Question

$f\left(x\right)\{\begin{array}{cc}=\frac{|x^2-x|}{x^2-x},& x\ne 0,x\ne 1\\ =1,& x=0\\ =-1,& x=1\end{array}$ Discuss its continuity in $0<x\le 1.$

• A. continuous at $x=0$
• B. dis-continuous at $x=1$
• C. dis-continuous at $x=0$
• D. continuous at $x=1$

Answer 1

Answer: (B), (C)

The correct options are
B dis-continuous at $x=1$
C dis-continuous at $x=0$

$f\left(x\right)\{\begin{array}{cc}=1,& x<0orx>1\\ =-1,& 0<x<1\\ =1,& x=0\\ =-1,& x=1\end{array}$
At $x=1,RHL=1,LHL=-1,V=-1$ $\therefore$ Discontinuous.
At $x=0,RHL=-1,LHL=1,V=1$ $\therefore$ Discontinuous.
Hence f(x) is continuous $\forall x$ except at $x=0$ and $x=1$
Ans: B, C