Define $\operatorname{sgn}(x)= \begin{cases}1, & \text { if } x \gt 0, \\ -1, & \text { if } x \lt 0, \\ 0, & \text { if } x=0 .\end{cases}$
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be the function defined by $f(x)=(x-\sqrt{5}) \operatorname{sgn}\left(x^2-5\right)$. Then the number of discontinuities of $f$ is

• A

  $0$
• Correct answer
  B

  $1$
• C

  $2$
• D

  $3$

Hence, the correct answer is option: 'B'.