(cos^{-1}(\frac{2sin^{-1}(\frac{1}{4x^2-1})}{π})) is: | Mathematics | SolveeIt | Solveeit

Today, August 24

Question

$cos^{-1}(\frac{2sin^{-1}(\frac{1}{4x^2-1})}{π})$ is:

A

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

B

$(-∞,-1] ∪[1,∞)∪\left\\{{0}\right\\}$

C

$(-∞,\frac{-1}{2}] ∪[\frac{1}{2},∞)∪\left\\{{0}\right\\}$

D

$(-∞,\frac{-1}{\sqrt2}] ∪[\frac{1}{\sqrt2},∞)∪\left\\{{0}\right\\}$

Answer

$(-∞,\frac{-1}{\sqrt2}] ∪[\frac{1}{\sqrt2},∞)∪\left\\{{0}\right\\}$

Explanation

Solution

The correct option is(D): $(-∞,\frac{-1}{\sqrt2}] ∪[\frac{1}{\sqrt2},∞)∪\left\\{{0}\right\\}$ .