Question

Number of points of discontiuity of the function $f\left(x\right)=\frac{1}{2\sin x-1}$ in $[0,2021\pi ]$ is

• A. $2020$
• B. $2021$
• C. $2022$
• D. $2023$

Answer 1

The correct option is C $2022$

$f\left(x\right)=\frac{1}{2\sin x-1}$

$f\left(x\right)$ is discontinuous when
$2\sin x-1=0\Rightarrow \sin x=\frac{1}{2}\Rightarrow x=n\pi +(-1{)}^n\frac{\pi }{6},n\in Z$
For every $2\pi$ interval there are $2$ discontinuity points.
Now, length of domain $=2021\pi =1010\cdot 2\pi +\pi$
So, the total number of points of discontinuity $=1010\cdot 2+2(\because \text{there are two points of discontinuity in}[0,\pi ])=2022$