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

If $0 < x < \frac{\pi }{2},$ then

Let $f : R \rightarrow R$ be a function such that $f(x)=\frac{x^2+2 x+1}{x^2+1}$. Then

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

If $f(x + ay,\;x - ay) = axy$, then $f(x,\;y)$ is equal to

Show that the function $f : R \rightarrow R$ given by $f ( x )= x ^{3}$ is injective.