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Let $f:[0, \pi] \rightarrow R$ be defined as $f(x)=\left\{\begin{array}{ll}\sin x, & \text { if } x \text { is irrational and } x \in[0, \pi] \\ \tan ^{2} x, & \text { if } x \text { is rational and } x \in[0, \pi]\end{array}\right\}$
The number of points in $[0, \pi]$ at which the function $f$ is continuous is :
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The number of points in $[0, \pi]$ at which the function $f$ is continuous is :
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The correct answer is:
4
$\sin x=\tan ^{2} x$
$\sin ^{2} x=\sin x\left(1-\sin ^{2} x\right)$
$\sin x\left(\sin ^{2} x+\sin -1\right)=0$
$\sin x=0 \quad \sin x=\frac{\sqrt{5}-1}{2}$
$x=0, \pi \quad \sin ^{-1} \frac{\sqrt{5}-1}{2}, \pi-\sin ^{-1} \frac{\sqrt{5}-1}{2}$
$\sin ^{2} x=\sin x\left(1-\sin ^{2} x\right)$
$\sin x\left(\sin ^{2} x+\sin -1\right)=0$
$\sin x=0 \quad \sin x=\frac{\sqrt{5}-1}{2}$
$x=0, \pi \quad \sin ^{-1} \frac{\sqrt{5}-1}{2}, \pi-\sin ^{-1} \frac{\sqrt{5}-1}{2}$
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