| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Topic | Modulus function |
| Type | Interpret or complete given sketch of two |linear| functions |
| Difficulty | Moderate -0.3 This is a straightforward modulus equation problem with a helpful diagram provided. Part (a) requires solving |2x-3|=|x| by considering cases (typically 2-3 cases based on critical points), which is a standard C3 technique. Part (b) uses the graph to read off the solution to the inequality directly. The question is slightly easier than average because the diagram guides students and the algebraic manipulation is routine. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02t Solve modulus equations: graphically with modulus function |
1 The diagram below shows the graphs of $y = | 2 x - 3 |$ and $y = | x |$.\\
\includegraphics[max width=\textwidth, alt={}, center]{063bbfa5-df49-44a1-8143-5e076397f63f-02_579_1150_351_482}
\begin{enumerate}[label=(\alph*)]
\item Find the $x$-coordinates of the points of intersection of the graphs of $y = | 2 x - 3 |$ and $y = | x |$.\\
(3 marks)
\item Hence, or otherwise, solve the inequality
$$| 2 x - 3 | \geqslant | x |$$
(2 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2013 Q1 [5]}}