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

The range of the polynomial $P(x)=4 x^3-3 x$ as $x$ varies over the interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$ is

Let $f(x)=x^6-2 x^3+x^3+x^2-x-1$ and $g(x)=x^4-x^3-x^2-1$ be two polynomials. Let $a, b, c$ and $d$ be the roots of $g(x)=0$. Then, the value of $f(a)+f(b)+f(c)+f(d)$ is

Let a function $f : R \rightarrow  R$ is defined such that $3f(2x^2 -3x + 5) + 2f(3x^2 -2x + 4) = x^2 -7x + 9\ \ \  \forall  x \in R$, then the value of $f(5)$ is-

If non-zero real numbers $b$ and $c$ are such that $min \,f\left( x \right) > \max \,g\left( x \right)$, where $f\left( x \right) = {x^2} + 2bx + 2{c^2}$  and $g\left( x \right) = {-x^2} - 2cx + {b^2}$$\left( {x \in R} \right)$; then $\left| {\frac{c}{b}} \right|$ lies in the interval

Domain of the function $f(x)\,=\,\frac{1}{{\sqrt {(x + 1)({e^x} - 1)(x - 4)(x + 5)(x - 6)} }}$