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:

Let the functions $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined as $f(x) = \begin{cases} x+2, & x < 0 \\ x^2, & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} x^3, & x < 1 \\ 3x-2, & x \geq 1 \end{cases}$. Then,the number of points in $R$ where $(f \circ g)(x)$ is $NOT$ differentiable is equal to

Let $f : R \rightarrow R$ and $g : R \rightarrow R$ be defined as $f(x) = \begin{cases} x+a, & x < 0 \\ |x-1|, & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} x+1, & x < 0 \\ (x-1)^2+b, & x \geq 0 \end{cases}$ where $a, b$ are non-negative real numbers. If $(g \circ f)(x)$ is continuous for all $x \in R$,then $a+b$ is equal to ......

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=x-[x]$ and $g(x)=[x]$ for $x \in R$, where $[x]$ is the greatest integer not exceeding $x$, then for every $x \in R, f(g(x))$ is equal to

If $f:[-6,6] \rightarrow R$ is defined by $f(x)=x^2-3$ for $x \in R$, then $(f \circ f \circ f)(-1)+(f \circ f \circ f)(0)+(f \circ f \circ f)(1)$ is equal to

Let $f(x)=x^2$ and $g(x)=\sin x$ for all $x \in R$. Then the set of all $x$ satisfying $(f \circ g \circ g \circ f)(x)=(g \circ g \circ f)(x)$,where $(f \circ g)(x)=f(g(x))$,is