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Application of Derivatives: 1 mark Questions (CBSE Class 12)

9 different 1 mark questions on Application of Derivatives from CBSE Class 12 Maths board exams 2026, newest first.

1 mark (9)2 marks (12)3 marks (3)4 marks (4)5 marks (6)

Q71 markMCQApplication of Derivatives CBSE 2026 · 65/1/1

The least value of $f (x) = x^{3} - 12 x, x \in [0, 3]$ is

(A) $- 16$
(B) $- 9$
(C)0
(D)16

Show answer & solution

Answer: (A)

- 16

$f^{'} (x) = 3 x^{2} - 12 = 0 \Rightarrow x = 2$ (in $[0, 3]$ ).
$f (0) = 0$ , $f (2) = 8 - 24 = - 16$ , $f (3) = 27 - 36 = - 9$ .
Least value $= - 16$ .

Q181 markMCQApplication of Derivatives CBSE 2026 · 65/1/2

The absolute maximum value of $f (x) = x^{2} + 1$ in $[- 5, 2]$ is

(A)26
(B)1
(C)5
(D)2

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Answer: (A) 26

$f^{'} (x) = 2 x = 0 \Rightarrow x = 0$ .
$f (- 5) = 26$ , $f (0) = 1$ , $f (2) = 5$ .
Absolute maximum $= 26$ .

Q111 markMCQApplication of Derivatives CBSE 2026 · 65/1/3

The least value of $f (x) = e^{- x}$ in [0, 3] is

(A) $e^{- 3}$
(B) $- 1$
(C)1
(D) $- e^{3}$

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Answer: (A)

e^{- 3}

$f^{'} (x) = - e^{- x} < 0$ , so f is decreasing on [0, 3].
Least value $= f (3) = e^{- 3}$ .

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

Q61 markMCQApplication of Derivatives CBSE 2026 · 65/4/1

The surface area of a sphere when its volume changes at the same rate as its radius is :

(A) $4 π$ sq. units
(B)1 sq. unit
(C)4 sq. units
(D) $π$ sq. units

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Answer: (B) 1 sq. unit

$V = \frac{4}{3} π r^{3} \Rightarrow \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$ .
Given $\frac{d V}{d t} = \frac{d r}{d t}$ , so $4 π r^{2} = 1$ .
Surface area $= 4 π r^{2} = 1$ sq. unit.

Q61 markMCQApplication of Derivatives CBSE 2026 · 65/4/2

If the distance travelled by a particle in t seconds is given by $S = 72 t + 3 t^{2} - t^{3}$ , then time taken by the particle to come to rest is :

(A)4 seconds
(B)6 seconds
(C)3 seconds
(D)0 seconds

Show answer & solution

Answer: (B) 6 seconds

Velocity $v = \frac{d S}{d t} = 72 + 6 t - 3 t^{2}$ .
At rest $v = 0$ : $t^{2} - 2 t - 24 = 0 \Rightarrow (t - 6) (t + 4) = 0$ .
$t > 0$ , so $t = 6$ seconds.

Q61 markMCQApplication of Derivatives CBSE 2026 · 65/4/3

A cylindrical tank is being filled with sand at a rate of 314 m $^{3}$ /h. If the radius of the tank is 10 m, then the height of sand in the tank increases at the rate of :

(A)1.1 m/h
(B)1 m/h
(C) $π$ m/h
(D) $\frac{π}{2}$ m/h

Show answer & solution

Answer: (B) 1 m/h

$V = π r^{2} h$ with $r = 10$ fixed, so $\frac{d V}{d t} = 100 π \frac{d h}{d t}$ .
$314 = 100 π \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = \frac{3.14}{π}$ .
Taking $π = 3.14$ , $\frac{d h}{d t} = 1$ m/h.

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

Show answer & solution

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

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