Math Problem Statement

Utiliser la raisonnement par l'absurde avec des étapes détaillées

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

Solution

To prove that x12x2x12|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: x1|x| \leq 1 We want to show: 2x2x12|2x^2 - x - 1| \leq 2 Using proof by contradiction, we assume that this inequality does not hold. Thus, we assume: 2x2x1>2|2x^2 - x - 1| > 2

Step 2: Analyze the Assumption

The absolute value inequality 2x2x1>2|2x^2 - x - 1| > 2 implies two cases:

  1. 2x2x1>22x^2 - x - 1 > 2
  2. 2x2x1<22x^2 - x - 1 < -2

We'll analyze each case separately.

Case 1: 2x2x1>22x^2 - x - 1 > 2

Simplify the inequality: 2x2x12>02x^2 - x - 1 - 2 > 0 2x2x3>02x^2 - x - 3 > 0 Now, factorize 2x2x32x^2 - x - 3: 2x2x3=(2x+3)(x1)2x^2 - x - 3 = (2x + 3)(x - 1) So we have: (2x+3)(x1)>0(2x + 3)(x - 1) > 0 This inequality can be solved by analyzing the intervals defined by x=32x = -\frac{3}{2} and x=1x = 1.

Since x1|x| \leq 1, we know xx is restricted to the interval [1,1][-1, 1]. This interval includes x=1x = -1 to x=1x = 1 but does not include 32-\frac{3}{2}.

We can check values within this interval to verify if the inequality 2x2x1>22x^2 - x - 1 > 2 can hold, and we find that it does not for any xx in [1,1][-1, 1].

Case 2: 2x2x1<22x^2 - x - 1 < -2

Now we simplify this inequality: 2x2x1+2<02x^2 - x - 1 + 2 < 0 2x2x+1<02x^2 - x + 1 < 0 However, the expression 2x2x+12x^2 - x + 1 has no real roots and is always positive for real xx, meaning this inequality can never hold.

Conclusion

Since both cases lead to contradictions, we conclude that our assumption 2x2x1>2|2x^2 - x - 1| > 2 is false. Therefore: 2x2x12|2x^2 - x - 1| \leq 2 is true whenever x1|x| \leq 1.

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


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 2x2x+12x^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.

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Math Problem Analysis

Mathematical Concepts

Absolute Value
Proof by Contradiction
Quadratic Inequalities

Formulas

Absolute Value Inequality |a| > b implies a > b or a < -b

Theorems

Proof by Contradiction
Properties of Quadratic Functions

Suitable Grade Level

Grade 11-12