Question

Let $f:[0,\pi ]\to R$ be defined as
$f(x)=\{\begin{array}{ll}\sin x& \text{if x is irrational and x∈[0,π]}\\ y& \text{if x is rational and x∈[0,π]}\end{array}$
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,
$f(x)=\{\begin{array}{ll}\sin x& \text{if x is irrational and x∈[0,π]}\\ y& \text{if x is rational and x∈[0,π]}\end{array}$
$f(x)$ is continuous at $x=0$ and $\pi$
For other point, then
$\sin x={\tan }^{2}x$
$\sin x=\frac{{\sin }^{2}x}{{\cos }^{2}x}$
$\Rightarrow \sin x(1-{\sin }^{2}x)={\sin }^{2}x$
$\Rightarrow 1-{\sin }^{2}x=\sin x$
$\Rightarrow {\sin }^{2}x+\sin x-1=0$
$\Rightarrow \sin x=\frac{-1\ne \sqrt{5}}{2}$
$\sin x=\frac{\sqrt{5}-1}{2},\sin x\pm \frac{-1-\sqrt{5}}{2}$
$x={\sin }^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$
$x=\pi -{\sin }^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$
$\therefore f(x)$ 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.