For x in (2,4), the sign of x^2 - 6x + 5 is negative. For x in (-infinity, 2) union (4, infinity), the sign of x^2 - 6x + 5 is positive. Therefore, Both I and II statements are true.

To determine the validity of the statements about the sign of the quadratic function $f(x) = x^2 - 6x + 5$, let's analyze it step by step.

Step 1: Find the roots of the quadratic equation

The quadratic equation can be solved using the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

For the given equation $x^2 - 6x + 5$, we have:

• $a = 1$
• $b = -6$
• $c = 5$

Plugging these values into the formula gives:

$$x = \frac{6 \pm \sqrt{36 - 20}}{2}$$

Simplifying inside the square root:

$$x = \frac{6 \pm \sqrt{16}}{2}$$

Since $\sqrt{16} = 4$:

$$x = \frac{6 \pm 4}{2}$$

This yields two roots:

$$x = \frac{6 + 4}{2} = 5 \quad \text{and} \quad x = \frac{6 - 4}{2} = 1$$

Step 2: Write the quadratic function in factored form

Using the roots, we can factorize the quadratic as:

$$f(x) = (x - 1)(x - 5)$$

Step 3: Determine the sign of the function over different intervals

The roots $x = 1$ and $x = 5$ divide the number line into three intervals:

1. $(-\infty, 1)$
2. $(1, 5)$
3. $(5, \infty)$

To find the sign of $f(x)$ in these intervals, we need to test values from each interval in the factored form.

• For $x \in (-\infty, 1)$: Choose $x = 0$.

  $$f(0) = (0 - 1)(0 - 5) = (-1)(-5) = 5 \implies \text{positive}$$

• For $x \in (1, 5)$: Choose $x = 2$.

  $$f(2) = (2 - 1)(2 - 5) = (1)(-3) = -3 \implies \text{negative}$$

• For $x \in (5, \infty)$: Choose $x = 6$.

  $$f(6) = (6 - 1)(6 - 5) = (5)(1) = 5 \implies \text{positive}$$

Step 4: Evaluate the given statements

• Statement I: For $x \in (2, 4)$, the sign of $x^2 - 6x + 5$ is negative.

  • This is true as within the interval $(1, 5)$, $f(x)$ is negative.

• Statement II: For $x \in (-\infty, 2) \cup (4, \infty)$, the sign of $x^2 - 6x + 5$ is positive.

  • This is partially true. While $x \in (-\infty, 1) \cup (5, \infty)$ indeed gives a positive value, the interval $(2, 4) \subset (1, 5)$ does not align with the true sign change behavior around the roots.

Given this analysis, only Statement I is entirely true. Therefore, the correct answer is:

Final Answer: A