| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2015 |
| 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 straightforward modulus equation solved by considering cases or squaring both sides, yielding x = 7/6. Part (ii) requires the substitution x = 2^y and then solving 2^y = 7/6 using logarithms—a direct application of the first part with one additional logarithm step. This is slightly easier than average due to the explicit 'hence' guidance and routine techniques. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State or imply equation \((3x + 4)^2 = (3x - 11)^2\) or \(3x + 4 = -(3x - 11)\) | B1 | |
| Attempt solution of 'quadratic' equation or linear equation | M1 | |
| Obtain \(x = \frac{7}{6}\) or equivalent (and no other solutions) | A1 | [3] |
| (ii) Use logarithms to solve equation of form \(2^y =\) their answer to (i) (must be \(+\)ve) | M1 | |
| Obtain 0.222 (and no other solutions) | A1 | [2] |
(i) State or imply equation $(3x + 4)^2 = (3x - 11)^2$ or $3x + 4 = -(3x - 11)$ | B1 |
Attempt solution of 'quadratic' equation or linear equation | M1 |
Obtain $x = \frac{7}{6}$ or equivalent (and no other solutions) | A1 | [3]
(ii) Use logarithms to solve equation of form $2^y =$ their answer to (i) (must be $+$ve) | M1 |
Obtain 0.222 (and no other solutions) | A1 | [2]
---1 (i) Solve the equation $| 3 x + 4 | = | 3 x - 11 |$.\\
(ii) Hence, using logarithms, solve the equation $\left| 3 \times 2 ^ { y } + 4 \right| = \left| 3 \times 2 ^ { y } - 11 \right|$, giving the answer correct to 3 significant figures.
\hfill \mbox{\textit{CAIE P2 2015 Q1 [5]}}