| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve |linear| > |linear| |
| Difficulty | Moderate -0.3 This is a standard modulus inequality requiring case-by-case analysis based on critical points x = -7/5 and x = 3/2. While it involves multiple cases and algebraic manipulation, it follows a routine algorithmic approach taught in P2 with no novel insight required, making it slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Solve \(5x+7=2x-3\) to obtain \(-\frac{10}{3}\) | B1 | Or inequality |
| Attempt solution of linear equation where \(5x\) and \(2x\) have different signs | M1 | Or inequality |
| Obtain \(-\frac{4}{7}\) | A1 | |
| State \(x < -\frac{10}{3}\), \(x > -\frac{4}{7}\) | A1 | A0 if '… and …' used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply non-modulus equation \((5x+7)^2 = (2x-3)^2\) | (B1) | Or inequality |
| Attempt solution of three-term quadratic equation | (M1) | Or inequality |
| Obtain \(-\frac{10}{3}\) and \(-\frac{4}{7}\) | (A1) | |
| State \(x < -\frac{10}{3}\), \(x > -\frac{4}{7}\) | (A1) | A0 if '… and …' used |
## Question 1:
| Answer | Mark | Guidance |
|--------|------|----------|
| Solve $5x+7=2x-3$ to obtain $-\frac{10}{3}$ | **B1** | Or inequality |
| Attempt solution of linear equation where $5x$ and $2x$ have different signs | **M1** | Or inequality |
| Obtain $-\frac{4}{7}$ | **A1** | |
| State $x < -\frac{10}{3}$, $x > -\frac{4}{7}$ | **A1** | A0 if '… and …' used |
**Alternative Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply non-modulus equation $(5x+7)^2 = (2x-3)^2$ | **(B1)** | Or inequality |
| Attempt solution of three-term quadratic equation | **(M1)** | Or inequality |
| Obtain $-\frac{10}{3}$ and $-\frac{4}{7}$ | **(A1)** | |
| State $x < -\frac{10}{3}$, $x > -\frac{4}{7}$ | **(A1)** | A0 if '… and …' used |
**Total: 4 marks**
---1 Solve the inequality $| 5 x + 7 | > | 2 x - 3 |$.\\
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\includegraphics[max width=\textwidth, alt={}, center]{76df3465-9617-4f2b-a8b7-f474b2817504-02_2715_37_143_2010}
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\hfill \mbox{\textit{CAIE P2 2024 Q1 [4]}}