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Application of Derivatives: 2 marks Questions (CBSE Class 12)

12 different 2 marks 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)

Q222 marksVery Short AnswerApplication of Derivatives CBSE 2026 · 65/1/1

A room freshner bottle in the shape of an inverted cone sprays the perfume at regular intervals such that volume of the perfume in the bottle decreases at the steady rate of 1 mm $^{3}$ /min. Find the rate at which level of perfume is dropping at an instant when level of perfume in the bottle is 10 mm, if the semi-vertical angle of conical bottle is $\frac{π}{6}$ .

Diagram for CBSE 2026 Class 12 Maths question 22

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Answer: The level is dropping at

\frac{3}{100 π}

mm/min.

Let h be the level of perfume and r the radius of its surface. Then $r = h tan \frac{π}{6} = \frac{h}{3}$ .
$V = \frac{1}{3} π r^{2} h = \frac{π h ^{3}}{9}$ .
$\frac{d V}{d t} = \frac{π h ^{2}}{3} \frac{d h}{d t}$ .
With $\frac{d V}{d t} = - 1$ and $h = 10$ : $- 1 = \frac{100 π}{3} \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = - \frac{3}{100 π}$ .
So the level drops at $\frac{3}{100 π}$ mm/min.

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

Q222 marksVery Short AnswerApplication of Derivatives CBSE 2026 · 65/3/1

Find the sub-interval(s) of $(0, \frac{π}{2})$ in which $f (x) = tan x - 4 x$ is increasing.

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Answer:

(\frac{π}{3}, \frac{π}{2})

$f^{'} (x) = sec^{2} x - 4$ .
$f^{'} (x) > 0 ⟺ sec^{2} x > 4 ⟺ cos^{2} x < \frac{1}{4} ⟺ cos x < \frac{1}{2}$ (cos x > 0 here).
In $(0, \frac{π}{2})$ this means $x > \frac{π}{3}$ .
f is increasing on $(\frac{π}{3}, \frac{π}{2})$ (and decreasing on $(0, \frac{π}{3})$ ).

Q252 marksVery Short AnswerApplication of Derivatives CBSE 2026 · 65/3/2

Find the sub-interval of $(0, \infty)$ in which $f (x) = x^{2} e^{- x}$ is increasing.

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Answer:

(0, 2)

$f^{'} (x) = 2 x e^{- x} - x^{2} e^{- x} = x e^{- x} (2 - x)$ .
For $x > 0$ , $x e^{- x} > 0$ , so $f^{'} (x) > 0 ⟺ x < 2$ .
f is increasing on $(0, 2)$ (and decreasing on $(2, \infty)$ ).

Q232 marksVery Short AnswerApplication of Derivatives CBSE 2026 · 65/3/3

Find the sub-interval of $(0, \infty)$ in which $f (x) = x lo g x$ is increasing.

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Answer:

(\frac{1}{e}, \infty)

$f^{'} (x) = lo g x + 1$ .
$f^{'} (x) > 0 ⟺ lo g x > - 1 ⟺ x > \frac{1}{e}$ .
f is increasing on $(\frac{1}{e}, \infty)$ (and decreasing on $(0, \frac{1}{e})$ ).

Q232 marksVery Short AnswerApplication of Derivatives CBSE 2026 · 65/4/1

Find the values of x for which $f (x) = x^{x}$ , $x > 0$ is increasing.

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Answer:

x \geq \frac{1}{e}

, i.e. f is increasing on

[\frac{1}{e}, \infty)

$lo g f = x lo g x \Rightarrow \frac{f ^{'} ( x )}{f ( x )} = lo g x + 1$ , so $f^{'} (x) = x^{x} (1 + lo g x)$ .
$x^{x} > 0$ , so $f^{'} (x) > 0 ⟺ lo g x > - 1 ⟺ x > \frac{1}{e}$ .
$f^{'} (x) = 0$ only at $x = \frac{1}{e}$ , so f is increasing on $[\frac{1}{e}, \infty)$ .

