Question

Assertion :The number of points of discontinuity of $f\left(x\right)$ are $0$, where $f\left(x\right)={\int }_0^{x}t\sin \left(\frac{1}{t}\right)dt$ Reason: The function $g\left(x\right)=max\{-x,1,x^2\}\forall x\in R$ is not differentiable at two values of $x$.

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

A:

$f\left(x\right)={\int }_0^{x}t\sin \left(\frac{1}{t}\right)dt$

At $x=0$

${lim}_{x\to 0}f\left(x\right)={\int }_0^{0}t\sin \left(\frac{1}{t}\right)dt=0$

Continuous at $x=0$ and everywhere else

R:

$f\left(x\right)=\{\begin{array}{c}-xatx=-1\\ 1-1\le x\le 1\\ x^{2}xϵ[-\infty ,-1]\cup [1,\infty ]\end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{c}-1x=-1\\ 0-1\le x\le 1\\ 2xxϵ[-\infty ,-1]\cup [1,\infty ]\end{array}$

at $x=-1$

We have $2$ different values of $f^{\prime }\left(x\right)$

$f\left(x\right)$ is not differentiable at $x=-1$

Similarly at $x=1$