Question:
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
Solution:
(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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