Question

The function 

\[f\left( x \right) = \frac{4 - x^2}{4x - x^3}\]

 

Options

• discontinuous at only one point

• discontinuous exactly at two points

• discontinuous exactly at three points

• none of these

Answer 1

 discontinuous exactly at three points 

Given:

\[f\left( x \right) = \frac{4 - x^2}{4x - x^3}\]

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

\[\Rightarrow f\left( x \right) = \frac{1}{x}, x \neq 0 \text{ and }  4 - x^2 \neq 0 \text{ or } x \neq 0, \pm 2\]

Clearly,

\[f\left( x \right)\]  is defined and continuous for all real numbers except \[\left\{ 0, \pm 2 \right\}\] .

Therefore, 

\[f\left( x \right)\]  is discontinuous exactly at three points.