| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - direct evaluation |
| Difficulty | Moderate -0.3 This is a multi-part question requiring quotient rule differentiation, coordinate finding, and algebraic manipulation of exponential functions. While it has many parts (5 sub-questions), each individual step is routine for C3 level: finding y-intercepts, applying quotient rule, basic integration, and proving function properties. The algebraic manipulations are straightforward with no novel insights required, making it slightly easier than average despite its length. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02w Graph transformations: simple transformations of f(x)1.06a Exponential function: a^x and e^x graphs and properties1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((0, \frac{1}{2})\) | B1 [1] | allow \(y = \frac{1}{2}\), but not \((x=)\frac{1}{2}\) or \((\frac{1}{2}, 0)\) nor \(P = 1/2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{(1+e^{2x}) \cdot 2e^{2x} - e^{2x} \cdot 2e^{2x}}{(1+e^{2x})^2}\) | M1 | Quotient or product rule |
| A1 | correct expression – condone missing bracket | |
| \(= \frac{2e^{2x}}{(1+e^{2x})^2}\) | A1 | cao – mark final answer; product rule: \(\frac{dy}{dx} = e^{2x}\cdot 2e^{2x}(-1)(1+e^{2x})^{-2} + 2e^{2x}(1+e^{2x})^{-1}\) |
| When \(x=0\), \(dy/dx = 2e^0/(1+e^0)^2 = \frac{1}{2}\) | B1ft [4] | follow through their derivative; \(-\frac{2e^{2x}}{(1+e^{2x})^2}\) from \((udv - vdu)/v^2\) SC1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = \int_0^1 \frac{e^{2x}}{1+e^{2x}} \, dx\) | B1 | correct integral and limits (soi); condone no \(dx\) |
| \(= \left[\frac{1}{2}\ln(1+e^{2x})\right]_0^1\) | M1 | \(k\ln(1+e^{2x})\) |
| A1 | \(k = \frac{1}{2}\) | |
| or let \(u = 1+e^{2x}\), \(du/dx = 2e^{2x}\) o.e. | M1 | or \(v = e^{2x}\), \(dv/dx = 2e^{2x}\) o.e. |
| \(A = \int_2^{1+e^2} \frac{1/2}{u} \, du = \left[\frac{1}{2}\ln u\right]_2^{1+e^2}\) | A1 | \([\frac{1}{2}\ln u]\) or \([\frac{1}{2}\ln(v+1)]\) |
| \(= \frac{1}{2}\ln(1+e^2) - \frac{1}{2}\ln 2\) | M1 | substituting correct limits |
| \(= \frac{1}{2}\ln\left[\frac{1+e^2}{2}\right]\) * | E1 [5] | www; allow missing \(dx\)'s or incompatible limits, but penalise missing brackets |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(g(-x) = \frac{1}{2}\left[\frac{e^{-x}-e^x}{e^{-x}+e^x}\right] = -\frac{1}{2}\left[\frac{e^x - e^{-x}}{e^x+e^{-x}}\right] = -g(x)\) | M1 | substituting \(-x\) for \(x\) in \(g(x)\) |
| E1 | completion www – taking out \(-\)ve must be clear; not \(g(-x) \neq g(x)\); condone use of f for g | |
| Rotational symmetry of order 2 about O | B1 [3] | must have 'rotational', 'about O', 'order 2' (oe) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (A) \(g(x) + \frac{1}{2} = \frac{1}{2}\cdot\frac{e^x - e^{-x}}{e^x+e^{-x}} + \frac{1}{2} = \frac{1}{2}\cdot\frac{e^x-e^{-x}+e^x+e^{-x}}{e^x+e^{-x}}\) | M1 | combining fractions (correctly) |
| \(= \frac{1}{2}\cdot\frac{2e^x}{e^x+e^{-x}} = \frac{e^x \cdot e^x}{e^x(e^x+e^{-x})} = \frac{e^{2x}}{e^{2x}+1} = f(x)\) | A1, E1 | |
| (B) Translation \(\begin{pmatrix}0\\1/2\end{pmatrix}\) | M1 | translation in \(y\) direction; allow 'shift', 'move' in correct direction for M1 |
| A1 | up \(\frac{1}{2}\) unit dep 'translation' used; o.e. condone omission of 180°/order 2; \(\begin{pmatrix}0\\1/2\end{pmatrix}\) alone is SC1 | |
| (C) Rotational symmetry [of order 2] about P | B1 [6] |
## Question 9(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(0, \frac{1}{2})$ | B1 [1] | allow $y = \frac{1}{2}$, but not $(x=)\frac{1}{2}$ or $(\frac{1}{2}, 0)$ nor $P = 1/2$ |
---
## Question 9(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{(1+e^{2x}) \cdot 2e^{2x} - e^{2x} \cdot 2e^{2x}}{(1+e^{2x})^2}$ | M1 | Quotient or product rule |
| | A1 | correct expression – condone missing bracket |
