Question

Assertion :If $f\left(x\right)=0$ has two distinct positive real roots then number of non-differentiable points of $y=\left|f\right(-\left|x\right|\left)\right|$ is $1$ Reason: Graph of $y=f\left(\left|x\right|\right)$ is symmetrical about y-axis

• 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 B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

A.$y=\left|f(-\left|x\right|)\right|$

$y^{{}^{\prime }}=\frac{\left|f(-\left|x\right|)\right|}{f(-\left|x\right|)}\times \frac{-\left|x\right|}{x}$

given that $f\left(x\right)$ has only $2$ roots and they are positive.

$-\left|x\right|$ will give us only negative values

$\Rightarrow f(-\left|x\right|)$ can not be $0$

$\therefore y^{{}^{\prime }}$ is not differentiable at only $1$ point i.e., $x=0$ where $y^{{}^{\prime }}$ is not defined.

R: $y=f\left(\left|x\right|\right)\Rightarrow f\left(|-n|\right)=f\left(\left|n\right|\right)$

$x=\pm n$ will have same y value

The y-axis acts as a mirror

$\Rightarrow f\left(\left|x\right|\right)$ is symmetrical about y-axis.