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Chapter 1 · Class 12 Mathematics

Application of Derivatives — Important Questions

34 questions With answers CBSE format

SUMMARY: This chapter focuses on the various applications of derivatives in different fields such as rate of change, increasing and decreasing functions, tangents and normals, and optimization problems.
KEY TOPICS: rate of change of quantities, increasing and decreasing functions, tangents and normals, maxima and minima, first derivative test, second derivative test, Rolle's Theorem, Lagrange's Mean Value Theorem, optimization problems.

Next chapter Application of Integrals
Multiple Choice Questions (MCQs) 5 questions
Q1 1 Mark

The slope of the tangent to y = x² at x = 2 is:

A2
B4
C8
D1
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Correct answer: Option 2 — 4
Q2 1 Mark

The function f(x) = x³ − 3x is decreasing in:

A(−1, 1)
B(−∞, −1) ∪ (1, ∞)
C(0, ∞)
D(−∞, 0)
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Correct answer: Option 1 — (−1, 1)
Q3 1 Mark

The maximum value of f(x) = sin x + cos x on R is:

A1
B√2
C2
D√3
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Correct answer: Option 2 — √2
Q4 1 Mark

If the radius of a circle is increasing at 0.5 cm/s, the rate of change of area when r = 4 cm is:

A2π cm²/s
B4π cm²/s
C8π cm²/s
D16π cm²/s
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Correct answer: Option 2 — 4π cm²/s
Q5 1 Mark

By the first derivative test, f(x) has a local maximum at x = c if:

Af'(c) > 0
Bf'(c) < 0
Cf'(c) = 0 and f' changes from + to − at c
Df'(c) = 0 and f' changes from − to + at c
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Correct answer: Option 3 — f'(c) = 0 and f' changes from + to − at c
Short Answer Questions 5 questions
Q6 3 Marks

Find the equation of the tangent to y = x² at the point (1, 1).

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dy/dx = 2x. Slope at x = 1 is 2. Tangent line: y − 1 = 2(x − 1) ⇒ y = 2x − 1.
Q7 3 Marks

Find the intervals in which f(x) = x² − 4x + 5 is increasing and decreasing.

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f'(x) = 2x − 4. f'(x) > 0 ⇔ x > 2 (increasing on (2, ∞)). f'(x) < 0 ⇔ x < 2 (decreasing on (−∞, 2)). Minimum at x = 2.
Q8 3 Marks

Find the absolute maximum and minimum of f(x) = 2x³ − 15x² + 36x + 1 on [1, 5].

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f'(x) = 6x² − 30x + 36 = 6(x − 2)(x − 3). Critical points x = 2, 3. Values: f(1) = 24, f(2) = 29, f(3) = 28, f(5) = 56. Absolute max = 56 at x = 5; absolute min = 24 at x = 1.
Q9 3 Marks

The radius of a balloon is increasing at 0.5 cm/s. Find the rate of change of volume when r = 5 cm.

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V = (4/3) π r³. dV/dt = 4 π r² (dr/dt). At r = 5, dr/dt = 0.5: dV/dt = 4π · 25 · 0.5 = 50π cm³/s.
Q10 3 Marks

State Rolle's theorem.

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If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there exists at least one c ∈ (a, b) such that f'(c) = 0. Geometrically: somewhere on the curve there is a horizontal tangent.
Long Answer Questions 6 questions
Q11 6 Marks

Find the equation of the tangent and normal to the curve y = x³ at the point (1, 1).

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dy/dx = 3x². Slope of tangent at (1,1) is 3. Tangent equation: y − 1 = 3(x − 1) ⇒ y = 3x − 2. Slope of normal = −1/3. Normal equation: y − 1 = (−1/3)(x − 1) ⇒ 3y − 3 = −(x − 1) ⇒ x + 3y = 4.
Q12 6 Marks

Find the maxima and minima of f(x) = x³ − 6x² + 9x + 15. Identify intervals of increase and decrease.

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f'(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3). Critical points: x = 1, x = 3. f''(x) = 6x − 12. At x = 1: f''(1) = −6 < 0 → local maximum, f(1) = 1 − 6 + 9 + 15 = 19. At x = 3: f''(3) = 6 > 0 → local minimum, f(3) = 27 − 54 + 27 + 15 = 15. Sign of f'(x): positive on (−∞, 1), negative on (1, 3), positive on (3, ∞). So f is increasing on (−∞, 1) ∪ (3, ∞) and decreasing on (1, 3).
Q13 6 Marks

A man 2 m tall walks at 5 km/h away from a 6-m-tall lamp post. Find the rate at which his shadow lengthens.

