Question

The domain of the function $f\left(x\right)=\sqrt{{\sec }^{-1}\left\{\frac{1-|x|}{2}\right\}}$ is

• A. $(-\infty ,-3]\cup [3,+\infty )$
• B. $[3,+\infty )$
• C. $\Phi$
• D. $R$

Answer 1

The correct option is A $(-\infty ,-3]\cup [3,+\infty )$

${\sec }^{-1}\left\{\frac{1-|x|}{2}\right\}>0$

$\frac{1-|x|}{2}\le -1$ and $\frac{1-|x|}{2}\ge 1$
$3\le |x|$ and $-1\ge |x|$
Hence $3\le |x|$
$x\in (-\infty ,-3]\cup [3,\infty )$