| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve inequality with reciprocal in modulus |
| Difficulty | Standard +0.8 This Further Maths question requires solving a rational inequality by considering sign cases and critical points, then extending to modulus cases. Part (a) is moderately challenging (finding where x-5 < 9/(x+3) requires multiplying by (x+3)² to avoid sign issues), and part (b) requires systematic consideration of |x+3| = ±(x+3), doubling the casework. The reciprocal within the modulus adds algebraic complexity beyond standard modulus inequalities. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02l Modulus function: notation, relations, equations and inequalities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((x+3)(x-5)=9 \Rightarrow x^2-2x-24=0\) or equivalent cubic method | M1 | Correct algebraic method to find intersection points. Must reach at least a quadratic. |
| CVs: \(6,\ -4\ ;\ -3\) | A1; B1 | \(6, -4\) via valid algebraic method; CV \(-3\) seen anywhere. |
| \(x < -4,\ -3 < x < 6\) | dM1A1A1 (6) | Obtaining inequalities using all CVs; at least one correct interval; both correct ranges. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x < 6,\ x \neq -3\) or equivalent | B1ftB1 (2) | B1ft for "\(x<6\)"; B1 for fully correct answer. |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x+3)(x-5)=9 \Rightarrow x^2-2x-24=0$ or equivalent cubic method | M1 | Correct algebraic method to find intersection points. Must reach at least a quadratic. |
| CVs: $6,\ -4\ ;\ -3$ | A1; B1 | $6, -4$ via valid algebraic method; CV $-3$ seen anywhere. |
| $x < -4,\ -3 < x < 6$ | dM1A1A1 (6) | Obtaining inequalities using all CVs; at least one correct interval; both correct ranges. |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x < 6,\ x \neq -3$ or equivalent | B1ftB1 (2) | B1ft for "$x<6$"; B1 for fully correct answer. |
---\begin{enumerate}
\item (a) Use algebra to determine the set of values of $x$ for which
\end{enumerate}
$$x - 5 < \frac { 9 } { x + 3 }$$
(b) Hence, or otherwise, determine the set of values of $x$ for which
$$x - 5 < \frac { 9 } { | x + 3 | }$$
\hfill \mbox{\textit{Edexcel F2 2022 Q2 [8]}}