A) continuous
B) discontinuous
C) differentiable
D) non-zero
Correct Answer: B
Solution :
Given, \[f(x)=\frac{|2x-3|}{2x-3}\] \[\left\{ \begin{matrix} \frac{2x-3}{2x-3}, & if & x\ge \frac{3}{2} \\ \frac{-(2x-3)}{2x-3}, & if & x<\frac{3}{2} \\ \end{matrix} \right.\] \[\left\{ \begin{matrix} 1, & if & x\ge \frac{3}{2} \\ -1, & if & x<\frac{3}{2} \\ \end{matrix} \right.\] Now, \[RHL=\underset{x\to \frac{{{3}^{+}}}{2}}{\mathop{\lim }}\,\,f(x)=\underset{x\to \frac{{{3}^{+}}}{2}}{\mathop{\lim }}\,1=1\] and \[LHL=\underset{x\to \frac{{{3}^{-}}}{2}}{\mathop{\lim }}\,\,f(x)=\underset{x\to \frac{{{3}^{-}}}{2}}{\mathop{\lim }}\,(-1)=-1\] \[\because \] \[RHL\ne LHL\] \[\therefore \] \[f(x)\] is discontinuous at \[x=\frac{3}{2}\].
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