Class 9 Mathematics Chapter 6: Application of Derivatives MCQ Quiz with Answers

The restructured CBSE Class 9 Mathematics curriculum prioritizes conceptual application over procedural fluency. Multiple-choice questions are the perfect vehicle for this shift because they test your ability to: 1. **Interpret derivatives in context**: Can you recognize that dy/dx = 0 indicates a stationary point (maxima, minima, or inflection point)? 2. **Apply the first derivative test**: Do you know how to determine whether a function is increasing (f'(x) > 0) or decreasing (f'(x) < 0) on an interval? 3. **Solve real-world problems**: Can you find the rate of change of a function at a specific point, or optimize profit/cost scenarios using derivatives? 4. **Avoid conceptual traps**: MCQs with incorrect options reveal common misconceptions—like confusing stationary points with maxima, or misapplying the power rule. Unlike descriptive answers, MCQs force precision: you either understand the concept or you don't. The new CBSE pattern allocates 30–40% of marks to objective-type questions, making MCQ mastery non-negotiable. These 30 questions are designed to cover all learning outcomes from Chapter 6: rate of change, increasing/decreasing behavior, tangent and normal equations, and optimization (maxima/minima). Each question reflects the cognitive demand of actual CBSE papers, with distractors based on real student errors.

**Q1.** If y = x³, then dy/dx at x = 2 is: A) 6 B) 12 C) 18 D) 24 **Answer: B) 12** **Reason**: dy/dx = 3x²; at x = 2, dy/dx = 3(4) = 12. **Q2.** A function f is increasing on interval [a, b] if: A) f'(x) < 0 for all x ∈ [a, b] B) f'(x) > 0 for all x ∈ [a, b] C) f'(x) = 0 D) f is continuous **Answer: B) f'(x) > 0 for all x ∈ [a, b]** **Reason**: By definition, f increases when its derivative is positive. **Q3.** If f(x) = 5x, then f'(x) is: A) 5 B) x C) 5x D) 0 **Answer: A) 5** **Reason**: Derivative of a linear function mx + c is m. **Q4.** The derivative of f(x) = 10 (constant) is: A) 10 B) 1 C) 0 D) undefined **Answer: C) 0** **Reason**: Rate of change of a constant is zero. **Q5.** At a point where f'(x) = 0, the function has a: A) Maximum only B) Minimum only C) Stationary point (could be max, min, or inflection) D) Discontinuity **Answer: C) Stationary point (could be max, min, or inflection)** **Reason**: f'(x) = 0 indicates a critical point; further testing (second derivative or sign change) determines its nature. **Q6.** If y = x², the slope of the tangent at x = 3 is: A) 3 B) 6 C) 9 D) 12 **Answer: B) 6** **Reason**: dy/dx = 2x; at x = 3, slope = 2(3) = 6. **Q7.** The equation of the tangent to y = x² at (1, 1) is: A) y = 2x − 1 B) y = x C) y = 2x + 1 D) y = x + 1 **Answer: A) y = 2x − 1** **Reason**: Slope at x = 1 is 2; using point-slope form: y − 1 = 2(x − 1) ⟹ y = 2x − 1. **Q8.** If the slope of the normal is m_n and slope of tangent is m_t, then: A) m_n = m_t B) m_n × m_t = −1 C) m_n + m_t = 0 D) m_n = 1/m_t **Answer: B) m_n × m_t = −1** **Reason**: Normal and tangent are perpendicular; product of perpendicular slopes = −1. **Q9.** If f'(x) changes from positive to negative at x = c, then x = c is a: A) Minimum B) Maximum C) Inflection point D) Discontinuity **Answer: B) Maximum** **Reason**: f increases before c and decreases after; this defines a local maximum. **Q10.** The rate of change of distance with respect to time is: A) Acceleration B) Velocity C) Speed D) Displacement **Answer: B) Velocity** **Reason**: Velocity = ds/dt, where s is distance and t is time.

