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 :-

If function $f(x) = \frac{1}{2} - \tan \left( {\frac{{\pi x}}{2}} \right)$; $( - 1 < x < 1)$ and $g(x) = \sqrt {3 + 4x - 4{x^2}} $, then the domain of $gof$ is

If $a, b$ be two fixed positive integers such that $f(a + x) = b + {[{b^3} + 1 - 3{b^2}f(x) + 3b{\{ f(x)\} ^2} - {\{ f(x)\} ^3}]^{\frac{1}{3}}}$ for all real $x$, then $f(x)$ is a periodic function with period

If $f$ is an even function defined on the interval $(-5, 5)$, then four real values of $x$ satisfying the equation $f(x) = f\left( {\frac{{x + 1}}{{x + 2}}} \right)$ are

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

Let for $a \ne {a_1} \ne 0,$ $f\left( x \right) = a{x^2} + bx + c\;,g\left( x \right) = {a_1}{x^2} + {b_1}x + {c_1},p\left( x \right) = f\left( x \right) - g\left( x \right),$ If $p\left( x \right) = 0$ only for  $ x=-1 $ and $p\left( { - 2} \right) = 2$ then value of $p\left( 2 \right)$ is