| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Rational inequality algebraically |
| Difficulty | Challenging +1.2 This is a structured Further Maths rational inequality requiring algebraic manipulation to a standard form, then sign analysis. Part (a) guides students through the algebraic rearrangement with scaffolding, and part (b) applies this to a specific case. While it requires careful handling of critical points and sign changes, the heavy scaffolding and systematic approach make it more accessible than unguided rational inequalities. Slightly above average difficulty due to the algebraic complexity and need for rigorous inequality reasoning. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\times k(x+4)^2(x-1)^2 \Rightarrow k(x+4)(x-1)^2(x+2) \leqslant x(x+4)^2(x-1)\) | M1 | Multiply both sides by \(k(x+4)^2(x-1)^2\) or bring to common denominator; slips allowed but intention clear |
| \((x+4)(x-1)\left[kx^2+kx-2k-x^2-4x\right] \leqslant 0\) | dM1 | Factorise out \((x+4)(x-1)\) and/or multiply by \((x+4)^2(x-1)^2\); dependent on previous M |
| \((x+4)(x-1)\left[(k-1)x^2+(k-4)x-2k\right] \leqslant 0\) | A1cso | Correct form; accept positive multiples of coefficients |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(k=3 \Rightarrow 2x^2-x-6=0\); \([(2x+3)(x-2)=0] \Rightarrow x=-\frac{3}{2}, 2\) | M1 | Use \(k=3\) in quadratic from (a) and solve |
| \(-4 < x \leqslant -\frac{3}{2}\) and \(1 < x \leqslant 2\) | dM1 A1 A1 | Critical values \(-4\), \(1\) and two solutions from quadratic; correct regions; both regions completely correct |
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\times k(x+4)^2(x-1)^2 \Rightarrow k(x+4)(x-1)^2(x+2) \leqslant x(x+4)^2(x-1)$ | M1 | Multiply both sides by $k(x+4)^2(x-1)^2$ or bring to common denominator; slips allowed but intention clear |
| $(x+4)(x-1)\left[kx^2+kx-2k-x^2-4x\right] \leqslant 0$ | dM1 | Factorise out $(x+4)(x-1)$ and/or multiply by $(x+4)^2(x-1)^2$; dependent on previous M |
| $(x+4)(x-1)\left[(k-1)x^2+(k-4)x-2k\right] \leqslant 0$ | A1cso | Correct form; accept positive multiples of coefficients |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $k=3 \Rightarrow 2x^2-x-6=0$; $[(2x+3)(x-2)=0] \Rightarrow x=-\frac{3}{2}, 2$ | M1 | Use $k=3$ in quadratic from (a) and solve |
| $-4 < x \leqslant -\frac{3}{2}$ and $1 < x \leqslant 2$ | dM1 A1 A1 | Critical values $-4$, $1$ and two solutions from quadratic; correct regions; both regions completely correct |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
\section*{Solutions relying on calculator technology are not acceptable.}
Given that
$$\frac { x + 2 } { x + 4 } \leqslant \frac { x } { k ( x - 1 ) }$$
where $k$ is a positive constant,\\
(a) show that
$$( x + 4 ) ( x - 1 ) \left( p x ^ { 2 } + q x + r \right) \leqslant 0$$
where $p , q$ and $r$ are expressions in terms of $k$ to be determined.\\
(b) Hence, or otherwise, determine the values for $x$ for which
$$\frac { x + 2 } { x + 4 } \leqslant \frac { x } { 3 ( x - 1 ) }$$
\hfill \mbox{\textit{Edexcel F2 2023 Q3 [7]}}