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Let x = distance of man from post, y = length of his shadow. Triangles similar: (man height)/y = (post height)/(x + y) ⇒ 2/y = 6/(x+y) ⇒ 2(x + y) = 6y ⇒ 2x + 2y = 6y ⇒ x = 2y, i.e. y = x/2. Differentiate: dy/dt = (1/2)(dx/dt) = (1/2)(5) = 2.5 km/h.
Q14 6 Marks

Find the absolute max and min of f(x) = sin x + cos x on [0, 2π].

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f'(x) = cos x − sin x = 0 ⇒ tan x = 1 ⇒ x = π/4, 5π/4 in [0, 2π]. Values: f(0) = 1, f(π/4) = √2 (max), f(5π/4) = −√2 (min), f(2π) = 1. Absolute maximum is √2 at x = π/4; absolute minimum is −√2 at x = 5π/4.
Q15 6 Marks

Prove that the function f(x) = x³ − 3x² + 6x − 100 is increasing on R.

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f'(x) = 3x² − 6x + 6 = 3(x² − 2x + 2) = 3((x − 1)² + 1) > 0 for all x ∈ R. Since f' is positive everywhere, f is strictly increasing on R. Hence proved.
Q16 6 Marks

Differentiate between local maxima and local minima of a function in tabular form.

Assertion–Reason Questions 5 questions
Q17 1 Mark

Assertion (A): The slope of the tangent to a curve y = f(x) at a point is f'(x) at that point.

Reason (R): The derivative gives the instantaneous rate of change of y with respect to x.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q18 1 Mark

Assertion (A): A function is strictly increasing on an interval if its derivative is positive there.

Reason (R): A positive derivative implies the function value rises with x.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q19 1 Mark

Assertion (A): At a local maximum the first derivative is zero and changes sign from positive to negative.

Reason (R): A maximum corresponds to the function turning from rising to falling.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q20 1 Mark

Assertion (A): If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then f'(c) = 0 for some c in (a, b).

Reason (R): The function takes the same value at the endpoints, so it must turn somewhere in between.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q21 1 Mark

Assertion (A): At a critical point f'(c) = 0 or f'(c) does not exist.

Reason (R): Local extrema and inflection candidates occur only at critical points.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Statement-Based Questions 5 questions
Q22 1 Mark

Statement 1: The slope of the tangent to a curve at a point equals the derivative at that point.

Statement 2: The slope of the normal at the same point equals the negative reciprocal of the derivative.

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Correct answer: Option 1 — Both statements are true.
Q23 1 Mark

Statement 1: A function is increasing on an interval where its derivative is positive.

Statement 2: A function is decreasing on an interval where its derivative is negative.

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Correct answer: Option 1 — Both statements are true.
Q24 1 Mark

Statement 1: At a local maximum the derivative is zero and changes from positive to negative.

Statement 2: At a local minimum the derivative is zero and changes from negative to positive.

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Correct answer: Option 1 — Both statements are true.
Q25 1 Mark

Statement 1: Rolle's theorem requires continuity on a closed interval and differentiability on the open interval.

Statement 2: The endpoints of the interval must give equal function values.

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Correct answer: Option 1 — Both statements are true.
Q26 1 Mark

Statement 1: The mean value theorem is a generalisation of Rolle's theorem.

Statement 2: Rolle's theorem is the special case where f(a) = f(b).

