| 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 transformations (horizontal stretch and vertical translation), which is routine. Parts (ii)-(iii) involve straightforward integration and differentiation of sin 2x. Parts (iv)-(v) on inverse functions are standard bookwork. While comprehensive, each component is a textbook exercise requiring no novel insight, making it slightly easier than average. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.02w Graph transformations: simple transformations of f(x)1.07b Gradient as rate of change: dy/dx notation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Stretch in \(x\)-direction, scale factor... | M1 | (in either order) — allow 'contraction', dep 'stretch' |
| ...s.f. [implied] | 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 [4] | or \(\begin{pmatrix}0\\1\end{pmatrix}\) dep 'translation'. \(\begin{pmatrix}0\\1\end{pmatrix}\) alone is M1 A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = \int_{-\pi/4}^{\pi/4}(1 + \sin 2x)\,dx\) | 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 | |
| \(= \pi/4 - \frac{1}{2}\cos\frac{\pi}{2} + \pi/4 + \frac{1}{2}\cos\left(-\frac{\pi}{2}\right)\) | M1 | Substituting their limits (if zero lower limit used, must show evidence of substitution) |
| \(= \pi/2\) | A1 [4] | or 1.57 or better — cao (www) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = 1 + \sin 2x \Rightarrow \frac{dy}{dx} = 2\cos 2x\) | M1 | Differentiating — allow 1 error (but not \(x + 2\cos 2x\)) |
| A1 | ||
| When \(x=0\), \(\frac{dy}{dx} = 2\); so gradient at \((0,1)\) on \(f(x)\) is 2 | A1ft | |
| \(\Rightarrow\) gradient at \((1,0)\) on \(f^{-1}(x) = \frac{1}{2}\) | B1ft [4] | If 1, then must show evidence of using reciprocal, e.g. \(1/1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Domain is \(0 \leq x \leq 2\) | B1 | Allow 0 to 2, but not \(0 < x < 2\) or \(y\) instead of \(x\) |
| [Graph: reflection of \(f\) in \(y=x\), passing through \((0,-\pi/4)\) and \((2,\pi/4)\)] | M1 | Clear attempt to reflect in \(y = x\) |
| A1 [3] | Correct domain indicated (0 to 2), and reasonable shape |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = 1 + \sin 2x \; x \leftrightarrow y \Rightarrow x = 1 + \sin 2y\) | M1 | or \(\sin 2x = y - 1\) |
| \(\Rightarrow \sin 2y = x - 1\) | ||
| \(\Rightarrow 2y = \arcsin(x-1)\) | ||
| \(\Rightarrow y = \frac{1}{2}\arcsin(x-1)\) | A1 [2] | cao |
## Question 3:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Stretch in $x$-direction, scale factor... | M1 | (in either order) — allow 'contraction', dep 'stretch' |
| ...s.f. [implied] | 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 [4] | or $\begin{pmatrix}0\\1\end{pmatrix}$ dep 'translation'. $\begin{pmatrix}0\\1\end{pmatrix}$ alone is M1 A0 |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = \int_{-\pi/4}^{\pi/4}(1 + \sin 2x)\,dx$ | 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 | |
| $= \pi/4 - \frac{1}{2}\cos\frac{\pi}{2} + \pi/4 + \frac{1}{2}\cos\left(-\frac{\pi}{2}\right)$ | M1 | Substituting their limits (if zero lower limit used, must show evidence of substitution) |
| $= \pi/2$ | A1 [4] | or 1.57 or better — cao (www) |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = 1 + \sin 2x \Rightarrow \frac{dy}{dx} = 2\cos 2x$ | M1 | Differentiating — allow 1 error (but not $x + 2\cos 2x$) |
| | A1 | |
| When $x=0$, $\frac{dy}{dx} = 2$; so gradient at $(0,1)$ on $f(x)$ is 2 | A1ft | |
| $\Rightarrow$ gradient at $(1,0)$ on $f^{-1}(x) = \frac{1}{2}$ | B1ft [4] | If 1, then must show evidence of using reciprocal, e.g. $1/1$ |
### Part (iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Domain is $0 \leq x \leq 2$ | B1 | Allow 0 to 2, but not $0 < x < 2$ or $y$ instead of $x$ |
| [Graph: reflection of $f$ in $y=x$, passing through $(0,-\pi/4)$ and $(2,\pi/4)$] | M1 | Clear attempt to reflect in $y = x$ |
| | A1 [3] | Correct domain indicated (0 to 2), and reasonable shape |
### Part (v):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = 1 + \sin 2x \; x \leftrightarrow y \Rightarrow x = 1 + \sin 2y$ | M1 | or $\sin 2x = y - 1$ |
| $\Rightarrow \sin 2y = x - 1$ | | |
| $\Rightarrow 2y = \arcsin(x-1)$ | | |
| $\Rightarrow y = \frac{1}{2}\arcsin(x-1)$ | A1 [2] | cao |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]{b7588524-8a5e-42af-8b52-29cdddc09eeb-2_577_820_1114_675}
\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]}}