$f: R \rightarrow R$ and $g:[0, \infty) \rightarrow R$ are defined by $f(x)=x^2$ and $g(x)=\sqrt{x}$. Which one of the following is not true?

If the function is $f(x)=\frac{1}{x+2}$,then the point of discontinuity of the composite function $y=f(f(x))$ is

The graph of $y = f(x)$ is shown. The number of solutions of the equation $f(f(x)) = 2$ is:

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x) = \begin{cases} x+2, & x>0 \\ 2-x, & x \leq 0 \end{cases}$ and $g(x) = \begin{cases} x^2-2x-2, & 1 \leq x < 2 \\ x-7, & x \geq 2 \\ x+5, & x < 1 \end{cases}$,then $\lim _{x \rightarrow 0} g(f(x))$

Let $(g \circ f)(x) = \sin x$ and $(f \circ g)(x) = (\sin \sqrt{x})^2$. Then,

Let the functions $f:(-1,1) \rightarrow R$ and $g:(-1,1) \rightarrow(-1,1)$ be defined by $f(x)=|2 x-1|+|2 x+1|$ and $g(x)=x-[x]$,where $[x]$ denotes the greatest integer less than or equal to $x$. Let $f \circ g:(-1,1) \rightarrow R$ be the composite function defined by $(f \circ g)(x)=f(g(x))$. Suppose $c$ is the number of points in the interval $(-1,1)$ at which $f \circ g$ is $NOT$ continuous,and suppose $d$ is the number of points in the interval $(-1,1)$ at which $f \circ g$ is $NOT$ differentiable. Then the value of $c+d$ is.