Show that the function
Question:

Show that the function $f(x)= \begin{cases}|2 x-3|[x], & x \geq 1 \\ \sin \left(\frac{\pi x}{2}\right), & x<1\end{cases}$  is continuous but not differentiable at x = 1.

Solution:

Given: $f(x)=\left\{\begin{array}{l}|2 x-3|[x], \quad x \geq 1 \\ \sin \left(\frac{\pi \mathrm{x}}{2}\right), \quad x<1\end{array}\right.$

Continuity at = 1:

$(\mathrm{LHL}$ at $x=1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} \sin \left(\frac{\pi(1-h)}{2}\right)=\sin \frac{\pi}{2}=1$

$(\mathrm{RHL}$ at $x=1)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}|2(1+h)-3|[1+h]=\lim _{h \rightarrow 0}|2(1+h)-3|=1$

Hence, (LHL at = 1) = (RHL at = 1)

Differentiability at = 1:

$(\mathrm{LHD}$ at $x=1)=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}$

$(\mathrm{LHD}$ at $x=1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{1-h-1}$

$(\mathrm{LHD}$ at $x=1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}$

$(\mathrm{LHD}$ at $x=1)=\lim _{h \rightarrow 0} \frac{\sin \left(\frac{\pi(1-h)}{2}\right)-1}{-h}$

$(\mathrm{LHD}$ at $x=1)=\lim _{h \rightarrow 0} \frac{\cos \frac{\pi h}{2}-1}{-h}$

$(\mathrm{LHD}$ at $x=1)=-\frac{\pi}{2} \lim _{h \rightarrow 0} \frac{\cos \frac{\pi h}{2}-1}{\frac{\pi}{2} h}=0$

$(\mathrm{RHD}$ at $x=1)=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}$

$(\mathrm{RHD}$ at $x=1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{1+h-1}$

$(\mathrm{RHD}$ at $x=1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$

$(\mathrm{RHD}$ at $x=1)=\lim _{h \rightarrow 0} \frac{-(2(1+h)-3)-1}{h}$

$(\mathrm{RHD}$ at $x=1)=\lim _{h \rightarrow 0} \frac{-2 h}{h}=-2$

LHD ≠ RHD

Hence, the function is continuous but not differentiable at x = 1.