| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Stationary points after transformation |
| Difficulty | Moderate -0.3 This is a standard C3 transformations question testing routine application of transformation rules (modulus, stretches) and basic function properties. Part (a) requires direct application of transformation formulas, part (b) is a standard sketch of f(|x|), part (c) involves finding a translated parabola from given points (straightforward algebra), and part (d) tests understanding of one-to-one functions. While multi-part, each component is textbook-standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.02w Graph transformations: simple transformations of f(x)1.02x Combinations of transformations: multiple transformations |
| Answer | Marks |
|---|---|
| \((3, 4)\) | B1 B1 |
| Answer | Marks |
|---|---|
| \((6, -8)\) | B1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Correct sketch with turning points \((-3, -4)\) and \((3, -4)\), \(y\)-intercept at \((0, 5)\) | B1 B1 B1 | B1: correct shape for \(x \geq 0\), curve meets positive \(y\)-axis, turning point below \(x\)-axis; B1: symmetrical about \(y\)-axis or correct shape for \(x < 0\); B1: correct turning points \((-3,-4)\) and \((3,-4)\), and \((0,5)\) marked |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = (x-3)^2 - 4\) or \(f(x) = x^2 - 6x + 5\) | M1A1 | M1: states \(f(x)\) in form \((x \pm \alpha)^2 \pm \beta\), \(\alpha, \beta \neq 0\); A1: either \((x-3)^2 - 4\) or \(x^2 - 6x + 5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Either: the function f is a many-one mapping. Or: the function f is not a one-one mapping. | B1 | Also accept: inverse is one-many (not a function); \(f(0)=5\) and \(f(6)=5\); one \(y\)-coordinate has 2 corresponding \(x\)-coordinates |
## Question 6:
### Part (a)(i)
| $(3, 4)$ | B1 B1 | |
### Part (a)(ii)
| $(6, -8)$ | B1 B1 | |
### Part (b)
| Correct sketch with turning points $(-3, -4)$ and $(3, -4)$, $y$-intercept at $(0, 5)$ | B1 B1 B1 | B1: correct shape for $x \geq 0$, curve meets positive $y$-axis, turning point below $x$-axis; B1: symmetrical about $y$-axis or correct shape for $x < 0$; B1: correct turning points $(-3,-4)$ and $(3,-4)$, and $(0,5)$ marked |
### Part (c)
| $f(x) = (x-3)^2 - 4$ or $f(x) = x^2 - 6x + 5$ | M1A1 | M1: states $f(x)$ in form $(x \pm \alpha)^2 \pm \beta$, $\alpha, \beta \neq 0$; A1: either $(x-3)^2 - 4$ or $x^2 - 6x + 5$ |
### Part (d)
| Either: the function f is a many-one mapping. Or: the function f is not a one-one mapping. | B1 | Also accept: inverse is one-many (not a function); $f(0)=5$ and $f(6)=5$; one $y$-coordinate has 2 corresponding $x$-coordinates |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{52f73407-14c5-46e6-b911-aa096b9b5893-10_781_858_239_575}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve with the equation $y = \mathrm { f } ( x ) , x \in \mathbb { R }$. The curve has a turning point at $A ( 3 , - 4 )$ and also passes through the point $( 0,5 )$.
\begin{enumerate}[label=(\alph*)]
\item Write down the coordinates of the point to which $A$ is transformed on the curve with equation
\begin{enumerate}[label=(\roman*)]
\item $y = | \mathrm { f } ( x ) |$,
\item $y = 2 f \left( \frac { 1 } { 2 } x \right)$.
\end{enumerate}\item Sketch the curve with equation
$$y = \mathrm { f } ( | x | )$$
On your sketch show the coordinates of all turning points and the coordinates of the point at which the curve cuts the $y$-axis.
The curve with equation $y = \mathrm { f } ( x )$ is a translation of the curve with equation $y = x ^ { 2 }$.
\item Find $\mathrm { f } ( x )$.
\item Explain why the function f does not have an inverse.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2010 Q6 [10]}}