| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Composite transformation sketch |
| Difficulty | Moderate -0.8 This is a straightforward C1 transformations question requiring only recall of standard transformation rules (horizontal translation, horizontal stretch, reflection) applied to key features. Part (d) requires minimal geometric reasoning about where a horizontal line intersects the curve once. All parts are direct applications of memorized transformation formulas with no problem-solving or novel insight required. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((4, 7)\) | B1 | Accept \((4,7)\) or \(x=4, y=7\) or sketch of \(y=f(x-2)\) with maximum marked at \((4,7)\). No other coordinates. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x =) 2.5\) | B1 | Allow \((2.5, 0)\); no other values. Allow sketch of \(f(2x)\) with only \(x\)-intercept at \(x=2.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 1\) (e.g. \(y - 1 = 0\)) | B1 | Must be an equation, not just '1'; no other asymptotes stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k \leq 1\) or \(k = 7\) | B1 | Either \(k \leq 1\) or \(k = 7\). Note \(k=7\) may be embedded in e.g. \(k=0,1,7\) |
| \(k \leq 1 \quad k = 7\) | B1 | Both correct and in terms of \(k\) with no other solutions |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(4, 7)$ | B1 | Accept $(4,7)$ or $x=4, y=7$ or sketch of $y=f(x-2)$ with maximum marked at $(4,7)$. No other coordinates. |
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x =) 2.5$ | B1 | Allow $(2.5, 0)$; no other values. Allow sketch of $f(2x)$ with only $x$-intercept at $x=2.5$ |
## Question 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 1$ (e.g. $y - 1 = 0$) | B1 | Must be an equation, not just '1'; no other asymptotes stated |
## Question 5(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k \leq 1$ **or** $k = 7$ | B1 | Either $k \leq 1$ or $k = 7$. Note $k=7$ may be embedded in e.g. $k=0,1,7$ |
| $k \leq 1 \quad k = 7$ | B1 | Both correct and in terms of $k$ with no other solutions |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{937246f9-2b6a-48df-b919-c6db3d6f863b-12_963_1239_255_354}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the sketch of a curve with equation $y = \mathrm { f } ( x ) , x \in \mathbb { R }$.
The curve crosses the $y$-axis at $( 0,4 )$ and crosses the $x$-axis at $( 5,0 )$.
The curve has a single turning point, a maximum, at (2, 7).
The line with equation $y = 1$ is the only asymptote to the curve.
\begin{enumerate}[label=(\alph*)]
\item State the coordinates of the turning point on the curve with equation $y = \mathrm { f } ( x - 2 )$.
\item State the solution of the equation f( $2 x$ ) $= 0$
\item State the equation of the asymptote to the curve with equation $y = \mathrm { f } ( - x )$.
Given that the line with equation $y = k$, where $k$ is a constant, meets the curve $y = \mathrm { f } ( x )$ at only one point,
\item state the set of possible values for $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2018 Q5 [5]}}