Application of Derivatives: CBSE Class 12 Previous Year Questions

34 different questions from Application of Derivatives (NCERT Chapter 6) asked in CBSE Class 12 Maths board exams 2026. Pick a mark group to practise, each with answers and step-by-step solutions.

1 mark questions9 questions 2 marks questions12 questions 3 marks questions3 questions 4 marks questions4 questions 5 marks questions6 questions

Most asked Application of Derivatives questions

Q81 markMCQApplication of Derivatives CBSE 2026 · 65/2/1

For $f (x) = x + \frac{1}{x}$ $(x \neq = 0)$

(A)local maximum value is 2
(B)local minimum value is $- 2$
(C)local maximum value is $- 2$
(D)local minimum value < local maximum value

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Answer: (C) local maximum value is

- 2

$f^{'} (x) = 1 - \frac{1}{x ^{2}} = 0 \Rightarrow x = \pm 1$ ; $f^{''} (x) = \frac{2}{x ^{3}}$ .
$f^{''} (- 1) = - 2 < 0$ : local maximum at $x = - 1$ , value $f (- 1) = - 2$ .
$f^{''} (1) = 2 > 0$ : local minimum at $x = 1$ , value $f (1) = 2$ .
So the local maximum value is $- 2$ (and it is less than the local minimum value 2).

Also asked in: 2026 65/2/2, 2026 65/2/3

Q91 markMCQApplication of Derivatives CBSE 2026 · 65/3/1

The rate of change of volume of a sphere with respect to its diameter, when its radius is 5 cm, is :

(A) $400 π$ cm $^{3}$ /cm
(B) $100 π$ cm $^{3}$ /cm
(C) $50 π$ cm $^{3}$ /cm
(D) $25 π$ cm $^{3}$ /cm

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Answer: (C)

50 π

^{3}

/cm

With diameter D, $V = \frac{4}{3} π (\frac{D}{2})^{3} = \frac{π D ^{3}}{6}$ .
$\frac{d V}{d D} = \frac{π D ^{2}}{2}$ .
At $r = 5$ , $D = 10$ : $\frac{d V}{d D} = \frac{100 π}{2} = 50 π$ cm $^{3}$ /cm.

Also asked in: 2026 65/3/2, 2026 65/3/3

Q91 markMCQApplication of Derivatives CBSE 2026 · 65/5/1

Absolute minimum value of $f (x) = (x - 2)^{2} + 5$ in the interval $[- 3, 2]$ is :

(A) $- 3$
(B)2
(C)5
(D)30

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Answer: (C) 5

$f^{'} (x) = 2 (x - 2) \leq 0$ on $[- 3, 2]$ , so f is decreasing there.
Minimum is at $x = 2$ : $f (2) = 5$ (maximum $f (- 3) = 30$ ).

Also asked in: 2026 65/5/2, 2026 65/5/3

Q232 marksVery Short AnswerApplication of Derivatives CBSE 2026 · 65/2/1

Find the absolute maximum value of $f (x) = cos x + sin^{2} x$ , $x \in [0, π]$

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Answer:

\frac{5}{4}

(at

x = \frac{π}{3}

)

$f^{'} (x) = - sin x + 2 sin x cos x = sin x (2 cos x - 1)$ .
$f^{'} (x) = 0 \Rightarrow x = 0, π$ or $cos x = \frac{1}{2} \Rightarrow x = \frac{π}{3}$ .
$f (0) = 1$ , $f (\frac{π}{3}) = \frac{1}{2} + \frac{3}{4} = \frac{5}{4}$ , $f (π) = - 1$ .
Absolute maximum value $= \frac{5}{4}$ .

Also asked in: 2026 65/2/2, 2026 65/2/3

Q23 (OR) (OR)2 marksVery Short AnswerApplication of Derivatives CBSE 2026 · 65/2/1

If the volume of a solid hemisphere increases at a uniform rate, prove that its surface area varies inversely as its radius.

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Answer: Proved.

Let r be the radius. $V = \frac{2}{3} π r^{3}$ and $\frac{d V}{d t} = k$ (constant).
$\frac{d V}{d t} = 2 π r^{2} \frac{d r}{d t} = k \Rightarrow \frac{d r}{d t} = \frac{k}{2 π r ^{2}}$ .
Total surface area of a solid hemisphere $S = 2 π r^{2} + π r^{2} = 3 π r^{2}$ .
$\frac{d S}{d t} = 6 π r \frac{d r}{d t} = 6 π r \cdot \frac{k}{2 π r ^{2}} = \frac{3 k}{r}$ .
So the rate of change of the surface area varies inversely as the radius.

