Application of Derivatives: 5 marks Questions (CBSE Class 12)

6 different 5 marks questions on Application of Derivatives from CBSE Class 12 Maths board exams 2026, newest first.

Find the sub intervals in which $f (x) = cot^{- 1} (sin x + cos x)$ , $x \in (0, π)$ is increasing and decreasing.

Show answer & solution

Answer: Increasing on

(\frac{π}{4}, π)

, decreasing on

(0, \frac{π}{4})

$f^{'} (x) = - \frac{c o s x - s i n x}{1 + ( s i n x + c o s x ) ^{2}} = \frac{s i n x - c o s x}{1 + ( s i n x + c o s x ) ^{2}}$ .
The denominator is positive, so the sign of $f^{'}$ is that of $sin x - cos x$ .
$f^{'} (x) = 0 \Rightarrow tan x = 1 \Rightarrow x = \frac{π}{4}$ in $(0, π)$ .
For $0 < x < \frac{π}{4}$ , $sin x < cos x$ : $f^{'} < 0$ , f is decreasing.
For $\frac{π}{4} < x < π$ , $sin x > cos x$ : $f^{'} > 0$ , f is increasing.

A rectangle of perimeter 36 cm is revolved around one of its sides to sweep out a cylinder of maximum volume.
Find the dimensions of the rectangle.

Show answer & solution

Answer: 12 cm by 6 cm (revolved about the 6 cm side)

Let the side revolved about be $h$ (height) and the other side $r$ (radius). $2 (r + h) = 36 \Rightarrow h = 18 - r$ .
$V = π r^{2} h = π (18 r^{2} - r^{3})$ .
$\frac{d V}{d r} = π (36 r - 3 r^{2}) = 0 \Rightarrow r = 12$ ( $r \neq = 0$ ).
$\frac{d ^{2} V}{d r ^{2}} = π (36 - 6 r) = - 36 π < 0$ at $r = 12$ , so V is maximum.
Dimensions: $12$ cm $\times$ $6$ cm, the rectangle being revolved about its 6 cm side.

Find the sub-interval of $(0, π)$ in which $f (x) = tan^{- 1} (sin x - cos x)$ is increasing and decreasing.

Show answer & solution

Answer: Increasing on

(0, \frac{3 π}{4})

, decreasing on

(\frac{3 π}{4}, π)

$f^{'} (x) = \frac{c o s x + s i n x}{1 + ( s i n x - c o s x ) ^{2}}$ ; the denominator is positive.
$f^{'} (x) = 0 \Rightarrow sin x + cos x = 0 \Rightarrow tan x = - 1 \Rightarrow x = \frac{3 π}{4}$ in $(0, π)$ .
$sin x + cos x = 2 sin (x + \frac{π}{4}) > 0$ for $0 < x < \frac{3 π}{4}$ : f is increasing there.
It is $< 0$ for $\frac{3 π}{4} < x < π$ : f is decreasing there.

A rectangle of perimeter 24 cm is revolved along one of its sides to sweep out a cylinder of maximum volume. Find the dimensions of the rectangle.

Show answer & solution

Answer: 8 cm by 4 cm (revolved about the 4 cm side)

Let the side revolved about be $h$ (height) and the other side $r$ (radius). $2 (r + h) = 24 \Rightarrow h = 12 - r$ .
$V = π r^{2} h = π (12 r^{2} - r^{3})$ .
$\frac{d V}{d r} = π (24 r - 3 r^{2}) = 0 \Rightarrow r = 8$ ( $r \neq = 0$ ).
$\frac{d ^{2} V}{d r ^{2}} = π (24 - 6 r) = - 24 π < 0$ at $r = 8$ , so V is maximum.
Dimensions: $8$ cm $\times$ $4$ cm, the rectangle being revolved about its 4 cm side.

Find the sub-interval of $(0, \frac{π}{2})$ in which $f (x) = lo g (sin x + cos x)$ is increasing and decreasing.

Show answer & solution

Answer: Increasing on

(0, \frac{π}{4})

, decreasing on

(\frac{π}{4}, \frac{π}{2})

$f^{'} (x) = \frac{c o s x - s i n x}{s i n x + c o s x}$ ; on $(0, \frac{π}{2})$ the denominator is positive.
$f^{'} (x) = 0 \Rightarrow tan x = 1 \Rightarrow x = \frac{π}{4}$ .
For $0 < x < \frac{π}{4}$ , $cos x > sin x$ : $f^{'} > 0$ , f is increasing.
For $\frac{π}{4} < x < \frac{π}{2}$ , $cos x < sin x$ : $f^{'} < 0$ , f is decreasing.

A rectangle of perimeter 30 cm is revolved along one of its sides to sweep out a cylinder of maximum volume. Find the dimensions of the rectangle.

Show answer & solution

Answer: 10 cm by 5 cm (revolved about the 5 cm side)

Let the side revolved about be $h$ (height) and the other side $r$ (radius). $2 (r + h) = 30 \Rightarrow h = 15 - r$ .
$V = π r^{2} h = π (15 r^{2} - r^{3})$ .
$\frac{d V}{d r} = π (30 r - 3 r^{2}) = 0 \Rightarrow r = 10$ ( $r \neq = 0$ ).
$\frac{d ^{2} V}{d r ^{2}} = π (30 - 6 r) = - 30 π < 0$ at $r = 10$ , so V is maximum.
Dimensions: $10$ cm $\times$ $5$ cm, the rectangle being revolved about its 5 cm side.

Practise smarter: chapter-wise revision, formula sheets and step-by-step NCERT solutions on MonoMath CBSE →