| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Specific function transformation description |
| Difficulty | Standard +0.3 This is a multi-part question covering standard C3 topics (transformations, quotient rule differentiation, substitution integration) with clear guidance at each step. Part (i) is routine transformation identification, part (ii) applies quotient rule with verification, part (iii) provides the substitution explicitly, and part (iv) uses previous results. While requiring multiple techniques, each step is scaffolded and represents typical textbook exercises rather than requiring novel insight. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08e Area between curve and x-axis: using definite integrals1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Translation in the \(x\)-direction | M1 | Allow 'shift', 'move'. If just vectors given withhold one 'A' mark only |
| of \(\pi/4\) to the right | A1 | oe (e.g. using vector). 'Translate \(\begin{pmatrix}\pi/4\\1\end{pmatrix}\)' is 4 marks; if followed by additional incorrect transformation, SC M1M1A1A0 |
| Translation in the \(y\)-direction | M1 | Allow 'shift', 'move'. \(\begin{pmatrix}\pi/4\\1\end{pmatrix}\) only is M2A1A0 |
| of 1 unit up | A1 | oe (e.g. using vector) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(g(x) = \dfrac{2\sin x}{\sin x + \cos x}\) | Can deal with numerator and denominator separately | |
| \(g'(x) = \dfrac{(\sin x + \cos x)2\cos x - 2\sin x(\cos x - \sin x)}{(\sin x + \cos x)^2}\) | M1 | Quotient (or product) rule consistent with their derivs; \(\dfrac{vu'-uv'}{v^2}\); allow one slip, missing brackets |
| \(= \dfrac{2\sin x\cos x + 2\cos^2 x - 2\sin x\cos x + 2\sin^2 x}{(\sin x + \cos x)^2}\) | A1 | Correct expanded expression (could leave the '2' as a factor); \(\dfrac{uv'-vu'}{v^2}\) is M0; condone \(\cos^2 x\), \(\sin^2 x\) |
| \(= \dfrac{2\cos^2 x + 2\sin^2 x}{(\sin x + \cos x)^2} = \dfrac{2(\cos^2 x + \sin^2 x)}{(\sin x + \cos x)^2}\) | ||
| \(= \dfrac{2}{(\sin x + \cos x)^2}\) * | A1 | NB AG; must take out 2 as a factor or state \(\sin^2 x + \cos^2 x = 1\) |
| When \(x = \pi/4\), \(g'(\pi/4) = 2/(1/\sqrt{2}+1/\sqrt{2})^2\) | M1 | Substituting \(\pi/4\) into correct deriv |
| \(= 1\) | A1 | |
| \(f'(x) = \sec^2 x\) | M1 | oe, e.g. \(1/\cos^2 x\) |
| \(f'(0) = \sec^2(0) = 1\), [so gradient the same here] | A1 | |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^{\pi/4} f(x)\,dx = \int_0^{\pi/4} \dfrac{\sin x}{\cos x}\,dx\) | ||
| Let \(u = \cos x\), \(du = -\sin x\,dx\) | ||
| When \(x=0\), \(u=1\); when \(x=\pi/4\), \(u=1/\sqrt{2}\) | ||
| \(= \int_1^{1/\sqrt{2}} -\dfrac{1}{u}\,du\) | M1 | Substituting to get \(\int -1/u\,(du)\); ignore limits here, condone no \(du\) but not \(dx\); allow \(\int 1/u \cdot -du\) |
| \(= \int_{1/\sqrt{2}}^{1} \dfrac{1}{u}\,du\) * | A1 | NB AG; but for A1 must deal correctly with the \(-\)ve sign by interchanging limits |
| \(= \Big[\ln u\Big]_{1/\sqrt{2}}^{1}\) | M1 | \([\ln u]\) |
| \(= \ln 1 - \ln(1/\sqrt{2})\) | ||
| \(= \ln\sqrt{2} = \ln 2^{1/2} = \frac{1}{2}\ln 2\) | A1 | \(\ln\sqrt{2}\), \(\frac{1}{2}\ln 2\) or \(-\ln(1/\sqrt{2})\); mark final answer |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area = area in part (iii) translated up 1 unit | M1 | soi from \(\pi/4\) added; or \(\int_{\pi/4}^{\pi/2}(1+\tan(x-\pi/4))\,dx = \Big[x + \ln\sec(x-\pi/4)\Big]_{\pi/4}^{\pi/2}\) \(= \pi/2 + \ln\sqrt{2} - \pi/4 = \pi/4 + \ln\sqrt{2}\) B2 |
| So \(= \frac{1}{2}\ln 2 + 1 \times \pi/4 = \frac{1}{2}\ln 2 + \pi/4\) | A1cao | oe (as above) |
| [2] |
## Question 4:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Translation in the $x$-direction | M1 | Allow 'shift', 'move'. If just vectors given withhold **one** 'A' mark only |
| of $\pi/4$ to the right | A1 | oe (e.g. using vector). 'Translate $\begin{pmatrix}\pi/4\\1\end{pmatrix}$' is 4 marks; if followed by additional incorrect transformation, SC M1M1A1A0 |
