Solve this


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

(a) discontinuous at only one point

(b) discontinuous exactly at two points

(c) discontinuous exactly at three points

(d) none of these


(C) discontinuous exactly at three points

Given: $f(x)=\frac{4-x^{2}}{4 x-x^{3}}$

$\Rightarrow f(x)=\frac{4-x^{2}}{x\left(4-x^{2}\right)}$

$\Rightarrow f(x)=\frac{1}{x}, x \neq 0$ and $4-x^{2} \neq 0$ or $x \neq 0, \pm 2$

Clearly, $f(x)$ is defined and continuous for all real numbers except $\{0, \pm 2\}$.

Therefore, $f(x)$ is discontinuous exactly at three points.

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