Question

The function $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}|2x-3|.\left[x\right],& x\ge 1\end{array}\\ \begin{array}{cc}\sin \left(\frac{\pi x}{2}\right),& x<1\end{array}\end{array}$ $($ where $\left[x\right]$ denotes the greatest integer $\le x)$ is

• A. continuous at $x=2$
• B. differentiable at $x=1$
• C. continous but not differentiable at $x=1$
• D. none of these

Answer 1

Answer: (C)

continous but not differentiable at $x=1$

At $x=1,f\left(x\right)=1$

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}|2x-3|\left[x\right]=1$

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}\sin \frac{\pi x}{2}=1$

$\Rightarrow f\left(x\right)$ is continuous at $x=1$

$\underset{h\to 0^{+}}{lim}\frac{f(1+h)-1}{h}=\underset{h\to 0^{+}}{lim}\frac{|2h-1|-1}{h}=\underset{h\to 0^{+}}{lim}\frac{1-2h-1}{h}=-2$

$\underset{h\to 0^{-}}{lim}\frac{f(1+h)-1}{h}=\underset{h\to 0^{-}}{lim}\frac{\sin (\frac{\pi }{2}+\frac{\pi h}{2})-1}{h}=\underset{h\to 0^{-}}{lim}\frac{\cos \frac{\pi h}{2}-1}{h}=0$

$\Rightarrow f\left(x\right)$ is not differentiable at $x=1$