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WMA12/01 Jun 2024, Question 5: Proof by Exhaustion and Counter-example

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 6 marks

The question, in short

Prove that positive x and y satisfying (x − y)³ > x³ − y³ must have y > x, then use a counter example to show this fails for all reals.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 4 marks

Show by detailed reasoning that if x and y are positive and (x − y)³ > x³ − y³, then y > x.

  1. 1(x − y)³ = x³ − 3x²y + 3xy² − y³
  2. 2So x³ − 3x²y + 3xy² − y³ > x³ − y³
  3. 3−3x²y + 3xy² > 0
  4. 43xy(y − x) > 0
  5. 5x and y are positive, so 3xy > 0; hence y − x > 0, so y > x

AnswerShown: (x − y)³ > x³ − y³ simplifies to 3xy(y − x) > 0, and since x, y > 0 this gives y > x

Part (b) · 2 marks

Give a counter example with real numbers showing the part (a) result does not hold generally.

  1. 1Choose a positive x and a negative y, e.g. x = 3, y = −1
  2. 2(x − y)³ = 4³ = 64 and x³ − y³ = 27 − (−1) = 28
  3. 364 > 28 so the condition holds, but y = −1 is not greater than x = 3

Answere.g. x = 3, y = −1: (3 − (−1))³ = 64 > 27 − (−1) = 28, yet y < x

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