June 2023 Paper 2 Q10
10
(a) Expand and simplify \((a - b)^2\) [1 mark]
(b) Peter thinks that the sum of any rational number and its reciprocal is always greater than 2
Peter checks two examples:
\[\frac{2}{3} + \frac{3}{2} = 2.1\dot{6}\]\[2 + \frac{1}{2} = 2.5\]Use a counter example to show that Peter is incorrect. [2 marks](c) Given that \(a\) and \(b\) are distinct positive numbers, use proof by contradiction to prove that\[\frac{a}{b} + \frac{b}{a} \gt 2\] [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains \(a^2 - 2ab + b^2\) | B1 | 1.1b |
| (1) |
Typical solution
\[a^2 - 2ab + b^2\]| Scheme | Marks | AO |
|---|---|---|
| Forms a different sum of a non-zero rational and its reciprocal. | M1 | 1.1a |
| Finds a correct counter example and compares the result with 2 Must have used 1 or a negative value. | R1 | 2.3 |
| (2) |
Typical solution
\[-2 + \frac{1}{-2} = -\frac{5}{2}\]\[-\frac{5}{2} \lt 2\]| Scheme | Marks | AO |
|---|---|---|
| Forms the inequality \(\dfrac{a}{b} + \dfrac{b}{a} \leqslant 2\) (for a pair of distinct positive integers \(a\) and \(b\)) Condone \(\dfrac{a}{b} + \dfrac{b}{a} \lt 2\) | M1 | 2.1 |
| Rearranges and factorises to deduce \((a - b)^2 \leqslant 0\) Condone \((a - b)^2 \lt 0\) | A1 | 2.2a |
| Completes a reasoned argument to explain the contradiction. Must have started with \(\dfrac{a}{b} + \dfrac{b}{a} \leqslant 2\) and stated \(a \neq b\) or makes reference to them being distinct. | R1 | 2.1 |
| (3) | ||
| (6 marks) |
Typical solution
Assume
\[\frac{a}{b} + \frac{b}{a} \leqslant 2\]\[\frac{a^2 + b^2}{ab} \leqslant 2\]\[a^2 + b^2 \leqslant 2ab\]\[a^2 - 2ab + b^2 \leqslant 0\]\[(a - b)^2 \leqslant 0\]Since \(a \neq b\) this is a contradiction because \((a - b)^2 \gt 0\)
Hence \(\dfrac{a}{b} + \dfrac{b}{a} \gt 2\)