| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | October |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Algebraic inequality proof |
| Difficulty | Moderate -0.5 This is the classic AM-GM inequality proof, which is a standard textbook exercise at A-level. Part (a) requires squaring both sides and basic algebraic manipulation (a standard technique), while part (b) only asks for a counter-example (trivial - just substitute any two negative numbers). The question is easier than average because it's a well-known result with a routine proof method and minimal problem-solving required. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.01c Disproof by counter example |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((\sqrt{x} - \sqrt{y})^2 \geq 0 \Rightarrow x - 2\sqrt{xy} + y \geq 0\) | M1 A1 | Sets up correct inequality and expands to three terms; correct expanded equation |
| \(\Rightarrow \frac{x+y}{2} \geq \sqrt{xy}\) | A1* | Rearranges with no errors; if working in reverse allow first M and A but require minimal conclusion for final A |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. \(x = -8,\ y = -2\): \(\frac{x+y}{2} = -5\), \(\sqrt{xy} = 4\), so \(\frac{x+y}{2} < \sqrt{xy}\) | B1 | Suitable example with both sides evaluated correctly and minimal conclusion; no need to refer to \(x\) and \(y\) in conclusion as long as inequality shown not to hold |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{x+y}{2} \geq \sqrt{xy} \Rightarrow \frac{(x+y)^2}{4} \geq xy \Rightarrow \frac{x^2 + xy + y^2}{4} \geq xy\) | M1 | Assumes \(\frac{x+y}{2} \geq \sqrt{xy}\) true and attempts to square, obtaining at least three terms |
| \(\Rightarrow x^2 - 2xy + y^2 \geq 0 \Rightarrow (x-y)^2 \geq 0\) | A1 | Correct expansion and rearranges inequality correctly to factorise to a perfect square |
| \((x-y)^2 \geq 0\) as it is a square number, so \(\frac{x+y}{2} \geq \sqrt{xy}\) is true | A1* | A complete conclusion given |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| States \((x-y)^2 \geq 0 \Rightarrow x^2 - 2xy + y^2 \geq 0\) | M1 | Sets up an inequality using an appropriate perfect square and expands to at least three terms |
| \(\Rightarrow x^2 + 2xy + y^2 \geq 4xy \Rightarrow (x+y)^2 \geq 4xy\) | A1 | Makes a correct rearrangement and factors the left hand side to produce the equation shown |
| States that as \(x\), \(y\) positive, so \(x+y > 0\) (and \(xy > 0\)); \(\Rightarrow (x+y) \geq \sqrt{4xy} \Rightarrow \frac{x+y}{2} \geq \sqrt{xy}\) | A1* | Makes a full conclusion justifying why the square root gives \(x+y\) |
## Question 9:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\sqrt{x} - \sqrt{y})^2 \geq 0 \Rightarrow x - 2\sqrt{xy} + y \geq 0$ | M1 A1 | Sets up correct inequality and expands to three terms; correct expanded equation |
| $\Rightarrow \frac{x+y}{2} \geq \sqrt{xy}$ | A1* | Rearranges with no errors; if working in reverse allow first M and A but require minimal conclusion for final A |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. $x = -8,\ y = -2$: $\frac{x+y}{2} = -5$, $\sqrt{xy} = 4$, so $\frac{x+y}{2} < \sqrt{xy}$ | B1 | Suitable example with both sides evaluated correctly and minimal conclusion; no need to refer to $x$ and $y$ in conclusion as long as inequality shown not to hold |
## Question 9(a) Alt 1:
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{x+y}{2} \geq \sqrt{xy} \Rightarrow \frac{(x+y)^2}{4} \geq xy \Rightarrow \frac{x^2 + xy + y^2}{4} \geq xy$ | M1 | Assumes $\frac{x+y}{2} \geq \sqrt{xy}$ true and attempts to square, obtaining at least three terms |
| $\Rightarrow x^2 - 2xy + y^2 \geq 0 \Rightarrow (x-y)^2 \geq 0$ | A1 | Correct expansion and rearranges inequality correctly to factorise to a perfect square |
| $(x-y)^2 \geq 0$ as it is a square number, so $\frac{x+y}{2} \geq \sqrt{xy}$ is true | A1* | A complete conclusion given |
---
## Question 9(a) Alt 2:
| Working | Mark | Guidance |
|---------|------|----------|
| States $(x-y)^2 \geq 0 \Rightarrow x^2 - 2xy + y^2 \geq 0$ | M1 | Sets up an inequality using an appropriate perfect square and expands to at least three terms |
| $\Rightarrow x^2 + 2xy + y^2 \geq 4xy \Rightarrow (x+y)^2 \geq 4xy$ | A1 | Makes a correct rearrangement and factors the left hand side to produce the equation shown |
| States that as $x$, $y$ positive, so $x+y > 0$ (and $xy > 0$); $\Rightarrow (x+y) \geq \sqrt{4xy} \Rightarrow \frac{x+y}{2} \geq \sqrt{xy}$ | A1* | Makes a full conclusion justifying why the square root gives $x+y$ |
---9. (a) Prove that for all positive values of $x$ and $y$,
$$\frac { x + y } { 2 } \geqslant \sqrt { x y }$$
(b) Prove by counter-example that this inequality does not hold when $x$ and $y$ are both negative.\\
(1)\\
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\includegraphics[max width=\textwidth, alt={}, center]{124ee19f-8a49-42df-9f4b-5a1cc2139be9-29_61_54_2608_1852}\\
\hfill \mbox{\textit{Edexcel P2 2021 Q9 [4]}}