| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2015 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve modulus equation then apply exponential/log substitution |
| Difficulty | Standard +0.3 Part (i) is a standard modulus equation requiring case analysis (critical points at x=-3/2 and x=-8), yielding two linear equations to solve. Part (ii) applies the substitution 2^y = x and uses logarithms to find y, which is routine once part (i) is solved. This is slightly above average difficulty due to the two-part structure and exponential-logarithm connection, but both components use standard A-level techniques without requiring novel insight. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State or imply non-modulus equation \((2x+3)^2=(x+8)^2\) or corresponding pair of linear equations | B1 | Either method |
| Solve 3-term quadratic equation or 2 linear equations | M1 | |
| Obtain \(x=-\frac{11}{3}\) and \(x=5\) | A1 | |
| Obtain \(x=5\) from graphical method, inspection, equation... | B1 | Or method |
| Obtain \(x=-\frac{11}{3}\) similarly | B2 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use logarithms to solve equation of form \(2^y=k\) where \(k>0\) | M1 | |
| Obtain \(2.32\) | A1 | [2] |
## Question 2:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply non-modulus equation $(2x+3)^2=(x+8)^2$ or corresponding pair of linear equations | B1 | Either method |
| Solve 3-term quadratic equation or 2 linear equations | M1 | |
| Obtain $x=-\frac{11}{3}$ and $x=5$ | A1 | |
| Obtain $x=5$ from graphical method, inspection, equation... | B1 | Or method |
| Obtain $x=-\frac{11}{3}$ similarly | B2 | [3] |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use logarithms to solve equation of form $2^y=k$ where $k>0$ | M1 | |
| Obtain $2.32$ | A1 | [2] |
---2 (i) Solve the equation $| 2 x + 3 | = | x + 8 |$.\\
(ii) Hence, using logarithms, solve the equation $\left| 2 ^ { y + 1 } + 3 \right| = \left| 2 ^ { y } + 8 \right|$. Give the answer correct to 3 significant figures.
\hfill \mbox{\textit{CAIE P2 2015 Q2 [5]}}