The number of points in $[0, \pi]$ at which the function $f$ is continuous is
We have,
$f(x)=\left\{\begin{array}{ll} \sin x, & \text { if } x \text { is irrational and } x \in[0, \pi] \\\tan ^2 x, & \text { if } x \text { is rational and } x \in[0, \pi]\end{array}\right.$
$f(x)$ is continuous at $x=0$ and $\pi$
For other point, then
$\sin x=\tan ^2 x$
$\sin x=\frac{\sin ^2 x}{\cos ^2 x}$
$\Rightarrow \quad \sin x\left(1-\sin ^2 x\right)=\sin ^2 x$
$\Rightarrow \quad \quad \sin ^2 x+\sin x-1=0$
$\Rightarrow \quad \sin x =\frac{-1 \pm \sqrt{5}}{2}$
$\Rightarrow \quad \sin x =\frac{\sqrt{5}-1}{2}, \sin x \neq \frac{-1-\sqrt{5}}{2}$
$\quad x=\sin ^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$
$x=\pi-\sin ^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$
$\therefore f(x)$ is continuous at $x=0, \pi$,
$\sin ^{-1} \frac{\sqrt{5}-1}{2}$ and $\pi-\sin ^{-1} \frac{\sqrt{5}-1}{2}$
Hence, $f(x)$ is continuous at $4$ points.
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