| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve |linear| = linear (non-modulus) |
| Difficulty | Moderate -0.8 This is a straightforward modulus equation requiring students to consider two cases (x ≥ 0 and x < 0) and solve resulting linear equations. Part (a) is trivial coordinate reading. The algebraic manipulation is routine with no conceptual challenges beyond the basic definition of modulus. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02t Solve modulus equations: graphically with modulus function |
3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0f6fd881-4d4b-4f80-96cc-6da41cc33c60-05_654_967_244_507}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the graph with equation $y = 2 | x | - 5$.\\
The graph intersects the positive $x$-axis at the point $P$ and the negative $y$-axis at the point $Q$.
\begin{enumerate}[label=(\alph*)]
\item State the coordinates of $P$ and the coordinates of $Q$.
\item Solve the equation
$$2 | x | - 5 = 3 - x$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2013 Q3 [5]}}