Also asked in: 2026 65/2/2, 2026 65/2/3

Q364 marksCase StudyApplication of Derivatives CBSE 2026 · 65/1/1

An online delivery company in a city has 5000 subscribers and collects annual subscription fees of ₹ 300 per subscriber for unlimited free deliveries.
The company wishes to increase the annual subscription fee. It is predicted that, for every increase of ₹ 1, ten subscribers will discontinue. Assume that the company increased the annual fee by ₹ $x$ .
Based on the given information, answer the following questions :
(i) How many subscribers will discontinue after an increase of ₹ $x$ in annual fee ? (1)
(ii) If R(x) denotes the total revenue collected after the increase of ₹ $x$ in subscription fee, express R(x) as a function of $x$ . (1)
(iii) Find the value of $x$ for which R(x) is maximum. (2)
OR (iii) Find the sub-intervals of (0, 5000) in which R(x) is increasing and decreasing. (2)

Diagram for CBSE 2026 Class 12 Maths question 36

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Answer: (i)

10 x

(ii)

R (x) = (300 + x) (5000 - 10 x)

(iii)

x = 100

; OR increasing on (0, 100), decreasing on (100, 5000)

(i) 10 subscribers leave per ₹ 1 increase, so $10 x$ subscribers discontinue.
(ii) Fee $= 300 + x$ , subscribers $= 5000 - 10 x$ , so $R (x) = (300 + x) (5000 - 10 x) = 1500000 + 2000 x - 10 x^{2}$ .
(iii) $R^{'} (x) = 2000 - 20 x = 0 \Rightarrow x = 100$ . $R^{''} (x) = - 20 < 0$ , so R is maximum at $x = 100$ .
OR (iii) $R^{'} (x) = 20 (100 - x)$ : $R^{'} (x) > 0$ on (0, 100) and $R^{'} (x) < 0$ on (100, 5000).
So R is increasing on (0, 100) and decreasing on (100, 5000).

Also asked in: 2026 65/1/2, 2026 65/1/3

Q364 marksCase StudyApplication of Derivatives CBSE 2026 · 65/3/1

Two vertical light poles of height 22 m and 16 m stand on the opposite sides of a 20 m wide road as shown below in the figure.
Two ladders of length $l_{1}$ and $l_{2}$ are placed from a common point R on the road at a distance of x m from the smaller pole.
Based on the above information, answer the following questions :
(i) Express $p (x) = l_{1} + l_{2}$ in terms of x. (1)
(ii) Find $p^{'} (x)$ . (1)
(iii) (a) Find the value of x for which $l_{1}^{2} + l_{2}^{2}$ is minimum. (2)
OR (iii) (b) If the 22 m long pole is also replaced by a 16 m long pole, at what distance from either pole should the ladders be kept so that the sum of squares of lengths of ladders needed to reach the top of the pole is minimum ? (2)

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Answer: (i)

p (x) = x^{2} - 40 x + 884 + x^{2} + 256

(ii)

p^{'} (x) = \frac{x - 20}{x ^{2} - 40 x + 884} + \frac{x}{x ^{2} + 256}

(iii)(a)

x = 10

m (iii)(b) 10 m from each pole

(i) R is x m from the 16 m pole and $(20 - x)$ m from the 22 m pole.
$l_{1} = 2 2^{2} + (20 - x)^{2} = x^{2} - 40 x + 884$ , $l_{2} = 1 6^{2} + x^{2} = x^{2} + 256$ .
$p (x) = x^{2} - 40 x + 884 + x^{2} + 256$ .
(ii) $p^{'} (x) = \frac{x - 20}{x ^{2} - 40 x + 884} + \frac{x}{x ^{2} + 256}$ .
(iii)(a) $S = l_{1}^{2} + l_{2}^{2} = 2 x^{2} - 40 x + 1140$ ; $S^{'} = 4 x - 40 = 0$ gives $x = 10$ ; $S^{''} = 4 > 0$ , so minimum at $x = 10$ m.
(iii)(b) $S = 256 + (20 - x)^{2} + 256 + x^{2}$ ; $S^{'} = - 2 (20 - x) + 2 x = 4 x - 40 = 0$ gives $x = 10$ , $S^{''} = 4 > 0$ .
The ladders should be kept 10 m from each pole (the middle of the road).

