The domain of the function $\cos ^{-1}\left(\frac{2 \sin ^{-1}\left(\frac{1}{4 x^{2}-1}\right)}{\pi}\right)$ is :

(D) $\left(-\infty, \frac{-1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, \infty\right) \cup\{0\}$

(A) $R-\left\{-\frac{1}{2}, \frac{1}{2}\right\}$

(B) $(-\infty,-1] \cup[1, \infty) \cup\{0\}$

(C) $\left(-\infty, \frac{-1}{2}\right) \cup\left(\frac{1}{2}, \infty\right) \cup\{0\}$

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Inverse Trigonometric Functions

The domain of the function

cos^{- 1} (\frac{2 s i n ^{- 1} ( \frac{1}{4 x ^{2} - 1} )}{π})

is :

Answer: (D)

For

cos^{- 1}

to be defined, we need:

- 1 \leq \frac{2 sin ^{- 1} ( \frac{1}{4 x ^{2} - 1} )}{π} \leq 1

- \frac{π}{2} \leq sin^{- 1} (\frac{1}{4 x ^{2} - 1}) \leq \frac{π}{2}

For

sin^{- 1}

to be defined:

- 1 \leq \frac{1}{4 x ^{2} - 1} \leq 1

This requires

4 x^{2} - 1 \geq 1

4 x^{2} - 1 \leq - 1

.

From

4 x^{2} - 1 \geq 1

x^{2} \geq \frac{1}{2}

, giving

x \in (- \infty, - \frac{1}{2}] \cup [\frac{1}{2}, \infty)

.

From

4 x^{2} - 1 \leq - 1

x^{2} \leq 0

, giving

x = 0

.

Therefore, the domain is

(- \infty, \frac{- 1}{2}] \cup [\frac{1}{2}, \infty) \cup {0}

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option