| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - direct evaluation |
| Difficulty | Standard +0.3 This is a structured multi-part question requiring quotient rule differentiation, point evaluation, definite integration, and algebraic manipulation of exponential functions. While it involves several steps, each part uses standard C3 techniques with clear guidance ('show that', 'hence'). The algebraic simplification is routine for this level, making it slightly easier than average. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.06a Exponential function: a^x and e^x graphs and properties1.06d Natural logarithm: ln(x) function and properties1.07q Product and quotient rules: differentiation1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | 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 \(P = \frac{1}{2}\) |
| Answer | 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 |
| \(= \frac{2e^{2x}}{(1+e^{2x})^2}\) | A1 | Correct expression – condone missing bracket |
| When \(x = 0\): \(\frac{dy}{dx} = \frac{2e^0}{(1+e^0)^2} = \frac{1}{2}\) | A1, B1ft [4] | cao; follow through their derivative; product rule alternative: \(-\frac{2e^{2x}}{(1+e^{2x})^2}\) from \((udv - vdu)/v^2\) SC1 |
| Answer | 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 \(u = 1 + e^{2x}\), \(\frac{du}{dx} = 2e^{2x}\): \(A = \int_2^{1+e^2} \frac{1/2}{u}\, du = \left[\frac{1}{2}\ln u\right]_2^{1+e^2}\) | M1, 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 |
|---|---|---|
| \(g(-x) = \frac{1}{2}\left[\frac{e^{-x}-e^x}{e^{-x}+e^x}\right] = -\frac{1}{2}\left[\frac{e^x - e^{-x}}{e^x + e^{-x}}\right] = -g(x)\) | 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 |
|---|---|---|
| \(g(x) + \frac{1}{2} = \frac{1}{2}\cdot\frac{e^x - e^{-x}}{e^x + e^{-x}} + \frac{1}{2} = \frac{1}{2}\cdot\left(\frac{e^x - e^{-x} + e^x + e^{-x}}{e^x + e^{-x}}\right)\) | M1 | Combining fractions correctly |
| \(= \frac{1}{2}\cdot\frac{2e^x}{e^x + e^{-x}}\) | A1 | |
| \(= \frac{e^x \cdot e^x}{e^x(e^x + e^{-x})} = \frac{e^{2x}}{e^{2x}+1} = f(x)\) | E1 | |
| (B) Translation \(\begin{pmatrix}0\\1/2\end{pmatrix}\) | M1, A1 | Translation in \(y\) direction; up \(\frac{1}{2}\) unit dep 'translation' used; allow 'shift', 'move' in correct direction for M1; \(\begin{pmatrix}0\\1/2\end{pmatrix}\) alone is SC1 |
| (C) Rotational symmetry [of order 2] about P | B1 [6] | o.e.; condone omission of 180°/order 2 |
## Question 3:
### Part (i):
$(0, \frac{1}{2})$ | B1 [1] | Allow $y = \frac{1}{2}$, but not $(x=)\frac{1}{2}$ or $(\frac{1}{2}, 0)$, nor $P = \frac{1}{2}$
### Part (ii):
$\frac{dy}{dx} = \frac{(1+e^{2x})\cdot 2e^{2x} - e^{2x}\cdot 2e^{2x}}{(1+e^{2x})^2}$ | M1 | Quotient or product rule
$= \frac{2e^{2x}}{(1+e^{2x})^2}$ | A1 | Correct expression – condone missing bracket
When $x = 0$: $\frac{dy}{dx} = \frac{2e^0}{(1+e^0)^2} = \frac{1}{2}$ | A1, B1ft [4] | cao; follow through their derivative; product rule alternative: $-\frac{2e^{2x}}{(1+e^{2x})^2}$ from $(udv - vdu)/v^2$ SC1
### Part (iii):
$A = \int_0^1 \frac{e^{2x}}{1+e^{2x}}\, dx$ | 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 $u = 1 + e^{2x}$, $\frac{du}{dx} = 2e^{2x}$: $A = \int_2^{1+e^2} \frac{1/2}{u}\, du = \left[\frac{1}{2}\ln u\right]_2^{1+e^2}$ | M1, 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
### Part (iv):
$g(-x) = \frac{1}{2}\left[\frac{e^{-x}-e^x}{e^{-x}+e^x}\right] = -\frac{1}{2}\left[\frac{e^x - e^{-x}}{e^x + e^{-x}}\right] = -g(x)$ | 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)
### Part (v)(A):
$g(x) + \frac{1}{2} = \frac{1}{2}\cdot\frac{e^x - e^{-x}}{e^x + e^{-x}} + \frac{1}{2} = \frac{1}{2}\cdot\left(\frac{e^x - e^{-x} + e^x + e^{-x}}{e^x + e^{-x}}\right)$ | M1 | Combining fractions correctly
$= \frac{1}{2}\cdot\frac{2e^x}{e^x + e^{-x}}$ | A1 |
$= \frac{e^x \cdot e^x}{e^x(e^x + e^{-x})} = \frac{e^{2x}}{e^{2x}+1} = f(x)$ | E1 |
**(B)** Translation $\begin{pmatrix}0\\1/2\end{pmatrix}$ | M1, A1 | Translation in $y$ direction; up $\frac{1}{2}$ unit dep 'translation' used; allow 'shift', 'move' in correct direction for M1; $\begin{pmatrix}0\\1/2\end{pmatrix}$ alone is SC1
**(C)** Rotational symmetry [of order 2] about P | B1 [6] | o.e.; condone omission of 180°/order 23 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]{72893fd5-bc8e-433b-8358-f7979b2da636-3_594_1230_514_494}
\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 Q3 [19]}}