| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Multiple separate transformations (sketch-based, standard transformations) |
| Difficulty | Standard +0.3 This is a comprehensive but standard C3 transformations question covering routine topics: basic function transformations (stretch, reflection, translation), inverse function concepts, and gradient relationships. While multi-part with 5 sections, each individual part uses well-practiced techniques—sketching standard transformations, explaining why functions fail the horizontal line test, reflecting graphs in y=x, using the chain rule for differentiation, and solving equations. The most challenging aspect (part v) requires algebraic manipulation but follows a predictable pattern. This is slightly easier than average as it's mostly recall and application of standard procedures without requiring novel problem-solving insight. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| (i)(A) The transformation is a stretch with the \(x\)-axis invariant and of scale factor 2. | B1 | Same orientation |
| B1 | \(y\) values doubled | |
| 2 | ||
| (i)(B) The transformation is a reflection in the \(y\)-axis. | B1 | same shape |
| B2 | Inversion | |
| 3 | ||
| (i)(C) The transformation is a translation of 2 units parallel to the \(x\)-axis, i.e \(\begin{pmatrix} 2 \\ 0 \end{pmatrix}\) | B1 | Same shape |
| B2 | Moved 2 to the right | |
| 3 | ||
| (ii) There is a set of values of \(y\) (for example, \(y = 1\)) for which there are three corresponding values of \(x\) (so an inverse would be multivalued). | B1 B1 | |
| 2 | ||
| (iii) Sketch of restricted domain (e.g., \(x \geq 0\)) showing monotonic increasing curve | B1 | |
| 1 |
| Answer | Marks |
|---|---|
| \(\Rightarrow f'(x) = 3x^2+4x\) | M1 A1 |
| So the gradient at \((1,3)\) is \(7\). | M1 |
| The gradient on the inverse (which is a reflection of the original in \(y = x\)) is therefore \(\frac{1}{7}\). | A1 |
| 4 | |
| (v) The graph and its reflection must intersect on the axis of reflection, i.e \(y = x\), so solve | M1 |
| \(y = x\), \(y = x^2(x+2)\) | M1 |
| Answer | Marks |
|---|---|
| \(\Rightarrow x = 0, -1 \pm \sqrt{2}\) | E1 |
| The positive non-zero root is as given. | |
| 3 |
**(i)(A)** The transformation is a stretch with the $x$-axis invariant and of scale factor 2. | B1 | Same orientation
| B1 | $y$ values doubled
| 2 |
**(i)(B)** The transformation is a reflection in the $y$-axis. | B1 | same shape
| B2 | Inversion
| 3 |
**(i)(C)** The transformation is a translation of 2 units parallel to the $x$-axis, i.e $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$ | B1 | Same shape
| B2 | Moved 2 to the right
| 3 |
**(ii)** There is a set of values of $y$ (for example, $y = 1$) for which there are three corresponding values of $x$ (so an inverse would be multivalued). | B1 B1 |
| 2 |
**(iii)** Sketch of restricted domain (e.g., $x \geq 0$) showing monotonic increasing curve | B1 |
| 1 |
**(iv)** $f(x) = x^2(x+2)$
$\Rightarrow f'(x) = 3x^2+4x$ | M1 A1 |
So the gradient at $(1,3)$ is $7$. | M1 |
The gradient on the inverse (which is a reflection of the original in $y = x$) is therefore $\frac{1}{7}$. | A1 |
| 4 |
**(v)** The graph and its reflection must intersect on the axis of reflection, i.e $y = x$, so solve | M1 |
$y = x$, $y = x^2(x+2)$ | M1 |
$\Rightarrow x = x^2(x+2)$
$\Rightarrow 0 = x(x^2+2x-1)$
$\Rightarrow x = 0, -1 \pm \sqrt{2}$ | E1 |
The positive non-zero root is as given. | |
| 3 |9 Answer parts (i) and (iii) on the insert provided.
Fig. 9 shows a sketch graph of $y = \mathrm { f } ( x )$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3f8be5ab-d241-4027-af26-c49da9394adc-4_401_799_488_593}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item On the Insert sketch graphs of\\
(A) $y = 2 \mathrm { f } ( x )$,\\
(B) $y = \mathrm { f } ( - x )$,\\
(C) $y = \mathrm { f } ( x - 2 )$
In each case describe the transformations.
\item Explain why the function $y = \mathrm { f } ( x )$ does not have an inverse function.
\item The function $\mathrm { g } ( x )$ is defined as follows:
$$\mathrm { g } ( x ) = \mathrm { f } ( x ) \text { for } x \geq 0$$
On the Insert sketch the graph of $y = \mathrm { g } ^ { - 1 } ( x )$.
\item You are given that $\mathrm { f } ( x ) = x ^ { 2 } ( x + 2 )$.
Calculate the gradient of the curve $y = \mathrm { f } ( x )$ at the point $( 1,3 )$.\\
Deduce the gradient of the function $\mathrm { g } ^ { - 1 } ( x )$ at the point where $x = 3$.
\item Show that $\mathrm { g } ( x )$ and $\mathrm { g } ^ { - 1 } ( x )$ cross where $x = - 1 + \sqrt { 2 }$.
\section*{Insert for question 9.}
\item (A) On the axes below sketch the graph of $y = 2 \mathrm { f } ( x )$.
Describe the transformation.\\
\includegraphics[max width=\textwidth, alt={}, center]{3f8be5ab-d241-4027-af26-c49da9394adc-5_563_1102_484_395}
Description:
\item (B) On the axes below sketch the graph of $y = \mathrm { f } ( - x )$.
Describe the transformation.\\
\includegraphics[max width=\textwidth, alt={}, center]{3f8be5ab-d241-4027-af26-c49da9394adc-5_588_1154_1576_404}
Description:
\item (C) On the axes below sketch the graph of $y = \mathrm { f } ( x - 2 )$.
Describe the transformation.\\
\includegraphics[max width=\textwidth, alt={}, center]{3f8be5ab-d241-4027-af26-c49da9394adc-6_615_1230_402_406}
Description:
\item The function $\mathrm { g } ( x )$ is defined as follows:
$$\mathrm { g } ( x ) = \mathrm { f } ( x ) \text { for } x \geq 0$$
On the axes below sketch the graph of $y = g ^ { - 1 } ( x )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{3f8be5ab-d241-4027-af26-c49da9394adc-6_677_1356_1567_312}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 Q9 [18]}}