| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Composite transformation sketch |
| Difficulty | Moderate -0.3 This is a straightforward C1 transformation question requiring a horizontal translation of +1. Students must sketch the transformed curve, identify new asymptotes (x=3, y=1), and find axis intercepts by substituting x=0 and y=0. While it involves multiple steps, each is routine application of standard transformation rules with no novel problem-solving required. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks |
|---|---|
| Correct shape with a single crossing of each axis | B1 |
| \(y = 1\) labelled or stated | B1 |
| \(x = 3\) labelled or stated | B1 |
| (3 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Horizontal translation so crosses the \(x\)-axis at \((1, 0)\) | B1 | For point \((1,0)\) identified; may be marked on sketch as 1 on \(x\)-axis; accept \(x=1\) |
| New equation is \((y =)\ \dfrac{x \pm 1}{(x \pm 1)-2}\) | M1 | For attempt at new equation with either numerator or denominator correct |
| When \(x = 0\), \(y =\) | M1 | For setting \(x=0\) in their new equation and solving as far as \(y = \ldots\) |
| \(= \dfrac{1}{3}\) | A1 | For \(\frac{1}{3}\) or exact equivalent; must see \(y = \frac{1}{3}\) or \((0, \frac{1}{3})\) or point marked on \(y\)-axis |
| (4 marks total) |
## Question 5:
### Part (a):
| Correct shape with a single crossing of each axis | B1 | |
| $y = 1$ labelled or stated | B1 | |
| $x = 3$ labelled or stated | B1 | |
|(3 marks total)|||
### Part (b):
| Horizontal translation so crosses the $x$-axis at $(1, 0)$ | B1 | For point $(1,0)$ identified; may be marked on sketch as 1 on $x$-axis; accept $x=1$ |
| New equation is $(y =)\ \dfrac{x \pm 1}{(x \pm 1)-2}$ | M1 | For attempt at new equation with either numerator or denominator correct |
| When $x = 0$, $y =$ | M1 | For setting $x=0$ in their new equation and solving as far as $y = \ldots$ |
| $= \dfrac{1}{3}$ | A1 | For $\frac{1}{3}$ or exact equivalent; must see $y = \frac{1}{3}$ or $(0, \frac{1}{3})$ or point marked on $y$-axis |
|(4 marks total)|||
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{95e11fd7-765c-477d-800b-7574bc1af81f-06_640_1063_322_438}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curve with equation $y = \mathrm { f } ( x )$ where
$$\mathrm { f } ( x ) = \frac { x } { x - 2 } , \quad x \neq 2$$
The curve passes through the origin and has two asymptotes, with equations $y = 1$ and $x = 2$, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item In the space below, sketch the curve with equation $y = \mathrm { f } ( x - 1 )$ and state the equations of the asymptotes of this curve.
\item Find the coordinates of the points where the curve with equation $y = \mathrm { f } ( x - 1 )$ crosses the coordinate axes.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2011 Q5 [7]}}