$f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{\sqrt{x}}$,then:

Let $S, T, U$ be three non-void sets and $f: S \rightarrow T, g: T \rightarrow U$ and the composed mapping $g \circ f: S \rightarrow U$ be defined. If $g \circ f$ is an injective mapping,then:

Let $f$ and $g$ be functions defined by $f(x) = \frac{x}{x + 1}$ and $g(x) = \frac{x}{1 - x}$. Then $(fog)(x)$ is

For a suitably chosen real constant $a$,let a 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 $f\left(-\frac{1}{5}\right)$ is equal to

If $f(x)=e^x$ and $g(x)=\ln(x)$ for all $x \in [1, \infty)$,then $f \circ g$ is . . . . . .

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x) = x^{2} - 3x + 4$ and $g(x) = 2x + 1$,then the value of $x$ for which $f(x) = (f \circ g)(x)$ is