Question

Assertion :Let $f:R\to R$ be a function defined by $f\left(x\right)=max\{x,x^3\}$. Then, $f\left(x\right)$ is not differentiable at $x=-1,0,1$ Reason: $f\left(x\right)=\{\begin{array}{c}x,x\le -1\\ x^3,-1<x\le 0\\ x,0<x\le 1\\ x^3,x>1\end{array}$

• 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. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

If $x<-1,$ then $x>x^3.$ So, $f\left(x\right)=x$

If $x=-1,$ then $x=x^3.$ So, $f\left(x\right)$

If $-1<x<1,$ then $x<x^3.$So, $f\left(x\right)=x^3$

If $x=0,$ then $x=x^3.$ So, $f\left(x\right)=x^3$

If $0<x<1,$ then $x>x^3.$ So, $f\left(x\right)=x$

If $x=1,$ then $x=x^3.$ So, $f\left(x\right)=x$

If $x>1,$ then $x<x^3.$ So, $f\left(x\right)=x^3$

Thus, $f\left(x\right)=x,x\le -1$

$f\left(x\right)=x^3,-1<x\le 0$

$f\left(x\right)=x,0<x\le 1$

$f\left(x\right)=x^3,x>1$

Clearly, $f\left(x\right)$ is not differentiable at $x=-1,0,1.$