Question

If $f\left(x\right)$ defined by $f\left(x\right)=\{\begin{array}{c}\frac{|x^2-x|}{x^2-x},x\ne 0,1\\ 1,x=0\\ -1,x=1\end{array}$ then $f\left(x\right)$ 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: (D)

$x$ except at $x=0$ and $x=1$

We have: $f\left(x\right)=\{\begin{array}{c}\frac{x^2-x}{x^2-x}=1,\text{if}x<0orx>1\\ -\frac{(x^2-x)}{(x^2-x)}=-1,\text{if}0<x<1\\ 1,\text{if}x=0\\ -1,\text{if}x=1\end{array}$
$=\{\begin{array}{c}1,\text{if}x\le 0\text{or}x>1\\ -1,\text{if}0<x\le 1\end{array}$
Now, $\underset{x\to 0^{-}}{lim}1=1$ and $\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0}{lim}-1=-1$
Clearly, $\underset{x\to 0^{-}}{lim}f\left(x\right)\ne \underset{x\to 0^{+}}{lim}f\left(x\right)$
So, $f\left(x\right)$ is not continuous at $x=0$. It can be easily seen that it is not continuous at $x=1$ also.