^(-1)(1 - x) - 2^(-1)x = ()/2, than x is equal to (a) 0 (b) 1 (c) 1/2 (d) 2 Choose the most appropriate answer from the options given below: Option 1: (a) and (b) only Option 2: (a) and (c) only Option 3: (a) only Option 4: (c) only

Answer: (a) only. Solution: Domain requires x[0,1]. Maximum of ^(-1)(1-x) on [0,1] is ()/2 (at x=0), and ^(-1)x is non-negative on [0,1]. For LHS to equal ()/2, we need ^(-1)(1-x) = ()/2 and ^(-1)x = 0 simultaneously. Both occur only at x=0. So only (a) is valid.

Geometry

Trigonometry

Medium

$\sin^{-1}(1 - x) - 2\sin^{-1}x = \frac{\pi}{2}$ , than $x$ is equal to

(a) $0$

(b) $1$

(c) $\frac{1}{2}$

(d) $2$

Choose the most appropriate answer from the options given below:

(a) and (b) only

(a) and (c) only

(a) only

✅ Correct Option: 3

Domain requires $x\in[0,1]$ .

Maximum of $\sin^{-1}(1-x)$ on $[0,1]$ is $\frac{\pi}{2}$ (at $x=0$ ), and $\sin^{-1}x$ is non-negative on $[0,1]$ .

For LHS to equal $\frac{\pi}{2}$ , we need $\sin^{-1}(1-x) = \frac{\pi}{2}$ and $\sin^{-1}x = 0$ simultaneously. Both occur only at $x=0$ .

So only (a) is valid.

2025: 13 May Shift 2

Match List-I with List-II

List-I	List-II
(A) $\sin^{-1}(-1)$	(I) $\frac{5\pi}{6}$
(B) $\cot^{-1}(-1)$	(II) $\frac{-\pi}{2}$
(C) $\sec^{-1}\left(\frac{-2}{\sqrt{3}}\right)$	(III) $\frac{\pi}{4}$
(D) $\tan^{-1}(1)$	(IV) $\frac{3\pi}{4}$

Choose the correct answer from the options given below:

2025: 30 May Shift 1

Consider $f(x) = \sin(3x) + 4, \forall x \in \mathbb{R}$ . Then

(A) Maximum value of $f(x)$ is 5

(B) Minimum value of $f(x)$ is 3

(D) Minimum value of $f(x)$ is attained at $x = 0$

Choose the correct answer from the options given below:

2022: 4 Aug Shift 1

The value of $\sec^2(\tan^{-1}2) + cosec^2(\cot^{-1}3)$ is:

2026: 22 May Shift 1

Match List-I with List-II

List-I (Function)	List-II (Principal value of y)
(A) $y=\sin^{-1}\left(-\frac{1}{2}\right)$	(I) $\frac{3\pi}{4}$
(B) $y=\cot^{-1}\left(-\frac{1}{\sqrt3}\right)$	(II) $-\frac{\pi}{4}$
(C) $y=cosec^{-1}(-\sqrt2)$	(III) $-\frac{\pi}{6}$
(D) $y=\cos^{-1}\left(-\frac{1}{\sqrt2}\right)$	(IV) $\frac{2\pi}{3}$

Choose the correct answer from the options given below:

2025: 19 May Shift 1

The maximum value of $\sin x \cdot \cos x$ is:

2026: 18 May Shift 2

Match the LIST-I with LIST-II

LIST-I Functions	LIST-II Principal value
A. $\sin^{-1}\left(\dfrac{1}{2}\right)$	I. $\dfrac{\pi}{4}$
B. $\sin^{-1}\left(-\dfrac{1}{\sqrt{2}}\right)$	II. $\dfrac{\pi}{6}$
C. $\tan^{-1}(1)$	III. $-\dfrac{\pi}{4}$
D. $\cos^{-1}\left(-\dfrac{1}{2}\right)$	IV. $\dfrac{2\pi}{3}$

Choose the correct answer from the options given below:

sin⁡−1(1−x)−2sin⁡−1x=π2\sin^{-1}(1 - x) - 2\sin^{-1}x = \frac{\pi}{2}sin−1(1−x)−2sin−1x=2π​, than xxx is equal to (a) 000 (b) 111 (c) 12\frac{1}{2}21​ (d) 222 Choose the most appropriate answer from the options given below:

Solution

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Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

$\sin^{-1}(1 - x) - 2\sin^{-1}x = \frac{\pi}{2}$ , than $x$ is equal to

(a) $0$

(b) $1$

(c) $\frac{1}{2}$

(d) $2$

Choose the most appropriate answer from the options given below: