The function $f(x) = \frac{4 - x^2}{4x - x^3}$ is

discontinuous at exactly three points

The correct answer is Option (3) → discontinuous at exactly three points ##

We have, $f(x) = \frac{4 - x^2}{4x - x^3} = \frac{(4 - x^2)}{x(4 - x^2)}$

$= \frac{(4 - x^2)}{x(2^2 - x^2)} = \frac{4 - x^2}{x(2 + x)(2 - x)}$

Clearly, $f(x)$ is discontinuous at exactly three points $x = 0$, $x = -2$ and $x = 2$.