Question

The function $f(x)=\{\begin{array}{ll}x[x],& \text{ if }0\le x<2\\ (x-1)x,& \text{ if }2\le x<3\end{array}$ is is

• A. differentiable at $x=2$
• B. not differentiable at $x=2$
• C. continuous at $x=2$
• D. None of these

Answer 1

Answer: (B)

Here, $f(x)=\{\begin{array}{l}x[x],\text{ if }0\le x<2\\ (x-1)x,\text{ if }2\le x<3\end{array}$ at $x=2$
$LHD=L{f}^{\prime }(2)=\underset{h\to 0}{\lim }\frac{f(2-h)-f(2)}{-h}$
$=\underset{h\to 0}{\lim }\frac{(2-h)[2-h]-(2-1)2}{-h}$
$=\underset{h\to 0}{\lim }\frac{(2-h)(1)-2}{-h}[\because (2-h)=1]$
$=\underset{h\to 0}{\lim }\frac{2-h-2}{-h}=\underset{h\to 0}{\lim }\frac{-h}{-h}=1$
RHD $=R{f}^{\prime }(2)=\underset{h\to 0}{\lim }\frac{f(2+h)-f(2)}{h}$
$=\underset{h\to 0}{\lim }\frac{(2+h-1)(2+h)-(2-1)(2)}{h}$
$=\underset{h\to 0}{\lim }\frac{(h+1)(2+h)-2}{h}$
$=\underset{h\to 0}{\lim }\frac{h^2+3h+2-2}{h}$
$=\underset{h\to 0}{\lim }\frac{h^2+3h}{h}=\underset{h\to 0}{\lim }\frac{h(h+3)}{h}$
$=3$
$\because LHD\ne RHD$
$\therefore f(x)$ is not differentiable at $x=2$.