Question

Assertion :Consider the function $f\left(x\right)=x^2-|x^2-1|+2|\left|x\right|-1|+2\left|x\right|-7$.

$f$ is not differentiable at $x=1,-1$ and $0$.

Reason: $\left|x\right|$ is not differentiable at $x=0$ and $|x^2-1|$ is not differentiable at $x=1$ and $-1$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

$\left|x\right|$ is not differentiable at $x=0$ and $|x^2-1|$ is not differentiable at $x=\pm 1$

So reason is correct

Now $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}{x}^2-x^2+1-2x-2-2x-7;& x\le -1\end{array}\\ \begin{array}{cc}{x}^2+x^2-1+2x+2-2x-7;& -1<x\le 0\end{array}\\ \begin{array}{cc}{x}^2+x^2-1-2x+2+2x-7;& 0<x\le 1\end{array}\\ \begin{array}{cc}{x}^2-x^2+1+2x-2+2x-7;& x>1\end{array}\end{array}$

$\Rightarrow f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}-4x-8;& x\le -1\end{array}\\ \begin{array}{cc}2x^2-6;& -1<x\le 0\end{array}\\ \begin{array}{cc}2x^2-6;& 0<x\le 1\end{array}\\ \begin{array}{cc}4x-8;& x>1\end{array}\end{array}\Rightarrow f^{\prime }\left(x\right)=\{\begin{array}{c}\begin{array}{cc}-4;& x<-1\end{array}\\ \begin{array}{cc}4x;& -1<x<1\end{array}\\ \begin{array}{cc}4;& x>1\end{array}\end{array}$

Clearly $f\left(x\right)$ is continuous at $x\in R$

$\therefore f\left(x\right)$ is differentiable at $x=-1$ and $1$

Therefore assertion is wrong but reason is correct.