Question

The function$f\left(x\right)=$$\frac{4-\mathrm{x}}{4\mathrm{x}-{\mathrm{x}}^3}$ is,

• A. discontinous at one point
• B. discontinous at two points
• C. discontinous at three points
• D. none of these

Answer 1

The correct option is C

discontinous at three points

Continuous Function:

A function is said to be continuous at a point$\mathrm{x}=\mathrm{a}$, if $\mathrm{f}\left(\mathrm{x}\right)$ exists, and $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{a}\right)$. It implies that if the left-hand limit $(\mathrm{L}.\mathrm{H}.\mathrm{L})$, right-hand limit $(\mathrm{R}.\mathrm{H}.\mathrm{L})$, and the value of the function at $\mathrm{x}=\mathrm{a}$ exists and these parameters are equal to each other, then the function $\mathrm{f}$ is said to be continuous at$\mathrm{x}=\mathrm{a}$. If the function is undefined or does not exist, then we say that the function is discontinuous.

Continuity in open intervals $(\mathrm{a},\mathrm{b})$

$\mathrm{f}\left(\mathrm{x}\right)$ will be continuous in the open interval $(\mathrm{a},\mathrm{b})$ if at any point in the given interval the function is continuous.

$f\left(x\right)=\frac{4-\mathrm{x}}{4\mathrm{x}-{\mathrm{x}}^3}$

At $\mathrm{x}=0$, the value of denominator is 0. So the function is discontinuous at $\mathrm{x}=0$.

$\mathrm{f}\left(\mathrm{x}\right)$ is discontinuous where $4\mathrm{x}-{\mathrm{x}}^3=0$

$$\begin{gathered} \Rightarrow \mathrm{x}(4\mathrm{x}-{\mathrm{x}}^2)=0 \\ \Rightarrow \mathrm{x}=0\mathrm{or}4\mathrm{x}-{\mathrm{x}}^2=0 \\ \Rightarrow \mathrm{x}=0\mathrm{or}\mathrm{x}=\pm 2 \end{gathered}$$

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)$is discontinuous at exactly three points.

Hence, the correct answer is (C) discontinuous at three points.