Question

Discuss the continuity and differentiability of the function f(x) in(0,3) where $f\left(x\right)=\{\begin{array}{cc}|2x-3|\left[x\right],& 0\le x\le 2\\ \frac{x^2}{2},& 2<x\le 3\end{array}$

• A. continuous everywhere
• B. discontinuous at some points
• C. not differentiable at some points
• D. differentiable everywhere

Answer 1

Answer: (B), (C)

not differentiable at some points
discontinuous at some points

$f\left(x\right)=\{\begin{array}{cc}|2x-3|\left[x\right],& 0\le x\le 2\\ \frac{x^2}{2},& 2<x\le 3\end{array}$
$\Rightarrow f\left(x\right)=\{\begin{array}{cc}0,& 0\le x<1\\ -2(x-\frac{3}{2}),& 1\le x<\frac{3}{2}\\ 2(x-\frac{3}{2}),& \frac{3}{2}\le x<2\\ 2,& x=2\\ x^2/2,& 2<x<3\end{array}$
Since, $f(1-)=f(1+)$
Therefore, $f$ is discontinuous at some points.
Ans: B