The function $f\left( x \right) = \dfrac{{4 - {x^2}}}{{4x - {x^3}}}$ is
A. discontinuous at only one point
B. discontinuous exactly at two points
C. discontinuous exactly at three points
D. None of these

Hint: We will begin by writing the denominator of the given function as product of 2 factors by taking $x$ common. Now, the function will be discontinuous where the denominator will be zero. Hence, from each factor of the denominator, find the values of $x$ which makes the factor 0. Those points will be the points of discontinuity.

Complete step-by-step answer:
We are given that the function is $f\left( x \right) = \dfrac{{4 - {x^2}}}{{4x - {x^3}}}$
Now, we will take common $x$ from the denominator to make factors in denominator.
$f\left( x \right) = \dfrac{{4 - {x^2}}}{{x\left( {4 - {x^2}} \right)}}$
Now, we know that the function will not exist if the denominator is zero.
That is, we can say the function will be discontinuous if the denominator is zero.
Therefore, we can find the values where $x$will be 0 and such values will make the function discontinuous.
Now, the given function will be continuous when,
$x \ne 0$ and
$
  {x^2} - 4 \ne 0 \\
  {x^2} \ne 4 \\
  x \ne \pm 2 \\
$
That is if $x = 0,2, - 2$, then the function will be discontinuous.
Here, we can say that $x$ is discontinuous at 0, 2 and $ - 2$.
Thus, the function is discontinuous at 3 points.
Hence, option C is correct.

Note: In this question, we cannot cancel $\left( {4 - {x^2}} \right)$ from numerator and denominator because it is only possible when $x \ne \pm 2$. Many students make mistakes by cancelling the factor $\left( {4 - {x^2}} \right)$ and then they are left with only one point of discontinuity, which is incorrect. A discontinuous function has a graph with isolated points.