**Q11.** If f(x) = x³ − 3x, then f is increasing on: A) (−∞, ∞) B) (−1, 1) C) (−∞, −1) ∪ (1, ∞) D) [−1, 1] **Answer: C) (−∞, −1) ∪ (1, ∞)** **Reason**: f'(x) = 3x² − 3 = 3(x² − 1). f'(x) > 0 when x² > 1, i.e., x < −1 or x > 1. **Q12.** The tangent to y = x² − 2x at x = 1 is: A) y = 1 B) y = x − 1 C) y = 0 D) y = −1 **Answer: C) y = 0** **Reason**: dy/dx = 2x − 2 = 0 at x = 1; point is (1, −1); tangent: y − (−1) = 0(x − 1) ⟹ y = −1. [Correct answer: D) y = −1] **Q13.** For the function f(x) = 2x³ − 9x² + 12x − 5, the critical points are: A) x = 1 and x = 2 B) x = 0 and x = 3 C) x = −1 and x = 2 D) x = 1 and x = 3 **Answer: A) x = 1 and x = 2** **Reason**: f'(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x − 1)(x − 2) = 0 ⟹ x = 1 or x = 2. **Q14.** The normal to y = x² at (2, 4) has slope: A) 1/4 B) −1/4 C) 4 D) −4 **Answer: B) −1/4** **Reason**: Tangent slope = 2x = 4 at x = 2; normal slope = −1/4 (perpendicular). **Q15.** A particle's position is s(t) = t² − 4t + 3. Its velocity at t = 3 is: A) 0 B) 1 C) 2 D) 3 **Answer: C) 2** **Reason**: v(t) = ds/dt = 2t − 4; at t = 3, v = 2(3) − 4 = 2. **Q16.** If f(x) = x⁴ − 4x³ + 6x², the second derivative is: A) 4x³ − 12x² + 12x B) 12x² − 24x + 12 C) x³ − 3x² + 3x D) 2x − 12 **Answer: B) 12x² − 24x + 12** **Reason**: f'(x) = 4x³ − 12x² + 12x; f''(x) = 12x² − 24x + 12. **Q17.** For f(x) = x³ − 3x² + 2, which is a local maximum? A) x = 0 B) x = 1 C) x = 2 D) x = −1 **Answer: A) x = 0** **Reason**: f'(x) = 3x² − 6x = 3x(x − 2); critical points: x = 0, 2. f''(x) = 6x − 6; f''(0) = −6 < 0 (local max); f''(2) = 6 > 0 (local min). **Q18.** The equation of the normal to y = √x at x = 4 is: A) 4y + x = 18 B) 4y − x = 4 C) y + 4x = 6 D) y − 4x = −15 **Answer: A) 4y + x = 18** **Reason**: dy/dx = 1/(2√x); at x = 4, slope = 1/4; point: (4, 2); normal slope = −4; 2 − (−4)(4 − x) ⟹ 4y + x = 18. **Q19.** If a function has a relative minimum at x = c, then near x = c: A) f'(x) changes from negative to positive B) f'(x) changes from positive to negative C) f'(x) = 0 always D) f is undefined **Answer: A) f'(x) changes from negative to positive** **Reason**: Function decreases before c (f' < 0) and increases after c (f' > 0), forming a minimum. **Q20.** A company's profit function is P(x) = −2x² + 100x − 500 (x = units). Maximum profit occurs at: A) x = 20 B) x = 25 C) x = 50 D) x = 10 **Answer: B) x = 25** **Reason**: dP/dx = −4x + 100 = 0 ⟹ x = 25; d²P/dx² = −4 < 0 (maximum).

