Question

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

Options

• x 

•  x except at x = 0

• x except at x = 1

•  x except at x = 0 and x = 1.

Answer 1

x except at x = 0 and x = 1.
Given:

\[f\left( x \right) = \begin{cases}\frac{\left| x^2 - x \right|}{x^2 - x}, & x \neq 0, 1 \\ 1 , & x = 0 \\ - 1 , & x = 1\end{cases}\] 

\[\Rightarrow f\left( x \right) = \begin{cases}\frac{\left| x \right| \left| x - 1 \right|}{x\left( x - 1 \right)}, x \neq 0, 1 \\ 1 , x = 0 \\ - 1 , x = 1\end{cases}\]

\[\Rightarrow f\left( x \right) = \begin{cases}1, x > 1 \\ 1, x < 0 \\ \begin{array}- 1, 0 < x < 1 \\ 1, x = 0 \\ - 1, x = 1\end{array}\end{cases}\]

\[\Rightarrow f\left( x \right) = \begin{cases}1, x > 1 \\ 1, x ≤  0 \\ \begin{array} - -1, 0 < x ≤ 1 \\ \end{array}\end{cases}\]

so , 

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( - h \right) = 1\]

Also,

\[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( h \right) = - 1\]

\[\Rightarrow \lim_{x \to 0^+} f\left( x \right) \neq \lim_{x \to 0^-} f\left( x \right)\]

Thus,  

\[f\left( x \right)\] is discontinuous at  \[x = 0\] .

Now, 

\[\lim_{x \to 1^-} f\left( x \right) = \lim_{h \to 0} f\left( 1 - h \right) = - 1\]

\[\lim_{x \to 1^+} f\left( x \right) = \lim_{h \to 0} f\left( 1 + h \right) = 1\]

\[\Rightarrow \lim_{x \to 1^+} f\left( x \right) \neq \lim_{x \to 1^-} f\left( x \right)\]

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.