| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Sketch y=|linear| then solve equation or inequality (numeric coefficients) |
| Difficulty | Moderate -0.3 This is a straightforward modulus question requiring basic graph sketching and solving a simple modulus equation by considering two cases (x ≥ 0 and x < 0). The inequality in part (d) follows directly from the equation solution. While it requires multiple steps, each individual step uses standard techniques with no novel insight needed, making it slightly easier than average. |
| 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 |
|---|---|---|
| Answer | Mark | Guidance |
| Correct graph sketch | B1 | |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Shape: inverted V in all four quadrants | M1 | Shape inverted V in all four quadrants |
| Symmetrical about \(y\) axis | A1 | Symmetrical about \(y\) axis |
| Correct coordinates \((-2,0)\), \((2,0)\), \((0,4)\) | A1 | Coordinates |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4 - | 2x | = x\) |
| \(4 - 2x = x \Rightarrow x = \frac{4}{3}\) | M1, A1 | Attempt to solve |
| \(4 + 2x = x \Rightarrow x = -4\) | A1 | And no others |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(-4 < x < \frac{4}{3}\) | M1, A1 | Either correct; other solution and no extras |
| SC \(-4 \leq x \leq \frac{4}{3}\) B1 | ||
| Total: 2 |
# Question 7:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct graph sketch | B1 | |
| **Total: 1** | | |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Shape: inverted V in all four quadrants | M1 | Shape inverted V in all four quadrants |
| Symmetrical about $y$ axis | A1 | Symmetrical about $y$ axis |
| Correct coordinates $(-2,0)$, $(2,0)$, $(0,4)$ | A1 | Coordinates |
| **Total: 3** | | |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $4 - |2x| = x$ | | |
| $4 - 2x = x \Rightarrow x = \frac{4}{3}$ | M1, A1 | Attempt to solve |
| $4 + 2x = x \Rightarrow x = -4$ | A1 | And no others |
| **Total: 3** | | |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $-4 < x < \frac{4}{3}$ | M1, A1 | Either correct; other solution and no extras |
| | | **SC** $-4 \leq x \leq \frac{4}{3}$ B1 |
| **Total: 2** | | |
---7
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of $y = | 2 x |$.
\item On a separate diagram, sketch the graph of $y = 4 - | 2 x |$, indicating the coordinates of the points where the graph crosses the coordinate axes.
\item Solve $4 - | 2 x | = x$.
\item Hence, or otherwise, solve the inequality $4 - | 2 x | > x$.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2007 Q7 [9]}}