| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve |linear| compared to linear: algebraic only |
| Difficulty | Standard +0.3 Part (a) is straightforward substitution requiring recognition that |x|=3 gives x=±3. Part (b) is a standard modulus inequality requiring case analysis (2x-1≥0 and 2x-1<0), solving two linear inequalities, and expressing in set notation. This is slightly above average difficulty due to the case work and set notation requirement, but remains a textbook exercise with well-established methods. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities |
1
\begin{enumerate}[label=(\alph*)]
\item If $| x | = 3$, find the possible values of $| 2 x - 1 |$.
\item Find the set of values of $x$ for which $| 2 x - 1 | > x + 1$.
Give your answer in set notation.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 Q1 [7]}}