Question

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

(d) x except at x = 0 and x = 1.

Given: $f\left(x\right)=\left\{\begin{array}{c}\frac{\left|x^2-x\right|}{x^2-x},x\ne 0,1\\ 1,x=0\\ -1,x=1\end{array}\right.$


$\Rightarrow f\left(x\right)=\left\{\begin{array}{c}\frac{\left|x\right|\left|x-1\right|}{x\left(x-1\right)},x\ne 0,1\\ 1,x=0\\ -1,x=1\end{array}\right.$


$\Rightarrow f\left(x\right)=\left\{\begin{array}{c}1,x>1\\ 1,x<0\\ \begin{array}{c}-1,0<x<1\\ 1,x=0\\ -1,x=1\end{array}\end{array}\right.$


$\Rightarrow f\left(x\right)=\left\{\begin{array}{c}\begin{array}{c}\begin{array}{c}1,x>1\\ 1,x\le 0\\ \begin{array}{c}-1,0<x\le 1\end{array}\end{array}\end{array}\end{array}\right.$

So,

$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(-h\right)=1$

Also,

$\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(h\right)=-1$

$\Rightarrow \underset{x\to 0^{+}}{\lim }f\left(x\right)\ne \underset{x\to 0^{-}}{\lim }f\left(x\right)$

Thus, $f\left(x\right)$ is discontinuous at $x=0$.

Now,

$\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1-h\right)=-1$

$\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1+h\right)=1$

$\Rightarrow \underset{x\to 1^{+}}{\lim }f\left(x\right)\ne \underset{x\to 1^{-}}{\lim }f\left(x\right)$

So, $f\left(x\right)$ is discontinuous at $x=1$.

Hence, $f\left(x\right)$ is continuous for all x except at $x=0$ and x = 1.