Which of the following functions are continuous on $(0,\pi )$

$\mathrm{On}(0,\pi )$

$\text{ a) }f\left(x\right)=\tan x\Rightarrow$We know that tan x is discontinuous $\text{ at }x=\pi /2$

$\text{ b) }f\left(x\right)={\int }_0^x t\sin \left(\frac{1}{t}\right)dt\text{ or }{f}^{\mathrm{\prime }}\left(x\right)=x\sin \left(\frac{1}{x}\right)\text{, }$$f\left(x\right)$ differentiable, continuous on $(0,\pi )$

$$\begin{gathered} \text{ c) L.H.L. }=\underset{x\to \frac{3\pi }{4}}{lim}\left(1\right)=1 \\ R.H.L=\underset{x\to \frac{3\pi }{4}}{lim}2\sin \frac{2x}{9}=1 \end{gathered}$$

$\text{ Also, }f\left(\frac{3\pi }{4}\right)=1.f\left(x\right)$ is continuous on $(0,\pi )$

$$\begin{gathered} \text{ d) L.H.L. }=\underset{x\to \frac{\pi }{2}}{lim}x\sin x=\frac{\pi }{2} \\ R.H.L=\underset{x\to \frac{\pi }{2}}{lim} \frac{\pi }{2}\sin (\pi +x)=-\frac{\pi }{2} \end{gathered}$$

f(x) is  not continuous