| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Rational inequality algebraically |
| Difficulty | Standard +0.8 This is a Further Maths F2 rational inequality requiring algebraic manipulation to a common denominator, rearranging to standard form, factoring a quadratic, and careful sign analysis with critical points. The presence of two distinct denominators and the requirement to avoid calculator methods elevates this above a standard C1/C2 inequality, but it follows a systematic procedure without requiring novel insight. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = -2, \ 3\) | B1 | Critical values −2 and 3 stated or used. May be seen on a diagram. |
| \(\frac{x+1}{(x-3)(x+2)} = 1 - \frac{2}{x-3} \Rightarrow x+1 = x^2 - x - 6 - 2(x+2) \Rightarrow x^2 - 4x - 11 = 0\) or \(\frac{x+1}{(x-3)(x+2)} - 1 + \frac{2}{x-3} \leq 0 \Rightarrow \frac{4x - x^2 + 11}{(x-3)(x+2)} = 0\) or \((x+1)(x-3)(x+2) \leq (x-3)^2(x+2)^2 - 2(x-3)(x+2)^2 \Rightarrow (x-3)(x+2)[4x - x^2 + 11] = 0\) | M1 | Any complete algebraic method to reach at least an expanded quadratic for the other two critical values. |
| \(x^2 - 4x - 11 = 0 \Rightarrow x = \frac{4 \pm \sqrt{16+44}}{2} = \ldots\) | M1 | Correct method to solve the three term quadratic. Accept if correct solutions appear without working. |
| \(x = 2 \pm \sqrt{15}\) | A1 | Must be exact. |
| \(x < -2, \ 2 - \sqrt{15} \leq x < 3, \ x \geq 2 + \sqrt{15}\) | M1A1 | M1: correct shape \(x < \alpha, b \leq x < \gamma, x \geq \delta\). A1: fully correct, no extra regions, exact values. Accept \((-\infty,-2) \cup [2-\sqrt{15},3) \cup [2+\sqrt{15},\infty)\) |
# Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = -2, \ 3$ | B1 | Critical values −2 and 3 stated or used. May be seen on a diagram. |
| $\frac{x+1}{(x-3)(x+2)} = 1 - \frac{2}{x-3} \Rightarrow x+1 = x^2 - x - 6 - 2(x+2) \Rightarrow x^2 - 4x - 11 = 0$ **or** $\frac{x+1}{(x-3)(x+2)} - 1 + \frac{2}{x-3} \leq 0 \Rightarrow \frac{4x - x^2 + 11}{(x-3)(x+2)} = 0$ **or** $(x+1)(x-3)(x+2) \leq (x-3)^2(x+2)^2 - 2(x-3)(x+2)^2 \Rightarrow (x-3)(x+2)[4x - x^2 + 11] = 0$ | M1 | Any complete algebraic method to reach at least an expanded quadratic for the other two critical values. |
| $x^2 - 4x - 11 = 0 \Rightarrow x = \frac{4 \pm \sqrt{16+44}}{2} = \ldots$ | M1 | Correct method to solve the three term quadratic. Accept if correct solutions appear without working. |
| $x = 2 \pm \sqrt{15}$ | A1 | Must be exact. |
| $x < -2, \ 2 - \sqrt{15} \leq x < 3, \ x \geq 2 + \sqrt{15}$ | M1A1 | M1: correct shape $x < \alpha, b \leq x < \gamma, x \geq \delta$. A1: fully correct, no extra regions, exact values. Accept $(-\infty,-2) \cup [2-\sqrt{15},3) \cup [2+\sqrt{15},\infty)$ |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.\\
Use algebra to determine the values of $x$ for which
$$\frac { x + 1 } { ( x - 3 ) ( x + 2 ) } \leqslant 1 - \frac { 2 } { x - 3 }$$
\hfill \mbox{\textit{Edexcel F2 2024 Q5 [6]}}