| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve |linear| < |linear| |
| Difficulty | Standard +0.3 Part (i) requires squaring both sides to eliminate modulus signs, then solving a quadratic inequality—a standard technique for |linear| < |linear| problems. Part (ii) applies substitution x = 3^(0.1n) and uses logarithms to find n, which adds one extra step but remains procedural. This is slightly above average difficulty due to the two-part structure and the exponential substitution, but all techniques are standard A-level fare. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| \( | 3x-5 | < |
| \((3x-5)^2 < (x+3)^2\) | M1 | Squaring both sides |
| \(9x^2 - 30x + 25 < x^2 + 6x + 9\) | ||
| \(8x^2 - 36x + 16 < 0\) | A1 | Correct 3-term quadratic |
| \(2x^2 - 9x + 4 < 0\) | ||
| \((2x-1)(x-4) < 0\) | M1 | Attempt to solve quadratic = 0 |
| \(\dfrac{1}{2} < x < 4\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \( | 3^{0.1n+1} - 5 | < |
| Let \(x = 3^{0.1n}\), so \(3x = 3^{0.1n+1}\) | M1 | Connecting to part (i) result |
| \(\dfrac{1}{2} < 3^{0.1n} < 4\) | ||
| \(3^{0.1n} < 4 \Rightarrow 0.1n\ln 3 < \ln 4\) | M1 | Taking logarithms |
| \(n < \dfrac{\ln 4}{0.1\ln 3} = 12.6...\) | A1 | |
| Greatest integer \(n = 12\) | A1 |
# Question 2(i):
$|3x-5| < |x+3|$
| $(3x-5)^2 < (x+3)^2$ | M1 | Squaring both sides |
|---|---|---|
| $9x^2 - 30x + 25 < x^2 + 6x + 9$ | | |
| $8x^2 - 36x + 16 < 0$ | A1 | Correct 3-term quadratic |
| $2x^2 - 9x + 4 < 0$ | | |
| $(2x-1)(x-4) < 0$ | M1 | Attempt to solve quadratic = 0 |
| $\dfrac{1}{2} < x < 4$ | A1 | Correct answer |
---
# Question 2(ii):
$|3^{0.1n+1} - 5| < |3^{0.1n} + 3|$
| Let $x = 3^{0.1n}$, so $3x = 3^{0.1n+1}$ | M1 | Connecting to part (i) result |
|---|---|---|
| $\dfrac{1}{2} < 3^{0.1n} < 4$ | | |
| $3^{0.1n} < 4 \Rightarrow 0.1n\ln 3 < \ln 4$ | M1 | Taking logarithms |
| $n < \dfrac{\ln 4}{0.1\ln 3} = 12.6...$ | A1 | |
| Greatest integer $n = 12$ | A1 | |
---2 (i) Solve the inequality $| 3 x - 5 | < | x + 3 |$.\\
(ii) Hence find the greatest integer $n$ satisfying the inequality $\left| 3 ^ { 0.1 n + 1 } - 5 \right| < \left| 3 ^ { 0.1 n } + 3 \right|$.\\
\hfill \mbox{\textit{CAIE P2 2019 Q2 [6]}}