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(133, 11182, '# IF by Rudyard Kipling - Grade 10 English Notes\r\n## Unit 16: Success & Celebration - Complete Chapter with All Exercises - NEB Curriculum\r\n\r\n## 🎯 Chapter Overview\r\nWhat happens when **timeless wisdom** **meets** **poetic brilliance** to **create** a **moral compass** for **navigating** **life\'s** **greatest challenges**? How can **four stanzas** of **carefully crafted verse** **provide** a **complete blueprint** for **character development**, **personal integrity**, and **sustainable success** that **transcends** **cultural boundaries** and **generational differences**? Rudyard Kipling\'s **\"IF\"** **presents** one of **English literature\'s** **most beloved** and **influential** **didactic poems** - a **father\'s advice** to his **son** that has **inspired millions** to **pursue** **excellence** **through** **moral fortitude** and **balanced living**.\r\n\r\nThis **masterful poem** **demonstrates** how **literature** can **serve** as **practical philosophy**, **offering** **concrete guidance** for **developing** **resilience**, **maintaining** **composure**, **handling** **both** **success** and **failure**, and **building** **authentic** **character** that **withstands** **life\'s** **inevitable** **storms**. Through **conditional statements** that **build** toward a **powerful conclusion**, **Kipling** **creates** a **moral framework** that **remains** **remarkably relevant** for **contemporary readers**.\r\n\r\nThe **poem** **serves** as both **literary masterpiece** and **life manual**, **teaching** **students** to **appreciate** how **poetry** can **embody** **philosophical wisdom** while **providing** **practical tools** for **personal development**, **ethical decision-making**, and **meaningful** **success** that **extends** **beyond** **material achievement** to **encompass** **character**, **service**, and **human dignity**.\r\n\r\n**Skills Focus:** Didactic poetry analysis, moral philosophy, conditional structures, prepositional phrases, speech writing  \r\n\r\n## 📚 Learning Objectives\r\nAfter studying this chapter, you will be able to:\r\n- Analyze didactic poetry and understand how literature can convey moral and philosophical teachings\r\n- Interpret conditional structures in poetry and their function in building logical arguments\r\n- Use advanced vocabulary related to character traits, moral values, and personal development\r\n- Master prepositional phrases and their various functions in complex sentence construction\r\n- Write compelling speeches that combine personal reflection with universal themes\r\n- Evaluate the relationship between character development and sustainable success\r\n- Discuss how poetic wisdom applies to contemporary challenges and personal growth\r\n- Express informed opinions about moral values, integrity, and ethical leadership in modern contexts\r\n\r\n\r\n\r\n### Historical and Literary Context\r\n\r\n#### Biographical Background\r\n**Rudyard Kipling** (1865-1936) **wrote** **\"IF\"** in **1895** **during** the **height** of the **British Empire**, when **Victorian values** of **stoicism**, **duty**, and **moral character** **dominated** **cultural discourse**. **Born** in **British India** and **educated** in **England**, **Kipling** **experienced** **firsthand** the **complexities** of **imperial responsibility**, **cultural diversity**, and **personal challenge** that **inform** the **poem\'s** **moral framework**.\r\n\r\n**The poem** was **reportedly inspired** by **Dr. Leander Starr Jameson**, a **British colonial administrator** whose **failed raid** in **South Africa** **demonstrated** both **courage** and **questionable judgment**. **Kipling** **admired** **Jameson\'s** **resilience** **following** **public disgrace**, **seeing** in him **qualities** that **exemplified** **true** **character** **under** **pressure**.\r\n\r\n#### Literary Significance\r\n**\"IF\"** **became** one of **Britain\'s** **most popular** **poems**, **frequently** **voted** as the **nation\'s favorite** **verse**. Its **enduring appeal** **stems** from its **practical wisdom**, **memorable rhythm**, and **universal applicability** **across** **cultures** and **generations**. The **poem** **has been** **translated** into **dozens** of **languages** and **continues** to **inspire** **leaders**, **athletes**, **students**, and **individuals** **facing** **personal challenges**.\r\n\r\n### Detailed Stanza-by-Stanza Analysis\r\n\r\n#### Stanza 1: Foundation of Composure and Self-Trust\r\n\r\n**Lines 1-2: \"If you can keep your head when all about you / Are losing theirs and blaming it on you\"**\r\n\r\n**Enhanced Analysis:** **Kipling** **opens** with **perhaps** the **most famous** **lines** in **English poetry**, **establishing** the **fundamental** **requirement** for **leadership** and **character**: **maintaining** **composure** **during** **crisis**. **\"Keeping your head\"** **means** **remaining** **calm**, **thinking** **clearly**, and **making** **rational decisions** **when** **others** **panic** or **act** **irrationally**. The **phrase** **\"blaming it on you\"** **acknowledges** that **leaders** **often** **become** **scapegoats** **during** **difficult** **times**.\r\n\r\n**Contemporary Application:** In **modern contexts** - **business crises**, **team failures**, **family conflicts** - this **advice** **remains** **crucial**. **Effective leaders** **don\'t** **succumb** to **mass hysteria** or **defensive reactions** but **maintain** **perspective** and **focus** on **solutions**.\r\n\r\n**Lines 3-4: \"If you can trust yourself when all men doubt you, / But make allowance for their doubting too\"**\r\n\r\n**Enhanced Analysis:** This **couplet** **presents** a **sophisticated** **balance** between **self-confidence** and **humility**. **Trusting yourself** **despite** **universal doubt** **requires** **strong** **inner** **conviction** and **self-knowledge**. **However**, **\"making allowance for their doubting\"** **prevents** **arrogance** by **acknowledging** that **others\'** **skepticism** **might** **contain** **valid** **concerns** **worth** **considering**.\r\n\r\n**Psychological Insight:** This **balance** **prevents** both **paralysis** from **excessive** **self-doubt** and **dangerous** **overconfidence** from **ignoring** **legitimate** **feedback**.\r\n\r\n**Lines 5-8: Complex Moral Challenges**\r\n\r\n**\"If you can wait and not be tired by waiting\"**: **Patience** **without** **passive resignation** - **active**, **purposeful** **waiting** that **maintains** **readiness** for **action**.\r\n\r\n**\"Or being lied about, don\'t deal in lies\"**: **Maintaining** **truthfulness** **despite** **others\'** **dishonesty** **about** you. **Refusing** to **respond** to **lies** **with** **counter-lies**.\r\n\r\n**\"Or being hated, don\'t give way to hating\"**: **Perhaps** the **most** **challenging** **moral** **standard** - **refusing** to **reciprocate** **hatred** **even** when **targeted** **unfairly**.\r\n\r\n**\"And yet don\'t look too good, nor talk too wise\"**: **Avoiding** **self-righteousness** and **intellectual** **superiority** that **alienate** **others**.\r\n\r\n#### Stanza 2: Dreams, Thinking, and Resilience\r\n\r\n**Lines 9-10: \"If you can dream—and not make dreams your master; / If you can think—and not make thoughts your aim\"**\r\n\r\n**Enhanced Analysis:** **Kipling** **advocates** for **balanced** **approach** to **ambition** and **intellectualism**. **Dreams** **provide** **vision** and **motivation** but **shouldn\'t** **become** **escapist** **fantasies** that **prevent** **practical action**. **Similarly**, **thinking** **develops** **understanding** but **shouldn\'t** **become** **paralyzed** **analysis** that **prevents** **decisive** **action**.\r\n\r\n**Modern Relevance:** In our **information-saturated age**, this **advice** **warns** **against** **overthinking** that **leads** to **analysis paralysis** or **dreaming** that **substitutes** for **concrete** **effort**.\r\n\r\n**Lines 11-12: \"If you can meet with Triumph and Disaster / And treat those two impostors just the same\"**\r\n\r\n**Enhanced Analysis:** **Kipling** **personifies** **\"Triumph\"** and **\"Disaster\"** as **\"impostors\"** - **false** **representatives** that **deceive** us **about** our **true** **worth** and **circumstances**. **Success** can **create** **false** **sense** of **superiority** and **invulnerability**, while **failure** can **create** **false** **sense** of **worthlessness** and **hopelessness**.\r\n\r\n**Philosophical Depth:** This **concept** **anticipates** **modern** **psychological understanding** about **emotional regulation** and **the** **temporary nature** of **both** **positive** and **negative** **circumstances**.\r\n\r\n**Lines 13-16: Rebuilding After Destruction**\r\n\r\n**Enhanced Analysis:** These **lines** **address** **one** of **life\'s** **most** **painful** **experiences**: **watching** **life\'s work** **destroyed** or **corrupted** by **others**. **\"Knaves\"** (**dishonest people**) **twist** **truth** to **deceive** **\"fools\"** (**naive people**), while **life projects** **we\'ve** **devoted** **years** to **building** **collapse**.\r\n\r\n**The** **crucial** **test** **involves** **willingness** to **\"stoop and build \'em up with worn-out tools\"** - **humbling** ourselves to **start** **over** with **whatever** **resources** **remain**, **without** **bitterness** or **despair**.\r\n\r\n#### Stanza 3: Ultimate Risk and Perseverance\r\n\r\n**Lines 17-20: The Ultimate Gamble**\r\n\r\n**Enhanced Analysis:** **\"Pitch-and-toss\"** **refers** to a **gambling game** where **coins** are **tossed** toward a **target**. **Kipling** **uses** this **metaphor** for **ultimate** **entrepreneurial** **risk**: **being** **willing** to **stake** **everything** on a **single** **opportunity**.\r\n\r\n**The** **true** **test** **isn\'t** **the** **willingness** to **risk** but the **response** to **total loss**: **starting** **again** **without** **complaint**, **self-pity**, or **public** **lamentation** **about** the **setback**.\r\n\r\n**Modern Application:** This **applies** to **business** **ventures**, **career** **changes**, **relationship** **commitments**, and **any** **significant** **life** **decision** **involving** **substantial** **risk**.\r\n\r\n**Lines 21-24: Physical and Mental Exhaustion**\r\n\r\n**Enhanced Analysis:** These **lines** **address** **moments** when **physical strength** (**\"heart and nerve and sinew\"**) **fails** but **will** **remains**. **\"Force\"** **suggests** **conscious** **effort** to **push** **beyond** **natural** **limits**.\r\n\r\n**The** **personified** **\"Will\"** **becomes** a **commanding** **voice** that **overrides** **physical** and **emotional** **exhaustion**. This **reflects** **understanding** that **ultimate** **achievement** **often** **requires** **pushing** **through** **periods** when **only** **determination** **sustains** **effort**.\r\n\r\n#### Stanza 4: Social Balance and Time Management\r\n\r\n**Lines 25-26: Social Versatility Without Compromise**\r\n\r\n**Enhanced Analysis:** **\"Talk with crowds and keep your virtue\"** **addresses** **challenge** of **maintaining** **moral** **standards** in **group** **settings** where **pressure** **exists** to **compromise** **principles** for **acceptance**.\r\n\r\n**\"Walk with Kings—nor lose the common touch\"** **warns** **against** **elitism** that **disconnects** us from **ordinary** **people** and **their** **concerns**. **True** **character** **adapts** **communication** **style** **without** **changing** **core** **values**.\r\n\r\n**Lines 27-28: Emotional Independence**\r\n\r\n**Enhanced Analysis:** **\"Neither foes nor loving friends can hurt you\"** **doesn\'t** **advocate** **emotional** **numbness** but **emotional** **resilience** that **prevents** **others** from **controlling** your **self-worth** or **decision-making**.\r\n\r\n**\"All men count with you, but none too much\"** **suggests** **balanced** **relationships** where **everyone** is **valued** but **no** **individual** **has** **disproportionate** **influence** **over** your **choices**.\r\n\r\n**Lines 29-30: The Unforgiving Minute**\r\n\r\n**Enhanced Analysis:** **\"Fill the unforgiving minute / With sixty seconds\' worth of distance run\"** **uses** **track** **metaphor** to **emphasize** **maximum** **effort** **within** **time** **constraints**. **\"Distance run\"** **suggests** **measurable** **progress** rather than **busy work**.\r\n\r\n**This** **became** one of **Kipling\'s** **most** **quoted** **lines**, **inspiring** **athletes**, **students**, and **professionals** to **maximize** **productivity** and **avoid** **wasted** **time**.\r\n\r\n**Lines 31-32: The Ultimate Reward**\r\n\r\n**Enhanced Analysis:** The **concluding** **promise** **reveals** the **stakes**: **following** **these** **principles** **leads** to **having** **\"the Earth and everything that\'s in it\"** - **not** **through** **conquest** but through **character** that **opens** **opportunities** and **builds** **trust**.\r\n\r\n**\"And—which is more—you\'ll be a Man, my son!\"** **uses** **Victorian** **concept** of **\"manhood\"** to **represent** **complete** **human** **development**. **Modern** **readers** **interpret** this as **mature**, **responsible**, **ethical** **personhood** **regardless** of **gender**.\r\n\r\n### Literary Devices and Poetic Techniques\r\n\r\n#### Structural Analysis\r\n**Form:** **Four stanzas** of **eight lines** each (**octaves**)\r\n**Rhyme Scheme:** **ABABCDCD** **throughout**\r\n**Meter:** **Predominantly** **iambic pentameter** with **variations** for **emphasis**\r\n**Overall Structure:** **One** **extended** **conditional sentence** (**32 lines**) **building** to **final** **reward**\r\n\r\n#### Major Literary Devices\r\n\r\n**Anaphora:** **Repetition** of **\"If you can\"** **creates** **rhythmic** **emphasis** and **builds** **cumulative** **effect**\r\n\r\n**Personification:** \r\n- **\"Triumph and Disaster\"** as **\"impostors\"**\r\n- **\"Will\"** **speaking** to **exhausted** **body** **parts**\r\n- **\"Unforgiving minute\"** **as** **harsh** **taskmaster**\r\n\r\n**Metaphor:**\r\n- **\"Keep your head\"** = **maintain** **composure**\r\n- **\"Pitch-and-toss\"** = **life\'s** **ultimate** **gambles**\r\n- **\"Distance run\"** = **productive** **effort**\r\n\r\n**Alliteration:** **\"nerve and sinew,\"** **\"foes...friends,\"** **\"distance...distance\"**\r\n\r\n**Symbolism:** **\"Earth\"** **represents** **complete** **success** and **opportunity**\r\n\r\n#### Rhythm and Sound\r\n**Kipling\'s** **masterful** **use** of **rhythm** **creates** **sense** of **measured**, **authoritative** **advice**. **Variations** in **meter** **emphasize** **key** **concepts** while **maintaining** **overall** **musical** **quality** that **aids** **memorization**.\r\n\r\n### Philosophical and Moral Framework\r\n\r\n#### Core Values Promoted\r\n1. **Emotional Regulation:** **Composure** **during** **crisis**, **balanced** **response** to **success/failure**\r\n2. **Intellectual Balance:** **Dreams** **without** **fantasy**, **thinking** **without** **paralysis**\r\n3. **Moral Integrity:** **Truthfulness** **despite** **lies**, **love** **despite** **hatred**\r\n4. **Social Adaptability:** **Relating** to **all** **social** **levels** **without** **compromising** **values**\r\n5. **Resilience:** **Starting** **over** **after** **failure**, **persevering** **through** **exhaustion**\r\n6. **Time Management:** **Maximizing** **effort** **within** **available** **time**\r\n\r\n#### Relevance to Contemporary Success\r\n**Modern** **research** in **psychology**, **leadership**, and **personal** **development** **validates** **many** of **Kipling\'s** **insights**:\r\n- **Emotional intelligence** **and** **regulation**\r\n- **Growth mindset** **and** **resilience**\r\n- **Authentic leadership** **and** **values-based** **decision-making**\r\n- **Work-life balance** **and** **sustainable** **effort**\r\n\r\n## 📝 Advanced Vocabulary Development\r\n\r\n### Character and Moral Vocabulary\r\n| Word | Pronunciation | Definition | Enhanced Context in Character Development |\r\n|------|---------------|------------|------------------------------------------|\r\n| **Composure** | /kəmˈpoʊʒər/ | Calm and controlled behavior under pressure | Essential leadership quality demonstrated during crises |\r\n| **Adversity** | /ədˈvɜːrsəti/ | Difficult or challenging circumstances | Character-building experiences that test moral resolve |\r\n| **Integrity** | /ɪnˈteɡrəti/ | Moral uprightness and honest consistency | Foundation of trustworthy relationships and sustainable success |\r\n| **Aspirations** | /ˌæspəˈreɪʃənz/ | Strong desires to achieve worthy goals | Motivating force balanced with practical action |\r\n| **Virtue** | /ˈvɜːrtʃu/ | Moral excellence and righteous behavior | Character traits that create positive influence on others |\r\n| **Perseverance** | /ˌpɜːrsəˈvɪrəns/ | Continued effort despite difficulties | Key factor distinguishing achievers from those who quit |\r\n| **Resilience** | /rɪˈzɪljəns/ | Ability to recover from setbacks quickly | Mental toughness enabling comeback from failures |\r\n| **Humility** | /hjuˈmɪləti/ | Modest view of one\'s importance | Prevents arrogance while maintaining healthy self-confidence |\r\n\r\n### Success and Achievement Terms\r\n| Term | Definition | Application in Personal Development |\r\n|------|------------|-----------------------------------|\r\n| **Triumph** | Great victory or significant achievement | Temporary success requiring balanced perspective |\r\n| **Disaster** | Sudden catastrophic failure or loss | Temporary setback teaching valuable lessons |\r\n| **Impostor** | Person pretending to be something they\'re not | Metaphor for false nature of extreme success/failure |\r\n| **Fortitude** | Courage and strength during difficult times | Mental resource enabling persistence through challenges |\r\n| **Stoicism** | Philosophy emphasizing reason over emotion | Approach to life maintaining equilibrium during ups/downs |\r\n| **Equanimity** | Mental calmness and composure | Emotional state enabling clear thinking during stress |\r\n\r\n### Action and Effort Vocabulary\r\n- **Stoop**: Lower oneself physically or morally\r\n- **Bear**: Endure or tolerate difficult circumstances\r\n- **Force**: Compel through willpower and determination\r\n- **Serve**: Work diligently toward goals despite obstacles\r\n- **Risk**: Accept potential loss for possible gain\r\n- **Build**: Construct systematically with available resources\r\n- **Hold on**: Maintain grip or position despite pressure\r\n- **Fill**: Use completely and productively\r\n\r\n### Social and Relationship Terms\r\n- **Crowds**: Large groups of ordinary people\r\n- **Kings**: Powerful, influential, or elite individuals\r\n- **Foes**: Enemies or opponents\r\n- **Knaves**: Dishonest, untrustworthy people\r\n- **Fools**: Naive or easily deceived individuals\r\n- **Common touch**: Ability to relate to ordinary people\r\n- **Count**: Have importance or value in relationships\r\n\r\n## 📋 Exercise Solutions (Enhanced and Rewritten)\r\n\r\n### A. Vocabulary Matching (Enhanced Character Development Context)\r\n\r\n| Word/Phrase | Correct Match | Enhanced Analysis and Character Application |\r\n|-------------|---------------|---------------------------------------------|\r\n| **unforgiving** | **vii. not giving second chance; not happening again** | **Time** **is** **unforgiving** **because** **wasted** **moments** **cannot** **be** **recovered**, **emphasizing** **need** for **productivity** |\r\n| **virtue** | **iii. behavior showing high moral standards** | **Virtue** **maintains** **consistency** **across** **all** **social** **situations**, **from** **crowds** to **kings** **to** **personal** **solitude** |\r\n| **sinew** | **vi. strong band of tissues in body** | **Physical** **strength** **supporting** **mental** **determination**, **representing** **complete** **human** **resources** **in** **challenging** **times** |\r\n| **pitch-and-toss** | **v. game of skill and chance** | **Metaphor** for **life\'s** **ultimate** **risks** where **everything** **is** **staked** on **uncertain** **outcomes** |\r\n| **worn-out** | **ii. badly damaged and no longer useful** | **Starting** **over** with **limited** **resources** **after** **major** **setbacks**, **demonstrating** **resilience** **and** **resourcefulness** |\r\n| **stoop** | **ix. bend body forwards and downwards** | **Humbling** **oneself** to **rebuild** **from** **ruins**, **showing** **willingness** to **do** **whatever** **necessary** |\r\n| **knave** | **viii. person who cheats by pretending** | **Dishonest** **individuals** who **twist** **truth** for **personal** **advantage**, **testing** **others\'** **integrity** |\r\n| **triumph** | **iv. great success, achievement, or victory** | **Major** **success** that **can** **create** **false** **sense** of **superiority** if **not** **handled** **with** **perspective** |\r\n| **impostor** | **vi. dishonest person** | **Both** **extreme** **success** and **failure** **are** **impostors** **because** they **misrepresent** **true** **worth** |\r\n\r\n### B. Summary Completion (Enhanced Thematic Analysis)\r\n\r\n**Enhanced Summary with Character Development Focus:**\r\n\r\nIn **\'IF,\'** **Rudyard Kipling** **provides** **inspirational** **insights** into **character-based** **living**. In the **first stanza**, the **poet** **suggests** **maintaining** **calm** **composure**, **developing** **self-trust**, and **learning** to **bear** **failure** **without** **becoming** **bitter** or **defensive**. **According** to the **poet**, it is **crucial** to be **thoughtful** and **strategic** about **actions** **before** **undertaking** them, **avoiding** **impulsive** **decisions** that **compromise** **long-term** **goals**. \r\n\r\n**Similarly**, **balancing** **between** **success** and **failure** **represents** another **essential** **quality** of **successful**, **well-rounded** **individuals**. In the **third stanza**, the **poet** **advises** us to **value** our **winnings** **appropriately** **without** **becoming** **attached** to **material** **possessions**. We **must** **devote** our **intellect**, **heart**, and **physical** **strength** in the **work** we **undertake**, and we **should** **always** **maintain** **strong** **virtue** to **achieve** **sustainable** **success**.\r\n\r\nIn the **final stanza**, the **poet** **emphasizes** **maintaining** **truthfulness** **regardless** of **social** **context** or **company**. **Most** **importantly**, we **must** **recognize** the **importance** of **time** **management** and **make** **every** **minute** **count** **toward** **meaningful** **accomplishment**.\r\n\r\n### C. Comprehensive Reading Questions (Enhanced Analysis)\r\n\r\n**a. What does the poet suggest about dealing with lies?**  \r\n**Enhanced Answer:** **Kipling** **advocates** for **maintaining** **absolute** **truthfulness** **even** **when** **others** **lie** **about** **you** or **use** **dishonesty** **in** **their** **interactions**. **The** **poet** **specifically** **states** **\"Or being lied about, don\'t deal in lies,\"** **establishing** a **moral** **standard** that **refuses** to **respond** to **dishonesty** with **counter-dishonesty**. **This** **principle** **requires** **tremendous** **self-discipline** because the **natural** **human** **tendency** **involves** **defending** **oneself** **against** **false** **accusations** **through** **whatever** **means** **necessary**. **However**, **Kipling** **argues** that **compromising** **truthfulness** **damages** **personal** **integrity** **more** than **enduring** **false** **accusations**. **By** **maintaining** **honesty** **despite** **others\'** **lies**, **individuals** **build** **reputation** for **reliability** that **ultimately** **proves** **more** **valuable** than **short-term** **defensive** **victories**. **This** **approach** **also** **prevents** the **moral** **deterioration** that **occurs** when **people** **justify** **dishonesty** as **response** to **others\'** **bad** **behavior**.\r\n\r\n**b. What do you mean by \'not be tired by waiting\'?**  \r\n**Enhanced Answer:** **\"Not be tired by waiting\"** **represents** **active**, **purposeful** **patience** that **maintains** **readiness** and **engagement** **rather** than **passive** **resignation** or **frustrated** **impatience**. **This** **concept** **goes** **beyond** **simple** **patience** to **encompass** **strategic** **waiting** that **uses** **delay** **periods** **productively**. **True** **patience** **involves** **maintaining** **mental** **alertness**, **emotional** **stability**, and **physical** **preparation** **while** **circumstances** **develop** **according** to **their** **natural** **timeline**. **Kipling** **suggests** that **mature** **individuals** **understand** that **many** **worthwhile** **achievements** **require** **extended** **preparation** **periods** **before** **action** **becomes** **appropriate** or **effective**. **During** these **waiting** **periods**, **character** **is** **tested** through **temptation** to **abandon** **goals**, **compromise** **standards**, or **act** **prematurely**. **\"Not being tired\"** **means** **sustaining** **energy**, **focus**, and **commitment** **throughout** **necessary** **delay** **periods** **without** **losing** **enthusiasm** or **direction**.\r\n\r\n**c. Why should we not aim only to dream in life?**  \r\n**Enhanced Answer:** **Kipling** **warns** **against** **making** **dreams** **our** **\"master\"** **because** **excessive** **focus** on **idealistic** **visions** **can** **prevent** **practical** **action** **necessary** for **actual** **achievement**. **Dreams** **serve** **crucial** **functions** as **motivation**, **inspiration**, and **goal-setting** **tools**, but they **become** **counterproductive** when they **substitute** for **concrete** **effort** or **create** **unrealistic** **expectations** that **discourage** **practical** **steps**. **People** who **become** **slaves** to **dreams** **often** **spend** **excessive** **time** **fantasizing** **about** **future** **success** **without** **developing** **skills**, **building** **relationships**, or **taking** **incremental** **actions** **required** for **real** **progress**. **Additionally**, **over-attachment** to **specific** **dream** **outcomes** **can** **create** **inflexibility** that **prevents** **adaptation** when **circumstances** **change** or **alternative** **opportunities** **arise**. **Balanced** **individuals** **use** **dreams** as **directional** **guides** while **remaining** **grounded** in **present** **reality** and **responsive** to **changing** **conditions**. **This** **approach** **combines** **visionary** **thinking** with **practical** **execution**.\r\n\r\n**d. How do you think we can achieve our dreams?**  \r\n**Enhanced Answer:** **Based** on **Kipling\'s** **framework**, **achieving** **dreams** **requires** **integrating** **multiple** **character** **qualities** and **strategic** **approaches**. **First**, **maintain** **balance** by **using** **dreams** as **motivation** **without** **becoming** **enslaved** to **rigid** **expectations**. **Second**, **develop** **resilience** through **treating** **both** **success** and **failure** as **temporary** **experiences** that **don\'t** **define** **permanent** **worth**. **Third**, **build** **persistence** by **continuing** **effort** **even** **when** **physical** and **emotional** **resources** **feel** **depleted**. **Fourth**, **maintain** **integrity** by **refusing** to **compromise** **moral** **standards** for **short-term** **advantages**. **Fifth**, **cultivate** **patience** **during** **necessary** **preparation** **and** **waiting** **periods**. **Sixth**, **develop** **social** **skills** that **enable** **effective** **relationships** **across** **diverse** **groups** **without** **losing** **personal** **authenticity**. **Finally**, **maximize** **time** **usage** by **filling** **every** **minute** with **purposeful** **activity** **that** **advances** **meaningful** **goals**. **Most** **importantly**, **be** **prepared** to **start** **over** **after** **major** **setbacks** **without** **complaint** or **self-pity**, **using** **whatever** **resources** **remain** **available**.\r\n\r\n**e. What can knaves do to us?**  \r\n**Enhanced Answer:** **According** to **Kipling**, **knaves** (**dishonest**, **unscrupulous** **individuals**) **can** **\"twist\"** the **truth** we\'ve **spoken** **to** **\"make a trap for fools.\"** **This** **means** **dishonest** **people** **deliberately** **distort**, **misrepresent**, or **take** **out** of **context** our **honest** **words** and **actions** to **create** **deceptive** **situations** that **harm** **naive** or **trusting** **individuals**. **Knaves** **might** **quote** us **selectively**, **change** **context**, **exaggerate** **meanings**, or **completely** **fabricate** **statements** **attributed** to us. **They** **use** our **reputation** for **honesty** to **lend** **credibility** to their **deceptions**, **making** **others** **more** **likely** to **believe** **false** **information**. **This** **creates** **particularly** **painful** **situations** because our **own** **integrity** **becomes** **weapon** **used** **against** **innocent** **people**. **The** **challenge** **involves** **maintaining** **truthfulness** **despite** **knowing** that **dishonest** **individuals** **might** **exploit** our **words**. **Kipling** **suggests** that **people** of **character** **must** **\"bear to hear\"** this **distortion** **without** **becoming** **cynical** or **reducing** **their** **own** **honesty** in **response**.\r\n\r\n**f. Why should we forget our failures in life?**  \r\n**Enhanced Answer:** **Kipling** **advocates** **\"forgetting\"** **failures** in the **sense** of **not** **dwelling** on **past** **setbacks** or **allowing** them to **define** **future** **possibilities**, **rather** than **completely** **erasing** **lessons** **learned** from **those** **experiences**. **The** **poem** **suggests** **starting** **\"again at your beginnings / And never breathe a word about your loss\"** - **focusing** **forward** **rather** than **backward**. **Excessive** **focus** on **past** **failures** **creates** **several** **problems**: it **wastes** **mental** **energy** that **could** be **directed** toward **current** **opportunities**; it **creates** **self-doubt** that **undermines** **confidence** **needed** for **future** **attempts**; it **can** **become** **excuse** for **avoiding** **new** **risks** or **challenges**; and it **might** **generate** **sympathy-seeking** **behavior** that **weakens** **personal** **accountability**. **Healthy** **approach** **involves** **learning** from **failures** **without** **being** **paralyzed** by them. **Extract** **valuable** **lessons** about **what** **didn\'t** **work**, **then** **focus** **energy** on **improved** **strategies** **going** **forward**. **This** **doesn\'t** **mean** **ignoring** **mistakes** but **rather** **processing** them **quickly** and **moving** **toward** **constructive** **action**.\r\n\r\n**g. What should we do if we lose the \'pitch-and-toss\'?**  \r\n**Enhanced Answer:** **When** we **lose** **everything** in **life\'s** **ultimate** **gamble** (**\"pitch-and-toss\"**), **Kipling** **advises** **starting** **\"again at your beginnings / And never breathe a word about your loss.\"** **This** **response** **requires** **extraordinary** **mental** **toughness** and **character**. **\"Starting again at your beginnings\"** **means** **returning** to **fundamental** **principles** and **basic** **building** **blocks** **without** **expecting** **shortcuts** or **special** **treatment** **based** on **past** **achievements**. **It** **requires** **humility** to **accept** **reduced** **circumstances** and **begin** **rebuilding** **from** **ground** **level**. **\"Never breathe a word about your loss\"** **doesn\'t** **mean** **denying** **reality** but **avoiding** **complaint**, **self-pity**, or **attempts** to **gain** **sympathy** that **might** **undermine** **future** **relationships** and **opportunities**. **This** **approach** **maintains** **dignity**, **prevents** **victim** **mentality**, and **focuses** **energy** on **solutions** **rather** than **problems**. **People** **who** **respond** to **total** **loss** **this** **way** **often** **discover** **inner** **strength** they **didn\'t** **know** **existed** and **build** **second** **success** on **firmer** **foundation** **than** the **first**.\r\n\r\n### D. Personal Reflection on Success Advice (Enhanced Application)\r\n\r\n**Enhanced Personal Response:**\r\n\r\n**Common** **suggestions** that **parents** and **teachers** **provide** for **life** **success** **often** **align** **remarkably** **well** with **Kipling\'s** **timeless** **wisdom** in **\"IF.\"**\r\n\r\n**Goal Setting and Vision:** **Adults** **consistently** **emphasize** **setting** **clear**, **achievable** **goals** that **provide** **direction** and **purpose**. **This** **connects** to **Kipling\'s** **advice** about **dreaming** **without** **making** **dreams** our **master** - **having** **vision** while **remaining** **grounded** in **practical** **reality**.\r\n\r\n**Work Ethic and Discipline:** **Parents** and **teachers** **stress** **hard** **work**, **dedication**, and **maintaining** **discipline** in **daily** **routines**. **This** **parallels** **Kipling\'s** **emphasis** on **filling** **\"the unforgiving minute / With sixty seconds\' worth of distance run\"** - **maximizing** **effort** **within** **available** **time**.\r\n\r\n**Positive Attitude and Self-Belief:** **Adults** **encourage** **maintaining** **positive** **attitude**, **believing** in **personal** **abilities**, and **developing** **confidence**. **Kipling\'s** **advice** to **\"trust yourself when all men doubt you\"** **while** **making** **\"allowance for their doubting too\"** **provides** **sophisticated** **balance** between **self-confidence** and **humility**.\r\n\r\n**Continuous Learning:** **Educators** **emphasize** **acquiring** **new** **knowledge**, **developing** **skills**, and **staying** **current** in **respective** **fields**. **This** **connects** to **Kipling\'s** **advice** about **thinking** **without** **making** **thoughts** our **aim** - **using** **intellect** **productively** **without** **becoming** **paralyzed** by **analysis**.\r\n\r\n**Calculated Risk-Taking:** **Adults** **teach** **willingness** to **take** **reasonable** **risks** for **growth** and **learning**. **Kipling\'s** **\"pitch-and-toss\"** **metaphor** **represents** **ultimate** **risk-taking** while **emphasizing** **resilient** **response** to **potential** **loss**.\r\n\r\n**Relationship Skills:** **Parents** and **teachers** **stress** **building** **positive** **relationships**, **effective** **communication**, and **collaboration**. **Kipling\'s** **advice** to **\"talk with crowds and keep your virtue, / Or walk with Kings—nor lose the common touch\"** **provides** **framework** for **authentic** **relationships** **across** **social** **boundaries**.\r\n\r\n**Resilience and Perseverance:** **Adults** **emphasize** **bouncing** **back** from **setbacks** and **maintaining** **effort** **despite** **difficulties**. **Kipling\'s** **entire** **poem** **focuses** on **character** **traits** that **enable** **persistence** **through** **challenges**.\r\n\r\n**The** **convergence** **between** **modern** **advice** and **Kipling\'s** **Victorian** **wisdom** **suggests** that **fundamental** **principles** of **character** **development** and **sustainable** **success** **remain** **constant** **across** **generations** and **cultures**.\r\n\r\n## 🎯 Grammar Focus: Mastering Prepositional Phrases\r\n\r\n### Understanding Prepositional Phrases in Complex Communication\r\n**Prepositional phrases** **consist** of **prepositions** **followed** by **objects** (**nouns** or **pronouns**) and **any** **modifying** **words**. **These** **phrases** **function** as **adjectives** or **adverbs**, **providing** **essential** **information** about **location**, **time**, **manner**, **purpose**, or **relationship** **between** **sentence** **elements**.\r\n\r\n### Prepositional Phrase Categories\r\n\r\n#### Location and Direction\r\n- **In/on/at** specific places\r\n- **Across/through/around** movement patterns\r\n- **Above/below/beside** relative positions\r\n- **Into/onto/out of** directional changes\r\n\r\n#### Time and Duration  \r\n- **During/throughout** time periods\r\n- **Before/after/since** temporal relationships\r\n- **By/until/from** time boundaries\r\n- **For/in/within** duration specifications\r\n\r\n#### Manner and Method\r\n- **With/without** accompaniment or tools\r\n- **By means of/through** methods\r\n- **In spite of/despite** contrasting circumstances\r\n- **According to/instead of** alternative approaches\r\n\r\n### A. Prepositional Phrase Completion (Enhanced Contextual Analysis)\r\n\r\n**Enhanced Analysis with Character Development Applications:**\r\n\r\n| Sentence | Correct Phrase | Enhanced Context | Character Development Application |\r\n|----------|----------------|------------------|----------------------------------|\r\n| **Bus and lorry collided _____** | **with each other** | **Physical collision** **requiring** **preposition** **showing** **mutual** **action** | **Conflicts** **require** **people** **working** **with each other** **for** **resolution** |\r\n| **Amin is brilliant _____** | **at playing** | **Skill** **demonstration** **requiring** **\"at\"** **with** **gerund** | **Excellence** **at playing** **any** **role** **requires** **dedicated** **practice** |\r\n| **Mr. Sharma not _____** | **interested in** | **\"Interested in\"** **standard** **collocation** **with** **topics** | **Character** **development** **means** **being** **interested in** **others\'** **welfare** |\r\n| **Criminal _____** | **at large** | **Idiomatic** **expression** **meaning** **free** **and** **potentially** **dangerous** | **Problems** **left** **at large** **without** **solutions** **become** **bigger** **issues** |\r\n| **His phone is _____** | **out of reach** | **Physical** **and** **metaphorical** **inaccessibility** | **Success** **shouldn\'t** **put** **us** **out of reach** **of** **ordinary** **people** |\r\n| **Concert went _____** | **out of hand** | **Lost** **control** **situation** **requiring** **intervention** | **Emotions** **going** **out of hand** **compromise** **decision-making** **ability** |\r\n| **Sanskriti is _____** | **on a diet** | **Temporary** **behavioral** **change** **for** **health** **goals** | **Self-discipline** **on a diet** **builds** **willpower** **for** **other** **challenges** |\r\n| **I am _____** | **in a hurry** | **Time** **pressure** **affecting** **current** **availability** | **Always** **being** **in a hurry** **prevents** **thoughtful** **decision-making** |\r\n| **All passengers _____** | **on board** | **Location** **within** **transportation** **vehicle** | **Getting** **everyone** **on board** **with** **plans** **requires** **good** **communication** |\r\n| **Payment _____** | **in cash** | **Method** **of** **financial** **transaction** | **Paying** **in cash** **for** **character** **development** **means** **immediate** **effort** |\r\n\r\n### Advanced Prepositional Phrase Functions\r\n\r\n#### Adjectival Function (Modifying Nouns)\r\n- **The book** **on the table** **belongs** **to Sarah**\r\n- **Students** **with strong character** **overcome** **challenges**\r\n- **People** **of integrity** **inspire** **trust**\r\n\r\n#### Adverbial Function (Modifying Verbs)\r\n- **She** **spoke** **with confidence** **during** **the presentation**\r\n- **They** **worked** **through the night** **to meet** **deadlines**\r\n- **He** **succeeded** **despite many obstacles**\r\n\r\n#### Complex Prepositional Constructions\r\n- **\"In spite of\"** **challenges**, **he** **maintained** **composure**\r\n- **\"According to\"** **the plan**, **we\'ll** **finish** **by Friday**\r\n- **\"Instead of\"** **complaining**, **she** **found** **solutions**\r\n\r\n### Common Prepositional Phrase Errors\r\n\r\n#### Incorrect Preposition Choice\r\n**Wrong:** \"He\'s good in mathematics\" → **Correct:** \"He\'s good **at** mathematics\"\r\n**Wrong:** \"Interested on politics\" → **Correct:** \"Interested **in** politics\"\r\n\r\n#### Missing Prepositions\r\n**Wrong:** \"She arrived 9 AM\" → **Correct:** \"She arrived **at** 9 AM\"\r\n**Wrong:** \"Different his brother\" → **Correct:** \"Different **from** his brother\"\r\n\r\n#### Redundant Prepositions\r\n**Wrong:** \"Where is he at?\" → **Correct:** \"Where is he?\"\r\n**Wrong:** \"Outside of the building\" → **Correct:** \"Outside the building\"\r\n\r\n### Prepositional Phrases in Character Development\r\n\r\n#### Describing Personal Qualities\r\n- **Person** **of integrity** **maintains** **honesty** **under pressure**\r\n- **Individual** **with perseverance** **continues** **despite** **setbacks**\r\n- **Leader** **of character** **inspires** **through** **example**\r\n\r\n#### Expressing Relationships\r\n- **Working** **with others** **requires** **empathy** and **communication**\r\n- **Learning** **from failure** **builds** **resilience** and **wisdom**\r\n- **Succeeding** **through effort** **creates** **sustainable** **achievement**\r\n\r\n#### Indicating Circumstances\r\n- **Acting** **under pressure** **reveals** **true** **character**\r\n- **Maintaining** **virtue** **among crowds** **requires** **strong** **principles**\r\n- **Building** **with worn-out tools** **demonstrates** **resourcefulness**\r\n\r\n## ✍️ Writing Practice: Inspirational Speech Composition\r\n\r\n### Model Speech: \"Failure as the Foundation of Success\"\r\n\r\n**FAILURE AS THE PILLAR OF SUCCESS**  \r\n*Transforming Setbacks into Stepping Stones*\r\n\r\n---\r\n\r\n**Distinguished** **guests**, **fellow** **students**, **and** **esteemed** **educators**,\r\n\r\n**Today**, I **stand** **before** **you** to **challenge** a **fundamental** **misconception** that **pervades** our **society** and **limits** our **potential**. **We** have been **conditioned** to **view** **failure** as **something** to **fear**, **avoid**, and **hide** from **others**. **But** I **propose** a **radically** **different** **perspective**: **failure** is **not** the **enemy** of **success** - it is **success\'s** **most** **essential** **foundation**.\r\n\r\n**This** **morning**, I **want** to **share** **with** **you** **compelling** **evidence** that **failure**, **when** **approached** with the **right** **mindset** and **character**, **becomes** the **most** **powerful** **teacher**, **the** **strongest** **motivator**, and the **surest** **path** to **authentic**, **lasting** **achievement**.\r\n\r\n### The Universal Nature of Failure\r\n\r\n**Let** me **begin** with a **fundamental** **truth**: **every** **person** in this **room** has **experienced** **failure**. **Every** **successful** **individual** **throughout** **history** has **faced** **setbacks**, **disappointments**, and **moments** when their **dreams** **seemed** **impossible**. **The** **difference** between **those** who **ultimately** **succeed** and **those** who **remain** **stuck** lies **not** in **avoiding** **failure** but in **transforming** **failure** **into** **wisdom**, **strength**, and **renewed** **determination**.\r\n\r\n**Consider** **Thomas Edison**, the **brilliant** **inventor** who **gave** us the **incandescent** **light bulb**, **phonograph**, and **hundreds** of **other** **innovations** that **revolutionized** **modern** **life**. **Edison** **didn\'t** **succeed** **despite** **failure** - he **succeeded** **because** of **failure**. **When** **critics** **pointed** to his **thousands** of **unsuccessful** **experiments**, **Edison** **famously** **replied**, **\"I have not failed. I\'ve just found 10,000 ways that won\'t work.\"**\r\n\r\n**This** **perspective** **transformed** **every** **setback** **into** **valuable** **data**, **every** **disappointment** **into** **education**, **every** **\"failure\"** **into** a **step** **closer** to **breakthrough**. **Edison\'s** **laboratory** **notebooks** **contained** **detailed** **records** of **experiments** that **didn\'t** **work** - **not** as **monuments** to **failure** but as **maps** **guiding** him **away** from **dead** **ends** toward **successful** **solutions**.\r\n\r\n### Modern Examples of Triumph Through Adversity\r\n\r\n**Let** me **share** **another** **inspiring** **example** from our **contemporary** **world**. **Oprah Winfrey**, one of the **most** **influential** **media** **personalities** in **history**, was **born** into **poverty**, **faced** **childhood** **abuse**, **experienced** **teenage** **pregnancy**, and was **told** by **early** **employers** that she was **\"unfit for television news.\"** **Rather** than **accepting** these **setbacks** as **permanent** **limitations**, **Oprah** **used** **each** **challenge** as **fuel** for **personal** **growth** and **professional** **development**.\r\n\r\n**Her** **early** **career** **included** **being** **fired** from her **first** **television** **job** as **co-anchor** because **producers** **considered** her **\"too emotionally invested\"** in **stories**. **This** **\"failure\"** **led** her to **discover** her **true** **strength**: **connecting** **authentically** with **people** through **empathy** and **genuine** **concern**. **What** **seemed** like a **career-ending** **weakness** **became** the **foundation** of her **unprecedented** **success**.\r\n\r\n**Similarly**, **consider** **Michael Jordan**, **widely** **regarded** as **basketball\'s** **greatest** **player**. **Jordan** was **cut** from his **high school** **varsity** **basketball** **team** as a **sophomore**. **His** **coach** **considered** him **too** **short** and **insufficiently** **skilled** for **competitive** **play**. **Rather** than **abandoning** his **dreams**, **Jordan** **used** this **rejection** as **motivation** for **relentless** **improvement**.\r\n\r\n**He** **later** **said**, **\"I\'ve missed more than 9,000 shots in my career. I\'ve lost almost 300 games. 26 times, I\'ve been trusted to take the game-winning shot and missed. I\'ve failed over and over and over again in my life. And that is why I succeed.\"**\r\n\r\n### The Science Behind Failure\'s Value\r\n\r\n**Modern** **psychological research** **validates** what **great** **achievers** have **always** **known**: **failure** **provides** **unique** **learning** **opportunities** **impossible** to **gain** through **success** **alone**. **When** we **fail**, our **brains** **activate** **different** **neural** **pathways**, **forcing** us to **analyze** **what** **went** **wrong**, **consider** **alternative** **approaches**, and **develop** **creative** **solutions**.\r\n\r\n**Failure** **also** **builds** **psychological** **resilience** - what **researchers** call **\"failure** **tolerance\"** or **\"grit.\"** **Each** **time** we **survive** a **setback** and **continue** **pursuing** our **goals**, we **develop** **confidence** that we **can** **handle** **future** **challenges**. **This** **emotional** **strength** **becomes** **invaluable** **during** **subsequent** **difficulties**.\r\n\r\n**Moreover**, **failure** **prevents** **complacency**. **People** who **experience** **early**, **easy** **success** **often** **develop** **overconfidence** that **makes** them **unprepared** for **eventual** **challenges**. **Those** who **learn** through **failure** **develop** **humility**, **preparation** **habits**, and **contingency** **thinking** that **sustain** **long-term** **achievement**.\r\n\r\n### Failure as Character Development\r\n\r\n**But** **failure\'s** **greatest** **value** **lies** not in **what** it **teaches** **about** **strategy** or **technique** but in **what** it **reveals** and **develops** **about** **character**. **As** **Rudyard Kipling** **wisely** **wrote** in his **immortal** **poem** **\"IF,\"** **true** **success** **requires** **treating** **\"Triumph and Disaster\"** as **\"two impostors just the same.\"**\r\n\r\n**Failure** **strips** **away** **pretense** and **forces** us to **confront** **fundamental** **questions**: **Do** we **have** **genuine** **passion** for our **goals**, or are we **motivated** **only** by **external** **recognition**? **Can** we **maintain** **integrity** **under** **pressure**? **Will** we **blame** **others** for our **setbacks**, or **take** **responsibility** for **improvement**? **Do** we **have** the **perseverance** to **start** **over** when **everything** **collapses**?\r\n\r\n**These** **character** **tests** **separate** **those** who **achieve** **temporary** **success** from **those** who **build** **lasting** **legacies**. **Failure** **reveals** whether we **possess** the **moral** **foundation** **necessary** for **sustainable** **achievement**.\r\n\r\n### The Innovation Connection\r\n\r\n**History** **demonstrates** that **breakthrough** **innovations** **emerge** **from** **failure** **more** **often** than **from** **success**. **The** **Post-it Note**, now **essential** in **offices** **worldwide**, **resulted** from a **\"failed\"** **experiment** to **create** **super-strong** **adhesive**. **Instead**, **3M** **scientist** **Spencer Silver** **created** **weak**, **repositionable** **adhesive** that **seemed** **useless** until **colleague** **Art Fry** **realized** it **solved** his **problem** of **bookmarks** **falling** **out** of **hymn** **books**.\r\n\r\n**Penicillin**, the **life-saving** **antibiotic**, was **discovered** when **Alexander Fleming** **accidentally** **contaminated** **bacterial** **cultures** in his **laboratory**. **What** **appeared** to be **ruined** **experiment** **revealed** that **mold** **killed** **bacteria**, **leading** to one of **medicine\'s** **greatest** **advances**.\r\n\r\n**These** **examples** **illustrate** **crucial** **principle**: **failure** **often** **opens** **doors** to **possibilities** we **never** **considered**. **When** our **original** **plans** **collapse**, we\'re **forced** to **explore** **new** **territories**, **develop** **creative** **solutions**, and **discover** **opportunities** **invisible** to **those** who **succeed** **immediately**.\r\n\r\n### Reframing Our Relationship with Failure\r\n\r\n**So** **how** do we **transform** our **relationship** with **failure**? **First**, we **must** **change** our **language**. **Instead** of **saying** **\"I failed,\"** **say** **\"I learned.\"** **Instead** of **\"This didn\'t work,\"** **say** **\"This approach needs modification.\"** **Language** **shapes** **thought**, and **thought** **shapes** **action**.\r\n\r\n**Second**, we **must** **develop** **systematic** **approaches** to **learning** from **setbacks**. **After** **every** **disappointment**, **ask**: **What** **specific** **factors** **contributed** to this **outcome**? **What** **could** I **have** **done** **differently**? **What** **resources** or **skills** do I **need** to **develop**? **What** **assumptions** **proved** **incorrect**? **How** can I **test** **new** **approaches** **safely**?\r\n\r\n**Third**, we **must** **build** **support** **systems** that **encourage** **intelligent** **risk-taking**. **Surround** **yourself** with **people** who **understand** that **failure** is **part** of **growth**, not **evidence** of **inadequacy**. **Share** your **setbacks** **honestly** with **trusted** **mentors** who **can** **provide** **perspective** and **guidance**.\r\n\r\n### The Courage to Begin Again\r\n\r\n**Perhaps** **failure\'s** **greatest** **gift** is the **courage** it **develops** to **begin** **again**. **Every** **time** we **survive** a **setback** and **choose** to **continue**, we **prove** to **ourselves** that we\'re **stronger** than our **circumstances**. **This** **self-knowledge** **becomes** **unshakeable** **foundation** for **future** **endeavors**.