Question

Let $f:R\to R$ be differentiable at $c\in R$ and $f\left(c\right)=0$. If $g\left(x\right)=\left|f\right(x\left)\right|$, then at $x=c,g$ is :

• A. differentiable if $f^{\prime }\left(c\right)\ne 0$
• B. differentiable if $f^{\prime }\left(c\right)=0$
• C. not differentiable if $f^{\prime }\left(c\right)=0$
• D. not differentiable

Answer 1

The correct option is B differentiable if $f^{\prime }\left(c\right)=0$

$g\left(x\right)=\left|f\right(x\left)\right|\Rightarrow g^{\prime }\left(c^{-}\right)=\underset{x\to c^{-}}{lim}\frac{\left|f\right(x\left)\right|-f\left(c\right)}{x-c}=\pm f^{\prime }\left(c\right)$
and
$\Rightarrow g^{\prime }\left(c^{+}\right)=\underset{x\to c^{+}}{lim}\frac{\left|f\right(x\left)\right|-f\left(c\right)}{x-c}=\pm f^{\prime }\left(c\right)$

For $g\left(x\right)$ to be differentiable at $x=c$.
$\text{R.H.D}=\text{L.H.D}$ i.e., $g^{\prime }\left(c^{-}\right)=g^{\prime }\left(c^{+}\right)$
So, $f^{\prime }\left(c\right)$ must be $0$, else it is non-differentiable.