Properties of Differentiable Function Derived from Strictly Decreasing Base Function

To determine the correct statement(s), we analyze the properties of the functions $F(x)$ and $f(x) = xF(x)$ given the conditions:

1. $F(1) = 0$ and $F(3) = -4$
2. $F'(x) < 0$ for all $x \in \left(\frac{1}{2}, 3\right)$

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1. Analysis of Statement (A):

The derivative of $f(x) = xF(x)$ with respect to $x$ is: $f'(x) = F(x) + xF'(x)$

Evaluating at $x = 1$: $f'(1) = F(1) + 1 \cdot F'(1)$

Since $F(1) = 0$, we have: $f'(1) = F'(1)$

Given that $1 \in \left(\frac{1}{2}, 3\right)$ and $F'(x) < 0$ for all $x \in \left(\frac{1}{2}, 3\right)$, it follows that: $F'(1) < 0 \implies f'(1) < 0$

Thus, Statement (A) is correct.

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2. Analysis of Statement (B):

Since $F'(x) < 0$ for all $x \in \left(\frac{1}{2}, 3\right)$, the function $F(x)$ is strictly decreasing on $[1, 3]$.

For $x = 2 \in (1, 3)$: $2 > 1 \implies F(2) < F(1) = 0 \implies F(2) < 0$

Now, evaluate $f(2)$: $f(2) = 2F(2)$

Since $2 > 0$ and $F(2) < 0$, we obtain: $f(2) < 0$

Thus, Statement (B) is correct.

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3. Analysis of Statements (C) and (D):

For any $x \in (1, 3)$:

• $F(x) < F(1) = 0$ (since $F$ is strictly decreasing), so $F(x) < 0$.
• $x > 0$ and $F'(x) < 0$, so $xF'(x) < 0$.

Therefore, the derivative $f'(x)$ on $(1, 3)$ is: $f'(x) = F(x) + xF'(x) < 0 \quad \text{for all } x \in (1, 3)$

Since $f'(x) < 0$ for every $x \in (1, 3)$, $f'(x)$ cannot be zero anywhere in $(1, 3)$. Hence, $f'(x) \neq 0$ for any $x \in (1, 3)$.

Thus, Statement (C) is correct and Statement (D) is incorrect.

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Conclusion:

The correct statements are (A), (B), and (C).