\r\n\r\n**J.K. Rowling**, **author** of the **beloved** **Harry Potter** **series**, **was** a **single** **mother** **living** on **welfare** when she **began** **writing** her **first** **novel**. **Twelve** **publishers** **rejected** **her** **manuscript** **before** **one** **finally** **accepted** it. **She** **later** **said**, **\"Rock bottom became the solid foundation on which I rebuilt my life.\"**\r\n\r\n**Rowling\'s** **early** **struggles** with **poverty**, **depression**, and **rejection** **didn\'t** **destroy** her **dreams** - they **clarified** her **priorities**, **strengthened** her **resolve**, and **provided** **authentic** **material** for **stories** that **resonate** with **millions** **worldwide**.\r\n\r\n### Failure in the Digital Age\r\n\r\n**In** our **social media** **age**, we\'re **constantly** **exposed** to **curated** **highlight** **reels** of **others\'** **lives**, **creating** **illusion** that **success** **comes** **easily** and **failure** **indicates** **personal** **inadequacy**. **This** **perspective** **is** **not** **only** **false** but **dangerous** - it **prevents** us from **taking** **necessary** **risks** for **growth** and **innovation**.\r\n\r\n**We** **must** **remember** that **behind** **every** **overnight** **success** **lies** **years** of **preparation**, **countless** **setbacks**, and **persistent** **effort** **invisible** to **casual** **observers**. **The** **most** **successful** **entrepreneurs**, **artists**, **athletes**, and **leaders** **have** **failure** **stories** **they** **rarely** **share** **publicly** but **consider** **essential** to their **development**.\r\n\r\n### Practical Steps Forward\r\n\r\n**As** you **leave** here **today**, I **challenge** you to **embrace** **failure** as your **teacher**, **not** your **enemy**. **Set** **ambitious** **goals** that **might** **result** in **setbacks**. **Take** **calculated** **risks** in **pursuit** of **meaningful** **objectives**. **When** **disappointment** **comes** - and it **will** - **extract** **every** **lesson**, **adjust** your **approach**, and **begin** **again** with **renewed** **wisdom**.\r\n\r\n**Remember** **Edison\'s** **laboratory**, **Jordan\'s** **high school** **rejection**, **Oprah\'s** **early** **career** **struggles**, and **Rowling\'s** **twelve** **rejections**. **Each** **setback** **prepared** them for **greater** **achievement** **than** they **originally** **imagined**.\r\n\r\n**Most** **importantly**, **develop** the **character** **traits** that **transform** **failure** into **fuel**: **persistence** **despite** **disappointment**, **integrity** **under** **pressure**, **humility** in **both** **success** and **failure**, and **courage** to **begin** **again** when **everything** **collapses**.\r\n\r\n### Conclusion\r\n\r\n**Ladies** and **gentlemen**, **failure** is **not** the **opposite** of **success** - it is the **pathway** to **success**. **Every** **failure** **contains** **seeds** of **future** **triumph** **if** we **have** the **wisdom** to **plant** them and the **patience** to **nurture** their **growth**.\r\n\r\n**As** you **pursue** your **dreams**, **remember** that **setbacks** are **not** **evidence** that you\'re **on** the **wrong** **path** - they\'re **proof** that you\'re **on** a **path** **worth** **traveling**. **The** **easy** **roads** **lead** to **ordinary** **destinations**. **The** **difficult** **roads**, **marked** by **failure** and **challenge**, **lead** to **extraordinary** **achievement**.\r\n\r\n**Embrace** **failure**. **Learn** from it. **Grow** **through** it. **And** **let** it **become** the **unshakeable** **pillar** **upon** which you **build** your **success**.\r\n\r\n**Thank** **you**.\r\n\r\n---\r\n\r\n### Speech Writing Guidelines\r\n\r\n#### Effective Speech Structure\r\n1. **Compelling opening** that **challenges** **conventional** **thinking**\r\n2. **Clear thesis statement** **establishing** **main** **argument**\r\n3. **Multiple examples** **supporting** **central** **theme**\r\n4. **Research-based evidence** **validating** **key** **points**\r\n5. **Practical applications** **for** **audience** **implementation**\r\n6. **Inspirational conclusion** **motivating** **action**\r\n\r\n#### Rhetorical Techniques\r\n- **Repetition** for **emphasis** and **memorability**\r\n- **Parallel structure** **creating** **rhythmic** **flow**\r\n- **Specific examples** **making** **abstract** **concepts** **concrete**\r\n- **Questions** **engaging** **audience** **participation**\r\n- **Personal stories** **creating** **emotional** **connection**\r\n- **Call to action** **providing** **clear** **next** **steps**\r\n\r\n#### Language Features\r\n- **Conversational tone** **appropriate** for **audience**\r\n- **Active voice** **creating** **dynamic**, **engaging delivery**\r\n- **Varied sentence length** **maintaining** **audience** **interest**\r\n- **Metaphors** and **analogies** **clarifying** **complex** **ideas**\r\n- **Emotional language** **inspiring** **rather** than **manipulating**\r\n\r\n## ✅ Chapter Assessment Checklist\r\n\r\n**Poetry Analysis Excellence:**\r\n- [ ] I can analyze didactic poetry and understand how literature conveys moral teachings\r\n- [ ] I understand conditional structures in poetry and their function in building logical arguments\r\n- [ ] I can identify and interpret literary devices including metaphor, personification, and symbolism\r\n- [ ] I recognize how rhythm and structure support thematic content in motivational poetry\r\n\r\n**Advanced Vocabulary Mastery:**\r\n- [ ] I know 25+ character traits and moral vocabulary terms with appropriate usage\r\n- [ ] I can use success and achievement terminology in discussions and writing\r\n- [ ] I understand the nuances between related concepts like confidence vs. arrogance\r\n- [ ] I can express complex moral and philosophical ideas using precise vocabulary\r\n\r\n**Grammar Sophistication:**\r\n- [ ] I can use prepositional phrases accurately across various contexts and functions\r\n- [ ] I understand how prepositional phrases modify nouns and verbs differently\r\n- [ ] I can identify and correct common prepositional phrase errors\r\n- [ ] I recognize complex prepositional constructions in formal writing\r\n\r\n**Communication Excellence:**\r\n- [ ] I can write compelling speeches that combine personal reflection with universal themes\r\n- [ ] I can structure arguments logically with supporting evidence and examples\r\n- [ ] I can adapt language and tone for different audiences and occasions\r\n- [ ] I can deliver inspirational messages that motivate action rather than just inform\r\n\r\n**Character and Personal Development:**\r\n- [ ] I can discuss the relationship between character traits and sustainable success\r\n- [ ] I understand how moral principles apply to contemporary challenges\r\n- [ ] I can evaluate my own character development needs and create improvement plans\r\n- [ ] I appreciate how literature provides practical wisdom for personal growth\r\n\r\n', 1, '2025-07-25 01:11:38', '2025-09-17 06:10:43');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(134, 11183, '# London vs Paris - Grade 10 English Complete Guide\r\n## Unit 17: Countries & Towns - NEB English Class 10\r\n\r\n## 🎯 Chapter Introduction\r\nThis chapter compares two famous European cities - **London** (capital of England/UK) and **Paris** (capital of France). Both cities are world-famous tourist destinations with rich history, culture, and attractions. Students will learn to compare cities, understand cultural differences, and practice comparative writing skills.\r\n\r\n**Main Topics Covered:**\r\n- City comparison and contrasts\r\n- Cultural and historical differences  \r\n- Tourist attractions and facilities\r\n- Transportation and lifestyle\r\n- Grammar: Adjectives vs Adverbs\r\n- Writing: Comparative essays\r\n\r\n## 📖 Text Summary: London vs Paris Comparison\r\n\r\n### Basic Information\r\n**London** and **Paris** are two outstanding European cities, each with unique features and attractions. **London** combines modern and historical architecture with famous sites like **Tower of London** and **Buckingham Palace**. **Paris** is known for romantic architecture including the **Eiffel Tower** and **Notre-Dame Cathedral**.\r\n\r\n### Key Differences and Similarities\r\n\r\n| Aspect | London | Paris |\r\n|--------|--------|--------|\r\n| **Country** | United Kingdom | France |\r\n| **River** | Thames | Seine |\r\n| **Area** | 1,572 km² | 105 km² |\r\n| **Famous For** | Multicultural, Financial hub | Fashion, Art, Cuisine |\r\n| **Transport** | Extensive network | Good metro system |\r\n| **Green Spaces** | Hyde Park (large parks) | Luxembourg Gardens (designed gardens) |\r\n\r\n### Cultural Contributions\r\nBoth cities have significantly influenced **art**, **literature**, **music**, and **theater** worldwide. **Paris** excels in **culinary arts** and **gourmet cuisine**, while **London** offers **diverse international food** and **vibrant street food culture**.\r\n\r\n### Tourism and Shopping\r\nBoth cities provide excellent **shopping opportunities** and have **extensive public transportation**. **London** serves as a **global financial center**, while **Paris** leads in **fashion** and **artistic creativity**.\r\n\r\n## 📝 Vocabulary Building\r\n\r\n### A. Crossword Puzzle Solutions\r\n\r\n#### DOWN Clues:\r\n1. **urbanite** - a person who lives in a city or town\r\n4. **showcase** - an impressive display or exhibition of something\r\n6. **impeccable** - perfect according to highest standards; without faults  \r\n7. **rusty** - skills that have deteriorated due to lack of practice\r\n\r\n#### ACROSS Clues:\r\n2. **chugger** - person who approaches people on streets asking for charity donations\r\n3. **tramway** - railway tracks designed for trams to travel on\r\n5. **vibrant** - full of energy, life, and activity\r\n8. **baguette** - traditional long, narrow French bread loaf\r\n9. **gastronomy** - the art and science of good eating and cooking\r\n10. **clique** - small, exclusive group of people with shared interests\r\n\r\n### Key Vocabulary with Meanings:\r\n- **iconic** - widely recognized and representative\r\n- **bustling** - full of energetic activity  \r\n- **cosmopolitan** - including people from many different countries\r\n- **metropolitan** - relating to a large city and surrounding areas\r\n- **culinary** - related to cooking and food preparation\r\n- **architecture** - design and construction of buildings\r\n- **heritage** - traditions and culture passed down through generations\r\n\r\n## 📋 Reading Comprehension Exercises\r\n\r\n### B. Multiple Choice Questions (with explanations)\r\n\r\n**a. London\'s transport network is:**\r\n- Answer: **ii. widespread** ✓\r\n- *Explanation: London has extensive transport covering large metropolitan area*\r\n\r\n**b. Number of museums in London:**\r\n- Answer: **iii. 240** ✓  \r\n- *Explanation: London has significantly more museums than Paris*\r\n\r\n**c. Paris excels over London in:**\r\n- Answer: **i. variety in cuisine** ✓\r\n- *Explanation: Paris is renowned for French culinary excellence*\r\n\r\n**d. NOT a major attraction of Paris:**\r\n- Answer: **iv. public library** ✓\r\n- *Explanation: Paris is famous for fashion, cuisine, and art, not libraries*\r\n\r\n**e. NOT a similarity between both cities:**\r\n- Answer: **ii. Both can be traveled year-round with nice weather** ✓\r\n- *Explanation: Both cities have variable weather, not consistently nice*\r\n\r\n### C. True/False Statements (with corrections)\r\n\r\n**a. French prefer handshakes to kisses - FALSE**\r\n- *Correction: French people commonly greet with air kisses (bisous)*\r\n\r\n**b. Londoners are easy to get along with - FALSE** \r\n- *Correction: British people can be reserved initially, though polite*\r\n\r\n**c. French enjoy informal talks in espresso bars - TRUE**\r\n- *Explanation: Café culture is central to French social life*\r\n\r\n**d. London is fifteen times larger than Paris geographically - FALSE**\r\n- *Correction: London is about 15 times larger, but statement says \"almost fifteen\"*\r\n\r\n**e. Cost of living is similar in both cities - FALSE**\r\n- *Correction: London is generally more expensive than Paris*\r\n\r\n**f. Paris is less populous - FALSE**\r\n- *Correction: Paris has lower total population but higher density*\r\n\r\n### D. Detailed Reading Questions & Answers\r\n\r\n**a. What are the famous rivers in London and Paris?**\r\n**Answer:** The **River Thames** flows through London, serving as the city\'s main waterway and historic trading route. The **River Seine** flows through Paris, dividing the city into Left Bank and Right Bank areas. Both rivers are central to each city\'s development and tourism.\r\n\r\n**b. Why do football fans prefer London over Paris?**\r\n**Answer:** Football enthusiasts favor London because it hosts **six major Premier League clubs** (Arsenal, Chelsea, Tottenham, West Ham, Fulham, Brentford) competing in England\'s top division. This provides multiple match options and rivalries. Paris has only **one major club - Paris Saint-Germain (PSG)**, limiting football tourism opportunities.\r\n\r\n**c. Which city received more tourists in 2018?**\r\n**Answer:** According to the **2018 MasterCard Global Cities Index**, **London** attracted more international visitors, ranking as the **world\'s second most-visited city** after Bangkok. **Paris** ranked **third globally**. London\'s English-speaking advantage and diverse attractions contribute to higher tourist numbers.\r\n\r\n**d. When is the best time to visit London?**\r\n**Answer:** **Autumn (September-November)** is considered optimal for visiting London. During this period, temperatures remain **comfortable and pleasant**, the city hosts various **British cultural events and festivals**, and tourist crowds are smaller than summer peak season. Weather is mild enough for extensive walking and sightseeing.\r\n\r\n**e. Which city is more affordable for travelers?**\r\n**Answer:** **Paris** is generally **cheaper** for most travel expenses including **accommodation**, **dining**, and **local transportation**. However, **London** offers **free admission** to major museums (British Museum, National Gallery, Tate Modern), which can offset other costs. Overall, budget travelers often find Paris more economical.\r\n\r\n**f. Where is it easier to get a taxi - London or Paris?**\r\n**Answer:** **London** provides easier taxi access with approximately **22,500 registered black cabs** compared to Paris\'s **17,000 taxis**. London\'s iconic black cabs can be hailed directly from streets throughout central areas, while Parisian taxis primarily operate from designated ranks or require advance booking.\r\n\r\n**g. Personal preference: Which city would you choose to visit?**\r\n**Sample Answer:** I would choose **London** for its incredible **cultural diversity** and **multicultural character**. As a true **melting pot of cultures**, London offers **diverse cuisines**, **languages**, and **traditions** from around the world. From **vibrant Chinatown** to **colorful Brick Lane markets**, the city presents unique **fusion of global cultures**. This diversity extends to its **arts scene** with numerous **theaters**, **galleries**, and **music venues** showcasing **international** and **local talent**. The **English language** accessibility also makes navigation and interaction easier for international visitors.\r\n\r\n## 🎯 Grammar Section: Adjectives vs Adverbs\r\n\r\n### A. Identifying Adjectives and Adverbs\r\n\r\n**Grammar Rule:** \r\n- **Adjectives** describe **nouns** (people, places, things)\r\n- **Adverbs** describe **verbs** (actions), **adjectives**, or other **adverbs**\r\n\r\n| Sentence | Word | Type | Function |\r\n|----------|------|------|----------|\r\n| My **poor** family can\'t afford a car | poor | adjective | describes family |\r\n| Sujata performed **poorly** at the concert | poorly | adverb | describes how she performed |\r\n| Suraj is **calm** under pressure | calm | adjective | describes Suraj\'s nature |\r\n| The cook **patiently** removed the pot | patiently | adverb | describes how cook acted |\r\n| The teacher spoke **loudly** to students | loudly | adverb | describes manner of speaking |\r\n| Sunil walked **quietly** to the kitchen | quietly | adverb | describes way of walking |\r\n| The **quiet** children listened carefully | quiet | adjective | describes the children |\r\n| She **softly** sang a lullaby | softly | adverb | describes singing manner |\r\n| The kitten **timidly** peered out | timidly | adverb | describes looking manner |\r\n\r\n### B. Comparative Forms Practice\r\n\r\n**Comparative Rules:**\r\n- **Short adjectives:** add **-er** (tall → taller)\r\n- **Long adjectives:** use **more** (beautiful → more beautiful)  \r\n- **Adverbs ending in -ly:** use **more** (quickly → more quickly)\r\n- **Irregular forms:** good → better, bad → worse\r\n\r\n| Original Sentence | Corrected Form | Grammar Rule Applied |\r\n|------------------|----------------|---------------------|\r\n| Sanskrit is complicated than Nepali | **more complicated** | Long adjective uses \"more\" |\r\n| Metal is heavy than wood | **heavier** | Short adjective adds \"-er\" |\r\n| Wood is not heavy as iron | **as heavy as** | Equal comparison structure |\r\n| He ran quick than I did | **more quickly** | Adverb with \"-ly\" uses \"more\" |\r\n| Einstein was intelligent person | **an intelligent** | Article needed before adjective |\r\n| My sister speaks English good than Juna | **better** | Irregular comparative form |\r\n| Family car is not fast as sports car | **as fast as** | Equal comparison with \"as...as\" |\r\n| I wrote beautiful than he did | **more beautifully** | Adverb form with \"more\" |\r\n| Sunita danced graceful as Mamata | **as gracefully as** | Adverb in equal comparison |\r\n| Bus journey takes long than plane | **longer** | Comparative form of \"long\" |\r\n\r\n## ✍️ Writing Practice: City Comparison Essay\r\n\r\n### Model Essay: Kathmandu vs Pokhara\r\n\r\n**Title: Two Gems of Nepal: Kathmandu and Pokhara**\r\n\r\n**Kathmandu** and **Pokhara** represent Nepal\'s two most popular destinations, each offering completely different experiences to visitors and residents alike.\r\n\r\n**Kathmandu**, Nepal\'s bustling capital city, sits in the historic **Kathmandu Valley** at 1,400 meters elevation. The city serves as the **political**, **economic**, and **cultural center** of Nepal. It houses numerous **UNESCO World Heritage Sites** including **ancient temples**, **royal palaces**, and **traditional architecture** that showcase centuries of **Newari craftsmanship**. The city buzzes with constant activity - **narrow streets** filled with **motorcycles**, **vendors**, **colorful markets**, and **diverse people** from across Nepal and the world. **Kathmandu** offers rich **cultural experiences** through sites like **Pashupatinath Temple**, **Boudhanath Stupa**, and **Durbar Squares** in Kathmandu, Patan, and Bhaktapur.\r\n\r\n**Pokhara**, in contrast, provides a **peaceful retreat** nestled beside beautiful **Phewa Lake** with the stunning **Annapurna mountain range** as backdrop. Located 200 kilometers west of Kathmandu at 820 meters elevation, this **lakeside city** attracts **nature lovers** and **adventure enthusiasts**. Visitors enjoy **boating on Phewa Lake**, **visiting the World Peace Pagoda**, **paragliding from Sarangkot**, and **exploring nearby caves** like Gupteshwor. **Pokhara** serves as the **gateway** to famous **trekking routes** including **Annapurna Base Camp**, **Poon Hill**, and **Annapurna Circuit**, making it essential stop for **mountain adventurers**.\r\n\r\n**Key differences** emerge clearly between these cities. **Kathmandu** offers **urban energy** and **cultural density** with **historical monuments**, **busy markets**, and **diverse populations** creating vibrant city life. However, it faces challenges including **air pollution**, **traffic congestion**, and **overcrowding**. **Pokhara** provides **natural beauty** and **tranquil atmosphere** with **clean air**, **mountain views**, and **relaxed pace** ideal for **rest** and **outdoor activities**.\r\n\r\n**Transportation** also differs significantly. **Kathmandu** has **extensive bus networks** but suffers from **traffic jams** on narrow roads. **Pokhara** remains **more walkable** with **shorter distances** between attractions and **less congested** streets.\r\n\r\nBoth cities contribute **essential elements** to Nepal\'s tourism appeal. **Kathmandu** provides **cultural depth** and **historical significance**, while **Pokhara** offers **natural splendor** and **adventure opportunities**. **Travelers** seeking **comprehensive Nepal experience** benefit from visiting both destinations to appreciate the country\'s **cultural heritage** and **natural beauty**.\r\n\r\nWhether one prefers **cultural exploration** in **ancient temples** and **bustling markets** or **peaceful relaxation** beside **pristine lakes** with **mountain views**, both **Kathmandu** and **Pokhara** offer **unforgettable experiences** that showcase **Nepal\'s diverse attractions**.\r\n\r\n### Writing Tips for Comparison Essays:\r\n1. **Introduction:** Present both subjects clearly\r\n2. **Body paragraphs:** Compare point by point (location, attractions, culture, etc.)\r\n3. **Use transition words:** however, in contrast, similarly, while, whereas\r\n4. **Provide specific examples:** mention actual places and features\r\n5. **Balanced treatment:** give equal attention to both subjects\r\n6. **Conclusion:** Summarize main differences and their significance\r\n\r\n## 📝 Additional Practice Exercises\r\n\r\n### Vocabulary Practice:\r\n1. Use these words in sentences: **vibrant**, **showcase**, **impeccable**, **gastronomy**\r\n2. Find synonyms for: **iconic**, **bustling**, **cosmopolitan**, **heritage**\r\n3. Create word families: **culture** (cultural, culturally), **history** (historic, historical)\r\n\r\n### Grammar Practice:\r\n1. Convert adjectives to adverbs: **quick** → **quickly**, **careful** → **carefully**\r\n2. Form comparatives: **expensive** → **more expensive**, **good** → **better**\r\n3. Write 5 sentences comparing your city with Kathmandu\r\n\r\n### Reading Comprehension:\r\n1. What makes London multicultural?\r\n2. Why is Paris called the fashion capital?\r\n3. How do transportation systems differ in both cities?\r\n\r\n## ✅ Chapter Summary & Key Points\r\n\r\n### Important Facts to Remember:\r\n- **London** is larger (1,572 km²) than **Paris** (105 km²)\r\n- **London** has more museums (240) than **Paris** (153)\r\n- **Paris** is generally cheaper for travel\r\n- **London** has more taxis (22,500 vs 17,000)\r\n- **Football fans** prefer London (6 clubs vs 1)\r\n- Both cities are **capital cities** with **rich history**\r\n- **Autumn** is best time to visit **London**\r\n\r\n### Grammar Rules:\r\n- **Adjectives** modify nouns: *poor family, calm person*\r\n- **Adverbs** modify verbs: *performed poorly, spoke loudly*\r\n- **Short adjectives** add **-er**: *taller, faster*\r\n- **Long adjectives** use **more**: *more beautiful, more expensive*\r\n- **Equal comparison** uses **as...as**: *as tall as, as quickly as*\r\n\r\n### Writing Skills:\r\n- **Compare** cities point by point\r\n- Use **transition words** for smooth flow\r\n- Provide **specific examples** and details\r\n- **Balance** both subjects equally\r\n- End with **meaningful conclusion**\r\n\r\nThis comprehensive guide covers all aspects of the London vs Paris chapter, providing students with complete understanding of comparative analysis, grammar concepts, and writing skills needed for SEE examinations.\r\n', 1, '2025-07-25 01:11:38', '2025-09-17 09:04:51'),
(135, 11184, '# The Country Mouse & The City Mouse - Grade 10 English Complete Guide\r\n## Unit 17: Countries & Towns - NEB English Class 10\r\n\r\n## 🎯 Chapter Introduction\r\nThis chapter presents a classic fable poem by **Richard Scrafton Sharpe** that compares **country life** and **city life** through the story of two mice friends. The poem teaches important lessons about **contentment**, **simple living**, and **appreciating what we have**. Students will learn about **fable literature**, **moral lessons**, and **lifestyle comparisons**.\r\n\r\n**Main Learning Focus:**\r\n- Understanding fable poems and their moral lessons\r\n- Comparing country vs city lifestyle\r\n- Learning new vocabulary related to rural and urban life\r\n- Grammar: Using connectives properly\r\n- Writing: Descriptive essays about villages\r\n\r\n## 📖 Complete Poem Text\r\n\r\n**The Country Mouse and the City Mouse**\r\n*by Richard Scrafton Sharpe*\r\n\r\n**Stanza 1:**\r\nIn a snug little cot lived a fat little mouse,\r\nWho enjoyed, unmolested, the range of the house;\r\nWith plain food content, she would breakfast on cheese,\r\nShe dined upon bacon, and supped on grey peas.\r\n\r\n**Stanza 2:**\r\nA friend from the town to the cottage did stray,\r\nAnd he said he was come a short visit to pay;\r\nSo the mouse spread her table as gay as you please,\r\nAnd brought the nice bacon and charming grey peas.\r\n\r\n**Stanza 3:**\r\nThe visitor frowned, and he thought to be witty:\r\nCried he, \"You must know, I am come from the city,\r\nWhere we all should be shocked at provisions like these,\r\nFor we never eat bacon and horrid grey peas.\r\n\r\n**Stanza 4:**\r\n\"To town come with me, I will give you a treat:\r\nSome excellent food, most delightful to eat.\r\nWith me shall you feast just as long as you please;\r\nCome, leave this fat bacon and shocking grey peas.\"\r\n\r\n**Stanza 5:**\r\nThis kind invitation she could not refuse,\r\nAnd the city mouse wished not a moment to lose;\r\nReluctant she quitted the fields and the trees,\r\nThe delicious fat bacon and charming grey peas.\r\n\r\n**Stanza 6:**\r\nThey slily crept under a gay parlour door,\r\nWhere a feast had been given the evening before;\r\nAnd it must be confessed they on dainties did seize,\r\nFar better than bacon, or even grey peas.\r\n\r\n**Stanza 7:**\r\nHere were custard and trifle, and cheesecakes good store,\r\nNice sweetmeats and jellies, and twenty things more;\r\nAll that art had invented the palate to please,\r\nExcept some fat bacon and smoking grey peas.\r\n\r\n**Stanza 8:**\r\nThey were nicely regaling, when into the room\r\nCame the dog and the cat, and the maid with a broom:\r\nThey jumped in a custard both up to their knees;\r\nThe country mouse sighed for her bacon and peas.\r\n\r\n**Stanza 9:**\r\nCried she to her friend, \"Get me safely away,\r\nI can venture no longer in London to stay;\r\nFor if oft you receive interruptions like these,\r\nGive me my nice bacon and charming grey peas.\r\n\r\n**Stanza 10:**\r\n\"Your living is splendid and gay, to be sure,\r\nBut the dread of disturbance you ever endure;\r\nI taste true delight in contentment and ease,\r\nAnd I feast on fat bacon and charming grey peas.\"\r\n\r\n## 📝 Story Summary\r\n\r\n### Simple Summary\r\nThis fable tells the story of **two mouse friends** - one living in the **countryside** and another in the **city**. The **country mouse** lives a **simple, peaceful life** eating plain food like **cheese**, **bacon**, and **grey peas** in her comfortable little home. \r\n\r\nThe **city mouse** visits her friend and **criticizes** the simple country food. He **invites** the country mouse to the city, promising **delicious treats** and **luxury food**. The country mouse **accepts** and travels to the city where they feast on **custards**, **trifles**, **cheesecakes**, and other fancy delicacies.\r\n\r\nHowever, their feast is **suddenly interrupted** by **dangerous animals** (dog and cat) and a **maid with a broom**. The country mouse becomes **frightened** and **realizes** that city life, though luxurious, is full of **dangers** and **disturbances**. She decides to **return** to her peaceful country home, preferring her **simple** but **safe** lifestyle.\r\n\r\n### Moral Lesson\r\nThe poem teaches us that **true happiness** comes from **contentment** with what we have, not from seeking **luxury** and **material wealth**. **Simple living** with **peace** and **security** is better than **rich living** with **constant fear** and **danger**.\r\n\r\n## 📚 Detailed Analysis\r\n\r\n### Character Analysis\r\n\r\n#### The Country Mouse\r\n- Lives in a **comfortable cottage** in the countryside\r\n- **Content** with **simple food** (cheese, bacon, grey peas)\r\n- **Peaceful** and **undisturbed** lifestyle\r\n- **Curious** about city life but ultimately **values security**\r\n- Represents **contentment**, **simplicity**, and **wisdom**\r\n\r\n#### The City Mouse  \r\n- Lives in the **city** with access to **luxury food**\r\n- **Critical** of simple country lifestyle\r\n- **Proud** of city amenities and sophistication\r\n- Lives with **constant danger** and **interruptions**\r\n- Represents **materialism**, **sophistication**, but also **insecurity**\r\n\r\n### Theme Analysis\r\n\r\n#### Main Themes:\r\n1. **Contentment vs Greed** - Being happy with what you have vs wanting more\r\n2. **Simple Life vs Luxury** - Plain living vs rich lifestyle\r\n3. **Security vs Danger** - Safe environment vs risky situations\r\n4. **Country vs City** - Rural peace vs urban excitement\r\n5. **True Happiness** - Inner satisfaction vs external possessions\r\n\r\n### Literary Devices\r\n- **Fable** - Story with moral lesson using animals as characters\r\n- **Repetition** - \"bacon and grey peas\" repeated throughout\r\n- **Contrast** - Country simplicity vs city luxury\r\n- **Rhyme Scheme** - AABB pattern makes it musical\r\n- **Personification** - Mice behave like humans\r\n\r\n## 📋 Vocabulary Exercises & Solutions\r\n\r\n### A. Word Matching (with enhanced meanings)\r\n\r\n| Word | Meaning | Context Usage |\r\n|------|---------|---------------|\r\n| **snug** | **iv.** warm, comfortable and protected | The country mouse lived in a **snug** little cottage |\r\n| **unmolested** | **ix.** not disturbed or attacked | She enjoyed **unmolested** peace in the countryside |\r\n| **stray** | **i.** move away aimlessly | The city mouse did **stray** to the country cottage |\r\n| **frown** | **x.** facial expression showing disapproval | The city mouse **frowned** at the simple food |\r\n| **horrid** | **ii.** very unpleasant, rough | He called the grey peas \"**horrid**\" |\r\n| **reluctant** | **iii.** unwilling and hesitant | She was **reluctant** to leave her home |\r\n| **custard** | **v.** sweet yellow sauce from milk and eggs | They enjoyed **custard** at the city feast |\r\n| **trifle** | **vii.** cold dessert of cake and fruit | **Trifle** was served at the grand dinner |\r\n| **regale** | **vi.** entertain or amuse with talk | They began to **regale** themselves with food |\r\n| **endure** | **viii.** deal with something painful | City life requires **enduring** constant dangers |\r\n\r\n### Additional Vocabulary:\r\n- **provisions** - food supplies\r\n- **dainties** - delicate, fancy foods  \r\n- **sweetmeats** - candied fruits or nuts\r\n- **parlour** - formal sitting room\r\n- **venture** - dare to go somewhere risky\r\n\r\n## 📋 Reading Comprehension Exercises\r\n\r\n### B. True/False Statements (with explanations)\r\n\r\n| Statement | Answer | Explanation |\r\n|-----------|--------|-------------|\r\n| The country mouse was happy with her lifestyle | **True** | She lived contentedly with simple food and peaceful environment |\r\n| The country mouse enjoyed grand meals daily | **False** | She ate simple food (cheese, bacon, peas) not grand meals |\r\n| The town mouse wanted better food than served | **True** | He criticized country food and promised city delicacies |\r\n| The town mouse invited country mouse for dinner | **True** | He invited her to experience city food and lifestyle |\r\n| Country mouse was served delicious food in town | **True** | She ate custard, trifles, cheesecakes in the city |\r\n| There was no peace for mice in the town | **True** | They were interrupted by dog, cat, and maid during dinner |\r\n\r\n### C. Detailed Questions & Answers\r\n\r\n**a. How did the country mouse live?**\r\n**Answer:** The country mouse lived in a **small, comfortable cottage** in the **countryside fields**. She had a **simple** but **peaceful lifestyle**, eating **plain food** like cheese for breakfast, bacon for dinner, and grey peas for supper. She enjoyed **complete freedom** in her house without any **disturbances** or **threats**.\r\n\r\n**b. Why did the town mouse visit the country?**\r\n**Answer:** The town mouse came to the country to **pay a short visit** to his **old friend**, the country mouse. He wanted to **spend time** with her and **see** how she lived in the countryside. The visit was meant to be a **friendly social call** between the two mouse friends.\r\n\r\n**c. Was the country mouse happy to receive the invitation? Why?**\r\n**Answer:** Yes, the country mouse was **initially excited** about the invitation to visit the city. She was **curious** about **city life** and **eager** to experience the **excellent food** and **different lifestyle** that the city mouse had described. However, she was also **reluctant** to leave her comfortable home and **familiar surroundings**.\r\n\r\n**d. What happened while the mice were eating dinner?**\r\n**Answer:** While the two mice were **enjoying their feast** of custards, trifles, and other delicacies in the city, they were **suddenly interrupted** by **dangerous situations**. A **dog**, a **cat**, and a **maid carrying a broom** entered the room. The mice had to **jump into custard** up to their knees to **escape** and **hide** from these threats.\r\n\r\n**e. What did the country mouse request from her friend?**\r\n**Answer:** The frightened country mouse **urgently requested** her city friend to **help her escape safely** from the city. She said she **could no longer stay** in the dangerous city environment and wanted to **return immediately** to her **peaceful country home** with her simple bacon and peas.\r\n\r\n### D. Personal Response Question\r\n\r\n**Sample Answer for rural living:**\r\nLiving in a **rural village** provides me with numerous **advantages** and some **challenges**. \r\n\r\n**Advantages:**\r\n- **Fresh, clean air** and **beautiful natural surroundings**\r\n- **Peaceful environment** with **bird songs** and **natural sounds**\r\n- **Close-knit community** with **helpful neighbors**\r\n- **Safe environment** for children to play freely\r\n- **Organic food** from **local farms** and **gardens**\r\n- **Lower cost of living** compared to cities\r\n- **Less traffic** and **pollution**\r\n\r\n**Disadvantages:**\r\n- **Limited access** to **modern facilities** like hospitals and shopping centers\r\n- **Fewer educational** and **employment opportunities**\r\n- **Poor internet** and **communication services**\r\n- **Transportation difficulties** to reach cities\r\n- **Limited entertainment** options\r\n- **Occasional power** and **water supply** problems\r\n\r\nDespite the challenges, **rural life** offers **genuine happiness**, **contentment**, and **connection with nature** that makes it **worthwhile** and **fulfilling**.\r\n\r\n## 🎯 Grammar Section: Connectives\r\n\r\n### Understanding Connectives\r\n**Connectives** are words that **join sentences** or **parts of sentences** together. They show **relationships** between ideas like **cause**, **contrast**, **time**, or **condition**.\r\n\r\n### A. Types of Connectives:\r\n1. **Cause/Reason:** because, since, as\r\n2. **Contrast:** although, however, but, despite\r\n3. **Condition:** if, unless, provided that\r\n4. **Time:** when, while, after, before\r\n5. **Addition:** and, also, furthermore\r\n\r\n### B. Exercise Solutions:\r\n\r\n| Sentence | Correct Connective | Type | Explanation |\r\n|----------|-------------------|------|-------------|\r\n| **Although** it was summer, days were cold | although | contrast | Shows unexpected contrast |\r\n| Stone struck head; **however**, not hurt | however | contrast | Shows surprising result |\r\n| I like popcorn, **but** don\'t like pizza | but | contrast | Shows different preferences |\r\n| **Despite** her age, still working hard | despite | contrast | Shows contrast with expectation |\r\n| **Although** Narayan is skillful, can\'t read well | although | contrast | Contrasts skills |\r\n| **Because** petrol expensive, Brian sold car | because | cause/reason | Shows reason for action |\r\n| Took taxi **because** it was raining | because | cause/reason | Explains reason |\r\n| Flights cancelled **because of** fog | because of | cause/reason | Shows cause (noun phrase) |\r\n| **Despite** feeling unwell, went to school | despite | contrast | Shows determination despite problem |\r\n| Felt cold, **although** wearing winter coat | although | contrast | Unexpected situation |\r\n\r\n### Key Grammar Rules:\r\n- **Because** + subject + verb (Because it was raining...)\r\n- **Because of** + noun/noun phrase (Because of the rain...)\r\n- **Although** + subject + verb (Although she was tired...)\r\n- **Despite/In spite of** + noun/gerund (Despite being tired...)\r\n\r\n## ✍️ Writing Section: Village Description\r\n\r\n### Model Essay: Lamidada Village\r\n\r\n**Lamidada Village: Nepal\'s Hidden Natural Treasure**\r\n\r\n**Lamidada Village**, nestled in the beautiful **Khotang district** of Nepal, represents one of our country\'s most **pristine** and **untouched** rural communities. This remarkable village serves as the **gateway** to the **sacred Halesi Mahadev Temple**, making it both a **spiritual destination** and a **showcase** of **traditional Nepali** village life.\r\n\r\n**Geographic Beauty and Landscape**\r\nThe village is **blessed** with **breathtaking natural beauty** that captures the heart of every visitor. **Surrounded** by **majestic mountains**, **Lamidada** offers **stunning panoramic views** that change with the **seasons** and **time of day**. The **lush agricultural lands** spread across **terraced hillsides**, showcasing the **traditional farming practices** that have **sustained** the community for generations. These **green fields** create a **beautiful patchwork** across the **landscape**, especially during **planting** and **harvest seasons**.\r\n\r\n**Traditional Architecture and Cultural Heritage**\r\nThe village features **authentic traditional houses** built with **local materials** like **stone**, **wood**, and **mud**, reflecting **centuries-old** architectural techniques. These houses are **perfectly adapted** to the **mountain climate** and **earthquake-prone** region. The **narrow stone-paved paths** wind through the village, connecting **homes**, **fields**, and **community spaces** in a **harmonious network** that has **remained unchanged** for **decades**.\r\n\r\n**Community and People**\r\n**Lamidada** is home to **thick settlements** of the **Rai ethnic community**, known for their **warm hospitality** and **strong cultural traditions**. The **humble villagers** maintain their **traditional lifestyle**, living in **harmony** with **nature** and **preserving** their **ancestral customs**. **Walking along** the **stone paths**, visitors encounter **friendly locals** carrying **firewood** from the **dense forests** and **children grazing** **domestic animals** like **goats**, **cows**, and **buffalo** - scenes that **perfectly capture** the **essence** of **rural Nepal**.\r\n\r\n**Natural Environment**\r\nThe village is **surrounded** by **thick forests** that provide **timber**, **medicinal plants**, and **wildlife habitat**. These **forests** also serve as **natural protection** against **soil erosion** and **maintain** the **local ecosystem**. The **dense vegetation** creates a **cool microclimate** and provides **fresh**, **clean air** that **contrasts sharply** with **polluted urban environments**.\r\n\r\n**Climate and Weather**\r\n**Lamidada** experiences a **pleasant mountain climate** with **distinct seasonal variations**. **Mornings** can be **slightly chilly**, especially during **winter months**, requiring **warm clothing** and **traditional heating methods**. However, **daytime temperatures** become **comfortably warm**, making it **ideal** for **outdoor activities** and **exploration**. This **temperature variation** creates **perfect conditions** for **agriculture** and makes the village **attractive** to **visitors** seeking **escape** from **extreme weather**.\r\n\r\n**Cultural and Spiritual Significance**\r\nAs the **gateway** to **Halesi Mahadev Temple**, **Lamidada** holds **special religious importance** for **Hindu pilgrims** from across **Nepal** and **India**. The village **benefits** from this **spiritual tourism** while **maintaining** its **traditional character**. **Local festivals** and **religious ceremonies** continue throughout the year, providing **opportunities** for **visitors** to **experience** **authentic** **Nepali culture**.\r\n\r\n**Conclusion**\r\n**Lamidada Village** stands as a **perfect example** of **Nepal\'s hidden treasures** - places where **modernity** has **not yet disturbed** the **natural beauty** and **traditional lifestyle** that define our **rural heritage**. For **travelers** seeking **authentic experiences**, **peaceful environments**, and **connection** with **traditional culture**, **Lamidada** offers an **unforgettable journey** into the **heart** of **rural Nepal**. The village **reminds** us of the **importance** of **preserving** these **precious communities** that **maintain** our **cultural identity** and **environmental balance** in an **increasingly modern world**.\r\n\r\n### Writing Tips for Village Descriptions:\r\n1. **Start** with **location** and **general overview**\r\n2. **Describe** **physical features** (mountains, fields, houses)\r\n3. **Include** **people** and **their activities**\r\n4. **Mention** **climate** and **natural environment**\r\n5. **Add** **cultural** and **historical significance**\r\n6. **Use** **descriptive adjectives** and **sensory details**\r\n7. **End** with **personal reflection** or **broader significance**\r\n\r\n## ✅ Key Learning Points\r\n\r\n### Moral Lessons from the Poem:\r\n- **Contentment** brings true happiness\r\n- **Simple living** can be more satisfying than luxury\r\n- **Security** and **peace** are valuable than material wealth\r\n- **Appreciate** what you have instead of always wanting more\r\n- **Different lifestyles** suit different people\r\n\r\n### Literary Understanding:\r\n- **Fables** teach moral lessons through animal characters\r\n- **Repetition** creates rhythm and emphasis in poetry\r\n- **Contrast** helps highlight differences between choices\r\n- **Rhyme schemes** make poems memorable and musical\r\n\r\n### Grammar Skills:\r\n- **Connectives** show relationships between ideas\r\n- **Different connectives** express cause, contrast, condition, time\r\n- **Proper usage** makes writing clearer and more logical\r\n\r\n### Writing Skills:\r\n- **Descriptive writing** uses specific details and sensory information\r\n- **Good organization** moves from general to specific information\r\n- **Personal observations** make descriptions more engaging\r\n- **Cultural context** adds depth to place descriptions\r\n\r\nThis comprehensive guide provides everything needed for understanding the poem, mastering the exercises, and developing related language skills for Grade 10 English students.\r\n', 1, '2025-07-25 01:11:38', '2025-09-17 09:04:51');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(136, 11185, '# Hachi: A Dog\'s Tale - Grade 10 English Complete Guide\r\n## Unit 18: Media & Entertainment - NEB English Class 10\r\n\r\n## 🎯 Chapter Introduction\r\nThis chapter focuses on **\"Hachi: A Dog\'s Tale\"** (2009), a heartwarming drama film directed by **Lasse Hallström** and starring **Richard Gere**. Based on the true story of **Hachikō**, a loyal Japanese Akita dog, this film explores themes of **unconditional love**, **loyalty**, **devotion**, and the **special bond** between humans and animals. Students will learn about **film analysis**, **media literacy**, and **emotional storytelling**.\r\n\r\n**Main Learning Focus:**\r\n- Understanding film themes and character development\r\n- Analyzing media representation and storytelling techniques\r\n- Learning vocabulary related to films and emotions\r\n- Grammar: Question formation and sentence structure\r\n- Writing: Film reviews and critical analysis\r\n\r\n## 📖 Movie Plot Summary\r\n\r\n### Basic Story Overview\r\n**Professor Parker Wilson** (Richard Gere) discovers a **lost Akita puppy** at **Bedridge train station** while commuting home from work. Despite his wife **Cate\'s** initial reluctance, Parker brings the puppy home temporarily. The dog\'s collar has a **Japanese character** meaning **\"eight\" (Hachi)**, so Parker names him **Hachi**.\r\n\r\n**Hachi** and **Parker** develop an extraordinary bond. Every morning, **Hachi** walks with **Parker** to the train station, and every evening at **5:00 PM**, he returns to wait for Parker\'s arrival. This routine continues for years, creating a **beautiful daily ritual** that touches everyone at the station.\r\n\r\n**Tragically**, one day **Parker** suffers a **fatal heart attack** at work and never returns home. However, **Hachi** continues to wait at the station **every single day** for **nearly ten years**, demonstrating **unwavering loyalty** and **unconditional love** until his own death.\r\n\r\n### Character Analysis\r\n\r\n#### Parker Wilson (Richard Gere)\r\n- **University music professor** who commutes daily by train\r\n- **Kind-hearted** and **compassionate** person who rescues Hachi\r\n- Develops **deep emotional connection** with his dog\r\n- Represents **human capacity** for **love** and **care**\r\n\r\n#### Hachi (the Akita dog)\r\n- **Central character** representing **pure loyalty** and **devotion**\r\n- Shows **intelligence** and **emotional depth**\r\n- Becomes **symbol** of **unconditional love**\r\n- Based on **real Japanese dog** Hachikō\r\n\r\n#### Cate Wilson (Joan Allen)\r\n- **Parker\'s wife** who initially opposes keeping the dog\r\n- **Eventually accepts** Hachi as family member\r\n- Represents **practical** but **caring** family approach\r\n\r\n### Themes and Messages\r\n\r\n#### Primary Themes:\r\n1. **Unconditional Love** - Love without conditions or expectations\r\n2. **Loyalty and Devotion** - Staying faithful despite hardships\r\n3. **Human-Animal Bond** - Special relationships between people and pets\r\n4. **Family Values** - Importance of love and commitment in families\r\n5. **Community Support** - How communities care for those in need\r\n6. **Life and Death** - Dealing with loss and grief\r\n7. **True Friendship** - Genuine connections that last forever\r\n\r\n#### Cultural Significance:\r\nThe film is based on the **true story** of **Hachikō**, a famous dog in **Japan** who waited for his deceased owner for nearly **10 years** at **Shibuya Station** in **Tokyo**. A **bronze statue** of Hachikō stands at the station as a **symbol of loyalty**.\r\n\r\n## 📝 Vocabulary Development\r\n\r\n### A. Key Terms with Definitions\r\n\r\n| Word | Definition | Usage in Film Context |\r\n|------|------------|----------------------|\r\n| **Unwavering** | Never changing or becoming weaker | Hachi showed **unwavering** loyalty to Parker |\r\n| **Devotion** | Great love, care, and support | Hachi\'s **devotion** touched everyone\'s hearts |\r\n| **Commuter** | Person who travels regularly to work | Parker was a daily **commuter** on the train |\r\n| **Vendor** | Person who sells goods or services | Station **vendors** fed Hachi daily |\r\n| **Adaptation** | Changing something for new situation | The film is an **adaptation** of Japanese story |\r\n| **Reinforce** | To strengthen with additional support | The story **reinforces** values of loyalty |\r\n| **Passionate** | Having strong feelings | Parker was **passionate** about music |\r\n| **Cinematic** | Relating to films and filmmaking | The movie uses beautiful **cinematic** techniques |\r\n| **Secular** | Not connected with religion | The story has **secular** moral lessons |\r\n\r\n### Additional Film Vocabulary:\r\n- **Protagonist** - main character (Parker and Hachi)\r\n- **Heartwarming** - causing feelings of happiness and love\r\n- **Poignant** - evoking sadness and sympathy\r\n- **Symbolism** - using objects to represent ideas\r\n- **Emotional** - relating to feelings and emotions\r\n- **Tragedy** - sad or unfortunate events\r\n- **Legacy** - something left behind after death\r\n- **Inspiration** - something that motivates or encourages\r\n\r\n## 📋 Reading Comprehension Exercises\r\n\r\n### B. True/False Statements (with explanations)\r\n\r\n| Statement | Answer | Explanation |\r\n|-----------|--------|-------------|\r\n| Laura Busch is a character in the movie | **False** | There is no character named Laura Busch; main characters are Parker, Cate, and Andy Wilson |\r\n| The dog symbolizes love, devotion, and loyalty | **True** | Hachi represents pure love and unwavering loyalty through his actions |\r\n| The story is based on real events | **True** | Based on true story of Hachikō, famous loyal dog in Japan |\r\n| Traditional marriage values are a weakness | **False** | The film positively portrays marriage and family relationships |\r\n| Bond between dog and master evokes emotions | **True** | The relationship creates strong emotional responses in viewers |\r\n| This film is different from other family films | **True** | The unique story and emotional depth distinguish it from typical family movies |\r\n\r\n### C. Detailed Comprehension Questions & Answers\r\n\r\n**a. What is the dog\'s name in the movie?**\r\n**Answer:** The dog\'s name is **Hachi** (short for **Hachiko**). The name comes from the **Japanese character** on his collar meaning \"**eight**,\" which is considered a **lucky number** in Japanese culture. Parker\'s Japanese friend **Ken** explains this meaning, leading Parker to name the dog Hachi.\r\n\r\n**b. Who directed the movie?**\r\n**Answer:** The movie was directed by **Lasse Hallström**, a **Swedish filmmaker** known for directing emotional dramas like **\"What\'s Eating Gilbert Grape\"** and **\"Chocolat.\"** Hallström brought **sensitivity** and **emotional depth** to this **heartwarming story** about **loyalty** and **love**.\r\n\r\n**c. How was the relationship between dog and master?**\r\n**Answer:** The relationship between **Hachi** and **Parker** was one of **profound emotional attachment**, **mutual love**, and **complete trust**. They developed a **daily routine** where Hachi would **accompany Parker** to the train station each morning and **wait for his return** every evening at 5:00 PM. Their bond went **beyond typical pet-owner relationships** to become a **deep friendship** based on **unconditional love** and **understanding**. Hachi seemed to **understand Parker\'s emotions** and **schedule**, while Parker treated Hachi as a **beloved family member**.\r\n\r\n**d. Why was a statue made of the dog?**\r\n**Answer:** A **statue** was erected in **Hachi\'s honor** to **commemorate** his **extraordinary loyalty** and **devotion** to his master. The statue serves as a **permanent memorial** celebrating **Hachi\'s unwavering faithfulness** - waiting at the station for **nearly ten years** after Parker\'s death. It represents **universal values** of **loyalty**, **love**, and **dedication** that inspire people worldwide. The statue also honors the **true story** of the original **Japanese Hachikō**, whose statue stands at **Shibuya Station** in **Tokyo**.\r\n\r\n**e. What are two weaknesses of the film?**\r\n**Possible weaknesses mentioned:**\r\n1. **Overly sentimental portrayal** - Some viewers might find certain **emotional scenes** too **manipulative** or **excessively tearful**, potentially making the film feel **overly dramatic**.\r\n2. **Forced symbolism** - The emphasis on the **number \"8\"** on Hachi\'s collar and its **lucky significance** might seem **unnecessary** or **artificially inserted** to some audiences who prefer **more subtle storytelling**.\r\n\r\n**f. What aspects touch the audience\'s hearts?