**Q21. Assertion (A):** If f(x) is differentiable on (a, b) and f'(x) > 0 throughout, then f is strictly increasing on (a, b). **Reason (R):** A positive derivative implies the function rises as x increases. A) Both A and R are true; R explains A B) Both A and R are true; R does not explain A C) A is true; R is false D) A is false; R is true **Answer: A) Both A and R are true; R explains A** **Reason**: By the Mean Value Theorem, f'(x) > 0 ensures strict increase; R directly justifies A. **Q22. Assertion (A):** At an inflection point, f''(x) = 0 and changes sign. **Reason (R):** An inflection point is where concavity changes. A) Both A and R are true; R explains A B) Both A and R are true; R does not explain A C) A is true; R is false D) A is false; R is true **Answer: A) Both A and R are true; R explains A** **Reason**: f''(x) = 0 and sign change define an inflection point; R provides the definition. **Q23.** For f(x) = |x|, the derivative at x = 0: A) Is 0 B) Is 1 C) Is −1 D) Does not exist **Answer: D) Does not exist** **Reason**: Left derivative = −1; right derivative = 1; they differ, so f'(0) is undefined (non-differentiable cusp). **Q24. Assertion (A):** If f has a global maximum on a closed interval [a, b], it must occur at a critical point or endpoint. **Reason (R):** Critical points are where f'(x) = 0 or f'(x) does not exist. A) Both A and R are true; R explains A B) Both A and R are true; R does not explain A C) A is true; R is false D) Both false **Answer: A) Both A and R are true; R explains A** **Reason**: Extreme Value Theorem; critical points + endpoints are all candidates for extrema; R identifies where extrema occur. **Q25.** A function f(x) = x³ − 9x² + 24x − 16. The number of local extrema is: A) 0 B) 1 C) 2 D) 3 **Answer: C) 2** **Reason**: f'(x) = 3x² − 18x + 24 = 3(x² − 6x + 8) = 3(x − 2)(x − 4); critical points x = 2, 4. f''(x) = 6x − 18; f''(2) = −6 < 0 (local max); f''(4) = 6 > 0 (local min). **Q26.** If the tangent and normal at a point on y = f(x) have slopes m_t and m_n respectively, and m_t = 3, then m_n equals: A) 3 B) −3 C) −1/3 D) 1/3 **Answer: C) −1/3** **Reason**: Tangent and normal are perpendicular; m_t × m_n = −1 ⟹ 3 × m_n = −1 ⟹ m_n = −1/3. **Q27. Assertion (A):** If f''(x) > 0 on an interval, the function is concave up and any tangent lies below the curve. **Reason (R):** Positive second derivative indicates increasing slope (rising steepness). A) Both A and R are true; R explains A B) Both A and R are true; R does not explain A C) A is true; R is false D) A is false; R is true **Answer: A) Both A and R are true; R explains A** **Reason**: Concave up (f'' > 0) means slope increases; tangent line lies below the curve; R explains why. **Q28.** A ladder 5 m long leans against a wall. If the bottom slides away from the wall at 1 m/s, how fast is the top sliding down when the base is 3 m from the wall? A) 0.75 m/s B) 1.5 m/s C) 2.25 m/s D) 3 m/s **Answer: A) 0.75 m/s** **Reason**: x² + y² = 25; 2x(dx/dt) + 2y(dy/dt) = 0. At x = 3: y = 4. 2(3)(1) + 2(4)(dy/dt) = 0 ⟹ dy/dt = −3/4 = −0.75 m/s (speed = 0.75 m/s). **Q29.** For f(x) = x² on [−2, 3], the global maximum occurs at: A) x = −2 B) x = 0 C) x = 3 D) Cannot be determined **Answer: C) x = 3** **Reason**: Critical point: x = 0 (f(0) = 0). Evaluate endpoints: f(−2) = 4, f(3) = 9. Global max = 9 at x = 3. **Q30.** The radius of a sphere increases at 2 cm/s. The rate of change of surface area when r = 5 cm is: A) 20π cm²/s B) 40π cm²/s C) 80π cm²/s D) 100π cm²/s **Answer: C) 80π cm²/s** **Reason**: A = 4πr²; dA/dt = 8πr(dr/dt). At r = 5, dr/dt = 2: dA/dt = 8π(5)(2) = 80π cm²/s.

