Question

The points of discontinuity of the function

f(x) = ${\displaystyle \frac{1}{x-1}}$ if 0 ≤ x ≤ 2

= ${\displaystyle \frac{x+5}{x+3}if2<x\le 4}$ in its domain are.

Options

• x = 1, x = 2

• x = 0, x = 2

• x = 2 only

• x = 4 only

Answer 1

x = 1, x = 2

Explanation:

f(x) = ${\displaystyle \{\begin{array}{l}\frac{1}{x-1}\text{,}if0\le x\le 2\\ \frac{x+5}{x+3}\text{,}if2<x\le 4\end{array}}$

Clearly, f(x) is not defined at x = 1

Hence, f(x) is discontinuous at x = 1

At x = 2

${\displaystyle \underset{x\to 2^{-}}{lim}\text{f}\left(x\right)=\underset{x\to 2^{-}}{lim}\frac{1}{x-1}=\frac{1}{2-1}}$ = 1

${\displaystyle \underset{x\to 2^{+}}{lim}\text{f}\left(x\right)=\underset{x\to 2^{+}}{lim}\frac{x+5}{x+3}=\frac{2+5}{2+3}=\frac{7}{5}}$

∴ ${\displaystyle \underset{x\to 2^{-}}{lim}\text{f}\left(x\right)\ne \underset{x\to 2^{+}}{lim} \text{f}\left(x\right)}$

∴ f(x) is discontinuous at x = 2