CBSE Class 12 · Chapter 6 · Calculus · 2026-27

Application of Derivatives Class 12: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceRate of change

The radius of a circle is increasing at $2$ cm/s. The rate at which its area is increasing when the radius is $5$ cm is

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Answer: (a)

20 π {cm}^{2} / s

Why: dA/dt = 2πr dr/dt = 2π × 5 × 2.

$A = π r^{2} \Rightarrow \frac{d A}{d t} = 2 π r \frac{d r}{d t} = 2 π (5) (2) = 20 π {cm}^{2} / s$ .

Q2

·1 mark·Multiple choiceRate of change

The volume of a cube is increasing at $12 {cm}^{3} / s$ . How fast is its edge increasing when the edge is $2$ cm?

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Answer: (b)

1

cm/s

Why: dV/dt = 3a² da/dt, so 12 = 3 × 4 × da/dt.

$V = a^{3} \Rightarrow \frac{d V}{d t} = 3 a^{2} \frac{d a}{d t}$ . $12 = 3 (2)^{2} \frac{d a}{d t} \Rightarrow \frac{d a}{d t} = 1$ cm/s.

Q3

·1 mark·Multiple choiceIncreasing and decreasing functions

The function $f (x) = x^{2} - 6 x + 5$ is strictly increasing on

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Answer: (b)

(3, \infty)

Why: f′(x) = 2x − 6 > 0 exactly when x > 3.

$f^{'} (x) = 2 x - 6 > 0 \Leftrightarrow x > 3$ . So $f$ is strictly increasing on $(3, \infty)$ .

Q4

·1 mark·Multiple choiceIncreasing and decreasing functions

The function $f (x) = e^{2 x}$ is

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Answer: (a) strictly increasing on

R

Why: f′(x) = 2e^2x > 0 for every x.

$f^{'} (x) = 2 e^{2 x} > 0$ for all real $x$ , so $f$ is strictly increasing on $R$ .

Q5

·1 mark·Multiple choiceMaxima and minima

The local maximum value of $f (x) = - x^{2} + 4 x + 1$ is

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Answer: (c)

5

Why: f′(x) = −2x + 4 = 0 at x = 2, and f(2) = 5.

$f^{'} (x) = - 2 x + 4 = 0 \Rightarrow x = 2$ ; $f^{″} (x) = - 2 < 0$ , so a local maximum. Value $f (2) = - 4 + 8 + 1 = 5$ . ( $2$ is where it occurs, not the value.)

Q6

·1 mark·Multiple choiceMaxima and minima

The critical points of $f (x) = 2 x^{3} - 6 x$ are

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Answer: (d)

x = \pm 1

Why: f′(x) = 6x² − 6 = 0 gives x = ±1.

$f^{'} (x) = 6 x^{2} - 6 = 6 (x - 1) (x + 1) = 0 \Rightarrow x = \pm 1$ . ( $\pm \sqrt{3}$ and $0$ are the zeros of $f$ , not of $f^{'}$ .)

Q7

·1 mark·Multiple choiceMaxima and minima

The local minimum value of $f (x) = 2 x^{3} - 6 x$ is

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Answer: (a)

- 4

Why: f″(1) = 12 > 0, so x = 1 gives the local minimum f(1) = −4.

Critical points $x = \pm 1$ ; $f^{″} (x) = 12 x$ . $f^{″} (1) = 12 > 0$ : local minimum $f (1) = 2 - 6 = - 4$ . ( $f (- 1) = 4$ is the local maximum.)

Q8

·1 mark·Multiple choiceMaxima and minima

The greatest value of $f (x) = \sin x + \sqrt{3} \cos x$ for $0 \leq x \leq \frac{π}{2}$ is

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Answer: (b)

2

Why: f′(x) = cos x − √3 sin x = 0 at x = π/6, where f = 1/2 + 3/2.

$f^{'} (x) = \cos x - \sqrt{3} \sin x = 0 \Rightarrow \tan x = \frac{1}{\sqrt{3}} \Rightarrow x = \frac{π}{6}$ . $f (\frac{π}{6}) = \frac{1}{2} + \frac{3}{2} = 2$ ; $f (0) = \sqrt{3}$ , $f (\frac{π}{2}) = 1$ . Greatest value $= 2$ .

Q9

·1 mark·Multiple choiceMaxima and minima

The absolute maximum value of $f (x) = x^{2} - 4 x$ on $[0, 5]$ is

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Answer: (c)

5

Why: Compare f at the critical point x = 2 and at the end points 0 and 5.

$f^{'} (x) = 2 x - 4 = 0 \Rightarrow x = 2$ . $f (0) = 0$ , $f (2) = - 4$ , $f (5) = 5$ . Absolute maximum $= 5$ (at $x = 5$ ); absolute minimum $= - 4$ .

Q10

·1 mark·Multiple choiceMaxima and minima

Two positive numbers have sum $16$ . The greatest possible value of their product is

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Answer: (d)

64

Why: P = x(16 − x) is greatest at x = 8.

$P (x) = x (16 - x)$ , $P^{'} (x) = 16 - 2 x = 0 \Rightarrow x = 8$ ; $P^{″} (x) = - 2 < 0$ . Maximum $P = 8 \times 8 = 64$ .

Q11

·1 mark·Multiple choiceMaxima and minima

The minimum value of $f (x) = x e^{x}$ is

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Answer: (a)

- \frac{1}{e}

Why: f′(x) = e^x(1 + x) = 0 at x = −1, and f″(−1) > 0.