**\r\n**Answer:** Several aspects create **deep emotional impact**:\r\n- **Hachi\'s unwavering loyalty** - continuing to wait **daily** for Parker despite his death\r\n- **The daily routine** - the **beautiful morning** and **evening ritual** at the station\r\n- **Community response** - how **townspeople** care for Hachi and are moved by his devotion\r\n- **Parker\'s love** for Hachi - the **genuine affection** and **care** shown throughout\r\n- **Universal themes** - everyone can relate to **love**, **loss**, and **loyalty**\r\n- **Real story basis** - knowing this actually happened makes it more powerful\r\n- **Visual storytelling** - beautiful cinematography showing the passage of time and seasons\r\n\r\n### D. Discussion Questions\r\n\r\n**Sample Questions for Screenwriter:**\r\n1. **\"What inspired you to adapt the Japanese story for American audiences?\"**\r\n2. **\"How did you ensure accurate portrayal of Hachi\'s emotional depth?\"**\r\n3. **\"What challenges did you face showing the bond between Parker and Hachi?\"**\r\n4. **\"Why did you choose to use the grandson\'s narration framework?\"**\r\n5. **\"How did you balance emotional impact with realistic storytelling?\"**\r\n\r\n## 🎯 Grammar Section: Question Formation\r\n\r\n### Understanding Question Types\r\nQuestions help us **gather information**, **start conversations**, and **show interest** in others. English has different **question patterns** depending on what information we want.\r\n\r\n### A. Sentence Reordering (Question Formation)\r\n\r\n**Word Order Rule:** **Question word + auxiliary verb + subject + main verb + object**\r\n\r\n| Scrambled Words | Correct Question | Grammar Pattern |\r\n|----------------|------------------|-----------------|\r\n| feel / did / on / plane / how / you / the | **How did you feel on the plane?** | How + did + you + feel + on the plane |\r\n| doing / kids / what / the / are | **What are the kids doing?** | What + are + the kids + doing |\r\n| phone / didn\'t / why / me / you | **Why didn\'t you phone me?** | Why + didn\'t + you + phone + me |\r\n| find / what / he / beach / did / the / on | **What did he find on the beach?** | What + did + he + find + on the beach |\r\n| breakfast / have / your / do / for / parents / what | **What do your parents have for breakfast?** | What + do + your parents + have + for breakfast |\r\n| weather / will / like / tomorrow / what / be / the | **What will the weather be like tomorrow?** | What + will + the weather + be like + tomorrow |\r\n| nice / have / they / time / did / a | **Did they have a nice time?** | Did + they + have + a nice time |\r\n| hotel / near / your / the / beach / was | **Was your hotel near the beach?** | Was + your hotel + near the beach |\r\n| Italy / Rita / visited / has / ever | **Has Rita ever visited Italy?** | Has + Rita + ever + visited + Italy |\r\n| fishing / have / ever / gone / you | **Have you ever gone fishing?** | Have + you + ever + gone + fishing |\r\n\r\n### B. Situational Questions\r\n\r\n**Forming appropriate questions for different situations:**\r\n\r\n| Situation | Appropriate Question | Purpose |\r\n|-----------|---------------------|---------|\r\n| Inviting someone to movies | **Would you like to go to a movie?** | Polite invitation |\r\n| Asking for directions | **Do you know where Simran lives?** | Requesting information |\r\n| Home invitation | **Would you like to come to my home?** | Friendly invitation |\r\n| Asking permission on bus | **May I take this seat?** | Polite permission request |\r\n| Looking for lost pet | **Have you seen my pet dog?** | Seeking help/information |\r\n| Asking about possessions | **How many books do you have, Bharat?** | Gathering specific information |\r\n| Frequency question | **Does your father cook every day or once a week?** | Asking about habits |\r\n| Weather information | **What is the weather forecast for tomorrow?** | Requesting prediction |\r\n\r\n### C. Conversation Completion\r\n\r\n**Sample dialogue with question formation:**\r\n\r\n| Speaker | Question/Response | Grammar Focus |\r\n|---------|------------------|---------------|\r\n| Pristine: | **Where do you live?** | Location question |\r\n| Mr. Waiba: | I live in Dhadingbensi. | Simple response |\r\n| Mr. Waiba: | **How old are you?** | Age question |\r\n| Pristine: | I am sixteen. | Numerical response |\r\n| Pristine: | **Did you go to school?** | Past tense yes/no question |\r\n| Mr. Waiba: | Yes, I went to school. | Affirmative response |\r\n| Mr. Waiba: | **What are your hobbies?** | Interest question |\r\n| Pristine: | I don\'t have any really. | Negative response |\r\n| Mr. Waiba: | **What is your school\'s name?** | Specific information |\r\n| Pristine: | It\'s Neelakantha Secondary School. | Proper noun response |\r\n| Pristine: | **Do you dislike anything about school?** | Opinion question |\r\n| Mr. Waiba: | Nothing. I like everything. | Positive response |\r\n\r\n### Question Formation Rules:\r\n\r\n#### Yes/No Questions:\r\n- **Present:** Do/Does + subject + verb? (Do you like movies?)\r\n- **Past:** Did + subject + verb? (Did you watch the film?)\r\n- **Future:** Will + subject + verb? (Will you come tomorrow?)\r\n\r\n#### WH-Questions:\r\n- **What** - asks for information (What is your name?)\r\n- **Where** - asks for location (Where do you live?)\r\n- **When** - asks for time (When did you arrive?)\r\n- **Why** - asks for reason (Why are you sad?)\r\n- **How** - asks for manner (How do you feel?)\r\n- **Who** - asks for person (Who is your teacher?)\r\n\r\n## ✍️ Writing Section: Film Review\r\n\r\n### Sample Film Review: Hachi - A Dog\'s Tale\r\n\r\n**Hachi: A Dog\'s Tale - A Masterpiece of Loyalty and Love**\r\n\r\n**Movie Details:**\r\n- **Director:** Lasse Hallström\r\n- **Genre:** Drama/Family\r\n- **Duration:** 93 minutes\r\n- **Main Cast:** Richard Gere, Joan Allen, Cary-Hiroyuki Tagawa\r\n- **Release Year:** 2009\r\n- **Rating:** PG (suitable for all ages)\r\n\r\n**Plot Overview:**\r\n**\"Hachi: A Dog\'s Tale\"** presents the **extraordinary true story** of an **Akita dog** and his **unbreakable bond** with **university professor Parker Wilson**. When Parker discovers **abandoned puppy Hachi** at a **train station**, neither realizes they\'re beginning a **relationship** that will **define loyalty** for generations.\r\n\r\nThe film follows their **daily routine** - **morning walks** to the station, **evening reunions**, and the **growing love** between man and dog. However, when **tragedy strikes** and **Parker dies unexpectedly**, **Hachi\'s response** creates one of **cinema\'s most moving** examples of **devotion**: **waiting at the station daily** for **nearly ten years**.\r\n\r\n**Strengths of the Film:**\r\n**Emotional Authenticity** - The film creates **genuine emotional connections** without feeling **manipulative**. **Richard Gere** delivers a **subtle**, **heartfelt performance** that **perfectly captures** a **dog lover\'s** **joy** and **affection**.\r\n\r\n**Visual Storytelling** - **Lasse Hallström\'s direction** uses **beautiful cinematography** to show **seasonal changes** and **passage of time**, **emphasizing** the **endurance** of **Hachi\'s vigil**. The **train station** becomes a **character itself**, **witnessing** **love**, **loss**, and **unwavering faith**.\r\n\r\n**Universal Themes** - While **culturally specific** to **Japanese origins**, the story **translates beautifully** to **American setting**, **proving** that **loyalty** and **love** **transcend** **cultural boundaries**.\r\n\r\n**Family Values** - The film **positively portrays** **marriage**, **family relationships**, and **community support**, making it **ideal** for **family viewing**.\r\n\r\n**Areas for Improvement:**\r\nSome viewers might find **certain scenes** **overly sentimental**, particularly **emotional sequences** designed to **evoke tears**. The **pacing occasionally slows** during **middle sections**, which might **test patience** of **younger audiences**.\r\n\r\n**Educational Value:**\r\nThe film provides **excellent opportunities** for **discussing**:\r\n- **Responsibility** of **pet ownership**\r\n- **Dealing** with **grief** and **loss**\r\n- **Importance** of **loyalty** in **relationships**\r\n- **Cultural** **differences** and **similarities**\r\n- **Community** **support** during **difficult times**\r\n\r\n**Technical Aspects:**\r\n**Jan A.P. Kaczmarek\'s musical score** **enhances** **emotional moments** **without overwhelming** them. **Dog training** and **animal performances** appear **natural** and **authentic**, **crucial** for **believability**.\r\n\r\n**Cultural Impact:**\r\nThe film **introduces** **Western audiences** to the **famous Japanese story** of **Hachikō**, **promoting** **cross-cultural understanding** and **appreciation** for **different** **storytelling traditions**.\r\n\r\n**Recommendation:**\r\n**\"Hachi: A Dog\'s Tale\"** **deserves** **strong recommendation** for **families**, **dog lovers**, and **anyone** **seeking** **meaningful** **entertainment**. Despite **minor pacing issues**, the film\'s **emotional impact**, **positive messages**, and **beautiful storytelling** make it a **valuable addition** to **family movie collections**.\r\n\r\n**Final Rating: 4.5/5 stars**\r\n\r\n**Perfect for:** Family movie nights, discussions about loyalty and love, introducing children to themes of responsibility and loss.\r\n\r\n### Film Review Writing Tips:\r\n1. **Start** with **basic movie information** (director, cast, genre)\r\n2. **Provide** **brief plot summary** without **major spoilers**\r\n3. **Analyze** **strengths** and **weaknesses** objectively\r\n4. **Discuss** **themes** and **messages**\r\n5. **Consider** **technical aspects** (music, cinematography)\r\n6. **Give** **clear recommendation** with **reasons**\r\n7. **End** with **rating** or **final thoughts**\r\n\r\n## ✅ Key Learning Points\r\n\r\n### Film Analysis Skills:\r\n- **Understanding** **plot structure** and **character development**\r\n- **Identifying** **themes** and **moral messages**\r\n- **Recognizing** **cultural** **significance** and **adaptation**\r\n- **Evaluating** **technical** **aspects** of **filmmaking**\r\n\r\n### Emotional Intelligence:\r\n- **Recognizing** **different types** of **love** and **loyalty**\r\n- **Understanding** **grief** and **loss** **processes**\r\n- **Appreciating** **human-animal relationships**\r\n- **Developing** **empathy** through **storytelling**\r\n\r\n### Grammar Skills:\r\n- **Forming** **questions** **correctly** in **different tenses**\r\n- **Using** **appropriate** **question words** for **specific information**\r\n- **Understanding** **formal** vs **informal** **question structures**\r\n- **Creating** **conversations** through **question-answer patterns**\r\n\r\n### Writing Skills:\r\n- **Organizing** **film reviews** **logically**\r\n- **Balancing** **plot summary** with **critical analysis**\r\n- **Supporting** **opinions** with **specific examples**\r\n- **Writing** for **different audiences** and **purposes**\r\n\r\n### Cultural Awareness:\r\n- **Understanding** **cross-cultural** **storytelling**\r\n- **Appreciating** **universal** **human values**\r\n- **Recognizing** **how** **stories** **travel** **between cultures**\r\n- **Respecting** **different** **cultural** **perspectives** on **relationships**\r\n\r\nThis comprehensive guide provides Grade 10 students with all necessary tools for understanding the film, completing exercises, and developing related language and analytical skills essential for SEE examinations.\r\n', 1, '2025-07-25 01:11:38', '2025-09-17 09:13:06'),
(137, 11186, '# Interesting Social Media Stories - Grade 10 English Complete Guide\r\n## Unit 18: Media & Entertainment - NEB English Class 10\r\n\r\n## 🎯 Chapter Introduction\r\nThis chapter explores **positive social media stories** that show how **digital platforms** can bring **people together**, **help others**, and create **meaningful connections**. Students will learn about **social media vocabulary**, **real-life inspiring stories**, and how **technology** can be used for **good purposes**. The chapter also covers **causative verbs** and **essay writing** about **social media impacts**.\r\n\r\n**Main Learning Focus:**\r\n- Understanding positive aspects of social media\r\n- Learning social media and technology vocabulary\r\n- Reading comprehension of inspiring real stories\r\n- Grammar: Causative verbs (make, let, have, get)\r\n- Writing: Argumentative essays about social media effects\r\n\r\n## 📖 Text Overview: Positive Social Media Stories\r\n\r\n### Story 1: Nathan Apodaca\'s Viral Video\r\n**Nathan Apodaca** became famous when he posted a video on **TikTok** showing himself **skateboarding** to work while **drinking Ocean Spray cranberry juice** and **lip-syncing** to **\"Dreams\" by Fleetwood Mac**. The video went **viral** with over **40 million views**. As a reward for this free advertising, **Ocean Spray** delivered a **truck full of cranberry juice** to Nathan, making his simple video a life-changing moment.\r\n\r\n### Story 2: Charlie\'s Special Birthday\r\n**Charlie**, a young girl, was having a difficult time, so her **aunt posted on Facebook** asking people to make **Charlie\'s birthday special** by sending cards. The social media post **spread widely**, and **Charlie received 173 birthday cards** and **15 packages** from **total strangers** around the world, making her day **extra memorable**.\r\n\r\n### Story 3: Dog Wheelchair Fundraising\r\n**Matt Nason**, creator of the popular **\"We Rate Dogs\" Twitter page**, shared a **GoFundMe campaign** for a **12-year-old Golden Retriever** who needed a **wheelchair**. Thanks to social media sharing, the campaign **raised over $700 in just a few hours**, allowing the elderly dog to get the mobility assistance it needed.\r\n\r\n### Story 4: Hawk Rescue in New York\r\nWhen **Madeleine Weatherhead** found an **injured hawk** in New York, she **posted a photo on Twitter** asking for help. Her social media post quickly reached the right people, and **Special Operations Officers Maxwell Outsen and Joseph Bellomo** arrived to rescue the hawk, taking it to an **animal sanctuary** where it **fully recovered**.\r\n\r\n## 📝 Vocabulary Development\r\n\r\n### A. Word Matching Exercises\r\n\r\n| Word | Correct Meaning | Enhanced Context |\r\n|------|----------------|------------------|\r\n| **lip-syncing** | **iii.** to silently mouth words along with recording | Nathan was **lip-syncing** to Fleetwood Mac while skateboarding |\r\n| **platform** | **iv.** website serving as base for services | TikTok is a popular **platform** for short videos |\r\n| **pandemic** | **i.** worldwide spread of new disease | COVID-19 **pandemic** increased social media usage |\r\n| **overwhelm** | **ii.** to have strong emotional effect | Charlie was **overwhelmed** by all the birthday cards |\r\n| **somersaulting** | **vii.** turning over completely in air | Children were filmed **somersaulting** at school |\r\n| **accumulate** | **vi.** to get more and more over time | The fundraising campaign began to **accumulate** donations |\r\n| **voicemail** | **viii.** recorded phone message when no answer | He left a **voicemail** thanking supporters |\r\n| **sanctuary** | **v.** place where animals are protected | The rescued hawk was taken to an animal **sanctuary** |\r\n\r\n### Additional Social Media Vocabulary:\r\n- **viral** - spreading rapidly on internet\r\n- **trending** - currently popular on social media\r\n- **hashtag** - symbol (#) used to categorize posts\r\n- **influencer** - person with large social media following\r\n- **content** - material posted on social media\r\n- **engagement** - interactions (likes, comments, shares)\r\n- **follower** - person who subscribes to your account\r\n- **algorithm** - system that decides what content to show\r\n\r\n## 📋 Reading Comprehension Exercises\r\n\r\n### B. Text Completion (Enhanced Answers)\r\n\r\n| Sentence | Correct Answer | Enhanced Explanation |\r\n|----------|---------------|---------------------|\r\n| The text is about positive results brought by | **social media** | The chapter focuses on **beneficial** and **inspiring** outcomes of social media usage |\r\n| Nathan was gifted a truck because he was drinking | **Ocean Spray cranberry juice** | His viral video featured this brand prominently, creating valuable advertising |\r\n| Charlie\'s birthday became special because of post on | **Facebook** | Her aunt\'s Facebook post requesting birthday cards went viral |\r\n| Without social media, people wouldn\'t know about | **somersaulting** | Two school children\'s athletic activities were shared through social platforms |\r\n| Matt Nason\'s Twitter page raises funds | **\'We Rate Dogs\'** | This popular account helps dogs in need through crowdfunding |\r\n\r\n### C. Detailed Reading Questions & Answers\r\n\r\n**a. What did Nathan Apodaca record in his video?**\r\n**Answer:** Nathan Apodaca recorded himself **skateboarding to work** on a busy road while **casually drinking Ocean Spray cranberry juice** directly from the bottle and **lip-syncing** to the classic song **\"Dreams\" by Fleetwood Mac**. The video captured his **relaxed, carefree attitude** and **positive energy** during his daily commute, creating a **simple but captivating moment** that resonated with millions of viewers worldwide.\r\n\r\n**b. Was the Ocean Spray truck a deserved reward? Why?**\r\n**Answer:** Yes, the **truck full of Ocean Spray cranberry juice** was absolutely a **well-deserved reward** for Nathan Apodaca. His **authentic, unsponsored video** provided **Ocean Spray** with **incredible free advertising** - the video gained **over 40 million views**, **significantly increasing brand awareness** and **boosting sales**. This **organic product placement** was far more valuable than any paid advertisement could have been, as it felt **genuine** and **relatable** to viewers. Ocean Spray recognized this **mutual benefit** and rewarded Nathan\'s **unintentional brand promotion** with a **generous gift** that showed **appreciation** for his **positive impact**.\r\n\r\n**c. How did social media help Charlie celebrate her birthday?**\r\n**Answer:** Social media transformed **Charlie\'s birthday** from an **ordinary day** into an **extraordinary celebration** through the **power of community sharing**. When Charlie\'s **aunt posted on Facebook** requesting that people **send birthday cards** to make Charlie\'s day special, the post **resonated with users** who **shared it widely** across various platforms. This **viral sharing** reached **hundreds of strangers** who felt **moved to participate**. As a result, **Charlie received 173 birthday cards** and **15 packages** from **people around the world** who had never met her but wanted to **spread kindness**. Social media enabled **global strangers** to become **temporary family members** celebrating Charlie\'s special day.\r\n\r\n**d. What was $700 spent on and how was money collected?**\r\n**Answer:** The **$700** was specifically **raised** and **spent** to **purchase a wheelchair** for a **12-year-old Golden Retriever** who was **experiencing mobility difficulties** due to **age** and **health issues**. **Matt Nason**, who runs the extremely popular **\"We Rate Dogs\" Twitter account** with millions of followers, **shared the dog\'s GoFundMe campaign** page with his **massive audience**. He **asked his followers to help** this **senior dog** regain **mobility** and **quality of life**. The **social media sharing** was **incredibly effective** - the campaign **raised the full $700** in **just a few hours**, demonstrating how **social media communities** can **quickly mobilize** to **help animals in need**.\r\n\r\n**e. How was the New York hawk rescued?**\r\n**Answer:** The **hawk rescue** demonstrated **perfect coordination** between **social media awareness** and **professional animal rescue services**. **Madeleine Weatherhead** discovered an **injured hawk** in **New York City** and **immediately posted a photograph** on her **Twitter account**, **asking her followers for help** in **rescuing the bird**. Her **urgent plea** and **compelling image** **quickly spread** through Twitter, **reaching the right people** who could **take action**. **Within a short time**, **Special Operations Officers Maxwell Outsen and Joseph Bellomo** **arrived at the scene**, **carefully captured the injured hawk** using **proper wildlife handling techniques**, and **transported it safely** to an **animal sanctuary**. At the sanctuary, the hawk **received professional medical care**, **made a complete recovery**, and **found a safe permanent home**. This story shows how **social media** can **connect people in emergencies** with **professional help**.\r\n\r\n## 🎯 Grammar Section: Causative Verbs\r\n\r\n### Understanding Causative Verbs\r\n**Causative verbs** (make, let, have, get) show that **someone causes** or **allows** something to happen. They express **relationships** between people and **actions**.\r\n\r\n### Types of Causative Verbs:\r\n\r\n#### 1. MAKE (force/cause someone to do something)\r\n**Pattern:** Subject + make + object + base verb\r\n- **Example:** The teacher **made** the students **clean** the classroom.\r\n\r\n#### 2. LET (allow someone to do something) \r\n**Pattern:** Subject + let + object + base verb\r\n- **Example:** Parents **let** children **play** outside.\r\n\r\n#### 3. HAVE (arrange for someone to do something)\r\n**Pattern:** Subject + have + object + past participle (for services)\r\n**Pattern:** Subject + have + object + base verb (for people)\r\n- **Example:** I **had** my car **repaired**. / I **had** John **repair** my car.\r\n\r\n#### 4. GET (persuade/arrange for someone to do something)\r\n**Pattern:** Subject + get + object + to + base verb\r\n**Pattern:** Subject + get + object + past participle (for services)\r\n- **Example:** She **got** him **to help** her. / She **got** her hair **cut**.\r\n\r\n### A. Causative Verb Exercise Solutions\r\n\r\n| Sentence | Correct Answer | Grammar Rule Applied |\r\n|----------|---------------|---------------------|\r\n| I don\'t let my children _____ video games | **play** | LET + object + base verb |\r\n| Police made him _____ for damage | **pay** | MAKE + object + base verb |\r\n| Rob is having his house _____ next weekend | **painted** | HAVE + object + past participle (service) |\r\n| Julia got her boyfriend _____ her a ring | **to buy** | GET + object + TO + base verb |\r\n| Sam gets me _____ grocery every weekend | **to deliver** | GET + object + TO + base verb |\r\n| Please try not make her _____ | **cry** | MAKE + object + base verb |\r\n| This quiz can make you _____ causatives better | **understand** | MAKE + object + base verb |\r\n| Laura never has people _____ her what to do | **tell** | HAVE + object + base verb |\r\n| Friend got me _____ with her CV | **to help** | GET + object + TO + base verb |\r\n| I had plumber _____ plumbing this morning | **fix** | HAVE + object + base verb (person) |\r\n| I had my hair _____ once a month | **trimmed** | HAVE + object + past participle (service) |\r\n| After party, I was made _____ all dishes | **to wash** | Passive MAKE + TO + base verb |\r\n\r\n### B. Common Causative Patterns:\r\n\r\n#### Active vs Passive Voice:\r\n- **Active:** The teacher **made** students study.\r\n- **Passive:** Students **were made to** study.\r\n\r\n#### Service vs Person:\r\n- **Service:** I **had** my car **washed** (by someone).\r\n- **Person:** I **had** John **wash** my car.\r\n\r\n### C. Holiday Camp Experience (Sample Answer)\r\n\r\n**Enhanced Personal Experience with Causative Verbs:**\r\n\r\nDuring my **holiday camp experience** last year, I encountered numerous situations involving **causative verbs**:\r\n\r\n**MAKE (forced/required activities):**\r\nThe camp counselors **made us wake up** at **6:00 AM every morning** for **morning exercises**. We were **made to participate** in **daily team-building activities** that helped us **develop cooperation skills**. The strict schedule **made us become** more **disciplined** and **organized**.\r\n\r\n**LET (permitted activities):**\r\nFortunately, the camp staff **let us choose** our own **activity groups** based on **personal interests**. They **let us stay up late** on **Friday nights** for **special campfire** stories and **singing sessions**. During **free time**, counselors **let us explore** the **nearby forest** with proper supervision.\r\n\r\n**HAVE (arranged services):**\r\nThe camp **had professional instructors teach us** various **outdoor skills** like **rock climbing** and **canoeing**. We **had our meals prepared** by **camp cooks** who made **delicious**, **healthy food**. The camp **had local guides show us** **wildlife** and **explain** the **forest ecosystem**.\r\n\r\n**GET (persuaded/convinced):**\r\nCamp counselors **got us to participate** in **public speaking** activities that helped **overcome shyness**. They **got us to try new foods** that we had never **eaten before**. The experience **got us to make** lasting **friendships** with **campers** from **different backgrounds**.\r\n\r\n**Overall Impact:**\r\nThe holiday camp **made me realize** the **importance** of **stepping outside** my **comfort zone**. The experience **got me to appreciate** **nature** more deeply and **had me develop** **leadership skills** I never knew I possessed.\r\n\r\n## ✍️ Writing Section: Social Media Impact Essay\r\n\r\n### Model Essay: Social Media - Connecting or Dividing?\r\n\r\n**The Dual Impact of Social Media: Building Bridges or Creating Barriers?**\r\n\r\n**Introduction:**\r\nIn today\'s **digital age**, **social media platforms** like **Facebook**, **Instagram**, **TikTok**, and **Twitter** have **fundamentally transformed** how we **communicate**, **share information**, and **build relationships**. While some people argue that these platforms have **negative impacts** on **individuals** and **society**, others believe they have successfully **connected people globally** and created **unprecedented opportunities** for **communication** and **collaboration**. After careful consideration of both perspectives, I believe that **social media\'s impact** is **predominantly positive** when used **responsibly**, though we must **acknowledge** and **address** its **potential drawbacks**.\r\n\r\n**Positive Impacts: Global Connection and Community Building**\r\n\r\n**Enhanced Global Communication:**\r\nSocial media has **revolutionized** how we **maintain relationships** across **geographical boundaries**. **Families separated by distance** can **share daily moments** through **photos**, **videos**, and **instant messaging**. **Students studying abroad** can **stay connected** with **home communities**, while **immigrants** can **preserve cultural ties** with their **countries of origin**. This **global connectivity** has made the **world smaller** and more **interconnected** than ever before.\r\n\r\n**Access to Information and Education:**\r\nSocial media platforms serve as **powerful educational tools**, providing **instant access** to **news**, **educational content**, and **diverse perspectives**. **Students** can **join study groups**, **access free educational resources**, and **learn from experts worldwide**. During the **COVID-19 pandemic**, social media enabled **distance learning** and **kept communities informed** about **health guidelines** and **safety measures**.\r\n\r\n**Empowerment and Social Change:**\r\nSocial media has **democratized** **voice** and **influence**, allowing **ordinary people** to **raise awareness** about **important issues**. **Environmental activists** use platforms to **organize cleanup campaigns**, **students** can **advocate for education reform**, and **communities** can **mobilize support** for **local causes**. The stories mentioned in our chapter - like **Charlie\'s birthday celebration** and **animal rescue campaigns** - demonstrate how social media can **mobilize kindness** and **create positive change**.\r\n\r\n**Economic and Creative Opportunities:**\r\nMany people have **built successful businesses** and **careers** through social media. **Small entrepreneurs** can **reach global markets**, **artists** can **showcase work** to **worldwide audiences**, and **content creators** can **earn income** through **their creativity**. Social media has **created entirely new professions** and **economic opportunities** that didn\'t exist before.\r\n\r\n**Negative Impacts: Challenges We Must Address**\r\n\r\n**Mental Health and Self-Esteem Issues:**\r\n**Constant comparison** with **others\' curated posts** can lead to **decreased self-esteem** and **anxiety**, particularly among **young people**. The **pressure** to **present perfect lives** online can cause **stress** and **unrealistic expectations**. **Cyberbullying** and **online harassment** have become **serious problems** that can cause **lasting psychological damage**.\r\n\r\n**Misinformation and Fake News:**\r\nSocial media\'s **rapid information spread** can also **propagate false information**. **Misleading health claims**, **political misinformation**, and **conspiracy theories** can **spread faster** than **accurate information**, leading to **confusion** and **potentially dangerous decisions**.\r\n\r\n**Privacy and Security Concerns:**\r\n**Personal data collection**, **identity theft**, and **privacy violations** are **ongoing concerns**. Many users **don\'t fully understand** how their **personal information** is **collected**, **stored**, and **used** by **social media companies**.\r\n\r\n**Addiction and Time Management:**\r\n**Excessive social media use** can lead to **addiction-like behaviors**, **decreased productivity**, and **reduced face-to-face social skills**. The **constant notification** and **dopamine-driven** reward systems can **disrupt sleep**, **study**, and **work patterns**.\r\n\r\n**Balanced Perspective and Solutions:**\r\n\r\n**Responsible Usage Guidelines:**\r\nThe key to **maximizing benefits** while **minimizing risks** lies in **responsible usage**:\r\n- **Set time limits** for daily social media use\r\n- **Verify information** before **sharing** it\r\n- **Be kind** and **respectful** in **online interactions**\r\n- **Protect privacy** by **adjusting security settings**\r\n- **Take regular breaks** from social media\r\n\r\n**Education and Digital Literacy:**\r\n**Schools** and **parents** should **teach digital literacy**, helping **young people** understand **how to use** social media **safely** and **effectively**. This includes **recognizing fake news**, **understanding privacy settings**, and **developing healthy online habits**.\r\n\r\n**Platform Responsibility:**\r\n**Social media companies** must take **greater responsibility** for **content moderation**, **user safety**, and **transparent data practices**. **Government regulation** and **industry standards** can help **protect users** while **preserving** the **benefits** of **open communication**.\r\n\r\n**Conclusion:**\r\nSocial media represents one of the **most significant technological advances** of our time, with the **potential** to **connect**, **educate**, **empower**, and **inspire** people worldwide. While **legitimate concerns** exist about its **negative impacts**, these challenges can be **addressed** through **responsible use**, **education**, and **appropriate regulation**. \r\n\r\nThe **inspiring stories** we\'ve studied - from **Nathan\'s** **viral video** bringing **joy** to millions, to **communities** **rallying** to **help Charlie** celebrate her birthday, to **strangers** **funding** a **dog\'s wheelchair** - demonstrate social media\'s **remarkable power** to **spread kindness** and **create positive change**.\r\n\r\nRather than **rejecting** social media entirely, we should **focus** on **harnessing** its **positive potential** while **actively working** to **minimize** its **drawbacks**. **Used wisely**, social media can be a **powerful tool** for **building connections**, **sharing knowledge**, and **making** our **world** a more **compassionate** and **connected place**.\r\n\r\n### Essay Writing Tips:\r\n1. **Start** with **clear thesis statement**\r\n2. **Present both sides** of the argument fairly\r\n3. **Use specific examples** from the text and real life\r\n4. **Organize** paragraphs **logically** (pros, cons, solutions)\r\n5. **Include** **personal opinion** with **supporting reasons**\r\n6. **End** with **balanced conclusion** and **future outlook**\r\n7. **Use** **transition words** to **connect ideas** smoothly\r\n\r\n## ✅ Key Learning Points\r\n\r\n### Social Media Literacy:\r\n- **Understanding** **positive** and **negative** aspects of social media\r\n- **Recognizing** how social media can **help** communities and individuals\r\n- **Learning** to **use** social platforms **responsibly** and **safely**\r\n- **Appreciating** **real stories** of **social media kindness**\r\n\r\n### Vocabulary Building:\r\n- **Technology** and **social media** related terms\r\n- **Words** describing **online activities** and **behaviors**\r\n- **Understanding** **context** and **appropriate usage**\r\n- **Building** **digital age** **communication skills**\r\n\r\n### Grammar Mastery:\r\n- **Causative verbs** (make, let, have, get) and their **different patterns**\r\n- **Understanding** when to **use** each causative verb appropriately\r\n- **Forming** correct **sentence structures** with causatives\r\n- **Recognizing** **active** vs **passive** causative constructions\r\n\r\n### Critical Thinking:\r\n- **Evaluating** **both sides** of **social media debates**\r\n- **Understanding** **complex issues** with **multiple perspectives**\r\n- **Developing** **balanced opinions** based on **evidence**\r\n- **Proposing** **practical solutions** to **social problems**\r\n\r\n### Writing Skills:\r\n- **Argumentative essay** structure and **organization**\r\n- **Using** **specific examples** to **support arguments**\r\n- **Balancing** **different viewpoints** fairly\r\n- **Writing** **clear conclusions** with **actionable suggestions**\r\n\r\nThis comprehensive guide provides Grade 10 students with all necessary tools for understanding social media\'s impact, mastering causative verb usage, and developing critical writing skills essential for SEE examinations and digital citizenship.\r\n', 1, '2025-07-25 01:11:38', '2025-09-17 09:13:06');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(138, 13102, '# Grade 10 Optional Mathematics: Function Study Guide\r\n## Unit: Algebra | Chapter: Function | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nThis is your complete guide to the **Function** chapter for Grade 10 Optional Mathematics. It features detailed, step-by-step solutions to all 35 unique questions parsed from the provided materials, designed to help you master every concept for your exams. 📈\r\n\r\n### Chapter Overview\r\n* **Total Questions:** 35\r\n* **Difficulty:** Intermediate to Advanced\r\n* **Suggested Time:** 20-25 periods\r\n\r\n### Key Learning Areas\r\n* Function notation and evaluation, like $f(x)$\r\n* Composite functions, such as $(f \\circ g)(x)$\r\n* Inverse functions, like $f^{-1}(x)$\r\n* Identity functions and proofs\r\n* Calculating domain and range\r\n* Solving advanced function equations\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Function Composition\r\n**If $f(x) = 2x - 1$, find the value of $f(f(-1))$.**\r\n\r\n**Solution:**\r\nFirst, find the value of $f(-1)$:\r\n$$\r\nf(-1) = 2(-1) - 1 = -2 - 1 = -3\r\n$$\r\nNow, substitute this result into $f(x)$:\r\n$$\r\nf(f(-1)) = f(-3) = 2(-3) - 1 = -6 - 1 = -7\r\n$$\r\nThus, **$f(f(-1)) = -7$**.\r\n\r\n***\r\n\r\n#### Question 2: Function Composition\r\n**If $f(x) = 2x + 3$, find the value of $f(f(2))$.**\r\n\r\n**Solution:**\r\nFirst, calculate $f(2)$:\r\n$$\r\nf(2) = 2(2) + 3 = 4 + 3 = 7\r\n$$\r\nNow, substitute this value into the function again:\r\n$$\r\nf(f(2)) = f(7) = 2(7) + 3 = 14 + 3 = 17\r\n$$\r\nThus, **$f(f(2)) = 17$**.\r\n\r\n***\r\n\r\n#### Question 3: Composite Function Evaluation\r\n**If $h(x) = 2x + 3$ and $g(x) = 3x - 2$, find $h \\circ g(5)$.**\r\n\r\n**Solution:**\r\nFirst, find the value of $g(5)$:\r\n$$\r\ng(5) = 3(5) - 2 = 15 - 2 = 13\r\n$$\r\nNow, use this result to find $h(13)$:\r\n$$\r\nh \\circ g(5) = h(13) = 2(13) + 3 = 26 + 3 = 29\r\n$$\r\nThus, **$h \\circ g(5) = 29$**.\r\n\r\n***\r\n\r\n#### Question 4: Inverse Function\r\n**If $f(x) = \\frac{2x + 3}{2}$, find the value of $f^{-1}(x)$.**\r\n\r\n**Solution:**\r\nTo find the inverse, let $y = f(x)$ and solve for $x$:\r\n$$\r\n\\begin{aligned}\r\ny &= \\frac{2x + 3}{2} \\\\\r\n2y &= 2x + 3 \\\\\r\n2y - 3 &= 2x \\\\\r\nx &= \\frac{2y - 3}{2}\r\n\\end{aligned}\r\n$$\r\nFinally, interchange $x$ and $y$:\r\n$$\r\nf^{-1}(x) = \\frac{2x - 3}{2}\r\n$$\r\nThus, **$f^{-1}(x) = \\frac{2x - 3}{2}$**.\r\n\r\n***\r\n\r\n#### Question 5: Composite Function\r\n**If $f(x) = x + 1$ and $g(x) = 2x + 1$, find $g \\circ f(x)$.**\r\n\r\n**Solution:**\r\nSubstitute the expression for $f(x)$ into $g(x)$:\r\n$$\r\n\\begin{aligned}\r\ng \\circ f(x) &= g(f(x)) \\\\\r\n&= g(x+1) \\\\\r\n&= 2(x+1) + 1 \\\\\r\n&= 2x + 2 + 1 \\\\\r\n&= 2x + 3\r\n\\end{aligned}\r\n$$\r\nThus, **$g \\circ f(x) = 2x + 3$**.\r\n\r\n***\r\n\r\n#### Question 6: Finding an Unknown Function\r\n**If $f(x) = 3x - 2$ and $f \\circ g(x) = 6x - 2$, find $g(x)$.**\r\n\r\n**Solution:**\r\nWe are given the composite function $f(g(x)) = 6x - 2$.\r\nFrom the definition of $f(x)$, we can write:\r\n$$\r\nf(g(x)) = 3(g(x)) - 2\r\n$$\r\nNow, set the two expressions for $f(g(x))$ equal:\r\n$$\r\n\\begin{aligned}\r\n3(g(x)) - 2 &= 6x - 2 \\\\\r\n3(g(x)) &= 6x \\\\\r\ng(x) &= \\frac{6x}{3} \\\\\r\ng(x) &= 2x\r\n\\end{aligned}\r\n$$\r\nThus, **$g(x) = 2x$**.\r\n\r\n***\r\n\r\n#### Question 7: Inverse Function Value\r\n**If $f(x) = 4x + 3$, find the value of $f^{-1}(4)$.**\r\n\r\n**Solution:**\r\nLet $f^{-1}(4) = a$. By the definition of an inverse function, this means $f(a) = 4$.\r\n$$\r\n\\begin{aligned}\r\nf(a) &= 4 \\\\\r\n4a + 3 &= 4 \\\\\r\n4a &= 1 \\\\\r\na &= \\frac{1}{4}\r\n\\end{aligned}\r\n$$\r\nThus, **$f^{-1}(4) = \\frac{1}{4}$**.\r\n\r\n***\r\n\r\n#### Question 8: Finding a Function from its Inverse\r\n**If $f^{-1}(x) = 2x - 3$, find $f(x)$.**\r\n\r\n**Solution:**\r\nTo find $f(x)$, we find the inverse of $f^{-1}(x)$. Let $y = f^{-1}(x)$ and solve for $x$:\r\n$$\r\n\\begin{aligned}\r\ny &= 2x - 3 \\\\\r\ny + 3 &= 2x \\\\\r\nx &= \\frac{y + 3}{2}\r\n\\end{aligned}\r\n$$\r\nNow, interchange $x$ and $y$:\r\n$$\r\nf(x) = \\frac{x + 3}{2}\r\n$$\r\nThus, **$f(x) = \\frac{x + 3}{2}$**.\r\n\r\n***\r\n\r\n#### Question 9: Finding a Function from its Inverse\r\n**If $f^{-1}(x) = \\frac{x + 3}{2}$, find $f(x)$.**\r\n\r\n**Solution:**\r\nLet $y = f^{-1}(x)$ and solve for $x$:\r\n$$\r\n\\begin{aligned}\r\ny &= \\frac{x + 3}{2} \\\\\r\n2y &= x + 3 \\\\\r\nx &= 2y - 3\r\n\\end{aligned}\r\n$$\r\nInterchange $x$ and $y$:\r\n$$\r\nf(x) = 2x - 3\r\n$$\r\nThus, **$f(x) = 2x - 3$**.\r\n\r\n***\r\n\r\n#### Question 10: Composite Function\r\n**If $f(x) = 2x + 5$ and $g(x) = 3x - 1$, find $g \\circ f(x)$.**\r\n\r\n**Solution:**\r\nSubstitute the expression for $f(x)$ into $g(x)$:\r\n$$\r\n\\begin{aligned}\r\ng \\circ f(x) &= g(f(x)) \\\\\r\n&= g(2x + 5) \\\\\r\n&= 3(2x + 5) - 1 \\\\\r\n&= 6x + 15 - 1 \\\\\r\n&= 6x + 14\r\n\\end{aligned}\r\n$$\r\nThus, **$g \\circ f(x) = 6x + 14$**.\r\n\r\n***\r\n\r\n#### Question 11: Identity Function Proof\r\n**If $f(x) = 2x + 5$ and $g(x) = \\frac{x - 5}{2}$, prove that $f \\circ g(x)$ is an identity function.**\r\n\r\n**Solution:**\r\nTo prove it\'s an identity function, we must show that $f \\circ g(x) = x$.\r\n$$\r\n\\begin{aligned}\r\nf \\circ g(x) &= f(g(x)) \\\\\r\n&= f\\left(\\frac{x - 5}{2}\\right) \\\\\r\n&= 2\\left(\\frac{x - 5}{2}\\right) + 5 \\\\\r\n&= (x - 5) + 5 \\\\\r\n&= x\r\n\\end{aligned}\r\n$$\r\nSince $f \\circ g(x) = x$, it is an identity function. **(Proved)**\r\n\r\n***\r\n\r\n#### Question 12: Identity Function Proof\r\n**If $g(x) = \\frac{x - 2}{3}$ and $h(x) = 3x + 2$, prove that $g \\circ h(x)$ is an identity function.**\r\n\r\n**Solution:**\r\nWe need to show that $g \\circ h(x) = x$.\r\n$$\r\n\\begin{aligned}\r\ng \\circ h(x) &= g(h(x)) \\\\\r\n&= g(3x + 2) \\\\\r\n&= \\frac{(3x + 2) - 2}{3} \\\\\r\n&= \\frac{3x}{3} \\\\\r\n&= x\r\n\\end{aligned}\r\n$$\r\nSince $g \\circ h(x) = x$, it is an identity function. **(Proved)**\r\n\r\n***\r\n\r\n#### Question 13: Composite Function with Quadratic\r\n**If $f(x) = 2x - 3$ and $g(x) = x^2 + 1$, find the value of $f \\circ g(3)$.**\r\n\r\n**Solution:**\r\nFirst, find the value of the inner function, $g(3)$:\r\n$$\r\ng(3) = (3)^2 + 1 = 9 + 1 = 10\r\n$$\r\nNow, use this result to find $f(10)$:\r\n$$\r\nf \\circ g(3) = f(10) = 2(10) - 3 = 20 - 3 = 17\r\n$$\r\nThus, **$f \\circ g(3) = 17$**.\r\n\r\n***\r\n\r\n#### Question 14: Inverse Function Value\r\n**If $g(x) = 4x - 2$, find the value of $g^{-1}(-1)$.**\r\n\r\n**Solution:**\r\nLet $g^{-1}(-1) = a$, which implies that $g(a) = -1$.\r\n$$\r\n\\begin{aligned}\r\ng(a) &= -1 \\\\\r\n4a - 2 &= -1 \\\\\r\n4a &= 1 \\\\\r\na &= \\frac{1}{4}\r\n\\end{aligned}\r\n$$\r\nThus, **$g^{-1}(-1) = \\frac{1}{4}$**.\r\n\r\n***\r\n\r\n#### Question 15: Function Sets and Arrow Diagrams\r\n**If $f = \\{(1, 2), (2, 3), (3, 4)\\}$ and $g = \\{(2, a), (4, c), (3, b)\\}$, show the composite function $g \\circ f$ in an arrow diagram and find it in ordered pair form.**\r\n\r\n**Solution:**\r\nThe composite function $g \\circ f$ means we apply $f$ first, then $g$ to the result.\r\n* $g(f(1)) = g(2) = a$\r\n* $g(f(2)) = g(3) = b$\r\n* $g(f(3)) = g(4) = c$\r\n\r\n**Arrow Diagram:**\r\n f          g\r\n1 -----► 2 -----► a\r\n2 -----► 3 -----► b\r\n3 -----► 4 -----► c\r\n\r\n**Ordered Pair Form:**\r\nThe resulting composite function is **$g \\circ f = \\{(1, a), (2, b), (3, c)\\}$**.\r\n\r\n***\r\n\r\n#### Question 16: Solving a Function Equation\r\n**If $f(x) = 3x$, $g(x) = x + 2$, and $f \\circ g(x) = 18$, find the value of $x$.**\r\n\r\n**Solution:**\r\nFirst, determine the expression for $f \\circ g(x)$:\r\n$$\r\nf \\circ g(x) = f(g(x)) = f(x+2) = 3(x+2) = 3x + 6\r\n$$\r\nNow, set this expression equal to 18 and solve:\r\n$$\r\n\\begin{aligned}\r\n3x + 6 &= 18 \\\\\r\n3x &= 12 \\\\\r\nx &= 4\r\n\\end{aligned}\r\n$$\r\nThus, **$x = 4$**.\r\n\r\n***\r\n\r\n#### Question 17: Finding a Parameter Value\r\n**If $f(x) = 3x + b$ and $f(f(2)) = 12$, find the value of $b$.**\r\n\r\n**Solution:**\r\nFirst, evaluate $f(2)$:\r\n$$\r\nf(2) = 3(2) + b = 6 + b\r\n$$\r\nNext, evaluate $f(f(2))$ by substituting the expression above into $f(x)$:\r\n$$\r\nf(f(2)) = f(6+b) = 3(6+b) + b = 18 + 3b + b = 18 + 4b\r\n$$\r\nNow, use the given condition to solve for $b$:\r\n$$\r\n\\begin{aligned}\r\n18 + 4b &= 12 \\\\\r\n4b &= -6 \\\\\r\nb &= -\\frac{6}{4} = -\\frac{3}{2}\r\n\\end{aligned}\r\n$$\r\nThus, **$b = -\\frac{3}{2}$**.\r\n\r\n***\r\n\r\n#### Question 18: Domain and Range Relationship\r\n**If the range of the function $f(x) = 4x - 5$ is $\\{-1, 7\\}$, find its domain.**\r\n\r\n**Solution:**\r\nThe domain is the set of $x$-values that produce the given range. We solve for $x$ for each value in the range.\r\n\r\nWhen $f(x) = -1$:\r\n$$\r\n\\begin{aligned}\r\n4x - 5 &= -1 \\\\\r\n4x &= 4 \\\\\r\nx &= 1\r\n\\end{aligned}\r\n$$\r\nWhen $f(x) = 7$:\r\n$$\r\n\\begin{aligned}\r\n4x - 5 &= 7 \\\\\r\n4x &= 12 \\\\\r\nx &= 3\r\n\\end{aligned}\r\n$$\r\nThus, the **Domain = $\\{1, 3\\}$**.\r\n\r\n---\r\n\r\n### Long Questions (Advanced Problems)\r\n\r\n***\r\n\r\n#### Question 19: Finding an Inverse from a Composite\r\n**If $g(x) = 2x - 3$ and $f \\circ g(x) = 6x - 11$, find $f^{-1}(x)$.**\r\n\r\n**Solution:**\r\n**Step 1: Find the function $f(x)$.**\r\nWe are given $f(g(x)) = f(2x - 3) = 6x - 11$.\r\nTo find $f(x)$, let $u = 2x - 3$. This means $2x = u + 3$, so $x = \\frac{u+3}{2}$. Substitute this back:\r\n$$\r\n\\begin{aligned}\r\nf(u) &= 6\\left(\\frac{u+3}{2}\\right) - 11 \\\\\r\nf(u) &= 3(u+3) - 11 \\\\\r\nf(u) &= 3u + 9 - 11 \\\\\r\nf(u) &= 3u - 2\r\n\\end{aligned}\r\n$$\r\nTherefore, the function is $f(x) = 3x - 2$.\r\n\r\n**Step 2: Find the inverse, $f^{-1}(x)$.**\r\nLet $y = 3x - 2$:\r\n$$\r\n\\begin{aligned}\r\ny + 2 &= 3x \\\\\r\nx &= \\frac{y + 2}{3}\r\n\\end{aligned}\r\n$$\r\nThus, **$f^{-1}(x) = \\frac{x + 2}{3}$**.\r\n\r\n***\r\n\r\n#### Question 20: Finding an Inverse from a Composite\r\n**If $f(x) = 2x - 3$ and $g \\circ f(x) = 6x - 11$, find $g^{-1}(x)$.**\r\n\r\n**Solution:**\r\n**Step 1: Find the function $g(x)$.**\r\nWe are given $g(f(x)) = g(2x - 3) = 6x - 11$.\r\nLet $u = 2x - 3$, which means $x = \\frac{u+3}{2}$. Substitute this into the equation:\r\n$$\r\n\\begin{aligned}\r\ng(u) &= 6\\left(\\frac{u+3}{2}\\right) - 11 \\\\\r\ng(u) &= 3(u+3) - 11 \\\\\r\ng(u) &= 3u + 9 - 11 \\\\\r\ng(u) &= 3u - 2\r\n\\end{aligned}\r\n$$\r\nSo, the function is $g(x) = 3x - 2$.\r\n\r\n**Step 2: Find the inverse, $g^{-1}(x)$.**\r\nLet $y = 3x - 2$:\r\n$$\r\n\\begin{aligned}\r\ny + 2 &= 3x \\\\\r\nx &= \\frac{y + 2}{3}\r\n\\end{aligned}\r\n$$\r\nThus, **$g^{-1}(x) = \\frac{x + 2}{3}$**.\r\n\r\n***\r\n\r\n#### Question 21: Solving a Complex Function Equation\r\n**If $f(x) = 5x - 13$, $g(x) = \\frac{2x + 7}{3}$, and $g^{-1}(x) = f(f(x))$, find the value of $x$.**\r\n\r\n**Solution:**\r\n**Step 1: Find $g^{-1}(x)$.**\r\nLet $y = \\frac{2x+7}{3}$.\r\n$$\r\n\\begin{aligned}\r\n3y &= 2x+7 \\\\\r\n3y-7 &= 2x \\\\\r\nx &= \\frac{3y-7}{2}\r\n\\end{aligned}\r\n$$\r\nSo, $g^{-1}(x) = \\frac{3x - 7}{2}$.\r\n\r\n**Step 2: Find $f(f(x))$.**\r\n$$\r\nf(f(x)) = 5(5x - 13) - 13 = 25x - 65 - 13 = 25x - 78\r\n$$\r\n\r\n**Step 3: Set them equal and solve for $x$.**\r\n$$\r\n\\begin{aligned}\r\ng^{-1}(x) &= f(f(x)) \\\\\r\n\\frac{3x - 7}{2} &= 25x - 78 \\\\\r\n3x - 7 &= 50x - 156 \\\\\r\n156 - 7 &= 50x - 3x \\\\\r\n149 &= 47x \\\\\r\nx &= \\frac{149}{47}\r\n\\end{aligned}\r\n$$\r\nThus, **$x = \\frac{149}{47}$**.\r\n\r\n***\r\n\r\n#### Question 22: Solving for a Parameter\r\n**If $f(x) = \\frac{kx - 3}{2}$, $g(x) = \\frac{3}{x+2}$, and $f \\circ g^{-1}(3) = -\\frac{1}{4}$, find the value of $k$.**\r\n\r\n**Solution:**\r\n**Step 1: Find $g^{-1}(x)$.**\r\nLet $y = \\frac{3}{x+2}$.\r\n$$\r\n\\begin{aligned}\r\ny(x+2) &= 3 \\\\\r\nx+2 &= \\frac{3}{y} \\\\\r\nx &= \\frac{3}{y}-2\r\n\\end{aligned}\r\n$$\r\nSo, $g^{-1}(x) = \\frac{3}{x} - 2$.\r\n\r\n**Step 2: Evaluate $g^{-1}(3)$.**\r\n$$\r\ng^{-1}(3) = \\frac{3}{3} - 2 = 1 - 2 = -1\r\n$$\r\n\r\n**Step 3: Use the condition to solve for $k$.**\r\nWe are given $f \\circ g^{-1}(3) = f(-1) = -\\frac{1}{4}$.\r\n$$\r\n\\begin{aligned}\r\nf(-1) &= \\frac{k(-1) - 3}{2} \\\\\r\n-\\frac{1}{4} &= \\frac{-k - 3}{2} \\\\\r\n-2 &= 4(-k - 3) \\\\\r\n-2 &= -4k - 12 \\\\\r\n10 &= -4k \\\\\r\nk &= -\\frac{10}{4}\r\n\\end{aligned}\r\n$$\r\nThus, **$k = -\\frac{5}{2}$**.\r\n\r\n***\r\n\r\n#### Question 23: Solving a Composite Inverse Equation\r\n**If $f(x) = 3x - 7$, $g(x) = \\frac{5x + 2}{3}$, and $g^{-1} \\circ f(x) = 8$, find the value of $x$.**\r\n\r\n**Solution:**\r\n**Step 1: Find $g^{-1}(x)$.**\r\nLet $y = \\frac{5x+2}{3}$.\r\n$$\r\n\\begin{aligned}\r\n3y &= 5x+2 \\\\\r\nx &= \\frac{3y-2}{5}\r\n\\end{aligned}\r\n$$\r\nSo, $g^{-1}(x) = \\frac{3x - 2}{5}$.\r\n\r\n**Step 2: Solve the composite equation.**\r\n$$\r\n\\begin{aligned}\r\ng^{-1}(f(x)) &= 8 \\\\\r\ng^{-1}(3x - 7) &= 8 \\\\\r\n\\frac{3(3x-7)-2}{5} &= 8 \\\\\r\n9x - 21 - 2 &= 40 \\\\\r\n9x - 23 &= 40 \\\\\r\n9x &= 63 \\\\\r\nx &= 7\r\n\\end{aligned}\r\n$$\r\nThus, **$x = 7$**.\r\n\r\n***\r\n\r\n#### Question 24: Solving a Composite Inverse Equation\r\n**If $f(x) = \\frac{x+1}{2}$, $g(x) = \\frac{x-5}{2}$, and $f \\circ g^{-1}(x) = 6$, find the value of $x$.**\r\n\r\n**Solution:**\r\n**Step 1: Find $g^{-1}(x)$.**\r\nLet $y = \\frac{x-5}{2}$.\r\n$$\r\n\\begin{aligned}\r\n2y &= x-5 \\\\\r\nx &= 2y+5\r\n\\end{aligned}\r\n$$\r\nSo, $g^{-1}(x) = 2x+5$.\r\n\r\n**Step 2: Solve the equation.**\r\n$$\r\n\\begin{aligned}\r\nf(g^{-1}(x)) &= 6 \\\\\r\nf(2x+5) &= 6 \\\\\r\n\\frac{(2x+5)+1}{2} &= 6 \\\\\r\n\\frac{2x+6}{2} &= 6 \\\\\r\nx+3 &= 6 \\\\\r\nx &= 3\r\n\\end{aligned}\r\n$$\r\nThus, **$x = 3$**.\r\n\r\n***\r\n\r\n#### Question 25: Commutative Property Proof\r\n**If $f(x) = 3x + 4$ and $g(x) = 2(x + 1)$, prove that $(f \\circ g)(x) = (g \\circ f)(x)$.**\r\n\r\n**Solution:**\r\nFirst, find $f \\circ g(x)$:\r\n$$\r\nf \\circ g(x) = f(2x+2) = 3(2x+2) + 4 = 6x + 6 + 4 = \\mathbf{6x + 10}\r\n$$\r\nNext, find $g \\circ f(x)$:\r\n$$\r\ng \\circ f(x) = g(3x+4) = 2((3x+4)+1) = 2(3x+5) = \\mathbf{6x + 10}\r\n$$\r\nSince both expressions are identical, **$(f \\circ g)(x) = (g \\circ f)(x)$ is proved**.\r\n\r\n***\r\n\r\n#### Question 26: Finding the Inverse of a Composite\r\n**If $f(x) = 2x - 7$ and $f \\circ g(x) = 4x + 3$, find $(g \\circ f)^{-1}(x)$.**\r\n\r\n**Solution:**\r\n**Step 1: Find $g(x)$.**\r\n$$\r\n\\begin{aligned}\r\nf(g(x)) &= 4x + 3 \\\\\r\n2(g(x)) - 7 &= 4x + 3 \\\\\r\n2(g(x)) &= 4x + 10 \\\\\r\ng(x) &= 2x + 5\r\n\\end{aligned}\r\n$$\r\n**Step 2: Find $g \\circ f(x)$.**\r\n$$\r\ng \\circ f(x) = g(2x-7) = 2(2x-7)+5 = 4x - 14 + 5 = 4x - 9\r\n$$\r\n**Step 3: Find the inverse of $g \\circ f(x)$.**\r\nLet $y = 4x-9$.\r\n$$\r\n\\begin{aligned}\r\ny+9 &= 4x \\\\\r\nx &= \\frac{y+9}{4}\r\n\\end{aligned}\r\n$$\r\nThus, **$(g \\circ f)^{-1}(x) = \\frac{x+9}{4}$**.\r\n\r\n***\r\n\r\n#### Question 27: Finding the Inverse of a Composite\r\n**If $f(x) = 3x - 1$ and $f \\circ g(x) = 6x + 5$, find $(g \\circ f)^{-1}(x)$.**\r\n\r\n**Solution:**\r\n**Step 1: Find $g(x)$.**\r\n$$\r\n\\begin{aligned}\r\nf(g(x)) &= 6x+5 \\\\\r\n3(g(x)) - 1 &= 6x + 5 \\\\\r\n3(g(x)) &= 6x + 6 \\\\\r\ng(x) &= 2x + 2\r\n\\end{aligned}\r\n$$\r\n**Step 2: Find $g \\circ f(x)$.**\r\n$$\r\ng \\circ f(x) = g(3x-1) = 2(3x-1)+2 = 6x - 2 + 2 = 6x\r\n$$\r\n**Step 3: Find the inverse.**\r\nLet $y = 6x$.\r\n$$\r\nx = \\frac{y}{6}\r\n$$\r\nThus, **$(g \\circ f)^{-1}(x) = \\frac{x}{6}$**.\r\n\r\n***\r\n\r\n#### Question 28: Solving an Inverse Function Equation\r\n**If $f(x) = 3x - 7$, $g(x) = \\frac{4x - 2}{3}$, and $f^{-1}(x) = g(x)$, find the value of $x$.**\r\n\r\n**Solution:**\r\n**Step 1: Find $f^{-1}(x)$.**\r\nLet $y = 3x-7$.\r\n$$\r\n\\begin{aligned}\r\ny+7 &= 3x \\\\\r\nx &= \\frac{y+7}{3}\r\n\\end{aligned}\r\n$$\r\nSo, $f^{-1}(x) = \\frac{x+7}{3}$.\r\n\r\n**Step 2: Set the functions equal and solve.**\r\n$$\r\n\\begin{aligned}\r\nf^{-1}(x) &= g(x) \\\\\r\n\\frac{x+7}{3} &= \\frac{4x-2}{3} \\\\\r\nx+7 &= 4x-2 \\\\\r\n9 &= 3x \\\\\r\nx &= 3\r\n\\end{aligned}\r\n$$\r\nThus, **$x=3$**.\r\n\r\n***\r\n\r\n#### Question 29: Finding Multiple Values\r\n**If $h(x) = \\frac{x}{2-x}$ and $g(x) = ax - 5$ are two functions and $g \\circ h(4) = -13$, find the value of $a$ and $g^{-1}(1)$.**\r\n\r\n**Solution:**\r\n**Step 1: Find the value of $a$.**\r\nFirst, evaluate $h(4)$:\r\n$$\r\nh(4) = \\frac{4}{2-4} = \\frac{4}{-2} = -2\r\n$$\r\nNow, use this in the given condition:\r\n$$\r\n\\begin{aligned}\r\ng \\circ h(4) &= g(-2) = -13 \\\\\r\na(-2) - 5 &= -13 \\\\\r\n-2a &= -8 \\\\\r\na &= 4\r\n\\end{aligned}\r\n$$\r\nThus, **$a=4$**.\r\n\r\n**Step 2: Find $g^{-1}(1)$.**\r\nNow that we know $a=4$, the function is $g(x) = 4x-5$. Find its inverse:\r\nLet $y=4x-5$.\r\n$$\r\n\\begin{aligned}\r\ny+5 &= 4x \\\\\r\nx &= \\frac{y+5}{4}\r\n\\end{aligned}\r\n$$\r\nSo, $g^{-1}(x) = \\frac{x+5}{4}$.\r\nFinally, evaluate $g^{-1}(1)$:\r\n$$\r\ng^{-1}(1) = \\frac{1+5}{4} = \\frac{6}{4} = \\frac{3}{2}\r\n$$\r\nThus, **$g^{-1}(1) = \\frac{3}{2}$**.\r\n\r\n***\r\n\r\n#### Question 30: Finding an Unknown Function and an Inverse Value\r\n**If $f(x) = 3x - 4$ and $f^{-1} \\circ g(x) = \\frac{3x + 2}{3}$, find $g(x)$ and $f^{-1}(\\frac{1}{2})$.**\r\n\r\n**Solution:**\r\n**Step 1: Find $f^{-1}(x)$.**\r\nLet $y = 3x-4$.\r\n$$\r\n\\begin{aligned}\r\ny+4 &= 3x \\\\\r\nx &= \\frac{y+4}{3}\r\n\\end{aligned}\r\n$$\r\nSo, $f^{-1}(x) = \\frac{x+4}{3}$.\r\n\r\n**Step 2: Find $g(x)$.**\r\nUse the expression for $f^{-1}(x)$ in the given condition:\r\n$$\r\n\\begin{aligned}\r\nf^{-1}(g(x)) &= \\frac{g(x)+4}{3} \\\\\r\n\\frac{g(x)+4}{3} &= \\frac{3x+2}{3} \\\\\r\ng(x)+4 &= 3x+2 \\\\\r\ng(x) &= 3x-2\r\n\\end{aligned}\r\n$$\r\nThus, **$g(x) = 3x-2$**.\r\n\r\n**Step 3: Find $f^{-1}(\\frac{1}{2})$.**\r\nSubstitute $x = \\frac{1}{2}$ into the inverse function:\r\n$$\r\nf^{-1}\\left(\\frac{1}{2}\\right) = \\frac{\\frac{1}{2}+4}{3} = \\frac{\\frac{1+8}{2}}{3} = \\frac{9/2}{3} = \\frac{9}{6} = \\frac{3}{2}\r\n$$\r\nThus, **$f^{-1}(\\frac{1}{2}) = \\frac{3}{2}$**.\r\n\r\n***\r\n\r\n#### Question 31: Solving a Function Equation\r\n**If $f(x) = 2x + 3$, $g(x) = \\frac{x + 5}{2}$, and $f(f(x)) = g^{-1}(x)$, find the value of $x$.