| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Sketch y=|linear| then solve equation or inequality (numeric coefficients) |
| Difficulty | Moderate -0.8 This is a straightforward modulus question requiring a basic V-shaped graph sketch and solving by considering two cases (3x-6 ≥ 0 and 3x-6 < 0). The algebraic manipulation is simple linear equations, and this is a standard textbook exercise testing routine application of modulus concepts with no problem-solving insight required. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02s Modulus graphs: sketch graph of |ax+b|1.02t Solve modulus equations: graphically with modulus function |
| Answer | Marks | Guidance |
|---|---|---|
| Graph showing two half-lines meeting at \((2, 0)\) with correct orientation | B1, B1 | One for two half-lines; one for correct orientation and meeting at \((2, 0)\). 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| At P: \(-3x + 6 = x + 4 \Rightarrow x = \frac{1}{2}\) | M1, A1 | |
| At Q: \(3x - 6 = x + 4 \Rightarrow x = 5\) | (implied) | |
| The solution is \(x = \frac{1}{2}, 5\). As shown on graph | A1, E1 | 4 marks |
### (i)
Graph showing two half-lines meeting at $(2, 0)$ with correct orientation | B1, B1 | One for two half-lines; one for correct orientation and meeting at $(2, 0)$. 2 marks
### (ii)
At P: $-3x + 6 = x + 4 \Rightarrow x = \frac{1}{2}$ | M1, A1
At Q: $3x - 6 = x + 4 \Rightarrow x = 5$ | (implied)
The solution is $x = \frac{1}{2}, 5$. As shown on graph | A1, E1 | 4 marks
---\begin{enumerate}[label=(\roman*)]
\item Sketch the graph of $y = |3x - 6|$. [2]
\item Solve the equation $|3x - 6| = x + 4$ and illustrate your answer on your graph. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 Q3 [6]}}