$f^{'} (x) = e^{x} (1 + x) = 0 \Rightarrow x = - 1$ ; $f^{″} (x) = e^{x} (2 + x)$ , $f^{″} (- 1) = e^{- 1} > 0$ . Minimum $= f (- 1) = - \frac{1}{e}$ .

Q12

·1 mark·Multiple choiceRate of change

The total cost of producing $x$ units of an item is $C (x) = 2 x^{2} + 10 x + 100$ rupees. The marginal cost when $5$ units are produced is

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Answer: (b) ₹

30

Why: Marginal cost = C′(x) = 4x + 10.

$C^{'} (x) = 4 x + 10$ , so $C^{'} (5) = 30$ : about ₹ $30$ for the next unit. ( $C (5) = 200$ is the total cost.)

Q13

·1 mark·Multiple choiceIncreasing and decreasing functions

The function $f (x) = x^{3} - 6 x^{2} + 9 x$ is strictly decreasing on

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Answer: (c)

(1, 3)

Why: f′(x) = 3(x − 1)(x − 3) is negative between 1 and 3.

$f^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3) < 0$ for $1 < x < 3$ . So $f$ is strictly decreasing on $(1, 3)$ .

Q14

·1 mark·Multiple choiceMaxima and minima

For $f (x) = x^{4} - 2 x^{2}$ , which statement about $x = 0$ is true?

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Answer: (d)

f

has a local maximum at

x = 0

Why: f′(0) = 0 and f″(0) = −4 < 0.

$f^{'} (x) = 4 x^{3} - 4 x$ , $f^{'} (0) = 0$ ; $f^{″} (x) = 12 x^{2} - 4$ , $f^{″} (0) = - 4 < 0$ . By the second derivative test, $x = 0$ is a point of local maximum.

Q15

·1 mark·Multiple choiceMaxima and minima

A rectangle has perimeter $36$ cm. Its greatest possible area is

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Answer: (b)

81 {cm}^{2}

Why: A = x(18 − x) is greatest at x = 9 (a square).

Sides $x$ and $18 - x$ : $A = 18 x - x^{2}$ , $A^{'} = 18 - 2 x = 0 \Rightarrow x = 9$ , $A^{″} = - 2 < 0$ . Maximum area $= 81 {cm}^{2}$ .

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 12 case study questions by chapter.

Case study 1: Fencing a kitchen garden (4 marks)

Anil has $40$ m of fencing to enclose a rectangular kitchen garden next to the long straight wall of his house. The wall forms one side, so only the other three sides need fencing. Let each side perpendicular to the wall be $x$ m.

(i) Express the area

A

of the garden in terms of

x

. [1 mark]

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Answer:

A (x) = x (40 - 2 x)

The side parallel to the wall is

40 - 2 x

, so

A (x) = x (40 - 2 x) = 40 x - 2 x^{2}

. A1

(ii) Find

A^{'} (x)

. [1 mark]

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Answer:

40 - 4 x

A^{'} (x) = 40 - 4 x

. A1

(iii) Find the dimensions that give the largest area, and that area. [2 marks]

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Answer:

10

m by

20

200 m^{2}

A^{'} (x) = 0 \Rightarrow x = 10

;

A^{″} (x) = - 4 < 0

, a maximum M1. Garden

10

\times

20

m, area

200 m^{2}

. A1

OR Show that

A

is increasing for

0 < x < 10

and decreasing for

10 < x < 20

. [2 marks]

Show answer

Answer: Shown

A^{'} (x) = 40 - 4 x = 4 (10 - x)

M1: positive for

x < 10

, negative for

x > 10

. A1

Case study 2: Inflating a balloon (4 marks)

For a school fete, a spherical balloon is pumped up with a small electric pump so that its radius increases at a steady $0.5$ cm/s. (Volume $V = \frac{4}{3} π r^{3}$ , surface area $S = 4 π r^{2}$ .)

(i) Find the rate at which the volume increases when

r = 10

cm. [1 mark]

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Answer:

200 π {cm}^{3} / s

\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} = 4 π (100) (0.5) = 200 π {cm}^{3} / s

. A1

(ii) Find the rate at which the surface area increases when

r = 10

cm. [1 mark]

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Answer:

40 π {cm}^{2} / s

\frac{d S}{d t} = 8 π r \frac{d r}{d t} = 8 π (10) (0.5) = 40 π {cm}^{2} / s

. A1

(iii) At what radius is the volume increasing at

50 π {cm}^{3} / s

? [2 marks]

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Answer:

5

4 π r^{2} (0.5) = 50 π

\Rightarrow r^{2} = 25 \Rightarrow r = 5

cm. A1

OR Find the radius at which the volume is increasing at

72 π {cm}^{3} / s

, and the rate at which the surface area is increasing at that moment. [2 marks]

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Answer:

6

cm;

24 π {cm}^{2} / s

4 π r^{2} (0.5) = 72 π \Rightarrow r^{2} = 36 \Rightarrow r = 6

cm M1;

\frac{d S}{d t} = 8 π r \frac{d r}{d t} = 8 π (6) (0.5) = 24 π {cm}^{2} / s

. A1

Next steps for Application of Derivatives

This free set is separate from the chapter's question bank. On the Application of Derivatives chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 43 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Application of Derivatives: important questions What Pro includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.