If $f(x)$ defined by $f(x)=\left\{\begin{array}{ll}\frac{\left|x^{2}-x\right|}{x^{2}-x}, & x \neq 0,1 \\ 1 & , \quad x=0 \\ -1 & , \quad x=1\end{array}\right.$ 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 $x=1$.
(d) $x$ except at $x=0$ and $x=1$.
Given: $f(x)=\left\{\begin{array}{cc}\frac{\left|x^{2}-x\right|}{x^{2}-x} & , x \neq 0,1 \\ 1 & , x=0 \\ -1 & , x=1\end{array}\right.$
$\Rightarrow f(x)=\left\{\begin{array}{cc}\frac{|x||x-1|}{x(x-1)}, & x \neq 0,1 \\ 1 & , x=0 \\ -1 & , x=1\end{array}\right.$
$\Rightarrow f(x)=\left\{\begin{array}{c}1, x>1 \\ 1, x<0 \\ -1,0 $\Rightarrow f(x)=\left\{\begin{array}{c}1, x>1 \\ 1, x \leq 0 \\ -1,0 So, $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(-h)=1$ Also, $\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(h)=-1$ $\Rightarrow \lim _{x \rightarrow 0^{+}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x)$ Thus, $f(x)$ is discontinuous at $x=0$. Now, $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=-1$ $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=1$ $\Rightarrow \lim _{x \rightarrow 1^{+}} f(x) \neq \lim _{x \rightarrow 1^{-}} f(x)$ So, $f(x)$ is discontinuous at $x=1$. Hence, $f(x)$ is continuous for all $x$ except at $x=0$ and $x=1$.