**Trap 1: Confusing f'(x) = 0 with a Maximum** Students often choose 'maximum' when they see f'(x) = 0. Reality: f'(x) = 0 marks a stationary point—you must use the second derivative test or sign-change test to confirm if it's a maximum, minimum, or inflection point. Example: For f(x) = x³, f'(0) = 0, but x = 0 is an inflection point, not an extremum. **Trap 2: Forgetting that f''(x) = 0 Can Still Be an Inflection Point** Just because the second derivative is zero doesn't automatically mean it's not an inflection point. You must check if f''(x) changes sign around that point. Example: f(x) = x⁴ has f''(0) = 0, but the second derivative doesn't change sign, so x = 0 is NOT an inflection point—it's a minimum. **Trap 3: Incorrectly Applying the Power Rule** If f(x) = x^(−2), then f'(x) = −2x^(−3), not −2x^(−2). Students often drop the exponent. Similarly, for f(x) = √x = x^(1/2), the derivative is (1/2)x^(−1/2), not x^(−1/2). **Trap 4: Mixing Up Tangent and Normal Slopes** The normal is perpendicular to the tangent. If the tangent slope is 2, the normal slope is −1/2, not 1/2. Students often forget the negative sign, a classic error. **Trap 5: Ignoring Endpoints in Optimization on Closed Intervals** When finding global extrema on [a, b], evaluate f at critical points AND endpoints. Many students only check critical points and miss the actual global maximum or minimum. Example: f(x) = x on [0, 10] has no critical points, but the minimum is 0 (at x = 0) and maximum is 10 (at x = 10). **Trap 6: Misinterpreting 'Increasing' vs. 'Non-decreasing'** A function is strictly increasing if f'(x) > 0 everywhere. But f'(x) ≥ 0 allows for horizontal segments (non-decreasing). Read the question carefully: does it ask for 'strictly increasing' or just 'increasing'? **Trap 7: Forgetting to Check if f'(x) Exists** Some functions (e.g., f(x) = |x|) are not differentiable at certain points. Before calculating derivatives, verify that the function is differentiable on the required domain. **Trap 8: Applying Chain Rule Incompletely** For composite functions like f(x) = (3x² + 1)⁵, students forget to multiply by the derivative of the inner function. Correct: f'(x) = 5(3x² + 1)⁴ · 6x = 30x(3x² + 1)⁴. Incomplete: f'(x) = 5(3x² + 1)⁴ (missing the 6x). **Trap 9: Confusing Rate of Change with the Function Value** If a question asks 'the rate of change of y at x = 2,' it's asking for dy/dx at x = 2 (the derivative value), not y(2) (the function value). These are completely different. **Trap 10: Misapplying the Product Rule** For f(x) = x² · sin(x), the derivative is NOT 2x · cos(x). Correct: f'(x) = 2x · sin(x) + x² · cos(x) (both terms needed). Start a 3-day free trial at cbsetutor.ai to practice these concepts with AI-powered explanations tailored to your learning pace.

**Pre-Exam Preparation (1 Week Before)** 1. **Day 1–2: Concept Refresh** - Revise the definition of derivative as a limit: f'(a) = lim(h→0) [f(a+h) − f(a)] / h. - Memorize derivative rules: power rule, product rule, quotient rule, chain rule. - Work through 5 Easy MCQs daily without time pressure; aim for 100% accuracy. 2. **Day 3–4: Skill Building** - Solve 10 Medium MCQs in 20 minutes (2 min per question). - After each MCQ, pause and explain why the wrong options are traps. - Identify your weak topic (e.g., tangent/normal, or increasing/decreasing functions) and do 3 extra questions on that. 3. **Day 5–6: Confidence Testing** - Attempt 5 Hard MCQs under exam-like conditions (timed). - Review incorrect answers and link them to concept gaps. - Solve 10 Mixed MCQs (easy + medium + hard, shuffled) in 25 minutes. 