Step-by-step Derivations

Let's analyze the given functions and options. We are given f(x) = (x - 1)(x - 2)(x - 5) and $F(x) = \int_0^x f(t)dt$ for x > 0.

1. Find the derivative of F(x) and its critical points.

Using the Fundamental Theorem of Calculus (Leibniz rule), the derivative of F(x) is: $F'(x) = \frac{d}{dx} \left( \int_0^x f(t)dt \right) = f(x)$

So, F'(x) = (x - 1)(x - 2)(x - 5).

The critical points of F(x) are the values of x where F'(x) = 0. $F'(x) = 0 \implies (x - 1)(x - 2)(x - 5) = 0$ The critical points for x > 0 are x = 1, x = 2, and x = 5.

2. Analyze the sign of F'(x) to determine local extrema.

We analyze the sign of F'(x) = f(x) in the intervals defined by the critical points:

• For $x \in (0, 1)$: f(x) = (negative)(negative)(negative) = negative. So F'(x) < 0, and F(x) is decreasing.
• For $x \in (1, 2)$: f(x) = (positive)(negative)(negative) = positive. So F'(x) > 0, and F(x) is increasing.
• For $x \in (2, 5)$: f(x) = (positive)(positive)(negative) = negative. So F'(x) < 0, and F(x) is decreasing.
• For $x \in (5, \infty)$: f(x) = (positive)(positive)(positive) = positive. So F'(x) > 0, and F(x) is increasing.

3. Evaluate options related to local extrema (B, C, D).

• At x = 1: F'(x) changes from negative to positive. This indicates a local minimum at x = 1. Therefore, Option D is correct.

• At x = 2: F'(x) changes from positive to negative. This indicates a local maximum at x = 2. Therefore, Option B is correct.

• At x = 5: F'(x) changes from negative to positive. This indicates a local minimum at x = 5.

• Option C: States that F has two local maxima and one local minimum in $(0, \infty)$. Our analysis shows that F has one local maximum (at x=2) and two local minima (at x=1 and x=5). Therefore, Option C is incorrect.

4. Evaluate Option A: F(x) ≠ 0 for all x ∈ (0, 5).

To analyze the sign of F(x), let's find the explicit expression for F(x). First, expand f(t): $f(t) = (t - 1)(t - 2)(t - 5) = (t^2 - 3t + 2)(t - 5) = t^3 - 5t^2 - 3t^2 + 15t + 2t - 10 = t^3 - 8t^2 + 17t - 10$

Now, integrate f(t) from 0 to x: $F(x) = \int_0^x (t^3 - 8t^2 + 17t - 10) dt = \left[ \frac{t^4}{4} - \frac{8t^3}{3} + \frac{17t^2}{2} - 10t \right]_0^x$ $F(x) = \frac{x^4}{4} - \frac{8x^3}{3} + \frac{17x^2}{2} - 10x$

Let's evaluate F(x) at the boundaries and critical points:

• F(0) = 0.
• For $x \in (0, 1)$, we know f(t) < 0. Thus, $F(x) = \int_0^x f(t)dt < 0$.
• At the local minimum x = 1: $F(1) = \frac{1}{4} - \frac{8}{3} + \frac{17}{2} - 10 = \frac{3 - 32 + 102 - 120}{12} = -\frac{47}{12} < 0$.
• At the local maximum x = 2: $F(2) = \frac{2^4}{4} - \frac{8(2^3)}{3} + \frac{17(2^2)}{2} - 10(2) = 4 - \frac{64}{3} + 34 - 20 = 18 - \frac{64}{3} = \frac{54 - 64}{3} = -\frac{10}{3} < 0$.

Let's summarize the behavior of F(x) in (0, 5):

• F(0) = 0.
• F(x) decreases on (0, 1) to F(1) = -47/12.
• F(x) increases on (1, 2) from F(1) = -47/12 to F(2) = -10/3. Since both values are negative, F(x) remains negative throughout (1, 2).
• F(x) decreases on (2, 5). Since it starts from a negative value F(2) = -10/3, it will become even more negative. The value at x=5 is $F(5) = \frac{5^4}{4} - \frac{8(5^3)}{3} + \frac{17(5^2)}{2} - 10(5) = \frac{625}{4} - \frac{1000}{3} + \frac{425}{2} - 50 = \frac{1875 - 4000 + 2550 - 600}{12} = -\frac{175}{12} < 0$.

Since F(x) is strictly negative for all $x \in (0, 5]$, it is never zero in the interval (0, 5). Therefore, Option A is correct.

Conclusion

Based on the analysis:

• Option A is correct.
• Option B is correct.
• Option C is incorrect.
• Option D is correct.

The correct options are A, B, and D.