| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve modulus equation then apply exponential/log substitution |
| Difficulty | Moderate -0.3 Part (i) is a standard modulus equation solved by considering cases or squaring both sides - routine A-level technique. Part (ii) adds a simple exponential substitution (let u = e^(3y)) followed by taking logarithms, but requires no novel insight beyond applying the standard method twice. This is slightly easier than average due to being a straightforward two-part question with clear methodology. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply non-modular equation \((4+2x)^2 = (3-5x)^2\) or pair of linear equations | B1 | |
| Attempt solution of 3-term quadratic eqn or pair of linear equations | M1 | |
| Obtain \(-\frac{1}{7},\ \frac{7}{3}\) | A1 | SC B1 for \(x = -\frac{1}{7}\) from one linear equation |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt correct process to solve \(e^{3y} = k\) where \(k > 0\) from (i) | M1 | |
| Obtain \(0.282\) and no others | A1 | |
| Total | 2 |
**Question 2(i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply non-modular equation $(4+2x)^2 = (3-5x)^2$ or pair of linear equations | B1 | |
| Attempt solution of 3-term quadratic eqn or pair of linear equations | M1 | |
| Obtain $-\frac{1}{7},\ \frac{7}{3}$ | A1 | SC B1 for $x = -\frac{1}{7}$ from one linear equation |
| **Total** | **3** | |
---
**Question 2(ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt correct process to solve $e^{3y} = k$ where $k > 0$ from **(i)** | M1 | |
| Obtain $0.282$ and no others | A1 | |
| **Total** | **2** | |2 (i) Solve the equation $| 4 + 2 x | = | 3 - 5 x |$.\\
(ii) Hence solve the equation $\left| 4 + 2 e ^ { 3 y } \right| = \left| 3 - 5 e ^ { 3 y } \right|$, giving the answer correct to 3 significant figures.\\
\hfill \mbox{\textit{CAIE P2 2019 Q2 [5]}}