Edexcel P4 2018 Specimen — Question 6 4 marks

Exam BoardEdexcel
ModuleP4 (Pure Mathematics 4)
Year2018
SessionSpecimen
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof
TypeContradiction proof of inequality
DifficultyStandard +0.3 This is a standard proof by contradiction of the AM-GM inequality for two variables. The setup is straightforward: assume a + b < 2√(ab), square both sides, and arrive at the contradiction (√a - √b)² < 0. While it requires understanding proof by contradiction and algebraic manipulation, this is a well-known result with a routine proof structure that students at this level would have practiced. It's slightly easier than average due to its familiarity and limited number of steps.
Spec1.01d Proof by contradiction
6. Prove by contradiction that, if \(a , b\) are positive real numbers, then \(a + b \geqslant 2 \sqrt { a b }\) \includegraphics[max width=\textwidth, alt={}, center]{4de08317-5fb9-4789-8d57-ccf463224c78-20_2655_1943_114_118}
Question 6:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Assumption: there exist positive real numbers \(a\), \(b\) such that \(a + b < 2\sqrt{ab}\)B1 Must assume the contrary for proof by contradiction
Method 1: \(a + b - 2\sqrt{ab} < 0 \Rightarrow (\sqrt{a} - \sqrt{b})^2 < 0\)M1A1 Complete method for creating \((f(a,b))^2 < 0\)
Method 2: \((a+b)^2 = (2\sqrt{ab})^2 \Rightarrow a^2 + 2ab + b^2 < 4ab \Rightarrow a^2 - 2ab + b^2 < 0 \Rightarrow (a-b)^2 < 0\)M1A1 All algebra correct
This is a contradiction, therefore if \(a\), \(b\) are positive real numbers, then \(a + b \geqslant 2\sqrt{ab}\)A1 Must state \((a-b)^2 < 0\) or \((\sqrt{a}-\sqrt{b})^2 < 0\) is a contradiction and conclude \(a+b \geqslant 2\sqrt{ab}\) 
## Question 6:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Assumption: there exist positive real numbers $a$, $b$ such that $a + b < 2\sqrt{ab}$ | B1 | Must assume the contrary for proof by contradiction |
| **Method 1:** $a + b - 2\sqrt{ab} < 0 \Rightarrow (\sqrt{a} - \sqrt{b})^2 < 0$ | M1A1 | Complete method for creating $(f(a,b))^2 < 0$ |
| **Method 2:** $(a+b)^2 = (2\sqrt{ab})^2 \Rightarrow a^2 + 2ab + b^2 < 4ab \Rightarrow a^2 - 2ab + b^2 < 0 \Rightarrow (a-b)^2 < 0$ | M1A1 | All algebra correct |
| This is a contradiction, therefore if $a$, $b$ are positive real numbers, then $a + b \geqslant 2\sqrt{ab}$ | A1 | Must state $(a-b)^2 < 0$ or $(\sqrt{a}-\sqrt{b})^2 < 0$ is a contradiction and conclude $a+b \geqslant 2\sqrt{ab}$ |

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6. Prove by contradiction that, if $a , b$ are positive real numbers, then $a + b \geqslant 2 \sqrt { a b }$\\
\includegraphics[max width=\textwidth, alt={}, center]{4de08317-5fb9-4789-8d57-ccf463224c78-20_2655_1943_114_118}\\

\hfill \mbox{\textit{Edexcel P4 2018 Q6 [4]}}
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