1. Given function

We have $f(x)=\frac{1}{2}\,[g(x)+g(2-x)]$ where $g$ is twice differentiable and $g'\left(\frac12\right)=g'\left(\frac32\right).$

We need to analyze the options.

2. Differentiate $f(x)$

Using chain rule, $\frac{d}{dx}g(2-x)=g'(2-x)\cdot(-1).$ So, $f'(x)=\frac12\left[g'(x)-g'(2-x)\right].$

Differentiate again: $\frac{d}{dx}[-g'(2-x)] = -g''(2-x)(-1)=g''(2-x).$ Hence, $f''(x)=\frac12\left[g''(x)+g''(2-x)\right].$

3. Use the condition on $g'$

From $g'\left(\frac12\right)=g'\left(\frac32\right),$ consider the function $g'$, which is differentiable on $\mathbb R$.

By Rolle's theorem applied to $g'$ on $\left[\frac12,\frac32\right]$, there exists some $c\in\left(\frac12,\frac32\right)$ such that $g''(c)=0.$

Now observe symmetry in $f''$: $f''(x)=\frac12\left[g''(x)+g''(2-x)\right].$ Also, $f''(2-x)=\frac12\left[g''(2-x)+g''(x)\right]=f''(x).$ So $f''$ is symmetric about $x=1$.

4. A more useful relation from symmetry of $f'$

Since $f'(x)=\frac12[g'(x)-g'(2-x)],$ we get $f'(1)=\frac12[g'(1)-g'(1)]=0.$

Also, $f'\left(\frac12\right)=\frac12\left[g'\left(\frac12\right)-g'\left(\frac32\right)\right]=0$ by the given condition.

Similarly, $f'\left(\frac32\right)=\frac12\left[g'\left(\frac32\right)-g'\left(\frac12\right)\right]=0.$

So $f'$ has equal values at three points: $f'\left(\frac12\right)=f'(1)=f'\left(\frac32\right)=0.$

Now apply Rolle's theorem to $f'$:

• on $\left(\frac12,1\right)$, there exists $c_1\in\left(\frac12,1\right)$ such that $f''(c_1)=0;$
• on $\left(1,\frac32\right)$, there exists $c_2\in\left(1,\frac32\right)$ such that $f''(c_2)=0.$

Thus there are at least two points in $(0,2)$ where $f''(x)=0$.

Therefore, Option A is true.

5. Check remaining options

Option B

We already found $f'\left(\frac12\right)=0, \qquad f'\left(\frac32\right)=0.$ Therefore, $f'\left(\frac32\right)+f'\left(\frac12\right)=0 \ne 1.$ So B is false.

Option C

Option C says $f''(x)=0$ for no $x$ in $(0,1)$. But from Rolle's theorem on $\left(\frac12,1\right)$, there exists $c_1\in\left(\frac12,1\right)\subset(0,1)$ with $f''(c_1)=0$. So C is false.

Option D

Option D says exactly one $x$ in $(0,1)$ satisfies $f''(x)=0$. We can only guarantee at least one in $(0,1)$, not exactly one. There may be more. So D is false.

6. Final answer

The correct option is $\boxed{A}.$