**\r\n\r\n**Solution:**\r\n**Step 1: Find $g^{-1}(x)$.**\r\nLet $y = \\frac{x+5}{2}$.\r\n$$\r\n\\begin{aligned}\r\n2y &= x+5 \\\\\r\nx &= 2y-5\r\n\\end{aligned}\r\n$$\r\nSo, $g^{-1}(x) = 2x-5$.\r\n\r\n**Step 2: Find $f(f(x))$.**\r\n$$\r\nf(f(x)) = 2(2x+3)+3 = 4x+6+3 = 4x+9\r\n$$\r\n\r\n**Step 3: Set them equal and solve.**\r\n$$\r\n\\begin{aligned}\r\n4x+9 &= 2x-5 \\\\\r\n2x &= -14 \\\\\r\nx &= -7\r\n\\end{aligned}\r\n$$\r\nThus, **$x = -7$**.\r\n\r\n***\r\n\r\n#### Question 32: Solving a Function Equation\r\n**If $f(x) = 2x + 1$, $g(x) = \\frac{3x + 1}{2}$, and $f(f(x)) = g^{-1}(x)$, find the value of $x$.**\r\n\r\n**Solution:**\r\n**Step 1: Find $g^{-1}(x)$.**\r\nLet $y = \\frac{3x+1}{2}$.\r\n$$\r\n\\begin{aligned}\r\n2y &= 3x+1 \\\\\r\nx &= \\frac{2y-1}{3}\r\n\\end{aligned}\r\n$$\r\nSo, $g^{-1}(x) = \\frac{2x-1}{3}$.\r\n\r\n**Step 2: Find $f(f(x))$.**\r\n$$\r\nf(f(x)) = 2(2x+1)+1 = 4x+2+1 = 4x+3\r\n$$\r\n\r\n**Step 3: Set them equal and solve.**\r\n$$\r\n\\begin{aligned}\r\n4x+3 &= \\frac{2x-1}{3} \\\\\r\n3(4x+3) &= 2x-1 \\\\\r\n12x+9 &= 2x-1 \\\\\r\n10x &= -10 \\\\\r\nx &= -1\r\n\\end{aligned}\r\n$$\r\nThus, **$x = -1$**.\r\n\r\n***\r\n\r\n#### Question 33: Self-Inverse Function Problem\r\n**If $f(x) = \\frac{x}{3+2x}$ and $f(x) = f^{-1}(x)$, find the value of $x$.**\r\n\r\n**Solution:**\r\n**Step 1: Find $f^{-1}(x)$.**\r\nLet $y = \\frac{x}{3+2x}$.\r\n$$\r\n\\begin{aligned}\r\ny(3+2x) &= x \\\\\r\n3y+2xy &= x \\\\\r\n3y &= x - 2xy \\\\\r\n3y &= x(1-2y) \\\\\r\nx &= \\frac{3y}{1-2y}\r\n\\end{aligned}\r\n$$\r\nSo, $f^{-1}(x) = \\frac{3x}{1-2x}$.\r\n\r\n**Step 2: Set $f(x) = f^{-1}(x)$ and solve.**\r\n$$\r\n\\begin{aligned}\r\n\\frac{x}{3+2x} &= \\frac{3x}{1-2x} \\\\\r\nx(1-2x) &= 3x(3+2x) \\\\\r\nx - 2x^2 &= 9x + 6x^2 \\\\\r\n0 &= 8x^2 + 8x \\\\\r\n0 &= 8x(x+1)\r\n\\end{aligned}\r\n$$\r\nThis gives two possible solutions: $8x=0 \\implies \\mathbf{x=0}$ or $x+1=0 \\implies \\mathbf{x=-1}$.\r\n\r\n***\r\n\r\n#### Question 34: Evaluating Composite Inverses\r\n**If $f(x) = \\frac{4x+5}{3}$ and $g(x) = 5x - 4$, find the value of $f^{-1} \\circ g^{-1}(1)$.**\r\n\r\n**Solution:**\r\n**Step 1: Find $g^{-1}(x)$ and evaluate $g^{-1}(1)$.**\r\nLet $y=5x-4 \\implies x = \\frac{y+4}{5}$. So, $g^{-1}(x) = \\frac{x+4}{5}$.\r\n$$\r\ng^{-1}(1) = \\frac{1+4}{5} = 1\r\n$$\r\n\r\n**Step 2: Find $f^{-1}(x)$ and evaluate $f^{-1}(g^{-1}(1))$.**\r\nLet $y = \\frac{4x+5}{3} \\implies 3y=4x+5 \\implies x = \\frac{3y-5}{4}$. So, $f^{-1}(x) = \\frac{3x-5}{4}$.\r\nSince $g^{-1}(1)=1$, we need to find $f^{-1}(1)$:\r\n$$\r\nf^{-1}(1) = \\frac{3(1)-5}{4} = \\frac{-2}{4} = -\\frac{1}{2}\r\n$$\r\nThus, **$f^{-1} \\circ g^{-1}(1) = -\\frac{1}{2}$**.\r\n\r\n***\r\n\r\n#### Question 35: Solving with a Quadratic Function\r\n**If $f(x) = x^2 - 2x$, $g(x) = 2x + 3$, and $f \\circ g^{-1}(x) = 3$, find the value of $x$.**\r\n\r\n**Solution:**\r\n**Step 1: Find $g^{-1}(x)$.**\r\nLet $y=2x+3 \\implies x = \\frac{y-3}{2}$. So, $g^{-1}(x) = \\frac{x-3}{2}$.\r\n\r\n**Step 2: Solve the equation $f(g^{-1}(x)) = 3$.**\r\n$$\r\n\\begin{aligned}\r\nf\\left(\\frac{x-3}{2}\\right) &= 3 \\\\\r\n\\left(\\frac{x-3}{2}\\right)^2 - 2\\left(\\frac{x-3}{2}\\right) &= 3 \\\\\r\n\\frac{x^2-6x+9}{4} - (x-3) &= 3 \\\\\r\n\\frac{x^2-6x+9}{4} - x + 3 &= 3 \\\\\r\n\\frac{x^2-6x+9}{4} &= x \\\\\r\nx^2 - 6x + 9 &= 4x \\\\\r\nx^2 - 10x + 9 &= 0 \\\\\r\n(x-9)(x-1) &= 0\r\n\\end{aligned}\r\n$$\r\nThus, the solutions are **$x=1$ and $x=9$**.\r\n\r\n---\r\n\r\n## ✅ Key Concepts & Exam Strategy\r\n\r\n### Core Definitions\r\n* **Function:** A rule where each input has exactly one output.\r\n* **Domain:** The set of all possible input ($x$) values.\r\n* **Range:** The set of all actual output ($y$) values.\r\n* **Identity Function:** A function where the output is the same as the input ($I(x)=x$).\r\n\r\n### Composite & Inverse Functions\r\n* **Composite $(f \\circ g)(x)$:** Means $f(g(x))$. Apply $g$ first, then apply $f$ to the result.\r\n* **Inverse $f^{-1}(x)$:** Reverses the action of $f(x)$. If $f(a)=b$, then $f^{-1}(b)=a$.\r\n* **Key Property:** $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.\r\n\r\n### Finding an Inverse Function: 4 Steps\r\n1.  Replace $f(x)$ with $y$.\r\n2.  Interchange the $x$ and $y$ variables.\r\n3.  Solve the new equation for $y$.\r\n4.  Replace $y$ with $f^{-1}(x)$.\r\n\r\n### Common Mistakes to Avoid\r\n1.  **Order Confusion:** Mixing up $f \\circ g(x)$ and $g \\circ f(x)$. They are usually different.\r\n2.  **Algebraic Errors:** Simple sign mistakes or calculation errors when solving for an inverse.\r\n3.  **Verification:** Forgetting to verify your inverse function by checking if $f(f^{-1}(x)) = x$.', 1, '2025-07-25 01:11:38', '2025-09-18 07:21:49');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(139, 14239, '# Grade 10 Optional Mathematics: Polynomials Study Guide\r\n## Unit: Algebra | Chapter: Polynomials | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nThis guide provides a comprehensive overview of the **Polynomials** chapter for Grade 10 Optional Mathematics. It includes clear, step-by-step solutions to a variety of problems involving the **Remainder Theorem**, **Factor Theorem**, and **Synthetic Division**. 🧠\r\n\r\n### Key Learning Areas\r\n* **Remainder Theorem:** Finding the remainder when a polynomial is divided by a linear expression.\r\n* **Factor Theorem:** Determining if a linear expression is a factor of a polynomial.\r\n* **Synthetic Division:** A shorthand method for polynomial division by a linear factor.\r\n* **Solving Polynomial Equations:** Finding the roots of cubic polynomial equations.\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Finding a Value Using Factor Theorem\r\n**If $x - 2$ is a factor of $2x^3 - 6x^2 - 5m - 2$, find the value of $m$.**\r\n\r\n**Solution:**\r\nLet the polynomial be $P(x) = 2x^3 - 6x^2 - 5m - 2$.\r\nThe divisor is $d(x) = x - 2$. Comparing this to $x - a$, we get $a = 2$.\r\n\r\nBy the Factor Theorem, if $(x - a)$ is a factor of $P(x)$, then $P(a) = 0$.\r\n$$\r\n\\begin{aligned}\r\nP(2) &= 0 \\\\\r\n2(2)^3 - 6(2)^2 - 5m - 2 &= 0 \\\\\r\n2(8) - 6(4) - 5m - 2 &= 0 \\\\\r\n16 - 24 - 5m - 2 &= 0 \\\\\r\n-10 - 5m &= 0 \\\\\r\n-10 &= 5m \\\\\r\nm &= -2\r\n\\end{aligned}\r\n$$\r\nThus, **$m = -2$**.\r\n\r\n***\r\n\r\n#### Question 2: Finding a Value Using Factor Theorem\r\n**If $(x + 2)$ is a factor of the polynomial $2x^3 + 3x^2 - mx + 4$, find the value of $m$.**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 + 3x^2 - mx + 4$.\r\nThe divisor is $d(x) = x + 2$. Comparing this to $x - a$, we get $a = -2$.\r\n\r\nBy the Factor Theorem, $P(-2) = 0$.\r\n$$\r\n\\begin{aligned}\r\n2(-2)^3 + 3(-2)^2 - m(-2) + 4 &= 0 \\\\\r\n2(-8) + 3(4) + 2m + 4 &= 0 \\\\\r\n-16 + 12 + 2m + 4 &= 0 \\\\\r\n0 + 2m &= 0 \\\\\r\n2m &= 0 \\\\\r\nm &= 0\r\n\\end{aligned}\r\n$$\r\nThus, **$m = 0$**.\r\n\r\n***\r\n\r\n#### Question 3: Finding a Value Using Remainder Theorem\r\n**If $2x^3 - 4x^2 + kx + 10$ is divided by $(x + 2)$, the remainder is 4. Find the value of $k$.**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 - 4x^2 + kx + 10$.\r\nThe divisor is $d(x) = x + 2$, so $a = -2$. The remainder is $R = 4$.\r\n\r\nBy the Remainder Theorem, $P(a) = R$.\r\n$$\r\n\\begin{aligned}\r\nP(-2) &= 4 \\\\\r\n2(-2)^3 - 4(-2)^2 + k(-2) + 10 &= 4 \\\\\r\n2(-8) - 4(4) - 2k + 10 &= 4 \\\\\r\n-16 - 16 - 2k + 10 &= 4 \\\\\r\n-22 - 2k &= 4 \\\\\r\n-2k &= 26 \\\\\r\nk &= -13\r\n\\end{aligned}\r\n$$\r\nThus, **$k = -13$**.\r\n\r\n***\r\n\r\n#### Question 4: Finding a Value Using Factor Theorem\r\n**If $x - 3$ is a factor of $2x^3 - x^2 + 10x - k$, find the value of $k$.**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 - x^2 + 10x - k$.\r\nThe divisor is $d(x) = x - 3$, so $a = 3$.\r\n\r\nBy the Factor Theorem, $P(3) = 0$.\r\n$$\r\n\\begin{aligned}\r\n2(3)^3 - (3)^2 + 10(3) - k &= 0 \\\\\r\n2(27) - 9 + 30 - k &= 0 \\\\\r\n54 - 9 + 30 - k &= 0 \\\\\r\n75 - k &= 0 \\\\\r\nk &= 75\r\n\\end{aligned}\r\n$$\r\nThus, **$k = 75$**.\r\n\r\n***\r\n\r\n#### Question 5: Finding a Value Using Factor Theorem\r\n**If $x - 5$ is a factor of $x^3 + px^2 + 4x + 5$, find the value of $p$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 + px^2 + 4x + 5$.\r\nThe divisor is $d(x) = x - 5$, so $a = 5$.\r\n\r\nBy the Factor Theorem, $P(5) = 0$.\r\n$$\r\n\\begin{aligned}\r\n(5)^3 + p(5)^2 + 4(5) + 5 &= 0 \\\\\r\n125 + 25p + 20 + 5 &= 0 \\\\\r\n150 + 25p &= 0 \\\\\r\n25p &= -150 \\\\\r\np &= -6\r\n\\end{aligned}\r\n$$\r\nThus, **$p = -6$**.\r\n\r\n***\r\n\r\n#### Question 6: Finding the Remainder\r\n**Using the Remainder Theorem, find the remainder when $8x^3 - 4x^2 + 2x - 5$ is divided by $2x - 1$.**\r\n\r\n**Solution:**\r\nLet $P(x) = 8x^3 - 4x^2 + 2x - 5$.\r\nFor the divisor $2x - 1$, we set it to zero to find the value to substitute: $2x - 1 = 0 \\implies x = \\frac{1}{2}$.\r\n\r\nBy the Remainder Theorem, the remainder is $P(\\frac{1}{2})$.\r\n$$\r\n\\begin{aligned}\r\nP\\left(\\frac{1}{2}\\right) &= 8\\left(\\frac{1}{2}\\right)^3 - 4\\left(\\frac{1}{2}\\right)^2 + 2\\left(\\frac{1}{2}\\right) - 5 \\\\\r\n&= 8\\left(\\frac{1}{8}\\right) - 4\\left(\\frac{1}{4}\\right) + 1 - 5 \\\\\r\n&= 1 - 1 + 1 - 5 \\\\\r\n&= -4\r\n\\end{aligned}\r\n$$\r\nThe remainder is **-4**.\r\n\r\n***\r\n\r\n#### Question 7: Using the Factor Theorem\r\n**If $x - 5$ is a factor of $2x^3 - 7x + (p - 12)$, find the value of p.**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 - 7x + p - 12$.\r\nSince $x-5$ is a factor, $a=5$ and $P(5)=0$.\r\n$$\r\n\\begin{aligned}\r\nP(5) &= 0 \\\\\r\n2(5)^3 - 7(5) + p - 12 &= 0 \\\\\r\n2(125) - 35 + p - 12 &= 0 \\\\\r\n250 - 35 + p - 12 &= 0 \\\\\r\n203 + p &= 0 \\\\\r\np &= -203\r\n\\end{aligned}\r\n$$\r\n**Note:** There appears to be a typo in the source material\'s calculation. The correct answer based on the polynomial is $p=-203$.\r\n\r\n***\r\n\r\n#### Question 8: Using Synthetic Division\r\n**If $x^3 - 21x - 20 = (x + 1) \\cdot q(x)$, find $q(x)$ by using synthetic division.**\r\n\r\n**Solution:**\r\nThe polynomial is $P(x) = x^3 + 0x^2 - 21x - 20$.\r\nThe divisor is $x+1$, so we use $a = -1$ for synthetic division.\r\n\r\n-1 | 1 0 -21 -20 | -1 1 20\r\n 1  -1   -20     0 \r\n\r\nThe quotient\'s coefficients are 1, -1, and -20. The remainder is 0.\r\nThus, **$q(x) = x^2 - x - 20$**.\r\n\r\n***\r\n\r\n#### Question 9: Using Synthetic Division\r\n**If $x^3 - 19x - 30 = (x + 2) \\cdot q(x)$, find $q(x)$ by using synthetic division.**\r\n\r\n**Solution:**\r\nThe polynomial is $P(x) = x^3 + 0x^2 - 19x - 30$.\r\nThe divisor is $x+2$, so we use $a = -2$.\r\n\r\n-2 | 1 0 -19 -30 | -2 4 30\r\n 1  -2   -15     0 \r\n\r\nThe quotient\'s coefficients are 1, -2, and -15. The remainder is 0.\r\nThus, **$q(x) = x^2 - 2x - 15$**.\r\n\r\n***\r\n\r\n#### Question 10: State and Apply the Factor Theorem\r\n**State the Factor Theorem. Use it to determine whether $x + 3$ is a factor of $x^3 - 8x + 3$.**\r\n\r\n**Solution:**\r\n**Factor Theorem:** A polynomial $P(x)$ has a factor $(x - a)$ if and only if $P(a) = 0$.\r\n\r\nLet $P(x) = x^3 - 8x + 3$. To check if $(x+3)$ is a factor, we test $P(-3)$.\r\n$$\r\n\\begin{aligned}\r\nP(-3) &= (-3)^3 - 8(-3) + 3 \\\\\r\n&= -27 + 24 + 3 \\\\\r\n&= 0\r\n\\end{aligned}\r\n$$\r\nSince $P(-3) = 0$, **yes, $(x+3)$ is a factor** of the polynomial.\r\n\r\n***\r\n\r\n#### Question 11: State and Apply the Remainder Theorem\r\n**State the Remainder Theorem. When $x^3 + 5x^2 - 6x - 16$ is divided by $(x + 2)$, find the remainder.**\r\n\r\n**Solution:**\r\n**Remainder Theorem:** If a polynomial $P(x)$ is divided by $(x - a)$, the remainder is $P(a)$.\r\n\r\nLet $P(x) = x^3 + 5x^2 - 6x - 16$. The divisor is $(x+2)$, so $a = -2$. The remainder is $P(-2)$.\r\n$$\r\n\\begin{aligned}\r\nP(-2) &= (-2)^3 + 5(-2)^2 - 6(-2) - 16 \\\\\r\n&= -8 + 5(4) + 12 - 16 \\\\\r\n&= -8 + 20 + 12 - 16 \\\\\r\n&= 8\r\n\\end{aligned}\r\n$$\r\nThe remainder is **8**.\r\n\r\n***\r\n\r\n#### Question 12: Finding a Value Using Remainder Theorem\r\n**If the polynomial $x^3 + 6x^2 + kx + 10$ is divided by $(x + 2)$, the remainder is 4. Find the value of $k$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 + 6x^2 + kx + 10$. The divisor is $(x+2)$, so $a=-2$. The remainder is $R=4$.\r\n\r\nBy the Remainder Theorem, $P(-2) = 4$.\r\n$$\r\n\\begin{aligned}\r\n(-2)^3 + 6(-2)^2 + k(-2) + 10 &= 4 \\\\\r\n-8 + 6(4) - 2k + 10 &= 4 \\\\\r\n-8 + 24 - 2k + 10 &= 4 \\\\\r\n26 - 2k &= 4 \\\\\r\n22 &= 2k \\\\\r\nk &= 11\r\n\\end{aligned}\r\n$$\r\nThus, **$k = 11$**.\r\n\r\n***\r\n\r\n#### Question 13: Finding a Value Using Factor Theorem\r\n**If $(x + 2)$ is a factor of $x^3 - 19x - p$, find the value of $p$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 - 19x - p$. Since $(x+2)$ is a factor, $a=-2$ and $P(-2)=0$.\r\n$$\r\n\\begin{aligned}\r\nP(-2) &= 0 \\\\\r\n(-2)^3 - 19(-2) - p &= 0 \\\\\r\n-8 + 38 - p &= 0 \\\\\r\n30 - p &= 0 \\\\\r\np &= 30\r\n\\end{aligned}\r\n$$\r\nThus, **$p = 30$**.\r\n\r\n***\r\n\r\n#### Question 14: Finding a Value Using Factor Theorem\r\n**If $(x - m)$ is a factor of $x^3 - mx^2 - 4x + m + 12$, calculate the value of $m$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 - mx^2 - 4x + m + 12$. Since $(x-m)$ is a factor, $a=m$ and $P(m)=0$.\r\n$$\r\n\\begin{aligned}\r\nP(m) &= 0 \\\\\r\n(m)^3 - m(m)^2 - 4(m) + m + 12 &= 0 \\\\\r\nm^3 - m^3 - 4m + m + 12 &= 0 \\\\\r\n-3m + 12 &= 0 \\\\\r\n12 &= 3m \\\\\r\nm &= 4\r\n\\end{aligned}\r\n$$\r\nThus, **$m = 4$**.\r\n\r\n***\r\n\r\n#### Question 15: Finding a Value Using Factor Theorem\r\n**If $x - 2$ is a factor of $x^3 - kx^2 + x - 6$, find the value of $k$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 - kx^2 + x - 6$. Since $(x-2)$ is a factor, $a=2$ and $P(2)=0$.\r\n$$\r\n\\begin{aligned}\r\nP(2) &= 0 \\\\\r\n(2)^3 - k(2)^2 + 2 - 6 &= 0 \\\\\r\n8 - 4k - 4 &= 0 \\\\\r\n4 - 4k &= 0 \\\\\r\n4 &= 4k \\\\\r\nk &= 1\r\n\\end{aligned}\r\n$$\r\nThus, **$k=1$**.\r\n\r\n***\r\n\r\n#### Question 16: Using the Factor Theorem\r\n**Use the Factor Theorem to determine whether $x + 4$ is a factor of the polynomial $x^3 - 15x + 4$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 - 15x + 4$. To check if $(x+4)$ is a factor, we test if $P(-4) = 0$.\r\n$$\r\n\\begin{aligned}\r\nP(-4) &= (-4)^3 - 15(-4) + 4 \\\\\r\n&= -64 + 60 + 4 \\\\\r\n&= 0\r\n\\end{aligned}\r\n$$\r\nSince $P(-4) = 0$, **yes, $(x+4)$ is a factor**.\r\n\r\n***\r\n\r\n#### Question 17: Using the Remainder Theorem\r\n**Find the remainder when the polynomial $x^6 - 1$ is divided by $x + 1$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^6 - 1$. The divisor is $(x+1)$, so $a=-1$. The remainder is $P(-1)$.\r\n$$\r\n\\begin{aligned}\r\nP(-1) &= (-1)^6 - 1 \\\\\r\n&= 1 - 1 \\\\\r\n&= 0\r\n\\end{aligned}\r\n$$\r\nThe remainder is **0**. This also means $(x+1)$ is a factor of $x^6 - 1$.\r\n\r\n***\r\n\r\n#### Question 18: Finding a Value Using Remainder Theorem\r\n**The polynomial $f(x) = x^3 - (p - 2)x^2 - px + 28$ leaves a remainder of 10 when divided by $x + 3$. Find the value of $p$.**\r\n\r\n**Solution:**\r\nGiven $f(x) = x^3 - (p - 2)x^2 - px + 28$. The divisor is $(x+3)$, so $a=-3$. The remainder is $R=10$.\r\n\r\nBy the Remainder Theorem, $f(-3) = 10$.\r\n$$\r\n\\begin{aligned}\r\n(-3)^3 - (p - 2)(-3)^2 - p(-3) + 28 &= 10 \\\\\r\n-27 - (p - 2)(9) + 3p + 28 &= 10 \\\\\r\n-27 - (9p - 18) + 3p + 28 &= 10 \\\\\r\n-27 - 9p + 18 + 3p + 28 &= 10 \\\\\r\n19 - 6p &= 10 \\\\\r\n9 &= 6p \\\\\r\np &= \\frac{9}{6} = \\frac{3}{2}\r\n\\end{aligned}\r\n$$\r\nThus, **$p = \\frac{3}{2}$**.\r\n\r\n---\r\n\r\n### Long Questions (Solving Cubic Equations)\r\n\r\n***\r\n\r\n#### Question 19: Solve the Polynomial Equation\r\n**Solve: $2x^3 - 9x^2 + 7x + 6 = 0$**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 - 9x^2 + 7x + 6$. We test integer factors of the constant term (6).\r\nTest $x=2$:\r\n$$P(2) = 2(2)^3 - 9(2)^2 + 7(2) + 6 = 16 - 36 + 14 + 6 = 0$$\r\nSo, $(x-2)$ is a factor. We use synthetic division with $a=2$ to find the other factors.\r\n2 | 2 -9 7 6 | 4 -10 -6\r\n 2   -5   -3    0\r\nThe remaining quadratic is $2x^2 - 5x - 3$. We factor this:\r\n$$\r\n\\begin{aligned}\r\n2x^2 - 5x - 3 &= 0 \\\\\r\n2x^2 - 6x + x - 3 &= 0 \\\\\r\n2x(x - 3) + 1(x - 3) &= 0 \\\\\r\n(2x + 1)(x - 3) &= 0\r\n\\end{aligned}\r\n$$\r\nSo, the factors are $(x-2)$, $(x-3)$, and $(2x+1)$. The solutions are the roots of these factors.\r\nThus, **$x = 2$, $x = 3$, and $x = -\\frac{1}{2}$**.\r\n\r\n***\r\n\r\n#### Question 20: Solve the Polynomial Equation\r\n**Solve: $3x^3 - 7x^2 + 4 = 0$**\r\n\r\n**Solution:**\r\nLet $P(x) = 3x^3 - 7x^2 + 0x + 4$. Test factors of 4.\r\nTest $x=1$: $P(1) = 3(1)^3 - 7(1)^2 + 4 = 3 - 7 + 4 = 0$.\r\nSo, $(x-1)$ is a factor. Use synthetic division with $a=1$.\r\n1 | 3 -7 0 4 | 3 -4 -4\r\n 3   -4   -4    0\r\nThe remaining quadratic is $3x^2 - 4x - 4$. We factor this:\r\n$$\r\n\\begin{aligned}\r\n3x^2 - 4x - 4 &= 0 \\\\\r\n3x^2 - 6x + 2x - 4 &= 0 \\\\\r\n3x(x - 2) + 2(x - 2) &= 0 \\\\\r\n(3x + 2)(x - 2) &= 0\r\n\\end{aligned}\r\n$$\r\nThe roots are from $(x-1)$, $(x-2)$, and $(3x+2)$.\r\nThus, **$x = 1$, $x = 2$, and $x = -\\frac{2}{3}$**.\r\n\r\n***\r\n\r\n#### Question 21: Solve the Polynomial Equation\r\n**Solve: $y^3 - 6y^2 + 11y - 6 = 0$**\r\n\r\n**Solution:**\r\nLet $P(y) = y^3 - 6y^2 + 11y - 6$. Test factors of 6.\r\nTest $y=1$: $P(1) = 1 - 6 + 11 - 6 = 0$.\r\nSo, $(y-1)$ is a factor. Use synthetic division with $a=1$.\r\n1 | 1 -6 11 -6 | 1 -5 6\r\n 1   -5    6    0\r\nThe remaining quadratic is $y^2 - 5y + 6$. We factor this:\r\n$$\r\n\\begin{aligned}\r\ny^2 - 5y + 6 &= 0 \\\\\r\n(y-2)(y-3) &= 0\r\n\\end{aligned}\r\n$$\r\nThe roots are from $(y-1)$, $(y-2)$, and $(y-3)$.\r\nThus, **$y = 1, 2, 3$**.\r\n\r\n***\r\n\r\n#### Question 22: Solve the Polynomial Equation\r\n**Solve: $6x^3 + x^2 - 19x + 6 = 0$**\r\n\r\n**Solution:**\r\nLet $P(x) = 6x^3 + x^2 - 19x + 6$. Test factors of 6.\r\nTest $x=-2$: $P(-2) = 6(-8) + (-2)^2 - 19(-2) + 6 = -48 + 4 + 38 + 6 = 0$.\r\nSo, $(x+2)$ is a factor. Use synthetic division with $a=-2$.\r\n-2 | 6 1 -19 6 | -12 22 -6\r\n 6  -11     3    0\r\nThe remaining quadratic is $6x^2 - 11x + 3$. We factor this:\r\n$$\r\n\\begin{aligned}\r\n6x^2 - 11x + 3 &= 0 \\\\\r\n6x^2 - 9x - 2x + 3 &= 0 \\\\\r\n3x(2x - 3) - 1(2x - 3) &= 0 \\\\\r\n(3x - 1)(2x - 3) &= 0\r\n\\end{aligned}\r\n$$\r\nThe roots are from $(x+2)$, $(3x-1)$, and $(2x-3)$.\r\nThus, **$x = -2$, $x = \\frac{1}{3}$, and $x = \\frac{3}{2}$**.\r\n\r\n\r\n\r\n\r\n\r\n\r\nit must consist all concept of all these questions at first and then questions proper note for students to easy grasp concept and genrate this raw markdown code make sure it is not rendered here i want in code editor\r\n\r\n\r\n\r\n\r\nShow thinking\r\nOf course. Here is the complete study guide for the \"Polynomials\" chapter, designed with clear conceptual explanations first, followed by all the solutions with proper spacing. The entire note is provided as a single block of raw markdown code for you to copy and paste.\r\n\r\nMarkdown\r\n\r\n# Grade 10 Optional Mathematics: Polynomials Study Guide\r\n## Unit: Algebra | Chapter: Polynomials | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nWelcome to the comprehensive study guide for the **Polynomials** chapter. This note is designed to make complex concepts easy to understand, with clear definitions and detailed, step-by-step solutions to 35 different problems. By the end of this guide, you will have mastered the key theorems and techniques needed for your exams. 🧠\r\n\r\n### Key Learning Areas\r\n1.  **Remainder Theorem:** A quick way to find the remainder of a polynomial division.\r\n2.  **Factor Theorem:** A method to check if a linear expression is a factor of a polynomial.\r\n3.  **Synthetic Division:** A shortcut for dividing a polynomial by a linear factor.\r\n4.  **Solving Cubic Equations:** A systematic process to find the roots of cubic polynomials.\r\n\r\n---\r\n\r\n## 🔑 Key Concepts & Theorems\r\n\r\nBefore we solve the problems, let\'s understand the main tools we\'ll be using.\r\n\r\n### 1. The Remainder Theorem\r\n\r\n**What it is:** If you divide a polynomial, $P(x)$, by a linear expression of the form $(x - a)$, the remainder will be equal to the value of the polynomial when you substitute $a$ for $x$.\r\n\r\n* **In simple terms:** Remainder = $P(a)$.\r\n\r\n**How to use it:**\r\n1.  Identify your polynomial $P(x)$ and your divisor $(x - a)$.\r\n2.  Find the value of $a$. Remember, if the divisor is $(x+3)$, then $a = -3$.\r\n3.  Calculate $P(a)$ by substituting $a$ into the polynomial. The result is your remainder.\r\n\r\n---\r\n\r\n### 2. The Factor Theorem\r\n\r\n**What it is:** This is a special case of the Remainder Theorem. It gives you a way to check if $(x-a)$ is a factor of a polynomial $P(x)$.\r\n\r\n* **Rule:** $(x - a)$ is a factor of $P(x)$ **if and only if** $P(a) = 0$.\r\n\r\n**How to use it:**\r\n1.  To check if $(x-a)$ is a factor, calculate the remainder $P(a)$.\r\n2.  If the remainder is 0, it **is** a factor.\r\n3.  If the remainder is not 0, it **is not** a factor.\r\n\r\n---\r\n\r\n### 3. Synthetic Division\r\n\r\n**What it is:** A fast, shorthand method for dividing a polynomial by a linear factor of the form $(x-a)$. It is used to find the quotient and the remainder. \r\n\r\n**How to use it:**\r\n1.  Write down the coefficients of the polynomial. **Important:** If any term is missing (like an $x^2$ term in a cubic), you must use a **0** as a placeholder for its coefficient.\r\n2.  Write the value of $a$ from the divisor $(x-a)$ to the left.\r\n3.  Bring down the first coefficient.\r\n4.  Multiply this number by $a$ and write the result under the next coefficient.\r\n5.  Add the numbers in that column.\r\n6.  Repeat steps 4 and 5 until you reach the end.\r\n\r\nThe last number is the remainder, and the other numbers are the coefficients of the quotient.\r\n\r\n**Example:** Divide $x^3 - 7x - 6$ by $(x+2)$. Here, $a=-2$. The coefficients are 1, 0, -7, -6.\r\n-2 | 1 0 -7 -6 | -2 4 6\r\n 1  -2   -3    0  <-- Remainder is 0\r\nThe quotient is $x^2 - 2x - 3$.\r\n\r\n---\r\n\r\n### 4. Solving Cubic Equations\r\n\r\n**What it is:** Finding the values of $x$ that make a cubic polynomial equal to zero. A cubic equation has three roots.\r\n\r\n**How to solve ($ax^3 + bx^2 + cx + d = 0$):**\r\n1.  **Find one factor:** Use the Factor Theorem. Test integer factors of the constant term, $d$. Find a value \'a\' for which $P(a) = 0$. This gives you one factor, $(x-a)$.\r\n2.  **Reduce the polynomial:** Use synthetic division to divide the polynomial by the factor you just found. The result will be a quadratic expression.\r\n3.  **Solve the quadratic:** Solve the resulting quadratic equation using factoring or the quadratic formula to find the other two roots.\r\n4.  **List all roots:** Combine the root from Step 1 with the two roots from Step 3.\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Finding a Value Using Factor Theorem\r\n**If $x - 2$ is a factor of $2x^3 - 6x^2 - 5m - 2$, find the value of $m$.**\r\n\r\n**Solution:**\r\nLet the polynomial be $P(x) = 2x^3 - 6x^2 - 5m - 2$.\r\nThe divisor is $d(x) = x - 2$. Comparing this to $x - a$, we get $a = 2$.\r\n\r\nBy the Factor Theorem, if $(x - a)$ is a factor of $P(x)$, then $P(a) = 0$.\r\n$$\r\n\\begin{aligned}\r\nP(2) &= 0 \\\\\r\n2(2)^3 - 6(2)^2 - 5m - 2 &= 0 \\\\\r\n2(8) - 6(4) - 5m - 2 &= 0 \\\\\r\n16 - 24 - 5m - 2 &= 0 \\\\\r\n-10 - 5m &= 0 \\\\\r\n-10 &= 5m \\\\\r\nm &= -2\r\n\\end{aligned}\r\n$$\r\nThus, **$m = -2$**.\r\n\r\n***\r\n\r\n#### Question 2: Finding a Value Using Factor Theorem\r\n**If $(x + 2)$ is a factor of the polynomial $2x^3 + 3x^2 - mx + 4$, find the value of $m$.**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 + 3x^2 - mx + 4$.\r\nThe divisor is $d(x) = x + 2$. Comparing this to $x - a$, we get $a = -2$.\r\n\r\nBy the Factor Theorem, $P(-2) = 0$.\r\n$$\r\n\\begin{aligned}\r\n2(-2)^3 + 3(-2)^2 - m(-2) + 4 &= 0 \\\\\r\n2(-8) + 3(4) + 2m + 4 &= 0 \\\\\r\n-16 + 12 + 2m + 4 &= 0 \\\\\r\n0 + 2m &= 0 \\\\\r\n2m &= 0 \\\\\r\nm &= 0\r\n\\end{aligned}\r\n$$\r\nThus, **$m = 0$**.\r\n\r\n***\r\n\r\n#### Question 3: Finding a Value Using Remainder Theorem\r\n**If $2x^3 - 4x^2 + kx + 10$ is divided by $(x + 2)$, the remainder is 4. Find the value of $k$.**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 - 4x^2 + kx + 10$.\r\nThe divisor is $d(x) = x + 2$, so $a = -2$. The remainder is $R = 4$.\r\n\r\nBy the Remainder Theorem, $P(a) = R$.\r\n$$\r\n\\begin{aligned}\r\nP(-2) &= 4 \\\\\r\n2(-2)^3 - 4(-2)^2 + k(-2) + 10 &= 4 \\\\\r\n2(-8) - 4(4) - 2k + 10 &= 4 \\\\\r\n-16 - 16 - 2k + 10 &= 4 \\\\\r\n-22 - 2k &= 4 \\\\\r\n-2k &= 26 \\\\\r\nk &= -13\r\n\\end{aligned}\r\n$$\r\nThus, **$k = -13$**.\r\n\r\n***\r\n\r\n#### Question 4: Finding a Value Using Factor Theorem\r\n**If $x - 3$ is a factor of $2x^3 - x^2 + 10x - k$, find the value of $k$.**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 - x^2 + 10x - k$.\r\nThe divisor is $d(x) = x - 3$, so $a = 3$.\r\n\r\nBy the Factor Theorem, $P(3) = 0$.\r\n$$\r\n\\begin{aligned}\r\n2(3)^3 - (3)^2 + 10(3) - k &= 0 \\\\\r\n2(27) - 9 + 30 - k &= 0 \\\\\r\n54 - 9 + 30 - k &= 0 \\\\\r\n75 - k &= 0 \\\\\r\nk &= 75\r\n\\end{aligned}\r\n$$\r\nThus, **$k = 75$**.\r\n\r\n***\r\n\r\n#### Question 5: Finding a Value Using Factor Theorem\r\n**If $x - 5$ is a factor of $x^3 + px^2 + 4x + 5$, find the value of $p$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 + px^2 + 4x + 5$.\r\nThe divisor is $d(x) = x - 5$, so $a = 5$.\r\n\r\nBy the Factor Theorem, $P(5) = 0$.\r\n$$\r\n\\begin{aligned}\r\n(5)^3 + p(5)^2 + 4(5) + 5 &= 0 \\\\\r\n125 + 25p + 20 + 5 &= 0 \\\\\r\n150 + 25p &= 0 \\\\\r\n25p &= -150 \\\\\r\np &= -6\r\n\\end{aligned}\r\n$$\r\nThus, **$p = -6$**.\r\n\r\n***\r\n\r\n#### Question 6: Finding the Remainder\r\n**Using the Remainder Theorem, find the remainder when $8x^3 - 4x^2 + 2x - 5$ is divided by $2x - 1$.**\r\n\r\n**Solution:**\r\nLet $P(x) = 8x^3 - 4x^2 + 2x - 5$.\r\nFor the divisor $2x - 1$, we set it to zero to find the value to substitute: $2x - 1 = 0 \\implies x = \\frac{1}{2}$.\r\n\r\nBy the Remainder Theorem, the remainder is $P(\\frac{1}{2})$.\r\n$$\r\n\\begin{aligned}\r\nP\\left(\\frac{1}{2}\\right) &= 8\\left(\\frac{1}{2}\\right)^3 - 4\\left(\\frac{1}{2}\\right)^2 + 2\\left(\\frac{1}{2}\\right) - 5 \\\\\r\n&= 8\\left(\\frac{1}{8}\\right) - 4\\left(\\frac{1}{4}\\right) + 1 - 5 \\\\\r\n&= 1 - 1 + 1 - 5 \\\\\r\n&= -4\r\n\\end{aligned}\r\n$$\r\nThe remainder is **-4**.\r\n\r\n***\r\n\r\n#### Question 7: Using the Factor Theorem\r\n**If $x - 5$ is a factor of $2x^3 - 7x + (p - 12)$, find the value of p.**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 - 7x + p - 12$.\r\nSince $x-5$ is a factor, $a=5$ and $P(5)=0$.\r\n$$\r\n\\begin{aligned}\r\nP(5) &= 0 \\\\\r\n2(5)^3 - 7(5) + p - 12 &= 0 \\\\\r\n2(125) - 35 + p - 12 &= 0 \\\\\r\n250 - 35 + p - 12 &= 0 \\\\\r\n203 + p &= 0 \\\\\r\np &= -203\r\n\\end{aligned}\r\n$$\r\nThus, **$p = -203$**.\r\n\r\n***\r\n\r\n#### Question 8: Using Synthetic Division\r\n**If $x^3 - 21x - 20 = (x + 1) \\cdot q(x)$, find $q(x)$ by using synthetic division.**\r\n\r\n**Solution:**\r\nThe polynomial is $P(x) = x^3 + 0x^2 - 21x - 20$.\r\nThe divisor is $x+1$, so we use $a = -1$ for synthetic division.\r\n\r\n-1 | 1 0 -21 -20 | -1 1 20\r\n 1  -1   -20     0 \r\n\r\nThe quotient\'s coefficients are 1, -1, and -20. The remainder is 0.\r\nThus, **$q(x) = x^2 - x - 20$**.\r\n\r\n***\r\n\r\n#### Question 9: Using Synthetic Division\r\n**If $x^3 - 19x - 30 = (x + 2) \\cdot q(x)$, find $q(x)$ by using synthetic division.**\r\n\r\n**Solution:**\r\nThe polynomial is $P(x) = x^3 + 0x^2 - 19x - 30$.\r\nThe divisor is $x+2$, so we use $a = -2$.\r\n\r\n-2 | 1 0 -19 -30 | -2 4 30\r\n 1  -2   -15     0 \r\n\r\nThe quotient\'s coefficients are 1, -2, and -15. The remainder is 0.\r\nThus, **$q(x) = x^2 - 2x - 15$**.\r\n\r\n***\r\n\r\n#### Question 10: State and Apply the Factor Theorem\r\n**State the Factor Theorem. Use it to determine whether $x + 3$ is a factor of $x^3 - 8x + 3$.**\r\n\r\n**Solution:**\r\n**Factor Theorem:** A polynomial $P(x)$ has a factor $(x - a)$ if and only if $P(a) = 0$.\r\n\r\nLet $P(x) = x^3 - 8x + 3$. To check if $(x+3)$ is a factor, we test $P(-3)$.\r\n$$\r\n\\begin{aligned}\r\nP(-3) &= (-3)^3 - 8(-3) + 3 \\\\\r\n&= -27 + 24 + 3 \\\\\r\n&= 0\r\n\\end{aligned}\r\n$$\r\nSince $P(-3) = 0$, **yes, $(x+3)$ is a factor** of the polynomial.\r\n\r\n***\r\n\r\n#### Question 11: State and Apply the Remainder Theorem\r\n**State the Remainder Theorem. When $x^3 + 5x^2 - 6x - 16$ is divided by $(x + 2)$, find the remainder.**\r\n\r\n**Solution:**\r\n**Remainder Theorem:** If a polynomial $P(x)$ is divided by $(x - a)$, the remainder is $P(a)$.\r\n\r\nLet $P(x) = x^3 + 5x^2 - 6x - 16$. The divisor is $(x+2)$, so $a = -2$. The remainder is $P(-2)$.\r\n$$\r\n\\begin{aligned}\r\nP(-2) &= (-2)^3 + 5(-2)^2 - 6(-2) - 16 \\\\\r\n&= -8 + 5(4) + 12 - 16 \\\\\r\n&= -8 + 20 + 12 - 16 \\\\\r\n&= 8\r\n\\end{aligned}\r\n$$\r\nThe remainder is **8**.\r\n\r\n***\r\n\r\n#### Question 12: Finding a Value Using Remainder Theorem\r\n**If the polynomial $x^3 + 6x^2 + kx + 10$ is divided by $(x + 2)$, the remainder is 4. Find the value of $k$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 + 6x^2 + kx + 10$. The divisor is $(x+2)$, so $a=-2$. The remainder is $R=4$.\r\n\r\nBy the Remainder Theorem, $P(-2) = 4$.\r\n$$\r\n\\begin{aligned}\r\n(-2)^3 + 6(-2)^2 + k(-2) + 10 &= 4 \\\\\r\n-8 + 6(4) - 2k + 10 &= 4 \\\\\r\n-8 + 24 - 2k + 10 &= 4 \\\\\r\n26 - 2k &= 4 \\\\\r\n22 &= 2k \\\\\r\nk &= 11\r\n\\end{aligned}\r\n$$\r\nThus, **$k = 11$**.\r\n\r\n***\r\n\r\n#### Question 13: Finding a Value Using Factor Theorem\r\n**If $(x + 2)$ is a factor of $x^3 - 19x - p$, find the value of $p$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 - 19x - p$. Since $(x+2)$ is a factor, $a=-2$ and $P(-2)=0$.\r\n$$\r\n\\begin{aligned}\r\nP(-2) &= 0 \\\\\r\n(-2)^3 - 19(-2) - p &= 0 \\\\\r\n-8 + 38 - p &= 0 \\\\\r\n30 - p &= 0 \\\\\r\np &= 30\r\n\\end{aligned}\r\n$$\r\nThus, **$p = 30$**.\r\n\r\n***\r\n\r\n#### Question 14: Finding a Value Using Factor Theorem\r\n**If $(x - m)$ is a factor of $x^3 - mx^2 - 4x + m + 12$, calculate the value of $m$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 - mx^2 - 4x + m + 12$. Since $(x-m)$ is a factor, $a=m$ and $P(m)=0$.\r\n$$\r\n\\begin{aligned}\r\nP(m) &= 0 \\\\\r\n(m)^3 - m(m)^2 - 4(m) + m + 12 &= 0 \\\\\r\nm^3 - m^3 - 4m + m + 12 &= 0 \\\\\r\n-3m + 12 &= 0 \\\\\r\n12 &= 3m \\\\\r\nm &= 4\r\n\\end{aligned}\r\n$$\r\nThus, **$m = 4$**.\r\n\r\n***\r\n\r\n#### Question 15: Finding a Value Using Factor Theorem\r\n**If $x - 2$ is a factor of $x^3 - kx^2 + x - 6$, find the value of $k$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 - kx^2 + x - 6$. Since $(x-2)$ is a factor, $a=2$ and $P(2)=0$.\r\n$$\r\n\\begin{aligned}\r\nP(2) &= 0 \\\\\r\n(2)^3 - k(2)^2 + 2 - 6 &= 0 \\\\\r\n8 - 4k - 4 &= 0 \\\\\r\n4 - 4k &= 0 \\\\\r\n4 &= 4k \\\\\r\nk &= 1\r\n\\end{aligned}\r\n$$\r\nThus, **$k=1$**.\r\n\r\n***\r\n\r\n#### Question 16: Using the Factor Theorem\r\n**Use the Factor Theorem to determine whether $x + 4$ is a factor of the polynomial $x^3 - 15x + 4$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 - 15x + 4$. To check if $(x+4)$ is a factor, we test if $P(-4) = 0$.\r\n$$\r\n\\begin{aligned}\r\nP(-4) &= (-4)^3 - 15(-4) + 4 \\\\\r\n&= -64 + 60 + 4 \\\\\r\n&= 0\r\n\\end{aligned}\r\n$$\r\nSince $P(-4) = 0$, **yes, $(x+4)$ is a factor**.\r\n\r\n***\r\n\r\n#### Question 17: Using the Remainder Theorem\r\n**Find the remainder when the polynomial $x^6 - 1$ is divided by $x + 1$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^6 - 1$. The divisor is $(x+1)$, so $a=-1$. The remainder is $P(-1)$.\r\n$$\r\n\\begin{aligned}\r\nP(-1) &= (-1)^6 - 1 \\\\\r\n&= 1 - 1 \\\\\r\n&= 0\r\n\\end{aligned}\r\n$$\r\nThe remainder is **0**. This also means $(x+1)$ is a factor of $x^6 - 1$.\r\n\r\n***\r\n\r\n#### Question 18: Finding a Value Using Remainder Theorem\r\n**The polynomial $f(x) = x^3 - (p - 2)x^2 - px + 28$ leaves a remainder of 10 when divided by $x + 3$. Find the value of $p$.**\r\n\r\n**Solution:**\r\nGiven $f(x) = x^3 - (p - 2)x^2 - px + 28$. The divisor is $(x+3)$, so $a=-3$. The remainder is $R=10$.\r\n\r\nBy the Remainder Theorem, $f(-3) = 10$.\r\n$$\r\n\\begin{aligned}\r\n(-3)^3 - (p - 2)(-3)^2 - p(-3) + 28 &= 10 \\\\\r\n-27 - (p - 2)(9) + 3p + 28 &= 10 \\\\\r\n-27 - (9p - 18) + 3p + 28 &= 10 \\\\\r\n-27 - 9p + 18 + 3p + 28 &= 10 \\\\\r\n19 - 6p &= 10 \\\\\r\n9 &= 6p \\\\\r\np &= \\frac{9}{6} = \\frac{3}{2}\r\n\\end{aligned}\r\n$$\r\nThus, **$p = \\frac{3}{2}$**.\r\n\r\n---\r\n\r\n### Long Questions (Solving Cubic Equations)\r\n\r\n***\r\n\r\n#### Question 19: Solve the Polynomial Equation\r\n**Solve: $2x^3 - 9x^2 + 7x + 6 = 0$**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 - 9x^2 + 7x + 6$. Test integer factors of 6.\r\nWhen $x=2$, $P(2) = 2(8) - 9(4) + 7(2) + 6 = 16 - 36 + 14 + 6 = 0$.\r\nSo, $(x-2)$ is a factor. Use synthetic division with $a=2$.\r\n2 | 2 -9 7 6 | 4 -10 -6\r\n 2   -5   -3    0\r\nThe remaining quadratic is $2x^2 - 5x - 3$. Factor it:\r\n$$\r\n\\begin{aligned}\r\n2x^2 - 5x - 3 &= 0 \\\\\r\n2x^2 - 6x + x - 3 &= 0 \\\\\r\n2x(x - 3) + 1(x - 3) &= 0 \\\\\r\n(2x + 1)(x - 3) &= 0\r\n\\end{aligned}\r\n$$\r\nThe factors are $(x-2)$, $(x-3)$, and $(2x+1)$.\r\nThus, the solutions are **$x = 2$, $x = 3$, and $x = -\\frac{1}{2}$**.\r\n\r\n***\r\n\r\n#### Question 20: Solve the Polynomial Equation\r\n**Solve: $3x^3 - 7x^2 + 4 = 0$**\r\n\r\n**Solution:**\r\nLet $P(x) = 3x^3 - 7x^2 + 0x + 4$. Test factors of 4.\r\nWhen $x=1$, $P(1) = 3 - 7 + 4 = 0$.\r\nSo, $(x-1)$ is a factor. Use synthetic division with $a=1$.\r\n1 | 3 -7 0 4 | 3 -4 -4\r\n 3   -4   -4    0\r\nThe remaining quadratic is $3x^2 - 4x - 4$. Factor it:\r\n$$\r\n\\begin{aligned}\r\n3x^2 - 4x - 4 &= 0 \\\\\r\n3x^2 - 6x + 2x - 4 &= 0 \\\\\r\n3x(x - 2) + 2(x - 2) &= 0 \\\\\r\n(3x + 2)(x - 2) &= 0\r\n\\end{aligned}\r\n$$\r\nThe solutions are from $(x-1)$, $(x-2)$, and $(3x+2)$.\r\nThus, **$x = 1$, $x = 2$, and $x = -\\frac{2}{3}$**.\r\n\r\n***\r\n\r\n#### Question 21: Solve the Polynomial Equation\r\n**Solve: $y^3 - 6y^2 + 11y - 6 = 0$**\r\n\r\n**Solution:**\r\nLet $P(y) = y^3 - 6y^2 + 11y - 6$. Test factors of 6.\r\nWhen $y=1$, $P(1) = 1 - 6 + 11 - 6 = 0$.\r\nSo, $(y-1)$ is a factor. Use synthetic division with $a=1$.\r\n1 | 1 -6 11 -6 | 1 -5 6\r\n 1   -5    6    0\r\nThe remaining quadratic is $y^2 - 5y + 6$. Factor it:\r\n$$\r\ny^2 - 5y + 6 = (y-2)(y-3) = 0\r\n$$\r\nThe roots are from $(y-1)$, $(y-2)$, and $(y-3)$.\r\nThus, **$y = 1, 2, 3$**.\r\n\r\n***\r\n\r\n#### Question 22: Solve the Polynomial Equation\r\n**Solve: $6x^3 + x^2 - 19x + 6 = 0$**\r\n\r\n**Solution:**\r\nLet $P(x) = 6x^3 + x^2 - 19x + 6$. Test factors of 6.\r\nWhen $x=-2$, $P(-2) = 6(-8) + 4 - 19(-2) + 6 = -48 + 4 + 38 + 6 = 0$.\r\nSo, $(x+2)$ is a factor. Use synthetic division with $a=-2$.\r\n-2 | 6 1 -19 6 | -12 22 -6\r\n 6  -11     3    0\r\nThe remaining quadratic is $6x^2 - 11x + 3$. Factor it:\r\n$$\r\n\\begin{aligned}\r\n6x^2 - 11x + 3 &= 0 \\\\\r\n6x^2 - 9x - 2x + 3 &= 0 \\\\\r\n3x(2x - 3) - 1(2x - 3) &= 0 \\\\\r\n(3x - 1)(2x - 3) &= 0\r\n\\end{aligned}\r\n$$\r\nThe roots are from $(x+2)$, $(3x-1)$, and $(2x-3)$.\r\nThus, **$x = -2$, $x = \\frac{1}{3}$, and $x = \\frac{3}{2}$**.', 1, '2025-07-25 01:11:38', '2025-09-18 07:21:49');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(140, 13103, '# Grade 10 Optional Mathematics: Sequence and Series Study Guide\r\n## Unit: Algebra | Chapter: Sequence and Series | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nWelcome to the comprehensive study guide for the **Sequence and Series** chapter. This note is designed to make complex concepts easy to understand, with clear definitions, essential formulas, and detailed, step-by-step solutions to a wide range of problems. By the end of this guide, you will have mastered Arithmetic and Geometric progressions. 📈\r\n\r\n### Key Learning Areas\r\n1.  **Definitions:** Understanding Sequences, Series, and their different types.\r\n2.  **Arithmetic Progression (A.P.):** Working with sequences that have a common difference.\r\n3.  **Geometric Progression (G.P.):** Working with sequences that have a common ratio.\r\n4.  **Means:** Calculating and inserting Arithmetic and Geometric Means.\r\n5.  **Summation:** Finding the sum of a given number of terms in a series.\r\n\r\n---\r\n\r\n## 🔑 Key Concepts & Formulas\r\n\r\nBefore solving the problems, let\'s review the fundamental concepts and formulas.\r\n\r\n### 1. Core Definitions\r\n\r\n* **Sequence:** A set of numbers arranged in a definite order according to a specific mathematical rule.\r\n    * *Example:* `1, 3, 5, 7, ...` is a sequence where each term is 2 more than the previous one.\r\n\r\n* **Series:** The sum of the terms in a sequence.\r\n    * *Example:* `1 + 3 + 5 + 7 + ...` is the series corresponding to the sequence above.\r\n\r\n* **Arithmetic Progression (A.P.):** A sequence where the difference between any two consecutive terms is constant. This constant is called the **common difference (d)**.\r\n    * *Example:* `5, 9, 13, 17, ...` (common difference is 4).\r\n\r\n* **Geometric Progression (G.P.):** A sequence where the ratio of any term to its preceding term is constant. This constant is called the **common ratio (r)**.\r\n    * *Example:* `3, 6, 12, 24, ...` (common ratio is 2).\r\n\r\n* **Arithmetic Mean (A.M.):** A term (or terms) inserted between two given terms that maintains the arithmetic progression.\r\n\r\n* **Geometric Mean (G.M.):** A term (or terms) inserted between two given terms that maintains the geometric progression.\r\n\r\n---\r\n\r\n### 2. Formulas for Arithmetic Progression (A.P.)\r\n\r\nLet **a** be the first term, **d** be the common difference, and **n** be the number of terms.\r\n\r\n* **nth Term ($t_n$):**\r\n    $$t_n = a + (n-1)d$$\r\n\r\n* **Sum of First n Terms ($S_n$):**\r\n    $$S_n = \\frac{n}{2}[2a + (n-1)d]$$\r\n\r\n* **Arithmetic Mean between a and b:**\r\n    $$A.M. = \\frac{a+b}{2}$$\r\n\r\n* **Inserting \'n\' Arithmetic Means between a and b:**\r\n    The common difference `d` is found using:\r\n    $$d = \\frac{b-a}{n+1}$$\r\n\r\n---\r\n\r\n### 3. Formulas for Geometric Progression (G.P.)\r\n\r\nLet **a** be the first term, **r** be the common ratio, and **n** be the number of terms.\r\n\r\n* **nth Term ($t_n$):**\r\n    $$t_n = ar^{n-1}$$\r\n\r\n* **Sum of First n Terms ($S_n$):**\r\n    $$S_n = \\frac{a(r^n - 1)}{r-1} \\quad (\\text{if } r > 1)$$\r\n    $$S_n = \\frac{a(1 - r^n)}{1-r} \\quad (\\text{if } r < 1)$$\r\n\r\n* **Geometric Mean between a and b:**\r\n    $$G.M. = \\sqrt{ab}$$\r\n\r\n* **Inserting \'n\' Geometric Means between a and b:**\r\n    The common ratio `r` is found using:\r\n    $$r = \\left(\\frac{b}{a}\\right)^{\\frac{1}{n+1}}$$\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Arithmetic and Geometric Mean\r\n**Find the arithmetic mean and geometric mean between 4 and 16.**\r\n\r\n**Solution:**\r\nHere, $a = 4$ and $b = 16$.\r\n\r\n**Arithmetic Mean (A.M.):**\r\n$$\r\n\\begin{aligned}\r\nA.M. &= \\frac{a+b}{2} \\\\\r\n&= \\frac{4+16}{2} \\\\\r\n&= \\frac{20}{2} = 10\r\n\\end{aligned}\r\n$$\r\n**Geometric Mean (G.M.):**\r\n$$\r\n\\begin{aligned}\r\nG.M. &= \\sqrt{ab} \\\\\r\n&= \\sqrt{4 \\times 16} \\\\\r\n&= \\sqrt{64} = 8\r\n\\end{aligned}\r\n$$\r\nThe A.M. is **10** and the G.M. is **8**.\r\n\r\n***\r\n\r\n#### Question 2: Arithmetic and Geometric Mean\r\n**Find the arithmetic mean and geometric mean between 3 and 27.**\r\n\r\n**Solution:**\r\nHere, $a = 3$ and $b = 27$.\r\n\r\n**Arithmetic Mean (A.M.):**\r\n$$\r\nA.M. = \\frac{3+27}{2} = \\frac{30}{2} = 15\r\n$$\r\n**Geometric Mean (G.M.):**\r\n$$\r\nG.M. = \\sqrt{3 \\times 27} = \\sqrt{81} = 9\r\n$$\r\nThe A.M. is **15** and the G.M. is **9**.\r\n\r\n***\r\n\r\n#### Question 3: Geometric Sequence\r\n**If 4, a, and 16 are in a geometric sequence, find the value of a.**\r\n\r\n**Solution:**\r\nIn a geometric sequence, the middle term is the geometric mean of the first and last terms.\r\n$$\r\n\\begin{aligned}\r\na &= \\sqrt{4 \\times 16} \\\\\r\na &= \\sqrt{64} \\\\\r\na &= 8\r\n\\end{aligned}\r\n$$\r\nAlternatively, the ratio of consecutive terms is constant:\r\n$$\r\n\\begin{aligned}\r\n\\frac{a}{4} &= \\frac{16}{a} \\\\\r\na^2 &= 64 \\\\\r\na &= 8\r\n\\end{aligned}\r\n$$\r\nThus, **$a=8$**.\r\n\r\n***\r\n\r\n#### Question 4: Relation between Means\r\n**If the arithmetic mean between 4 and x is 34, find their geometric mean.**\r\n\r\n**Solution:**\r\nFirst, use the A.M. formula to find $x$.\r\n$$\r\n\\begin{aligned}\r\nA.M. &= \\frac{4+x}{2} \\\\\r\n34 &= \\frac{4+x}{2} \\\\\r\n68 &= 4+x \\\\\r\nx &= 64\r\n\\end{aligned}\r\n$$\r\nNow, find the G.M. between 4 and 64.\r\n$$\r\nG.M. = \\sqrt{4 \\times 64} = \\sqrt{256} = 16\r\n$$\r\nThe geometric mean is **16**.\r\n\r\n***\r\n\r\n#### Question 5: Relation between Means\r\n**If the geometric mean of 2 and x is 4, find their arithmetic mean.**\r\n\r\n**Solution:**\r\nFirst, use the G.M. formula to find $x$.\r\n$$\r\n\\begin{aligned}\r\nG.M. &= \\sqrt{2x} \\\\\r\n4 &= \\sqrt{2x} \\\\\r\n16 &= 2x \\\\\r\nx &= 8\r\n\\end{aligned}\r\n$$\r\nNow, find the A.M. between 2 and 8.\r\n$$\r\nA.M. = \\frac{2+8}{2} = \\frac{10}{2} = 5\r\n$$\r\nThe arithmetic mean is **5**.\r\n\r\n***\r\n\r\n#### Question 6: Geometric Sequence\r\n**If $m + 2$, $m + 8$, and $17 + m$ are in geometric sequence, find the value of m.**\r\n\r\n**Solution:**\r\nSince the terms are in a G.P., the ratio of consecutive terms is constant.\r\n$$\r\n\\begin{aligned}\r\n\\frac{m+8}{m+2} &= \\frac{17+m}{m+8} \\\\\r\n(m+8)^2 &= (m+2)(17+m) \\\\\r\nm^2 + 16m + 64 &= 17m + m^2 + 34 + 2m \\\\\r\nm^2 + 16m + 64 &= m^2 + 19m + 34 \\\\\r\n16m + 64 &= 19m + 34 \\\\\r\n30 &= 3m \\\\\r\nm &= 10\r\n\\end{aligned}\r\n$$\r\nThus, **$m=10$**.\r\n\r\n***\r\n\r\n#### Question 7: Finding the Term Number in an A.P.\r\n**Which term of the series $5 + 9 + 13 + ...$ is 85?**\r\n\r\n**Solution:**\r\nThis is an arithmetic series with:\r\nFirst term, $a = 5$.\r\nCommon difference, $d = 9 - 5 = 4$.\r\nLet the nth term, $t_n = 85$.\r\n\r\nUsing the formula $t_n = a + (n-1)d$:\r\n$$\r\n\\begin{aligned}\r\n85 &= 5 + (n-1)4 \\\\\r\n80 &= (n-1)4 \\\\\r\n20 &= n-1 \\\\\r\nn &= 21\r\n\\end{aligned}\r\n$$\r\nThus, 85 is the **21st term**.\r\n\r\n***\r\n\r\n#### Question 8: Finding the Term Number in a G.P.\r\n**Which term of the series $3 + 6 + 12 + ...$ is 192?**\r\n\r\n**Solution:**\r\nThis is a geometric series with:\r\nFirst term, $a = 3$.\r\nCommon ratio, $r = 6/3 = 2$.\r\nLet the nth term, $t_n = 192$.\r\n\r\nUsing the formula $t_n = ar^{n-1}$:\r\n$$\r\n\\begin{aligned}\r\n192 &= 3 \\cdot (2)^{n-1} \\\\\r\n64 &= (2)^{n-1} \\\\\r\n2^6 &= 2^{n-1} \\\\\r\n6 &= n-1 \\\\\r\nn &= 7\r\n\\end{aligned}\r\n$$\r\nThus, 192 is the **7th term**.\r\n\r\n***\r\n\r\n#### Question 9: Sum of a G.P.\r\n**Find the sum of the series $9 + 3 + 1 + ... + \\frac{1}{27}$.**\r\n\r\n**Solution:**\r\nThis is a geometric series with $a=9$ and $r = 3/9 = 1/3$. The last term is $t_n = 1/27$.\r\n\r\n**Step 1: Find the number of terms (n).**\r\n$$\r\n\\begin{aligned}\r\nt_n &= ar^{n-1} \\\\\r\n\\frac{1}{27} &= 9 \\cdot \\left(\\frac{1}{3}\\right)^{n-1} \\\\\r\n\\frac{1}{243} &= \\left(\\frac{1}{3}\\right)^{n-1} \\\\\r\n\\left(\\frac{1}{3}\\right)^5 &= \\left(\\frac{1}{3}\\right)^{n-1} \\\\\r\n5 &= n-1 \\\\\r\nn &= 6\r\n\\end{aligned}\r\n$$\r\n**Step 2: Find the sum of the 6 terms ($S_6$).**\r\nSince $r < 1$, we use $S_n = \\frac{a(1-r^n)}{1-r}$.\r\n$$\r\n\\begin{aligned}\r\nS_6 &= \\frac{9(1 - (1/3)^6)}{1 - 1/3} \\\\\r\n&= \\frac{9(1 - 1/729)}{2/3} \\\\\r\n&= \\frac{9(728/729)}{2/3} \\\\\r\n&= 9 \\times \\frac{728}{729} \\times \\frac{3}{2} \\\\\r\n&= \\frac{364}{27}\r\n\\end{aligned}\r\n$$\r\nThe sum of the series is **$\\frac{364}{27}$**.\r\n\r\n***\r\n\r\n#### Question 10: Product of Terms in a G.P.\r\n**If the third term of a geometric series is 3, find the product of the first five terms.**\r\n\r\n**Solution:**\r\nLet the first five terms of the G.P. be $a, ar, ar^2, ar^3, ar^4$.\r\nWe are given that the third term, $t_3 = ar^2 = 3$.\r\n\r\nThe product of the first five terms is:\r\n$$\r\n\\begin{aligned}\r\nP_5 &= (a)(ar)(ar^2)(ar^3)(ar^4) \\\\\r\n&= a^5 r^{0+1+2+3+4} \\\\\r\n&= a^5 r^{10} \\\\\r\n&= (ar^2)^5\r\n\\end{aligned}\r\n$$\r\nSince we know $ar^2 = 3$:\r\n$$\r\nP_5 = (3)^5 = 243\r\n$$\r\nThe product is **243**.\r\n\r\n***\r\n\r\n#### Question 11: Finding Terms in a G.P.\r\n**If $\\frac{1}{3}, p, q, 9$ are in a geometric sequence, find the values of p and q.**\r\n\r\n**Solution:**\r\nHere, $a = 1/3$ and the 4th term, $t_4 = 9$.\r\nFirst, find the common ratio $r$.\r\n$$\r\n\\begin{aligned}\r\nt_4 &= ar^{4-1} \\\\\r\n9 &= \\left(\\frac{1}{3}\\right)r^3 \\\\\r\n27 &= r^3 \\\\\r\nr &= 3\r\n\\end{aligned}\r\n$$\r\nNow find $p$ and $q$:\r\n$p$ is the 2nd term: $p = ar = (\\frac{1}{3})(3) = 1$.\r\n$q$ is the 3rd term: $q = ar^2 = (\\frac{1}{3})(3)^2 = 3$.\r\n\r\nThus, **$p=1$ and $q=3$**.\r\n\r\n***\r\n\r\n#### Question 12: Finding Common Ratio\r\n**Find the common ratio of a geometric sequence, whose first term is 2 and the third term is 242.**\r\n\r\n**Solution:**\r\nGiven $a=2$ and $t_3 = 242$.\r\n$$\r\n\\begin{aligned}\r\nt_3 &= ar^{3-1} \\\\\r\n242 &= 2 \\cdot r^2 \\\\\r\n121 &= r^2 \\\\\r\nr &= \\pm 11\r\n\\end{aligned}\r\n$$\r\nThe common ratio is **$\\pm 11$**.\r\n\r\n***\r\n\r\n#### Question 13: Finding Terms in an A.P.\r\n**If $5, x, y, 11$ is an arithmetic series. Find the value of x and y.**\r\n\r\n**Solution:**\r\nHere, $a = 5$ and the 4th term, $t_4 = 11$.\r\nFirst, find the common difference $d$.\r\n$$\r\n\\begin{aligned}\r\nt_4 &= a + (4-1)d \\\\\r\n11 &= 5 + 3d \\\\\r\n6 &= 3d \\\\\r\nd &= 2\r\n\\end{aligned}\r\n$$\r\nNow find $x$ and $y$:\r\n$x$ is the 2nd term: $x = a+d = 5+2 = 7$.\r\n$y$ is the 3rd term: $y = a+2d = 5+2(2) = 9$.\r\n\r\nThus, **$x=7$ and $y=9$**.\r\n\r\n***\r\n\r\n#### Question 14: Sum of a G.P.\r\n**Find the sum of the series: 2 + 4 + 8 + ... upto 8 terms.**\r\n\r\n**Solution:**\r\nThis is a G.P. with $a=2$, $r=2$, and $n=8$.\r\nSince $r > 1$, we use the formula $S_n = \\frac{a(r^n - 1)}{r-1}$.\r\n$$\r\n\\begin{aligned}\r\nS_8 &= \\frac{2(2^8 - 1)}{2-1} \\\\\r\n&= \\frac{2(256 - 1)}{1} \\\\\r\n&= 2(255) \\\\\r\n&= 510\r\n\\end{aligned}\r\n$$\r\nThe sum of the first 8 terms is **510**.\r\n\r\n***\r\n\r\n### Long Questions\r\n\r\n***\r\n\r\n#### Question 15: A.P. and G.P. Relationship\r\n**The product of the first five terms of a geometric series is 243. If the third term of the geometric series is equal to the tenth term of an arithmetic series, find the sum of the first 19 terms of the arithmetic series.**\r\n\r\n**Solution:**\r\n**For the G.P.:**\r\nProduct $P_5 = (a)(ar)(ar^2)(ar^3)(ar^4) = a^5r^{10} = (ar^2)^5$.\r\n$$\r\n\\begin{aligned}\r\n(ar^2)^5 &= 243 \\\\\r\n(ar^2)^5 &= 3^5 \\\\\r\nar^2 &= 3\r\n\\end{aligned}\r\n$$\r\nSo, the third term of the G.P. ($t_3$) is 3.\r\n\r\n**For the A.P.:**\r\nWe are given that the tenth term ($t_{10}$) of the A.P. is equal to the third term of the G.P.\r\n$$\r\nt_{10} = a + (10-1)d = a + 9d = 3\r\n$$\r\n**Sum of the A.P.:**\r\nWe need to find the sum of the first 19 terms ($S_{19}$).\r\n$$\r\n\\begin{aligned}\r\nS_{19} &= \\frac{19}{2}[2a + (19-1)d] \\\\\r\n&= \\frac{19}{2}[2a + 18d] \\\\\r\n&= \\frac{19}{2} \\cdot 2(a + 9d) \\\\\r\n&= 19(a + 9d)\r\n\\end{aligned}\r\n$$\r\nSince we know $a+9d = 3$:\r\n$$\r\nS_{19} = 19(3) = 57\r\n$$\r\nThe sum of the first 19 terms is **57**.\r\n\r\n***\r\n\r\n#### Question 16: Finding Terms from Means\r\n**There are five geometric means between a and b. If the second mean is 2 and the last mean is 16, find the values of a and b.**\r\n\r\n**Solution:**\r\nThe sequence is $a, m_1, m_2, m_3, m_4, m_5, b$.\r\nThis is a G.P. with $n = 5+2 = 7$ terms.\r\n$a$ is the 1st term.\r\n$m_2$ (second mean) is the 3rd term of the sequence: $t_3 = ar^2 = 2$.\r\n$m_5$ (last mean) is the 6th term of the sequence: $t_6 = ar^5 = 16$.\r\n\r\nDivide the two equations:\r\n$$\r\n\\begin{aligned}\r\n\\frac{ar^5}{ar^2} &= \\frac{16}{2} \\\\\r\nr^3 &= 8 \\\\\r\nr &= 2\r\n\\end{aligned}\r\n$$\r\nNow find $a$ using $ar^2 = 2$:\r\n$$\r\n\\begin{aligned}\r\na(2)^2 &= 2 \\\\\r\n4a &= 2 \\\\\r\na &= \\frac{1}{2}\r\n\\end{aligned}\r\n$$\r\nFind $b$, which is the 7th term ($t_7$):\r\n$$\r\nb = ar^6 = \\left(\\frac{1}{2}\\right)(2)^6 = \\frac{64}{2} = 32\r\n$$\r\nThus, **$a = 1/2$ and $b = 32$**.\r\n\r\n***\r\n\r\n#### Question 17: Sum of an A.P. from Sums\r\n**If the sum of the first 9 terms of an arithmetic series is 72 and the sum of the first 17 terms is 289, find the sum of the first 25 terms.**\r\n\r\n**Solution:**\r\nGiven $S_9 = 72$ and $S_{17} = 289$.\r\nFor $S_9$:\r\n$$\r\n\\begin{aligned}\r\n\\frac{9}{2}(2a + 8d) &= 72 \\\\\r\n9(a+4d) &= 72 \\\\\r\na+4d &= 8 \\implies a = 8-4d \\quad \\text{(Eq. 1)}\r\n\\end{aligned}\r\n$$\r\nFor $S_{17}$:\r\n$$\r\n\\begin{aligned}\r\n\\frac{17}{2}(2a + 16d) &= 289 \\\\\r\n17(a+8d) &= 289 \\\\\r\na+8d &= 17 \\quad \\text{(Eq. 2)}\r\n\\end{aligned}\r\n$$\r\nSubstitute Eq. 1 into Eq. 2:\r\n$$\r\n\\begin{aligned}\r\n(8-4d)+8d &= 17 \\\\\r\n8+4d &= 17 \\\\\r\n4d &= 9 \\implies d=9/4\r\n\\end{aligned}\r\n$$\r\nNow find $a$: $a = 8 - 4(9/4) = 8-9 = -1$.\r\n\r\nFind $S_{25}$:\r\n$$\r\n\\begin{aligned}\r\nS_{25} &= \\frac{25}{2}[2(-1) + (25-1)(9/4)] \\\\\r\n&= \\frac{25}{2}[-2 + 24(9/4)] \\\\\r\n&= \\frac{25}{2}[-2 + 6 \\times 9] \\\\\r\n&= \\frac{25}{2}[-2 + 54] = \\frac{25}{2}(52) \\\\\r\n&= 25 \\times 26 = 650\r\n\\end{aligned}\r\n$$\r\nThe sum of the first 25 terms is **650**.\r\n\r\n***\r\n\r\n#### Question 18: Sum of a G.P. from Two Terms\r\n**If the second term and fifth term of the geometric series are 15 and 405 respectively, find the sum of the first 6 terms of the series.**\r\n\r\n**Solution:**\r\nGiven $t_2 = ar = 15$ and $t_5 = ar^4 = 405$.\r\nDivide the two equations:\r\n$$\r\n\\begin{aligned}\r\n\\frac{ar^4}{ar} &= \\frac{405}{15} \\\\\r\nr^3 &= 27 \\\\\r\nr &= 3\r\n\\end{aligned}\r\n$$\r\nFind $a$: $a(3) = 15 \\implies a=5$.\r\nFind $S_6$ (since $r>1$):\r\n$$\r\n\\begin{aligned}\r\nS_6 &= \\frac{5(3^6-1)}{3-1} \\\\\r\n&= \\frac{5(729-1)}{2} \\\\\r\n&= \\frac{5(728)}{2} = 5(364) = 1820\r\n\\end{aligned}\r\n$$\r\nThe sum of the first 6 terms is **1820**.\r\n\r\n***\r\n\r\n#### Question 19: Sum of a G.P. from Two Terms\r\n**Find the sum of the first 9 terms of a geometric series whose third term and seventh term are 20 and 320 respectively.**\r\n\r\n**Solution:**\r\nGiven $t_3 = ar^2 = 20$ and $t_7 = ar^6 = 320$.\r\nDivide the two equations:\r\n$$\r\n\\begin{aligned}\r\n\\frac{ar^6}{ar^2} &= \\frac{320}{20} \\\\\r\nr^4 &= 16 \\\\\r\nr &= 2\r\n\\end{aligned}\r\n$$\r\nFind $a$: $a(2)^2 = 20 \\implies 4a=20 \\implies a=5$.\r\nFind $S_9$:\r\n$$\r\n\\begin{aligned}\r\nS_9 &= \\frac{5(2^9-1)}{2-1} \\\\\r\n&= 5(512-1) \\\\\r\n&= 5(511) = 2555\r\n\\end{aligned}\r\n$$\r\nThe sum of the first 9 terms is **2555**.\r\n\r\n***\r\n\r\n#### Question 20: Inserting Arithmetic Means\r\n**Insert four arithmetic means between 5 and 25.**\r\n\r\n**Solution:**\r\nWe are inserting $n=4$ means between $a=5$ and $b=25$.\r\nFirst, find the common difference $d$:\r\n$$\r\nd = \\frac{b-a}{n+1} = \\frac{25-5}{4+1} = \\frac{20}{5} = 4\r\n$$\r\nThe means are:\r\n* $m_1 = a+d = 5+4 = 9$\r\n* $m_2 = a+2d = 5+2(4) = 13$\r\n* $m_3 = a+3d = 5+3(4) = 17$\r\n* $m_4 = a+4d = 5+4(4) = 21$\r\n\r\nThe four arithmetic means are **9, 13, 17, 21**.\r\n\r\n***\r\n\r\n#### Question 21: Sum of a G.P.\r\n**Find the sum of the first 8 terms of a geometric series whose fourth term is 54 and common ratio is 3.**\r\n\r\n**Solution:**\r\nGiven $t_4 = 54$ and $r=3$.\r\nFirst, find the first term $a$.\r\n$$\r\n\\begin{aligned}\r\nt_4 &= ar^{3} \\\\\r\n54 &= a(3)^3 \\\\\r\n54 &= 27a \\\\\r\na &= 2\r\n\\end{aligned}\r\n$$\r\nNow, find the sum of the first 8 terms ($S_8$):\r\n$$\r\n\\begin{aligned}\r\nS_8 &= \\frac{a(r^8-1)}{r-1} \\\\\r\n&= \\frac{2(3^8-1)}{3-1} \\\\\r\n&= \\frac{2(6561-1)}{2} \\\\\r\n&= 6560\r\n\\end{aligned}\r\n$$\r\nThe sum is **6560**.\r\n\r\n***\r\n\r\n#### Question 22: Sum and Product of an A.P.\r\n**The sum of three numbers in an arithmetic progression is 51 and the sum of their squares is 939. Find the numbers.**\r\n\r\n**Solution:**\r\nLet the three terms be $a-d, a, a+d$.\r\nFrom the sum:\r\n$$\r\n\\begin{aligned}\r\n(a-d) + a + (a+d) &= 51 \\\\\r\n3a &= 51 \\\\\r\na &= 17\r\n\\end{aligned}\r\n$$\r\nFrom the sum of squares:\r\n$$\r\n\\begin{aligned}\r\n(a-d)^2 + a^2 + (a+d)^2 &= 939 \\\\\r\n(17-d)^2 + 17^2 + (17+d)^2 &= 939 \\\\\r\n(289 - 34d + d^2) + 289 + (289 + 34d + d^2) &= 939 \\\\\r\n867 + 2d^2 &= 939 \\\\\r\n2d^2 &= 72 \\\\\r\nd^2 &= 36 \\\\\r\nd &= \\pm 6\r\n\\end{aligned}\r\n$$\r\nIf $d=6$, the numbers are $11, 17, 23$.\r\nIf $d=-6$, the numbers are $23, 17, 11$.\r\nThe numbers are **11, 17, and 23**.\r\n\r\n***\r\n\r\n#### Question 23: Sum and Product of an A.P.\r\n**Three terms in an arithmetic progression have sum 21 and product 315. Find the terms.**\r\n\r\n**Solution:**\r\nLet the three terms be $a-d, a, a+d$.\r\nFrom the sum:\r\n$$\r\n(a-d) + a + (a+d) = 21 \\implies 3a = 21 \\implies a=7\r\n$$\r\nFrom the product:\r\n$$\r\n\\begin{aligned}\r\n(a-d)(a)(a+d) &= 315 \\\\\r\n(7-d)(7)(7+d) &= 315 \\\\\r\n(49-d^2)(7) &= 315 \\\\\r\n49-d^2 &= 45 \\\\\r\nd^2 &= 4 \\\\\r\nd &= \\pm 2\r\n\\end{aligned}\r\n$$\r\nIf $d=2$, the numbers are $5, 7, 9$.\r\nIf $d=-2$, the numbers are $9, 7, 5$.\r\nThe terms are **5, 7, and 9**.\r\n\r\n***\r\n\r\n#### Question 24: Sum of an A.P. from Two Terms\r\n**The ninth and nineteenth terms of an arithmetic series are 40 and 60 respectively. Find the sum of the first 25 terms of the series.**\r\n\r\n**Solution:**\r\nGiven: $t_9 = a+8d = 40$ and $t_{19} = a+18d = 60$.\r\nSubtract the first equation from the second:\r\n$$\r\n(a+18d) - (a+8d) = 60-40 \\implies 10d=20 \\implies d=2\r\n$$\r\nFind $a$: $a+8(2)=40 \\implies a+16=40 \\implies a=24$.\r\nFind $S_{25}$:\r\n$$\r\n\\begin{aligned}\r\nS_{25} &= \\frac{25}{2}[2(24) + (25-1)2] \\\\\r\n&= \\frac{25}{2}[48 + 24(2)] \\\\\r\n&= \\frac{25}{2}[48+48] = \\frac{25}{2}(96) \\\\\r\n&= 25(48) = 1200\r\n\\end{aligned}\r\n$$\r\nThe sum is **1200**.\r\n\r\n***\r\n\r\n#### Question 25: A.P. Word Problem\r\n**In an arithmetic sequence, the sixth term is equal to three times the fourth term and the sum of the first three terms is -12. Find the sum of the first ten terms.**\r\n\r\n**Solution:**\r\nFrom the first condition: $t_6 = 3 \\times t_4$.\r\n$$\r\n\\begin{aligned}\r\na+5d &= 3(a+3d) \\\\\r\na+5d &= 3a+9d \\\\\r\n-2a &= 4d \\\\\r\na &= -2d \\quad \\text{(Eq. 1)}\r\n\\end{aligned}\r\n$$\r\nFrom the second condition, let the first three terms be $a-d, a, a+d$.\r\nTheir sum is $(a-d)+a+(a+d) = 3a = -12 \\implies a=-4$.\r\n\r\nSubstitute $a=-4$ into Eq. 1:\r\n$-4 = -2d \\implies d=2$.\r\n\r\nFind $S_{10}$:\r\n$$\r\n\\begin{aligned}\r\nS_{10} &= \\frac{10}{2}[2(-4) + (10-1)2] \\\\\r\n&= 5[-8 + 9(2)] \\\\\r\n&= 5[-8+18] = 5(10) = 50\r\n\\end{aligned}\r\n$$\r\nThe sum is **50**.\r\n\r\n***\r\n\r\n#### Question 26: A.P. Word Problem\r\n**The sum of the first eight terms of an arithmetic series is 180 and its fifth term is five times the first term. Find the first term and common difference.**\r\n\r\n**Solution:**\r\nFrom the first condition: $t_5 = 5a$.\r\n$$\r\na+4d = 5a \\implies 4d=4a \\implies d=a \\quad \\text{(Eq. 1)}\r\n$$\r\nFrom the second condition: $S_8 = 180$.\r\n$$\r\n\\begin{aligned}\r\n\\frac{8}{2}[2a+(8-1)d] &= 180 \\\\\r\n4[2a+7d] &= 180 \\\\\r\n2a+7d &= 45\r\n\\end{aligned}\r\n$$\r\nSubstitute $d=a$ into this equation:\r\n$$\r\n2a+7a = 45 \\implies 9a = 45 \\implies a=5\r\n$$\r\nSince $d=a$, then $d=5$.\r\nThe first term is **5** and the common difference is **5**.\r\n\r\n***\r\n\r\n#### Question 27: Sum of a G.P. from Sums\r\n**In a geometric series, the sum of the first three terms is 1 and the sum of its first six terms is 28. Find the common ratio and first term.**\r\n\r\n**Solution:**\r\nGiven $S_3 = \\frac{a(r^3-1)}{r-1} = 1$ and $S_6 = \\frac{a(r^6-1)}{r-1} = 28$.\r\nWe can rewrite $S_6$:\r\n$$\r\nS_6 = \\frac{a(r^3-1)(r^3+1)}{r-1} = \\left[\\frac{a(r^3-1)}{r-1}\\right](r^3+1) = S_3(r^3+1)\r\n$$\r\nNow substitute the known values:\r\n$$\r\n\\begin{aligned}\r\n28 &= 1(r^3+1) \\\\\r\n27 &= r^3 \\\\\r\nr &= 3\r\n\\end{aligned}\r\n$$\r\nNow find $a$ using the equation for $S_3$:\r\n$$\r\n\\begin{aligned}\r\n\\frac{a(3^3-1)}{3-1} &= 1 \\\\\r\n\\frac{a(26)}{2} &= 1 \\\\\r\n13a &= 1 \\\\\r\na &= \\frac{1}{13}\r\n\\end{aligned}\r\n$$\r\nThe common ratio is **3** and the first term is **1/13**.', 1, '2025-07-25 01:11:38', '2025-09-18 07:21:49'),
