Question

The points of discontinuity of $\lim_{x \rightarrow 3^{-}}f(x) = 0$ where $\lim_{x \rightarrow 3^{+}}f(x) \neq \lim_{x \rightarrow 3^{-}}f(x)$ is

(1) $f(x) = \left\{ \begin{matrix} 3x,\text{if}0 \leq x \leq 1 \\ 5 - 3x,\text{if}1 < x \leq 2 \end{matrix} \right.$ (2) $\lim_{x \rightarrow 1}f(x) = f(1)$ (3)$\lim_{x \rightarrow 1}f(x) = 3$ (4) None

• A. $f(x) = \left\{ \begin{matrix} 3x,\text{if}0 \leq x \leq 1 \\ 5 - 3x,\text{if}1 < x \leq 2 \end{matrix} \right.$
• B. $\lim_{x \rightarrow 1}f(x) = f(1)$
• C. $\lim_{x \rightarrow 1}f(x) = 3$
• D. None

Answer 1

Answer: (A)

$f(x) = \left\{ \begin{matrix} 3x,\text{if}0 \leq x \leq 1 \\ 5 - 3x,\text{if}1 < x \leq 2 \end{matrix} \right.$

Answer 2

The function $u = f ( x ) = \frac { 1 } { x - 1 }$ is discontinuous at the point The function $y = g ( x ) = \frac { 1 } { u ^ { 2 } + u - 2 } = \frac { 1 } { ( u + 2 ) ( u - 1 ) }$ is discontinuous at $u = - 2$ and $u = 1$

when $u = - 2 \Rightarrow \frac { 1 } { x - 1 } = - 2 \Rightarrow x = \frac { 1 } { 2 }$ , when

$u = 1 \Rightarrow \frac { 1 } { x - 1 } = 1 \Rightarrow x = 2$

Hence, the composite $y = g ( f ( x ) )$ is discontinuous at three points $= \frac { 1 } { 2 } , 1,2$.