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Utiliser la raisonnement par l'absurde avec des étapes détaillées

To prove that \( |x| \leq 1 \Rightarrow |2x^2 - x - 1| \leq 2 \) using proof by contradiction, let's go through the steps in detail.

### Step 1: Assumption and Goal
We are given:
\[
|x| \leq 1
\]
We want to show:
\[
|2x^2 - x - 1| \leq 2
\]
Using proof by contradiction, we assume that this inequality **does not hold**. Thus, we assume:
\[
|2x^2 - x - 1| > 2
\]

### Step 2: Analyze the Assumption
The absolute value inequality \( |2x^2 - x - 1| > 2 \) implies two cases:
1. \( 2x^2 - x - 1 > 2 \)
2. \( 2x^2 - x - 1 < -2 \)

We'll analyze each case separately.

### Case 1: \( 2x^2 - x - 1 > 2 \)
Simplify the inequality:
\[
2x^2 - x - 1 - 2 > 0
\]
\[
2x^2 - x - 3 > 0
\]
Now, factorize \( 2x^2 - x - 3 \):
\[
2x^2 - x - 3 = (2x + 3)(x - 1)
\]
So we have:
\[
(2x + 3)(x - 1) > 0
\]
This inequality can be solved by analyzing the intervals defined by \( x = -\frac{3}{2} \) and \( x = 1 \).

Since \( |x| \leq 1 \), we know \( x \) is restricted to the interval \([-1, 1]\). This interval includes \( x = -1 \) to \( x = 1 \) but does **not** include \( -\frac{3}{2} \). 

We can check values within this interval to verify if the inequality \( 2x^2 - x - 1 > 2 \) can hold, and we find that it does not for any \( x \) in \([-1, 1]\).

### Case 2: \( 2x^2 - x - 1 < -2 \)
Now we simplify this inequality:
\[
2x^2 - x - 1 + 2 < 0
\]
\[
2x^2 - x + 1 < 0
\]
However, the expression \( 2x^2 - x + 1 \) has no real roots and is always positive for real \( x \), meaning this inequality can never hold.

### Conclusion
Since both cases lead to contradictions, we conclude that our assumption \( |2x^2 - x - 1| > 2 \) is false. Therefore:
\[
|2x^2 - x - 1| \leq 2
\]
is true whenever \( |x| \leq 1 \).

Would you like additional details on any specific step or a similar example?

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Here are 5 related questions to expand understanding:

1. How can we approach solving absolute value inequalities in general?
2. How does the method of proof by contradiction work in general?
3. What are alternative ways to solve this inequality, aside from contradiction?
4. How would we verify if the expression \( 2x^2 - x + 1 \) is always positive?
5. How would we solve similar inequalities involving different polynomials?

**Tip:** When using proof by contradiction, always clearly state the assumption you're negating and ensure each logical step aligns with that assumption.

Montrer que : |x| ≤ 1 => |2x^2 - x - 1| ≤ 2. Utiliser le raisonnement par l'absurde avec des étapes détaillées.

Explore a proof by contradiction to demonstrate that |x| ≤ 1 implies |2x^2 - x - 1| ≤ 2. This problem involves absolute values, quadratic inequalities, and step-by-step reasoning suitable for advanced high school students.

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