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.
Question
 For $x \in (2,4)$, the sign of $x^{2}  6x + 5$ is negative.
 For $x \in (\infty, 2) \cup (4, \infty)$, the sign of $x^{2}  6x + 5$ is positive.
Therefore, Both I and II statements are true.
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Answer
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:
 ( (\infty, 1) )
 ( (1, 5) )
 ( (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
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