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Correct answer: Option 1 — Both statements are true.
Case Study / Passage Questions 3 questions
Q27 3 Marks
The profit function (in lakhs of rupees) of a small firm is given by P(x) = −x² + 10x − 16 where x is the number of items produced (in thousands). Find the production level that maximises profit and the maximum profit.
  1. The production level that maximises profit is:
    A2 thousand
    B5 thousand
    C8 thousand
    D10 thousand
  2. The maximum profit equals:
    A₹5 lakh
    B₹9 lakh
    C₹16 lakh
    D₹25 lakh
  3. Find the critical point and maximum profit using the second derivative test.
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1. Option 2 — 5 thousand
2. Option 2 — ₹9 lakh
3. P'(x) = −2x + 10 = 0 gives x = 5 (critical point). P''(x) = −2 < 0 confirms maximum. P(5) = −25 + 50 − 16 = 9 lakh rupees.
Q28 3 Marks
A spherical balloon is being inflated at a rate of 100 cm³/s. Find the rate at which the radius is increasing when the radius is 5 cm. Use V = (4/3)πr³.
  1. Differentiating V = (4/3)πr³ with respect to t gives dV/dt =
    A4πr² · dr/dt
    B4πr · dr/dt
    C(4/3)πr³ · dr/dt
    D3r² · dr/dt
  2. At r = 5 cm dr/dt equals:
    A1/π cm/s
    B1/(4π) cm/s
    C1/(25π) cm/s
    D1/(100π) cm/s
  3. Calculate dr/dt step by step at the moment when r = 5 cm.
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1. Option 1 — 4πr² · dr/dt
2. Option 1 — 1/π cm/s
3. dV/dt = 4πr² (dr/dt) so 100 = 4π(25)(dr/dt) giving dr/dt = 1/π cm/s. The radius increases at 1/π ≈ 0.318 cm per second.
Q29 3 Marks
A curve is given by y = x² + 1. Find the equations of the tangent and normal to the curve at the point (1, 2).
  1. The slope of the tangent at (1, 2) is dy/dx |₁ which equals:
    A1
    B2
    C4
    D5
  2. The slope of the normal at the same point is:
    A−1/2
    B1/2
    C−2
    D2
  3. Write the equations of tangent and normal at (1, 2) and verify they pass through this point.
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1. Option 2 — 2
2. Option 1 — −1/2
3. dy/dx = 2x; at x = 1 the slope of tangent is 2; equation y − 2 = 2(x − 1) i.e. y = 2x. Normal slope = −1/2 so normal equation is y − 2 = (−1/2)(x − 1) i.e. 2y + x = 5.
Table-Based Questions 4 questions
Q30 3 Marks

Study the increasing/decreasing nature of f(x) = x² − 4x:

Intervalf'(x) = 2x − 4SignNature
x < 2Negative−Decreasing
x = 200Stationary
x > 2Positive+Increasing
  1. The critical point of f is at x =
    A1
    B2
    C3
    D4
  2. At the critical point the function has a:
    ALocal maximum
    BLocal minimum
    CInflection point
    DAsymptote
  3. Use the first-derivative test to identify the nature of the critical point.
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1. Option 2 — 2
2. Option 2 — Local minimum
3. The first-derivative test: if f' changes from negative to positive at x₀ there is a local minimum at x₀; if it changes from positive to negative there is a local maximum. Here f'(x) goes from − to + at x = 2 hence local minimum.
Q31 3 Marks

Study the second derivative test on f(x) = x³ − 3x:

StepResult
f'(x)3x² − 3
f'(x) = 0 atx = 1, x = −1
f''(x)6x
f''(1)6 > 0 (Local minimum)
f''(−1)−6 < 0 (Local maximum)
  1. At x = 1 the function has:
    ALocal minimum
    BLocal maximum
    CInflection
    DSaddle
  2. The critical points of f(x) = x³ − 3x are:
    A2 and −2
    B1 and −1
    C0 only
    DNo critical points
  3. State the second derivative test for classifying critical points.
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1. Option 1 — Local minimum
2. Option 2 — 1 and −1
3. f''(x) at a critical point determines its nature: if positive ⇒ local min if negative ⇒ local max. If zero we use the first-derivative test or higher derivatives. Here f''(1) = 6 > 0 (min) and f''(−1) = −6 < 0 (max).
Q32 6 Marks

For f(x) = x³ − 6x² + 9x + 5, find the intervals where f is increasing and decreasing using the sign of f'(x). Also list local maxima and minima.

IntervalSign of f'(x)Nature
x < 1??
1 < x < 3??
x > 3??
Q33 6 Marks

A spherical balloon is being inflated. Use the table to find the rate of change of volume when r = 4 cm and the rate of change of surface area when r = 4 cm given dr/dt = 2 cm/s.

VariableFormulaValue
r—4 cm
dr/dt—2 cm/s
V(4/3)πr³?
S4πr²?
Picture-Based Questions 1 question
Q34 3 Marks

Study the graph of f(x) = x³ − 3x and answer:

Application of Derivatives figure
  1. The critical points of f are at:
    Ax = −1 only
    Bx = 1 only
    Cx = ±1
    DNo critical points
  2. At x = 1 the function has a:
    ALocal maximum
    BLocal minimum
    CInflection
    DSaddle point
  3. Use the second-derivative test to classify the two critical points.
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1. Option 3 — x = ±1
2. Option 2 — Local minimum
3. f'(x) = 3x² − 3 = 0 ⇒ x = ±1. f''(x) = 6x. At x = −1, f''(−1) = −6 < 0 ⇒ local maximum (value 2). At x = 1, f''(1) = 6 > 0 ⇒ local minimum (value −2). The second-derivative test classifies critical points by curvature.
Next chapter Application of Integrals

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