| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Transformations of modulus graphs from given f(x) sketch |
| Difficulty | Moderate -0.3 This question tests standard transformations of modulus graphs (reflection in x-axis and y-axis) and solving equations with modulus functions. Parts (a)-(c) are routine graphical work requiring recall of transformation rules. Part (d) involves solving a linear equation with modulus, requiring case-by-case analysis but following a standard method. The multi-part structure adds some length, but each component is a textbook exercise with no novel problem-solving required. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02m Graphs of functions: difference between plotting and sketching1.02s Modulus graphs: sketch graph of |ax+b|1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\bigwedge\!\bigwedge\) shape | B1 | |
| Vertices correctly placed | B1 | (2) |
| (b) Shape | B1 | |
| Vertex and intersections with axes correctly placed | B1 | (2) |
| (c) \(P:(-1, 2)\) | B1 | |
| \(Q:(0, 1)\) | B1 | |
| \(R:(1, 0)\) | B1 | (3) |
| (d) For \(x > -1\): \(2 - x - 1 = \frac{1}{2}x\) | M1 A1 | |
| Leading to \(x = \frac{2}{3}\) | A1 | |
| For \(x < -1\): \(2 + x + 1 = \frac{1}{2}x\) | M1 | |
| Leading to \(x = -6\) | A1 | (5) [12] |
**(a)** $\bigwedge\!\bigwedge$ shape | B1 |
Vertices correctly placed | B1 | (2)
**(b)** Shape | B1 |
Vertex and intersections with axes correctly placed | B1 | (2)
**(c)** $P:(-1, 2)$ | B1 |
$Q:(0, 1)$ | B1 |
$R:(1, 0)$ | B1 | (3)
**(d)** For $x > -1$: $2 - x - 1 = \frac{1}{2}x$ | M1 A1 |
Leading to $x = \frac{2}{3}$ | A1 |
For $x < -1$: $2 + x + 1 = \frac{1}{2}x$ | M1 |
Leading to $x = -6$ | A1 | (5) [12]
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f47675f8-a2c2-4c4c-b878-ffe15a95c19d-05_623_977_207_479}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the graph of $y = f ( x ) , x \in \mathbb { R }$.\\
The graph consists of two line segments that meet at the point $P$.\\
The graph cuts the $y$-axis at the point $Q$ and the $x$-axis at the points $( - 3,0 )$ and $R$. Sketch, on separate diagrams, the graphs of
\begin{enumerate}[label=(\alph*)]
\item $y = | f ( x ) |$,
\item $y = \mathrm { f } ( - x )$.
Given that $\mathrm { f } ( x ) = 2 - | x + 1 |$,
\item find the coordinates of the points $P , Q$ and $R$,
\item solve $\mathrm { f } ( x ) = \frac { 1 } { 2 } x$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2008 Q3 [12]}}