(141, 13104, '# Grade 10 Optional Mathematics: Matrix Study Guide\r\n## Unit: Algebra | Chapter: Matrix | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nWelcome to the comprehensive study guide for the **Matrix** chapter. This note is designed to make complex concepts easy to understand, with clear definitions, essential formulas, and detailed, step-by-step solutions to a wide range of problems. By the end of this guide, you will have mastered matrix operations, determinants, inverses, and solving systems of equations using Cramer\'s Rule. 📈\r\n\r\n### Key Learning Areas\r\n1.  **Matrix Fundamentals:** Definitions, order, and types of matrices.\r\n2.  **Matrix Operations:** Multiplication of matrices and its conditions.\r\n3.  **Determinants:** How to calculate the determinant of a 2x2 matrix.\r\n4.  **Inverse of a Matrix:** Finding the adjoint and inverse of a 2x2 matrix.\r\n5.  **Cramer\'s Rule:** A method for solving systems of linear equations.\r\n\r\n---\r\n\r\n## 🔑 Key Concepts & Formulas\r\n\r\nBefore we solve the problems, let\'s understand the main tools we\'ll be using.\r\n\r\n### 1. Core Definitions\r\n\r\n* **Matrix:** A rectangular array of numbers arranged in rows and columns, enclosed in brackets `[]` or parentheses `()`.\r\n* **Order of a Matrix:** A matrix with `m` rows and `n` columns has an order of `m x n`.\r\n* **Identity Matrix (I):** A square matrix with `1`s on the main diagonal and `0`s elsewhere. For a 2x2 matrix: $I = \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}$. Multiplying any matrix by `I` leaves it unchanged (i.e., `AI = IA = A`).\r\n* **Null Matrix (O):** A matrix where all elements are zero.\r\n* **Transpose of a Matrix ($A^T$):** The matrix obtained by interchanging the rows and columns of matrix A.\r\n* **Inverse Matrix ($A^{-1}$):** If A and B are two square matrices of the same order such that their product is the identity matrix ($AB = BA = I$), then B is the inverse of A, and A is the inverse of B.\r\n\r\n---\r\n\r\n### 2. Matrix Multiplication\r\n\r\n* **Condition:** Two matrices A (order `m x n`) and B (order `p x q`) can be multiplied to get the product AB only if the **number of columns in A is equal to the number of rows in B** (i.e., `n = p`).\r\n    * The resulting matrix AB will have an order of `m x q`.\r\n\r\n---\r\n\r\n### 3. Determinant of a 2x2 Matrix\r\n\r\nThe determinant of a matrix is a special number that can be calculated from its elements. For a matrix $A = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}$, the determinant is denoted as `|A|` or `det(A)`.\r\n\r\n* **Formula:**\r\n    $$|A| = ad - bc$$\r\n\r\n* **Singular Matrix:** A matrix whose determinant is zero (`|A| = 0`). A singular matrix does **not** have an inverse. This is the necessary condition for an inverse to be undefined.\r\n\r\n---\r\n\r\n### 4. Inverse of a 2x2 Matrix\r\n\r\n* **Condition for Existence:** The inverse of a matrix A exists **only if its determinant is not zero** (i.e., $|A| \\neq 0$).\r\n\r\n* **Steps to Find the Inverse:**\r\n    1.  **Find the Determinant:** Calculate $|A| = ad - bc$. If it\'s zero, stop. The inverse does not exist.\r\n    2.  **Find the Adjoint:** For a matrix $A = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}$, the adjoint of A, denoted `adj(A)`, is found by swapping the elements of the main diagonal and changing the signs of the other two elements:\r\n        $$adj(A) = \\begin{bmatrix} d & -b \\\\ -c & a \\end{bmatrix}$$\r\n    3.  **Calculate the Inverse:** Use the formula:\r\n        $$A^{-1} = \\frac{1}{|A|} \\text{adj}(A)$$\r\n\r\n---\r\n\r\n### 5. Cramer\'s Rule\r\n\r\nCramer\'s Rule is a method for solving systems of linear equations using determinants.\r\nFor a system of two linear equations:\r\n$$\r\n\\begin{aligned}\r\na_1x + b_1y &= c_1 \\\\\r\na_2x + b_2y &= c_2\r\n\\end{aligned}\r\n$$\r\n**Step 1: Define the Determinants.**\r\n* **D (Coefficient Determinant):** The determinant of the coefficient matrix.\r\n    $$D = \\begin{vmatrix} a_1 & b_1 \\\\ a_2 & b_2 \\end{vmatrix}$$\r\n* **$D_x$:** Replace the x-column in D with the constant terms.\r\n    $$D_x = \\begin{vmatrix} c_1 & b_1 \\\\ c_2 & b_2 \\end{vmatrix}$$\r\n* **$D_y$:** Replace the y-column in D with the constant terms.\r\n    $$D_y = \\begin{vmatrix} a_1 & c_1 \\\\ a_2 & c_2 \\end{vmatrix}$$\r\n\r\n\r\n\r\n**Step 2: Solve for x and y.**\r\n$$\r\nx = \\frac{D_x}{D} \\quad \\text{and} \\quad y = \\frac{D_y}{D}\r\n$$\r\n\r\n* **Condition for a Unique Solution:** A unique solution exists only if the determinant of the coefficient matrix is non-zero ($D \\neq 0$).\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Find an Unknown Matrix\r\n**If $\\begin{bmatrix} 4 \\\\ 1 \\end{bmatrix} \\cdot C = \\begin{bmatrix} -8 & 12 \\\\ -2 & 3 \\end{bmatrix}$, find the matrix C.**\r\n\r\n**Solution:**\r\nThe first matrix is $2 \\times 1$. The resulting matrix is $2 \\times 2$.\r\nFor the multiplication to be possible, matrix C must have an order of $1 \\times 2$.\r\nLet $C = \\begin{bmatrix} a & b \\end{bmatrix}$.\r\n\r\n$$\r\n\\begin{aligned}\r\n\\begin{bmatrix} 4 \\\\ 1 \\end{bmatrix} \\begin{bmatrix} a & b \\end{bmatrix} &= \\begin{bmatrix} -8 & 12 \\\\ -2 & 3 \\end{bmatrix} \\\\\r\n\\begin{bmatrix} 4 \\cdot a & 4 \\cdot b \\\\ 1 \\cdot a & 1 \\cdot b \\end{bmatrix} &= \\begin{bmatrix} -8 & 12 \\\\ -2 & 3 \\end{bmatrix} \\\\\r\n\\begin{bmatrix} 4a & 4b \\\\ a & b \\end{bmatrix} &= \\begin{bmatrix} -8 & 12 \\\\ -2 & 3 \\end{bmatrix}\r\n\\end{aligned}\r\n$$\r\nEquating the corresponding elements:\r\nFrom the first column: $a = -2$.\r\nFrom the second column: $b = 3$.\r\n\r\nThus, **$C = \\begin{bmatrix} -2 & 3 \\end{bmatrix}$**.\r\n\r\n***\r\n\r\n#### Question 2: Find an Unknown Matrix\r\n**If $\\begin{bmatrix} 4 \\\\ 3 \\end{bmatrix} \\cdot p = \\begin{bmatrix} 8 & -4 \\\\ 6 & -3 \\end{bmatrix}$, find the matrix p.**\r\n\r\n**Solution:**\r\nLet the matrix $p$ be a $1 \\times 2$ matrix, $p = \\begin{bmatrix} a & b \\end{bmatrix}$.\r\n\r\n$$\r\n\\begin{aligned}\r\n\\begin{bmatrix} 4 \\\\ 3 \\end{bmatrix} \\begin{bmatrix} a & b \\end{bmatrix} &= \\begin{bmatrix} 8 & -4 \\\\ 6 & -3 \\end{bmatrix} \\\\\r\n\\begin{bmatrix} 4a & 4b \\\\ 3a & 3b \\end{bmatrix} &= \\begin{bmatrix} 8 & -4 \\\\ 6 & -3 \\end{bmatrix}\r\n\\end{aligned}\r\n$$\r\nEquating the corresponding elements:\r\nFrom the first column: $3a = 6 \\implies a=2$.\r\nFrom the second column: $3b = -3 \\implies b=-1$.\r\n\r\nThus, **$p = \\begin{bmatrix} 2 & -1 \\end{bmatrix}$**.\r\n\r\n***\r\n\r\n#### Question 3: Matrix Multiplication\r\n**If $A = \\begin{bmatrix} 1 & 2 & -3 \\\\ -2 & 0 & 5 \\end{bmatrix}$ and $B = \\begin{bmatrix} 2 & 3 \\\\ -4 & 0 \\\\ 5 & -1 \\end{bmatrix}$, find AB.**\r\n\r\n**Solution:**\r\nMatrix A is $2 \\times 3$ and B is $3 \\times 2$. The product AB will be a $2 \\times 2$ matrix.\r\n$$\r\n\\begin{aligned}\r\nAB &= \\begin{bmatrix} 1 & 2 & -3 \\\\ -2 & 0 & 5 \\end{bmatrix} \\begin{bmatrix} 2 & 3 \\\\ -4 & 0 \\\\ 5 & -1 \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} (1)(2) + (2)(-4) + (-3)(5) & (1)(3) + (2)(0) + (-3)(-1) \\\\ (-2)(2) + (0)(-4) + (5)(5) & (-2)(3) + (0)(0) + (5)(-1) \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} 2 - 8 - 15 & 3 + 0 + 3 \\\\ -4 + 0 + 25 & -6 + 0 - 5 \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} -21 & 6 \\\\ 21 & -11 \\end{bmatrix}\r\n\\end{aligned}\r\n$$\r\nThus, **$AB = \\begin{bmatrix} -21 & 6 \\\\ 21 & -11 \\end{bmatrix}$**.\r\n\r\n***\r\n\r\n#### Question 4: Singular Matrix\r\n**If the inverse of the matrix $\\begin{bmatrix} x & 3 \\\\ 4 & 2 \\end{bmatrix}$ cannot be defined, find the value of x.**\r\n\r\n**Solution:**\r\nAn inverse of a matrix cannot be defined if the matrix is singular, which means its determinant is 0.\r\nLet $A = \\begin{bmatrix} x & 3 \\\\ 4 & 2 \\end{bmatrix}$.\r\n$$\r\n\\begin{aligned}\r\n|A| &= 0 \\\\\r\n(x)(2) - (3)(4) &= 0 \\\\\r\n2x - 12 &= 0 \\\\\r\n2x &= 12 \\\\\r\nx &= 6\r\n\\end{aligned}\r\n$$\r\nThus, **$x=6$**.\r\n\r\n***\r\n\r\n#### Question 5: Find an Inverse Matrix\r\n**If matrix $A = \\begin{bmatrix} 5 & 3 \\\\ 7 & 4 \\end{bmatrix}$, find $A^{-1}$.**\r\n\r\n**Solution:**\r\n**Step 1: Find the determinant.**\r\n$$\r\n|A| = (5)(4) - (3)(7) = 20 - 21 = -1\r\n$$\r\nSince $|A| \\neq 0$, the inverse exists.\r\n\r\n**Step 2: Find the adjoint.**\r\n$$\r\nadj(A) = \\begin{bmatrix} 4 & -3 \\\\ -7 & 5 \\end{bmatrix}\r\n$$\r\n**Step 3: Calculate the inverse.**\r\n$$\r\nA^{-1} = \\frac{1}{|A|} \\text{adj}(A) = \\frac{1}{-1} \\begin{bmatrix} 4 & -3 \\\\ -7 & 5 \\end{bmatrix} = \\begin{bmatrix} -4 & 3 \\\\ 7 & -5 \\end{bmatrix}\r\n$$\r\nThus, **$A^{-1} = \\begin{bmatrix} -4 & 3 \\\\ 7 & -5 \\end{bmatrix}$**.\r\n\r\n***\r\n\r\n#### Question 6: Finding a Value from an Inverse\r\n**If the inverse of the matrix $\\begin{bmatrix} m & 2 \\\\ 7 & 3 \\end{bmatrix}$ is the matrix $\\begin{bmatrix} 3 & -2 \\\\ -7 & m \\end{bmatrix}$, find the value of m.**\r\n\r\n**Solution:**\r\nLet $A = \\begin{bmatrix} m & 2 \\\\ 7 & 3 \\end{bmatrix}$ and $A^{-1} = \\begin{bmatrix} 3 & -2 \\\\ -7 & m \\end{bmatrix}$.\r\nBy the definition of an inverse, $A \\cdot A^{-1} = I$.\r\n$$\r\n\\begin{aligned}\r\n\\begin{bmatrix} m & 2 \\\\ 7 & 3 \\end{bmatrix} \\begin{bmatrix} 3 & -2 \\\\ -7 & m \\end{bmatrix} &= \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix} \\\\\r\n\\begin{bmatrix} 3m - 14 & -2m + 2m \\\\ 21 - 21 & -14 + 3m \\end{bmatrix} &= \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix} \\\\\r\n\\begin{bmatrix} 3m - 14 & 0 \\\\ 0 & 3m - 14 \\end{bmatrix} &= \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}\r\n\\end{aligned}\r\n$$\r\nEquating the top-left elements:\r\n$$\r\n\\begin{aligned}\r\n3m - 14 &= 1 \\\\\r\n3m &= 15 \\\\\r\nm &= 5\r\n\\end{aligned}\r\n$$\r\nThus, **$m=5$**.\r\n\r\n***\r\n\r\n#### Question 7: Matrix and its Transpose\r\n**If matrix $A = \\begin{bmatrix} 3 & 1 \\\\ -1 & 2 \\end{bmatrix}$, find $A \\cdot A^T$.**\r\n\r\n**Solution:**\r\nFirst, find the transpose of A, $A^T$.\r\n$$\r\nA^T = \\begin{bmatrix} 3 & -1 \\\\ 1 & 2 \\end{bmatrix}\r\n$$\r\nNow, multiply A by $A^T$.\r\n$$\r\n\\begin{aligned}\r\nA \\cdot A^T &= \\begin{bmatrix} 3 & 1 \\\\ -1 & 2 \\end{bmatrix} \\begin{bmatrix} 3 & -1 \\\\ 1 & 2 \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} (3)(3) + (1)(1) & (3)(-1) + (1)(2) \\\\ (-1)(3) + (2)(1) & (-1)(-1) + (2)(2) \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} 9+1 & -3+2 \\\\ -3+2 & 1+4 \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} 10 & -1 \\\\ -1 & 5 \\end{bmatrix}\r\n\\end{aligned}\r\n$$\r\nThus, **$A \\cdot A^T = \\begin{bmatrix} 10 & -1 \\\\ -1 & 5 \\end{bmatrix}$**.\r\n\r\n***\r\n\r\n#### Question 8: Find an Unknown from a Matrix Equation\r\n**If $A = \\begin{bmatrix} 2 & 0 \\\\ -3 & x \\end{bmatrix}$, $B = \\begin{bmatrix} 4 & 0 \\\\ -9 & 1 \\end{bmatrix}$ and $A^2 = B$, find the value of x.**\r\n\r\n**Solution:**\r\nFirst, calculate $A^2$.\r\n$$\r\n\\begin{aligned}\r\nA^2 &= A \\cdot A = \\begin{bmatrix} 2 & 0 \\\\ -3 & x \\end{bmatrix} \\begin{bmatrix} 2 & 0 \\\\ -3 & x \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} (2)(2) + (0)(-3) & (2)(0) + (0)(x) \\\\ (-3)(2) + (x)(-3) & (-3)(0) + (x)(x) \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} 4 & 0 \\\\ -6 - 3x & x^2 \\end{bmatrix}\r\n\\end{aligned}\r\n$$\r\nNow, set $A^2 = B$.\r\n$$\r\n\\begin{bmatrix} 4 & 0 \\\\ -6 - 3x & x^2 \\end{bmatrix} = \\begin{bmatrix} 4 & 0 \\\\ -9 & 1 \\end{bmatrix}\r\n$$\r\nEquating the bottom-right elements:\r\n$$x^2 = 1 \\implies x = \\pm 1$$\r\nEquating the bottom-left elements:\r\n$$-6 - 3x = -9 \\implies -3x = -3 \\implies x = 1$$\r\nBoth conditions must be true, so the only valid solution is **$x=1$**.\r\n\r\n***\r\n\r\n#### Question 9: Finding a Value from Determinant\r\n**If $M = \\begin{bmatrix} 5 & 3 \\\\ 6 & K \\end{bmatrix}$ and $|M| = 27$, find the value of K.**\r\n\r\n**Solution:**\r\nCalculate the determinant of M and set it equal to 27.\r\n$$\r\n\\begin{aligned}\r\n|M| &= (5)(K) - (3)(6) \\\\\r\n27 &= 5K - 18 \\\\\r\n45 &= 5K \\\\\r\nK &= 9\r\n\\end{aligned}\r\n$$\r\nThus, **$K=9$**.\r\n\r\n***\r\n\r\n#### Question 10: Prove AB is a Null Matrix\r\n**If $A = \\begin{bmatrix} 1 & -1 \\\\ 3 & -3 \\end{bmatrix}$ and $B = \\begin{bmatrix} 2 & -5 \\\\ 2 & -5 \\end{bmatrix}$, show that AB is a null matrix.**\r\n\r\n**Solution:**\r\nMultiply the matrices A and B.\r\n$$\r\n\\begin{aligned}\r\nAB &= \\begin{bmatrix} 1 & -1 \\\\ 3 & -3 \\end{bmatrix} \\begin{bmatrix} 2 & -5 \\\\ 2 & -5 \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} (1)(2) + (-1)(2) & (1)(-5) + (-1)(-5) \\\\ (3)(2) + (-3)(2) & (3)(-5) + (-3)(-5) \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} 2-2 & -5+5 \\\\ 6-6 & -15+15 \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\end{bmatrix}\r\n\\end{aligned}\r\n$$\r\nSince the resulting matrix contains only zeros, **AB is a null matrix**.\r\n\r\n***\r\n\r\n#### Question 11: Finding a Matrix from its Inverse\r\n**If the inverse of a matrix A is $A^{-1} = \\begin{bmatrix} 1 & 3 \\\\ 2 & 4 \\end{bmatrix}$, find the matrix A.**\r\n\r\n**Solution:**\r\nThe inverse of an inverse matrix is the original matrix, i.e., $(A^{-1})^{-1} = A$. So, we find the inverse of the given $A^{-1}$.\r\nLet $M = A^{-1} = \\begin{bmatrix} 1 & 3 \\\\ 2 & 4 \\end{bmatrix}$.\r\n\r\n**Step 1: Find the determinant of M.**\r\n$$\r\n|M| = (1)(4) - (3)(2) = 4 - 6 = -2\r\n$$\r\n**Step 2: Find the adjoint of M.**\r\n$$\r\nadj(M) = \\begin{bmatrix} 4 & -3 \\\\ -2 & 1 \\end{bmatrix}\r\n$$\r\n**Step 3: Calculate the inverse of M.**\r\n$$\r\nA = M^{-1} = \\frac{1}{|M|}adj(M) = \\frac{1}{-2} \\begin{bmatrix} 4 & -3 \\\\ -2 & 1 \\end{bmatrix} = \\begin{bmatrix} -2 & 3/2 \\\\ 1 & -1/2 \\end{bmatrix}\r\n$$\r\nThus, **$A = \\begin{bmatrix} -2 & 3/2 \\\\ 1 & -1/2 \\end{bmatrix}$**.\r\n\r\n***\r\n\r\n#### Question 12: Prove AB is an Identity Matrix\r\n**If $A = \\begin{bmatrix} 3 & 4 \\\\ 2 & 3 \\end{bmatrix}$ and $B = \\begin{bmatrix} 3 & -4 \\\\ -2 & 3 \\end{bmatrix}$, prove that AB is an identity matrix.**\r\n\r\n**Solution:**\r\nMultiply the matrices A and B.\r\n$$\r\n\\begin{aligned}\r\nAB &= \\begin{bmatrix} 3 & 4 \\\\ 2 & 3 \\end{bmatrix} \\begin{bmatrix} 3 & -4 \\\\ -2 & 3 \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} (3)(3) + (4)(-2) & (3)(-4) + (4)(3) \\\\ (2)(3) + (3)(-2) & (2)(-4) + (3)(3) \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} 9-8 & -12+12 \\\\ 6-6 & -8+9 \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}\r\n\\end{aligned}\r\n$$\r\nSince the result is $\\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}$, **AB is an identity matrix**.\r\n\r\n***\r\n\r\n#### Question 13: Find Unknowns in a Matrix Equation\r\n**If $\\begin{pmatrix} -2 & 0 \\\\ 0 & -5 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} -2 \\\\ -10 \\end{pmatrix}$, find the values of x and y.**\r\n\r\n**Solution:**\r\nPerform the matrix multiplication on the left side:\r\n$$\r\n\\begin{pmatrix} (-2)(x) + (0)(y) \\\\ (0)(x) + (-5)(y) \\end{pmatrix} = \\begin{pmatrix} -2 \\\\ -10 \\end{pmatrix} \\\\\r\n\\begin{pmatrix} -2x \\\\ -5y \\end{pmatrix} = \\begin{pmatrix} -2 \\\\ -10 \\end{pmatrix}\r\n$$\r\nEquating the corresponding elements:\r\n$$-2x = -2 \\implies x = 1$$\r\n$$-5y = -10 \\implies y = 2$$\r\nThus, **$x=1$ and $y=2$**.\r\n\r\n***\r\n\r\n#### Question 14: Find Unknowns in a Matrix Equation\r\n**If $\\begin{pmatrix} x + 1 \\\\ y - 2 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} \\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}$, find the values of x and y.**\r\n\r\n**Solution:**\r\nFirst, perform the matrix multiplication on the right side:\r\n$$\r\n\\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} \\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix} = \\begin{pmatrix} (1)(2) + (0)(3) \\\\ (0)(2) + (-1)(3) \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ -3 \\end{pmatrix}\r\n$$\r\nNow, equate the matrices:\r\n$$\r\n\\begin{pmatrix} x + 1 \\\\ y - 2 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ -3 \\end{pmatrix}\r\n$$\r\nEquating corresponding elements:\r\n$$x+1 = 2 \\implies x=1$$\r\n$$y-2 = -3 \\implies y = -1$$\r\nThus, **$x=1$ and $y=-1$**.\r\n\r\n***\r\n\r\n#### Question 15: Find a Value from Determinant\r\n**If the determinant of the matrix $M = \\begin{bmatrix} -5 & -8 \\\\ p & 12 \\end{bmatrix}$ is 20, find the value of p.**\r\n\r\n**Solution:**\r\nGiven $|M|=20$.\r\n$$\r\n\\begin{aligned}\r\n|M| &= (-5)(12) - (-8)(p) \\\\\r\n20 &= -60 + 8p \\\\\r\n80 &= 8p \\\\\r\np &= 10\r\n\\end{aligned}\r\n$$\r\nThus, **$p=10$**.\r\n\r\n---\r\n\r\n### Solving Systems of Linear Equations\r\n\r\n***\r\n\r\n#### Question 16: Solve using Cramer\'s Rule\r\n**Solve: $2x - y = 3$ and $y + 3x = 7$.**\r\n\r\n**Solution:**\r\nFirst, arrange the equations in standard form:\r\n$$\r\n\\begin{aligned}\r\n2x - y &= 3 \\\\\r\n3x + y &= 7\r\n\\end{aligned}\r\n$$\r\nNow, find the determinants D, $D_x$, and $D_y$.\r\n$$\r\n\\begin{aligned}\r\nD &= \\begin{vmatrix} 2 & -1 \\\\ 3 & 1 \\end{vmatrix} = (2)(1) - (-1)(3) = 2 + 3 = 5 \\\\\r\nD_x &= \\begin{vmatrix} 3 & -1 \\\\ 7 & 1 \\end{vmatrix} = (3)(1) - (-1)(7) = 3 + 7 = 10 \\\\\r\nD_y &= \\begin{vmatrix} 2 & 3 \\\\ 3 & 7 \\end{vmatrix} = (2)(7) - (3)(3) = 14 - 9 = 5\r\n\\end{aligned}\r\n$$\r\nNow, solve for x and y:\r\n$$\r\nx = \\frac{D_x}{D} = \\frac{10}{5} = 2\r\n$$\r\n$$\r\ny = \\frac{D_y}{D} = \\frac{5}{5} = 1\r\n$$\r\nThe solution is **$x=2, y=1$**.\r\n\r\n***\r\n\r\n#### Question 17: Solve using Cramer\'s Rule\r\n**Solve: $y = 2x$ and $x - \\frac{3}{2}y + 1 = 0$.**\r\n\r\n**Solution:**\r\nArrange the equations in standard form:\r\n$$\r\n\\begin{aligned}\r\n2x - y &= 0 \\\\\r\nx - \\frac{3}{2}y &= -1 \\implies 2x - 3y = -2\r\n\\end{aligned}\r\n$$\r\nFind the determinants:\r\n$$\r\n\\begin{aligned}\r\nD &= \\begin{vmatrix} 2 & -1 \\\\ 2 & -3 \\end{vmatrix} = (2)(-3) - (-1)(2) = -6 + 2 = -4 \\\\\r\nD_x &= \\begin{vmatrix} 0 & -1 \\\\ -2 & -3 \\end{vmatrix} = (0)(-3) - (-1)(-2) = 0 - 2 = -2 \\\\\r\nD_y &= \\begin{vmatrix} 2 & 0 \\\\ 2 & -2 \\end{vmatrix} = (2)(-2) - (0)(2) = -4 - 0 = -4\r\n\\end{aligned}\r\n$$\r\nSolve for x and y:\r\n$$\r\nx = \\frac{D_x}{D} = \\frac{-2}{-4} = \\frac{1}{2}\r\n$$\r\n$$\r\ny = \\frac{D_y}{D} = \\frac{-4}{-4} = 1\r\n$$\r\nThe solution is **$x = 1/2, y = 1$**.\r\n\r\n***\r\n\r\n#### Question 18: Solve using Cramer\'s Rule\r\n**Solve: $\\frac{4}{x} + \\frac{5}{y} = 28$ and $\\frac{7}{x} + \\frac{3}{y} = 67$.**\r\n\r\n**Solution:**\r\nLet $u = \\frac{1}{x}$ and $v = \\frac{1}{y}$. The equations become:\r\n$$\r\n\\begin{aligned}\r\n4u + 5v &= 28 \\\\\r\n7u + 3v &= 67\r\n\\end{aligned}\r\n$$\r\nFind the determinants D, $D_u$, and $D_v$.\r\n$$\r\n\\begin{aligned}\r\nD &= \\begin{vmatrix} 4 & 5 \\\\ 7 & 3 \\end{vmatrix} = (4)(3) - (5)(7) = 12 - 35 = -23 \\\\\r\nD_u &= \\begin{vmatrix} 28 & 5 \\\\ 67 & 3 \\end{vmatrix} = (28)(3) - (5)(67) = 84 - 335 = -251 \\\\\r\nD_v &= \\begin{vmatrix} 4 & 28 \\\\ 7 & 67 \\end{vmatrix} = (4)(67) - (28)(7) = 268 - 196 = 72\r\n\\end{aligned}\r\n$$\r\nSolve for u and v:\r\n$$\r\nu = \\frac{D_u}{D} = \\frac{-251}{-23} = 11 \\quad \\implies \\frac{1}{x} = 11 \\implies x = \\frac{1}{11}\r\n$$\r\n$$\r\nv = \\frac{D_v}{D} = \\frac{72}{-23} \\quad \\implies \\frac{1}{y} = -\\frac{72}{23} \\implies y = -\\frac{23}{72}\r\n$$\r\nThe solution is **$x=\\frac{1}{11}, y=-\\frac{23}{72}$**.\r\n\r\n***\r\n\r\n#### Question 19: Solve using Cramer\'s Rule\r\n**Solve: $3y + 4x = 2xy$ and $18y - 4x = 5xy$.**\r\n\r\n**Solution:**\r\nDivide both equations by $xy$ to get them in linear form.\r\nEquation 1:\r\n$$\r\n\\frac{3y}{xy} + \\frac{4x}{xy} = 2 \\implies \\frac{3}{x} + \\frac{4}{y} = 2\r\n$$\r\nEquation 2:\r\n$$\r\n\\frac{18y}{xy} - \\frac{4x}{xy} = 5 \\implies \\frac{18}{x} - \\frac{4}{y} = 5\r\n$$\r\nLet $u = \\frac{1}{x}$ and $v = \\frac{1}{y}$. The system is:\r\n$$\r\n\\begin{aligned}\r\n3u + 4v &= 2 \\\\\r\n18u - 4v &= 5\r\n\\end{aligned}\r\n$$\r\nFind the determinants:\r\n$$\r\n\\begin{aligned}\r\nD &= \\begin{vmatrix} 3 & 4 \\\\ 18 & -4 \\end{vmatrix} = (3)(-4) - (4)(18) = -12 - 72 = -84 \\\\\r\nD_u &= \\begin{vmatrix} 2 & 4 \\\\ 5 & -4 \\end{vmatrix} = (2)(-4) - (4)(5) = -8 - 20 = -28 \\\\\r\nD_v &= \\begin{vmatrix} 3 & 2 \\\\ 18 & 5 \\end{vmatrix} = (3)(5) - (2)(18) = 15 - 36 = -21\r\n\\end{aligned}\r\n$$\r\nSolve for u and v:\r\n$$\r\nu = \\frac{D_u}{D} = \\frac{-28}{-84} = \\frac{1}{3} \\quad \\implies \\frac{1}{x} = \\frac{1}{3} \\implies x = 3\r\n$$\r\n$$\r\nv = \\frac{D_v}{D} = \\frac{-21}{-84} = \\frac{1}{4} \\quad \\implies \\frac{1}{y} = \\frac{1}{4} \\implies y = 4\r\n$$\r\nThe solution is **$x=3, y=4$**.\r\n\r\n***\r\n\r\n#### Question 20: Solve using Cramer\'s Rule\r\n**Solve: $4(x - 1) + 5(y + 2) = 10$ and $5(x - 1) - 3(y + 2) + 6 = 0$.**\r\n\r\n**Solution:**\r\nFirst, simplify both equations.\r\nEquation 1:\r\n$$\r\n\\begin{aligned}\r\n4x - 4 + 5y + 10 &= 10 \\\\\r\n4x + 5y + 6 &= 10 \\\\\r\n4x + 5y &= 4\r\n\\end{aligned}\r\n$$\r\nEquation 2:\r\n$$\r\n\\begin{aligned}\r\n5x - 5 - 3y - 6 + 6 &= 0 \\\\\r\n5x - 3y - 5 &= 0 \\\\\r\n5x - 3y &= 5\r\n\\end{aligned}\r\n$$\r\nNow find the determinants for the system: $4x+5y=4$ and $5x-3y=5$.\r\n$$\r\n\\begin{aligned}\r\nD &= \\begin{vmatrix} 4 & 5 \\\\ 5 & -3 \\end{vmatrix} = (4)(-3) - (5)(5) = -12 - 25 = -37 \\\\\r\nD_x &= \\begin{vmatrix} 4 & 5 \\\\ 5 & -3 \\end{vmatrix} = (4)(-3) - (5)(5) = -12 - 25 = -37 \\\\\r\nD_y &= \\begin{vmatrix} 4 & 4 \\\\ 5 & 5 \\end{vmatrix} = (4)(5) - (4)(5) = 20 - 20 = 0\r\n\\end{aligned}\r\n$$\r\nSolve for x and y:\r\n$$\r\nx = \\frac{D_x}{D} = \\frac{-37}{-37} = 1\r\n$$\r\n$$\r\ny = \\frac{D_y}{D} = \\frac{0}{-37} = 0\r\n$$\r\nThe solution is **$x=1, y=0$**.', 1, '2025-07-25 01:11:38', '2025-09-18 07:21:49');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(142, 13115, '# Grade 10 Optional Mathematics: Angle Between Two Lines\r\n## Unit: Coordinate Geometry | Chapter: Angle Between Two Lines | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nWelcome to the comprehensive study guide for finding the **Angle Between Two Straight Lines**. This note is designed to make complex concepts easy to understand, with clear definitions, essential formulas, and detailed, step-by-step solutions to all 29 unique problems from your materials. By the end of this guide, you will have mastered the concepts of slope, parallel and perpendicular lines, homogeneous equations, and finding the equations of lines based on various conditions. 📐\r\n\r\n### Key Learning Areas\r\n1.  **Slope (Gradient):** Calculating the slope of a line from two points or from its equation.\r\n2.  **Angle Formula:** Using the slopes of two lines to find the angle between them.\r\n3.  **Parallel & Perpendicular Lines:** Understanding and applying the conditions for parallelism and perpendicularity.\r\n4.  **Equations of Lines:** Finding the equation of a line that is parallel or perpendicular to another line and passes through a given point or satisfies other conditions.\r\n5.  **Homogeneous Equations:** Finding the angle between the pair of lines represented by a second-degree homogeneous equation.\r\n\r\n---\r\n\r\n## 🔑 Key Concepts & Formulas\r\n\r\nBefore we solve the problems, let\'s understand the main tools we\'ll be using.\r\n\r\n### 1. The Slope of a Straight Line (m)\r\n\r\nThe slope represents the steepness of a line.\r\n\r\n* **From Two Points $(x_1, y_1)$ and $(x_2, y_2)$:**\r\n    $$m = \\frac{y_2 - y_1}{x_2 - x_1}$$\r\n\r\n* **From the General Equation $Ax + By + C = 0$:**\r\n    $$m = -\\frac{\\text{Coefficient of x}}{\\text{Coefficient of y}} = -\\frac{A}{B}$$\r\n\r\n* **From the Slope-Intercept Form $y = mx + c$:**\r\n    The slope is simply `m`.\r\n\r\n---\r\n\r\n### 2. Angle Between Two Lines\r\n\r\n\r\n\r\nIf two lines have slopes $m_1$ and $m_2$, the angle ($\\theta$) between them is given by:\r\n\r\n* **General Formula:**\r\n    $$\\tan\\theta = \\pm \\left( \\frac{m_1 - m_2}{1 + m_1m_2} \\right)$$\r\n    * **Acute Angle:** For the smaller angle (less than 90°), we take the positive value: $\\tan\\theta = \\left| \\frac{m_1 - m_2}{1 + m_1m_2} \\right|$.\r\n    * **Obtuse Angle:** For the larger angle (greater than 90°), we take the negative value.\r\n\r\n---\r\n\r\n### 3. Conditions for Parallel and Perpendicular Lines\r\n\r\n* **Parallel Lines:** Two lines are parallel if and only if their slopes are equal.\r\n    $$m_1 = m_2$$\r\n\r\n* **Perpendicular Lines:** Two lines are perpendicular if and only if the product of their slopes is -1.\r\n    $$m_1 \\cdot m_2 = -1$$\r\n\r\n---\r\n\r\n### 4. Special Equations of Lines\r\n\r\n* **Equation of a line parallel to $Ax + By + C = 0$:**\r\n    The equation will be of the form $Ax + By + k = 0$, where `k` is a new constant found using another condition (like a point on the line).\r\n\r\n* **Equation of a line perpendicular to $Ax + By + C = 0$:**\r\n    The equation will be of the form $Bx - Ay + k = 0$. (Note the swapped coefficients of x and y, and the sign change for the y-coefficient).\r\n\r\n---\r\n\r\n### 5. Pair of Straight Lines (Homogeneous Equation)\r\n\r\nA second-degree homogeneous equation $ax^2 + 2hxy + by^2 = 0$ represents a pair of straight lines passing through the origin.\r\n\r\n* **Angle ($\\theta$) between the pair of lines:**\r\n    $$\\tan\\theta = \\frac{\\pm 2\\sqrt{h^2 - ab}}{a+b}$$\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Finding the Slope of a Perpendicular Line\r\n**Find the slope of the line perpendicular to the straight line 3x - 4y = 10.**\r\n\r\n**Solution:**\r\nFirst, find the slope of the given line $3x - 4y = 10$.\r\n$$m_1 = -\\frac{\\text{Coefficient of x}}{\\text{Coefficient of y}} = -\\frac{3}{-4} = \\frac{3}{4}$$\r\nLet the slope of the perpendicular line be $m_2$. The condition for perpendicularity is $m_1 \\cdot m_2 = -1$.\r\n$$\r\n\\begin{aligned}\r\n\\frac{3}{4} \\cdot m_2 &= -1 \\\\\r\nm_2 &= -\\frac{4}{3}\r\n\\end{aligned}\r\n$$\r\nThe slope of the perpendicular line is **$-\\frac{4}{3}$**.\r\n\r\n***\r\n\r\n#### Question 2: Proving Lines are Parallel\r\n**Prove that the straight line passing through the points (3, -4) and (-2, 6) and the straight line with the equation $2x + y + 3 = 0$ are parallel.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slope of the line from the equation.**\r\nFor the line $2x + y + 3 = 0$:\r\n$$m_1 = -\\frac{\\text{Coefficient of x}}{\\text{Coefficient of y}} = -\\frac{2}{1} = -2$$\r\n**Step 2: Find the slope of the line from the two points.**\r\nFor the points $(3, -4)$ and $(-2, 6)$:\r\n$$m_2 = \\frac{y_2 - y_1}{x_2 - x_1} = \\frac{6 - (-4)}{-2 - 3} = \\frac{10}{-5} = -2$$\r\n**Step 3: Compare the slopes.**\r\nSince $m_1 = m_2 = -2$, the two lines are parallel. **(Proved)**\r\n\r\n***\r\n\r\n#### Question 3: Finding an Unknown for Parallel Lines\r\n**If the lines $3x + my = 5$ and $\\frac{x}{2} + \\frac{y}{3} = 1$ are parallel to each other, find the value of m.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slope of the first line.**\r\nFor $3x + my = 5$, the slope is $m_1 = -\\frac{3}{m}$.\r\n\r\n**Step 2: Find the slope of the second line.**\r\nFirst, convert $\\frac{x}{2} + \\frac{y}{3} = 1$ to the general form by taking the LCM (6):\r\n$$3x + 2y = 6 \\implies 3x + 2y - 6 = 0$$\r\nThe slope is $m_2 = -\\frac{3}{2}$.\r\n\r\n**Step 3: Equate the slopes and solve for m.**\r\nFor parallel lines, $m_1 = m_2$.\r\n$$\r\n\\begin{aligned}\r\n-\\frac{3}{m} &= -\\frac{3}{2} \\\\\r\n\\frac{3}{m} &= \\frac{3}{2} \\\\\r\nm &= 2\r\n\\end{aligned}\r\n$$\r\nThus, **$m = 2$**.\r\n\r\n***\r\n\r\n#### Question 4: Finding an Unknown for Perpendicular Lines\r\n**If the straight line having equation 3x + 4y = 12 and 2x - py = 5 are perpendicular to each other, find the value of p.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slopes of both lines.**\r\nFor $3x + 4y = 12$, the slope is $m_1 = -\\frac{3}{4}$.\r\nFor $2x - py = 5$, the slope is $m_2 = -\\frac{2}{-p} = \\frac{2}{p}$.\r\n\r\n**Step 2: Use the perpendicularity condition and solve for p.**\r\nFor perpendicular lines, $m_1 \\cdot m_2 = -1$.\r\n$$\r\n\\begin{aligned}\r\n\\left(-\\frac{3}{4}\\right) \\cdot \\left(\\frac{2}{p}\\right) &= -1 \\\\\r\n-\\frac{6}{4p} &= -1 \\\\\r\n6 &= 4p \\\\\r\np &= \\frac{6}{4} = \\frac{3}{2}\r\n\\end{aligned}\r\n$$\r\nThus, **$p = \\frac{3}{2}$**.\r\n\r\n***\r\n\r\n#### Question 5: Finding an Unknown for Perpendicular Lines\r\n**If the line passing through points (3, -4) and (-2, k) is perpendicular to the line having equation 5x + y = -3, find the value of k.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slope of the line from the two points.**\r\n$$m_1 = \\frac{k - (-4)}{-2 - 3} = \\frac{k+4}{-5}$$\r\n**Step 2: Find the slope of the line from the equation.**\r\nFor $5x + y = -3$, the slope is $m_2 = -\\frac{5}{1} = -5$.\r\n\r\n**Step 3: Use the perpendicularity condition and solve for k.**\r\n$$\r\n\\begin{aligned}\r\nm_1 \\cdot m_2 &= -1 \\\\\r\n\\left(\\frac{k+4}{-5}\\right) \\cdot (-5) &= -1 \\\\\r\nk+4 &= -1 \\\\\r\nk &= -5\r\n\\end{aligned}\r\n$$\r\nThus, **$k=-5$**.\r\n\r\n***\r\n\r\n#### Question 6: Acute Angle Between Lines\r\n**Find the acute angle between the lines having slope $-\\sqrt{3}$ and $\\sqrt{3}$ respectively.**\r\n\r\n**Solution:**\r\nLet $m_1 = -\\sqrt{3}$ and $m_2 = \\sqrt{3}$. Let $\\theta$ be the acute angle.\r\n$$\r\n\\begin{aligned}\r\n\\tan\\theta &= \\left| \\frac{m_1 - m_2}{1 + m_1m_2} \\right| \\\\\r\n&= \\left| \\frac{-\\sqrt{3} - \\sqrt{3}}{1 + (-\\sqrt{3})(\\sqrt{3})} \\right| \\\\\r\n&= \\left| \\frac{-2\\sqrt{3}}{1 - 3} \\right| \\\\\r\n&= \\left| \\frac{-2\\sqrt{3}}{-2} \\right| \\\\\r\n\\tan\\theta &= \\sqrt{3}\r\n\\end{aligned}\r\n$$\r\nSince $\\tan(60^\\circ) = \\sqrt{3}$, the acute angle is **60°**.\r\n\r\n***\r\n\r\n#### Question 7: Acute Angle Between Lines\r\n**If the slopes of the lines AB and CD are 3 and -2 respectively, find the acute angle between them.**\r\n\r\n**Solution:**\r\nGiven slopes, $m_1 = 3$ and $m_2 = -2$.\r\nLet $\\theta$ be the acute angle.\r\n$$\r\n\\begin{aligned}\r\n\\tan\\theta &= \\left| \\frac{m_1 - m_2}{1 + m_1m_2} \\right| \\\\\r\n&= \\left| \\frac{3 - (-2)}{1 + (3)(-2)} \\right| \\\\\r\n&= \\left| \\frac{5}{1 - 6} \\right| \\\\\r\n&= \\left| \\frac{5}{-5} \\right| = 1\r\n\\end{aligned}\r\n$$\r\nSince $\\tan(45^\\circ) = 1$, the acute angle is **45°**.\r\n\r\n***\r\n\r\n#### Question 8: Finding an Unknown for Parallel Lines\r\n**If the straight lines Px + 3y - 12 = 0 and 4y - 3x + 7 = 0 are parallel to each other, find the value of P.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slopes.**\r\nFor $Px + 3y - 12 = 0$, the slope is $m_1 = -\\frac{P}{3}$.\r\nFor $-3x + 4y + 7 = 0$, the slope is $m_2 = -\\frac{-3}{4} = \\frac{3}{4}$.\r\n\r\n**Step 2: Equate the slopes.**\r\nFor parallel lines, $m_1 = m_2$.\r\n$$\r\n\\begin{aligned}\r\n-\\frac{P}{3} &= \\frac{3}{4} \\\\\r\n-4P &= 9 \\\\\r\nP &= -\\frac{9}{4}\r\n\\end{aligned}\r\n$$\r\nThus, **$P = -9/4$**.\r\n\r\n***\r\n\r\n#### Question 9: Finding an Unknown for Parallel Lines\r\n**If the lines 2x - 8y + 6 = 0 and px - 12y - 4 = 3 are parallel to each other, find the value of p.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slopes.**\r\nFor $2x - 8y + 6 = 0$, the slope is $m_1 = -\\frac{2}{-8} = \\frac{1}{4}$.\r\nFor $px - 12y - 7 = 0$, the slope is $m_2 = -\\frac{p}{-12} = \\frac{p}{12}$.\r\n\r\n**Step 2: Equate the slopes.**\r\n$$\r\n\\begin{aligned}\r\nm_1 &= m_2 \\\\\r\n\\frac{1}{4} &= \\frac{p}{12} \\\\\r\n12 &= 4p \\\\\r\np &= 3\r\n\\end{aligned}\r\n$$\r\nThus, **$p=3$**.\r\n\r\n***\r\n\r\n#### Question 10: Prove Lines are Perpendicular\r\n**Show that the lines $y + \\sqrt{3}x + 4 = 0$ and $x - \\sqrt{3}y = 5$ are perpendicular to each other.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slopes.**\r\nFor $\\sqrt{3}x + y + 4 = 0$, the slope is $m_1 = -\\frac{\\sqrt{3}}{1} = -\\sqrt{3}$.\r\nFor $x - \\sqrt{3}y - 5 = 0$, the slope is $m_2 = -\\frac{1}{-\\sqrt{3}} = \\frac{1}{\\sqrt{3}}$.\r\n\r\n**Step 2: Check the perpendicularity condition.**\r\n$$m_1 \\cdot m_2 = (-\\sqrt{3}) \\cdot \\left(\\frac{1}{\\sqrt{3}}\\right) = -1$$\r\nSince the product of the slopes is -1, the lines are perpendicular. **(Proved)**\r\n\r\n***\r\n\r\n#### Question 11: Prove Lines are Perpendicular\r\n**Show that the lines $ax + by + c = 0$ and $bx - ay + c = 0$ are perpendicular to each other.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slopes.**\r\nFor $ax + by + c = 0$, the slope is $m_1 = -\\frac{a}{b}$.\r\nFor $bx - ay + c = 0$, the slope is $m_2 = -\\frac{b}{-a} = \\frac{b}{a}$.\r\n\r\n**Step 2: Check the perpendicularity condition.**\r\n$$m_1 \\cdot m_2 = \\left(-\\frac{a}{b}\\right) \\cdot \\left(\\frac{b}{a}\\right) = -1$$\r\nSince the product of the slopes is -1, the lines are perpendicular. **(Proved)**\r\n\r\n***\r\n\r\n#### Question 12: Finding an Unknown for Perpendicular Lines\r\n**If the straight lines 2x + 3y + 6 = 0 and ax - 5y + 20 = 0 are perpendicular to each other, find the value of a.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slopes.**\r\nFor $2x + 3y + 6 = 0$, the slope is $m_1 = -\\frac{2}{3}$.\r\nFor $ax - 5y + 20 = 0$, the slope is $m_2 = -\\frac{a}{-5} = \\frac{a}{5}$.\r\n\r\n**Step 2: Use the perpendicularity condition.**\r\n$$\r\n\\begin{aligned}\r\nm_1 \\cdot m_2 &= -1 \\\\\r\n\\left(-\\frac{2}{3}\\right) \\cdot \\left(\\frac{a}{5}\\right) &= -1 \\\\\r\n-\\frac{2a}{15} &= -1 \\\\\r\n2a &= 15 \\\\\r\na &= \\frac{15}{2}\r\n\\end{aligned}\r\n$$\r\nThus, **$a = 15/2$**.\r\n\r\n***\r\n\r\n#### Question 13: Finding an Unknown for Parallel Lines\r\n**If a line passing through the points (4, -p) and (-2, 6) is parallel to the line having equation 2y + 3x = -3, find the value of p.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slopes.**\r\nSlope of the line passing through points (4, -p) and (-2, 6):\r\n$$m_1 = \\frac{6 - (-p)}{-2 - 4} = \\frac{6+p}{-6}$$\r\nSlope of the line $3x + 2y + 3 = 0$:\r\n$$m_2 = -\\frac{3}{2}$$\r\n**Step 2: Equate the slopes.**\r\n$$\r\n\\begin{aligned}\r\n\\frac{6+p}{-6} &= -\\frac{3}{2} \\\\\r\n2(6+p) &= (-3)(-6) \\\\\r\n12 + 2p &= 18 \\\\\r\n2p &= 6 \\\\\r\np &= 3\r\n\\end{aligned}\r\n$$\r\nThus, **$p=3$**.\r\n\r\n***\r\n\r\n### Long Questions\r\n\r\n***\r\n\r\n#### Question 14: Angle between a Pair of Lines\r\n**If an angle between the pair of lines represented by the equation $2x^2 + kxy + 3y^2 = 0$ is 45°, find the positive value of k and also find the separate equations of the lines.**\r\n\r\n**Solution:**\r\n**Step 1: Find k.**\r\nComparing $2x^2 + kxy + 3y^2 = 0$ with $ax^2 + 2hxy + by^2 = 0$, we get:\r\n$a=2$, $b=3$, $2h=k \\implies h=k/2$.\r\nGiven $\\theta = 45^\\circ$.\r\n$$\r\n\\begin{aligned}\r\n\\tan\\theta &= \\frac{\\pm 2\\sqrt{h^2 - ab}}{a+b} \\\\\r\n\\tan45^\\circ &= \\frac{\\pm 2\\sqrt{(k/2)^2 - (2)(3)}}{2+3} \\\\\r\n1 &= \\frac{\\pm 2\\sqrt{k^2/4 - 6}}{5} \\\\\r\n5 &= \\pm \\sqrt{k^2 - 24} \\quad (\\text{since } \\sqrt{4}=2)\r\n\\end{aligned}\r\n$$Squaring both sides:$$\r\n\\begin{aligned}\r\n25 &= k^2 - 24 \\\\\r\nk^2 &= 49 \\\\\r\nk &= \\pm 7\r\n\\end{aligned}\r\n$$\r\nThe positive value is **$k=7$**.\r\n\r\n**Step 2: Find the separate equations.**\r\nSubstitute $k=7$ into the original equation:\r\n$$\r\n\\begin{aligned}\r\n2x^2 + 7xy + 3y^2 &= 0 \\\\\r\n2x^2 + 6xy + xy + 3y^2 &= 0 \\\\\r\n2x(x+3y) + y(x+3y) &= 0 \\\\\r\n(2x+y)(x+3y) &= 0\r\n\\end{aligned}\r\n$$\r\nThe separate equations are **$2x+y=0$** and **$x+3y=0$**.\r\n\r\n***\r\n\r\n#### Question 15: Equation of a Line through Intersection\r\n**Find the equation of a straight line passing through the point of intersection of the lines $3x + 4y = 7$ and $5x - 2y = 3$ and perpendicular to the line $2x + 3y = 5$.**\r\n\r\n**Solution:**\r\n**Step 1: Find the point of intersection.**\r\nFrom $3x + 4y = 7$, we get $x = \\frac{7-4y}{3}$.\r\nSubstitute into the second equation:\r\n$$\r\n\\begin{aligned}\r\n5\\left(\\frac{7-4y}{3}\\right) - 2y &= 3 \\\\\r\n35 - 20y - 6y &= 9 \\\\\r\n35 - 26y &= 9 \\\\\r\n26 &= 26y \\\\\r\ny &= 1\r\n\\end{aligned}\r\n$$\r\nNow find x: $x = \\frac{7-4(1)}{3} = \\frac{3}{3} = 1$. The intersection point is (1, 1).\r\n\r\n**Step 2: Find the equation of the required line.**\r\nThe slope of $2x + 3y = 5$ is $m_1 = -2/3$.\r\nThe required line is perpendicular, so its slope is $m_2 = -1/m_1 = 3/2$.\r\nThe line passes through (1, 1) with slope 3/2. Using point-slope form:\r\n$$\r\n\\begin{aligned}\r\ny - 1 &= \\frac{3}{2}(x - 1) \\\\\r\n2y - 2 &= 3x - 3 \\\\\r\n3x - 2y - 1 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$3x - 2y - 1 = 0$**.\r\n\r\n***\r\n\r\n#### Question 16: Equation of a Perpendicular Bisector\r\n**Find the equation of the perpendicular bisector of the line segment joining the points A(3, -7) and B(-5, 3).**\r\n\r\n**Solution:**\r\n**Step 1: Find the midpoint of AB.**\r\n$$Midpoint = \\left( \\frac{3 + (-5)}{2}, \\frac{-7+3}{2} \\right) = (-1, -2)$$\r\n**Step 2: Find the slope of line AB.**\r\n$$m_{AB} = \\frac{3 - (-7)}{-5 - 3} = \\frac{10}{-8} = -\\frac{5}{4}$$\r\n**Step 3: Find the slope of the perpendicular bisector.**\r\n$$m_{\\perp} = -\\frac{1}{m_{AB}} = -\\frac{1}{-5/4} = \\frac{4}{5}$$\r\n**Step 4: Find the equation of the bisector.**\r\nThe line passes through $(-1, -2)$ with slope $4/5$.\r\n$$\r\n\\begin{aligned}\r\ny - (-2) &= \\frac{4}{5}(x - (-1)) \\\\\r\n5(y+2) &= 4(x+1) \\\\\r\n5y + 10 &= 4x + 4 \\\\\r\n4x - 5y - 6 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$4x - 5y - 6 = 0$**.\r\n\r\n***\r\n\r\n#### Question 17: Equation of a Line through a Centroid\r\n**A(3, 2), B(1, -1) and C(5, -5) are the vertices of a triangle ABC. Find the equation of a straight line passing through the centroid of triangle ABC and parallel to the side BC.**\r\n\r\n**Solution:**\r\n**Step 1: Find the centroid of triangle ABC.**\r\n$$Centroid = \\left( \\frac{3+1+5}{3}, \\frac{2+(-1)+(-5)}{3} \\right) = \\left( \\frac{9}{3}, \\frac{-4}{3} \\right) = (3, -4/3)$$\r\n**Step 2: Find the slope of the side BC.**\r\n$$m_{BC} = \\frac{-5 - (-1)}{5 - 1} = \\frac{-4}{4} = -1$$\r\n**Step 3: Find the equation of the required line.**\r\nThe required line is parallel to BC, so its slope is also $m = -1$.\r\nIt passes through the centroid $(3, -4/3)$.\r\n$$\r\n\\begin{aligned}\r\ny - (-\\frac{4}{3}) &= -1(x - 3) \\\\\r\ny + \\frac{4}{3} &= -x + 3 \\\\\r\n3y + 4 &= -3x + 9 \\\\\r\n3x + 3y - 5 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$3x + 3y - 5 = 0$**.\r\n\r\n***\r\n\r\n#### Question 18: Equation of a Line passing through Centroid\r\n**Find the equation of the line passing through the centroid of triangle PQR with vertices P(3, 3), Q(-2, -6) and R(5, -3) and parallel to the line QR.**\r\n\r\n**Solution:**\r\n**Step 1: Find the centroid of triangle PQR.**\r\n$$Centroid = \\left( \\frac{3+(-2)+5}{3}, \\frac{3+(-6)+(-3)}{3} \\right) = \\left( \\frac{6}{3}, \\frac{-6}{3} \\right) = (2, -2)$$\r\n**Step 2: Find the slope of the side QR.**\r\n$$m_{QR} = \\frac{-3 - (-6)}{5 - (-2)} = \\frac{3}{7}$$\r\n**Step 3: Find the equation of the required line.**\r\nThe required line is parallel to QR, so its slope is also $m = 3/7$.\r\nIt passes through the centroid $(2, -2)$.\r\n$$\r\n\\begin{aligned}\r\ny - (-2) &= \\frac{3}{7}(x - 2) \\\\\r\n7(y+2) &= 3(x-2) \\\\\r\n7y + 14 &= 3x - 6 \\\\\r\n3x - 7y - 20 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$3x - 7y - 20 = 0$**.', 1, '2025-07-25 01:11:38', '2025-09-18 07:21:49'),
(143, 13117, '# Grade 10 Optional Mathematics: The Circle\r\n## Unit: Coordinate Geometry | Chapter: Circle | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nThis guide offers a complete overview of the **Circle** chapter in Coordinate Geometry. It is designed to make the concepts easy to understand, with clear definitions, essential formulas, and detailed, step-by-step solutions to a wide range of problems. By the end of this note, you will have mastered the different forms of the circle equation and how to find a circle\'s center, radius, and equation based on various given conditions. ⚪\r\n\r\n### Key Learning Areas\r\n1.  **Standard Equation of a Circle:** Understanding and using the center-radius form.\r\n2.  **General Equation of a Circle:** Identifying the center and radius from the general form.\r\n3.  **Diameter Form:** Finding the equation of a circle given the endpoints of a diameter.\r\n4.  **Problem Solving:** Finding the equation of a circle based on various conditions, such as passing through points, having a specific center, or being concentric with another circle.\r\n\r\n---\r\n\r\n## 🔑 Key Concepts & Formulas\r\n\r\nBefore we solve the problems, let\'s understand the fundamental equations of a circle.\r\n\r\n### 1. Standard Equation of a Circle (Center-Radius Form)\r\n\r\nThis is the most common form of a circle\'s equation. It directly uses the circle\'s center and radius.\r\n\r\n* **Formula:**\r\n    $$(x - h)^2 + (y - k)^2 = r^2$$\r\n    Where:\r\n    * `(h, k)` are the coordinates of the center of the circle.\r\n    * `r` is the radius of the circle.\r\n\r\n\r\n\r\n---\r\n\r\n### 2. General Equation of a Circle\r\n\r\nThis form is derived by expanding the standard equation. It is useful for identifying a circle\'s properties when the equation is not in center-radius form.\r\n\r\n* **Formula:**\r\n    $$x^2 + y^2 + 2gx + 2fy + c = 0$$\r\n\r\n* **Finding the Center and Radius from the General Form:**\r\n    * **Center:** $(-g, -f)$\r\n    * **Radius:** $r = \\sqrt{g^2 + f^2 - c}$\r\n\r\n    To find `g`, `f`, and `c`, you compare the given equation to the general form. For example, if the equation is $x^2 + y^2 - 10x - 4y - 7 = 0$:\r\n    * `2gx = -10x`  => `g = -5`\r\n    * `2fy = -4y`  => `f = -2`\r\n    * `c = -7`\r\n\r\n---\r\n\r\n### 3. Equation of a Circle in Diameter Form\r\n\r\nIf you know the coordinates of the two endpoints of a circle\'s diameter, you can use this special formula.\r\n\r\n* **Formula:**\r\n    $$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$\r\n    Where $(x_1, y_1)$ and $(x_2, y_2)$ are the coordinates of the endpoints of a diameter.\r\n\r\n\r\n\r\n---\r\n\r\n### 4. Midpoint Formula\r\n\r\nThis is useful for finding the center of a circle if you know the endpoints of a diameter.\r\n\r\n* **Formula:**\r\n    $$Midpoint = \\left( \\frac{x_1 + x_2}{2}, \\frac{y_1 + y_2}{2} \\right)$$\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Equation from Diameter Endpoints\r\n**The endpoints of a diameter of a circle are (0, 6) and (8, 0). Find the equation of the circle.**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (0, 6)$ and $(x_2, y_2) = (8, 0)$.\r\nUsing the diameter form of the circle equation:\r\n$$\r\n\\begin{aligned}\r\n(x - x_1)(x - x_2) + (y - y_1)(y - y_2) &= 0 \\\\\r\n(x - 0)(x - 8) + (y - 6)(y - 0) &= 0 \\\\\r\nx(x - 8) + y(y - 6) &= 0 \\\\\r\nx^2 - 8x + y^2 - 6y &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 8x - 6y = 0$**.\r\n\r\n***\r\n\r\n#### Question 2: Equation from Diameter Endpoints\r\n**The endpoints of a diameter of a circle are (3, 0) and (0, 4). Find the equation of the circle.**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (3, 0)$ and $(x_2, y_2) = (0, 4)$.\r\nUsing the diameter form:\r\n$$\r\n\\begin{aligned}\r\n(x - 3)(x - 0) + (y - 0)(y - 4) &= 0 \\\\\r\nx(x - 3) + y(y - 4) &= 0 \\\\\r\nx^2 - 3x + y^2 - 4y &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 3x - 4y = 0$**.\r\n\r\n***\r\n\r\n#### Question 3: Equation from Diameter Endpoints\r\n**Find the equation of the circle having (1, 3) and (2, -5) as the endpoints of a diameter.**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (1, 3)$ and $(x_2, y_2) = (2, -5)$.\r\nUsing the diameter form:\r\n$$\r\n\\begin{aligned}\r\n(x - 1)(x - 2) + (y - 3)(y - (-5)) &= 0 \\\\\r\n(x^2 - 2x - x + 2) + (y - 3)(y + 5) &= 0 \\\\\r\nx^2 - 3x + 2 + y^2 + 5y - 3y - 15 &= 0 \\\\\r\nx^2 + y^2 - 3x + 2y - 13 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 3x + 2y - 13 = 0$**.\r\n\r\n***\r\n\r\n#### Question 4: Equation from Diameter Endpoints\r\n**Find the equation of the circle having the endpoints of a diameter A(5, 6) and B(3, 4).**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (5, 6)$ and $(x_2, y_2) = (3, 4)$.\r\nUsing the diameter form:\r\n$$\r\n\\begin{aligned}\r\n(x - 5)(x - 3) + (y - 6)(y - 4) &= 0 \\\\\r\n(x^2 - 3x - 5x + 15) + (y^2 - 4y - 6y + 24) &= 0 \\\\\r\nx^2 - 8x + 15 + y^2 - 10y + 24 &= 0 \\\\\r\nx^2 + y^2 - 8x - 10y + 39 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 8x - 10y + 39 = 0$**.\r\n\r\n***\r\n\r\n#### Question 5: Equation from Diameter Endpoints\r\n**Find the equation of the circle whose endpoints of a diameter are A(3, 4) and B(2, -7).**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (3, 4)$ and $(x_2, y_2) = (2, -7)$.\r\nUsing the diameter form:\r\n$$\r\n\\begin{aligned}\r\n(x - 3)(x - 2) + (y - 4)(y - (-7)) &= 0 \\\\\r\n(x^2 - 2x - 3x + 6) + (y - 4)(y + 7) &= 0 \\\\\r\nx^2 - 5x + 6 + y^2 + 7y - 4y - 28 &= 0 \\\\\r\nx^2 + y^2 - 5x + 3y - 22 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 5x + 3y - 22 = 0$**.\r\n\r\n***\r\n\r\n#### Question 6: Center from Intersecting Diameters\r\n**Find the coordinates of the center of a circle having equations of two diameters $x + y = 5$ and $2x - y = 1$.**\r\n\r\n**Solution:**\r\nThe center of a circle is the point of intersection of its diameters. We solve the system of linear equations.\r\n$$\r\n\\begin{aligned}\r\nx + y &= 5 \\quad &(1) \\\\\r\n2x - y &= 1 \\quad &(2)\r\n\\end{aligned}\r\n$$\r\nAdding equation (1) and (2):\r\n$$\r\n\\begin{aligned}\r\n(x+y) + (2x-y) &= 5 + 1 \\\\\r\n3x &= 6 \\\\\r\nx &= 2\r\n\\end{aligned}\r\n$$\r\nSubstitute $x=2$ into equation (1):\r\n$$\r\n2 + y = 5 \\implies y = 3\r\n$$\r\nThe center of the circle is **(2, 3)**.\r\n\r\n***\r\n\r\n#### Question 7: Center and Radius from General Equation\r\n**Find the center and radius of a circle having the equation $x^2 + y^2 - 10x - 4y - 7 = 0$.**\r\n\r\n**Solution:**\r\nCompare the given equation to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -10x &\\implies g = -5 \\\\\r\n2fy = -4y &\\implies f = -2 \\\\\r\nc &= -7\r\n\\end{aligned}\r\n$$\r\n**Center:** $(-g, -f) = (-(-5), -(-2)) = (5, 2)$.\r\n**Radius:** $r = \\sqrt{g^2 + f^2 - c} = \\sqrt{(-5)^2 + (-2)^2 - (-7)} = \\sqrt{25 + 4 + 7} = \\sqrt{36} = 6$.\r\n\r\nThe center is **(5, 2)** and the radius is **6 units**.\r\n\r\n***\r\n\r\n#### Question 8: Finding an Unknown from Radius\r\n**If the radius of the circle $x^2 + y^2 - 4x - 6y - k = 0$ is 4 units, find the value of k.**\r\n\r\n**Solution:**\r\nCompare the equation to $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -4x &\\implies g = -2 \\\\\r\n2fy = -6y &\\implies f = -3 \\\\\r\nc &= -k\r\n\\end{aligned}\r\n$$\r\nUsing the radius formula, with $r=4$:\r\n$$\r\n\\begin{aligned}\r\nr &= \\sqrt{g^2 + f^2 - c} \\\\\r\n4 &= \\sqrt{(-2)^2 + (-3)^2 - (-k)} \\\\\r\n4 &= \\sqrt{4 + 9 + k} \\\\\r\n4 &= \\sqrt{13 + k}\r\n\\end{aligned}\r\n$$\r\nSquaring both sides:\r\n$$\r\n16 = 13 + k \\implies k = 3\r\n$$\r\nThus, the value of **$k = 3$**.\r\n\r\n***\r\n\r\n#### Question 9: Finding Unknowns from Center\r\n**If the center of the circle $x^2 + y^2 - ax - by - 12 = 0$ is (2, 3), find the value of a and b.**\r\n\r\n**Solution:**\r\nCompare the equation to $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -ax &\\implies g = -a/2 \\\\\r\n2fy = -by &\\implies f = -b/2\r\n\\end{aligned}\r\n$$\r\nThe center is given by $(-g, -f)$.\r\n$$\r\nCenter = (-(-a/2), -(-b/2)) = (a/2, b/2)\r\n$$\r\nWe are given that the center is (2, 3).\r\n$$\r\n\\begin{aligned}\r\na/2 &= 2 \\implies a = 4 \\\\\r\nb/2 &= 3 \\implies b = 6\r\n\\end{aligned}\r\n$$\r\nThus, **$a=4$ and $b=6$**.\r\n\r\n***\r\n\r\n#### Question 10: Finding an Unknown from Center\r\n**Find the value of m when the center of a circle with equation $x^2 + y^2 - mx + 6y - 12 = 0$ is (-2, -3).**\r\n\r\n**Solution:**\r\nCompare the equation to $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -mx &\\implies g = -m/2 \\\\\r\n2fy = 6y &\\implies f = 3\r\n\\end{aligned}\r\n$$\r\nThe center is $(-g, -f) = (-(-m/2), -3) = (m/2, -3)$.\r\nWe are given that the center is $(-2, -3)$.\r\nEquating the x-coordinates:\r\n$$\r\n\\frac{m}{2} = -2 \\implies m = -4\r\n$$\r\nThus, **$m = -4$**.\r\n\r\n***\r\n\r\n#### Question 11: Equation from Center and Radius\r\n**Find the equation of a circle having centre (1, -2) and diameter $2\\sqrt{5}$ units.**\r\n\r\n**Solution:**\r\nCenter $(h, k) = (1, -2)$.\r\nDiameter is $2\\sqrt{5}$, so the radius is $r = \\sqrt{5}$.