| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find stationary points - polynomial/exponential products |
| Difficulty | Moderate -0.3 This is a multi-part question involving standard C3 techniques: finding intercepts by substitution, differentiating a product (polynomial × exponential) using the product rule to find stationary points, integration by parts for area, and applying transformations. While it has multiple parts (5 marks worth), each individual step is routine and follows textbook procedures without requiring novel insight or complex problem-solving. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((1, 0)\) and \((0, 1)\) | B1B1 | \(x=0, y=1\); \(y=0, x=1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f'(x) = 2(1-x)e^{2x} - e^{2x}\) | B1 | \(\frac{d}{dx}(e^{2x}) = 2e^{2x}\) |
| M1 | product rule consistent with their derivatives | |
| \(= e^{2x}(1-2x)\) | A1 | correct expression, so \((1-x)e^{2x} - e^{2x}\) is B0M1A0 |
| \(f'(x) = 0\) when \(x = \frac{1}{2}\) | M1dep | setting their derivative to 0 dep 1st M1 |
| A1cao | \(x = \frac{1}{2}\) | |
| \(y = \frac{1}{2}e\) | B1 | allow \(\frac{1}{2}e^1\) isw |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(A = \int_0^1 (1-x)e^{2x}\,dx\) | B1 | correct integral and limits; condone no \(dx\) (limits may be seen later) |
| \(u = (1-x),\; u' = -1,\; v' = e^{2x},\; v = \frac{1}{2}e^{2x}\) | M1 | \(u\), \(u'\), \(v'\), \(v\), all correct; or if split up \(u=x\), \(u'=1\), \(v'=e^{2x}\), \(v=\frac{1}{2}e^{2x}\) |
| \(\Rightarrow A = \left[\frac{1}{2}(1-x)e^{2x}\right]_0^1 - \int_0^1 \frac{1}{2}e^{2x}\cdot(-1)\,dx\) | A1 | condone incorrect limits; or from above \(\ldots\left[\frac{1}{2}xe^{2x}\right]_0^1 - \int_0^1 \frac{1}{2}e^{2x}\,dx\) |
| \(= \left[\frac{1}{2}(1-x)e^{2x} + \frac{1}{4}e^{2x}\right]_0^1\) | A1 | o.e. if integral split up; condone incorrect limits |
| \(= \frac{1}{4}e^2 - \frac{1}{2} - \frac{1}{4}\) | ||
| \(= \frac{1}{4}(e^2 - 3)\) * | A1cao | NB AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(g(x) = 3f(\frac{1}{2}x) = 3(1-\frac{1}{2}x)e^x\) | B1 | o.e.; mark final answer |
| [Graph] through \((2, 0)\) and \((0, 3)\) | B1 | through \((2,0)\) and \((0,3)\) – condone errors in writing coordinates (e.g. \((0,2)\)) |
| [Graph] reasonable shape | B1dep | reasonable shape, dep previous B1 |
| TP at \((1, \frac{3e}{2})\) or \((1, 4.1)\) (or better) | B1 | Must be evidence that \(x=1\), \(y=4.1\) is indeed the TP – appearing in a table of values is not enough on its own. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(6 \times \frac{1}{4}(e^2-3)\; [= \frac{3(e^2-3)}{2}]\) | B1 | o.e. mark final answer |
## Question 8:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(1, 0)$ and $(0, 1)$ | B1B1 | $x=0, y=1$; $y=0, x=1$ |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f'(x) = 2(1-x)e^{2x} - e^{2x}$ | B1 | $\frac{d}{dx}(e^{2x}) = 2e^{2x}$ |
| | M1 | product rule consistent with their derivatives |
| $= e^{2x}(1-2x)$ | A1 | correct expression, so $(1-x)e^{2x} - e^{2x}$ is B0M1A0 |
| $f'(x) = 0$ when $x = \frac{1}{2}$ | M1dep | setting their derivative to 0 dep 1st M1 |
| | A1cao | $x = \frac{1}{2}$ |
| $y = \frac{1}{2}e$ | B1 | allow $\frac{1}{2}e^1$ isw |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $A = \int_0^1 (1-x)e^{2x}\,dx$ | B1 | correct integral and limits; condone no $dx$ (limits may be seen later) |
| $u = (1-x),\; u' = -1,\; v' = e^{2x},\; v = \frac{1}{2}e^{2x}$ | M1 | $u$, $u'$, $v'$, $v$, all correct; or if split up $u=x$, $u'=1$, $v'=e^{2x}$, $v=\frac{1}{2}e^{2x}$ |
| $\Rightarrow A = \left[\frac{1}{2}(1-x)e^{2x}\right]_0^1 - \int_0^1 \frac{1}{2}e^{2x}\cdot(-1)\,dx$ | A1 | condone incorrect limits; or from above $\ldots\left[\frac{1}{2}xe^{2x}\right]_0^1 - \int_0^1 \frac{1}{2}e^{2x}\,dx$ |
| $= \left[\frac{1}{2}(1-x)e^{2x} + \frac{1}{4}e^{2x}\right]_0^1$ | A1 | o.e. if integral split up; condone incorrect limits |
| $= \frac{1}{4}e^2 - \frac{1}{2} - \frac{1}{4}$ | | |
| $= \frac{1}{4}(e^2 - 3)$ * | A1cao | **NB AG** |
### Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $g(x) = 3f(\frac{1}{2}x) = 3(1-\frac{1}{2}x)e^x$ | B1 | o.e.; mark final answer |
| [Graph] through $(2, 0)$ and $(0, 3)$ | B1 | through $(2,0)$ and $(0,3)$ – condone errors in writing coordinates (e.g. $(0,2)$) |
| [Graph] reasonable shape | B1dep | reasonable shape, dep previous B1 |
| TP at $(1, \frac{3e}{2})$ or $(1, 4.1)$ (or better) | B1 | Must be evidence that $x=1$, $y=4.1$ is indeed the TP – appearing in a table of values is not enough on its own. |
### Part (v):
| Answer | Mark | Guidance |
|--------|------|----------|
| $6 \times \frac{1}{4}(e^2-3)\; [= \frac{3(e^2-3)}{2}]$ | B1 | o.e. mark final answer |8 Fig. 8 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = ( 1 - x ) \mathrm { e } ^ { 2 x }$, with its turning point P .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{28ce1bcc-e9d5-4ae6-98c0-67b5b8c50bc6-5_716_810_404_609}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Write down the coordinates of the intercepts of $y = \mathrm { f } ( x )$ with the $x$ - and $y$-axes.\\
(ii) Find the exact coordinates of the turning point P .\\
(iii) Show that the exact area of the region enclosed by the curve and the $x$ - and $y$-axes is $\frac { 1 } { 4 } \left( e ^ { 2 } - 3 \right)$. The function $\mathrm { g } ( x )$ is defined by $\mathrm { g } ( x ) = 3 \mathrm { f } \left( \frac { 1 } { 2 } x \right)$.\\
(iv) Express $\mathrm { g } ( x )$ in terms of $x$.
Sketch the curve $y = \mathrm { g } ( x )$ on the copy of Fig. 8, indicating the coordinates of its intercepts with the $x$ - and $y$-axes and of its turning point.\\
(v) Write down the exact area of the region enclosed by the curve $y = \mathrm { g } ( x )$ and the $x$-and $y$-axes.
\hfill \mbox{\textit{OCR MEI C3 2013 Q8 [18]}}