| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Algebraic inequality proof |
| Difficulty | Standard +0.3 This is a structured algebraic proof with clear guidance. Part (i) involves routine algebraic expansion/verification that any C3 student should handle mechanically. Part (ii) requires recognizing that the given identities show x³-y³ = (x-y)(positive expression) when x>y, but this insight is heavily scaffolded by the question structure. Slightly easier than average due to the step-by-step guidance. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| \((x - y)(x^2 + xy + y^2) = x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3 = x^3 - y^3\) | M1, E1 | Expanding – allow tabulation www |
| Answer | Marks | Guidance |
|---|---|---|
| \((x + \frac{1}{2}y)^2 + \frac{3}{4}y^2 = x^2 + xy + \frac{1}{4}y^2 + \frac{3}{4}y^2 = x^2 + xy + y^2\) | M1, E1 cao www | \((x + \frac{1}{2}y)^2 = x^2 + \frac{1}{2}xy + \frac{1}{4}y^2\) o.e. [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^3 - y^3 = (x - y)[(x + \frac{1}{2}y)^2 + \frac{3}{4}y^2]\) | M1 | Substituting results of (i) |
| \((x + \frac{1}{2}y)^2 + \frac{3}{4}y^2 > 0\) [as squares \(\geq 0\)] | M1 | |
| \(\Rightarrow\) if \(x - y > 0\) then \(x^3 - y^3 > 0\) \(\Rightarrow\) if \(x > y\) then \(x^3 > y^3\) * | E1 | [3] |
**Part (i)**
**Part (A)**
| $(x - y)(x^2 + xy + y^2) = x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3 = x^3 - y^3$ | M1, E1 | Expanding – allow tabulation www |
**Part (B)**
| $(x + \frac{1}{2}y)^2 + \frac{3}{4}y^2 = x^2 + xy + \frac{1}{4}y^2 + \frac{3}{4}y^2 = x^2 + xy + y^2$ | M1, E1 cao www | $(x + \frac{1}{2}y)^2 = x^2 + \frac{1}{2}xy + \frac{1}{4}y^2$ o.e. [4] |
**Part (ii)**
| $x^3 - y^3 = (x - y)[(x + \frac{1}{2}y)^2 + \frac{3}{4}y^2]$ | M1 | Substituting results of (i) |
| $(x + \frac{1}{2}y)^2 + \frac{3}{4}y^2 > 0$ [as squares $\geq 0$] | M1 | |
| $\Rightarrow$ if $x - y > 0$ then $x^3 - y^3 > 0$ $\Rightarrow$ if $x > y$ then $x^3 > y^3$ * | E1 | [3] |
---7
\begin{enumerate}[label=(\roman*)]
\item Show that\\
(A) $( x - y ) \left( x ^ { 2 } + x y + y ^ { 2 } \right) = x ^ { 3 } - y ^ { 3 }$,\\
(B) $\left( x + \frac { 1 } { 2 } y \right) ^ { 2 } + \frac { 3 } { 4 } y ^ { 2 } = x ^ { 2 } + x y + y ^ { 2 }$.
\item Hence prove that, for all real numbers $x$ and $y$, if $x > y$ then $x ^ { 3 } > y ^ { 3 }$.
Section B (36 marks)
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 2009 Q7 [7]}}