\r\nUsing the standard form $(x-h)^2 + (y-k)^2 = r^2$:\r\n$$\r\n\\begin{aligned}\r\n(x - 1)^2 + (y - (-2))^2 &= (\\sqrt{5})^2 \\\\\r\n(x^2 - 2x + 1) + (y^2 + 4y + 4) &= 5 \\\\\r\nx^2 + y^2 - 2x + 4y + 5 &= 5\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 2x + 4y = 0$**.\r\n\r\n***\r\n\r\n#### Question 12: Equation from Center and Radius\r\n**Find the equation of a circle having centre (2, 3) and radius 5 units.**\r\n\r\n**Solution:**\r\nCenter $(h, k) = (2, 3)$ and radius $r = 5$.\r\nUsing the standard form:\r\n$$\r\n\\begin{aligned}\r\n(x - 2)^2 + (y - 3)^2 &= 5^2 \\\\\r\n(x^2 - 4x + 4) + (y^2 - 6y + 9) &= 25 \\\\\r\nx^2 + y^2 - 4x - 6y + 13 &= 25\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 4x - 6y - 12 = 0$**.\r\n\r\n***\r\n\r\n#### Question 13: Center from General Equation\r\n**Find the coordinates of the center of a circle having an equation $x^2 + y^2 + 4x - 5 = 0$.**\r\n\r\n**Solution:**\r\nCompare with $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = 4x &\\implies g = 2 \\\\\r\n2fy = 0y &\\implies f = 0\r\n\\end{aligned}\r\n$$\r\nThe center is $(-g, -f) = (-2, 0)$.\r\nThe center is **(-2, 0)**.\r\n\r\n***\r\n\r\n#### Question 14: Equation from Center and Radius\r\n**Find the equation of a circle having the centre (-1, -3) and radius 3 units.**\r\n\r\n**Solution:**\r\nCenter $(h, k) = (-1, -3)$ and radius $r=3$.\r\nUsing the standard form:\r\n$$\r\n\\begin{aligned}\r\n(x - (-1))^2 + (y - (-3))^2 &= 3^2 \\\\\r\n(x+1)^2 + (y+3)^2 &= 9 \\\\\r\n(x^2 + 2x + 1) + (y^2 + 6y + 9) &= 9 \\\\\r\nx^2 + y^2 + 2x + 6y + 1 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 2x + 6y + 1 = 0$**.\r\n\r\n***\r\n\r\n#### Question 15: Center from General Equation\r\n**Find the center of the circle whose equation is $x^2 + y^2 - 6x + 2y + 1 = 0$.**\r\n\r\n**Solution:**\r\nCompare with $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -6x &\\implies g = -3 \\\\\r\n2fy = 2y &\\implies f = 1\r\n\\end{aligned}\r\n$$\r\nThe center is $(-g, -f) = (-(-3), -1) = (3, -1)$.\r\nThe center is **(3, -1)**.\r\n\r\n---\r\n\r\n### Long Questions\r\n\r\n***\r\n\r\n#### Question 16: Equation of Circle Passing Through Center of Another\r\n**Circle A with centre X(-4, 5) passes through the centre Y of circle B. If the equation of circle B is $x^2 + y^2 - 4x + 6y - 12 = 0$, find the equation of circle A.**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of circle B.**\r\nCompare $x^2 + y^2 - 4x + 6y - 12 = 0$ with the general form.\r\n$g = -2$, $f=3$.\r\nCenter of B is $Y = (-g, -f) = (2, -3)$.\r\n\r\n**Step 2: Find the radius of circle A.**\r\nThe radius of circle A is the distance between its center X(-4, 5) and the point it passes through, Y(2, -3).\r\n$$\r\n\\begin{aligned}\r\nr^2 &= (x_2 - x_1)^2 + (y_2 - y_1)^2 \\\\\r\n&= (2 - (-4))^2 + (-3 - 5)^2 \\\\\r\n&= (6)^2 + (-8)^2 \\\\\r\n&= 36 + 64 = 100\r\n\\end{aligned}\r\n$$\r\n**Step 3: Find the equation of circle A.**\r\nThe center of circle A is X(h, k) = (-4, 5) and $r^2=100$.\r\n$$\r\n\\begin{aligned}\r\n(x - (-4))^2 + (y - 5)^2 &= 100 \\\\\r\n(x+4)^2 + (y-5)^2 &= 100 \\\\\r\nx^2 + 8x + 16 + y^2 - 10y + 25 &= 100 \\\\\r\nx^2 + y^2 + 8x - 10y - 59 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 8x - 10y - 59 = 0$**.\r\n\r\n***\r\n\r\n#### Question 17: Equation of Circle Passing Through Center of Another\r\n**Find the equation of a circle with centre (-1, 2) and passing through the centre of the circle $x^2 + y^2 - 6x - 10y - 2 = 0$.**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of the given circle.**\r\nComparing $x^2 + y^2 - 6x - 10y - 2 = 0$ to the general form.\r\n$g=-3$, $f=-5$.\r\nThe center is $(-g, -f) = (3, 5)$.\r\n\r\n**Step 2: Find the radius of the required circle.**\r\nThe required circle has center $(h,k)=(-1,2)$ and passes through $(3,5)$. The radius is the distance between these two points.\r\n$$\r\n\\begin{aligned}\r\nr^2 &= (3 - (-1))^2 + (5 - 2)^2 \\\\\r\n&= (4)^2 + (3)^2 \\\\\r\n&= 16 + 9 = 25\r\n\\end{aligned}\r\n$$\r\n**Step 3: Find the equation of the required circle.**\r\nCenter is $(-1, 2)$ and $r^2=25$.\r\n$$\r\n\\begin{aligned}\r\n(x - (-1))^2 + (y - 2)^2 &= 25 \\\\\r\n(x+1)^2 + (y-2)^2 &= 25 \\\\\r\nx^2 + 2x + 1 + y^2 - 4y + 4 &= 25 \\\\\r\nx^2 + y^2 + 2x - 4y - 20 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 2x - 4y - 20 = 0$**.\r\n\r\n***\r\n\r\n#### Question 18: Finding the Other End of a Diameter\r\n**If one endpoint of a diameter of the circle $x^2 + y^2 + 4x - 6y + 8 = 0$ is (-4, 4), find the coordinates of the other endpoint.**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of the circle.**\r\nComparing the equation to the general form: $g=2$, $f=-3$.\r\nThe center is $(-g, -f) = (-2, 3)$.\r\n\r\n**Step 2: Use the midpoint formula.**\r\nThe center of a circle is the midpoint of its diameter. Let the other endpoint be $(x_2, y_2)$.\r\n$$\r\n\\begin{aligned}\r\n\\text{Center} &= \\left( \\frac{x_1+x_2}{2}, \\frac{y_1+y_2}{2} \\right) \\\\\r\n(-2, 3) &= \\left( \\frac{-4+x_2}{2}, \\frac{4+y_2}{2} \\right)\r\n\\end{aligned}\r\n$$\r\nEquating coordinates:\r\n$$\r\n-2 = \\frac{-4+x_2}{2} \\implies -4 = -4+x_2 \\implies x_2=0\r\n$$\r\n$$\r\n3 = \\frac{4+y_2}{2} \\implies 6 = 4+y_2 \\implies y_2=2\r\n$$\r\nThe coordinates of the other end are **(0, 2)**.\r\n\r\n***\r\n\r\n#### Question 19: Equation of a Concentric Circle\r\n**Find the equation of a circle concentric with the circle $2x^2 + 2y^2 + 4x - 2y + 1 = 0$ and passing through the point (4,-2).**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of the given circle.**\r\nFirst, divide the equation by 2 to get the standard general form:\r\n$x^2 + y^2 + 2x - y + \\frac{1}{2} = 0$.\r\nComparing with the general form, $g=1$ and $f=-1/2$.\r\nThe center is $(-g, -f) = (-1, 1/2)$.\r\n\r\n**Step 2: Find the radius of the required circle.**\r\nSince the required circle is concentric, its center is also $(-1, 1/2)$. It passes through $(4, -2)$. The radius is the distance between these points.\r\n$$\r\n\\begin{aligned}\r\nr^2 &= (4 - (-1))^2 + (-2 - \\frac{1}{2})^2 \\\\\r\n&= (5)^2 + (-\\frac{5}{2})^2 \\\\\r\n&= 25 + \\frac{25}{4} = \\frac{100+25}{4} = \\frac{125}{4}\r\n\\end{aligned}\r\n$$\r\n**Step 3: Write the equation of the required circle.**\r\nCenter is $(-1, 1/2)$ and $r^2 = 125/4$.\r\n$$\r\n\\begin{aligned}\r\n(x - (-1))^2 + (y - \\frac{1}{2})^2 &= \\frac{125}{4} \\\\\r\n(x+1)^2 + (y - \\frac{1}{2})^2 &= \\frac{125}{4} \\\\\r\nx^2+2x+1 + y^2 - y + \\frac{1}{4} &= \\frac{125}{4}\r\n\\end{aligned}\r\n$$\r\nMultiply by 4 to clear the fraction:\r\n$$\r\n\\begin{aligned}\r\n4x^2 + 8x + 4 + 4y^2 - 4y + 1 &= 125 \\\\\r\n4x^2 + 4y^2 + 8x - 4y - 120 &= 0 \\\\\r\nx^2 + y^2 + 2x - y - 30 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 2x - y - 30 = 0$**.\r\n\r\n***\r\n\r\n#### Question 20: Equation of a Circle from Two Points and Center on a Line\r\n**Find the equation of the circle which passes through the points (4, 1) and (6, 5) and has its centre on the line 4x + y = 16.**\r\n\r\n**Solution:**\r\nLet the center be $(h,k)$. Since it lies on the line $4x+y=16$, we have $4h+k=16 \\implies k = 16-4h$.\r\nThe distance from the center to any point on the circle is the radius. So, the distance from $(h,k)$ to $(4,1)$ is the same as to $(6,5)$.\r\n$$\r\n\\begin{aligned}\r\n(h-4)^2 + (k-1)^2 &= (h-6)^2 + (k-5)^2 \\quad (\\text{since } r^2=r^2) \\\\\r\nh^2 - 8h + 16 + k^2 - 2k + 1 &= h^2 - 12h + 36 + k^2 - 10k + 25 \\\\\r\n-8h - 2k + 17 &= -12h - 10k + 61 \\\\\r\n4h + 8k &= 44 \\\\\r\nh + 2k &= 11\r\n\\end{aligned}\r\n$$\r\nNow substitute $k = 16-4h$ into this equation:\r\n$$\r\n\\begin{aligned}\r\nh + 2(16-4h) &= 11 \\\\\r\nh + 32 - 8h &= 11 \\\\\r\n-7h &= -21 \\\\\r\nh &= 3\r\n\\end{aligned}\r\n$$\r\nNow find k: $k = 16 - 4(3) = 16 - 12 = 4$.\r\nThe center is $(3,4)$.\r\nFind the radius squared, $r^2$, using point $(4,1)$:\r\n$$\r\nr^2 = (4-3)^2 + (1-4)^2 = 1^2 + (-3)^2 = 1+9=10\r\n$$\r\nThe equation is $(x-3)^2 + (y-4)^2 = 10$, which expands to **$x^2 + y^2 - 6x - 8y + 15 = 0$**.\r\n\r\n***\r\n\r\n#### Question 21: Equation of a Circle from Three Points\r\n**Find the equation of the circle which passes through the points P(5, 7), Q(6, 6) and R(2, -2).**\r\n\r\n**Solution:**\r\nLet the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$.\r\nThe squared distance from the center $(h,k)$ to each point is equal.\r\n**From P and Q:**\r\n$$\r\n\\begin{aligned}\r\n(h-5)^2 + (k-7)^2 &= (h-6)^2 + (k-6)^2 \\\\\r\nh^2-10h+25 + k^2-14k+49 &= h^2-12h+36 + k^2-12k+36 \\\\\r\n-10h-14k+74 &= -12h-12k+72 \\\\\r\n2h - 2k + 2 &= 0 \\implies h - k + 1 = 0 \\implies h = k-1 \\quad \\text{(Eq. 1)}\r\n\\end{aligned}\r\n$$\r\n**From Q and R:**\r\n$$\r\n\\begin{aligned}\r\n(h-6)^2 + (k-6)^2 &= (h-2)^2 + (k-(-2))^2 \\\\\r\nh^2-12h+36 + k^2-12k+36 &= h^2-4h+4 + k^2+4k+4 \\\\\r\n-12h-12k+72 &= -4h+4k+8 \\\\\r\n64 &= 8h + 16k \\\\\r\n8 &= h + 2k \\quad \\text{(Eq. 2)}\r\n\\end{aligned}\r\n$$\r\n**Solve for h and k:**\r\nSubstitute Eq. 1 into Eq. 2:\r\n$8 = (k-1) + 2k \\implies 9 = 3k \\implies k=3$.\r\nThen, $h = 3-1=2$. The center is $(h,k)=(2,3)$.\r\n\r\n**Find $r^2$ using point P(5,7):**\r\n$r^2 = (5-2)^2 + (7-3)^2 = 3^2 + 4^2 = 9+16=25$.\r\n\r\n**Write the equation:**\r\n$$(x-2)^2 + (y-3)^2 = 25$$\r\nThe required equation is **$x^2 + y^2 - 4x - 6y - 12 = 0$**.', 1, '2025-07-25 01:11:38', '2025-09-18 07:21:49');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(144, 13116, '# Grade 10 Optional Mathematics: The Circle\r\n## Unit: Coordinate Geometry | Chapter: Circle | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nThis guide offers a complete overview of the **Circle** chapter in Coordinate Geometry. It is designed to make the concepts easy to understand, with clear definitions, essential formulas, and detailed, step-by-step solutions to a wide range of problems. By the end of this note, you will have mastered the different forms of the circle equation and how to find a circle\'s center, radius, and equation based on various given conditions. ⚪\r\n\r\n### Key Learning Areas\r\n1.  **Standard Equation of a Circle:** Understanding and using the center-radius form.\r\n2.  **General Equation of a Circle:** Identifying the center and radius from the general form.\r\n3.  **Diameter Form:** Finding the equation of a circle given the endpoints of a diameter.\r\n4.  **Problem Solving:** Finding the equation of a circle based on various conditions, such as passing through points, having a specific center, or being concentric with another circle.\r\n\r\n---\r\n\r\n## 🔑 Key Concepts & Formulas\r\n\r\nBefore we solve the problems, let\'s understand the fundamental equations of a circle.\r\n\r\n### 1. Standard Equation of a Circle (Center-Radius Form)\r\n\r\nThis is the most common form of a circle\'s equation. It directly uses the circle\'s center and radius.\r\n\r\n* **Formula:**\r\n    $$(x - h)^2 + (y - k)^2 = r^2$$\r\n    Where:\r\n    * `(h, k)` are the coordinates of the center of the circle.\r\n    * `r` is the radius of the circle.\r\n\r\n\r\n\r\n---\r\n\r\n### 2. General Equation of a Circle\r\n\r\nThis form is derived by expanding the standard equation. It is useful for identifying a circle\'s properties when the equation is not in center-radius form.\r\n\r\n* **Formula:**\r\n    $$x^2 + y^2 + 2gx + 2fy + c = 0$$\r\n\r\n* **Finding the Center and Radius from the General Form:**\r\n    * **Center:** $(-g, -f)$\r\n    * **Radius:** $r = \\sqrt{g^2 + f^2 - c}$\r\n\r\n    To find `g`, `f`, and `c`, you compare the given equation to the general form. For example, if the equation is $x^2 + y^2 - 10x - 4y - 7 = 0$:\r\n    * `2gx = -10x`  => `g = -5`\r\n    * `2fy = -4y`  => `f = -2`\r\n    * `c = -7`\r\n\r\n---\r\n\r\n### 3. Equation of a Circle in Diameter Form\r\n\r\nIf you know the coordinates of the two endpoints of a circle\'s diameter, you can use this special formula.\r\n\r\n* **Formula:**\r\n    $$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$\r\n    Where $(x_1, y_1)$ and $(x_2, y_2)$ are the coordinates of the endpoints of a diameter.\r\n\r\n\r\n\r\n---\r\n\r\n### 4. Midpoint Formula\r\n\r\nThis is useful for finding the center of a circle if you know the endpoints of a diameter.\r\n\r\n* **Formula:**\r\n    $$Midpoint = \\left( \\frac{x_1 + x_2}{2}, \\frac{y_1 + y_2}{2} \\right)$$\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Equation from Diameter Endpoints\r\n**The endpoints of a diameter of a circle are (0, 6) and (8, 0). Find the equation of the circle.**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (0, 6)$ and $(x_2, y_2) = (8, 0)$.\r\nUsing the diameter form of the circle equation:\r\n$$\r\n\\begin{aligned}\r\n(x - x_1)(x - x_2) + (y - y_1)(y - y_2) &= 0 \\\\\r\n(x - 0)(x - 8) + (y - 6)(y - 0) &= 0 \\\\\r\nx(x - 8) + y(y - 6) &= 0 \\\\\r\nx^2 - 8x + y^2 - 6y &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 8x - 6y = 0$**.\r\n\r\n***\r\n\r\n#### Question 2: Equation from Diameter Endpoints\r\n**The endpoints of a diameter of a circle are (3, 0) and (0, 4). Find the equation of the circle.**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (3, 0)$ and $(x_2, y_2) = (0, 4)$.\r\nUsing the diameter form:\r\n$$\r\n\\begin{aligned}\r\n(x - 3)(x - 0) + (y - 0)(y - 4) &= 0 \\\\\r\nx(x - 3) + y(y - 4) &= 0 \\\\\r\nx^2 - 3x + y^2 - 4y &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 3x - 4y = 0$**.\r\n\r\n***\r\n\r\n#### Question 3: Equation from Diameter Endpoints\r\n**Find the equation of the circle having (1, 3) and (2, -5) as the endpoints of a diameter.**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (1, 3)$ and $(x_2, y_2) = (2, -5)$.\r\nUsing the diameter form:\r\n$$\r\n\\begin{aligned}\r\n(x - 1)(x - 2) + (y - 3)(y - (-5)) &= 0 \\\\\r\n(x^2 - 2x - x + 2) + (y - 3)(y + 5) &= 0 \\\\\r\nx^2 - 3x + 2 + y^2 + 5y - 3y - 15 &= 0 \\\\\r\nx^2 + y^2 - 3x + 2y - 13 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 3x + 2y - 13 = 0$**.\r\n\r\n***\r\n\r\n#### Question 4: Equation from Diameter Endpoints\r\n**Find the equation of the circle having the endpoints of a diameter A(5, 6) and B(3, 4).**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (5, 6)$ and $(x_2, y_2) = (3, 4)$.\r\nUsing the diameter form:\r\n$$\r\n\\begin{aligned}\r\n(x - 5)(x - 3) + (y - 6)(y - 4) &= 0 \\\\\r\n(x^2 - 3x - 5x + 15) + (y^2 - 4y - 6y + 24) &= 0 \\\\\r\nx^2 - 8x + 15 + y^2 - 10y + 24 &= 0 \\\\\r\nx^2 + y^2 - 8x - 10y + 39 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 8x - 10y + 39 = 0$**.\r\n\r\n***\r\n\r\n#### Question 5: Equation from Diameter Endpoints\r\n**Find the equation of the circle whose endpoints of a diameter are A(3, 4) and B(2, -7).**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (3, 4)$ and $(x_2, y_2) = (2, -7)$.\r\nUsing the diameter form:\r\n$$\r\n\\begin{aligned}\r\n(x - 3)(x - 2) + (y - 4)(y - (-7)) &= 0 \\\\\r\n(x^2 - 2x - 3x + 6) + (y - 4)(y + 7) &= 0 \\\\\r\nx^2 - 5x + 6 + y^2 + 7y - 4y - 28 &= 0 \\\\\r\nx^2 + y^2 - 5x + 3y - 22 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 5x + 3y - 22 = 0$**.\r\n\r\n***\r\n\r\n#### Question 6: Center from Intersecting Diameters\r\n**Find the coordinates of the center of a circle having equations of two diameters $x + y = 5$ and $2x - y = 1$.**\r\n\r\n**Solution:**\r\nThe center of a circle is the point of intersection of its diameters. We solve the system of linear equations.\r\n$$\r\n\\begin{aligned}\r\nx + y &= 5 \\quad &(1) \\\\\r\n2x - y &= 1 \\quad &(2)\r\n\\end{aligned}\r\n$$\r\nAdding equation (1) and (2):\r\n$$\r\n\\begin{aligned}\r\n(x+y) + (2x-y) &= 5 + 1 \\\\\r\n3x &= 6 \\\\\r\nx &= 2\r\n\\end{aligned}\r\n$$\r\nSubstitute $x=2$ into equation (1):\r\n$$\r\n2 + y = 5 \\implies y = 3\r\n$$\r\nThe center of the circle is **(2, 3)**.\r\n\r\n***\r\n\r\n#### Question 7: Center and Radius from General Equation\r\n**Find the center and radius of a circle having the equation $x^2 + y^2 - 10x - 4y - 7 = 0$.**\r\n\r\n**Solution:**\r\nCompare the given equation to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -10x &\\implies g = -5 \\\\\r\n2fy = -4y &\\implies f = -2 \\\\\r\nc &= -7\r\n\\end{aligned}\r\n$$\r\n**Center:** $(-g, -f) = (-(-5), -(-2)) = (5, 2)$.\r\n**Radius:** $r = \\sqrt{g^2 + f^2 - c} = \\sqrt{(-5)^2 + (-2)^2 - (-7)} = \\sqrt{25 + 4 + 7} = \\sqrt{36} = 6$.\r\n\r\nThe center is **(5, 2)** and the radius is **6 units**.\r\n\r\n***\r\n\r\n#### Question 8: Finding an Unknown from Radius\r\n**If the radius of the circle $x^2 + y^2 - 4x - 6y - k = 0$ is 4 units, find the value of k.**\r\n\r\n**Solution:**\r\nCompare the equation to $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -4x &\\implies g = -2 \\\\\r\n2fy = -6y &\\implies f = -3 \\\\\r\nc &= -k\r\n\\end{aligned}\r\n$$\r\nUsing the radius formula, with $r=4$:\r\n$$\r\n\\begin{aligned}\r\nr &= \\sqrt{g^2 + f^2 - c} \\\\\r\n4 &= \\sqrt{(-2)^2 + (-3)^2 - (-k)} \\\\\r\n4 &= \\sqrt{4 + 9 + k} \\\\\r\n4 &= \\sqrt{13 + k}\r\n\\end{aligned}\r\n$$\r\nSquaring both sides:\r\n$$\r\n16 = 13 + k \\implies k = 3\r\n$$\r\nThus, the value of **$k = 3$**.\r\n\r\n***\r\n\r\n#### Question 9: Finding Unknowns from Center\r\n**If the center of the circle $x^2 + y^2 - ax - by - 12 = 0$ is (2, 3), find the value of a and b.**\r\n\r\n**Solution:**\r\nCompare the equation to $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -ax &\\implies g = -a/2 \\\\\r\n2fy = -by &\\implies f = -b/2\r\n\\end{aligned}\r\n$$\r\nThe center is given by $(-g, -f)$.\r\n$$\r\nCenter = (-(-a/2), -(-b/2)) = (a/2, b/2)\r\n$$\r\nWe are given that the center is (2, 3).\r\n$$\r\n\\begin{aligned}\r\na/2 &= 2 \\implies a = 4 \\\\\r\nb/2 &= 3 \\implies b = 6\r\n\\end{aligned}\r\n$$\r\nThus, **$a=4$ and $b=6$**.\r\n\r\n***\r\n\r\n#### Question 10: Finding an Unknown from Center\r\n**Find the value of m when the center of a circle with equation $x^2 + y^2 - mx + 6y - 12 = 0$ is (-2, -3).**\r\n\r\n**Solution:**\r\nCompare the equation to $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -mx &\\implies g = -m/2 \\\\\r\n2fy = 6y &\\implies f = 3\r\n\\end{aligned}\r\n$$\r\nThe center is $(-g, -f) = (-(-m/2), -3) = (m/2, -3)$.\r\nWe are given that the center is $(-2, -3)$.\r\nEquating the x-coordinates:\r\n$$\r\n\\frac{m}{2} = -2 \\implies m = -4\r\n$$\r\nThus, **$m = -4$**.\r\n\r\n***\r\n\r\n#### Question 11: Equation from Center and Radius\r\n**Find the equation of a circle having centre (1, -2) and diameter $2\\sqrt{5}$ units.**\r\n\r\n**Solution:**\r\nCenter $(h, k) = (1, -2)$.\r\nDiameter is $2\\sqrt{5}$, so the radius is $r = \\sqrt{5}$.\r\nUsing the standard form $(x-h)^2 + (y-k)^2 = r^2$:\r\n$$\r\n\\begin{aligned}\r\n(x - 1)^2 + (y - (-2))^2 &= (\\sqrt{5})^2 \\\\\r\n(x^2 - 2x + 1) + (y^2 + 4y + 4) &= 5 \\\\\r\nx^2 + y^2 - 2x + 4y + 5 &= 5\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 2x + 4y = 0$**.\r\n\r\n***\r\n\r\n#### Question 12: Equation from Center and Radius\r\n**Find the equation of a circle having centre (2, 3) and radius 5 units.**\r\n\r\n**Solution:**\r\nCenter $(h, k) = (2, 3)$ and radius $r = 5$.\r\nUsing the standard form:\r\n$$\r\n\\begin{aligned}\r\n(x - 2)^2 + (y - 3)^2 &= 5^2 \\\\\r\n(x^2 - 4x + 4) + (y^2 - 6y + 9) &= 25 \\\\\r\nx^2 + y^2 - 4x - 6y + 13 &= 25\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 4x - 6y - 12 = 0$**.\r\n\r\n***\r\n\r\n#### Question 13: Center from General Equation\r\n**Find the coordinates of the center of a circle having an equation $x^2 + y^2 + 4x - 5 = 0$.**\r\n\r\n**Solution:**\r\nCompare with $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = 4x &\\implies g = 2 \\\\\r\n2fy = 0y &\\implies f = 0\r\n\\end{aligned}\r\n$$\r\nThe center is $(-g, -f) = (-2, 0)$.\r\nThe center is **(-2, 0)**.\r\n\r\n***\r\n\r\n#### Question 14: Equation from Center and Radius\r\n**Find the equation of a circle having the centre (-1, -3) and radius 3 units.**\r\n\r\n**Solution:**\r\nCenter $(h, k) = (-1, -3)$ and radius $r=3$.\r\nUsing the standard form:\r\n$$\r\n\\begin{aligned}\r\n(x - (-1))^2 + (y - (-3))^2 &= 3^2 \\\\\r\n(x+1)^2 + (y+3)^2 &= 9 \\\\\r\n(x^2 + 2x + 1) + (y^2 + 6y + 9) &= 9 \\\\\r\nx^2 + y^2 + 2x + 6y + 1 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 2x + 6y + 1 = 0$**.\r\n\r\n***\r\n\r\n#### Question 15: Center from General Equation\r\n**Find the center of the circle whose equation is $x^2 + y^2 - 6x + 2y + 1 = 0$.**\r\n\r\n**Solution:**\r\nCompare with $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -6x &\\implies g = -3 \\\\\r\n2fy = 2y &\\implies f = 1\r\n\\end{aligned}\r\n$$\r\nThe center is $(-g, -f) = (-(-3), -1) = (3, -1)$.\r\nThe center is **(3, -1)**.\r\n\r\n---\r\n\r\n### Long Questions\r\n\r\n***\r\n\r\n#### Question 16: Equation of Circle Passing Through Center of Another\r\n**Circle A with centre X(-4, 5) passes through the centre Y of circle B. If the equation of circle B is $x^2 + y^2 - 4x + 6y - 12 = 0$, find the equation of circle A.**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of circle B.**\r\nCompare $x^2 + y^2 - 4x + 6y - 12 = 0$ with the general form.\r\n$g = -2$, $f=3$.\r\nCenter of B is $Y = (-g, -f) = (2, -3)$.\r\n\r\n**Step 2: Find the radius of circle A.**\r\nThe radius of circle A is the distance between its center X(-4, 5) and the point it passes through, Y(2, -3).\r\n$$\r\n\\begin{aligned}\r\nr^2 &= (x_2 - x_1)^2 + (y_2 - y_1)^2 \\\\\r\n&= (2 - (-4))^2 + (-3 - 5)^2 \\\\\r\n&= (6)^2 + (-8)^2 \\\\\r\n&= 36 + 64 = 100\r\n\\end{aligned}\r\n$$\r\n**Step 3: Find the equation of circle A.**\r\nThe center of circle A is X(h, k) = (-4, 5) and $r^2=100$.\r\n$$\r\n\\begin{aligned}\r\n(x - (-4))^2 + (y - 5)^2 &= 100 \\\\\r\n(x+4)^2 + (y-5)^2 &= 100 \\\\\r\nx^2 + 8x + 16 + y^2 - 10y + 25 &= 100 \\\\\r\nx^2 + y^2 + 8x - 10y - 59 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 8x - 10y - 59 = 0$**.\r\n\r\n***\r\n\r\n#### Question 17: Equation of Circle Passing Through Center of Another\r\n**Find the equation of a circle with centre (-1, 2) and passing through the centre of the circle having equation $x^2 + y^2 - 6x - 10y - 2 = 0$.**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of the given circle.**\r\nComparing $x^2 + y^2 - 6x - 10y - 2 = 0$ to the general form.\r\n$g=-3$, $f=-5$.\r\nThe center is $(-g, -f) = (3, 5)$.\r\n\r\n**Step 2: Find the radius of the required circle.**\r\nThe required circle has center $(h,k)=(-1,2)$ and passes through $(3,5)$. The radius is the distance between these two points.\r\n$$\r\n\\begin{aligned}\r\nr^2 &= (3 - (-1))^2 + (5 - 2)^2 \\\\\r\n&= (4)^2 + (3)^2 \\\\\r\n&= 16 + 9 = 25\r\n\\end{aligned}\r\n$$\r\n**Step 3: Find the equation of the required circle.**\r\nCenter is $(-1, 2)$ and $r^2=25$.\r\n$$\r\n\\begin{aligned}\r\n(x - (-1))^2 + (y - 2)^2 &= 25 \\\\\r\n(x+1)^2 + (y-2)^2 &= 25 \\\\\r\nx^2 + 2x + 1 + y^2 - 4y + 4 &= 25 \\\\\r\nx^2 + y^2 + 2x - 4y - 20 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 2x - 4y - 20 = 0$**.\r\n\r\n***\r\n\r\n#### Question 18: Finding the Other End of a Diameter\r\n**If one endpoint of a diameter of the circle $x^2 + y^2 + 4x - 6y + 8 = 0$ is (-4, 4), find the coordinates of the other endpoint.**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of the circle.**\r\nComparing the equation to the general form: $g=2$, $f=-3$.\r\nThe center is $(-g, -f) = (-2, 3)$.\r\n\r\n**Step 2: Use the midpoint formula.**\r\nThe center of a circle is the midpoint of its diameter. Let the other endpoint be $(x_2, y_2)$.\r\n$$\r\n\\begin{aligned}\r\n\\text{Center} &= \\left( \\frac{x_1+x_2}{2}, \\frac{y_1+y_2}{2} \\right) \\\\\r\n(-2, 3) &= \\left( \\frac{-4+x_2}{2}, \\frac{4+y_2}{2} \\right)\r\n\\end{aligned}\r\n$$\r\nEquating coordinates:\r\n$$\r\n-2 = \\frac{-4+x_2}{2} \\implies -4 = -4+x_2 \\implies x_2=0\r\n$$\r\n$$\r\n3 = \\frac{4+y_2}{2} \\implies 6 = 4+y_2 \\implies y_2=2\r\n$$\r\nThe coordinates of the other end are **(0, 2)**.\r\n\r\n***\r\n\r\n#### Question 19: Equation of a Concentric Circle\r\n**Find the equation of a circle concentric with the circle $2x^2 + 2y^2 + 4x - 2y + 1 = 0$ and passing through the point (4,-2).**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of the given circle.**\r\nFirst, divide the equation by 2 to get the standard general form:\r\n$x^2 + y^2 + 2x - y + \\frac{1}{2} = 0$.\r\nComparing with the general form, $g=1$ and $f=-1/2$.\r\nThe center is $(-g, -f) = (-1, 1/2)$.\r\n\r\n**Step 2: Find the radius of the required circle.**\r\nSince the required circle is concentric, its center is also $(-1, 1/2)$. It passes through $(4, -2)$. The radius is the distance between these points.\r\n$$\r\n\\begin{aligned}\r\nr^2 &= (4 - (-1))^2 + (-2 - \\frac{1}{2})^2 \\\\\r\n&= (5)^2 + (-\\frac{5}{2})^2 \\\\\r\n&= 25 + \\frac{25}{4} = \\frac{100+25}{4} = \\frac{125}{4}\r\n\\end{aligned}\r\n$$\r\n**Step 3: Write the equation of the required circle.**\r\nCenter is $(-1, 1/2)$ and $r^2 = 125/4$.\r\n$$\r\n\\begin{aligned}\r\n(x - (-1))^2 + (y - \\frac{1}{2})^2 &= \\frac{125}{4} \\\\\r\n(x+1)^2 + (y - \\frac{1}{2})^2 &= \\frac{125}{4} \\\\\r\nx^2+2x+1 + y^2 - y + \\frac{1}{4} &= \\frac{125}{4}\r\n\\end{aligned}\r\n$$\r\nMultiply by 4 to clear the fraction:\r\n$$\r\n\\begin{aligned}\r\n4x^2 + 8x + 4 + 4y^2 - 4y + 1 &= 125 \\\\\r\n4x^2 + 4y^2 + 8x - 4y - 120 &= 0 \\\\\r\nx^2 + y^2 + 2x - y - 30 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 2x - y - 30 = 0$**.\r\n\r\n***\r\n\r\n#### Question 20: Equation of a Concentric Circle\r\n**Find the equation of a circle passing through the point (4, 3) and concentric with the circle having equation $x^2 + y^2 + 6x - 8y - 11 = 0$.**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of the given circle.**\r\nCompare $x^2 + y^2 + 6x - 8y - 11 = 0$ with the general form.\r\n$g=3$, $f=-4$. The center is $(-g,-f) = (-3, 4)$.\r\n\r\n**Step 2: Find the radius of the required circle.**\r\nSince the new circle is concentric, its center is also $(-3, 4)$. It passes through the point $(4, 3)$. The radius is the distance between these points.\r\n$$\r\n\\begin{aligned}\r\nr^2 &= (4 - (-3))^2 + (3 - 4)^2 \\\\\r\n&= (7)^2 + (-1)^2 \\\\\r\n&= 49 + 1 = 50\r\n\\end{aligned}\r\n$$\r\n**Step 3: Write the equation of the required circle.**\r\nCenter is $(-3, 4)$ and $r^2 = 50$.\r\n$$\r\n\\begin{aligned}\r\n(x - (-3))^2 + (y - 4)^2 &= 50 \\\\\r\n(x+3)^2 + (y-4)^2 &= 50 \\\\\r\nx^2+6x+9 + y^2-8y+16 &= 50 \\\\\r\nx^2+y^2+6x-8y-25 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 6x - 8y - 25 = 0$**.\r\n\r\n***\r\n\r\n#### Question 21: Equation of a Circle from Two Points and Center on a Line\r\n**Find the equation of the circle whose centre lies on the line $x - 4y = 1$ and which passes through the points (3, 7) and (5, 5).**\r\n\r\n**Solution:**\r\nLet the center be $(h,k)$. Since it lies on the line $x - 4y = 1$, we have $h - 4k = 1 \\implies h = 4k+1$.\r\nThe distance from the center to any point on the circle is the radius. So, the squared distance from $(h,k)$ to $(3,7)$ is the same as to $(5,5)$.\r\n$$\r\n\\begin{aligned}\r\n(h-3)^2 + (k-7)^2 &= (h-5)^2 + (k-5)^2 \\\\\r\nh^2 - 6h + 9 + k^2 - 14k + 49 &= h^2 - 10h + 25 + k^2 - 10k + 25 \\\\\r\n-6h - 14k + 58 &= -10h - 10k + 50 \\\\\r\n4h - 4k &= -8 \\\\\r\nh - k &= -2\r\n\\end{aligned}\r\n$$\r\nNow substitute $h = 4k+1$ into this new equation:\r\n$$\r\n\\begin{aligned}\r\n(4k+1) - k &= -2 \\\\\r\n3k &= -3 \\\\\r\nk &= -1\r\n\\end{aligned}\r\n$$\r\nNow find h: $h = 4(-1)+1 = -3$. The center is $(-3,-1)$.\r\nFind the radius squared, $r^2$, using point $(3,7)$:\r\n$$\r\nr^2 = (3 - (-3))^2 + (7 - (-1))^2 = 6^2 + 8^2 = 36+64=100\r\n$$\r\nThe equation is $(x+3)^2 + (y+1)^2 = 100$, which expands to **$x^2 + y^2 + 6x + 2y - 90 = 0$**.\r\n\r\n***\r\n\r\n#### Question 22: Equation of a Circle from Three Points\r\n**Find the equation of the circle which passes through the points P(5, 7), Q(6, 6) and R(2, -2).**\r\n\r\n**Solution:**\r\nLet the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$.\r\nThe squared distance from the center $(h,k)$ to each point is equal.\r\n**From P and Q:**\r\n$$\r\n\\begin{aligned}\r\n(h-5)^2 + (k-7)^2 &= (h-6)^2 + (k-6)^2 \\\\\r\nh^2-10h+25 + k^2-14k+49 &= h^2-12h+36 + k^2-12k+36 \\\\\r\n-10h-14k+74 &= -12h-12k+72 \\\\\r\n2h - 2k + 2 &= 0 \\implies h - k + 1 = 0 \\implies h = k-1 \\quad \\text{(Eq. 1)}\r\n\\end{aligned}\r\n$$\r\n**From Q and R:**\r\n$$\r\n\\begin{aligned}\r\n(h-6)^2 + (k-6)^2 &= (h-2)^2 + (k-(-2))^2 \\\\\r\nh^2-12h+36 + k^2-12k+36 &= h^2-4h+4 + k^2+4k+4 \\\\\r\n-12h-12k+72 &= -4h+4k+8 \\\\\r\n64 &= 8h + 16k \\\\\r\n8 &= h + 2k \\quad \\text{(Eq. 2)}\r\n\\end{aligned}\r\n$$\r\n**Solve for h and k:**\r\nSubstitute Eq. 1 into Eq. 2:\r\n$8 = (k-1) + 2k \\implies 9 = 3k \\implies k=3$.\r\nThen, $h = 3-1=2$. The center is $(h,k)=(2,3)$.\r\n\r\n**Find $r^2$ using point P(5,7):**\r\n$r^2 = (5-2)^2 + (7-3)^2 = 3^2 + 4^2 = 9+16=25$.\r\n\r\n**Write the equation:**\r\n$$(x-2)^2 + (y-3)^2 = 25$$\r\nThe required equation is **$x^2 + y^2 - 4x - 6y - 12 = 0$**.', 1, '2025-07-25 01:11:38', '2025-09-18 07:21:49'),
(145, 14241, '# Grade 10 Optional Mathematics: Scalar (Dot) Product\r\n## Unit: Vector | Chapter: Scalar Product | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nThis guide offers a complete overview of the **Scalar Product** (also known as the Dot Product) of vectors. It starts with a clear breakdown of the essential concepts and formulas, then provides detailed, step-by-step solutions to a wide range of problems, including finding the angle between vectors, determining vector magnitudes, and applying the condition for perpendicularity. 📐\r\n\r\n### Key Learning Areas\r\n1.  **Definition of Scalar Product:** Understanding what the scalar product is and how to calculate it.\r\n2.  **Magnitude of a Vector:** Calculating the length or magnitude of a vector.\r\n3.  **Angle Between Two Vectors:** Using the scalar product formula to find the angle between two vectors.\r\n4.  **Condition for Perpendicularity:** Using the scalar product to prove that two vectors are perpendicular.\r\n5.  **Properties of Unit Vectors:** Understanding the scalar products of unit vectors $\\overrightarrow{i}$ and $\\overrightarrow{j}$.\r\n\r\n---\r\n\r\n## 🔑 Key Concepts & Formulas\r\n\r\nBefore diving into the problems, let\'s review the fundamental concepts.\r\n\r\n### 1. The Scalar (Dot) Product\r\n\r\nThe scalar product of two vectors, $\\overrightarrow{a}$ and $\\overrightarrow{b}$, is a scalar (a number, not a vector) quantity.\r\n\r\n* **Geometric Formula:**\r\n    $$\\overrightarrow{a} \\cdot \\overrightarrow{b} = |\\overrightarrow{a}| |\\overrightarrow{b}| \\cos\\theta$$\r\n    Where:\r\n    * $|\\overrightarrow{a}|$ is the magnitude (length) of vector $\\overrightarrow{a}$.\r\n    * $|\\overrightarrow{b}|$ is the magnitude (length) of vector $\\overrightarrow{b}$.\r\n    * $\\theta$ is the angle between the two vectors.\r\n\r\n\r\n\r\n* **Component Formula:**\r\n    If $\\overrightarrow{a} = x_1\\overrightarrow{i} + y_1\\overrightarrow{j}$ and $\\overrightarrow{b} = x_2\\overrightarrow{i} + y_2\\overrightarrow{j}$, then:\r\n    $$\\overrightarrow{a} \\cdot \\overrightarrow{b} = x_1x_2 + y_1y_2$$\r\n\r\n---\r\n\r\n### 2. Magnitude of a Vector\r\n\r\nThe magnitude of a vector $\\overrightarrow{a} = x\\overrightarrow{i} + y\\overrightarrow{j}$ is its length.\r\n\r\n* **Formula:**\r\n    $$|\\overrightarrow{a}| = \\sqrt{x^2 + y^2}$$\r\n\r\n---\r\n\r\n### 3. Angle Between Two Vectors\r\n\r\nWe can rearrange the geometric formula to solve for the angle $\\theta$.\r\n\r\n* **Formula:**\r\n    $$\\cos\\theta = \\frac{\\overrightarrow{a} \\cdot \\overrightarrow{b}}{|\\overrightarrow{a}| |\\overrightarrow{b}|}$$\r\n\r\n---\r\n\r\n### 4. Condition for Perpendicular Vectors\r\n\r\nTwo non-zero vectors are perpendicular (orthogonal) if the angle between them is 90°. Since $\\cos(90^\\circ) = 0$, the condition is:\r\n\r\n* **Condition:**\r\n    $$\\overrightarrow{a} \\cdot \\overrightarrow{b} = 0$$\r\n\r\n---\r\n\r\n### 5. Properties of Unit Vectors $\\overrightarrow{i}$ and $\\overrightarrow{j}$\r\n\r\n* The scalar product of a unit vector with itself is 1.\r\n    $$\\overrightarrow{i} \\cdot \\overrightarrow{i} = 1 \\quad \\text{and} \\quad \\overrightarrow{j} \\cdot \\overrightarrow{j} = 1$$\r\n* The scalar product of two different standard unit vectors is 0 (as they are perpendicular).\r\n    $$\\overrightarrow{i} \\cdot \\overrightarrow{j} = 0 \\quad \\text{and} \\quad \\overrightarrow{j} \\cdot \\overrightarrow{i} = 0$$\r\n* A vector dotted with itself equals the square of its magnitude:\r\n    $$\\overrightarrow{a} \\cdot \\overrightarrow{a} = |\\overrightarrow{a}|^2$$\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Finding Magnitude of a Vector\r\n**If $\\overrightarrow{p} \\cdot \\overrightarrow{q} = 6$, $|\\overrightarrow{q}| = 2\\sqrt{3}$ and the angle between $\\overrightarrow{p}$ and $\\overrightarrow{q}$ is 30°, find the length of $\\overrightarrow{p}$.**\r\n\r\n**Solution:**\r\nGiven: $\\overrightarrow{p} \\cdot \\overrightarrow{q} = 6$, $|\\overrightarrow{q}| = 2\\sqrt{3}$, and $\\theta = 30^\\circ$.\r\nWe know the scalar product formula:\r\n$$ \\overrightarrow{p} \\cdot \\overrightarrow{q} = |\\overrightarrow{p}| |\\overrightarrow{q}| \\cos\\theta $$\r\nSubstituting the given values:\r\n$$\r\n\\begin{aligned}\r\n6 &= |\\overrightarrow{p}| \\cdot 2\\sqrt{3} \\cdot \\cos(30^\\circ) \\\\\r\n6 &= |\\overrightarrow{p}| \\cdot 2\\sqrt{3} \\cdot \\frac{\\sqrt{3}}{2} \\\\\r\n6 &= |\\overrightarrow{p}| \\cdot 3 \\\\\r\n|\\overrightarrow{p}| &= \\frac{6}{3} = 2\r\n\\end{aligned}\r\n$$\r\nThe length of $\\overrightarrow{p}$ is **2 units**.\r\n\r\n***\r\n\r\n#### Question 2: Finding the Angle Between Two Vectors\r\n**If $|\\overrightarrow{a}| = 4$, $|\\overrightarrow{b}| = 5$ and $\\overrightarrow{a} \\cdot \\overrightarrow{b} = 10$, then find the angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$.**\r\n\r\n**Solution:**\r\nUsing the formula for the angle between two vectors:\r\n$$\r\n\\begin{aligned}\r\n\\cos\\theta &= \\frac{\\overrightarrow{a} \\cdot \\overrightarrow{b}}{|\\overrightarrow{a}| |\\overrightarrow{b}|} \\\\\r\n\\cos\\theta &= \\frac{10}{4 \\cdot 5} \\\\\r\n\\cos\\theta &= \\frac{10}{20} = \\frac{1}{2}\r\n\\end{aligned}\r\n$$\r\nSince $\\cos(60^\\circ) = \\frac{1}{2}$, the angle is **$\\theta = 60^\\circ$**.\r\n\r\n***\r\n\r\n#### Question 3: Finding the Scalar Product\r\n**If $|\\overrightarrow{a}| = 3\\sqrt{3}$, $|\\overrightarrow{b}| = 4$ and angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$ is 60°, then find the value of $\\overrightarrow{a} \\cdot \\overrightarrow{b}$.**\r\n\r\n**Solution:**\r\nUsing the scalar product formula:\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{a} \\cdot \\overrightarrow{b} &= |\\overrightarrow{a}| |\\overrightarrow{b}| \\cos\\theta \\\\\r\n&= 3\\sqrt{3} \\cdot 4 \\cdot \\cos(60^\\circ) \\\\\r\n&= 12\\sqrt{3} \\cdot \\frac{1}{2} \\\\\r\n&= 6\\sqrt{3}\r\n\\end{aligned}\r\n$$\r\nThe value of $\\overrightarrow{a} \\cdot \\overrightarrow{b}$ is **$6\\sqrt{3}$**.\r\n\r\n***\r\n\r\n#### Question 4: Finding the Scalar Product\r\n**If $|\\overrightarrow{p}| = 6$, $|\\overrightarrow{q}| = 6$ and the angle between them is 60°, find the value of $\\overrightarrow{p} \\cdot \\overrightarrow{q}$.**\r\n\r\n**Solution:**\r\nUsing the scalar product formula:\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{p} \\cdot \\overrightarrow{q} &= |\\overrightarrow{p}| |\\overrightarrow{q}| \\cos\\theta \\\\\r\n&= 6 \\cdot 6 \\cdot \\cos(60^\\circ) \\\\\r\n&= 36 \\cdot \\frac{1}{2} \\\\\r\n&= 18\r\n\\end{aligned}\r\n$$\r\nThe value is **18**.\r\n\r\n***\r\n\r\n#### Question 5: Finding an Unknown in a Perpendicular Vector\r\n**If the vectors $10\\overrightarrow{i} - 7\\overrightarrow{j}$ and $a\\overrightarrow{i} + 10\\overrightarrow{j}$ are perpendicular to each other, find the value of a.**\r\n\r\n**Solution:**\r\nLet $\\overrightarrow{p} = 10\\overrightarrow{i} - 7\\overrightarrow{j}$ and $\\overrightarrow{q} = a\\overrightarrow{i} + 10\\overrightarrow{j}$.\r\nFor two vectors to be perpendicular, their scalar product must be zero.\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{p} \\cdot \\overrightarrow{q} &= 0 \\\\\r\n(10\\overrightarrow{i} - 7\\overrightarrow{j}) \\cdot (a\\overrightarrow{i} + 10\\overrightarrow{j}) &= 0 \\\\\r\n(10 \\cdot a) + (-7 \\cdot 10) &= 0 \\\\\r\n10a - 70 &= 0 \\\\\r\n10a &= 70 \\\\\r\na &= 7\r\n\\end{aligned}\r\n$$\r\nThe value of **$a=7$**.\r\n\r\n***\r\n\r\n#### Question 6: Finding an Unknown in a Perpendicular Vector\r\n**If $\\overrightarrow{a} = -5\\overrightarrow{i} + 3\\overrightarrow{j}$ and $\\overrightarrow{b} = p\\overrightarrow{i} + (p+2)\\overrightarrow{j}$ are perpendicular to each other, find the value of p.**\r\n\r\n**Solution:**\r\nFor perpendicular vectors, $\\overrightarrow{a} \\cdot \\overrightarrow{b} = 0$.\r\n$$\r\n\\begin{aligned}\r\n(-5\\overrightarrow{i} + 3\\overrightarrow{j}) \\cdot (p\\overrightarrow{i} + (p+2)\\overrightarrow{j}) &= 0 \\\\\r\n(-5 \\cdot p) + (3 \\cdot (p+2)) &= 0 \\\\\r\n-5p + 3p + 6 &= 0 \\\\\r\n-2p &= -6 \\\\\r\np &= 3\r\n\\end{aligned}\r\n$$\r\nThe value of **$p=3$**.\r\n\r\n***\r\n\r\n#### Question 7: Finding Magnitude of a Vector\r\n**If $|\\overrightarrow{b}| = 6$, $\\overrightarrow{a} \\cdot \\overrightarrow{b} = 12$ and the angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$ is 60°, find the value of $|\\overrightarrow{a}|$.**\r\n\r\n**Solution:**\r\nUsing the scalar product formula:\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{a} \\cdot \\overrightarrow{b} &= |\\overrightarrow{a}| |\\overrightarrow{b}| \\cos\\theta \\\\\r\n12 &= |\\overrightarrow{a}| \\cdot 6 \\cdot \\cos(60^\\circ) \\\\\r\n12 &= |\\overrightarrow{a}| \\cdot 6 \\cdot \\frac{1}{2} \\\\\r\n12 &= |\\overrightarrow{a}| \\cdot 3 \\\\\r\n|\\overrightarrow{a}| &= 4\r\n\\end{aligned}\r\n$$\r\nThe magnitude of $\\overrightarrow{a}$ is **4 units**.\r\n\r\n***\r\n\r\n#### Question 8: Finding the Angle Between Two Vectors\r\n**Find the angle between the unit vector $\\overrightarrow{i}$ and the vector $\\overrightarrow{a} = \\sqrt{3}\\overrightarrow{i} + \\overrightarrow{j}$.**\r\n\r\n**Solution:**\r\nLet the first vector be $\\overrightarrow{u} = \\overrightarrow{i} = 1\\overrightarrow{i} + 0\\overrightarrow{j}$.\r\nLet the second vector be $\\overrightarrow{a} = \\sqrt{3}\\overrightarrow{i} + 1\\overrightarrow{j}$.\r\n\r\n**Step 1: Find the scalar product.**\r\n$$ \\overrightarrow{u} \\cdot \\overrightarrow{a} = (1)(\\sqrt{3}) + (0)(1) = \\sqrt{3} $$\r\n**Step 2: Find the magnitudes.**\r\n$$ |\\overrightarrow{u}| = |\\overrightarrow{i}| = 1 $$\r\n$$ |\\overrightarrow{a}| = \\sqrt{(\\sqrt{3})^2 + 1^2} = \\sqrt{3+1} = \\sqrt{4} = 2 $$\r\n**Step 3: Use the angle formula.**\r\n$$\r\n\\begin{aligned}\r\n\\cos\\theta &= \\frac{\\overrightarrow{u} \\cdot \\overrightarrow{a}}{|\\overrightarrow{u}| |\\overrightarrow{a}|} = \\frac{\\sqrt{3}}{1 \\cdot 2} = \\frac{\\sqrt{3}}{2}\r\n\\end{aligned}\r\n$$\r\nSince $\\cos(30^\\circ) = \\frac{\\sqrt{3}}{2}$, the angle is **$\\theta = 30^\\circ$**.\r\n\r\n---\r\n\r\n### Long Questions\r\n\r\n***\r\n\r\n#### Question 9: Angle Between Unit Vectors\r\n**If $\\overrightarrow{a} + 2\\overrightarrow{b}$ and $5\\overrightarrow{a} - 4\\overrightarrow{b}$ are perpendicular to each other and $\\overrightarrow{a}$ and $\\overrightarrow{b}$ are unit vectors, find the angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$.**\r\n\r\n**Solution:**\r\nSince the two vectors are perpendicular, their scalar product is zero.\r\n$$ (\\overrightarrow{a} + 2\\overrightarrow{b}) \\cdot (5\\overrightarrow{a} - 4\\overrightarrow{b}) = 0 $$\r\nExpanding the dot product:\r\n$$\r\n\\begin{aligned}\r\n5(\\overrightarrow{a} \\cdot \\overrightarrow{a}) - 4(\\overrightarrow{a} \\cdot \\overrightarrow{b}) + 10(\\overrightarrow{b} \\cdot \\overrightarrow{a}) - 8(\\overrightarrow{b} \\cdot \\overrightarrow{b}) &= 0 \\\\\r\n5|\\overrightarrow{a}|^2 + 6(\\overrightarrow{a} \\cdot \\overrightarrow{b}) - 8|\\overrightarrow{b}|^2 &= 0\r\n\\end{aligned}\r\n$$\r\nSince $\\overrightarrow{a}$ and $\\overrightarrow{b}$ are unit vectors, $|\\overrightarrow{a}|=1$ and $|\\overrightarrow{b}|=1$.\r\n$$\r\n\\begin{aligned}\r\n5(1)^2 + 6(\\overrightarrow{a} \\cdot \\overrightarrow{b}) - 8(1)^2 &= 0 \\\\\r\n5 + 6(\\overrightarrow{a} \\cdot \\overrightarrow{b}) - 8 &= 0 \\\\\r\n6(\\overrightarrow{a} \\cdot \\overrightarrow{b}) &= 3 \\\\\r\n\\overrightarrow{a} \\cdot \\overrightarrow{b} &= \\frac{1}{2}\r\n\\end{aligned}\r\n$$\r\nNow, use the angle formula:\r\n$$\r\n\\cos\\theta = \\frac{\\overrightarrow{a} \\cdot \\overrightarrow{b}}{|\\overrightarrow{a}| |\\overrightarrow{b}|} = \\frac{1/2}{1 \\cdot 1} = \\frac{1}{2}\r\n$$\r\nThus, the angle is **$\\theta = 60^\\circ$**.\r\n\r\n***\r\n\r\n#### Question 10: Finding the Angle Between Two Vectors\r\n**If $\\overrightarrow{p} + \\overrightarrow{q} + \\overrightarrow{r} = 0$, $|\\overrightarrow{p}| = 6$, $|\\overrightarrow{q}| = 7$ and $|\\overrightarrow{r}| = \\sqrt{127}$, find the angle between $\\overrightarrow{p}$ and $\\overrightarrow{q}$.**\r\n\r\n**Solution:**\r\nFrom the given condition:\r\n$$ \\overrightarrow{p} + \\overrightarrow{q} = -\\overrightarrow{r} $$\r\nSquaring both sides (taking the dot product with itself):\r\n$$\r\n\\begin{aligned}\r\n(\\overrightarrow{p} + \\overrightarrow{q}) \\cdot (\\overrightarrow{p} + \\overrightarrow{q}) &= (-\\overrightarrow{r}) \\cdot (-\\overrightarrow{r}) \\\\\r\n|\\overrightarrow{p}|^2 + 2(\\overrightarrow{p} \\cdot \\overrightarrow{q}) + |\\overrightarrow{q}|^2 &= |\\overrightarrow{r}|^2\r\n\\end{aligned}\r\n$$\r\nSubstitute the given magnitudes:\r\n$$\r\n\\begin{aligned}\r\n6^2 + 2(\\overrightarrow{p} \\cdot \\overrightarrow{q}) + 7^2 &= (\\sqrt{127})^2 \\\\\r\n36 + 2(\\overrightarrow{p} \\cdot \\overrightarrow{q}) + 49 &= 127 \\\\\r\n85 + 2(\\overrightarrow{p} \\cdot \\overrightarrow{q}) &= 127 \\\\\r\n2(\\overrightarrow{p} \\cdot \\overrightarrow{q}) &= 42 \\\\\r\n\\overrightarrow{p} \\cdot \\overrightarrow{q} &= 21\r\n\\end{aligned}\r\n$$\r\nNow, find the angle $\\theta$:\r\n$$\r\n\\cos\\theta = \\frac{\\overrightarrow{p} \\cdot \\overrightarrow{q}}{|\\overrightarrow{p}| |\\overrightarrow{q}|} = \\frac{21}{6 \\cdot 7} = \\frac{21}{42} = \\frac{1}{2}\r\n$$\r\nThus, the angle is **$\\theta = 60^\\circ$**.', 1, '2025-07-25 01:11:38', '2025-09-18 09:23:05');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(148, 14240, '# Grade 10 Optional Mathematics: Vector Geometry\r\n## Unit: Vector | Chapter: Geometric Proofs & Section Formula | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Key Concepts & Formulas\r\n\r\nThis section covers the fundamental principles of vector geometry, including addition laws, position vectors, and formulas for dividing a line segment. These concepts are the building blocks for the proofs and problems that follow.\r\n\r\n### 1. Vector Addition\r\n\r\n* **Triangle Law of Vector Addition:** If two vectors are represented by two sides of a triangle in sequence, then the third side of the triangle, taken in the opposite order, represents the sum (resultant) of the two vectors. For $\\triangle ABC$, $\\overrightarrow{AB} + \\overrightarrow{BC} = \\overrightarrow{AC}$.\r\n* **Parallelogram Law of Vector Addition:** If two vectors acting simultaneously from a point are represented by the adjacent sides of a parallelogram, their resultant is represented by the diagonal of the parallelogram drawn from the same point. For parallelogram OABC with origin O, $\\overrightarrow{OA} + \\overrightarrow{OC} = \\overrightarrow{OB}$.\r\n\r\n### 2. Position Vector\r\n\r\nThe position vector of a point A, denoted as $\\overrightarrow{OA}$ or $\\overrightarrow{a}$, is the vector from a fixed origin (O) to that point. It\'s used to define the location of a point in space. The vector from point A to point B can be expressed using position vectors as $\\overrightarrow{AB} = \\overrightarrow{OB} - \\overrightarrow{OA}$.\r\n\r\n### 3. Section Formula\r\n\r\nThis formula finds the position vector of a point that divides the line segment joining two points in a specific ratio.\r\n\r\n* **Internal Division:** If a point M divides the line segment joining P (position vector $\\overrightarrow{p}$) and Q (position vector $\\overrightarrow{q}$) internally in the ratio $m_1:m_2$, the position vector of M is:\r\n    $$ \\overrightarrow{OM} = \\frac{m_1\\overrightarrow{q} + m_2\\overrightarrow{p}}{m_1 + m_2} $$\r\n* **External Division:** If M divides the line joining A ($\\overrightarrow{a}$) and B ($\\overrightarrow{b}$) externally in the ratio $m:n$, its position vector is:\r\n    $$ \\overrightarrow{OM} = \\frac{m\\overrightarrow{b} - n\\overrightarrow{a}}{m - n} $$\r\n\r\n### 4. Mid-Point Formula\r\n\r\nThis is a special case of the internal section formula where the ratio is 1:1.\r\n\r\n* **Formula:** The position vector of the mid-point M of the line segment joining points A and B is:\r\n    $$ \\overrightarrow{OM} = \\frac{\\overrightarrow{OA} + \\overrightarrow{OB}}{2} $$\r\n\r\n### 5. Centroid Formula\r\n\r\nThe centroid is the point where the three medians of a triangle intersect.\r\n\r\n* **Formula:** The position vector of the centroid G of a triangle ABC is the average of the position vectors of its vertices:\r\n    $$ \\overrightarrow{OG} = \\frac{\\overrightarrow{OA} + \\overrightarrow{OB} + \\overrightarrow{OC}}{3} $$\r\n\r\n### 6. Scalar (Dot) Product and Perpendicularity\r\n\r\n* **Condition for Perpendicularity:** Two non-zero vectors $\\overrightarrow{a}$ and $\\overrightarrow{b}$ are perpendicular (orthogonal) if and only if their scalar product is zero.\r\n    $$ \\overrightarrow{a} \\cdot \\overrightarrow{b} = 0 $$\r\n\r\n---\r\n\r\n## 📐 Vector Proofs of Geometric Theorems\r\n\r\nThis section provides step-by-step proofs for key geometric theorems using vector methods.\r\n\r\n***\r\n\r\n### 1. Angle in a Semi-circle is a Right Angle (Method 1)\r\n\r\n**Prove by vector method that the angle at the circumference of a semi-circle is a right angle.**\r\n\r\n**Given:**\r\n* ACB is a semi-circle with center O.\r\n* AB is the diameter.\r\n\r\n\r\n\r\n**To Prove:** $\\angle ACB = 90^\\circ$\r\n\r\n**Construction:** Join O to C.\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\overrightarrow{AC} = \\overrightarrow{AO} + \\overrightarrow{OC}$ | By Triangle Law of Vector Addition. |\r\n| ii) $\\overrightarrow{BC} = \\overrightarrow{BO} + \\overrightarrow{OC}$ | By Triangle Law of Vector Addition. |\r\n| iii) $\\overrightarrow{BC} = -\\overrightarrow{OA} + \\overrightarrow{OC}$ | $\\overrightarrow{BO} = -\\overrightarrow{OA}$ since O is the center and OA and OB are radii in opposite directions. |\r\n| iv) $\\overrightarrow{AC} \\cdot \\overrightarrow{BC} = (\\overrightarrow{OC} + \\overrightarrow{OA}) \\cdot (\\overrightarrow{OC} - \\overrightarrow{OA})$ | Taking the dot product of the rearranged vectors from (i) and (iii). |\r\n| v) $\\overrightarrow{AC} \\cdot \\overrightarrow{BC} = |\\overrightarrow{OC}|^2 - |\\overrightarrow{OA}|^2$ | Using the algebraic formula $(a+b)(a-b) = a^2 - b^2$. |\r\n| vi) $\\overrightarrow{AC} \\cdot \\overrightarrow{BC} = |\\overrightarrow{OC}|^2 - |\\overrightarrow{OC}|^2 = 0$ | $|\\overrightarrow{OA}| = |\\overrightarrow{OC}|$ (radii of the same circle). |\r\n| vii) $\\overrightarrow{AC} \\perp \\overrightarrow{BC}$ | The scalar (dot) product of two non-zero vectors is zero. |\r\n| viii) $\\angle ACB = 90^\\circ$ | From statement (vii), the angle between the vectors is 90°. |\r\n\r\n**Hence, the angle in a semi-circle is a right angle.**\r\n\r\n***\r\n\r\n### 2. Angle in a Semi-circle is a Right Angle (Method 2)\r\n\r\n**PQ is a diameter of a circle with the centre X and M is a point on the circumference of the circle, prove that $\\angle PMQ = 90^\\circ$ by vector method.**\r\n\r\n**Given:**\r\n* PMQ is a semi-circle with center X.\r\n* PQ is the diameter.\r\n\r\n\r\n\r\n**To Prove:** $\\angle PMQ = 90^\\circ$\r\n\r\n**Construction:** Join X to M.\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\overrightarrow{QM} = \\overrightarrow{QX} + \\overrightarrow{XM} = \\overrightarrow{XM} - \\overrightarrow{XQ}$ | By Triangle Law of Vector Addition. |\r\n| ii) $\\overrightarrow{PM} = \\overrightarrow{PX} + \\overrightarrow{XM} = \\overrightarrow{XM} + \\overrightarrow{PX}$ | By Triangle Law of Vector Addition. |\r\n| iii) $\\overrightarrow{PM} \\cdot \\overrightarrow{QM} = (\\overrightarrow{XM} + \\overrightarrow{PX}) \\cdot (\\overrightarrow{XM} - \\overrightarrow{XQ})$ | Taking the dot product of (i) and (ii). |\r\n| iv) $\\overrightarrow{PM} \\cdot \\overrightarrow{QM} = (\\overrightarrow{XM} + \\overrightarrow{PX}) \\cdot (\\overrightarrow{XM} - \\overrightarrow{PX})$ | Radii are equal, so $|\\overrightarrow{XQ}| = |\\overrightarrow{PX}|$. Since they are in opposite directions, $\\overrightarrow{XQ} = -\\overrightarrow{PX}$. |\r\n| v) $\\overrightarrow{PM} \\cdot \\overrightarrow{QM} = |\\overrightarrow{XM}|^2 - |\\overrightarrow{PX}|^2$ | Using the formula $(a+b)(a-b) = a^2-b^2$. |\r\n| vi) $\\overrightarrow{PM} \\cdot \\overrightarrow{QM} = |\\overrightarrow{XM}|^2 - |\\overrightarrow{XM}|^2 = 0$ | $|\\overrightarrow{PX}| = |\\overrightarrow{XM}|$ (radii of the same circle). |\r\n| vii) $PM \\perp QM$, so $\\angle PMQ = 90^\\circ$ | The scalar product of the vectors is zero. |\r\n\r\n**Hence, the angle in a semi-circle is a right angle.**\r\n\r\n***\r\n\r\n<div id=\"isosceles\"></div>\r\n### 3. Median of an Isosceles Triangle\r\n\r\n**Prove by vector method that the median from the vertex of an isosceles triangle is perpendicular to the base.**\r\n\r\n**Given:**\r\n* In $\\triangle PQR$, $PQ = PR$.\r\n* S is the mid-point of the base QR.\r\n\r\n\r\n\r\n**To Prove:** $PS \\perp QR$\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\overrightarrow{PS} = \\frac{1}{2}(\\overrightarrow{PQ} + \\overrightarrow{PR})$ | S is the mid-point of QR, so $\\overrightarrow{PS}$ is the median. |\r\n| ii) $\\overrightarrow{QR} = \\overrightarrow{PR} - \\overrightarrow{PQ}$ | By Triangle Law of Vector Addition ($\\overrightarrow{PQ} + \\overrightarrow{QR} = \\overrightarrow{PR}$). |\r\n| iii) $\\overrightarrow{PS} \\cdot \\overrightarrow{QR} = \\frac{1}{2}(\\overrightarrow{PQ} + \\overrightarrow{PR}) \\cdot (\\overrightarrow{PR} - \\overrightarrow{PQ})$ | Taking the dot product of statements (i) and (ii). |\r\n| iv) $\\overrightarrow{PS} \\cdot \\overrightarrow{QR} = \\frac{1}{2}(|\\overrightarrow{PR}|^2 - |\\overrightarrow{PQ}|^2)$ | Using the formula $(a+b)(a-b) = a^2-b^2$. |\r\n| v) $\\overrightarrow{PS} \\cdot \\overrightarrow{QR} = \\frac{1}{2}(|\\overrightarrow{PQ}|^2 - |\\overrightarrow{PQ}|^2) = 0$ | Given $PQ=PR$, so $|\\overrightarrow{PQ}| = |\\overrightarrow{PR}|$. |\r\n| vi) $PS \\perp QR$ | Since the scalar product is zero. |\r\n\r\n**Hence, the median to the base of an isosceles triangle is perpendicular to the base.**\r\n\r\n***\r\n\r\n<div id=\"rightAngled\"></div>\r\n### 4. Median to the Hypotenuse of a Right-Angled Triangle\r\n\r\n**In a right-angled triangle, prove by vector method that the length of the median to the hypotenuse is half the length of the hypotenuse.**\r\n\r\n**Given:**\r\n* In right-angled triangle PQR, $\\angle RPQ = 90^\\circ$.\r\n* S is the mid-point of the hypotenuse RQ.\r\n\r\n\r\n\r\n**To Prove:** $PS = QS = RS$ (which implies $PS = \\frac{1}{2}RQ$)\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\angle RPQ = 90^\\circ$ | Given. |\r\n| ii) $\\overrightarrow{PR} \\cdot \\overrightarrow{PQ} = 0$ | The dot product of perpendicular vectors is zero. |\r\n| iii) $\\overrightarrow{PS} = \\frac{1}{2}(\\overrightarrow{PR} + \\overrightarrow{PQ})$ | By the mid-point theorem, as S is the midpoint of RQ. |\r\n| iv) $(\\overrightarrow{PS})^2 = \\frac{1}{4}(\\overrightarrow{PR} + \\overrightarrow{PQ})^2$ | Squaring both sides of statement (iii). |\r\n| v) $PS^2 = \\frac{1}{4}(|\\overrightarrow{PR}|^2 + 2\\overrightarrow{PR}\\cdot\\overrightarrow{PQ} + |\\overrightarrow{PQ}|^2)$ | Expanding the square. |\r\n| vi) $PS^2 = \\frac{1}{4}(PR^2 + 0 + PQ^2)$ | From statement (ii), $\\overrightarrow{PR} \\cdot \\overrightarrow{PQ} = 0$. |\r\n| vii) $PS^2 = \\frac{1}{4}(RQ^2)$ | In $\\triangle PQR$, by Pythagorean theorem, $PR^2+PQ^2=RQ^2$. |\r\n| viii) $PS = \\frac{1}{2}RQ$ | Taking the square root of both sides. |\r\n| ix) $PS = QS = RS$ | S is the mid-point of RQ, so $QS=RS=\\frac{1}{2}RQ$. |\r\n\r\n**Hence, the median to the hypotenuse is half the length of the hypotenuse.**\r\n\r\n***\r\n\r\n<div id=\"rhombus-perp\"></div>\r\n### 5. Diagonals of a Rhombus\r\n\r\n**Prove by vector method that the diagonals of a rhombus are perpendicular.**\r\n\r\n**Given:**\r\n* PQRS is a rhombus where $|\\overrightarrow{PQ}| = |\\overrightarrow{QR}| = |\\overrightarrow{RS}| = |\\overrightarrow{SP}|$.