| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2018 |
| Session | March |
| Marks | 6 |
| Topic | Modulus function |
| Type | Solve |linear| = |linear| (both linear inside) |
| Difficulty | Moderate -0.8 Part (i) requires simple substitution of n=±5 into an expression. Part (ii) is a standard modulus equation solved by considering cases (squaring both sides or critical points at x=2 and x=6), requiring routine algebraic manipulation but no novel insight. This is easier than average A-level content. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02t Solve modulus equations: graphically with modulus function |
| Answer | Marks | Guidance |
|---|---|---|
| \(n = \pm 5\); \( | 2(-5) - 3 | = 13\) and \( |
| Answer | Marks | Guidance |
|---|---|---|
| \(3x - 6 = x - 6\), so \(x = 0\) OR \(3x - 6 = -x + 6\), \(4x = 12\), \(x = 3\) | B1, M1, A1 | Obtain \(x = 0\) OR M1 square both sides, A1 Obtain \(x = 3\) |
**(i)**
| $n = \pm 5$; $|2(-5) - 3| = 13$ and $|2(5) - 3| = 7$; $13 > 7$ as $13 > 7$ | B1, M1, A1 | Identify that $n = \pm 5$; Substitute both $n = \pm 5$ into expression; Conclude with 13 and explanation |
**(ii)**
| $3x - 6 = x - 6$, so $x = 0$ OR $3x - 6 = -x + 6$, $4x = 12$, $x = 3$ | B1, M1, A1 | Obtain $x = 0$ OR M1 square both sides, A1 Obtain $x = 3$ |2 (i) Given that $| n | = 5$, find the greatest value of $| 2 n - 3 |$, justifying your answer.\\
(ii) Solve the equation $| 3 x - 6 | = | x - 6 |$.
\hfill \mbox{\textit{OCR H240/01 2018 Q2 [6]}}