Consider the function f: ( 0, 2 ) → R defined by f ( x ) = (x)/(2)...

Given,

$f: ( 0, 2 ) \to R; f ( x ) = \dfrac{x}{2} + \dfrac{2}{x}$

$⇒ {f}^{'}\left(x\right) = \dfrac{1}{2} - \dfrac{2}{{x}^{2}}$

$⇒ {f}^{'}\left(x\right) = \dfrac{{x}^{2} - 4}{2{x}^{2}}$

$∴f\left(x\right)$ is decreasing in domain \left(0, 2\right)$.</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/6d1d452d-21f6-4392-b5d7-bdf699a69082.png" style="width: 300px; height: 223px;"/></p><p>$⇒ g\left(x\right) = \left\{$$\begin{array}{ll}\dfrac{x}{2} + \dfrac{2}{x}0 < x \le 1 \\ \dfrac{3}{2} +x1 < x < 2\end{array}$$\right

$⇒ g\left(1\right) = \dfrac{1}{2} + \dfrac{2}{1} = \dfrac{5}{2}$

$⇒ g\left({1}^{+}\right) = \dfrac{3}{2} + 1 = \dfrac{5}{2}$

$⇒ g\left(x\right)$ is continuous in \left(0, 2\right)$.</p><p>$⇒ {g}^{'}\left(x\right) = \left\{$$\begin{array}{ll}{f}^{'}\left(x\right), 0 < x \le 1 \\ 1, 1 < x < 2\end{array}$$\right

$⇒ {g}^{'}\left(1\right) = {f}^{'}\left(1\right) = \dfrac{- 3}{2}$

$⇒ {g}^{'}\left(1\right) \ne {g}^{'}\left({1}^{+}\right)$

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