Mathematics - Continuity - Differentiability Question with Solution | TestHub

Given,

$\mathrm{f}:(0,2)\to \mathrm{R};\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{2}+\frac{2}{\mathrm{x}}$

$\Rightarrow {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=\frac{1}{2}-\frac{2}{{\mathrm{x}}^2}$

$\Rightarrow {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2-4}{2{\mathrm{x}}^2}$

$\therefore$$\mathrm{f}\left(\mathrm{x}\right)$ is decreasing in domain $\left(0,2\right)$.

$\Rightarrow \mathrm{g}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{\mathrm{x}}{2}+\frac{2}{\mathrm{x}} 0<\mathrm{x}\le 1\\ \frac{3}{2}+\mathrm{x} 1<\mathrm{x}<2\end{array}\right.$

$\Rightarrow \mathrm{g}\left(1\right)=\frac{1}{2}+\frac{2}{1}=\frac{5}{2}$

$\Rightarrow \mathrm{g}\left(1^{+}\right)=\frac{3}{2}+1=\frac{5}{2}$

$\Rightarrow \mathrm{g}\left(\mathrm{x}\right)$ is continuous in $\left(0,2\right)$.

$\Rightarrow {\mathrm{g}}^{\text{'}}\left(\mathrm{x}\right)=\left\{\begin{array}{l}{\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right), 0<\mathrm{x}\le 1\\ 1, 1<\mathrm{x}<2\end{array}\right.$

$\Rightarrow {\mathrm{g}}^{\text{'}}\left(1\right)={\mathrm{f}}^{\text{'}}\left(1\right)=\frac{-3}{2}$

$\Rightarrow {\mathrm{g}}^{\text{'}}\left(1\right)\ne {\mathrm{g}}^{\text{'}}\left(1^{+}\right)$

So, $\mathrm{g}\left(\mathrm{x}\right)$ is not differentiable at $\mathrm{x}=1$.