If $f(x) = x^2 + 1$,then $fof(x)$ is equal to

For a suitably chosen real constant $a$,let the function $f: R-\{-a\} \rightarrow R$ be defined by $f(x)=\frac{a-x}{a+x}$. Further,suppose that for any real number $x \neq-a$ and $f(x) \neq-a$,$(f \circ f)(x)=x$. Then,$f\left(-\frac{1}{2}\right)$ is equal to

Given the functions $f: R \rightarrow R$ defined by $f(x) = 2x^2 - 5$ and $g: R \rightarrow R$ defined by $g(x) = \frac{x}{x^2 + 1}$,find the composite function $(g \circ f)(x)$.

If $\phi(x) = x^2 + 1$ and $\psi(x) = 3^x$,then find $\phi \{ \psi(x) \}$ and $\psi \{ \phi(x) \}$.

If $f(x)=\frac{3x+4}{5x-7}$ and $g(x)=\frac{7x+4}{5x-3}$,then $f(g(x))=$

If $f: R - \{\frac{3}{7}\} \rightarrow R - \{\frac{3}{7}\}$ is given by $f(x) = \frac{3x+5}{7x-3}$,then the statement which is not true is: