| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2017 |
| Session | Specimen |
| Marks | 7 |
| Topic | Modulus function |
| Type | Solve |linear| compared to linear: algebraic only |
| Difficulty | Moderate -0.3 Part (a) is straightforward substitution into an absolute value expression. Part (b) requires solving an absolute value inequality by considering cases (2x-1 ≥ 0 and 2x-1 < 0), which is a standard technique taught in M1/C1. The question tests routine application of absolute value properties with minimal problem-solving demand, making it slightly easier than average but not trivial due to the case-work required in part (b). |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02l Modulus function: notation, relations, equations and inequalities |
\begin{enumerate}[label=(\alph*)]
\item If $|x| = 3$, find the possible values of $|2x - 1|$. [3]
\item Find the set of values of $x$ for which $|2x - 1| > x + 1$.
Give your answer in set notation. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2017 Q1 [7]}}