Q232 marksVery Short AnswerApplication of Derivatives CBSE 2026 · 65/4/2

Find the interval(s) in which the function $f (x) = \frac{x}{l o g x}$ , where $x \in (0, 1) \cup (1, \infty)$ , is increasing.

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Answer: f is increasing on

[e, \infty)

$f^{'} (x) = \frac{l o g x - x \cdot \frac{1}{x}}{( l o g x ) ^{2}} = \frac{l o g x - 1}{( l o g x ) ^{2}}$ .
$(lo g x)^{2} > 0$ in the domain, so $f^{'} (x) > 0 ⟺ lo g x > 1 ⟺ x > e$ .
$f^{'} (x) < 0$ on $(0, 1)$ and $(1, e)$ , and $f^{'} (e) = 0$ . So f is increasing on $[e, \infty)$ .

Q232 marksVery Short AnswerApplication of Derivatives CBSE 2026 · 65/4/3

Find the interval(s) for which the function $f (x) = \frac{x}{3} + \frac{3}{x}$ , $x \neq = 0$ is increasing.

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Answer:

(- \infty, - 3]

and

[3, \infty)

$f^{'} (x) = \frac{1}{3} - \frac{3}{x ^{2}} = \frac{x ^{2} - 9}{3 x ^{2}}$ .
$f^{'} (x) > 0 ⟺ x^{2} > 9 ⟺ x < - 3$ or $x > 3$ ; $f^{'} (x) = 0$ at $x = \pm 3$ .
So f is increasing on $(- \infty, - 3]$ and on $[3, \infty)$ .

Q232 marksVery Short AnswerApplication of Derivatives CBSE 2026 · 65/5/1

Determine the values of x for which $f (x) = \frac{x - 3}{x + 1}$ , $x \neq = - 1$ is an increasing function.

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Answer: f is increasing for all real

x \neq = - 1

, i.e. on

(- \infty, - 1)

and

(- 1, \infty)

$f^{'} (x) = \frac{( x + 1 ) - ( x - 3 )}{( x + 1 ) ^{2}} = \frac{4}{( x + 1 ) ^{2}}$ .
$f^{'} (x) > 0$ for every $x \neq = - 1$ .
So f is increasing on $(- \infty, - 1)$ and on $(- 1, \infty)$ .

Q232 marksVery Short AnswerApplication of Derivatives CBSE 2026 · 65/5/2

Determine the interval(s) in which $f (x) = 5 x^{3/2} - 3 x^{5/2}$ , $x > 0$ is increasing.

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Answer: f is increasing on

(0, 1)

$f^{'} (x) = \frac{15}{2} x^{1/2} - \frac{15}{2} x^{3/2} = \frac{15}{2} x (1 - x)$ .
For $x > 0$ , $x > 0$ , so $f^{'} (x) > 0$ when $0 < x < 1$ and $f^{'} (x) < 0$ when $x > 1$ .
So f is increasing on $(0, 1)$ .

Q232 marksVery Short AnswerApplication of Derivatives CBSE 2026 · 65/5/3

Determine the interval(s) in which $f (x) = \frac{x}{3} + \frac{3}{x}$ , $x \neq = 0$ is increasing.

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Answer: f is increasing on

(- \infty, - 3]

and on

[3, \infty)

$f^{'} (x) = \frac{1}{3} - \frac{3}{x ^{2}} = \frac{x ^{2} - 9}{3 x ^{2}} = \frac{( x - 3 ) ( x + 3 )}{3 x ^{2}}$ .
$f^{'} (x) > 0$ when $x < - 3$ or $x > 3$ , and $f^{'} (x) < 0$ when $- 3 < x < 0$ or $0 < x < 3$ .
So f is increasing on $(- \infty, - 3]$ and on $[3, \infty)$ .

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