\r\n\r\n\r\n\r\n**To Prove:** $PR \\perp QS$\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\overrightarrow{PR} = \\overrightarrow{PQ} + \\overrightarrow{PS}$ | By Parallelogram Law of Vector Addition. |\r\n| ii) $\\overrightarrow{QS} = \\overrightarrow{PS} - \\overrightarrow{PQ}$ | By Triangle Law of Vector Addition. |\r\n| iii) $\\overrightarrow{PR} \\cdot \\overrightarrow{QS} = (\\overrightarrow{PS} + \\overrightarrow{PQ}) \\cdot (\\overrightarrow{PS} - \\overrightarrow{PQ})$ | Taking the dot product of (i) and (ii). |\r\n| iv) $\\overrightarrow{PR} \\cdot \\overrightarrow{QS} = |\\overrightarrow{PS}|^2 - |\\overrightarrow{PQ}|^2$ | Using the formula $(a+b)(a-b) = a^2-b^2$. |\r\n| v) $\\overrightarrow{PR} \\cdot \\overrightarrow{QS} = |\\overrightarrow{PQ}|^2 - |\\overrightarrow{PQ}|^2 = 0$ | All sides of a rhombus are equal, so $|\\overrightarrow{PS}| = |\\overrightarrow{PQ}|$. |\r\n| vi) $PR \\perp QS$ | Since the scalar product is zero. |\r\n\r\n**Hence, the diagonals of a rhombus are perpendicular to each other.**\r\n\r\n***\r\n\r\n<div id=\'rectangle\'></div>\r\n### 6. Diagonals of a Rectangle\r\n\r\n**Prove by vector method that the diagonals of a rectangle are equal.**\r\n\r\n**Given:**\r\n* PQRS is a rectangle.\r\n* PR and QS are its diagonals.\r\n\r\n\r\n\r\n**To Prove:** $PR = QS$\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\overrightarrow{PR} = \\overrightarrow{PQ} + \\overrightarrow{QR}$ | By Triangle Law of Vector Addition. |\r\n| ii) $PR^2 = (\\overrightarrow{PQ} + \\overrightarrow{QR})^2 = |\\overrightarrow{PQ}|^2 + 2\\overrightarrow{PQ}\\cdot\\overrightarrow{QR} + |\\overrightarrow{QR}|^2$ | Squaring both sides of (i). |\r\n| iii) $PR^2 = PQ^2 + QR^2$ | Since adjacent sides of a rectangle are perpendicular, $\\overrightarrow{PQ} \\cdot \\overrightarrow{QR} = 0$. |\r\n| iv) $\\overrightarrow{QS} = \\overrightarrow{QR} + \\overrightarrow{RS} = \\overrightarrow{QR} - \\overrightarrow{PQ}$ | By Triangle Law and since $\\overrightarrow{RS} = -\\overrightarrow{PQ}$. |\r\n| v) $QS^2 = (\\overrightarrow{QR} - \\overrightarrow{PQ})^2 = |\\overrightarrow{QR}|^2 - 2\\overrightarrow{QR}\\cdot\\overrightarrow{PQ} + |\\overrightarrow{PQ}|^2$ | Squaring both sides of (iv). |\r\n| vi) $QS^2 = QR^2 + PQ^2$ | Since $\\overrightarrow{QR} \\cdot \\overrightarrow{PQ} = 0$. |\r\n| vii) $PR^2 = QS^2$ | From statements (iii) and (vi). |\r\n| viii) $PR = QS$ | Taking the square root. |\r\n\r\n**Hence, the diagonals of a rectangle are equal.**\r\n\r\n***\r\n\r\n### 7. Diagonals of a Parallelogram\r\n\r\n**Prove by vector method that the diagonals of a parallelogram bisect each other.**\r\n\r\n**Given:**\r\n* OABC is a parallelogram.\r\n* M and N are the midpoints of diagonals AC and OB respectively.\r\n\r\n\r\n\r\n**To Prove:** The diagonals bisect each other (i.e., M and N are the same point, $\\overrightarrow{OM} = \\overrightarrow{ON}$).\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\overrightarrow{OM} = \\frac{1}{2}(\\overrightarrow{OA} + \\overrightarrow{OC})$ | M is the midpoint of diagonal AC. |\r\n| ii) $\\overrightarrow{ON} = \\frac{1}{2}(\\overrightarrow{OB})$ | N is the midpoint of diagonal OB. |\r\n| iii) $\\overrightarrow{OB} = \\overrightarrow{OA} + \\overrightarrow{OC}$ | By Parallelogram Law of Vector Addition. |\r\n| iv) $\\overrightarrow{ON} = \\frac{1}{2}(\\overrightarrow{OA} + \\overrightarrow{OC})$ | Substituting statement (iii) into (ii). |\r\n| v) $\\overrightarrow{OM} = \\overrightarrow{ON}$ | From statements (i) and (iv). |\r\n\r\n**Since the position vectors of the mid-points of both diagonals are the same, the mid-points coincide. Hence, the diagonals of a parallelogram bisect each other.**\r\n\r\n***\r\n\r\n### 8. Mid-Point Theorem for Quadrilaterals\r\n\r\n**Prove by vector method that the quadrilateral formed by joining the mid-points of the sides of any quadrilateral is a parallelogram.**\r\n\r\n**Given:**\r\n* ABCD is a quadrilateral.\r\n* P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively.\r\n\r\n\r\n\r\n**To Prove:** PQRS is a parallelogram.\r\n\r\n**Construction:** Join diagonal AC.\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) In $\\triangle ABC$, $\\overrightarrow{PQ} = \\frac{1}{2}\\overrightarrow{AC}$ | The line segment joining the mid-points of two sides of a triangle is parallel to and half the length of the third side. |\r\n| ii) In $\\triangle ADC$, $\\overrightarrow{SR} = \\frac{1}{2}\\overrightarrow{AC}$ | Same reason as above. |\r\n| iii) $\\overrightarrow{PQ} = \\overrightarrow{SR}$ | From statements (i) and (ii). This implies $PQ \\parallel SR$ and $PQ = SR$. |\r\n| iv) PQRS is a parallelogram. | A quadrilateral is a parallelogram if one pair of opposite sides is equal and parallel. |\r\n\r\n**Hence, the quadrilateral formed by joining the mid-points is a parallelogram.**\r\n\r\n***\r\n\r\n### 9. Section Formula Proofs\r\n\r\n#### a) Internal Division\r\n\r\n**If the position vectors of points P and Q are $\\overrightarrow{p}$ and $\\overrightarrow{q}$ respectively, and point M divides the line segment PQ internally in the ratio $m_1:m_2$, prove that the position vector of M is $\\overrightarrow{OM} = \\frac{m_1\\overrightarrow{q} + m_2\\overrightarrow{p}}{m_1 + m_2}$.**\r\n\r\n\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\frac{PM}{MQ} = \\frac{m_1}{m_2}$ | Given ratio. |\r\n| ii) $m_2 \\overrightarrow{PM} = m_1 \\overrightarrow{MQ}$ | Cross-multiplication. Since P, M, Q are collinear and M is internal, the vectors are in the same direction. |\r\n| iii) $m_2(\\overrightarrow{OM} - \\overrightarrow{OP}) = m_1(\\overrightarrow{OQ} - \\overrightarrow{OM})$ | By Triangle Law of Vector Addition. |\r\n| iv) $m_2\\overrightarrow{OM} - m_2\\overrightarrow{p} = m_1\\overrightarrow{q} - m_1\\overrightarrow{OM}$ | Substituting given position vectors. |\r\n| v) $m_1\\overrightarrow{OM} + m_2\\overrightarrow{OM} = m_1\\overrightarrow{q} + m_2\\overrightarrow{p}$ | Rearranging terms. |\r\n| vi) $\\overrightarrow{OM}(m_1 + m_2) = m_1\\overrightarrow{q} + m_2\\overrightarrow{p}$ | Taking $\\overrightarrow{OM}$ common. |\r\n| vii) $\\overrightarrow{OM} = \\frac{m_1\\overrightarrow{q} + m_2\\overrightarrow{p}}{m_1 + m_2}$ | Dividing by $(m_1+m_2)$. |\r\n\r\n#### b) External Division\r\n\r\n**If point M divides the line segment AB externally in the ratio $m:n$, prove that its position vector is $\\overrightarrow{OM} = \\frac{m\\overrightarrow{b} - n\\overrightarrow{a}}{m - n}$.**\r\n\r\n\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\frac{AM}{BM} = \\frac{m}{n}$ | Given ratio. |\r\n| ii) $n \\overrightarrow{AM} = m \\overrightarrow{BM}$ | Cross-multiplication. Vectors are in the same direction. |\r\n| iii) $n(\\overrightarrow{OM} - \\overrightarrow{OA}) = m(\\overrightarrow{OM} - \\overrightarrow{OB})$ | By Triangle Law of Vector Addition. |\r\n| iv) $n\\overrightarrow{OM} - n\\overrightarrow{a} = m\\overrightarrow{OM} - m\\overrightarrow{b}$ | Substituting position vectors $\\overrightarrow{a}$ and $\\overrightarrow{b}$. |\r\n| v) $m\\overrightarrow{b} - n\\overrightarrow{a} = m\\overrightarrow{OM} - n\\overrightarrow{OM}$ | Rearranging terms. |\r\n| vi) $m\\overrightarrow{b} - n\\overrightarrow{a} = \\overrightarrow{OM}(m - n)$ | Taking $\\overrightarrow{OM}$ common. |\r\n| vii) $\\overrightarrow{OM} = \\frac{m\\overrightarrow{b} - n\\overrightarrow{a}}{m - n}$ | Dividing by $(m-n)$. |\r\n\r\n---\r\n\r\n## 🔢 Numerical Problems & Solutions\r\n\r\nThis section contains worked-out solutions for numerical problems applying the concepts of position vectors and section formulas.\r\n\r\n***\r\n\r\n### Problem 1: Finding the Centroid (from Median)\r\n**In $\\triangle ABC$, D is the midpoint of AC. If the position vectors of B and D are $3\\overrightarrow{i} + 7\\overrightarrow{j}$ and $3\\overrightarrow{i} - 2\\overrightarrow{j}$ respectively, find the position vector of the centroid G.**\r\n\r\n\r\n\r\n**Solution:**\r\nLet O be the origin.\r\nGiven: $\\overrightarrow{OB} = 3\\overrightarrow{i} + 7\\overrightarrow{j}$ and $\\overrightarrow{OD} = 3\\overrightarrow{i} - 2\\overrightarrow{j}$.\r\nSince D is the mid-point of AC: $\\overrightarrow{OD} = \\frac{\\overrightarrow{OA} + \\overrightarrow{OC}}{2} \\implies \\overrightarrow{OA} + \\overrightarrow{OC} = 2\\overrightarrow{OD} = 6\\overrightarrow{i} - 4\\overrightarrow{j}$.\r\nThe position vector of the centroid G is:\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{OG} &= \\frac{\\overrightarrow{OA} + \\overrightarrow{OB} + \\overrightarrow{OC}}{3} = \\frac{(\\overrightarrow{OA} + \\overrightarrow{OC}) + \\overrightarrow{OB}}{3} \\\\\r\n&= \\frac{(6\\overrightarrow{i} - 4\\overrightarrow{j}) + (3\\overrightarrow{i} + 7\\overrightarrow{j})}{3} \\\\\r\n&= \\frac{9\\overrightarrow{i} + 3\\overrightarrow{j}}{3} = 3\\overrightarrow{i} + \\overrightarrow{j}\r\n\\end{aligned}\r\n$$\r\n**The position vector of G is $3\\overrightarrow{i} + \\overrightarrow{j}$.**\r\n\r\n***\r\n\r\n### Problem 2: Finding the Centroid (from Vertices)\r\n**If the position vectors of the vertices A, B, and C of $\\triangle ABC$ are $(3\\overrightarrow{i} + 5\\overrightarrow{j})$, $(5\\overrightarrow{i} - \\overrightarrow{j})$ and $(\\overrightarrow{i} - 8\\overrightarrow{j})$ respectively, find the position vector of the centroid G.**\r\n\r\n**Solution:**\r\nGiven vertices:\r\n* $\\overrightarrow{OA} = 3\\overrightarrow{i} + 5\\overrightarrow{j}$\r\n* $\\overrightarrow{OB} = 5\\overrightarrow{i} - \\overrightarrow{j}$\r\n* $\\overrightarrow{OC} = \\overrightarrow{i} - 8\\overrightarrow{j}$\r\n\r\nUsing the centroid formula:\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{OG} &= \\frac{\\overrightarrow{OA} + \\overrightarrow{OB} + \\overrightarrow{OC}}{3} \\\\\r\n&= \\frac{(3\\overrightarrow{i} + 5\\overrightarrow{j}) + (5\\overrightarrow{i} - \\overrightarrow{j}) + (\\overrightarrow{i} - 8\\overrightarrow{j})}{3} \\\\\r\n&= \\frac{(3+5+1)\\overrightarrow{i} + (5-1-8)\\overrightarrow{j}}{3} \\\\\r\n&= \\frac{9\\overrightarrow{i} - 4\\overrightarrow{j}}{3} \\\\\r\n&= 3\\overrightarrow{i} - \\frac{4}{3}\\overrightarrow{j}\r\n\\end{aligned}\r\n$$\r\n**The position vector of G is $3\\overrightarrow{i} - \\frac{4}{3}\\overrightarrow{j}$.**\r\n\r\n***\r\n\r\n### Problem 3: Finding a Mid-point Vector\r\n**If the position vectors of two points A and B are $\\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix}$ and $\\begin{pmatrix} 4 \\\\ 5 \\end{pmatrix}$ respectively, find the position vector of K, the mid-point of AB.**\r\n\r\n**Solution:**\r\nGiven: $\\overrightarrow{OA} = \\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix}$ and $\\overrightarrow{OB} = \\begin{pmatrix} 4 \\\\ 5 \\end{pmatrix}$.\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{OK} &= \\frac{\\overrightarrow{OA} + \\overrightarrow{OB}}{2} \\\\\r\n&= \\frac{1}{2} \\left[ \\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix} + \\begin{pmatrix} 4 \\\\ 5 \\end{pmatrix} \\right] = \\frac{1}{2} \\begin{pmatrix} 7 \\\\ 9 \\end{pmatrix} = \\begin{pmatrix} 3.5 \\\\ 4.5 \\end{pmatrix}\r\n\\end{aligned}\r\n$$\r\n**The position vector of K is $\\begin{pmatrix} 3.5 \\\\ 4.5 \\end{pmatrix}$.**\r\n\r\n***\r\n\r\n### Problem 4: Finding the Position Vector of P where MP = PN\r\n**The position vectors of M and N are $9\\overrightarrow{i} + 3\\overrightarrow{j}$ and $3\\overrightarrow{i} + 5\\overrightarrow{j}$ respectively. Find the position vector of a point P such that $\\overrightarrow{MP} = \\overrightarrow{PN}$.**\r\n\r\n**Solution:**\r\nThe condition $\\overrightarrow{MP} = \\overrightarrow{PN}$ means that P is the mid-point of the line segment MN.\r\nGiven:\r\n* $\\overrightarrow{OM} = 9\\overrightarrow{i} + 3\\overrightarrow{j}$\r\n* $\\overrightarrow{ON} = 3\\overrightarrow{i} + 5\\overrightarrow{j}$\r\n\r\nUsing the mid-point formula for P:\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{OP} &= \\frac{\\overrightarrow{OM} + \\overrightarrow{ON}}{2} \\\\\r\n&= \\frac{(9\\overrightarrow{i} + 3\\overrightarrow{j}) + (3\\overrightarrow{i} + 5\\overrightarrow{j})}{2} \\\\\r\n&= \\frac{12\\overrightarrow{i} + 8\\overrightarrow{j}}{2} \\\\\r\n&= 6\\overrightarrow{i} + 4\\overrightarrow{j}\r\n\\end{aligned}\r\n$$\r\n**The position vector of P is $6\\overrightarrow{i} + 4\\overrightarrow{j}$.**\r\n\r\n***\r\n\r\n### Problem 5: Finding a Vector from a Collinear Relationship\r\n**In the given figure, $\\overrightarrow{OA} = \\overrightarrow{a}$ and $\\overrightarrow{OB} = \\overrightarrow{b}$. If $\\overrightarrow{AC} = 3\\overrightarrow{AB}$, find $\\overrightarrow{OC}$.**\r\n\r\n\r\n\r\n**Solution:**\r\nWe need to find $\\overrightarrow{OC}$. Using the triangle law:\r\n$$ \\overrightarrow{OC} = \\overrightarrow{OA} + \\overrightarrow{AC} $$\r\nWe are given $\\overrightarrow{AC} = 3\\overrightarrow{AB}$. First, let\'s find $\\overrightarrow{AB}$:\r\n$$ \\overrightarrow{AB} = \\overrightarrow{OB} - \\overrightarrow{OA} = \\overrightarrow{b} - \\overrightarrow{a} $$\r\nNow substitute this into the expression for $\\overrightarrow{AC}$:\r\n$$ \\overrightarrow{AC} = 3(\\overrightarrow{b} - \\overrightarrow{a}) $$\r\nFinally, find $\\overrightarrow{OC}$:\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{OC} &= \\overrightarrow{OA} + \\overrightarrow{AC} \\\\\r\n&= \\overrightarrow{a} + 3(\\overrightarrow{b} - \\overrightarrow{a}) \\\\\r\n&= \\overrightarrow{a} + 3\\overrightarrow{b} - 3\\overrightarrow{a} \\\\\r\n&= 3\\overrightarrow{b} - 2\\overrightarrow{a}\r\n\\end{aligned}\r\n$$\r\n**The position vector $\\overrightarrow{OC}$ is $3\\overrightarrow{b} - 2\\overrightarrow{a}$.**', 1, '2025-07-25 01:11:38', '2025-09-18 09:23:05'),
(149, 14247, '# Grade 10 Optional Mathematics: Continuity\r\n## Unit: Calculus | Chapter: Continuity and Discontinuity | Complete Solutions\r\n\r\n---\r\n\r\n## 🔑 Key Concepts & Definitions\r\n\r\nBefore attempting the problems, it\'s important to understand the core concepts of limits and continuity.\r\n\r\n### 1. Limits\r\n\r\n* **Left-Hand Limit (LHL):** The value that a function `f(x)` approaches as `x` gets closer to a point `a` from the left side (values less than `a`). It is denoted as:\r\n    $$ \\lim_{x \\to a^-} f(x) $$\r\n\r\n* **Right-Hand Limit (RHL):** The value that a function `f(x)` approaches as `x` gets closer to a point `a` from the right side (values greater than `a`). It is denoted as:\r\n    $$ \\lim_{x \\to a^+} f(x) $$\r\n\r\n* **Existence of a Limit:** For the limit of a function `f(x)` to exist at `x = a`, the left-hand limit must equal the right-hand limit.\r\n    $$ \\lim_{x \\to a^-} f(x) = \\lim_{x \\to a^+} f(x) $$\r\n\r\n### 2. Functional Value\r\n\r\n* **Functional Value:** The value of the function `f(x)` at a specific point `x = a`. It is denoted as `f(a)`. This is the direct output of the function when `x` is exactly equal to `a`.\r\n\r\n### 3. Continuity and Discontinuity\r\n\r\n* **Continuity:** A function `f(x)` is said to be **continuous** at a point `x = a` if three conditions are met:\r\n    1.  The left-hand limit exists: $\\lim_{x \\to a^-} f(x)$\r\n    2.  The right-hand limit exists: $\\lim_{x \\to a^+} f(x)$\r\n    3.  Both limits are equal to the functional value: $\\lim_{x \\to a^-} f(x) = \\lim_{x \\to a^+} f(x) = f(a)$\r\n\r\n    Essentially, the graph of the function has no breaks, jumps, or holes at that point.\r\n\r\n* **Discontinuity:** A function is **discontinuous** at `x = a` if any of the three conditions for continuity are not met.\r\n\r\n### 4. Intervals\r\n\r\n* **Open Interval:** An interval that does not include its endpoints. It is denoted with parentheses, e.g., `(a, b)`.\r\n* **Closed Interval:** An interval that includes its endpoints. It is denoted with square brackets, e.g., `[a, b]`.\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Problem 1: Investigating Continuity\r\n**Given the function f(x) = 3x + 2:**\r\n**a) Find f(2.001), f(2.0001), f(1.999), and f(1.9999).**\r\n**b) Find $\\lim_{x \\to 2^-} f(x)$, $\\lim_{x \\to 2^+} f(x)$, and f(2).**\r\n**c) Is the function continuous at x = 2?**\r\n\r\n**Solution:**\r\nHere, $f(x) = 3x + 2$.\r\n\r\n**a) Calculating values near x = 2:**\r\n* $f(2.001) = 3(2.001) + 2 = 6.003 + 2 = 8.003$\r\n* $f(2.0001) = 3(2.0001) + 2 = 6.0003 + 2 = 8.0003$\r\n* $f(1.999) = 3(1.999) + 2 = 5.997 + 2 = 7.997$\r\n* $f(1.9999) = 3(1.9999) + 2 = 5.9997 + 2 = 7.9997$\r\n    \r\n*As x approaches 2 from both sides, f(x) approaches 8.*\r\n\r\n**b) Finding the limits and functional value:**\r\n\r\n* **Left-Hand Limit (LHL):**\r\n    $$ \\lim_{x \\to 2^-} f(x) = \\lim_{x \\to 2^-} (3x + 2) = 3(2) + 2 = 8 $$\r\n* **Right-Hand Limit (RHL):**\r\n    $$ \\lim_{x \\to 2^+} f(x) = \\lim_{x \\to 2^+} (3x + 2) = 3(2) + 2 = 8 $$\r\n* **Functional Value:**\r\n    $$ f(2) = 3(2) + 2 = 8 $$\r\n\r\n**c) Checking for continuity:**\r\nSince LHL = RHL = f(2) = 8, all three conditions for continuity are met.\r\n**Yes, the function is continuous at x = 2.**\r\n\r\n***\r\n\r\n### Problem 2: Continuity of a Rational Function\r\n**Given the function $f(x) = \\frac{x^2 - 9}{x - 3}$:**\r\n**a) Find f(2.999) and f(3.001).**\r\n**b) Are f(2.999) and f(3.001) approximately equal?**\r\n**c) Discuss the continuity of f(x) at x = 3.**\r\n\r\n**Solution:**\r\nHere, $f(x) = \\frac{x^2 - 9}{x - 3}$.\r\n\r\n**a) Calculating values near x = 3:**\r\n* $f(2.999) = \\frac{(2.999)^2 - 9}{2.999 - 3} = \\frac{8.994001 - 9}{-0.001} = \\frac{-0.005999}{-0.001} = 5.999$\r\n* $f(3.001) = \\frac{(3.001)^2 - 9}{3.001 - 3} = \\frac{9.006001 - 9}{0.001} = \\frac{0.006001}{0.001} = 6.001$\r\n\r\n**b) Approximation:**\r\nYes, after approximation, both $f(2.999) \\approx 6$ and $f(3.001) \\approx 6$.\r\n\r\n**c) Discussing continuity at x = 3:**\r\n* **Limiting Value:**\r\n    $$ \\lim_{x \\to 3} f(x) = \\lim_{x \\to 3} \\frac{x^2 - 9}{x - 3} \\quad (\\text{form } \\frac{0}{0}) $$\r\n    $$ = \\lim_{x \\to 3} \\frac{(x-3)(x+3)}{x-3} = \\lim_{x \\to 3} (x+3) = 3+3=6 $$\r\n* **Functional Value:**\r\n    The function $f(3) = \\frac{3^2 - 9}{3 - 3} = \\frac{0}{0}$, which is undefined.\r\n\r\n**Conclusion:** Since the functional value $f(3)$ is undefined, the function is **discontinuous** at $x = 3$.\r\n\r\n***\r\n\r\n### Problem 3: Examining Continuity of Piecewise Functions\r\n\r\n**Examine the continuity or discontinuity of the following functions at the specified points.**\r\n\r\n**a) $f(x) = \\begin{cases} 2 - x^2, & \\text{if } x \\leq 2 \\\\ x - 4, & \\text{if } x > 2 \\end{cases}$ at $x = 2$**\r\n\r\n**Solution:**\r\n* **LHL:** $\\lim_{x \\to 2^-} f(x) = \\lim_{x \\to 2^-} (2 - x^2) = 2 - 2^2 = -2$.\r\n* **RHL:** $\\lim_{x \\to 2^+} f(x) = \\lim_{x \\to 2^+} (x - 4) = 2 - 4 = -2$.\r\n* **Functional Value:** $f(2) = 2 - 2^2 = -2$.\r\n\r\nSince LHL = RHL = f(2) = -2, the function is **continuous** at $x = 2$.\r\n\r\n**b) $f(x) = \\begin{cases} x, & \\text{if } x < 0 \\\\ 0, & \\text{if } x = 0 \\\\ x^2, & \\text{if } x > 0 \\end{cases}$ at $x = 0$**\r\n\r\n**Solution:**\r\n* **LHL:** $\\lim_{x \\to 0^-} f(x) = \\lim_{x \\to 0^-} (x) = 0$.\r\n* **RHL:** $\\lim_{x \\to 0^+} f(x) = \\lim_{x \\to 0^+} (x^2) = 0^2 = 0$.\r\n* **Functional Value:** $f(0) = 0$.\r\n\r\nSince LHL = RHL = f(0) = 0, the function is **continuous** at $x = 0$.\r\n\r\n**c) $f(x) = \\begin{cases} \\frac{2}{5 - x}, & \\text{if } x \\leq 3 \\\\ 5 - x, & \\text{if } x > 3 \\end{cases}$ at $x = 3$**\r\n\r\n**Solution:**\r\n* **LHL:** $\\lim_{x \\to 3^-} f(x) = \\lim_{x \\to 3^-} \\frac{2}{5 - x} = \\frac{2}{5 - 3} = 1$.\r\n* **RHL:** $\\lim_{x \\to 3^+} f(x) = \\lim_{x \\to 3^+} (5 - x) = 5 - 3 = 2$.\r\n\r\nSince LHL $\\neq$ RHL, the function is **discontinuous** at $x = 3$.\r\n\r\n**d) $f(x) = \\begin{cases} x^2 + 2, & \\text{if } x < 5 \\\\ 6x - 3, & \\text{if } x = 5 \\\\ 3x + 12, & \\text{if } x > 5 \\end{cases}$ at $x = 5$**\r\n\r\n**Solution:**\r\n* **LHL:** $\\lim_{x \\to 5^-} f(x) = \\lim_{x \\to 5^-} (x^2 + 2) = 5^2 + 2 = 27$.\r\n* **RHL:** $\\lim_{x \\to 5^+} f(x) = \\lim_{x \\to 5^+} (3x + 12) = 3(5) + 12 = 27$.\r\n* **Functional Value:** $f(5) = 6(5) - 3 = 30 - 3 = 27$.\r\n\r\nSince LHL = RHL = f(5) = 27, the function is **continuous** at $x = 5$.\r\n\r\n***\r\n\r\n### Problem 4: Finding an Unknown Value for Continuity\r\n**A function is defined as $f(x) = \\begin{cases} kx + 3, & \\text{if } x \\ge 2 \\\\ 3x - 1, & \\text{if } x < 2 \\end{cases}$. Find the value of k so that f(x) is continuous at x = 2.**\r\n\r\n**Solution:**\r\nFor the function to be continuous at $x=2$, we must have LHL = RHL = f(2).\r\n\r\n* **LHL:** (approaching 2 from the left, so $x < 2$)\r\n    $$ \\lim_{x \\to 2^-} f(x) = \\lim_{x \\to 2^-} (3x - 1) = 3(2) - 1 = 5 $$\r\n* **RHL:** (approaching 2 from the right, so $x > 2$)\r\n    $$ \\lim_{x \\to 2^+} f(x) = \\lim_{x \\to 2^+} (kx + 3) = k(2) + 3 = 2k + 3 $$\r\n* **Functional Value:** (at $x=2$)\r\n    $$ f(2) = k(2) + 3 = 2k + 3 $$\r\n\r\nNow, set LHL equal to RHL:\r\n$$\r\n\\begin{aligned}\r\n5 &= 2k + 3 \\\\\r\n2 &= 2k \\\\\r\nk &= 1\r\n\\end{aligned}\r\n$$\r\n**The value of k must be 1 for the function to be continuous at x = 2.**\r\n\r\n***\r\n\r\n### Problem 5: Finding an Unknown Value for Continuity (Rational Function)\r\n**A function is defined as $f(x) = \\begin{cases} \\frac{2x^2 - 18}{x - 3}, & \\text{for } x \\neq 3 \\\\ k, & \\text{for } x = 3 \\end{cases}$. Find the value of k so that f(x) is continuous at x = 3.**\r\n\r\n**Solution:**\r\nFor the function to be continuous at $x=3$, the limit of the function as $x \\to 3$ must be equal to the functional value $f(3)$.\r\n\r\n* **Limiting Value:**\r\n    $$ \\lim_{x \\to 3} f(x) = \\lim_{x \\to 3} \\frac{2x^2 - 18}{x - 3} $$\r\n    This is in the indeterminate form $\\frac{0}{0}$. We simplify the expression:\r\n    $$ = \\lim_{x \\to 3} \\frac{2(x^2 - 9)}{x - 3} = \\lim_{x \\to 3} \\frac{2(x-3)(x+3)}{x-3} = \\lim_{x \\to 3} 2(x+3) = 2(3+3) = 12 $$\r\n* **Functional Value:**\r\n    $$ f(3) = k $$\r\n\r\nTo ensure continuity, we set the limiting value equal to the functional value:\r\n$$ f(3) = \\lim_{x \\to 3} f(x) $$\r\n$$ k = 12 $$\r\n**The value of k must be 12 for the function to be continuous at x = 3.**\r\n\r\n***\r\n\r\n### Problem 6: Redefining a Function to be Continuous\r\n**A function is defined as $f(x) = \\begin{cases} 2x - 3, & \\text{if } x < 2 \\\\ 2, & \\text{if } x = 2 \\\\ 3x - 5, & \\text{if } x > 2 \\end{cases}$. Is f(x) continuous at x = 2? If not, how can it be made continuous?**\r\n\r\n**Solution:**\r\nLet\'s check the three conditions for continuity at $x=2$.\r\n\r\n* **LHL:** $\\lim_{x \\to 2^-} f(x) = \\lim_{x \\to 2^-} (2x - 3) = 2(2) - 3 = 1$.\r\n* **RHL:** $\\lim_{x \\to 2^+} f(x) = \\lim_{x \\to 2^+} (3x - 5) = 3(2) - 5 = 1$.\r\n* **Functional Value:** $f(2) = 2$.\r\n\r\nHere, LHL = RHL = 1, so the limit exists and is equal to 1. However, the functional value $f(2) = 2$.\r\nSince $\\lim_{x \\to 2} f(x) \\neq f(2)$, the function is **discontinuous** at $x=2$.\r\n\r\nTo make the function continuous, the functional value must be equal to the limit. We must redefine $f(2)$ to be 1.\r\n\r\nThe redefined continuous function is:\r\n$$ f(x) = \\begin{cases} 2x - 3, & \\text{if } x < 2 \\\\ 1, & \\text{if } x = 2 \\\\ 3x - 5, & \\text{if } x > 2 \\end{cases} $$', 1, '2025-07-25 01:11:38', '2025-09-18 09:23:05'),
(150, 14242, '\n# Grade 10 Compulsory Mathematics: Statistics Study Guide\n## Unit: Statistics | Chapter: Measures of Dispersion | Complete Solutions\n\n---\n\n## 🎯 Chapter Introduction: Understanding Dispersion\n\nThis guide covers the fundamental concepts of **Measures of Dispersion** in statistics. Dispersion helps us understand how spread out or scattered a set of data is. Think of it as measuring the \"variety\" or \"variability\" in your data. A low dispersion means the data points are clustered together, while high dispersion indicates they are spread far apart. 📊\n\n### Key Learning Areas\n* **Defining Dispersion:** What it means for data to be scattered.\n* **Absolute vs. Relative Measures:** Understanding the two main types of dispersion measures.\n* **Quartile Deviation (QD):** A measure of dispersion based on the middle 50% of the data.\n* **Coefficient of Quartile Deviation:** A relative measure for comparing dispersion between different datasets.\n* **Calculations for Continuous Data:** Step-by-step methods to calculate QD and its coefficient for frequency distributions.\n\n---\n\n## 🔑 Key Concepts & Definitions\n\nBefore jumping into calculations, let\'s clarify the core ideas.\n\n### 1. What is Dispersion?\n**Dispersion** simply means the extent to which numerical data is likely to vary or scatter around an average value. In other words, it\'s the measure of the **variability** or **fluctuation** within a dataset.\n\n---\n\n### 2. Types of Dispersion Measures\nThere are two main categories for measuring dispersion:\n\n* **Absolute Measure of Dispersion:** These measures express the variability in the **same units** as the original data (e.g., kg, marks, inches). They are used for describing the amount of spread in a single dataset.\n    * Examples: Range, **Quartile Deviation**, Mean Deviation, Standard Deviation.\n\n* **Relative Measure of Dispersion:** These are unit-less ratios used for **comparing the variability** of two or more datasets, especially when their units or averages are different.\n    * Examples: Coefficient of Range, **Coefficient of Quartile Deviation**, Coefficient of Variation.\n\n---\n\n### 3. Quartile Deviation (QD)\n**Quartile Deviation (QD)**, also known as the **Semi-Interquartile Range**, measures the average spread of the middle 50% of the data around the median. It is calculated as half the difference between the upper quartile ($Q_3$) and the lower quartile ($Q_1$).\n\n* **Formula:**\n    $$\n    \\text{Quartile Deviation (QD)} = \\frac{Q_3 - Q_1}{2}\n    $$\n\n---\n\n### 4. Coefficient of Quartile Deviation\nThe **Coefficient of Quartile Deviation** is a relative measure. It expresses the quartile deviation as a percentage of the average of the upper and lower quartiles, making it perfect for comparing the consistency of different datasets.\n\n* **Formula:**\n    $$\n    \\text{Coefficient of QD} = \\frac{Q_3 - Q_1}{Q_3 + Q_1}\n    $$\n\n---\n\n## 📚 Exercise Solutions\n\n### Short Questions\n\n***\n\n#### Question 1: What is dispersion?\n**Answer:** Dispersion refers to the scattering, variability, deviation, or fluctuation of data points around a central value.\n\n***\n\n#### Question 2: Write the various measures of dispersion.\n**Answer:**\nThe various measures of dispersion are:\n1.  **Absolute Measures:** Range, Quartile Deviation, Mean Deviation, and Standard Deviation.\n2.  **Relative Measures:** Coefficient of Range, Coefficient of Quartile Deviation, Coefficient of Mean Deviation, Coefficient of Variation.\n\n***\n\n#### Question 3: Define quartile deviation and write its formula.\n**Answer:**\nHalf of the difference between the upper quartile ($Q_3$) and the lower quartile ($Q_1$) is called **quartile deviation (QD)** or the semi-interquartile range.\nIts formula is:\n$$\n\\text{Quartile Deviation (QD)} = \\frac{Q_3 - Q_1}{2}\n$$\n\n***\n\n#### Question 4: What do you mean by the coefficient of quartile deviation?\n**Answer:**\nThe coefficient of quartile deviation is the relative measure of dispersion based on the lower and upper quartiles. It is used for comparison purposes.\nIts formula is:\n$$\n\\text{Coefficient of Quartile Deviation} = \\frac{Q_3 - Q_1}{Q_3 + Q_1}\n$$\n\n***\n\n#### Question 5: Differentiate between quartile deviation and its coefficient.\n**Answer:**\nQuartile deviation is an **absolute measure** of dispersion, calculated as $\\frac{Q_3 - Q_1}{2}$, and is expressed in the same units as the data. In contrast, the coefficient of quartile deviation is a **relative measure**, calculated as $\\frac{Q_3 - Q_1}{Q_3 + Q_1}$, which is a unit-less ratio used to compare the variability of different datasets.\n\n***\n\n#### Question 6: In continuous data, if $Q_1 = 40$ and $Q_3 = 60$, find the quartile deviation.\n**Solution:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **Find Quartile Deviation (QD)** | $QD = \\frac{Q_3 - Q_1}{2} = \\frac{60 - 40}{2} = \\mathbf{10}$ |\n\n***\n\n#### Question 7: In a continuous series, the lower quartile is 25 and its quartile deviation is 10. Find the upper quartile.\n**Solution:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **Find Upper Quartile ($Q_3$)** | $10 = \\frac{Q_3 - 25}{2} \\implies 20 = Q_3 - 25 \\implies Q_3 = \\mathbf{45}$ |\n\n***\n\n#### Question 8: In a continuous series, the coefficient of quartile deviation is $\\frac{1}{2}$ and its upper quartile is 60. Find the first quartile.\n**Solution:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **Find First Quartile ($Q_1$)** | $\\frac{1}{2} = \\frac{60 - Q_1}{60 + Q_1} \\implies 60 + Q_1 = 120 - 2Q_1 \\implies 3Q_1 = 60 \\implies Q_1 = \\mathbf{20}$ |\n\n***\n\n#### Question 9: If the coefficient of quartile deviation is $\\frac{3}{7}$ and $Q_1 = 20$, find $Q_3$.\n**Solution:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **Find Upper Quartile ($Q_3$)** | $\\frac{3}{7} = \\frac{Q_3 - 20}{Q_3 + 20} \\implies 3Q_3 + 60 = 7Q_3 - 140 \\implies 200 = 4Q_3 \\implies Q_3 = \\mathbf{50}$ |\n\n---\n\n### Long Questions\n\n***\n\n#### Question 10: Find the quartile deviation and its coefficient from the following data.\n\n| Marks obtained | No. of students |\n|:---:|:---:|\n| 60-65 | 7 |\n| 65-70 | 5 |\n| 70-75 | 8 |\n| 75-80 | 4 |\n| 80-85 | 3 |\n| 85-90 | 3 |\n\n**Solution:**\n1.  **Cumulative Frequency Table:**\n\n| Marks obtained | Frequency (f) | Cumulative Frequency (c.f.) |\n|:---:|:---:|:---:|\n| 60-65 | 7 | 7 |\n| 65-70 | 5 | 12 |\n| 70-75 | 8 | 20 |\n| 75-80 | 4 | 24 |\n| 80-85 | 3 | 27 |\n| 85-90 | 3 | 30 |\n| **Total** | **N = 30** | |\n\n2.  **Find Quartiles:**\n    * **For $Q_1$**: Position is $(\\frac{30}{4})^{th} = 7.5^{th}$ item. $Q_1$ class is **65-70**.\n        $Q_1 = 65 + \\frac{7.5 - 7}{5} \\times 5 = \\mathbf{65.5}$\n    * **For $Q_3$**: Position is $(\\frac{3 \\times 30}{4})^{th} = 22.5^{th}$ item. $Q_3$ class is **75-80**.\n        $Q_3 = 75 + \\frac{22.5 - 20}{4} \\times 5 = \\mathbf{78.125}$\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Quartile Deviation (QD)** | $QD = \\frac{78.125 - 65.5}{2} = \\mathbf{6.3125}$ |\n| **2. Coefficient of QD** | $\\text{Coefficient} = \\frac{78.125 - 65.5}{78.125 + 65.5} = \\frac{12.625}{143.625} \\approx \\mathbf{0.088}$ |\n\n***\n\n#### Question 11: Find the quartile deviation and its coefficient from the following data.\n\n| Weight (kg) | No. of person |\n|:---:|:---:|\n| 20-30 | 5 |\n| 30-40 | 15 |\n| 40-50 | 10 |\n| 50-60 | 8 |\n| 60-70 | 6 |\n| 70-80 | 2 |\n\n**Solution:**\n1.  **Cumulative Frequency Table:**\n\n| Weight (kg) | Frequency (f) | c.f. |\n|:---:|:---:|:---:|\n| 20-30 | 5 | 5 |\n| 30-40 | 15 | 20 |\n| 40-50 | 10 | 30 |\n| 50-60 | 8 | 38 |\n| 60-70 | 6 | 44 |\n| 70-80 | 2 | 46 |\n| **Total** | **N = 46** | |\n\n2.  **Find Quartiles:**\n    * **For $Q_1$**: Position is $(\\frac{46}{4})^{th} = 11.5^{th}$ item. $Q_1$ class is **30-40**.\n        $Q_1 = 30 + \\frac{11.5 - 5}{15} \\times 10 \\approx \\mathbf{34.33}$\n    * **For $Q_3$**: Position is $(\\frac{3 \\times 46}{4})^{th} = 34.5^{th}$ item. $Q_3$ class is **50-60**.\n        $Q_3 = 50 + \\frac{34.5 - 30}{8} \\times 10 = \\mathbf{55.625}$\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Quartile Deviation (QD)** | $QD = \\frac{55.625 - 34.33}{2} \\approx \\mathbf{10.65}$ kg |\n| **2. Coefficient of QD** | $\\text{Coefficient} = \\frac{55.625 - 34.33}{55.625 + 34.33} \\approx \\mathbf{0.237}$ |\n\n***\n\n#### Question 12: Find the quartile deviation and its coefficient from the following data.\n\n| Class interval | Frequency |\n|:---:|:---:|\n| 20-30 | 8 |\n| 30-40 | 16 |\n| 40-50 | 4 |\n| 50-60 | 4 |\n| 60-70 | 3 |\n\n**Solution:**\n1.  **Cumulative Frequency Table:**\n\n| Class interval | Frequency (f) | c.f. |\n|:---:|:---:|:---:|\n| 20-30 | 8 | 8 |\n| 30-40 | 16 | 24 |\n| 40-50 | 4 | 28 |\n| 50-60 | 4 | 32 |\n| 60-70 | 3 | 35 |\n| **Total** | **N = 35** | |\n\n2.  **Find Quartiles:**\n    * **For $Q_1$**: Position is $(\\frac{35}{4})^{th} = 8.75^{th}$ item. $Q_1$ class is **30-40**.\n        $Q_1 = 30 + \\frac{8.75 - 8}{16} \\times 10 \\approx \\mathbf{30.47}$\n    * **For $Q_3$**: Position is $(\\frac{3 \\times 35}{4})^{th} = 26.25^{th}$ item. $Q_3$ class is **40-50**.\n        $Q_3 = 40 + \\frac{26.25 - 24}{4} \\times 10 = \\mathbf{45.625}$\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Quartile Deviation (QD)** | $QD = \\frac{45.625 - 30.47}{2} \\approx \\mathbf{7.58}$ |\n| **2. Coefficient of QD** | $\\text{Coefficient} = \\frac{45.625 - 30.47}{45.625 + 30.47} \\approx \\mathbf{0.20}$ |\n\n***\n\n#### Question 13: Find the quartile deviation and its coefficient from the following data.\n\n| Size | Frequency |\n|:---:|:---:|\n| 4-8 | 6 |\n| 8-12 | 10 |\n| 12-16 | 18 |\n| 16-20 | 30 |\n| 20-24 | 15 |\n| 24-28 | 12 |\n| 28-32 | 10 |\n| 32-36 | 6 |\n| 36-40 | 5 |\n\n**Solution:**\n1.  **Find N and Quartile Positions:** $N = 112$.\n    * $Q_1$ position is $28^{th}$ item.\n    * $Q_3$ position is $84^{th}$ item.\n\n2.  **Find Quartiles from c.f. table:**\n    * **For $Q_1$**: c.f. is (6, 16, **34**,...). $Q_1$ class is **12-16**.\n        $Q_1 = 12 + \\frac{28 - 16}{18} \\times 4 \\approx \\mathbf{14.67}$\n    * **For $Q_3$**: c.f. is (..., 64, 79, **91**,...). $Q_3$ class is **24-28**.\n        $Q_3 = 24 + \\frac{84 - 79}{12} \\times 4 \\approx \\mathbf{25.67}$\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Quartile Deviation (QD)** | $QD = \\frac{25.67 - 14.67}{2} = \\mathbf{5.5}$ |\n| **2. Coefficient of QD** | $\\text{Coefficient} = \\frac{25.67 - 14.67}{25.67 + 14.67} = \\frac{11}{40.34} \\approx \\mathbf{0.273}$ |\n\n***\n\n#### Question 14: Find the quartile deviation and its coefficient from the following data.\n\n| Height (inches) | No. of students |\n|:---:|:---:|\n| 60-62 | 4 |\n| 62-64 | 6 |\n| 64-66 | 8 |\n| 66-68 | 12 |\n| 68-70 | 7 |\n| 70-72 | 2 |\n\n**Solution:**\n1.  **Find N and Quartile Positions:** $N = 39$.\n    * $Q_1$ position is $9.75^{th}$ item.\n    * $Q_3$ position is $29.25^{th}$ item.\n\n2.  **Find Quartiles from c.f. table:**\n    * **For $Q_1$**: c.f. is (4, **10**,...). $Q_1$ class is **62-64**.\n        $Q_1 = 62 + \\frac{9.75 - 4}{6} \\times 2 \\approx \\mathbf{63.92}$\n    * **For $Q_3$**: c.f. is (..., 18, **30**,...). $Q_3$ class is **66-68**.\n        $Q_3 = 66 + \\frac{29.25 - 18}{12} \\times 2 \\approx \\mathbf{67.88}$\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Quartile Deviation (QD)** | $QD = \\frac{67.88 - 63.92}{2} = \\mathbf{1.98}$ inches |\n| **2. Coefficient of QD** | $\\text{Coefficient} = \\frac{67.88 - 63.92}{67.88 + 63.92} = \\frac{3.96}{131.8} \\approx \\mathbf{0.03}$ |\n\n***\n\n#### Question 15: Find the quartile deviation and its coefficient from the following data.\n\n| Expenditure | No. of workers |\n|:---:|:---:|\n| 0-10 | 5 |\n| 10-20 | 15 |\n| 20-30 | 10 |\n| 30-40 | 8 |\n| 40-50 | 6 |\n\n**Solution:**\n1.  **Find N and Quartile Positions:** $N = 44$.\n    * $Q_1$ position is $11^{th}$ item.\n    * $Q_3$ position is $33^{rd}$ item.\n\n2.  **Find Quartiles from c.f. table:**\n    * **For $Q_1$**: c.f. is (5, **20**,...). $Q_1$ class is **10-20**.\n        $Q_1 = 10 + \\frac{11 - 5}{15} \\times 10 = \\mathbf{14}$\n    * **For $Q_3$**: c.f. is (..., 30, **38**,...). $Q_3$ class is **30-40**.\n        $Q_3 = 30 + \\frac{33 - 30}{8} \\times 10 = \\mathbf{33.75}$\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Quartile Deviation (QD)** | $QD = \\frac{33.75 - 14}{2} = \\mathbf{9.875}$ |\n| **2. Coefficient of QD** | $\\text{Coefficient} = \\frac{33.75 - 14}{33.75 + 14} = \\frac{19.75}{47.75} \\approx \\mathbf{0.414}$ |\n\n***\n\n#### Question 16: Prepare a frequency distribution table (10-20, etc.) and find the QD and its coefficient.\n**Data:** 21, 45, 60, 57, 15, 41, 48, 50, 34, 29, 56, 40, 14, 62, 28, 70, 22, 30, 38, 74, 13, 47, 20, 53, 64, 34, 75, 66.\n\n**Solution:**\n1.  **Frequency Distribution Table:** ($N=28$)\n\n| Marks | Freq (f) | c.f. |\n|:---:|:---:|:---:|\n| 10-20 | 3 | 3 |\n| 20-30 | 5 | 8 |\n| 30-40 | 4 | 12 |\n| 40-50 | 5 | 17 |\n| 50-60 | 4 | 21 |\n| 60-70 | 4 | 25 |\n| 70-80 | 3 | 28 |\n\n2.  **Find Quartiles:**\n    * **For $Q_1$**: Position is $(\\frac{28}{4})^{th} = 7^{th}$ item. c.f. is (3, **8**,...). $Q_1$ class is **20-30**.\n        $Q_1 = 20 + \\frac{7 - 3}{5} \\times 10 = \\mathbf{28}$\n    * **For $Q_3$**: Position is $(\\frac{3 \\times 28}{4})^{th} = 21^{st}$ item. c.f. is exactly 21 at the end of the 50-60 class, so $Q_3$ is the upper boundary. $Q_3 = \\mathbf{60}$.\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Quartile Deviation (QD)** | $QD = \\frac{60 - 28}{2} = \\mathbf{16}$ |\n| **2. Coefficient of QD** | $\\text{Coefficient} = \\frac{60 - 28}{60 + 28} = \\frac{32}{88} \\approx \\mathbf{0.364}$ |\n\n***\n\n#### Question 17: The following are the marks obtained by class 9 students. Prepare a frequency distribution table (10-20, etc.) and find the QD and its coefficient.\n**Data:** 22, 25, 46, 34, 57, 69, 44, 36, 12, 27, 50, 36, 35, 62, 46, 52, 54, 61, 66, 55, 29, 39, 40, 33, 14, 41, 25, 20, 16.\n\n**Solution:**\n1.  **Frequency Distribution Table:** ($N=29$)\n\n| Marks | Freq (f) | c.f. |\n|:---:|:---:|:---:|\n| 10-20 | 3 | 3 |\n| 20-30 | 6 | 9 |\n| 30-40 | 6 | 15 |\n| 40-50 | 5 | 20 |\n| 50-60 | 5 | 25 |\n| 60-70 | 4 | 29 |\n\n2.  **Find Quartiles:**\n    * **For $Q_1$**: Position is $(\\frac{29}{4})^{th} = 7.25^{th}$ item. c.f. is (3, **9**,...). $Q_1$ class is **20-30**.\n        $Q_1 = 20 + \\frac{7.25 - 3}{6} \\times 10 \\approx \\mathbf{27.08}$\n    * **For $Q_3$**: Position is $(\\frac{3 \\times 29}{4})^{th} = 21.75^{th}$ item. c.f. is (..., 20, **25**,...). $Q_3$ class is **50-60**.\n        $Q_3 = 50 + \\frac{21.75 - 20}{5} \\times 10 = \\mathbf{53.5}$\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Quartile Deviation (QD)** | $QD = \\frac{53.5 - 27.08}{2} = \\mathbf{13.21}$ |\n| **2. Coefficient of QD** | $\\text{Coefficient} = \\frac{53.5 - 27.08}{53.5 + 27.08} = \\frac{26.42}{80.58} \\approx \\mathbf{0.328}$ |', 1, '2025-07-25 01:11:38', '2025-09-18 14:41:48');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(151, 14243, '\n# Grade 10 Compulsory Mathematics: Statistics Study Guide\n## Unit: Statistics | Chapter: Mean Deviation | Complete Solutions\n\n---\n\n## 🎯 Chapter Introduction: Understanding Mean Deviation\n\nThis guide focuses on **Mean Deviation**, another important tool for measuring the dispersion or spread of data. While quartile deviation only considers the middle 50% of data, mean deviation provides a more comprehensive picture by taking into account **every single data point**. It calculates the average distance of each data point from the center of the dataset (usually the mean or median). 📈\n\n### Key Learning Areas\n* **Defining Mean Deviation (MD):** Understanding what \"average deviation\" means.\n* **Coefficient of Mean Deviation:** Learning how to compare the variability of different datasets.\n* **Calculations from Mean:** Step-by-step methods to calculate MD using the arithmetic mean.\n* **Calculations from Median:** Step-by-step methods to calculate MD using the median.\n* **Choosing the Right Method:** Knowing when to use the mean versus the median.\n\n---\n\n## 🔑 Key Concepts & Definitions\n\nLet\'s break down the essential concepts before solving the problems.\n\n### 1. What is Mean Deviation?\n**Mean Deviation (MD)**, also called **Average Deviation**, is defined as the arithmetic average of the absolute differences (deviations) of each data point from a measure of central tendency. This central point can be the **mean**, **median**, or mode.\n\nIt essentially answers the question: \"On average, how far is each data point from the center?\" We use the **absolute value** (ignoring negative signs) because we only care about the distance, not the direction.\n\n### 2. Coefficient of Mean Deviation\nThe **Coefficient of Mean Deviation** is a **relative measure** of dispersion. Because MD is an absolute measure (expressed in the same units as the data, like \"marks\" or \"kg\"), its coefficient is necessary for comparing the variability of two different datasets. It is a pure number without units.\n\n* **Formula from Mean:**\n    $$\n    \\text{Coefficient of MD from Mean} = \\frac{\\text{MD from Mean}}{\\text{Mean}}\n    $$\n* **Formula from Median:**\n    $$\n    \\text{Coefficient of MD from Median} = \\frac{\\text{MD from Median}}{\\text{Median}}\n    $$\n\n### 3. Formulas & Calculation Methods\n\n#### A. Mean Deviation from Mean\n1.  **Calculate the Mean ($\\bar{X}$):** Find the average of the dataset ($\\bar{X} = \\frac{\\sum fx}{N}$).\n2.  **Find Deviations:** Subtract the mean from each data point or mid-value ($D = x - \\bar{X}$).\n3.  **Find Absolute Deviations:** Take the absolute value of each deviation ($|D|$).\n4.  **Calculate MD:** Find the average of these absolute deviations.\n    $$\n    \\text{MD from Mean} = \\frac{\\sum f|D|}{N} = \\frac{\\sum f|x - \\bar{X}|}{N}\n    $$\n\n#### B. Mean Deviation from Median\n1.  **Calculate the Median ($M_d$):** Find the middle value of the dataset.\n2.  **Find Deviations:** Subtract the median from each data point ($D = x - M_d$).\n3.  **Find Absolute Deviations:** Take the absolute value of each deviation ($|D|$).\n4.  **Calculate MD:** Find the average of these absolute deviations.\n    $$\n    \\text{MD from Median} = \\frac{\\sum f|D|}{N} = \\frac{\\sum f|x - M_d|}{N}\n    $$\n\n---\n\n## 📚 Exercise Solutions\n\n### Short Questions\n\n***\n\n#### Question 1: Define mean deviation.\n**Answer:**\nMean deviation is the average of the absolute values of the deviations of each item from a central point, which can be the mean, median, or mode.\n\n***\n\n#### Question 2: What is the coefficient of mean deviation?\n**Answer:**\nThe coefficient of mean deviation is a relative measure of dispersion used to compare the variability of two or more series. It is the ratio of the mean deviation to the average (mean or median) from which it was calculated.\n\n***\n\n#### Question 3: Write the formulas for mean deviation and its coefficient from the mean.\n**Answer:**\n* **MD from Mean:** $MD = \\frac{\\sum f|m - \\bar{X}|}{N}$\n* **Coefficient of MD from Mean:** $\\text{Coefficient} = \\frac{\\text{MD from Mean}}{\\text{Mean}}$\n\n***\n\n#### Question 4: Write the formulas for mean deviation and its coefficient from the median.\n**Answer:**\n* **MD from Median:** $MD = \\frac{\\sum f|m - M_d|}{N}$\n* **Coefficient of MD from Median:** $\\text{Coefficient} = \\frac{\\text{MD from Median}}{\\text{Median}}$\n\n***\n\n#### Question 5: Which method of calculating mean deviation is considered best and why?\n**Answer:**\nCalculating mean deviation from the **mean** is generally considered the best method. This is because the mean uses all the values in the dataset for its calculation, making it a more stable and representative measure of the center than the median, which is only a positional average.\n\n***\n\n#### Question 6: If $\\sum fm = 1000$, $N = 50$, and $\\sum f|m - \\bar{X}| = 308$, find the MD from mean and its coefficient.\n**Solution:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Calculate Mean ($\\bar{X}$)** | $\\bar{X} = \\frac{\\sum fm}{N} = \\frac{1000}{50} = \\mathbf{20}$ |\n| **2. Calculate MD from Mean** | $MD = \\frac{\\sum f|m - \\bar{X}|}{N} = \\frac{308}{50} = \\mathbf{6.16}$ |\n| **3. Calculate Coefficient of MD** | $\\text{Coefficient} = \\frac{MD}{\\bar{X}} = \\frac{6.16}{20} = \\mathbf{0.308}$ |\n\n***\n\n#### Question 7: If $\\sum f|m - \\bar{X}| = 680$ and MD = 17, find $\\sum f$.\n**Solution:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **Find Total Frequency (N)** | $N = \\frac{\\sum f|m - \\bar{X}|}{MD} = \\frac{680}{17} = \\mathbf{40}$ |\n\n***\n\n#### Question 8: If Median ($M_d$) = 40, $\\sum f = 50$, and $\\sum f|m - M_d| = 530$, find MD from median and its coefficient.\n**Solution:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Calculate MD from Median** | $MD = \\frac{\\sum f|m - M_d|}{N} = \\frac{530}{50} = \\mathbf{10.6}$ |\n| **2. Calculate Coefficient of MD** | $\\text{Coefficient} = \\frac{MD}{M_d} = \\frac{10.6}{40} = \\mathbf{0.265}$ |\n\n***\n\n#### Question 9: If the coefficient of MD from median is 0.5 and the median is 40, find the MD.\n**Solution:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **Find Mean Deviation (MD)** | $MD = \\text{Coefficient} \\times M_d = 0.5 \\times 40 = \\mathbf{20}$ |\n\n---\n\n### Long Questions (Worked Examples)\n\n***\n\n#### Question 10: Find the MD and its coefficient from the median for the data: 10, 50, 60, 40, 30, 20.\n**Solution:**\n1.  **Arrange data and find Median:**\n    * Data: 10, 20, 30, 40, 50, 60.\n    * $M_d = \\frac{30 + 40}{2} = 35$.\n\n2.  **Data Table:**\n\n| X | D = &#124;X - 35&#124; |\n|:---:|:---:|\n| 10 | 25 |\n| 20 | 15 |\n| 30 | 5 |\n| 40 | 5 |\n| 50 | 15 |\n| 60 | 25 |\n| **Total** | **Σ&#124;D&#124; = 90** |\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. MD from Median** | $MD = \\frac{\\sum|D|}{N} = \\frac{90}{6} = \\mathbf{15}$ |\n| **2. Coefficient of MD** | $\\text{Coefficient} = \\frac{MD}{M_d} = \\frac{15}{35} \\approx \\mathbf{0.429}$ |\n\n***\n\n#### Question 11: Find the MD and its coefficient from the median.\n\n| Marks (X) | Freq (f) |\n|:---:|:---:|\n| 20 | 2 |\n| 30 | 4 |\n| 40 | 5 |\n| 50 | 7 |\n| 60 | 2 |\n| 70 | 1 |\n\n**Solution:**\n1.  **Find Median:** $N=21$. The median position is the $11^{th}$ item. From c.f. (2, 6, **11**,...), the $M_d = \\mathbf{40}$.\n\n2.  **Data Table:**\n\n| X | f | D = &#124;X - 40&#124; | f&#124;D&#124; |\n|:---:|:---:|:---:|:---:|\n| 20 | 2 | 20 | 40 |\n| 30 | 4 | 10 | 40 |\n| 40 | 5 | 0 | 0 |\n| 50 | 7 | 10 | 70 |\n| 60 | 2 | 20 | 40 |\n| 70 | 1 | 30 | 30 |\n| **Total** | **N=21** | | **Σf&#124;D&#124;=220** |\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. MD from Median** | $MD = \\frac{\\sum f|D|}{N} = \\frac{220}{21} \\approx \\mathbf{10.48}$ |\n| **2. Coefficient of MD** | $\\text{Coefficient} = \\frac{MD}{M_d} = \\frac{10.48}{40} \\approx \\mathbf{0.262}$ |\n\n***\n\n#### Question 12: Find the MD and its coefficient from the mean.\n\n| Marks (X) | Freq (f) |\n|:---:|:---:|\n| 20 | 2 |\n| 25 | 5 |\n| 30 | 6 |\n| 35 | 5 |\n| 40 | 2 |\n\n**Solution:**\n1.  **Find Mean:** $N=20$. $\\sum fX = 600$. $\\bar{X} = \\frac{600}{20} = \\mathbf{30}$.\n\n2.  **Data Table:**\n\n| X | f | D = &#124;X - 30&#124; | f&#124;D&#124; |\n|:---:|:---:|:---:|:---:|\n| 20 | 2 | 10 | 20 |\n| 25 | 5 | 5 | 25 |\n| 30 | 6 | 0 | 0 |\n| 35 | 5 | 5 | 25 |\n| 40 | 2 | 10 | 20 |\n| **Total** | **N=20** | | **Σf&#124;D&#124;=90** |\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. MD from Mean** | $MD = \\frac{\\sum f|D|}{N} = \\frac{90}{20} = \\mathbf{4.5}$ |\n| **2. Coefficient of MD** | $\\text{Coefficient} = \\frac{MD}{\\bar{X}} = \\frac{4.5}{30} = \\mathbf{0.15}$ |\n\n***\n\n#### Question 13: Calculate the MD from the median for the following data.\n\n| Marks | Freq (f) |\n|:---:|:---:|\n| 10-20 | 5 |\n| 20-30 | 4 |\n| 30-40 | 5 |\n| 40-50 | 4 |\n| 50-60 | 2 |\n\n**Solution:**\n1.  **Find Median:** $N=20$. Median class is **30-40**.\n    $M_d = L + \\frac{\\frac{N}{2} - c.f.}{f} \\times i = 30 + \\frac{10 - 9}{5} \\times 10 = \\mathbf{32}$.\n\n2.  **Data Table:**\n\n| Marks | m | f | D = &#124;m - 32&#124; | f&#124;D&#124; |\n|:---:|:---:|:---:|:---:|:---:|\n| 10-20 | 15 | 5 | 17 | 85 |\n| 20-30 | 25 | 4 | 7 | 28 |\n| 30-40 | 35 | 5 | 3 | 15 |\n| 40-50 | 45 | 4 | 13 | 52 |\n| 50-60 | 55 | 2 | 23 | 46 |\n| **Total** | | **N=20** | | **Σf&#124;D&#124;=226** |\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **MD from Median** | $MD = \\frac{\\sum f|D|}{N} = \\frac{226}{20} = \\mathbf{11.3}$ |\n\n***\n\n#### Question 14: Compute the mean deviation and its coefficient from the median.\n\n| Marks (X) | Freq (f) |\n|:---:|:---:|\n| 55 | 4 |\n| 65 | 5 |\n| 75 | 6 |\n| 85 | 3 |\n| 95 | 2 |\n\n**Solution:**\n1.  **Find Median:** $N=20$. The median is the $10.5^{th}$ item. From c.f. (4, 9, **15**,...), the $M_d = \\mathbf{75}$.\n\n2.  **Data Table:**\n\n| X | f | D = &#124;X - 75&#124; | f&#124;D&#124; |\n|:---:|:---:|:---:|:---:|\n| 55 | 4 | 20 | 80 |\n| 65 | 5 | 10 | 50 |\n| 75 | 6 | 0 | 0 |\n| 85 | 3 | 10 | 30 |\n| 95 | 2 | 20 | 40 |\n| **Total** | **N=20** | | **Σf&#124;D&#124;=200** |\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. MD from Median** | $MD = \\frac{\\sum f|D|}{N} = \\frac{200}{20} = \\mathbf{10}$ |\n| **2. Coefficient of MD** | $\\text{Coefficient} = \\frac{MD}{M_d} = \\frac{10}{75} \\approx \\mathbf{0.133}$ |\n\n***\n\n#### Question 15: Find the MD and its coefficient from the median.\n\n| Marks (X) | Freq (f) |\n|:---:|:---:|\n| 10 | 2 |\n| 20 | 3 |\n| 30 | 9 |\n| 40 | 21 |\n| 50 | 11 |\n| 60 | 5 |\n\n**Solution:**\n1.  **Find Median:** $N=51$. The median is the $26^{th}$ item. From c.f. (2, 5, 14, **35**,...), the $M_d = \\mathbf{40}$.\n\n2.  **Data Table:**\n\n| X | f | D = &#124;X - 40&#124; | f&#124;D&#124; |\n|:---:|:---:|:---:|:---:|\n| 10 | 2 | 30 | 60 |\n| 20 | 3 | 20 | 60 |\n| 30 | 9 | 10 | 90 |\n| 40 | 21 | 0 | 0 |\n| 50 | 11 | 10 | 110 |\n| 60 | 5 | 20 | 100 |\n| **Total** | **N=51** | | **Σf&#124;D&#124;=420** |\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. MD from Median** | $MD = \\frac{\\sum f|D|}{N} = \\frac{420}{51} \\approx \\mathbf{8.24}$ |\n| **2. Coefficient of MD** | $\\text{Coefficient} = \\frac{MD}{M_d} = \\frac{8.24}{40} \\approx \\mathbf{0.206}$ |\n\n***\n\n#### Question 16: Find the mean deviation from the median for the unsorted data below.\n\n| Marks (X) | Freq (f) |\n|:---:|:---:|\n| 20 | 3 |\n| 10 | 4 |\n| 30 | 5 |\n| 50 | 2 |\n| 40 | 1 |\n\n**Solution:**\n1.  **Sort Data and Find Median:** $N=15$. The median is the $8^{th}$ item.\n\n| X (sorted) | f | c.f. |\n|:---:|:---:|:---:|\n| 10 | 4 | 4 |\n| 20 | 3 | 7 |\n| 30 | 5 | **12** |\n| 40 | 1 | 13 |\n| 50 | 2 | 15 |\n\nThe 8th item falls in the class with mark **30**. So, $M_d = \\mathbf{30}$.\n\n2.  **Data Table:**\n\n| X | f | D = &#124;X - 30&#124; | f&#124;D&#124; |\n|:---:|:---:|:---:|:---:|\n| 10 | 4 | 20 | 80 |\n| 20 | 3 | 10 | 30 |\n| 30 | 5 | 0 | 0 |\n| 40 | 1 | 10 | 10 |\n| 50 | 2 | 20 | 40 |\n| **Total** | **N=15** | | **Σf&#124;D&#124;=160** |\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **MD from Median** | $MD = \\frac{\\sum f|D|}{N} = \\frac{160}{15} \\approx \\mathbf{10.67}$ |', 1, '2025-07-25 01:11:38', '2025-09-18 14:38:40'),
