Tho damnin of tho finction $\cos ^{-1}\left(\frac{2 \sin ^{-1}\left(\frac{1}{4 x^{2}-1}\right)}{\pi}\right)$ is
$R-\left\{-\frac{1}{2}, \frac{1}{2}\right\}$
$(-\infty,-1] \cup[1, \infty) \cup\{0\}$
$\left(-\infty, \frac{-1}{2}\right) \cup\left(\frac{1}{2}, \infty\right) \cup\{0\}$
$\left(-\infty, \frac{-1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, \infty\right) \cup\{0\}$
Which of the following function is even function
Let $A= \{1, 2, 3, 4\}$ and $R : A \to A$ be the relation defined by $R = \{ (1, 1), (2, 3), (3, 4), ( 4, 2) \}$. The correct statement is
If a function $g(x)$ is defined in $[-1, 1]$ and two vertices of an equilateral triangle are $(0, 0)$ and $(x, g(x))$ and its area is $\frac{\sqrt 3}{4}$ , then $g(x)$ equals :-
Minimum integral value of $\alpha$ for which graph of $f(x) = ||x -2| -\alpha|-5$ has exactly four $x-$intercepts-
The function $f(x) =$ ${x^{\frac{1}{{\ln \,x}}}}$