Question

If $f(x)$ defined by $f(x)=\begin{cases}\frac{\left|x^{2}-x\right|}{x^{2}-x}, & x \neq 0,1 \\ 1, & x=0 \text { then } \\ -1, & x=1\end{cases}$
$f(x)$ is continuous for all

• A. $ x $
• B. $ x $ except at $ x=0 $
• C. $ x $ except at $ x=1 $
• D. $ x $ except at $ x=0 $ and $ x=1 $

Answer 1

Answer: (C)

We have; $f(x)=\begin{cases}\frac{x^{2}-x}{x^{2}-x}=1, & \text { if } x< 0 \text { or } x > 1 \\ -\frac{\left(x^{2}-x\right)}{\left(x^{2}-x\right)}=-1, & \text { if } 0 < x < 1 \\ 1, & \text { if } x=0 \\ -1, & \text { if } x=1\end{cases}$
$=\begin{cases}1, & \text { if } & x \leq 0 \text { or } x > 1 \\ -1, & \text { if } & 0 < x \leq 1\end{cases}$
Now, $\displaystyle\lim _{x \rightarrow 0^{-}} 1=1$ and $\displaystyle\lim _{x \rightarrow 0^{+}} f(x)=\displaystyle\lim _{x \rightarrow 0}-1=-1$
Clearly, $\displaystyle\lim _{x \rightarrow 0^{-}} f(x) \neq \displaystyle\lim _{x \rightarrow 0^{+}} f(x)$.
So, $f(x)$ is not continuous at $x=0$.
It can be easily seen that it is not continuous at $x=1$ also.