1. Given data

We have a thrice differentiable function $F:\mathbb{R}\to\mathbb{R}$ such that $F(1)=0,\qquad F(3)=-4,$ and $F'(x)<0\quad \text{for all }x\in\left(\frac12,3\right).$ Also, $f(x)=xF(x).$

We must check each option.

2. Use the fact that $F'(x)<0$ on $\left(\frac12,3\right)$

Since $F'(x)<0$ on this interval, $F$ is strictly decreasing on $\left(\frac12,3\right)$.

Because $1<2<3$, we get $F(1)>F(2)>F(3).$ Using $F(1)=0$ and $F(3)=-4$, $0>F(2)>-4.$ Hence, $F(2)<0.$

Also, for every $x\in(1,3)$, since $F$ is strictly decreasing and $F(1)=0$, $F(x)<F(1)=0.$ So, $F(x)<0\quad \forall x\in(1,3).$

3. Find $f'(x)$

Since $f(x)=xF(x),$ by product rule, $f'(x)=F(x)+xF'(x).$

Now evaluate the options.

4. Option A: $f'(1)<0$

At $x=1$, $f'(1)=F(1)+1\cdot F'(1)=0+F'(1)=F'(1).$ Given $F'(x)<0$ for all $x\in\left(\frac12,3\right)$, in particular at $x=1$, $F'(1)<0.$ Therefore, $f'(1)<0.$

So A is correct.

5. Option B: $f(2)<0$

We have $f(2)=2F(2).$ From Step 2, $F(2)<0$. Since $2>0$, $2F(2)<0.$ Hence, $f(2)<0.$

So B is correct.

6. Option C: $f'(x)\neq 0$ for any $x\in(1,3)$

For $x\in(1,3)$, $f'(x)=F(x)+xF'(x).$ Now,

• $F(x)<0$ from Step 2,
• $F'(x)<0$ given,
• $x>0$ on $(1,3)$, so $xF'(x)<0$.

Thus, $f'(x)=\underbrace{F(x)}_{<0}+\underbrace{xF'(x)}_{<0}<0.$ Therefore, $f'(x)\neq 0\quad \forall x\in(1,3).$

So C is correct.

7. Option D: $f'(x)=0$ for some $x\in(1,3)$

From Step 6, for every $x\in(1,3)$, $f'(x)<0.$ So there is no such point where $f'(x)=0$ in $(1,3)$.

Hence D is false.

8. Final answer

The correct statements are: $\boxed{A,\ B,\ C}$

This matches the stored correct answer.