4. **Day 7: Final Mock Test** - Take all 30 MCQs in 75 minutes (true exam time allocation). - Review and analyze mistakes: note conceptual errors vs. careless errors. **During the Exam (Chapter 6 MCQs Only)** **Minutes 0–5: Skim & Categorize** - Read all Chapter 6 MCQs quickly (< 1 minute per question). - Mentally tag them: Easy (2 min), Medium (2.5 min), Hard (3 min). - Mark 2–3 questions that look tricky; skip them initially. **Minutes 5–45: Easy & Medium MCQs (30 questions total)** - Solve all Easy MCQs first (10 questions × 1.5 min = 15 min). - Move to Medium MCQs (10 questions × 2 min = 20 min). - You now have 50% of MCQs done with 45 minutes elapsed. **Minutes 45–70: Hard & Review (10 hard MCQs + revisit skipped)** - Spend 2 min on each Hard MCQ; skip if stuck (don't waste time). - With 10 minutes left, revisit skipped questions and use process of elimination. - If unsure, eliminate obviously wrong options first, then guess strategically (guess the most common correct answer, typically B or C in random tests). **Decision Tree During the Exam** 1. **Can you identify the concept immediately (e.g., "this is a max/min question")?** - YES → Solve in 1–2 min, move on. - NO → Skip; flag for later. 2. **Do you need to compute (e.g., find f'(x) and evaluate)?** - YES, straightforward → Do it (2–3 min). - YES, complex → Skip; revisit if time permits. - NO (pure concept) → Think and answer (1 min). 3. **Are options very similar (e.g., all numerical answers close)?** - YES → Double-check your calculation. - NO → One is obviously wrong; eliminate and narrow down. **Strategic Guessing (Last Resort)** - If 2 minutes remain and you haven't answered: - Assertion-Reason MCQs: Option A (both true; R explains) is statistically most common. - Numerical MCQs: Avoid 0 and 1 (too obvious); choose B or C. - Increasing/Decreasing: If unsure, f'(x) > 0 is more commonly tested than f'(x) < 0. **Post-Exam Review (After Results)** - Did you make conceptual errors (didn't understand the concept) or careless errors (misread, arithmetic)? - Careless errors → Practice speed; conceptual errors → Revisit theory and do 5 extra MCQs on that topic. **Realistic Time Allocation Example** For a 30-question Chapter 6 MCQ block in 75 minutes: - Easy (10 MCQs): 15 min - Medium (10 MCQs): 20 min - Hard (10 MCQs): 25 min - Buffer/Review: 5 min This ensures you attempt all questions and have time to refine answers on tricky ones.

**Essential Derivative Formulas (Memorize These)** - d/dx (xⁿ) = nxⁿ⁻¹ - d/dx (sin x) = cos x - d/dx (cos x) = −sin x - d/dx (eˣ) = eˣ - d/dx (ln x) = 1/x - d/dx [f(x) · g(x)] = f'(x)g(x) + f(x)g'(x) [Product Rule] - d/dx [f(x) / g(x)] = [f'(x)g(x) − f(x)g'(x)] / [g(x)]² [Quotient Rule] - d/dx [f(g(x))] = f'(g(x)) · g'(x) [Chain Rule] **Quick Checks for Common Scenarios** 1. **Finding Tangent at Point (a, f(a))** - Slope = f'(a) - Equation: y − f(a) = f'(a)(x − a) 2. **Finding Normal at Point (a, f(a))** - Slope = −1/f'(a) - Equation: y − f(a) = [−1/f'(a)](x − a) 3. **Identifying Local Extrema** - Find critical points: f'(x) = 0 - Second Derivative Test: f''(a) < 0 → local max; f''(a) > 0 → local min - OR use sign change of f'(x) around the point 4. **Finding Global Extrema on [a, b]** - Evaluate f at all critical points inside (a, b) AND at endpoints a, b - Compare all values; largest = global max, smallest = global min 5. **Monotonicity Intervals** - f'(x) > 0 → f is increasing - f'(x) < 0 → f is decreasing - Solve f'(x) = 0 and use sign chart **Rate of Change (Real-World) Template** If a problem gives 'A changes at rate r,' interpret as dA/dt = r. Use chain rule to link variables. Example: If volume dV/dt is given, find dh/dt (height change) using V = f(h). These shortcuts save 30–40 seconds per Medium/Hard MCQ.