If $f(x) = x^2$ and $g(x) = [x^2]$, where $[ \cdot ]$ represents the greatest integer function, then $(f \circ g)(\frac{3}{2}) + (g \circ f)(\frac{3}{2})$ is equal to ...

If $f:[0,3] \rightarrow [0,3]$ is defined by $f(x) = \begin{cases} 1+x, & 0 \leq x \leq 2 \\ 3-x, & 2 < x \leq 3 \end{cases}$,then $f(f(x))$ is:

If $g(x)=x^2+x-2$ and $\frac{1}{2}(g \circ f)(x)=2 x^2-5 x+2$,then one such function $f(x)=$

If $f: R \rightarrow R$ is defined by $f(x)= \begin{cases} |[x-5]|, & \text{for } x < 5 \\ [|x-5|], & \text{for } x \geq 5 \end{cases}$ Then,$(f \circ f)\left(-\frac{7}{2}\right) = ?$ (here,$[x]$ is the greatest integer function)

If $g(x)=x^{2}+x-1$ and $(g \circ f)(x)=4 x^{2}-10 x+5,$ then $f\left(\frac{5}{4}\right)$ is equal to

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