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:

  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


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