If $g(x) = x^2 + x - 2$ and $\frac{1}{2} (g \circ f)(x) = 2x^2 - 5x + 2$,then $f(x)$ is equal to:

If $f(x) = \frac{x - 3}{x + 1}$,then $f[f\{f(x)\}]$ equals

Let $f, g: R \rightarrow R$ be defined as: $f(x)=|x-1|$ and $g(x)=\begin{cases} e^x, & x \geq 0 \\ x+1, & x \leq 0 \end{cases}$. Then the function $f(g(x))$ is

If $f(x) = \begin{cases} 2+2x, & -1 \leq x < 0 \\ 1-\frac{x}{3}, & 0 \leq x \leq 3 \end{cases}$ and $g(x) = \begin{cases} -x, & -3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{cases}$,then the range of $(f \circ g)(x)$ is:

Let a relation $R$ be defined by $R = \{(4, 5), (1, 4), (4, 6), (7, 6), (3, 7)\}$. Then ${R^{ - 1}}oR$ is

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are two functions defined by $f(x)=2x-3$ and $g(x)=5x^2-2$,then the least value of the function $(g \circ f)(x)$ is