Also asked in: 2026 65/3/2, 2026 65/3/3

Q364 marksCase StudyApplication of Derivatives CBSE 2026 · 65/4/1

At a birthday party, children are being served orange juice in conical cups, as shown in the figure.
Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0.1 cm $^{3}$ /s.
On the basis of the above information, answer the following questions :
(i) Establish a relation between the height h of the juice in the cup and radius r of the surface of the juice in the cup, if the semi-vertical angle of the cone is $α$ . (1)
(ii) At what rate is the juice level in the cup rising when the juice is 6 cm deep ? (1)
(iii) When the juice is 6 cm deep, then find at what rate is the upper surface area of juice increasing ? (2)
OR
(iii) When the juice is 6 cm deep, then find the rate at which the wetted surface area of the cup is increasing. (2)

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Answer: (i)

r = h tan α

, here

tan α = \frac{5}{15}

r = \frac{h}{3}

(ii)

\frac{1}{40 π}

cm/s (iii)

\frac{1}{30}

^{2}

/s; OR

\frac{10}{30}

^{2}

(i) In the right triangle formed by the axis, a radius of the juice surface and the slant side, $tan α = \frac{r}{h}$ , so $r = h tan α$ . For the cup $tan α = \frac{5}{15} = \frac{1}{3}$ , so $r = \frac{h}{3}$ .
(ii) $V = \frac{1}{3} π r^{2} h = \frac{π h ^{3}}{27}$ , so $\frac{d V}{d t} = \frac{π h ^{2}}{9} \frac{d h}{d t}$ .
At $h = 6$ : $0.1 = 4 π \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = \frac{1}{40 π}$ cm/s.
(iii) Upper surface area $A = π r^{2} = \frac{π h ^{2}}{9}$ ; $\frac{d A}{d t} = \frac{2 π h}{9} \frac{d h}{d t} = \frac{12 π}{9} \cdot \frac{1}{40 π} = \frac{1}{30}$ cm $^{2}$ /s.
OR (iii) Wetted surface $S = π r l$ with $l = h^{2} + r^{2} = \frac{10}{3} h$ , so $S = \frac{10 π h ^{2}}{9}$ .
$\frac{d S}{d t} = \frac{2 10 π h}{9} \frac{d h}{d t} = \frac{12 10 π}{9} \cdot \frac{1}{40 π} = \frac{10}{30}$ cm $^{2}$ /s.

Also asked in: 2026 65/4/2, 2026 65/4/3

Q384 marksCase StudyApplication of Derivatives CBSE 2026 · 65/5/1

A company produces cylindrical tumblers, open from the top. Since they want uniformity in the product, they fix the surface area of the tumblers produced.
Based on the above information, answer the following questions :
If for a tumbler, V is its volume, h the height and r the radius of the circular base, then :
(i) Differentiate its volume with respect to radius of the base, where the surface area is constant. (2)
(ii) If the company wants to maximize the volume of each tumbler, then establish a relation between its height and the radius of the base. (2)

Diagram for CBSE 2026 Class 12 Maths question 38

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Answer: (i)

\frac{d V}{d r} = \frac{S - 3 π r ^{2}}{2}

, where S is the fixed surface area (ii) h = r

Let the fixed surface area be $S = π r^{2} + 2 π r h$ , so $h = \frac{S - π r ^{2}}{2 π r}$ .
$V = π r^{2} h = \frac{r ( S - π r ^{2} )}{2} = \frac{S r - π r ^{3}}{2}$ .
(i) $\frac{d V}{d r} = \frac{S - 3 π r ^{2}}{2}$ .
(ii) $\frac{d V}{d r} = 0 \Rightarrow S = 3 π r^{2} \Rightarrow π r^{2} + 2 π r h = 3 π r^{2} \Rightarrow h = r$ .
$\frac{d ^{2} V}{d r ^{2}} = - 3 π r < 0$ , so V is maximum when h = r (height equals radius of the base).

Also asked in: 2026 65/5/2, 2026 65/5/3

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