Let $f(x)$ and $g(x)$ be two functions given by $f\left( x \right) = \frac{{2\sin \pi x}}{x}$ and $g\left( x \right) = f\left( {1 – x} \right) + f\left( x \right).$ If $g\left( x \right) = kf(\frac{x}{2})f\left( {\frac{{1 – x}}{2}} \right)$,then the value of $k$ is

If the function $f\,:\,R – \,\{ 1, – 1\}  \to A$ defined by $f\,(x)\, = \frac{{{x^2}}}{{1 – {x^2}}},$ is surjective, then $A$ is equal to

Let $P(x)$ be a polynomial with real coefficients such that $P\left(\sin ^2 x\right)=P\left(\cos ^2 x\right)$ for all $x \in[0, \pi / 2)$. Consider the following statements:

$I.$ $P(x)$ is an even function.

$II.$ $P(x)$ can be expressed as a polynomial in $(2 x-1)^2$

$III.$ $P(x)$ is a polynomial of even degree.

Then,

The domain of the function $f(x) = \frac{{{{\sin }^{ – 1}}(3 – x)}}{{\ln (|x|\; – 2)}}$ is

If $f\left( x \right) + 2f\left( {\frac{1}{x}} \right) = 3x,x \ne 0$ and $S = \left\{ {x \in R:f\left( x \right) = f\left( { – x} \right)} \right\}$;then $S :$