Question

(\lambda) مجموعه جواب نامعادله زیر را بدست آورید:

\\\frac{x^2 - x}{x^2 - 2x + 2} \le 0\\

Explanation:

⚡ Using what you learned: solving two-step and multi-step inequalities

Step 1: Analyze the denominator

We examine the quadratic expression in the denominator:

$$ x^2 - 2x + 2 $$

To find its roots, we calculate the discriminant \( \Delta \):

$$ \Delta = b^2 - 4ac = (-2)^2 - 4(1)(2) = 4 - 8 = -4 $$

Since \( \Delta < 0 \) and the leading coefficient \( a = 1 > 0 \), the denominator \( x^2 - 2x + 2 \) is strictly positive for all real numbers \( x \):

$$ x^2 - 2x + 2 > 0 \quad \forall x \in \mathbb{R} $$

Step 2: Simplify the inequality

Since the denominator is always positive, the sign of the rational expression depends entirely on the numerator. The inequality:

$$ \frac{x^2 - x}{x^2 - 2x + 2} \le 0 $$

simplifies directly to:

$$ x^2 - x \le 0 $$

Step 3: Solve the simplified inequality

Factor the quadratic expression:

$$ x(x - 1) \le 0 $$

The roots of the equation \( x(x - 1) = 0 \) are \( x = 0 \) and \( x = 1 \).

We test the intervals determined by these roots:

• For \( x < 0 \): both \( x \) and \( x - 1 \) are negative, so their product is positive.
• For \( 0 \le x \le 1 \): \( x \) is non-negative and \( x - 1 \) is non-positive, so their product is non-positive (\( \le 0 \)).
• For \( x > 1 \): both \( x \) and \( x - 1 \) are positive, so their product is positive.

Thus, the inequality holds when \( x \) is between \( 0 \) and \( 1 \), inclusive.

Answer:

$$ [0, 1] \quad \text{or} \quad 0 \le x \le 1 $$