Question

Let, $f\left(x\right)=\{\begin{array}{c}\left[x\right],-2\le x\le -1\\ |x|+1,-1<x\le 2\end{array}$ and $g\left(x\right)=\{\begin{array}{c}\left[x\right],-\pi \le x\le 0\\ \sin x,0<x\le \pi \end{array},$ then the number of integral points in the range of $g\left(f\right(x\left)\right)$ is

Answer 1

$g\left(f\right(x\left)\right)=\{\begin{array}{c}\left[f\right(x\left)\right],-\pi \le f\left(x\right)\le 0\\ \sin f\left(x\right),0<f\left(x\right)\le \pi \end{array}$
Now, from graph of $f\left(x\right)$


$f\left(x\right)\in [-\pi ,0]\Rightarrow x\in [-2,-1]$
$f\left(x\right)\in (0,\pi ]\Rightarrow x\in (-1,2]$
$\therefore g\left(f\right(x\left)\right)=\{\begin{array}{c}\left[f\right(x\left)\right],x\in [-2,-1]\\ \sin f\left(x\right),x\in (-1,2]\end{array}$
$=\{\begin{array}{c}-2,x\in [-2,-1)\\ -1,x=-1\\ \sin (|x|+1),x\in (-1,2]\end{array}$
$\therefore$Range of $g\left(f\right(x\left)\right)=\{-2,-1\}\cup [\sin 1,\sin 3]$
$(\because \forall x\in (-1,2],|x|+1\in [1,3\left]\right)$
No. of integers in $[\sin 1,\sin 3]$ is $1$ ie. $\sin \frac{\pi }{2}$
$\therefore$ Total number of integers in the range $\{–2,–1,1\}$ is $3$