| 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 | Moderate -0.3 This is a structured multi-part question covering standard C3 techniques: finding coordinates by substitution, quotient rule differentiation, definite integration, and function transformations. While it requires multiple steps and some algebraic manipulation, each part follows routine procedures with clear guidance ('show that', 'hence'). The most challenging aspect is the algebraic simplification in part (ii), but overall this is slightly easier than a typical C3 question due to its heavily scaffolded nature. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.07q Product and quotient rules: differentiation1.08e Area between curve and x-axis: using definite integrals |
| 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 |
|---|---|---|
| \(\frac{dy}{dx} = \frac{(1+e^{2x})2e^{2x} - e^{2x}.2e^{2x}}{(1+e^{2x})^2}\) | M1 | Quotient or product rule, correct expression – condone missing bracket |
| \(= \frac{2e^{2x}}{(1+e^{2x})^2}\) | A1 | cao – mark final answer |
| When \(x=0\), \(dy/dx = \frac{2e^0}{(1+e^0)^2} = \frac{1}{2}\) | A1, B1ft [4] | Follow through their derivative. Product rule: \(\frac{dy}{dx} = e^{2x}.2e^{2x}(-1)(1+e^{2x})^{-2} + 2e^{2x}(1+e^{2x})^{-1}\); \(-\frac{2e^{2x}}{(1+e^{2x})^2}\) 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 \(u = 1+e^{2x}\), \(du/dx = 2e^{2x}\); \(A = \int_2^{1+e^2} \frac{1/2}{u}\,du = \left[\frac{1}{2}\ln u\right]_2^{1+e^2}\) | M1, A1 | Or \(v = e^{2x}\), \(dv/dx = 2e^{2x}\); \([\frac{1}{2}\ln u]\) or \([\frac{1}{2}\ln(v+1)]\) |
| \(= \frac{1}{2}\ln(1+e^2) - \frac{1}{2}\ln 2 = \frac{1}{2}\ln\left[\frac{1+e^2}{2}\right]\) | M1, E1 [5] | Substituting correct limits; 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 |
|---|---|---|
| \((A)\ g(x) + \frac{1}{2} = \frac{1}{2}\cdot\frac{e^x-e^{-x}}{e^x+e^{-x}} + \frac{1}{2} = \frac{1}{2}\cdot\left(\frac{e^x-e^{-x}+e^x+e^{-x}}{e^x+e^{-x}}\right) = \frac{1}{2}\cdot\left(\frac{2e^x}{e^x+e^{-x}}\right) = \frac{e^x \cdot e^x}{e^x(e^x+e^{-x})} = \frac{e^{2x}}{e^{2x}+1} = f(x)\) | M1, A1, E1 [part] | Combining fractions correctly |
| \((B)\) 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 2:
## 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)
| $\frac{dy}{dx} = \frac{(1+e^{2x})2e^{2x} - e^{2x}.2e^{2x}}{(1+e^{2x})^2}$ | M1 | Quotient or product rule, correct expression – condone missing bracket |
|---|---|---|
| $= \frac{2e^{2x}}{(1+e^{2x})^2}$ | A1 | cao – mark final answer |
| When $x=0$, $dy/dx = \frac{2e^0}{(1+e^0)^2} = \frac{1}{2}$ | A1, B1ft [4] | Follow through their derivative. Product rule: $\frac{dy}{dx} = e^{2x}.2e^{2x}(-1)(1+e^{2x})^{-2} + 2e^{2x}(1+e^{2x})^{-1}$; $-\frac{2e^{2x}}{(1+e^{2x})^2}$ 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 $u = 1+e^{2x}$, $du/dx = 2e^{2x}$; $A = \int_2^{1+e^2} \frac{1/2}{u}\,du = \left[\frac{1}{2}\ln u\right]_2^{1+e^2}$ | M1, A1 | Or $v = e^{2x}$, $dv/dx = 2e^{2x}$; $[\frac{1}{2}\ln u]$ or $[\frac{1}{2}\ln(v+1)]$ |
| $= \frac{1}{2}\ln(1+e^2) - \frac{1}{2}\ln 2 = \frac{1}{2}\ln\left[\frac{1+e^2}{2}\right]$ | M1, E1 [5] | Substituting correct limits; 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)
| $(A)\ g(x) + \frac{1}{2} = \frac{1}{2}\cdot\frac{e^x-e^{-x}}{e^x+e^{-x}} + \frac{1}{2} = \frac{1}{2}\cdot\left(\frac{e^x-e^{-x}+e^x+e^{-x}}{e^x+e^{-x}}\right) = \frac{1}{2}\cdot\left(\frac{2e^x}{e^x+e^{-x}}\right) = \frac{e^x \cdot e^x}{e^x(e^x+e^{-x})} = \frac{e^{2x}}{e^{2x}+1} = f(x)$ | M1, A1, E1 [part] | Combining fractions correctly |
|---|---|---|
| $(B)$ 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 |
---2 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]{65ac8807-cd93-450f-adb5-dc6864f8470c-2_595_1230_445_496}
\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 Q2 [19]}}