| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 4 |
| 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 critical points and cases, yielding x = 5.5. Part (ii) requires the substitution u = 3^y and then solving 3^y = 5.5 using logarithms. This is a routine two-part question testing basic modulus equation technique and exponential manipulation with no novel insight required, making it slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.06a Exponential function: a^x and e^x graphs and properties |
| Answer | Marks |
|---|---|
| (i) Either: Square both sides to obtain linear equation. Obtain \(x = \frac{165}{30}\) or \(\frac{33}{6}\) or \(\frac{11}{2}\) | M1, A1 [2] |
| Or: Solve linear equation in which, initially, signs of \(x\) are different. Obtain \(x + 2 = -x + 13\) or equivalent and hence \(\frac{11}{2}\) or equivalent | M1, A1 [2] |
| (ii) Apply logarithms and use power law. Obtain \(y \log 3 = \log \frac{11}{5}\) and hence \(y = 1.55\) | M1, A1 [2] |
(i) Either: Square both sides to obtain linear equation. Obtain $x = \frac{165}{30}$ or $\frac{33}{6}$ or $\frac{11}{2}$ | M1, A1 [2] |
Or: Solve linear equation in which, initially, signs of $x$ are different. Obtain $x + 2 = -x + 13$ or equivalent and hence $\frac{11}{2}$ or equivalent | M1, A1 [2] |
(ii) Apply logarithms and use power law. Obtain $y \log 3 = \log \frac{11}{5}$ and hence $y = 1.55$ | M1, A1 [2] |1 (i) Solve the equation $| x + 2 | = | x - 13 |$.\\
(ii) Hence solve the equation $\left| 3 ^ { y } + 2 \right| = \left| 3 ^ { y } - 13 \right|$, giving your answer correct to 3 significant figures.
\hfill \mbox{\textit{CAIE P2 2014 Q1 [4]}}