Extrema and Properties of Integral Function Defined by Cubic Polynomial

To determine the correct statements, we analyze the function $F(x) = \int_{0}^{x} f(t) \, dt$ for $x > 0$, where $f(x) = (x - 1)(x - 2)(x - 5)$.

Step 1: Derivative and Critical Points

By the Fundamental Theorem of Calculus, the derivative of $F(x)$ is: $F'(x) = f(x) = (x - 1)(x - 2)(x - 5)$

Setting $F'(x) = 0$ gives the critical points for $x > 0$: $x = 1, \quad x = 2, \quad x = 5$

Step 2: Sign of $F'(x)$ and Identification of Extrema

We check the sign of $F'(x)$ across the intervals in $(0, \infty)$:

1. For $x \in (0, 1)$: $F'(x) < 0 \implies F(x)$ is strictly decreasing.
2. For $x \in (1, 2)$: $F'(x) > 0 \implies F(x)$ is strictly increasing.
3. For $x \in (2, 5)$: $F'(x) < 0 \implies F(x)$ is strictly decreasing.
4. For $x \in (5, \infty)$: $F'(x) > 0 \implies F(x)$ is strictly increasing.

Using the First Derivative Test:

• At $x = 1$, $F'(x)$ changes from negative to positive, so $F(x)$ has a local minimum at $x = 1$. (Option A is correct)
• At $x = 2$, $F'(x)$ changes from positive to negative, so $F(x)$ has a local maximum at $x = 2$. (Option B is correct)
• At $x = 5$, $F'(x)$ changes from negative to positive, so $F(x)$ has a local minimum at $x = 5$.

Thus, $F(x)$ has one local maximum (at $x = 2$) and two local minima (at $x = 1$ and $x = 5$) on $(0, \infty)$. (Option C is incorrect)

Step 3: Behavior of $F(x)$ on $(0, 5)$

First, note that: $f(t) = (t - 1)(t - 2)(t - 5) = t^3 - 8t^2 + 17t - 10$

Evaluating the value of the local maximum at $x = 2$: $F(2) = \int_{0}^{2} (t^3 - 8t^2 + 17t - 10) \, dt = \left[ \frac{t^4}{4} - \frac{8t^3}{3} + \frac{17t^2}{2} - 10t \right]_{0}^{2}$ $F(2) = \frac{16}{4} - \frac{64}{3} + \frac{68}{2} - 20 = 4 - \frac{64}{3} + 34 - 20 = 18 - \frac{64}{3} = -\frac{10}{3} < 0$

Since:

• $F(0) = 0$ and $F(x)$ strictly decreases on $(0, 1]$, we have $F(x) < 0$ for all $x \in (0, 1]$.
• On $[1, 2]$, $F(x)$ increases to its local maximum $F(2) = -\frac{10}{3} < 0$, which means $F(x) \leq -\frac{10}{3} < 0$ for all $x \in [1, 2]$.
• On $[2, 5]$, $F(x)$ strictly decreases from $F(2) = -\frac{10}{3}$, which means $F(x) < -\frac{10}{3} < 0$ for all $x \in (2, 5]$.

Therefore, $F(x) < 0$ for all $x \in (0, 5)$, which implies: $F(x) \neq 0 \quad \text{for all } x \in (0, 5)$ (Option D is correct)

Conclusion:

The correct options are A, B, and D.