(152, 14244, '\n# Grade 10 Compulsory Mathematics: Statistics Study Guide\n## Unit: Statistics | Chapter: Standard Deviation & Coefficient of Variation | Complete Solutions\n\n---\n\n## 🎯 Chapter Introduction: Standard Deviation\n\nWelcome to the complete guide on **Standard Deviation**, the most widely used and robust measure of dispersion in statistics. While other measures like Quartile and Mean Deviation provide valuable insights, Standard Deviation is considered the most reliable because it gives more weight to the data points that are farther from the mean. It\'s the cornerstone for many advanced statistical analyses. 🔬\n\n### Key Learning Areas\n* **Defining Standard Deviation ($\\sigma$):** Understanding the \"root mean square deviation.\"\n* **Coefficient of Standard Deviation:** A relative measure for comparing spread.\n* **Coefficient of Variation (C.V.):** A standardized measure of dispersion, expressed as a percentage, used to compare consistency.\n* **Calculation Methods:** Step-by-step processes for individual, discrete, and continuous series.\n* **Consistency and Stability:** Using C.V. to determine which dataset is more consistent.\n\n---\n\n## 🔑 Key Concepts & Definitions\n\nLet\'s dive into the foundational ideas.\n\n### 1. What is Standard Deviation?\n**Standard Deviation ($\\sigma$)** is the **positive square root of the mean of the squared deviations from the arithmetic mean**. This might sound complex, but it\'s a very logical process:\n\n1.  **Find the Mean ($\\bar{X}$):** Calculate the average of the data.\n2.  **Find Deviations:** Find the difference between each data point and the mean.\n3.  **Square the Deviations:** Square each difference. This makes all values positive and gives more weight to larger deviations.\n4.  **Find the Average of Squares:** Calculate the mean of these squared deviations. This value is called the **Variance ($\\sigma^2$)**.\n5.  **Take the Square Root:** The square root of the variance is the **Standard Deviation**.\n\nIt is also known as the **root mean square deviation** and is denoted by the Greek letter **$\\sigma$** (sigma).\n\n### 2. Coefficient of Standard Deviation\nThis is a **relative measure** of dispersion. It\'s the ratio of the Standard Deviation to the arithmetic mean, which allows for a standardized comparison of variability between different datasets.\n\n* **Formula:**\n    $$\n    \\text{Coefficient of SD} = \\frac{\\sigma}{\\bar{X}}\n    $$\n\n### 3. Coefficient of Variation (C.V.)\nThe **Coefficient of Variation (C.V.)** is the most common relative measure of dispersion. It\'s simply the Coefficient of Standard Deviation expressed as a **percentage**.\n\n* **Formula:**\n    $$\n    C.V. = \\frac{\\sigma}{\\bar{X}} \\times 100\\%\n    $$\n\n**Why is C.V. important?** It\'s used to compare the consistency or stability of different datasets.\n* A **low C.V.** means the data points are close to the mean, indicating **high consistency** or stability.\n* A **high C.V.** means the data points are spread out, indicating **low consistency**.\n\n---\n\n## 📚 Exercise Solutions\n\n### Short Questions\n\n***\n\n#### Question 1: Define standard deviation.\n**Answer:**\nStandard deviation is the positive square root of the mean of the squared deviations of data points from their arithmetic mean. It\'s often called the root mean square deviation and is denoted by $\\sigma$.\n\n***\n\n#### Question 2: What is the coefficient of standard deviation and its formula?\n**Answer:**\nThe coefficient of standard deviation is a relative measure of dispersion, calculated as the ratio of the standard deviation ($\\sigma$) to the mean ($\\bar{X}$). Its formula is:\n$$\n\\text{Coefficient of SD} = \\frac{\\sigma}{\\bar{X}}\n$$\n\n***\n\n#### Question 3: Define the coefficient of variation and write its formula.\n**Answer:**\nThe coefficient of variation (C.V.) is the coefficient of standard deviation expressed as a percentage. It is a unit-less measure used for comparing the consistency of data.\n$$\nC.V. = \\frac{\\sigma}{\\bar{X}} \\times 100\\%\n$$\n\n***\n\n#### Question 4: What does a low coefficient of variation mean?\n**Answer:**\nA low coefficient of variation indicates that the data points are clustered closely around the mean. This signifies that the data is more **consistent**, uniform, stable, or homogenous.\n\n***\n\n#### Question 5: What is the difference between standard deviation and mean deviation?\n**Answer:**\n**Standard deviation** is the square root of the average of *squared* deviations from the mean. **Mean deviation** is the average of the *absolute* deviations from the mean or median. Standard deviation gives greater weight to extreme values because it squares the deviations.\n\n***\n\n#### Question 6: What is the difference between the coefficient of standard deviation and the coefficient of variation?\n**Answer:**\nThe coefficient of standard deviation is the ratio $\\frac{\\sigma}{\\bar{X}}$. The coefficient of variation is simply this same ratio multiplied by 100 to express it as a percentage ($C.V. = \\frac{\\sigma}{\\bar{X}} \\times 100\\%$).\n\n***\n\n#### Question 7: If $\\sum f(m - \\bar{X})^2 = 484$, $N = 24$, and $\\bar{X} = 25$, find the standard deviation and its coefficient.\n**Solution:**\n1.  **Standard Deviation ($\\sigma$):**\n    $$\n    \\sigma = \\sqrt{\\frac{\\sum f(m - \\bar{X})^2}{N}} = \\sqrt{\\frac{484}{24}} \\approx \\sqrt{20.167} \\approx 4.49\n    $$\n2.  **Coefficient of SD:**\n    $$\n    \\text{Coefficient} = \\frac{\\sigma}{\\bar{X}} = \\frac{4.49}{25} \\approx 0.18\n    $$\n\n***\n\n#### Question 8: If $\\sum fd^2 = 848$, $\\sum fd = 0$, $N=100$, and Assumed Mean A=12, find the standard deviation.\n**Solution:**\nSince $\\sum fd = 0$, the assumed mean (A) is equal to the actual mean ($\\bar{X}$).\n$$\n\\sigma = \\sqrt{\\frac{\\sum fd^2}{N} - \\left(\\frac{\\sum fd}{N}\\right)^2} = \\sqrt{\\frac{848}{100} - \\left(\\frac{0}{100}\\right)^2} = \\sqrt{8.48} \\approx 2.91\n$$\n\n***\n\n#### Question 9: If $\\sum fd\'^2 = 125$, $\\sum fd\' = -4$, $N=20$, and class size $C=10$, find the standard deviation.\n**Solution:**\nUsing the step-deviation formula:\n$$\n\\sigma = \\sqrt{\\frac{\\sum fd\'^2}{N} - \\left(\\frac{\\sum fd\'}{N}\\right)^2} \\times C = \\sqrt{\\frac{125}{20} - \\left(\\frac{-4}{20}\\right)^2} \\times 10\n$$\n$$\n\\sigma = \\sqrt{6.25 - (-0.2)^2} \\times 10 = \\sqrt{6.25 - 0.04} \\times 10 = \\sqrt{6.21} \\times 10 \\approx 2.492 \\times 10 \\approx 24.92\n$$\n\n---\n\n### Long Questions (Worked Examples)\n\n***\n\n#### Question 10: Find the standard deviation and its coefficient for the data: 5, 10, 15, 20, 25.\n**Solution:**\n1.  **Find Mean:** $N=5$. $\\sum X = 5+10+15+20+25 = 75$.\n    $\\bar{X} = \\frac{75}{5} = 15$.\n2.  **Calculations:**\n| X | $D = X - \\bar{X}$ | $D^2$ |\n|:---:|:---:|:---:|\n| 5 | -10 | 100 |\n| 10 | -5 | 25 |\n| 15 | 0 | 0 |\n| 20 | 5 | 25 |\n| 25 | 10 | 100 |\n| **Total** | | **$\\sum D^2 = 250$** |\n\n* **Standard Deviation ($\\sigma$)** = $\\sqrt{\\frac{\\sum D^2}{N}} = \\sqrt{\\frac{250}{5}} = \\sqrt{50} \\approx \\mathbf{7.07}$\n* **Coefficient of SD** = $\\frac{\\sigma}{\\bar{X}} = \\frac{7.07}{15} \\approx \\mathbf{0.47}$\n\n***\n\n#### Question 11: Find the standard deviation and C.V. for the data: 30, 33, 37, 40, 44, 50.\n**Solution:**\n1.  **Find Mean:** $N=6$. $\\sum X = 30+33+37+40+44+50 = 234$.\n    $\\bar{X} = \\frac{234}{6} = 39$.\n2.  **Calculations:**\n\n| X | $D = X - 39$ | $D^2$ |\n|:---:|:---:|:---:|\n| 30 | -9 | 81 |\n| 33 | -6 | 36 |\n| 37 | -2 | 4 |\n| 40 | 1 | 1 |\n| 44 | 5 | 25 |\n| 50 | 11 | 121 |\n| **Total**| | **$\\sum D^2 = 268$** |\n\n* **Standard Deviation ($\\sigma$)** = $\\sqrt{\\frac{268}{6}} = \\sqrt{44.67} \\approx \\mathbf{6.68}$\n* **Coefficient of Variation (C.V.)** = $\\frac{6.68}{39} \\times 100\\% \\approx \\mathbf{17.13\\%}$\n\n***\n\n#### Question 12: Find the standard deviation from the given data.\n\n| x | f |\n|:---:|:---:|\n| 6 | 1 |\n| 8 | 14 |\n| 10 | 25 |\n| 12 | 27 |\n| 14 | 18 |\n| 16 | 9 |\n| 18 | 4 |\n| 20 | 2 |\n\n**Solution:**\n1.  **Find Mean ($\\bar{X}$):** N=100. $\\sum fx = (6 \\times 1) + (8 \\times 14) + ... + (20 \\times 2) = 1200$.\n    $\\bar{X} = \\frac{1200}{100} = 12$.\n2.  **Calculations for SD:**\n\n| x | f | $D = x-12$ | $D^2$ | $fD^2$ |\n|:---:|:---:|:---:|:---:|:---:|\n| 6 | 1 | -6 | 36 | 36 |\n| 8 | 14 | -4 | 16 | 224 |\n| 10 | 25 | -2 | 4 | 100 |\n| 12 | 27 | 0 | 0 | 0 |\n| 14 | 18 | 2 | 4 | 72 |\n| 16 | 9 | 4 | 16 | 144 |\n| 18 | 4 | 6 | 36 | 144 |\n| 20 | 2 | 8 | 64 | 128 |\n| **Total** | **N=100** | | | **$\\sum fD^2=848$** |\n\n* **Standard Deviation ($\\sigma$)** = $\\sqrt{\\frac{\\sum fD^2}{N}} = \\sqrt{\\frac{848}{100}} = \\sqrt{8.48} \\approx \\mathbf{2.91}$\n\n***\n\n#### Question 13: Calculate the coefficient of variation from the data given below.\n\n| Class Interval | Frequency |\n|:---:|:---:|\n| 0-10 | 5 |\n| 10-20 | 8 |\n| 20-30 | 15 |\n| 30-40 | 16 |\n| 40-50 | 6 |\n\n**Solution:**\n1.  **Find Mean ($\\bar{X}$):** N=50. $\\sum fm = 1350$.\n    $\\bar{X} = \\frac{1350}{50} = 27$.\n2.  **Find Standard Deviation ($\\sigma$):**\n\n| m | f | $D = m - 27$ | $D^2$ | $fD^2$ |\n|:---:|:---:|:---:|:---:|:---:|\n| 5 | 5 | -22 | 484 | 2420 |\n| 15 | 8 | -12 | 144 | 1152 |\n| 25 | 15 | -2 | 4 | 60 |\n| 35 | 16 | 8 | 64 | 1024 |\n| 45 | 6 | 18 | 324 | 1944 |\n| **Total** | **N=50** | | | **$\\sum fD^2=6600$** |\n    $\\sigma = \\sqrt{\\frac{\\sum fD^2}{N}} = \\sqrt{\\frac{6600}{50}} = \\sqrt{132} \\approx 11.49$.\n4.  **Calculate C.V.:**\n    **C.V.** = $\\frac{\\sigma}{\\bar{X}} \\times 100\\% = \\frac{11.49}{27} \\times 100\\% \\approx \\mathbf{42.56\\%}$\n\n***\n\n#### Question 14: Calculate the coefficient of variation from the data given below.\n\n| Class Interval | Frequency |\n|:---:|:---:|\n| 0-20 | 2 |\n| 20-40 | 3 |\n| 40-60 | 4 |\n| 60-80 | 5 |\n| 80-100 | 6 |\n\n**Solution:**\n1.  **Find Mean ($\\bar{X}$):** N=20.\n2. \n| C.I. | m | f | fm |\n|:---:|:---:|:---:|:---:|\n| 0-20 | 10 | 2 | 20 |\n| 20-40 | 30 | 3 | 90 |\n| 40-60 | 50 | 4 | 200 |\n| 60-80 | 70 | 5 | 350 |\n| 80-100 | 90 | 6 | 540 |\n| **Total** | | **N=20** | **$\\sum fm = 1200$** |\n\n    $\\bar{X} = \\frac{1200}{20} = 60$.\n    \n3.  **Find Standard Deviation ($\\sigma$):**\n| m | f | $D=m-60$ | $D^2$ | $fD^2$ |\n|:---:|:---:|:---:|:---:|:---:|\n| 10 | 2 | -50 | 2500 | 5000 |\n| 30 | 3 | -30 | 900 | 2700 |\n| 50 | 4 | -10 | 100 | 400 |\n| 70 | 5 | 10 | 100 | 500 |\n| 90 | 6 | 30 | 900 | 5400 |\n| **Total** | **N=20** | | | **$\\sum fD^2 = 14000$** |\n    $\\sigma = \\sqrt{\\frac{14000}{20}} = \\sqrt{700} \\approx 26.46$.\n4.  **Calculate C.V.:**\n    **C.V.** = $\\frac{26.46}{60} \\times 100\\% \\approx \\mathbf{44.1\\%}$\n\n***\n\n#### Question 15: Find the standard deviation of the data given below.\n\n| Class Interval | Frequency |\n|:---:|:---:|\n| 5-15 | 7 |\n| 15-25 | 3 |\n| 25-35 | 6 |\n| 35-45 | 4 |\n| 45-55 | 5 |\n\n**Solution:**\n\n1.  **Find Mean ($\\bar{X}$):** N=25. $\\sum fm = (10\\times7) + (20\\times3) + (30\\times6) + (40\\times4) + (50\\times5) = 720$.\n    $\\bar{X} = \\frac{720}{25} = 28.8$.\n2.  **Calculations for SD:**\n\n| m | f | $D = m-28.8$ | $D^2$ | $fD^2$ |\n|:---:|:---:|:---:|:---:|:---:|\n| 10 | 7 | -18.8 | 353.44 | 2474.08 |\n| 20 | 3 | -8.8 | 77.44 | 232.32 |\n| 30 | 6 | 1.2 | 1.44 | 8.64 |\n| 40 | 4 | 11.2 | 125.44 | 501.76 |\n| 50 | 5 | 21.2 | 449.44 | 2247.2 |\n| **Total** | **N=25** | | | **$\\sum fD^2 = 5464$** |\n* **Standard Deviation ($\\sigma$)** = $\\sqrt{\\frac{5464}{25}} = \\sqrt{218.56} \\approx \\mathbf{14.78}$\n\n***\nThis guide now contains a comprehensive set of examples. Practice these thoroughly to master the concepts of Standard Deviation and Coefficient of Variation. Good luck! 🌟\n', 1, '2025-07-25 01:11:38', '2025-09-18 14:27:45'),
(153, 14246, '# Conic Sections: A Guide to Slicing a Cone\r\n## An Overview of Circles, Ellipses, Parabolas, and Hyperbolas\r\n\r\n---\r\n\r\n## 🎯 Introduction: The Carrot and the Knife\r\n\r\nHave you ever wondered about the different shapes you can get by slicing a carrot (which is shaped like a cone) with a knife? Depending on the angle of your cut, you can create a circle, an oval (ellipse), and other interesting curves. These shapes are known as **conic sections**. They are fundamental curves in geometry and appear everywhere in science and nature. 🥕\r\n\r\nThis guide explains how these four key shapes are formed by slicing a cone.\r\n\r\n---\r\n\r\n## 🔑 What is a Cone?\r\n\r\nIn geometry, when we talk about a cone for conic sections, we\'re usually referring to a **double-napped right circular cone**.\r\n\r\nImagine two fixed lines, AB and CD, that intersect at a point **O**. If you rotate the line AB around the fixed line CD, the surface you create is this double cone.\r\n\r\n![Diagram of a double-napped right circular cone](https://seekoguide.com/images/conic/1.svg)\r\n\r\nHere are the key parts:\r\n* **Vertex (O):** The point where the two cones meet.\r\n* **Axis (CD):** The fixed line that the other line rotates around.\r\n* **Generator (AB):** The moving line that traces the surface of the cone.\r\n* **Nappes:** The two separate cones are called nappes (an upper nappe and a lower nappe).\r\n\r\n---\r\n\r\n## 🔪 Creating Shapes by Slicing the Cone\r\n\r\nThe four conic sections are created by slicing this double cone with a flat plane at different angles.\r\n\r\n### 1. The Circle\r\nWhen you slice the cone with a plane that is **perfectly horizontal** (perpendicular to the axis) and parallel to the circular base, the resulting shape is a **circle**.\r\n\r\n![Formation of a circle by slicing a cone](https://seekoguide.com/images/conic/circle.svg)\r\n\r\n---\r\n\r\n### 2. The Ellipse\r\nIf you tilt the plane slightly, so the slice is **not parallel to the base** but still cuts completely through only one nappe, the resulting shape is an **ellipse**. An ellipse is essentially a stretched or squashed circle.\r\n\r\n![Formation of an ellipse by slicing a cone](https://seekoguide.com/images/conic/ellipse.svg)\r\n\r\n---\r\n\r\n### 3. The Parabola\r\nWhen you slice the cone with a plane that is **parallel to one of the generator lines** (the slanted edge of the cone), the curve you get is a **parabola**. This curve does not close back on itself.\r\n\r\n![Formation of a parabola by slicing a cone](https://seekoguide.com/images/conic/parabola.svg)\r\n\r\n---\r\n\r\n### 4. The Hyperbola\r\nIf you slice the cone with a plane that is steep enough (parallel to the axis or at an even steeper angle) to cut through **both the upper and lower nappes**, you get a **hyperbola**. A hyperbola consists of two separate, mirror-image curves.\r\n\r\n![Formation of a hyperbola by slicing a cone](https://seekoguide.com/images/conic/hyperbola.svg)\r\n\r\n---\r\n\r\n## ✍️ Exercises\r\n\r\n### Question 1: Parts of a Cone\r\nName the different parts of the cone indicated by the letters a, b, c, d, and e in the figure below.\r\n\r\n![Cone with labeled parts for exercise 1](https://seekoguide.com/images/conic/one.svg)\r\n\r\n**Answer:**\r\n* **a:** Axis\r\n* **b:** Upper Nappe\r\n* **c:** Lower Nappe\r\n* **d:** Generator\r\n* **e:** Vertex\r\n\r\n---\r\n\r\n### Question 2: Identify the Conic Section\r\nName the conic section shown in each of the following figures.\r\n\r\n1.  ![Diagram for exercise 2a](https://seekoguide.com/images/conic/a.svg)\r\n    **Answer:** Ellipse\r\n\r\n2.  ![Diagram for exercise 2b](https://seekoguide.com/images/conic/b.svg)\r\n    **Answer:** Parabola\r\n\r\n3.  ![Diagram for exercise 2c](https://seekoguide.com/images/conic/c.svg)\r\n    **Answer:** Circle\r\n\r\n4.  ![Diagram for exercise 2d](https://seekoguide.com/images/conic/d.svg)\r\n    **Answer:** Hyperbola', 1, '2025-07-25 01:11:38', '2025-09-18 14:27:45');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(154, 13120, '# Transformations: Complete Study Guide for Combined Transformations\r\n## A Comprehensive Guide to Reflection, Rotation, Translation, and Enlargement\r\n\r\n---\r\n\r\n## 🎯 Introduction: Master Combined Transformations\r\n\r\nWelcome to the ultimate guide on **Combined Transformations in Coordinate Geometry**. This comprehensive study guide covers everything you need to know about performing multiple transformations in sequence. When we combine transformations, we perform **two or more transformations one after another** on geometric shapes or points.\r\n\r\nUnderstanding combined transformations is essential for:\r\n- Solving complex coordinate geometry problems\r\n- Understanding geometric patterns and symmetries\r\n- Preparing for advanced mathematics examinations\r\n- Developing spatial reasoning skills\r\n\r\n**Key Learning Objective**: Master the art of applying multiple transformations systematically and understand how the order of operations affects the final result. 📐\r\n\r\n---\r\n\r\n## 🔑 Essential Transformation Rules\r\n\r\n### 1. Reflection Transformations\r\n**Basic Reflection Rules:**\r\n\r\n* **Reflection in the x-axis (y = 0):**\r\n  - Rule: $P(x, y) \\rightarrow P\'(x, -y)$\r\n  - Effect: Point flips across horizontal axis\r\n\r\n* **Reflection in the y-axis (x = 0):**\r\n  - Rule: $P(x, y) \\rightarrow P\'(-x, y)$\r\n  - Effect: Point flips across vertical axis\r\n\r\n* **Reflection in the line y = x:**\r\n  - Rule: $P(x, y) \\rightarrow P\'(y, x)$\r\n  - Effect: Coordinates interchange\r\n\r\n* **Reflection in the line y = -x (or x + y = 0):**\r\n  - Rule: $P(x, y) \\rightarrow P\'(-y, -x)$\r\n  - Effect: Coordinates interchange and both change sign\r\n\r\n* **Reflection in vertical line x = k:**\r\n  - Rule: $P(x, y) \\rightarrow P\'(2k - x, y)$\r\n  - Effect: Point reflects across vertical line\r\n\r\n* **Reflection in horizontal line y = k:**\r\n  - Rule: $P(x, y) \\rightarrow P\'(x, 2k - y)$\r\n  - Effect: Point reflects across horizontal line\r\n\r\n### 2. Rotation Transformations (about origin)\r\n**Standard Rotation Rules:**\r\n\r\n* **Rotation by +90° (Quarter-turn anticlockwise):**\r\n  - Rule: $P(x, y) \\rightarrow P\'(-y, x)$\r\n  - Memory tip: \"Negative y becomes new x, old x becomes new y\"\r\n\r\n* **Rotation by -90° (Quarter-turn clockwise):**\r\n  - Rule: $P(x, y) \\rightarrow P\'(y, -x)$\r\n  - Memory tip: \"y becomes new x, negative old x becomes new y\"\r\n\r\n* **Rotation by 180° (Half-turn):**\r\n  - Rule: $P(x, y) \\rightarrow P\'(-x, -y)$\r\n  - Memory tip: \"Both coordinates change sign\"\r\n\r\n### 3. Translation Transformations\r\n**Vector Translation Rule:**\r\n* **Translation by vector $\\begin{pmatrix} a \\\\ b \\end{pmatrix}$:**\r\n  - Rule: $P(x, y) \\rightarrow P\'(x+a, y+b)$\r\n  - Effect: Point moves a units horizontally, b units vertically\r\n\r\n### 4. Enlargement Transformations\r\n**Scaling Rules:**\r\n* **Enlargement from origin with scale factor k:**\r\n  - Rule: $P(x, y) \\rightarrow P\'(kx, ky)$\r\n  - Effect: Distance from origin multiplied by k\r\n\r\n* **Enlargement from center (a,b) with scale factor k:**\r\n  - Rule: $P(x, y) \\rightarrow P\'(k(x-a)+a, k(y-b)+b)$\r\n  - Effect: Point moves relative to center, scaled by k\r\n\r\n---\r\n\r\n## 📚 Worked Examples and Solutions\r\n\r\n### Problem 1: Triangle Translation and Enlargement\r\n**Question:** A triangle ABC with vertices A(2, 3), B(2, 6) and C(3, 4) is translated by T = $\\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$ and the image so obtained is enlarged by E[(0, 0), 2]. Write the co-ordinates of the vertices of the images so formed and represent △ABC and its both images in the same graph.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply translation**\r\nTranslation rule: $P(x, y) \\rightarrow P\'(x-2, y+3)$\r\n- A(2, 3) → A\'(0, 6)\r\n- B(2, 6) → B\'(0, 9) \r\n- C(3, 4) → C\'(1, 7)\r\n\r\n**Step 2: Apply enlargement**\r\nEnlargement rule: $P(x, y) \\rightarrow P\'\'(2x, 2y)$\r\n- A\'(0, 6) → A\'\'(0, 12)\r\n- B\'(0, 9) → B\'\'(0, 18)\r\n- C\'(1, 7) → C\'\'(2, 14)\r\n\r\n**Final Answer:**\r\n- After translation: **A\'(0, 6), B\'(0, 9), C\'(1, 7)**\r\n- After enlargement: **A\'\'(0, 12), B\'\'(0, 18), C\'\'(2, 14)**\r\n\r\n![Graph of Triangle ABC and its transformations](https://seekoguide.com/images/comb-of-transformation/transformation.svg)\r\n\r\n### Problem 2: Combined Transformation with Composition Notation\r\n**Question:** E denotes the enlargement about the centre (3, 1) with a scale factor of 2 and R denotes a reflection on the line y = x. Find the image of △ABC having the vertices A(2, 3), B(4, 5) and C(1, -2) under the combined transformation E∘R. Draw both figure on the same graph paper.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply R (Reflection on y = x)**\r\nReflection rule: $P(x, y) \\rightarrow P\'(y, x)$\r\n- A(2, 3) → A\'(3, 2)\r\n- B(4, 5) → B\'(5, 4) \r\n- C(1, -2) → C\'(-2, 1)\r\n\r\n**Step 2: Apply E (Enlargement from center (3, 1) with scale factor 2)**\r\nEnlargement rule: $P(x, y) \\rightarrow P\'\'(2(x-3)+3, 2(y-1)+1)$\r\n- A\'(3, 2) → A\'\'(2(3-3)+3, 2(2-1)+1) = A\'\'(3, 3)\r\n- B\'(5, 4) → B\'\'(2(5-3)+3, 2(4-1)+1) = B\'\'(7, 7)\r\n- C\'(-2, 1) → C\'\'(2(-2-3)+3, 2(1-1)+1) = C\'\'(-7, 1)\r\n\r\n**Final Answer:**\r\n- After reflection: **A\'(3, 2), B\'(5, 4), C\'(-2, 1)**\r\n- After enlargement: **A\'\'(3, 3), B\'\'(7, 7), C\'\'(-7, 1)**\r\n\r\n![Graph of Triangle ABC under E∘R transformation](https://seekoguide.com/images/comb-of-transformation/transformation-p2.svg)\r\n\r\n### Problem 3: Reflection followed by Rotation\r\n**Question:** Find the coordinates of the image of point A(3, -5) after reflection on the line x + y = 0 followed by rotation through +90° about the origin.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Identify the reflection line**\r\nThe line x + y = 0 can be rewritten as y = -x. This is reflection in the line y = -x.\r\n\r\n**Step 2: Apply reflection**\r\nFor reflection in line y = -x, the rule is: $P(x, y) \\rightarrow P\'(-y, -x)$\r\n- Apply to A(3, -5): $A(3, -5) \\rightarrow A\'(-(-5), -3) = A\'(5, -3)$\r\n\r\n**Step 3: Apply rotation**\r\nFor +90° rotation, the rule is: $P(x, y) \\rightarrow P\'(-y, x)$\r\n- Apply to A\'(5, -3): $A\'(5, -3) \\rightarrow A\'\'(-(-3), 5) = A\'\'(3, 5)$\r\n\r\n**Final Answer:**\r\n- Intermediate image after reflection: **A\'(5, -3)**\r\n- Final image after rotation: **A\'\'(3, 5)**\r\n\r\n### Problem 4: Triangle Reflection and Enlargement\r\n**Question:** Triangle PQR having vertices P(2, 1), Q(5, 3) and R(7, -1) is reflected on the y-axis. The image so formed is enlarged by E[(0, 0), 2]. Write the co-ordinates of the vertices images thus obtained and present the △PQR and its image in same graph paper.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply reflection on y-axis**\r\nReflection rule: $P(x, y) \\rightarrow P\'(-x, y)$\r\n- P(2, 1) → P\'(-2, 1)\r\n- Q(5, 3) → Q\'(-5, 3)\r\n- R(7, -1) → R\'(-7, -1)\r\n\r\n**Step 2: Apply enlargement**\r\nEnlargement rule: $P(x, y) \\rightarrow P\'\'(2x, 2y)$\r\n- P\'(-2, 1) → P\'\'(-4, 2)\r\n- Q\'(-5, 3) → Q\'\'(-10, 6)\r\n- R\'(-7, -1) → R\'\'(-14, -2)\r\n\r\n**Final Answer:**\r\n- After reflection: **P\'(-2, 1), Q\'(-5, 3), R\'(-7, -1)**\r\n- After enlargement: **P\'\'(-4, 2), Q\'\'(-10, 6), R\'\'(-14, -2)**\r\n\r\n![Graph of Triangle PQR and its transformations](https://seekoguide.com/images/comb-of-transformation/22.svg)\r\n\r\n### Problem 5: Parallelogram Rotation and Enlargement\r\n**Question:** L(2, 2), O(6, 2), V(7,4) and E(3, 4) are the vertices of a parallelogram LOVE. Find the coordinates of the vertices of the images of ▱LOVE under the rotation of positive 90° about origin followed by enlargement E[(0,0), 3]. Represent the object and image on the same graph paper.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply +90° rotation**\r\nRotation rule: $P(x, y) \\rightarrow P\'(-y, x)$\r\n- L(2, 2) → L\'(-2, 2)\r\n- O(6, 2) → O\'(-2, 6)\r\n- V(7, 4) → V\'(-4, 7)\r\n- E(3, 4) → E\'(-4, 3)\r\n\r\n**Step 2: Apply enlargement**\r\nEnlargement rule: $P(x, y) \\rightarrow P\'\'(3x, 3y)$\r\n- L\'(-2, 2) → L\'\'(-6, 6)\r\n- O\'(-2, 6) → O\'\'(-6, 18)\r\n- V\'(-4, 7) → V\'\'(-12, 21)\r\n- E\'(-4, 3) → E\'\'(-12, 9)\r\n\r\n**Final Answer:**\r\n- After rotation: **L\'(-2, 2), O\'(-2, 6), V\'(-4, 7), E\'(-4, 3)**\r\n- After enlargement: **L\'\'(-6, 6), O\'\'(-6, 18), V\'\'(-12, 21), E\'\'(-12, 9)**\r\n\r\n![Graph of Parallelogram LOVE and its transformations](https://seekoguide.com/images/comb-of-transformation/23.svg)\r\n\r\n### Problem 6: Triangle with Reflection in Line y = x\r\n**Question:** P(1, 2), Q(2, 1) and R(4, 3) are the vertices of the triangle PQR. Find the coordinates of the images of the triangle PQR under the reflection line y = x followed by the enlargement E[(0, 0), 2]. Present the △PQR and its images on the same graph.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply reflection in y = x**\r\nReflection rule: $P(x, y) \\rightarrow P\'(y, x)$\r\n- P(1, 2) → P\'(2, 1)\r\n- Q(2, 1) → Q\'(1, 2)\r\n- R(4, 3) → R\'(3, 4)\r\n\r\n**Step 2: Apply enlargement**\r\nEnlargement rule: $P(x, y) \\rightarrow P\'\'(2x, 2y)$\r\n- P\'(2, 1) → P\'\'(4, 2)\r\n- Q\'(1, 2) → Q\'\'(2, 4)\r\n- R\'(3, 4) → R\'\'(6, 8)\r\n\r\n**Final Answer:**\r\n- After reflection: **P\'(2, 1), Q\'(1, 2), R\'(3, 4)**\r\n- After enlargement: **P\'\'(4, 2), Q\'\'(2, 4), R\'\'(6, 8)**\r\n\r\n![Graph of Triangle PQR and its transformations](https://seekoguide.com/images/comb-of-transformation/24.svg)\r\n\r\n### Problem 7: Translation followed by Enlargement\r\n**Question:** The vertices of △ABC are A(2, 0), B(3, 1) and C(1, 1). △ABC is translated by vector $\\begin{pmatrix} 2 \\\\ -3 \\end{pmatrix}$. The image so formed is enlarged by E[(0, 0), 2]. Write the coordinates of the image thus obtained and represent the object and the image on the same graph paper.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply translation**\r\nTranslation rule: $P(x, y) \\rightarrow P\'(x+2, y-3)$\r\n- A(2, 0) → A\'(4, -3)\r\n- B(3, 1) → B\'(5, -2)\r\n- C(1, 1) → C\'(3, -2)\r\n\r\n**Step 2: Apply enlargement**\r\nEnlargement rule: $P(x, y) \\rightarrow P\'\'(2x, 2y)$\r\n- A\'(4, -3) → A\'\'(8, -6)\r\n- B\'(5, -2) → B\'\'(10, -4)\r\n- C\'(3, -2) → C\'\'(6, -4)\r\n\r\n**Final Answer:**\r\n- After translation: **A\'(4, -3), B\'(5, -2), C\'(3, -2)**\r\n- After enlargement: **A\'\'(8, -6), B\'\'(10, -4), C\'\'(6, -4)**\r\n\r\n![Graph of Triangle ABC and its transformations](https://seekoguide.com/images/comb-of-transformation/25.svg)\r\n\r\n### Problem 8: Rotation and Enlargement of Triangle CAT\r\n**Question:** C(2, 5), A(-1, 3) and T(4, 1) are the vertices Triangle CAT. Find the coordinates of the vertices of the image of △CAT under the rotation of positive 90° about origin followed by enlargement E[(0,0), 2]. Represent the object and images on the same graph paper.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply +90° rotation**\r\nRotation rule: $P(x, y) \\rightarrow P\'(-y, x)$\r\n- C(2, 5) → C\'(-5, 2)\r\n- A(-1, 3) → A\'(-3, -1)\r\n- T(4, 1) → T\'(-1, 4)\r\n\r\n**Step 2: Apply enlargement**\r\nEnlargement rule: $P(x, y) \\rightarrow P\'\'(2x, 2y)$\r\n- C\'(-5, 2) → C\'\'(-10, 4)\r\n- A\'(-3, -1) → A\'\'(-6, -2)\r\n- T\'(-1, 4) → T\'\'(-2, 8)\r\n\r\n**Final Answer:**\r\n- After rotation: **C\'(-5, 2), A\'(-3, -1), T\'(-1, 4)**\r\n- After enlargement: **C\'\'(-10, 4), A\'\'(-6, -2), T\'\'(-2, 8)**\r\n\r\n![Graph of Triangle CAT and its transformations](https://seekoguide.com/images/comb-of-transformation/26.svg)\r\n\r\n---\r\n\r\n## 🧮 Practice Problems with Solutions\r\n\r\n### Practice Set A: Equivalent Single Transformations\r\n\r\n**Problem A1:** State the single transformation equivalent to the combination of reflections on the X-axis and Y-axis respectively. Using this single transformation find the coordinates of the vertices of the image of △PQR having vertices P(4, 3), Q(1, 1) and R(5, -1). Also draw the object and image on the same graph.\r\n\r\n**Solution:**\r\n**Step 1: Find the equivalent transformation**\r\n- First reflection on x-axis: P(x, y) → P\'(x, -y)\r\n- Second reflection on y-axis: P\'(x, -y) → P\'\'(-x, -y)\r\n- Combined transformation: P(x, y) → P\'\'(-x, -y)\r\n- This is equivalent to **rotation of ±180° about the origin**\r\n\r\n**Step 2: Apply to triangle vertices**\r\n- P(4, 3) → P\'(-4, -3)\r\n- Q(1, 1) → Q\'(-1, -1)\r\n- R(5, -1) → R\'(-5, 1)\r\n\r\n![Graph showing triangle PQR and its image](https://seekoguide.com/images/comb-of-transformation/27.svg)\r\n\r\n**Answer: Single transformation = 180° rotation; Final vertices: P\'(-4, -3), Q\'(-1, -1), R\'(-5, 1)**\r\n\r\n**Problem A2:** A △ABC with vertices A(1, 2), B(4, -1) and C(2, 5) is reflected successively in the line y = x and the X-axis. Find the co-ordinates and graphically represent the image under these transformations. State also the single transformation given by the combinations of these transformations.\r\n\r\n**Solution:**\r\n**Step 1: Reflection in y = x**\r\nRule: P(x, y) → P\'(y, x)\r\n- A(1, 2) → A\'(2, 1)\r\n- B(4, -1) → B\'(-1, 4)\r\n- C(2, 5) → C\'(5, 2)\r\n\r\n**Step 2: Reflection in x-axis**\r\nRule: P(x, y) → P\'\'(x, -y)\r\n- A\'(2, 1) → A\'\'(2, -1)\r\n- B\'(-1, 4) → B\'\'(-1, -4)\r\n- C\'(5, 2) → C\'\'(5, -2)\r\n\r\n**Step 3: Find single equivalent transformation**\r\nCombined rule: P(x, y) → P\'\'(y, -x)\r\nThis is equivalent to **rotation of -90° about the origin**\r\n\r\n![Graph showing triangle ABC and its transformations](https://seekoguide.com/images/comb-of-transformation/29.svg)\r\n\r\n**Answer: Final vertices: A\'\'(2, -1), B\'\'(-1, -4), C\'\'(5, -2); Single transformation = -90° rotation**\r\n\r\n### Practice Set B: Complex Combined Transformations\r\n\r\n**Problem B1:** Translate the △PQR with vertices P(2, 4), Q(-1, 2) and R(5, -1) by translation vector T = $\\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$ and then reflect the image so obtained in the line x + y = 0. Write the coordinates of the vertices of the image and draw the object and images on the same graph paper.\r\n\r\n**Solution:**\r\n**Step 1: Apply translation**\r\nTranslation rule: P(x, y) → P\'(x-2, y+3)\r\n- P(2, 4) → P\'(0, 7)\r\n- Q(-1, 2) → Q\'(-3, 5)\r\n- R(5, -1) → R\'(3, 2)\r\n\r\n**Step 2: Apply reflection in x + y = 0**\r\nFor line x + y = 0 (y = -x), rule: P(x, y) → P\'\'(-y, -x)\r\n- P\'(0, 7) → P\'\'(-7, 0)\r\n- Q\'(-3, 5) → Q\'\'(-5, 3)\r\n- R\'(3, 2) → R\'\'(-2, -3)\r\n\r\n![Graph showing triangle PQR and its transformations](https://seekoguide.com/images/comb-of-transformation/32.svg)\r\n\r\n**Answer: Final vertices: P\'\'(-7, 0), Q\'\'(-5, 3), R\'\'(-2, -3)**\r\n\r\n**Problem B2:** Draw the image in graph that is formed when a triangle with vertices A(1, 0), B(2, 1) and C(3, -1) is translated by $\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix}$. Reflect the image in the line x = 2 and also draw the image in the same graph.\r\n\r\n**Solution:**\r\n**Step 1: Apply translation**\r\nTranslation rule: P(x, y) → P\'(x+1, y+2)\r\n- A(1, 0) → A\'(2, 2)\r\n- B(2, 1) → B\'(3, 3)\r\n- C(3, -1) → C\'(4, 1)\r\n\r\n**Step 2: Apply reflection in x = 2**\r\nRule: P(x, y) → P\'\'(4-x, y)\r\n- A\'(2, 2) → A\'\'(2, 2)\r\n- B\'(3, 3) → B\'\'(1, 3)\r\n- C\'(4, 1) → C\'\'(0, 1)\r\n\r\n![Graph showing triangle ABC and its transformations](https://seekoguide.com/images/comb-of-transformation/33.svg)\r\n\r\n**Answer: Final vertices: A\'\'(2, 2), B\'\'(1, 3), C\'\'(0, 1)**\r\n\r\n### Practice Set C: Advanced Reflection Problems\r\n\r\n**Problem C1:** A triangle with vertices N(1, 2), V(4, -1) and M(2, 5) is reflected successively in the lines x = -1 and y = 2. Find by stating coordinates and represent the images graphically under these transformations. Also state the single transformation given by these transformations.\r\n\r\n**Solution:**\r\n**Step 1: Reflection in x = -1**\r\nRule: P(x, y) → P\'(-2-x, y)\r\n- N(1, 2) → N\'(-3, 2)\r\n- V(4, -1) → V\'(-6, -1)\r\n- M(2, 5) → M\'(-4, 5)\r\n\r\n**Step 2: Reflection in y = 2**\r\nRule: P(x, y) → P\'\'(x, 4-y)\r\n- N\'(-3, 2) → N\'\'(-3, 2)\r\n- V\'(-6, -1) → V\'\'(-6, 5)\r\n- M\'(-4, 5) → M\'\'(-4, -1)\r\n\r\n**Step 3: Find single transformation**\r\nThe combined effect is reflection in line x = -1 (since reflecting in two perpendicular lines gives reflection in the line equidistant from them).\r\n\r\n![Graph showing triangle NVM and its transformations](https://seekoguide.com/images/comb-of-transformation/34.svg)\r\n\r\n**Answer: Final vertices: N\'\'(-3, 2), V\'\'(-6, 5), M\'\'(-4, -1); Single transformation = reflection in x = -1**\r\n\r\n### Practice Set D: Rotation and Enlargement Problems\r\n\r\n**Problem D1:** The vertices of △ABC are A(2, 5), B(-2, 3) and C(4, 1). △ABC is rotated through negative quarter turn about the origin and the image so obtained is enlarged by taking origin as the centre and scale factor 2. Hence, find the co-ordinates of the image of △ABC and show the object and image on the same graph.\r\n\r\n**Solution:**\r\n**Step 1: Apply -90° rotation (negative quarter turn)**\r\nRule: P(x, y) → P\'(y, -x)\r\n- A(2, 5) → A\'(5, -2)\r\n- B(-2, 3) → B\'(3, 2)\r\n- C(4, 1) → C\'(1, -4)\r\n\r\n**Step 2: Apply enlargement from origin with scale factor 2**\r\nRule: P(x, y) → P\'\'(2x, 2y)\r\n- A\'(5, -2) → A\'\'(10, -4)\r\n- B\'(3, 2) → B\'\'(6, 4)\r\n- C\'(1, -4) → C\'\'(2, -8)\r\n\r\n![Graph showing triangle ABC and its transformations](https://seekoguide.com/images/comb-of-transformation/30.svg)\r\n\r\n**Answer: Final vertices: A\'\'(10, -4), B\'\'(6, 4), C\'\'(2, -8)**\r\n\r\n**Problem D2:** K(2, 5), L(-1, 3) and M(4, 1) are the vertices of a triangle KLM. Find the co-ordinates of the vertices of the images of △KLM under the rotation of negative 90° about the origin followed by the enlargement E[(0, 0), 2]. Present object and its images on the same graph paper.\r\n\r\n**Solution:**\r\n**Step 1: Apply -90° rotation**\r\nRule: P(x, y) → P\'(y, -x)\r\n- K(2, 5) → K\'(5, -2)\r\n- L(-1, 3) → L\'(3, 1)\r\n- M(4, 1) → M\'(1, -4)\r\n\r\n**Step 2: Apply enlargement**\r\nRule: P(x, y) → P\'\'(2x, 2y)\r\n- K\'(5, -2) → K\'\'(10, -4)\r\n- L\'(3, 1) → L\'\'(6, 2)\r\n- M\'(1, -4) → M\'\'(2, -8)\r\n\r\n![Graph showing triangle KLM and its transformations](https://seekoguide.com/images/comb-of-transformation/31.svg)\r\n\r\n**Answer: Final vertices: K\'\'(10, -4), L\'\'(6, 2), M\'\'(2, -8)**\r\n\r\n---\r\n\r\n## 🎨 Understanding Order of Operations\r\n\r\n### Why Order Matters\r\n\r\nThe order in which transformations are applied is **crucial**. Different orders can produce completely different results.\r\n\r\n**Example Demonstration:**\r\nConsider point P(2, 1) with two transformations:\r\n- T₁: Reflection in x-axis\r\n- T₂: Translation by $\\begin{pmatrix} 3 \\\\ 0 \\end{pmatrix}$\r\n\r\n**Order 1: T₁ then T₂**\r\n- P(2, 1) → P\'(2, -1) → P\'\'(5, -1)\r\n\r\n**Order 2: T₂ then T₁**\r\n- P(2, 1) → P\'(5, 1) → P\'\'(5, -1)\r\n\r\n**Result:** Same final point, but this is not always the case!\r\n\r\n**Counter-example with different final results:**\r\nPoint P(1, 2) with transformations:\r\n- T₁: +90° rotation\r\n- T₂: Translation by $\\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}$\r\n\r\n**Order 1: T₁ then T₂**\r\n- P(1, 2) → (-2, 1) → (-1, 1)\r\n\r\n**Order 2: T₂ then T₁**\r\n- P(1, 2) → (2, 2) → (-2, 2)\r\n\r\n**Different results!** Order matters significantly.\r\n\r\n---\r\n\r\n## 💡 Problem-Solving Strategies\r\n\r\n### Strategy 1: Step-by-Step Method\r\n1. **Identify** all transformations and their order\r\n2. **Apply** each transformation sequentially\r\n3. **Check** your work by verifying intermediate steps\r\n4. **State** both intermediate and final results clearly\r\n\r\n### Strategy 2: Combined Rule Method\r\n1. **Derive** the single combined transformation rule\r\n2. **Apply** this rule directly to find the final image\r\n3. **Verify** by checking with step-by-step method\r\n\r\n### Strategy 3: Working Backwards\r\n1. **Start** with the final image\r\n2. **Reverse** each transformation in opposite order\r\n3. **Apply** inverse transformations to find original point\r\n\r\n---\r\n\r\n## 📊 Common Mistakes to Avoid\r\n\r\n### Mistake 1: Wrong Order of Operations\r\n❌ **Wrong:** Applying transformations in written order when composition notation indicates reverse order\r\n✅ **Correct:** For f ∘ g, apply g first, then f\r\n\r\n### Mistake 2: Sign Errors in Rotations\r\n❌ **Wrong:** Confusing +90° and -90° rotation rules\r\n✅ **Correct:** \r\n- +90°: (x, y) → (-y, x)\r\n- -90°: (x, y) → (y, -x)\r\n\r\n### Mistake 3: Reflection Line Confusion\r\n❌ **Wrong:** Using wrong reflection formulas\r\n✅ **Correct:** Always identify the reflection line clearly before applying rules\r\n\r\n---\r\n\r\n## 🎯 Examination Tips\r\n\r\n### Time-Saving Techniques\r\n1. **Master the basic rules** - know them by heart\r\n2. **Use systematic notation** - label each intermediate step\r\n3. **Double-check signs** - most errors occur with positive/negative signs\r\n4. **Verify with graphical thinking** - visualize transformations mentally\r\n\r\n### Common Exam Question Types\r\n1. **Direct application**: Apply given transformations in sequence\r\n2. **Finding unknowns**: Given final result, find original point or unknown parameters\r\n3. **Equivalent transformations**: Find single transformation equivalent to combination\r\n4. **Order dependency**: Compare results of different transformation orders\r\n\r\n---\r\n\r\n## 🏆 Advanced Applications\r\n\r\n### Composite Transformations in Real World\r\n- **Computer Graphics**: Rotating and scaling 3D models\r\n- **Engineering**: Coordinate system transformations\r\n- **Physics**: Reference frame changes\r\n- **Art and Design**: Creating patterns and symmetries\r\n\r\n### Extension Topics\r\n- Transformations in 3D space\r\n- Matrix representations of transformations\r\n- Inverse transformations\r\n- Transformation groups in advanced mathematics\r\n\r\n---\r\n\r\n## 📝 Summary Checklist\r\n\r\nBy the end of this guide, you should be able to:\r\n\r\n✅ **Apply basic transformations** (reflection, rotation, translation, enlargement) to points and shapes  \r\n✅ **Combine multiple transformations** in the correct order using proper notation  \r\n✅ **Work with triangles, parallelograms, and other polygons** in transformation problems  \r\n✅ **Handle reflection in various lines** (x-axis, y-axis, y=x, y=-x, x=k, y=k)  \r\n✅ **Execute rotations** (+90°, -90°, 180°) and understand quarter-turns  \r\n✅ **Apply translations** using vector notation and coordinate shifts  \r\n✅ **Perform enlargements** from origin and from other centers with various scale factors  \r\n✅ **Interpret composition notation** (E∘R, r₁∘r₂) correctly  \r\n✅ **Find equivalent single transformations** for combined operations  \r\n✅ **Understand why order matters** in combined transformations  \r\n✅ **Solve for unknown coordinates and parameters** in complex problems  \r\n✅ **Use efficient problem-solving strategies** for multi-step transformations  \r\n✅ **Avoid common mistakes** with signs, order, and transformation rules  \r\n✅ **Work with graphical representations** and coordinate plotting  \r\n✅ **Handle negative quarter-turns** and clockwise/counterclockwise rotations  \r\n✅ **Solve real examination-style problems** with complete step-by-step solutions  \r\n\r\n---\r\n\r\n## 🎓 Final Words\r\n\r\nMastering combined transformations requires **practice, patience, and systematic thinking**. The key to success is:\r\n\r\n1. **Understanding each basic transformation thoroughly**\r\n2. **Practicing the step-by-step approach consistently**\r\n3. **Checking your work at each stage**\r\n4. **Learning from mistakes and understanding why they occur**\r\n\r\nRemember: **Every complex transformation is just a series of simple steps applied correctly in the right order.**\r\n\r\nKeep practicing, stay organized in your approach, and soon you\'ll master even the most complex combined transformation problems! 🚀\r\n\r\n---\r\n\r\n', 1, '2025-07-25 01:11:38', '2025-09-18 14:27:45'),
(155, 14245, '# Circle Inversion: Complete Study Guide for Geometric Inversion\r\n## A Comprehensive Guide to Inversion with Respect to Circles\r\n\r\n---\r\n\r\n## 🎯 Introduction: Understanding Circle Inversion\r\n\r\nWelcome to the complete guide on **Circle Inversion** (also known as **Geometric Inversion**). Circle inversion is a powerful transformation in geometry that maps points inside a circle to points outside the circle and vice versa, while preserving certain geometric properties.\r\n\r\nThis transformation has fascinating applications in:\r\n- Complex analysis and conformal mapping\r\n- Geometric constructions and proofs\r\n- Non-Euclidean geometry\r\n- Physics and engineering problems\r\n- Advanced mathematics competitions\r\n\r\n**Key Learning Objective**: Master the concept of inversion, understand the fundamental inversion formula, and solve complex problems involving inverse points with respect to circles. 📐\r\n\r\n---\r\n\r\n## 🔑 Fundamental Concepts of Circle Inversion\r\n\r\n### What is Circle Inversion?\r\n\r\n**Circle Inversion** is a transformation that maps each point P in the plane (except the center) to a corresponding point P\' such that:\r\n- The center O, point P, and point P\' are **collinear**\r\n- The relationship **OP × OP\' = r²** holds (where r is the radius of the circle of inversion)\r\n\r\n### Basic Properties of Inversion\r\n\r\n1. **Reciprocal Relationship**: If P\' is the inverse of P, then P is also the inverse of P\'\r\n2. **Points on the Circle**: Points on the circle of inversion map to themselves\r\n3. **Inside ↔ Outside**: Points inside the circle map to points outside, and vice versa\r\n4. **Center Behavior**: The center of inversion has no inverse (maps to infinity)\r\n\r\n---\r\n\r\n## 📐 Essential Inversion Formulas\r\n\r\n### 1. For Circle Centered at Origin\r\n\r\n**Circle of Inversion**: $x^2 + y^2 = r^2$\r\n\r\n**Inversion Formula**:\r\n$$P\'(x\', y\') = \\left(\\frac{r^2 x}{x^2 + y^2}, \\frac{r^2 y}{x^2 + y^2}\\right)$$\r\n\r\n**Key Properties**:\r\n- Distance relationship: $|OP| \\times |OP\'| = r^2$\r\n- If $|OP| = d$, then $|OP\'| = \\frac{r^2}{d}$\r\n\r\n### 2. For Circle Centered at Point (h, k)\r\n\r\n**Circle of Inversion**: $(x - h)^2 + (y - k)^2 = r^2$\r\n\r\n**Inversion Formula**:\r\n$$P\'(x\', y\') = \\left(\\frac{r^2(x - h)}{(x - h)^2 + (y - k)^2} + h, \\frac{r^2(y - k)}{(x - h)^2 + (y - k)^2} + k\\right)$$\r\n\r\n**Key Properties**:\r\n- The center (h, k) is the center of inversion\r\n- Distance relationship: $|CP| \\times |CP\'| = r^2$ where C is the center\r\n\r\n---\r\n\r\n## 📚 Worked Examples and Solutions\r\n\r\n### Problem 1: Understanding the Basic Inversion Relationship\r\n**Question:** In the figure, O is centre of circle, OP = 4 units, OQ = 8 units and OP\' = 16 units, write the relation between P and P\'.\r\n\r\n![Circle inversion diagram](https://seekoguide.com/images/inversion/1.svg)\r\n\r\n**Complete Solution:**\r\n\r\n**Given Information:**\r\n- OP = 4 units\r\n- OQ = 8 units (radius of circle)\r\n- OP\' = 16 units\r\n\r\n**Step 1: Check the inversion relationship**\r\nFor inversion with respect to a circle, we must have: $OP \\times OP\' = r^2$\r\n\r\n**Step 2: Verify the relationship**\r\n- $OP \\times OP\' = 4 \\times 16 = 64$\r\n- $r^2 = OQ^2 = 8^2 = 64$\r\n- Since $64 = 64$, the relationship holds!\r\n\r\n**Step 3: State the conclusion**\r\nThe relationship is: $OP \\times OP\' = r^2$\r\n\r\n**Final Answer:** P\' is the **inversion point of P** (and vice versa) with respect to the circle centered at O with radius 8 units.\r\n\r\n### Problem 2: Finding Inverse with Respect to Circle Centered at Origin\r\n**Question:** Find the inverse image of the point (4, 5) with respect to the circle $x^2 + y^2 = 10$.\r\n\r\n**Complete Solution:**\r\n\r\n**Given Information:**\r\n- Circle: $x^2 + y^2 = 10$\r\n- Center of circle: (0, 0)\r\n- Radius: $r^2 = 10$ (so $r = \\sqrt{10}$)\r\n- Object point P(x, y) = (4, 5)\r\n- Inversion point P\'(x\', y\') = ?\r\n\r\n**Step 1: Apply the inversion formula**\r\nFor circle centered at origin:\r\n$$P\'(x\', y\') = \\left(\\frac{r^2 x}{x^2 + y^2}, \\frac{r^2 y}{x^2 + y^2}\\right)$$\r\n\r\n**Step 2: Substitute the values**\r\n$$P\'(x\', y\') = \\left(\\frac{10 \\times 4}{4^2 + 5^2}, \\frac{10 \\times 5}{4^2 + 5^2}\\right)$$\r\n\r\n**Step 3: Calculate the denominator**\r\n$4^2 + 5^2 = 16 + 25 = 41$\r\n\r\n**Step 4: Complete the calculation**\r\n$$P\'(x\', y\') = \\left(\\frac{40}{41}, \\frac{50}{41}\\right)$$\r\n\r\n**Final Answer:** The inverse of point (4, 5) is $\\left(\\frac{40}{41}, \\frac{50}{41}\\right)$\r\n\r\n**Verification:** $OP \\times OP\' = \\sqrt{41} \\times \\frac{\\sqrt{40^2 + 50^2}}{41} = \\sqrt{41} \\times \\frac{\\sqrt{4100}}{41} = \\sqrt{41} \\times \\frac{\\sqrt{41 \\times 100}}{41} = 10$ ✓\r\n\r\n### Problem 3: Finding Inverse with Respect to Circle with Different Center\r\n**Question:** Find the inverse image of the point (3, 4) about the circle $(x - 2)^2 + (y - 2)^2 = 36$.\r\n\r\n**Complete Solution:**\r\n\r\n**Given Information:**\r\n- Circle: $(x - 2)^2 + (y - 2)^2 = 36$\r\n- Center of circle (h, k) = (2, 2)\r\n- Radius: r = 6 (since $r^2 = 36$)\r\n- Object point (x, y) = (3, 4)\r\n- Inversion point (x\', y\') = ?\r\n\r\n**Step 1: Apply the inversion formula for non-origin center**\r\n$$P\'(x\', y\') = \\left(\\frac{r^2(x - h)}{(x - h)^2 + (y - k)^2} + h, \\frac{r^2(y - k)}{(x - h)^2 + (y - k)^2} + k\\right)$$\r\n\r\n**Step 2: Calculate the distance from point to center**\r\n- $(x - h) = 3 - 2 = 1$\r\n- $(y - k) = 4 - 2 = 2$\r\n- $(x - h)^2 + (y - k)^2 = 1^2 + 2^2 = 1 + 4 = 5$\r\n\r\n**Step 3: Substitute into the formula**\r\n$$P\'(x\', y\') = \\left(\\frac{36 \\times 1}{5} + 2, \\frac{36 \\times 2}{5} + 2\\right)$$\r\n\r\n**Step 4: Simplify**\r\n$$P\'(x\', y\') = \\left(\\frac{36}{5} + 2, \\frac{72}{5} + 2\\right)$$\r\n$$P\'(x\', y\') = \\left(\\frac{36 + 10}{5}, \\frac{72 + 10}{5}\\right)$$\r\n$$P\'(x\', y\') = \\left(\\frac{46}{5}, \\frac{82}{5}\\right)$$\r\n\r\n**Final Answer:** The inverse of point (3, 4) about the given circle is $\\left(\\frac{46}{5}, \\frac{82}{5}\\right)$\r\n\r\n---\r\n\r\n## 🧮 Practice Problems with Solutions\r\n\r\n### Practice Set A: Basic Inversion Problems\r\n\r\n**Problem A1:** Find the inverse of point (6, 8) with respect to the circle $x^2 + y^2 = 25$.\r\n\r\n**Solution:**\r\n- Center: (0, 0), $r^2 = 25$\r\n- $x^2 + y^2 = 6^2 + 8^2 = 36 + 64 = 100$\r\n- $P\'(x\', y\') = \\left(\\frac{25 \\times 6}{100}, \\frac{25 \\times 8}{100}\\right) = \\left(\\frac{3}{2}, 2\\right)$\r\n\r\n**Answer: $\\left(\\frac{3}{2}, 2\\right)$**\r\n\r\n**Problem A2:** If P(2, 3) and P\'(a, b) are inverse points with respect to circle $x^2 + y^2 = 26$, find a and b.\r\n\r\n**Solution:**\r\n- $r^2 = 26$, $x^2 + y^2 = 4 + 9 = 13$\r\n- $a = \\frac{26 \\times 2}{13} = 4$\r\n- $b = \\frac{26 \\times 3}{13} = 6$\r\n\r\n**Answer: a = 4, b = 6**\r\n\r\n### Practice Set B: Advanced Inversion Problems\r\n\r\n**Problem B1:** Find the inverse of point (5, 1) with respect to circle $(x - 3)^2 + (y - 2)^2 = 16$.\r\n\r\n**Solution:**\r\n- Center: (3, 2), $r^2 = 16$\r\n- $(x - h) = 5 - 3 = 2$, $(y - k) = 1 - 2 = -1$\r\n- $(x - h)^2 + (y - k)^2 = 4 + 1 = 5$\r\n- $x\' = \\frac{16 \\times 2}{5} + 3 = \\frac{32}{5} + 3 = \\frac{47}{5}$\r\n- $y\' = \\frac{16 \\times (-1)}{5} + 2 = -\\frac{16}{5} + 2 = -\\frac{6}{5}$\r\n\r\n**Answer: $\\left(\\frac{47}{5}, -\\frac{6}{5}\\right)$**\r\n\r\n**Problem B2:** Show that if two points are inverse with respect to a circle, then the circle is the perpendicular bisector of the line segment joining the two points when extended.\r\n\r\n**Solution:**\r\nThis is a theorem in inversion geometry. The proof involves showing that any circle orthogonal to the circle of inversion passes through both P and P\', making the inversion circle the locus of centers of all such circles.\r\n\r\n### Practice Set C: Applications and Properties\r\n\r\n**Problem C1:** Find the equation of the circle that passes through point (1, 2) and its inverse with respect to circle $x^2 + y^2 = 9$.\r\n\r\n**Solution:**\r\n**Step 1:** Find inverse of (1, 2)\r\n- $P\' = \\left(\\frac{9 \\times 1}{5}, \\frac{9 \\times 2}{5}\\right) = \\left(\\frac{9}{5}, \\frac{18}{5}\\right)$\r\n\r\n**Step 2:** Any circle passing through both points and orthogonal to $x^2 + y^2 = 9$\r\n- The required circle has the form: $x^2 + y^2 + Dx + Ey + F = 0$ where it\'s orthogonal to the inversion circle.\r\n\r\n**Problem C2:** Prove that inversion preserves angles.\r\n\r\n**Solution:**\r\nThis is a fundamental theorem: **Inversion is a conformal transformation**, meaning it preserves angles between curves at their intersection points.\r\n\r\n---\r\n\r\n## 🎨 Special Properties of Inversion\r\n\r\n### Geometric Properties Preserved Under Inversion\r\n\r\n1. **Angle Preservation**: Angles between curves are preserved\r\n2. **Circle/Line Mapping**: \r\n   - Circles through the center → Lines not through the center\r\n   - Circles not through the center → Circles not through the center\r\n   - Lines through the center → Lines through the center\r\n   - Lines not through the center → Circles through the center\r\n\r\n### Important Theorems\r\n\r\n**Theorem 1**: The inverse of a line passing through the center of inversion is the line itself.\r\n\r\n**Theorem 2**: The inverse of a line not passing through the center of inversion is a circle passing through the center of inversion.\r\n\r\n**Theorem 3**: The inverse of a circle passing through the center of inversion is a line not passing through the center.\r\n\r\n**Theorem 4**: The inverse of a circle not passing through the center of inversion is another circle not passing through the center.\r\n\r\n---\r\n\r\n## 💡 Problem-Solving Strategies\r\n\r\n### Strategy 1: Direct Formula Application\r\n1. **Identify** the circle of inversion (center and radius)\r\n2. **Determine** which formula to use (origin-centered or general)\r\n3. **Substitute** coordinates carefully\r\n4. **Simplify** step by step\r\n5. **Verify** using the fundamental relationship\r\n\r\n### Strategy 2: Geometric Approach\r\n1. **Draw** the configuration\r\n2. **Use** the property $OP \\times OP\' = r^2$\r\n3. **Apply** similar triangles when appropriate\r\n4. **Check** collinearity of O, P, and P\'\r\n\r\n### Strategy 3: Verification Method\r\n1. **Calculate** distances OP and OP\'\r\n2. **Verify** that $OP \\times OP\' = r^2$\r\n3. **Check** that O, P, P\' are collinear\r\n\r\n---\r\n\r\n## 📊 Common Mistakes to Avoid\r\n\r\n### Mistake 1: Wrong Formula Selection\r\n❌ **Wrong:** Using origin formula for non-origin centered circles\r\n✅ **Correct:** Always check the center of the circle of inversion\r\n\r\n### Mistake 2: Arithmetic Errors\r\n❌ **Wrong:** Making calculation errors in fractions\r\n✅ **Correct:** Double-check all arithmetic, especially with fractions\r\n\r\n### Mistake 3: Sign Confusion\r\n❌ **Wrong:** Missing negative signs in coordinates\r\n✅ **Correct:** Pay careful attention to coordinate signs\r\n\r\n### Mistake 4: Distance Calculation\r\n❌ **Wrong:** Using wrong distance in non-origin cases\r\n✅ **Correct:** Always calculate distance from the center of inversion\r\n\r\n---\r\n\r\n## 🎯 Examination Tips\r\n\r\n### Time-Saving Techniques\r\n1. **Master both formulas** - origin and general center cases\r\n2. **Practice fraction arithmetic** - most answers involve fractions\r\n3. **Use verification** - check your answer with $OP \\times OP\' = r^2$\r\n4. **Draw diagrams** when possible for geometric insight\r\n\r\n### Common Exam Question Types\r\n1. **Direct inversion**: Find inverse of given point\r\n2. **Finding unknowns**: Given one point and its inverse, find missing coordinates\r\n3. **Relationship verification**: Prove two points are inverse\r\n4. **Applications**: Using inversion in geometric constructions\r\n\r\n---\r\n\r\n## 🏆 Advanced Applications\r\n\r\n### Applications in Mathematics\r\n- **Complex Analysis**: Möbius transformations\r\n- **Hyperbolic Geometry**: Poincaré disk model\r\n- **Geometric Proofs**: Elegant solutions using inversion\r\n- **Conformal Mapping**: Preserving angles in transformations\r\n\r\n### Applications in Physics and Engineering\r\n- **Electrostatics**: Method of images\r\n- **Fluid Dynamics**: Flow around obstacles\r\n- **Optics**: Lens design and ray tracing\r\n- **Antenna Theory**: Radiation patterns\r\n\r\n---\r\n\r\n## 📝 Summary Checklist\r\n\r\nBy the end of this guide, you should be able to:\r\n\r\n✅ **Understand the concept** of circle inversion and its geometric meaning  \r\n✅ **Apply inversion formulas** for both origin-centered and general circles  \r\n✅ **Find inverse points** given any point and circle of inversion  \r\n✅ **Verify inversion relationships** using the fundamental property $OP \\times OP\' = r^2$  \r\n✅ **Work with fractional coordinates** accurately in inversion problems  \r\n✅ **Understand geometric properties** preserved under inversion  \r\n✅ **Solve complex inversion problems** involving multiple steps  \r\n✅ **Apply inversion** to geometric constructions and proofs  \r\n✅ **Recognize applications** of inversion in advanced mathematics  \r\n✅ **Use verification methods** to check solutions  \r\n\r\n---\r\n\r\n## 🎓 Final Words\r\n\r\nCircle inversion is one of the most elegant and powerful transformations in geometry. It provides:\r\n\r\n1. **Beautiful geometric insights** into the nature of circles and lines\r\n2. **Powerful problem-solving tools** for complex geometric constructions\r\n3. **Deep connections** to advanced areas of mathematics\r\n4. **Practical applications** in physics and engineering\r\n\r\n**Key to Success:**\r\n- **Practice the formulas** until they become second nature\r\n- **Understand the geometric meaning** behind the algebraic calculations  \r\n- **Visualize the transformation** - points inside map outside and vice versa\r\n- **Use verification** as a powerful check on your work\r\n\r\nRemember: **Inversion turns the inside out and the outside in, but angles stay the same!** \r\n\r\nThis transformation has fascinated mathematicians for centuries and continues to provide elegant solutions to seemingly difficult problems. Master it, and you\'ll have a powerful tool in your mathematical toolkit! 🚀\r\n\r\n---\r\n\r\n*This guide provides comprehensive coverage of circle inversion with detailed worked examples, practice problems, and advanced applications. Use it as your complete reference for mastering geometric inversion.*\r\n', 1, '2025-07-25 01:11:38', '2025-09-18 14:27:45');
