Question

If $f(x)$ defined by $f(x)=\{\begin{array}{ll}\frac{|x^2-x|}{x^2-x},& x\ne 0,1\\ 1,& x=0\text{ then }\\ -1,& x=1\end{array}$
$f(x)$ 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; $f(x)=\{\begin{array}{ll}\frac{x^2-x}{x^2-x}=1,& \text{ if }x<0\text{ or }x>1\\ -\frac{\left(x^2-x\right)}{\left(x^2-x\right)}=-1,& \text{ if }0<x<1\\ 1,& \text{ if }x=0\\ -1,& \text{ if }x=1\end{array}$
$=\{\begin{array}{llc}1,& \text{ if }& x\le 0\text{ or }x>1\\ -1,& \text{ if }& 0<x\le 1\end{array}$
Now, $\underset{x\to 0^{-}}{\lim }1=1$ and $\underset{x\to 0^{+}}{\lim }f(x)=\underset{x\to 0}{\lim }-1=-1$
Clearly, $\underset{x\to 0^{-}}{\lim }f(x)\ne \underset{x\to 0^{+}}{\lim }f(x)$.
So, $f(x)$ is not continuous at $x=0$.
It can be easily seen that it is not continuous at $x=1$ also.