| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Sequence of transformations order |
| Difficulty | Moderate -0.3 This is a multi-part question covering standard C3 transformations, integration, differentiation, and inverse functions. Part (i) requires identifying two basic transformations (horizontal stretch and vertical translation), which is routine. Parts (ii)-(v) involve standard techniques with no novel problem-solving required. The question is slightly easier than average because each part follows predictable patterns from the C3 syllabus, though the multi-part nature and inverse function work prevent it from being significantly below average. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.02w Graph transformations: simple transformations of f(x)1.05a Sine, cosine, tangent: definitions for all arguments1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Stretch in \(x\)-direction | M1 | (in either order) — allow 'contraction'; dep 'stretch' |
| s.f. \(\frac{1}{2}\) | A1 | |
| translation in \(y\)-direction | M1 | allow 'move', 'shift', etc — direction can be inferred from \(\begin{pmatrix}0\\1\end{pmatrix}\) |
| 1 unit up | A1 | or \(\begin{pmatrix}0\\1\end{pmatrix}\) dep 'translation'. \(\begin{pmatrix}0\\1\end{pmatrix}\) alone is M1 A0 |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = \displaystyle\int_{-\pi/4}^{\pi/4}(1+\sin 2x)\,dx\) | M1 | correct integral and limits; condone \(dx\) missing; limits may be implied from subsequent working |
| \(= \left[x - \frac{1}{2}\cos 2x\right]_{-\pi/4}^{\pi/4}\) | B1 | |
| \(= \pi/4 - \frac{1}{2}\cos\frac{\pi}{2} + \pi/4 + \frac{1}{2}\cos(-\pi/2)\) | M1 | substituting their limits |
| \(= \pi/2\) | A1 | or 1.57 or better — cao (www) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = 1 + \sin 2x \Rightarrow dy/dx = 2\cos 2x\) | M1 | differentiating — allow 1 error (but not \(x + 2\cos 2x\)) |
| When \(x=0\), \(dy/dx = 2\) | A1 | |
| So gradient at \((0,1)\) on \(f(x)\) is \(2\) | A1ft | |
| gradient at \((1,0)\) on \(f^{-1}(x) = \frac{1}{2}\) | B1ft | If 1, then must show evidence of using reciprocal, e.g. \(1/1\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Domain is \(0 \le x \le 2\) | B1 | Allow 0 to 2, but not \(0 < x < 2\) or \(y\) instead of \(x\) |
| [graph described: reflection of curve in \(y=x\)] | M1 | clear attempt to reflect in \(y=x\) |
| A1 | correct domain indicated (0 to 2), and reasonable shape | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = 1 + \sin 2x,\ x \leftrightarrow y\) | ||
| \(x = 1 + \sin 2y\) | M1 | or \(\sin 2x = y-1\) |
| \(\sin 2y = x - 1\) | ||
| \(2y = \arcsin(x-1)\) | ||
| \(\Rightarrow y = \frac{1}{2}\arcsin(x-1)\) | A1 | cao |
| [2] |
# Question 3:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Stretch in $x$-direction | M1 | (in either order) — allow 'contraction'; dep 'stretch' |
| s.f. $\frac{1}{2}$ | A1 | |
| translation in $y$-direction | M1 | allow 'move', 'shift', etc — direction can be inferred from $\begin{pmatrix}0\\1\end{pmatrix}$ |
| 1 unit up | A1 | or $\begin{pmatrix}0\\1\end{pmatrix}$ dep 'translation'. $\begin{pmatrix}0\\1\end{pmatrix}$ alone is M1 A0 |
| **[4]** | | |
---
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = \displaystyle\int_{-\pi/4}^{\pi/4}(1+\sin 2x)\,dx$ | M1 | correct integral and limits; condone $dx$ missing; limits may be implied from subsequent working |
| $= \left[x - \frac{1}{2}\cos 2x\right]_{-\pi/4}^{\pi/4}$ | B1 | |
| $= \pi/4 - \frac{1}{2}\cos\frac{\pi}{2} + \pi/4 + \frac{1}{2}\cos(-\pi/2)$ | M1 | substituting their limits |
| $= \pi/2$ | A1 | or 1.57 or better — cao (www) |
| **[4]** | | |
---
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = 1 + \sin 2x \Rightarrow dy/dx = 2\cos 2x$ | M1 | differentiating — allow 1 error (but not $x + 2\cos 2x$) |
| When $x=0$, $dy/dx = 2$ | A1 | |
| So gradient at $(0,1)$ on $f(x)$ is $2$ | A1ft | |
| gradient at $(1,0)$ on $f^{-1}(x) = \frac{1}{2}$ | B1ft | If 1, then must show evidence of using reciprocal, e.g. $1/1$ |
| **[4]** | | |
---
## Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Domain is $0 \le x \le 2$ | B1 | Allow 0 to 2, but not $0 < x < 2$ or $y$ instead of $x$ |
| [graph described: reflection of curve in $y=x$] | M1 | clear attempt to reflect in $y=x$ |
| | A1 | correct domain indicated (0 to 2), and reasonable shape |
| **[3]** | | |
---
## Part (v)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = 1 + \sin 2x,\ x \leftrightarrow y$ | | |
| $x = 1 + \sin 2y$ | M1 | or $\sin 2x = y-1$ |
| $\sin 2y = x - 1$ | | |
| $2y = \arcsin(x-1)$ | | |
| $\Rightarrow y = \frac{1}{2}\arcsin(x-1)$ | A1 | cao |
| **[2]** | | |
---3 Fig. 8 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = 1 + \sin 2 x$ for $- \frac { 1 } { 4 } \pi \leqslant x \leqslant \frac { 1 } { 4 } \pi$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1d12cd0d-07b0-429c-ad3b-e3bccb0fae18-3_577_815_392_719}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) State a sequence of two transformations that would map part of the curve $y = \sin x$ onto the curve $y = \mathrm { f } ( x )$.\\
(ii) Find the area of the region enclosed by the curve $y = \mathrm { f } ( x )$, the $x$-axis and the line $x = \frac { 1 } { 4 } \pi$.\\
(iii) Find the gradient of the curve $y = \mathrm { f } ( x )$ at the point ( 0,1 ). Hence write down the gradient of the curve $y = \mathrm { f } ^ { - 1 } ( x )$ at the point $( 1,0 )$.\\
(iv) State the domain of $\mathrm { f } ^ { - 1 } ( x )$. Add a sketch of $y = \mathrm { f } ^ { - 1 } ( x )$ to a copy of Fig. 8.\\
(v) Find an expression for $\mathrm { f } ^ { - 1 } ( x )$.
\hfill \mbox{\textit{OCR MEI C3 Q3 [17]}}