| $= \frac{2e^{2x}}{(1+e^{2x})^2}$ | A1 | cao – mark final answer; product rule: $\frac{dy}{dx} = e^{2x}\cdot 2e^{2x}(-1)(1+e^{2x})^{-2} + 2e^{2x}(1+e^{2x})^{-1}$ |
| When $x=0$, $dy/dx = 2e^0/(1+e^0)^2 = \frac{1}{2}$ | B1ft [4] | follow through their derivative; $-\frac{2e^{2x}}{(1+e^{2x})^2}$ from $(udv - vdu)/v^2$ SC1 |
---
## Question 9(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = \int_0^1 \frac{e^{2x}}{1+e^{2x}} \, dx$ | B1 | correct integral and limits (soi); condone no $dx$ |
| $= \left[\frac{1}{2}\ln(1+e^{2x})\right]_0^1$ | M1 | $k\ln(1+e^{2x})$ |
| | A1 | $k = \frac{1}{2}$ |
| or let $u = 1+e^{2x}$, $du/dx = 2e^{2x}$ o.e. | M1 | or $v = e^{2x}$, $dv/dx = 2e^{2x}$ o.e. |
| $A = \int_2^{1+e^2} \frac{1/2}{u} \, du = \left[\frac{1}{2}\ln u\right]_2^{1+e^2}$ | A1 | $[\frac{1}{2}\ln u]$ or $[\frac{1}{2}\ln(v+1)]$ |
| $= \frac{1}{2}\ln(1+e^2) - \frac{1}{2}\ln 2$ | M1 | substituting correct limits |
| $= \frac{1}{2}\ln\left[\frac{1+e^2}{2}\right]$ * | E1 [5] | www; allow missing $dx$'s or incompatible limits, but penalise missing brackets |
---
## Question 9(iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $g(-x) = \frac{1}{2}\left[\frac{e^{-x}-e^x}{e^{-x}+e^x}\right] = -\frac{1}{2}\left[\frac{e^x - e^{-x}}{e^x+e^{-x}}\right] = -g(x)$ | M1 | substituting $-x$ for $x$ in $g(x)$ |
| | E1 | completion www – taking out $-$ve must be clear; not $g(-x) \neq g(x)$; condone use of f for g |
| Rotational symmetry of order 2 about O | B1 [3] | must have 'rotational', 'about O', 'order 2' (oe) |
---
## Question 9(v):
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(A)** $g(x) + \frac{1}{2} = \frac{1}{2}\cdot\frac{e^x - e^{-x}}{e^x+e^{-x}} + \frac{1}{2} = \frac{1}{2}\cdot\frac{e^x-e^{-x}+e^x+e^{-x}}{e^x+e^{-x}}$ | M1 | combining fractions (correctly) |
| $= \frac{1}{2}\cdot\frac{2e^x}{e^x+e^{-x}} = \frac{e^x \cdot e^x}{e^x(e^x+e^{-x})} = \frac{e^{2x}}{e^{2x}+1} = f(x)$ | A1, E1 | |
| **(B)** Translation $\begin{pmatrix}0\\1/2\end{pmatrix}$ | M1 | translation in $y$ direction; allow 'shift', 'move' in correct direction for M1 |
| | A1 | up $\frac{1}{2}$ unit dep 'translation' used; o.e. condone omission of 180°/order 2; $\begin{pmatrix}0\\1/2\end{pmatrix}$ alone is SC1 |
| **(C)** Rotational symmetry [of order 2] about P | B1 [6] | |9 Fig. 9 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { \mathrm { e } ^ { 2 x } } { 1 + \mathrm { e } ^ { 2 x } }$. The curve crosses the $y$-axis at P.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{30d0d728-d6d6-4a54-baf9-a6df8646bf64-4_604_1233_358_456}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Find the coordinates of P .
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$, simplifying your answer.
Hence calculate the gradient of the curve at P .
\item Show that the area of the region enclosed by $y = \mathrm { f } ( x )$, the $x$-axis, the $y$-axis and the line $x = 1$ is\\
$\frac { 1 } { 2 } \ln \left( \frac { 1 + \mathrm { e } ^ { 2 } } { 2 } \right)$.
The function $\mathrm { g } ( x )$ is defined by $\mathrm { g } ( x ) = \frac { 1 } { 2 } \left( \frac { \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } } { \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } } \right)$.
\item Prove algebraically that $\mathrm { g } ( x )$ is an odd function.
Interpret this result graphically.
\item (A) Show that $\mathrm { g } ( x ) + \frac { 1 } { 2 } = \mathrm { f } ( x )$.\\
(B) Describe the transformation which maps the curve $y = \mathrm { g } ( x )$ onto the curve $y = \mathrm { f } ( x )$.\\
(C) What can you conclude about the symmetry of the curve $y = \mathrm { f } ( x )$ ?
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 2010 Q9 [19]}}