Question

If $f\left(x\right)=|1-x|,$ then the number of points (excluding the boundary points if exist) where $g\left(x\right)={\sin }^{-1}\left(f\right(|x|\left)\right)$ is non-differentiable, is

• A. $5$
• B. $2$
• C. $4$
• D. $3$

Answer 1

The correct option is D $3$

Domain of $g\left(x\right):$
$-1\le f\left(|x|\right)\le 1\Rightarrow -1\le |1-|x||\le 1\Rightarrow 0\le |1-|x||\le 1\Rightarrow -1\le 1-|x|\le 1\Rightarrow 0\le |x|\le 2\Rightarrow -2\le x\le 2$
$\therefore$ Domain of $g\left(x\right)$ is $[-2,2]$



$g\left(x\right)={\sin }^{-1}\left(f\right(|x|\left)\right)$
Clearly, $g\left(x\right)$ is non-differentiable at $x=-1,0,1$.
$\therefore$ Number of non-differentiable points is $3.$