| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Contradiction proof of inequality |
| Difficulty | Standard +0.8 This is a proof by contradiction of a classic inequality requiring students to manipulate algebraic expressions and recognize when they reach a contradiction. While the result is well-known and the algebraic manipulation is straightforward (multiplying through by ab, rearranging to get (a-b)² > 0), students must structure the proof correctly and recognize that the contradiction arises from the fact that distinct positive numbers cannot be equal. This is moderately challenging as it requires proof technique mastery beyond routine calculation, but the algebra itself is accessible to A-level students. |
| Spec | 1.01d Proof by contradiction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a^2 - 2ab + b^2\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Forms a different sum of a non-zero rational and its reciprocal, e.g. \(-2 + \frac{1}{-2} = -\frac{5}{2}\) | M1 | Must have used 1 or a negative value |
| \(-\frac{5}{2} < 2\), so disproves the statement | R1 | Finds correct counter example and compares result with 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Assumes \(\frac{a}{b} + \frac{b}{a} \leq 2\) for distinct positive integers \(a\) and \(b\); forms \(\frac{a^2+b^2}{ab} \leq 2\) | M1 | Condone \(\frac{a}{b}+\frac{b}{a} < 2\) |
| Rearranges and factorises to deduce \((a-b)^2 \leq 0\) | A1 | Condone \((a-b)^2 < 0\) |
| Since \(a \neq b\), \((a-b)^2 > 0\), this is a contradiction, hence \(\frac{a}{b}+\frac{b}{a} > 2\) | R1 | Must have started with \(\frac{a}{b}+\frac{b}{a} \leq 2\) and stated \(a \neq b\) or made reference to them being distinct |
## Question 10:
**Part 10(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $a^2 - 2ab + b^2$ | B1 | |
**Part 10(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Forms a different sum of a non-zero rational and its reciprocal, e.g. $-2 + \frac{1}{-2} = -\frac{5}{2}$ | M1 | Must have used 1 or a negative value |
| $-\frac{5}{2} < 2$, so disproves the statement | R1 | Finds correct counter example **and** compares result with 2 |
**Part 10(c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Assumes $\frac{a}{b} + \frac{b}{a} \leq 2$ for distinct positive integers $a$ and $b$; forms $\frac{a^2+b^2}{ab} \leq 2$ | M1 | Condone $\frac{a}{b}+\frac{b}{a} < 2$ |
| Rearranges and factorises to deduce $(a-b)^2 \leq 0$ | A1 | Condone $(a-b)^2 < 0$ |
| Since $a \neq b$, $(a-b)^2 > 0$, this is a contradiction, hence $\frac{a}{b}+\frac{b}{a} > 2$ | R1 | Must have started with $\frac{a}{b}+\frac{b}{a} \leq 2$ and stated $a \neq b$ or made reference to them being distinct |10
\begin{enumerate}[label=(\alph*)]
\item \\
10
\item \\
10
\item Given that $a$ and $b$ are distinct positive numbers, use proof by contradiction to prove that
$$\frac { a } { b } + \frac { b } { a } > 2$$
\section*{END OF SECTION A \\
TURN OVER FOR SECTION B}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2023 Q10 [6]}}