CBSE Maths › Class 12 PYQs › Inverse Trigonometric Functions

Inverse Trigonometric Functions: 5 marks Questions (CBSE Class 12)

3 different 5 marks questions on Inverse Trigonometric Functions from CBSE Class 12 Maths board exams 2026, newest first.

1 mark (11)2 marks (11)5 marks (3)

Q345 marksLong AnswerInverse Trigonometric Functions CBSE 2026 · 65/2/1

Find the domain of $g (x) = cos^{- 1} (x^{2} - 1)$ . Hence, find the value of $x$ for which $g (x) = \frac{π}{3}$ .
Also, write the range of $cos^{- 1} x$ other than its principal branch.

Show answer & solution

Answer: Domain

[- 2, 2]

;

x = \pm \frac{3}{2}

; e.g. range

[π, 2 π]

$cos^{- 1}$ needs $- 1 \leq x^{2} - 1 \leq 1 \Rightarrow 0 \leq x^{2} \leq 2 \Rightarrow - 2 \leq x \leq 2$ .
Domain $= [- 2, 2]$ .
$g (x) = \frac{π}{3} \Rightarrow x^{2} - 1 = cos \frac{π}{3} = \frac{1}{2} \Rightarrow x^{2} = \frac{3}{2} \Rightarrow x = \pm \frac{3}{2} = \pm \frac{6}{2}$ (both in the domain).
A branch other than the principal one: range $[π, 2 π]$ (or $[- π, 0]$ , etc.).

Q325 marksLong AnswerInverse Trigonometric Functions CBSE 2026 · 65/2/2

Find the domain of $p (x) = sin^{- 1} (1 - 2 x^{2})$ . Hence, find the value of $x$ for which $p (x) = \frac{π}{6}$ . Also, write the range of $2 p (x) + \frac{π}{2}$ .

Show answer & solution

Answer: Domain

[- 1, 1]

;

x = \pm \frac{1}{2}

; range

[- \frac{π}{2}, \frac{3 π}{2}]

$sin^{- 1}$ needs $- 1 \leq 1 - 2 x^{2} \leq 1 \Rightarrow 0 \leq x^{2} \leq 1 \Rightarrow - 1 \leq x \leq 1$ . Domain $= [- 1, 1]$ .
$p (x) = \frac{π}{6} \Rightarrow 1 - 2 x^{2} = sin \frac{π}{6} = \frac{1}{2} \Rightarrow x^{2} = \frac{1}{4} \Rightarrow x = \pm \frac{1}{2}$ .
On $[- 1, 1]$ , $1 - 2 x^{2}$ takes every value in $[- 1, 1]$ , so $p (x)$ takes every value in $[- \frac{π}{2}, \frac{π}{2}]$ .
Then $2 p (x) \in [- π, π]$ and $2 p (x) + \frac{π}{2} \in [- \frac{π}{2}, \frac{3 π}{2}]$ .

Q335 marksLong AnswerInverse Trigonometric Functions CBSE 2026 · 65/2/3

Find the domain of $q (x) = cos^{- 1} (4 x^{2} - 3)$ . Hence, find the value of $x$ for which $q (x) = 0$ . Also, write the range of $3 q (x) - π$ .

Show answer & solution

Answer: Domain

[- 1, - \frac{1}{2}] \cup [\frac{1}{2}, 1]

;

x = \pm 1

; range

[- π, 2 π]

$cos^{- 1}$ needs $- 1 \leq 4 x^{2} - 3 \leq 1 \Rightarrow 2 \leq 4 x^{2} \leq 4 \Rightarrow \frac{1}{2} \leq x^{2} \leq 1$ .
Domain $= [- 1, - \frac{1}{2}] \cup [\frac{1}{2}, 1]$ .
$q (x) = 0 \Rightarrow 4 x^{2} - 3 = cos 0 = 1 \Rightarrow x^{2} = 1 \Rightarrow x = \pm 1$ .
On the domain, $4 x^{2} - 3$ takes every value in $[- 1, 1]$ , so $q (x) \in [0, π]$ .
$3 q (x) - π \in [- π, 2 π]$ .

← Relations and Functions Matrices →

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