| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - at special curve features |
| Difficulty | Challenging +1.2 This is a substantial multi-part question requiring logarithm differentiation, inverse functions, and integration with substitution. However, each part follows standard C3 techniques with helpful hints provided. The most challenging aspect is part (iv) requiring reflection geometry to find the shaded area, but this is a known technique for inverse function problems. Overall, slightly above average difficulty for C3 due to length and the inverse function geometry, but not requiring novel insights. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = \ln\frac{2y}{1+y}\) | (\(x \leftrightarrow y\) here or at end to complete) | |
| \(\Rightarrow e^x = \frac{2y}{1+y}\) | B1 | |
| \(\Rightarrow e^x(1+y) = 2y\) | B1 | |
| \(\Rightarrow e^x = 2y - e^x y = y(2 - e^x)\) | B1 | |
| \(\Rightarrow y = \frac{e^x}{2-e^x}\) \([= g(x)]\) | B1 | completion |
| OR \(\text{gf}(x) = g\!\left(\frac{2x}{1+x}\right) = e^{\ln[2x/(1+x)]} / \{2 - e^{\ln[2x/(1+x)]}\}\) | M1 | forming gf or fg |
| \(= \dfrac{2x/(1+x)}{2 - 2x/(1+x)}\) | A1 | |
| \(= \dfrac{2x}{2+2x-2x} = \dfrac{2x}{2} = x\) | M1A1 | |
| gradient at \(R = 1/\frac{1}{2} = 2\) | B1 ft | 1/their ans in (ii) unless \(\pm 1\) or \(0\) |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| let \(u = 2 - e^x \Rightarrow du/dx = -e^x\) | ||
| \(x=0,\ u=1;\ x=\ln(4/3),\ u = 2-4/3 = 2/3\) | B1 | \(2-e^0 = 1\), and \(2 - e^{\ln(4/3)} = 2/3\) seen |
| \(\Rightarrow \displaystyle\int_0^{\ln(4/3)} g(x)\,dx = \int_1^{2/3} -\frac{1}{u}\,du\) | M1 | \(\int -1/u\,du\) — condone \(\int 1/u\,du\) |
| \(= [-\ln(u)]_1^{2/3}\) | A1 | \([-\ln(u)]\) (could be \([\ln u]\) if limits swapped) |
| \(= [-\ln(u)]_1^{2/3} = -\ln(2/3) + \ln 1 = \ln(3/2)^*\) | A1cao | NB AG |
| Shaded region \(=\) rectangle \(-\) integral | M1 | |
| \(= 2\ln(4/3) - \ln(3/2)\) | B1 | rectangle area \(= 2\ln(4/3)\) |
| \(= \ln(16/9 \times 2/3)\) | ||
| \(= \ln(32/27)^*\) | A1cao | NB AG must show at least one step from \(2\ln(4/3) - \ln(3/2)\) |
| [7] |
# Question 1:
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = \ln\frac{2y}{1+y}$ | | ($x \leftrightarrow y$ here or at end to complete) |
| $\Rightarrow e^x = \frac{2y}{1+y}$ | B1 | |
| $\Rightarrow e^x(1+y) = 2y$ | B1 | |
| $\Rightarrow e^x = 2y - e^x y = y(2 - e^x)$ | B1 | |
| $\Rightarrow y = \frac{e^x}{2-e^x}$ $[= g(x)]$ | B1 | completion |
| **OR** $\text{gf}(x) = g\!\left(\frac{2x}{1+x}\right) = e^{\ln[2x/(1+x)]} / \{2 - e^{\ln[2x/(1+x)]}\}$ | M1 | forming gf or fg |
| $= \dfrac{2x/(1+x)}{2 - 2x/(1+x)}$ | A1 | |
| $= \dfrac{2x}{2+2x-2x} = \dfrac{2x}{2} = x$ | M1A1 | |
| gradient at $R = 1/\frac{1}{2} = 2$ | B1 ft | 1/their ans in (ii) unless $\pm 1$ or $0$ |
| **[5]** | | |
---
## Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| let $u = 2 - e^x \Rightarrow du/dx = -e^x$ | | |
| $x=0,\ u=1;\ x=\ln(4/3),\ u = 2-4/3 = 2/3$ | B1 | $2-e^0 = 1$, and $2 - e^{\ln(4/3)} = 2/3$ seen |
| $\Rightarrow \displaystyle\int_0^{\ln(4/3)} g(x)\,dx = \int_1^{2/3} -\frac{1}{u}\,du$ | M1 | $\int -1/u\,du$ — condone $\int 1/u\,du$ |
| $= [-\ln(u)]_1^{2/3}$ | A1 | $[-\ln(u)]$ (could be $[\ln u]$ if limits swapped) |
| $= [-\ln(u)]_1^{2/3} = -\ln(2/3) + \ln 1 = \ln(3/2)^*$ | A1cao | NB AG |
| Shaded region $=$ rectangle $-$ integral | M1 | |
| $= 2\ln(4/3) - \ln(3/2)$ | B1 | rectangle area $= 2\ln(4/3)$ |
| $= \ln(16/9 \times 2/3)$ | | |
| $= \ln(32/27)^*$ | A1cao | NB AG must show at least one step from $2\ln(4/3) - \ln(3/2)$ |
| **[7]** | | |
---1 Fig. 9 shows the curves $y = \mathrm { f } ( x )$ and $y = \mathrm { g } ( x )$. The function $y = \mathrm { f } ( x )$ is given by
$$f ( x ) = \ln \left( \frac { 2 x } { 1 + x } \right) , x > 0$$
The curve $y = \mathrm { f } ( x )$ crosses the $x$-axis at P , and the line $x = 2$ at Q .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1d12cd0d-07b0-429c-ad3b-e3bccb0fae18-1_555_641_573_748}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
(i) Verify that the $x$-coordinate of P is 1 .
Find the exact $y$-coordinate of Q .\\
(ii) Find the gradient of the curve at P . [Hint: use $\ln \frac { a } { b } = \ln a - \ln b$.]
The function $\mathrm { g } ( x )$ is given by
$$\mathrm { g } ( x ) = \frac { \mathrm { e } ^ { x } } { 2 - \mathrm { e } ^ { x } } , \quad x < \ln 2 .$$
The curve $y = \mathrm { g } ( x )$ crosses the $y$-axis at the point R .\\
(iii) Show that $\mathrm { g } ( x )$ is the inverse function of $\mathrm { f } ( x )$.
Write down the gradient of $y = \mathrm { g } ( x )$ at R.\\
(iv) Show, using the substitution $u = 2 - \mathrm { e } ^ { x }$ or otherwise, that $\int _ { 0 } ^ { \ln \frac { 4 } { 3 } } \mathrm {~g} ( x ) \mathrm { d } x = \ln \frac { 3 } { 2 }$.
Using this result, show that the exact area of the shaded region shown in Fig. 9 is $\ln \frac { 32 } { 27 }$.\\[0pt]
[Hint: consider its reflection in $y = x$.]
\hfill \mbox{\textit{OCR MEI C3 Q1 [18]}}