| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 13 |
| 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 standard C3 modulus question covering routine techniques: sketching a V-shaped graph with transformations, solving by cases (splitting at x = -1/2), and describing transformations (reflection + translation). While multi-part, each component is textbook-standard with no novel insight required, making it slightly easier than average. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02s Modulus graphs: sketch graph of |ax+b|1.02t Solve modulus equations: graphically with modulus function1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Correct V-shape with vertex above x-axis | B1 | Correct shape |
| Vertex at \(\left(-\frac{1}{2}, 4\right)\) | B1 | Correct vertex coordinates marked |
| y-intercept at \((0, 3)\) | B1 | Correct y-intercept |
| x-intercepts at \(\left(\frac{3}{2}, 0\right)\) and \(\left(-\frac{5}{2}, 0\right)\) | B1 | Both x-intercepts correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Case 1: \(x = 4-(2x+1) = 3-2x \Rightarrow 3x = 3 \Rightarrow x=1\) | M1 | One valid case attempted |
| Case 2: \(x = 4-(-(2x+1)) = 4+2x+1 \Rightarrow -x = 5 \Rightarrow x=-5\) | M1 | Second case attempted |
| \(x = 1\) and \(x = -5\) | A1 | Both solutions correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-5 < x < 1\) | M1A1 | M1 for correct method using (b); A1 correct inequality |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Reflection in the x-axis | M1A1 | Correct transformation stated |
| Translation by \(\begin{pmatrix}0\\4\end{pmatrix}\) (translation 4 units up) | M1A1 | Correct second transformation |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct V-shape with vertex above x-axis | B1 | Correct shape |
| Vertex at $\left(-\frac{1}{2}, 4\right)$ | B1 | Correct vertex coordinates marked |
| y-intercept at $(0, 3)$ | B1 | Correct y-intercept |
| x-intercepts at $\left(\frac{3}{2}, 0\right)$ and $\left(-\frac{5}{2}, 0\right)$ | B1 | Both x-intercepts correct |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Case 1: $x = 4-(2x+1) = 3-2x \Rightarrow 3x = 3 \Rightarrow x=1$ | M1 | One valid case attempted |
| Case 2: $x = 4-(-(2x+1)) = 4+2x+1 \Rightarrow -x = 5 \Rightarrow x=-5$ | M1 | Second case attempted |
| $x = 1$ and $x = -5$ | A1 | Both solutions correct |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-5 < x < 1$ | M1A1 | M1 for correct method using (b); A1 correct inequality |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Reflection in the x-axis | M1A1 | Correct transformation stated |
| Translation by $\begin{pmatrix}0\\4\end{pmatrix}$ (translation 4 units up) | M1A1 | Correct second transformation |
---2
\begin{enumerate}[label=(\alph*)]
\item Sketch, on the axes below, the curve with equation $y = 4 - | 2 x + 1 |$, indicating the coordinates where the curve crosses the axes.
\item Solve the equation $x = 4 - | 2 x + 1 |$.
\item Solve the inequality $x < 4 - | 2 x + 1 |$.
\item Describe a sequence of two geometrical transformations that maps the graph of $y = | 2 x + 1 |$ onto the graph of $y = 4 - | 2 x + 1 |$.\\[0pt]
[4 marks]
\section*{Answer space for question 2}
(a)\\
\includegraphics[max width=\textwidth, alt={}, center]{2df59047-3bfe-4b9c-a77f-142bc7506cbc-04_851_1459_1000_319}
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2015 Q2 [13]}}