Let $\mathrm{f}=\mathbf{R} \rightarrow \mathbf{R}$ be differentiable at $\mathrm{c} \in \mathbf{R}$ and $\mathrm{f}(\mathrm{c})=0$. If $g(x)=|f(x)|$, then at $x=c, g$ is :

  1. (1) not differentiable if $f^{\prime}(c)=0$

  2. (2) differentiable if $f^{\prime \prime}(c) \neq 0$

  3. (3) differentiable if $\mathrm{f}^{\prime \prime}(\mathrm{c})=0$

  4. (4) not differentiable

Correct Option: , 3


$g^{\prime}(c)=\lim _{x \rightarrow c} \frac{g(x) \quad g(c)}{x \quad c}$

$\Rightarrow g^{\prime}(c)=\lim _{x \rightarrow c} \frac{|f(x)|-|f(c)|}{x-c}$

Since, $f(c)=0$

Then, $g^{\prime}(c)=\lim _{x \rightarrow c} \frac{|f(x)|}{x-c}$

$\Rightarrow g^{\prime}(c)=\lim _{x \rightarrow c} \frac{f(x)}{x-c} ;$ if $f(x)>0$

$\Rightarrow g^{\prime}(c)=f^{\prime}(c)=-f^{\prime}(c)$

$\Rightarrow 2 f^{\prime}(c)=0 \Rightarrow f^{\prime}(c)=0$

Hence, $g(x)$ is differentiable if $f^{\prime}(c)=0$


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now