| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Graph y = a|bx+c| + d: identify vertex and intercepts |
| Difficulty | Moderate -0.3 This is a straightforward modulus function question requiring students to use the vertex and y-intercept to find constants, then apply a standard horizontal stretch and vertical translation. The algebraic manipulation is routine and the transformations are standard C3 content, making it slightly easier than average but still requiring proper understanding of modulus graphs and transformations. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02s Modulus graphs: sketch graph of |ax+b|1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitute \((0,5)\): \(\ | b\ | = 5\) so \(b = \pm 5\) |
| Substitute \((\frac{1}{3}, 0)\): \(\ | \frac{1}{3}a + b\ | = 0\) so \(a = \mp 15\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \ | -15x + 5\ | \) or \(y = \ |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| V shape correct way up | B1 | V shape correct way up, any position but not on \(x\)-axis. Accept V's without symmetry |
| \(P\) at \(\left(\frac{2}{3}, 3\right)\) | B1 | Score if coordinates stated in text OR marked on axes |
| \(Q\) at \((0, 8)\) | B1 | For crossing \(y\)-axis at \((0,8)\). Accept 8 on correct axis. Condone \((8,0)\) on correct axis |
| (3) | There must be a sketch to score any marks |
# Question 7:
## Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitute $(0,5)$: $\|b\| = 5$ so $b = \pm 5$ | B1 | For both $b = \pm 5$, not just $\|b\| = 5$ |
| Substitute $(\frac{1}{3}, 0)$: $\|\frac{1}{3}a + b\| = 0$ so $a = \mp 15$ | M1 A1 | M1: Substitute $(\frac{1}{3}, 0)$ to give $\|\frac{1}{3}a + b\| = 0$. A1: $a = 15$ corresponding to $b = -5$ **and** $a = -15$ corresponding to $b = 5$ |
## Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \|-15x + 5\|$ or $y = \|15x - 5\|$ | B1 | Note this is an A1 mark on e-pen. If only one equation given scores B0 M1 A0. Both equations score B1 M1 A1. Linear equations without modulus score B0 M1 A0 |
| | **(4)** | |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| V shape correct way up | B1 | V shape correct way up, any position but not on $x$-axis. Accept V's without symmetry |
| $P$ at $\left(\frac{2}{3}, 3\right)$ | B1 | Score if coordinates stated in text OR marked on axes |
| $Q$ at $(0, 8)$ | B1 | For **crossing** $y$-axis at $(0,8)$. Accept 8 on correct axis. Condone $(8,0)$ on correct axis |
| | **(3)** | There must be a sketch to score any marks |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e30f0c28-1695-40a1-8e9a-6ea7e29042bf-12_458_433_264_781}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the graph of $y = \mathrm { f } ( x ) , x \in \mathbb { R }$.\\
The point $P \left( \frac { 1 } { 3 } , 0 \right)$ is the vertex of the graph.\\
The point $Q ( 0,5 )$ is the intercept with the $y$-axis.
Given that $\mathrm { f } ( x ) = | a x + b |$, where $a$ and $b$ are constants,
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item find all possible values for $a$ and $b$,
\item hence find an equation for the graph.
\end{enumerate}\item Sketch the graph with equation
$$y = \mathrm { f } \left( \frac { 1 } { 2 } x \right) + 3$$
showing the coordinates of its vertex and its intercept with the $y$-axis.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q7 [7]}}