Question

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

• A. $R-\left\{-\frac{1}{2}, \frac{1}{2}\right\}$
• B. $(-\infty,-1] \cup[1, \infty) \cup\{0\}$
• C. $\left(-\infty, \frac{-1}{2}\right) \cup\left(\frac{1}{2}, \infty\right) \cup\{0\}$
• D. $\left(-\infty, \frac{-1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, \infty\right) \cup\{0\}$

Answer 1

Answer: (D)

$-1 \leq \frac{2 \sin ^{-1}\left(\frac{1}{4 x^{2}-1}\right)}{\pi} \leq 1$
$-\pi / 2 \leq \sin ^{-1} \frac{1}{4 x^{2}-1} \leq \pi / 2$
Always $-1 \leq \frac{1}{4 x^{2}-1} \leq 1$
$x \in\left(\infty, \frac{1}{\sqrt{2}}\right) \cup\left[\frac{1}{\sqrt{2}}, \infty\right)$