Question

Let $f\left(x\right)={\int }_0^{x}t\sin \frac{1}{t}dt$. Then the number of points of discontinuity of the function $f\left(x\right)$ in the open interval $(0,\pi )$ is

• A. $0$
• B. $1$
• C. $2$
• D. infinite

Answer 1

The correct option is A $0$

$f\left(x\right)={\int }_0^{x}t\sin \frac{1}{t}dt$
$\Rightarrow f^{\prime }\left(x\right)=x\sin \frac{1}{x}=$ well defined and continuous in $(0,\pi )$
Thus in $(0,\pi ),$ $f^{\prime }\left(x\right)$ exist $\Rightarrow$ $f$ is differentiable function.
And we know if a function is differentiable in some interval then it must be continuous in that interval.