Question

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

• A. 6
• B. 4
• C. 2
• D. 0

Answer 1

Answer: (B)

We have,
$f(x) = \begin{cases} \sin\,x & \text{if $x$ is irrational and $x \in [0, \pi ]$} \\[2ex] y & \text{if $x$ is rational and $x\in [0, \pi]$} \end{cases}$
$f(x)$ is continuous at $x = 0$ and $\pi$
For other point, then
$\sin \,x = \tan^2 \,x$
$\sin\,x = \frac{\sin^2 \,x}{\cos^2\,x}$
$\Rightarrow \sin\,x ( 1 - \sin^2 \,x ) = \sin^2 \,x$
$\Rightarrow 1 - \sin^2 \,x = \sin\,x$
$\Rightarrow \sin^2 \,x + \sin\,x - 1 = 0$
$ \Rightarrow \sin\,x = \frac{-1 \neq \sqrt{5}}{2}$
$ \sin\,x = \frac{\sqrt{5} - 1}{2}, \sin \,x \pm \frac{-1-\sqrt{5}}{2}$
$ 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.