| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | October |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Contradiction proof of inequality |
| Difficulty | Moderate -0.3 This is a straightforward proof by contradiction requiring AM-GM inequality application (or completing the square), followed by a simple counterexample. The contradiction setup is routine, and part (b) requires only testing a negative value. Below average difficulty as it's a standard textbook-style proof with minimal problem-solving required. |
| Spec | 1.01d Proof by contradiction |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Assume that there exists a positive number \(k\) such that \(k + \frac{9}{k} < 6\) | B1 | Must include "assume"/"let", must be strict inequality. Accept "assume/let \(k + \frac{9}{k} < 6\)" |
| \(k + \frac{9}{k} < 6 \Rightarrow k^2 + 9 < 6k \Rightarrow k^2 - 6k + 9 < 0\) or \(k + \frac{9}{k} < 6 \Rightarrow k + \frac{9}{k} - 6 < 0 \Rightarrow (\sqrt{k}...)(\sqrt{k}...) < 0\) or \(k + \frac{9}{k} < 6 \Rightarrow \left(k + \frac{9}{k}\right)^2 < 36 \Rightarrow k^2 + 18 + \frac{81}{k^2} - 36 < 0\) | M1 | Starting from \(k+\frac{9}{k} < 6\) (or \(\leqslant\) or \(=\)): multiplies by \(k\) and reaches \(k^2...<0\); or collects terms and factorises to \((\sqrt{k}...)(\sqrt{k}...)\leqslant/ |
| \(\Rightarrow (k-3)^2 < 0\) or \(\Rightarrow \left(\sqrt{k} - \frac{3}{\sqrt{k}}\right)^2 < 0\) or \(\Rightarrow \left(k - \frac{9}{k}\right)^2 < 0\) | A1 | Reaches \((k-3)^2...0\) or equivalent squared expression |
| But numbers squared are \(\geqslant 0\), hence \(k + \frac{9}{k} \geqslant 6\) | A1* | Requires: correct algebra, reason ("numbers squared are \(\geqslant 0\)"/"this is impossible"), and minimal conclusion (e.g. "hence proven, QED" — not just "contradiction") |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| E.g. When \(k = -3\), \(-3 + \frac{9}{-3}(= -6)\) which is not \(\geqslant 6\) | B1 | Choose a negative number, substitute into \(k + \frac{9}{k}\), state result is not \(\geqslant 6\) or is \(< 6\). It is not necessary to evaluate: e.g. \(-3 + \frac{9}{-3}\) is sufficient |
| Alt (b): \(k < 0 \Rightarrow \frac{9}{k} < 0 \Rightarrow k + \frac{9}{k} < 0\) which is not \(\geqslant 6\) | B1 | States if \(k<0\) then \(k+\frac{9}{k}<0\), gives minimal conclusion |
## Question 4(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Assume that there exists a positive number $k$ such that $k + \frac{9}{k} < 6$ | B1 | Must include "assume"/"let", must be strict inequality. Accept "assume/let $k + \frac{9}{k} < 6$" |
| $k + \frac{9}{k} < 6 \Rightarrow k^2 + 9 < 6k \Rightarrow k^2 - 6k + 9 < 0$ **or** $k + \frac{9}{k} < 6 \Rightarrow k + \frac{9}{k} - 6 < 0 \Rightarrow (\sqrt{k}...)(\sqrt{k}...) < 0$ **or** $k + \frac{9}{k} < 6 \Rightarrow \left(k + \frac{9}{k}\right)^2 < 36 \Rightarrow k^2 + 18 + \frac{81}{k^2} - 36 < 0$ | M1 | Starting from $k+\frac{9}{k} < 6$ (or $\leqslant$ or $=$): multiplies by $k$ and reaches $k^2...<0$; or collects terms and factorises to $(\sqrt{k}...)(\sqrt{k}...)\leqslant/</ =0$; or squares both sides and collects terms |
| $\Rightarrow (k-3)^2 < 0$ **or** $\Rightarrow \left(\sqrt{k} - \frac{3}{\sqrt{k}}\right)^2 < 0$ **or** $\Rightarrow \left(k - \frac{9}{k}\right)^2 < 0$ | A1 | Reaches $(k-3)^2...0$ or equivalent squared expression |
| But numbers squared are $\geqslant 0$, hence $k + \frac{9}{k} \geqslant 6$ | A1* | Requires: correct algebra, reason ("numbers squared are $\geqslant 0$"/"this is impossible"), and minimal conclusion (e.g. "hence proven, QED" — not just "contradiction") |
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## Question 4(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| E.g. When $k = -3$, $-3 + \frac{9}{-3}(= -6)$ which is not $\geqslant 6$ | B1 | Choose a negative number, substitute into $k + \frac{9}{k}$, state result is not $\geqslant 6$ or is $< 6$. It is not necessary to evaluate: e.g. $-3 + \frac{9}{-3}$ is sufficient |
**Alt (b):** $k < 0 \Rightarrow \frac{9}{k} < 0 \Rightarrow k + \frac{9}{k} < 0$ which is not $\geqslant 6$ | B1 | States if $k<0$ then $k+\frac{9}{k}<0$, gives minimal conclusion
---\begin{enumerate}
\item (a) Prove by contradiction that for all positive numbers $k$
\end{enumerate}
$$k + \frac { 9 } { k } \geqslant 6$$
(b) Show that the result in part (a) is not true for all real numbers.
\hfill \mbox{\textit{Edexcel P4 2023 Q4 [5]}}