| Translation in the $y$-direction | M1 | Allow 'shift', 'move'. $\begin{pmatrix}\pi/4\\1\end{pmatrix}$ only is M2A1A0 |
| of 1 unit up | A1 | oe (e.g. using vector) |
| **[4]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $g(x) = \dfrac{2\sin x}{\sin x + \cos x}$ | | Can deal with numerator and denominator separately |
| $g'(x) = \dfrac{(\sin x + \cos x)2\cos x - 2\sin x(\cos x - \sin x)}{(\sin x + \cos x)^2}$ | M1 | Quotient (or product) rule consistent with their derivs; $\dfrac{vu'-uv'}{v^2}$; allow one slip, missing brackets |
| $= \dfrac{2\sin x\cos x + 2\cos^2 x - 2\sin x\cos x + 2\sin^2 x}{(\sin x + \cos x)^2}$ | A1 | Correct expanded expression (could leave the '2' as a factor); $\dfrac{uv'-vu'}{v^2}$ is M0; condone $\cos^2 x$, $\sin^2 x$ |
| $= \dfrac{2\cos^2 x + 2\sin^2 x}{(\sin x + \cos x)^2} = \dfrac{2(\cos^2 x + \sin^2 x)}{(\sin x + \cos x)^2}$ | | |
| $= \dfrac{2}{(\sin x + \cos x)^2}$ * | A1 | **NB AG**; must take out 2 as a factor or state $\sin^2 x + \cos^2 x = 1$ |
| When $x = \pi/4$, $g'(\pi/4) = 2/(1/\sqrt{2}+1/\sqrt{2})^2$ | M1 | Substituting $\pi/4$ into correct deriv |
| $= 1$ | A1 | |
| $f'(x) = \sec^2 x$ | M1 | oe, e.g. $1/\cos^2 x$ |
| $f'(0) = \sec^2(0) = 1$, [so gradient the same here] | A1 | |
| **[7]** | | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^{\pi/4} f(x)\,dx = \int_0^{\pi/4} \dfrac{\sin x}{\cos x}\,dx$ | | |
| Let $u = \cos x$, $du = -\sin x\,dx$ | | |
| When $x=0$, $u=1$; when $x=\pi/4$, $u=1/\sqrt{2}$ | | |
| $= \int_1^{1/\sqrt{2}} -\dfrac{1}{u}\,du$ | M1 | Substituting to get $\int -1/u\,(du)$; ignore limits here, condone no $du$ but not $dx$; allow $\int 1/u \cdot -du$ |
| $= \int_{1/\sqrt{2}}^{1} \dfrac{1}{u}\,du$ * | A1 | **NB AG**; but for A1 must deal correctly with the $-$ve sign by interchanging limits |
| $= \Big[\ln u\Big]_{1/\sqrt{2}}^{1}$ | M1 | $[\ln u]$ |
| $= \ln 1 - \ln(1/\sqrt{2})$ | | |
| $= \ln\sqrt{2} = \ln 2^{1/2} = \frac{1}{2}\ln 2$ | A1 | $\ln\sqrt{2}$, $\frac{1}{2}\ln 2$ or $-\ln(1/\sqrt{2})$; mark final answer |
| **[4]** | | |
### Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area = area in part **(iii)** translated up 1 unit | M1 | soi from $\pi/4$ added; or $\int_{\pi/4}^{\pi/2}(1+\tan(x-\pi/4))\,dx = \Big[x + \ln\sec(x-\pi/4)\Big]_{\pi/4}^{\pi/2}$ $= \pi/2 + \ln\sqrt{2} - \pi/4 = \pi/4 + \ln\sqrt{2}$ B2 |
| So $= \frac{1}{2}\ln 2 + 1 \times \pi/4 = \frac{1}{2}\ln 2 + \pi/4$ | A1cao | oe (as above) |
| **[2]** | | |
---4 Fig. 8 shows parts of the curves $y = \mathrm { f } ( x )$ and $y = \mathrm { g } ( x )$, where $\mathrm { f } ( x ) = \tan x$ and $\mathrm { g } ( x ) = 1 + \mathrm { f } \left( x - \frac { 1 } { 4 } \pi \right)$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{01bdea17-c698-44ae-a45a-7da4de631de4-2_687_888_419_609}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Describe a sequence of two transformations which maps the curve $y = \mathrm { f } ( x )$ to the curve $y = \mathrm { g } ( x )$. [4] It can be shown that $\mathrm { g } ( x ) = \frac { 2 \sin x } { \sin x + \cos x }$.\\
(ii) Show that $\mathrm { g } ^ { \prime } ( x ) = \frac { 2 } { ( \sin x + \cos x ) ^ { 2 } }$. Hence verify that the gradient of $y = \mathrm { g } ( x )$ at the point $\left( \frac { 1 } { 4 } \pi , 1 \right)$ is the same as that of $y = \mathrm { f } ( x )$ at the origin.\\
(iii) By writing $\tan x = \frac { \sin x } { \cos x }$ and using the substitution $u = \cos x$, show that $\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \mathrm { f } ( x ) \mathrm { d } x = \int _ { \frac { 1 } { \sqrt { 2 } } } ^ { 1 } \frac { 1 } { u } \mathrm {~d} u$. Evaluate this integral exactly.\\
(iv) Hence find the exact area of the region enclosed by the curve $y = \mathrm { g } ( x )$, the $x$-axis and the lines $x = \frac { 1 } { 4 } \pi$ and $x = \frac { 1 } { 2 } \pi$.
\hfill \mbox{\textit{OCR MEI C3 Q4 [17]}}