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

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.

$f(x) = (20 - x^4)^{1/4}$ for $0 < x < \sqrt{5}$, then $f(f(1/2))$ is equal to

Let $f(x) = \sin x$ and $g(x) = \ln |x|$. If the ranges of the composite functions $fog$ and $gof$ are $R_1$ and $R_2$ respectively,then:

For $x \in R, x \neq 0,$ let $f_0(x) = \frac{1}{1 - x}$ and $f_{n + 1}(x) = f_0(f_n(x)),$ $n = 0, 1, 2, ....$ Then the value of $f_{100}(3) + f_1\left( \frac{2}{3} \right) + f_2\left( \frac{3}{2} \right)$ is equal to

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