Question

Let $f\left(x\right)=\{\begin{array}{c}-1,-2\le x<0\\ x^2-1,0<x\le 2\end{array}$ and $g\left(x\right)=\left|f\left(x\right)\right|+f\left|x\right|$ then the number of points at which g(x) is non differentiable, is

• A. at most one point
• B. 2
• C. exactly one point
• D. infinite

Answer 1

Answer: (B)

2

$f\left(x\right)=\{\begin{array}{cc}-1,& -2\le x<0\\ x^2-1,& 0<x\le 2\end{array}$

$g\left(x\right)=|f\left(x\right)|+f\left(|x|\right)$

$g\left(x\right)=\{\begin{array}{cc}1+x^2-1,& -2\le x<0\\ |x^2-1|+x^2-1,& 0<x\le 2\end{array}g\left(x\right)=\{\begin{array}{cc}{x}^2& -2\le x<0\\ 0& 0<x\le 2\\ 2x^2-2& 1<x\le 2\end{array}$

At $x=0$ $g\left(x\right)$ is not defined.

At $x=1$

$LHD=1$

$RHD=4x(x=1)-4$

Not differentiable

Total $2$ points.

$B$ is correct.