Let f:[0, pi] rightarrow R be defined as f(x) | vedclass

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(B) function $f(x)$ defined as $f(x) = \begin{cases} g(x), & x \in \mathbb{Q} \ h(x), & x 
otin \mathbb{Q} \end{cases}$ is continuous at points wher

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$4$

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(B) function $f(x)$ defined as $f(x) = \begin{cases} g(x), & x \in \mathbb{Q} \ h(x), & x 
otin \mathbb{Q} \end{cases}$ is continuous at points where $g(x) = h(x)$.Here,$g(x) = 	an^2 x$ and $h(x) = \sin x$.We need to find points $x \in [0, \pi]$ such that $	an^2 x = \sin x$.$\frac{\sin^2 x}{\cos^2 x} = \sin x$$\sin^2 x = \sin x \cos^2 x = \sin x (1 - \sin^2 x)$$\sin^2 x + \sin x \sin^2 x - \sin x = 0$ is incorrect. Let's re-evaluate:$\sin^2 x = \sin x (1 - \sin^2 x) \implies \sin x (\sin x + \cos^2 x - 1) = 0$ is not correct.Correct approach: $\sin^2 x = \sin x \cos^2 x \implies \sin^2 x = \sin x (1 - \sin^2 x) \implies \sin^2 x + \sin^3 x - \sin x = 0$ is wrong.Correct: $\sin^2 x = \sin x (1 - \sin^2 x) \implies \sin^2 x = \sin x - \sin^3 x \implies \sin^3 x + \sin^2 x - \sin x = 0$.$\sin x (\sin^2 x + \sin x - 1) = 0$.Case $1$: $\sin x = 0 \implies x = 0, \pi$.Case $2$: $\sin^2 x + \sin x - 1 = 0$. Let $u = \sin x$. $u^2 + u - 1 = 0 \implies u = \frac{-1 \pm \sqrt{5}}{2}$.Since $u = \sin x \in [0, 1]$,we take $u = \frac{\sqrt{5}-1}{2}$.This gives two values of $x$ in $[0, \pi]$: $x_1 = \arcsin\left(\frac{\sqrt{5}-1}{2}ight)$ and $x_2 = \pi - \arcsin\left(\frac{\sqrt{5}-1}{2}ight)$.Total points are $0, \pi, x_1, x_2$,which are $4$ points.

Let $f:[0, \pi] \rightarrow R$ be defined as $f(x)=\begin{cases} \sin x, & \text{if } x \text{ is irrational and } x \in[0, \pi] \\ \tan^2 x, & \text{if } x \text{ is rational and } x \in[0, \pi] \end{cases}$. The number of points in $[0, \pi]$ at which the function $f$ is continuous is

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