Consider the function $$f: (0, 2) \to \mathbb{R}$$ defined by $$f(x) = \frac{x}{2} + \frac{2}{x}$$ and the function $$g(x)$$ defined by $$g(x) = \begin{cases} \min\{f(t)\},\ 0 < t \leq x & \text{and } 0 < x \leq 1 \\ \frac{3}{2} + x, & 1 < x < 2 \end{cases}$$. Then

For $$x \in (0,2)$$ the given function is $$f(x)=\dfrac{x}{2}+\dfrac{2}{x}\,.$$

Differentiate to study its monotonicity:
$$f'(x)=\dfrac{1}{2}-\dfrac{2}{x^{2}}\;.$$ $$-(1)$$

Put $$f'(x)=0$$:
$$\dfrac{1}{2}-\dfrac{2}{x^{2}}=0 \Longrightarrow x^{2}=4 \Longrightarrow x=2.$$
The point $$x=2$$ lies at the right end of the open interval $$(0,2)$$, so inside the domain we have
$$x\lt 2 \;\Rightarrow\; x^{2}\lt 4 \;\Rightarrow\; \dfrac{2}{x^{2}}\gt\dfrac{1}{2} \;\Rightarrow\; f'(x)\lt 0.$$

Thus $$f(x)$$ is strictly decreasing on $$(0,2).$$ For any $$x\in(0,1]$$ the minimum value of $$f(t)$$ on $$(0,t]\;$$ occurs at the right end point $$t=x$$ itself:

$$\min\{f(t):0\lt t\le x\}=f(x).$$

Hence the definition of $$g(x)$$ simplifies to
$$g(x)=f(x)=\dfrac{x}{2}+\dfrac{2}{x},\qquad 0\lt x\le 1.$$ $$-(2)$$

For $$1\lt x\lt 2$$ we are directly given
$$g(x)=\dfrac{3}{2}+x.$$ $$-(3)$$

CONTINUITY AT $$x=1$$

Left-hand limit:
Using $$(2)$$, $$\displaystyle \lim_{x\to1^{-}}g(x)=f(1)=\dfrac{1}{2}+2=\dfrac{5}{2}.$$

Right-hand limit:
Using $$(3)$$, $$\displaystyle \lim_{x\to1^{+}}g(x)=\dfrac{3}{2}+1=\dfrac{5}{2}.$$

Since both one-sided limits are equal and $$g(1)=\dfrac{5}{2},$$ $$g(x)$$ is continuous at $$x=1.$$ On every other point of $$(0,2)$$ the defining expressions are elementary, so $$g(x)$$ is continuous on the whole interval $$(0,2).$$

DIFFERENTIABILITY AT $$x=1$$

Left derivative:
From $$(2)$$, using $$(1)$$,
$$g'_{-}(1)=f'(1)=\dfrac{1}{2}-\dfrac{2}{1^{2}}=-\dfrac{3}{2}.$$

Right derivative:
From $$(3)$$,
$$g'_{+}(1)=\dfrac{d}{dx}\bigl(\dfrac{3}{2}+x\bigr)\big|_{x=1}=1.$$

Because $$g'_{-}(1)\neq g'_{+}(1),$$ the derivative does not exist at $$x=1.$$ Everywhere else in $$(0,2)$$ the expressions in $$(2)$$ and $$(3)$$ are differentiable, so $$g(x)$$ fails to be differentiable only at $$x=1.$$

Therefore $$g(x)$$ is continuous but not differentiable at $$x=1.$$ Hence, Option A is correct.