| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Prove inequality algebraically |
| Difficulty | Standard +0.3 This is a straightforward algebraic manipulation question requiring expansion of the cubic, simplification, and factoring to reach the conclusion. Part (a) involves routine algebra (expanding, collecting terms, factoring out xy) with a clear path once expanded. Part (b) simply asks for a counterexample, which can be found by testing negative values. While it requires careful algebraic reasoning, it's more accessible than average A-level questions as it follows a standard 'expand and simplify' approach with no novel insight needed. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.01c Disproof by counter example |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3\) | M1 | Attempts \((x-y)^3 = x^3 \pm Ax^2y \pm Bxy^2 \pm y^3\); condone at most one slip in index as long as middle terms have at least \(xy\) |
| \((x-y)^3 > x^3 - y^3 \Rightarrow -3x^2y + 3xy^2 > 0\) | A1 | Correct simplified inequality, cubed terms cancelled; terms need not all be gathered e.g. \(3xy^2 > 3x^2y\) |
| \(\Rightarrow 3xy(y-x) > 0\) | dM1 | Takes out common factor of \(kxy\) or divides by \(kxy\); must be clear division or stated as dividing |
| As \(x\) and \(y\) are positive numbers \((y-x) > 0 \Rightarrow y > x\) | A1* | Correct working before achieving \(y>x\); reasoning given e.g. as \(x\) and \(y\) are positive so \(3xy(y-x)>0 \Rightarrow (y-x)>0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \((x-y)^3 = (x-y)(x^2+axy+y^2)\) and \(x^3-y^3=(x-y)(x^2+bxy+y^2)\) | M1 | May be unsimplified |
| Both correct: \((x-y)^3=(x-y)(x^2-2xy+y^2)\) and \(x^3-y^3=(x-y)(x^2+xy+y^2)\) | A1 | |
| Takes to one side, takes out common factor \((x-y)\) and cancels square terms | dM1 | Dividing by \((x-y)\) is M0 |
| Correct working before \(y>x\); reasoning given in correct place | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Chooses suitable counter example e.g. \(x=3, y=-1\) | M1 | Any positive \(x\) and negative \(y\); substitution not needed for this mark; if both positive it is M0 |
| Shows \((3-{-1})^3 > (3)^3-(-1)^3\) as \(64>28\) (BUT \(-1<3\)) | A1 | Shows result is not true for their values; minimum accept substitution into both sides |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$ | M1 | Attempts $(x-y)^3 = x^3 \pm Ax^2y \pm Bxy^2 \pm y^3$; condone at most one slip in index as long as middle terms have at least $xy$ |
| $(x-y)^3 > x^3 - y^3 \Rightarrow -3x^2y + 3xy^2 > 0$ | A1 | Correct simplified inequality, cubed terms cancelled; terms need not all be gathered e.g. $3xy^2 > 3x^2y$ |
| $\Rightarrow 3xy(y-x) > 0$ | dM1 | Takes out common factor of $kxy$ or divides by $kxy$; must be clear division or stated as dividing |
| As $x$ and $y$ are positive numbers $(y-x) > 0 \Rightarrow y > x$ | A1* | Correct working before achieving $y>x$; reasoning given e.g. as $x$ and $y$ are positive so $3xy(y-x)>0 \Rightarrow (y-x)>0$ |
**Alternative method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $(x-y)^3 = (x-y)(x^2+axy+y^2)$ and $x^3-y^3=(x-y)(x^2+bxy+y^2)$ | M1 | May be unsimplified |
| Both correct: $(x-y)^3=(x-y)(x^2-2xy+y^2)$ and $x^3-y^3=(x-y)(x^2+xy+y^2)$ | A1 | |
| Takes to one side, takes out common factor $(x-y)$ and cancels square terms | dM1 | Dividing by $(x-y)$ is M0 |
| Correct working before $y>x$; reasoning given in correct place | A1* | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Chooses suitable counter example e.g. $x=3, y=-1$ | M1 | Any positive $x$ and negative $y$; substitution not needed for this mark; if both positive it is M0 |
| Shows $(3-{-1})^3 > (3)^3-(-1)^3$ as $64>28$ (BUT $-1<3$) | A1 | Shows result is not true for their values; minimum accept substitution into both sides |
---\begin{enumerate}
\item In this question you must show detailed reasoning.\\
(a) Given that $x$ and $y$ are positive numbers such that
\end{enumerate}
$$( x - y ) ^ { 3 } > x ^ { 3 } - y ^ { 3 }$$
prove that
$$y > x$$
(b) Using a counter example, show that the result in part (a) is not true for all real numbers.
\hfill \mbox{\textit{Edexcel P2 2024 Q5 [6]}}