Question

If $f\left(x\right)=\sqrt{{\sec }^{-1}\left(\frac{2-\left|x\right|}{4}\right)}$, then the domain of $f\left(x\right)$ is

• A. $[-2,2]$
• B. $[-6,6]$
• C. $(-\infty ,-6]\cup [6,-\infty )$
• D. $[-6,-6]\cup [2,6]$

Answer 1

Answer: (C)

$(-\infty ,-6]\cup [6,-\infty )$

$f\left(x\right)=\sqrt{{\sec }^{-1}\left(\frac{2-|x|}{4}\right)}$

$\frac{2-|x|}{4}\le -1$ or $\frac{2-|x|}{4}\ge 1$

$2-|x|\le -4$, or $2-|x|\ge 4$

$6\le |x|$, or $-2\ge |x|$

$\therefore |x|\le -2$ (Impossible)

$\therefore |x|\ge 6$

$\therefore xt\in [-\infty ,-6]U[6,\infty ]$