Question

If $f\left(x\right)=\{\begin{array}{cc}{x}^2+2,& x<0\\ -2e^{-x},& 0\le x<2,\\ \frac{-2}{e^2}(x-3),& x\ge 2\end{array}$ then

• A. $|f\left(x\right)|$ is discontinuous at $0$ points
• B. $|f\left(x\right)|$ is discontinuous at $2$ points
• C. $|f\left(x\right)|$ is not differentiable at $2$ points
• D. $|f\left(x\right)|$ is not differentiable at $3$ points

Answer 1

Answer: (A), (C)

The correct options are
A $|f\left(x\right)|$ is discontinuous at $0$ points
C $|f\left(x\right)|$ is not differentiable at $2$ points

Graph of $f\left(x\right)$



Graph of $|f\left(x\right)|$


The graph is discontinuous at $0$ points
Before $x=0$ the function is differentiable at all points because our function is a quadratic function.
At $x=2$ and $x=0$, the function changes definition.
So we need to check differentiability.

At $x=3$, a linear function changes direction (sharp turn). So the function is not differentiable.

At $x=0$
L.H.D. $=0$
R.H.D $=-2e^{-x}{|}_{x=0}=-2$
So, function is not differentiable.

At $x=2$
L.H.D. $=-2e^{-x}{|}_{x=2}=\frac{-2}{e^2}$
R.H.D. $=\frac{-2}{e^2}$

Therefore, $|f\left(x\right)|$ is not differentiable at $x=0$ and $x=3$
$\therefore$ There are only $2$ points where function is not differentiable.