| 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 manipulation, differentiation of ln functions using chain rule, inverse function verification, and integration with substitution. However, each individual step follows standard C3 techniques with helpful hints provided. The most challenging aspect is part (iv) connecting areas via reflection in y=x, but this is guided. Overall, it's moderately above average for C3 due to length and the inverse function/area relationship, but not exceptionally difficult. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.06d Natural logarithm: ln(x) function and properties1.06e Logarithm as inverse: ln(x) inverse of e^x1.06f Laws of logarithms: addition, subtraction, power rules1.07l Derivative of ln(x): and related functions1.08d Evaluate definite integrals: between limits1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| When \(x=1\), \(f(1)=\ln(2/2)=\ln 1=0\) so \(P\) is \((1,0)\) | B1 | or \(\ln(2x/(1+x))=0 \Rightarrow 2x/(1+x)=1 \Rightarrow 2x=1+x \Rightarrow x=1\) |
| \(f(2)=\ln(4/3)\) | B1 | if approximated, can isw after \(\ln(4/3)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y=\ln(2x)-\ln(1+x)\) | condone lack of brackets | |
| \(\Rightarrow \frac{dy}{dx}=\frac{2}{2x}-\frac{1}{1+x}\) | M1 | one term correct |
| A1cao | mark final ans; \(2/2x\) or \(-1/(1+x)\) | |
| OR \(\frac{d}{dx}\left(\frac{2x}{1+x}\right)=\frac{(1+x)2-2x\cdot1}{(1+x)^2}=\frac{2}{(1+x)^2}\) | B1 | correct quotient or product rule; need not be simplified |
| \(\frac{dy}{dx}=\frac{2}{(1+x)^2}\cdot\frac{2x/(1+x)}{1}=\frac{1}{x(1+x)}\) | M1 | chain rule attempted |
| A1 | o.e., but mark final ans; need not be simplified | |
| At \(P\), \(\frac{dy}{dx}=1-\frac{1}{2}=\frac{1}{2}\) | A1cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x=\ln\!\left[\frac{2y}{1+y}\right]\) or \(e^x=\frac{2y}{1+y}\) | B1 | (\(x\leftrightarrow y\) here or at end to complete) |
| \(e^x(1+y)=2y\) | B1 | |
| \(e^x=2y-e^xy=y(2-e^x)\) | B1 | |
| \(y=\frac{e^x}{2-e^x}\ [=g(x)]\) | B1 | completion |
| OR \(gf(x)=g(2x/(1+x))=e^{\ln[2x/(1+x)]}/\{2-e^{\ln[2x/(1+x)]}\}\) | M1 | forming \(gf\) or \(fg\) |
| \(=\frac{2x/(1+x)}{2-2x/(1+x)}=\frac{2x/(1+x)}{(2+2x-2x)/(1+x)}=\frac{2x}{2}=x\) | A1, M1A1 | |
| gradient at \(R = 1/\frac{1}{2}=2\) | B1ft | \(1/\text{their ans in (ii)}\) unless \(\pm1\) or \(0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Let \(u=2-e^x \Rightarrow du/dx=-e^x\) | ||
| \(x=0,\ u=1\);\ \(x=\ln(4/3),\ u=2-e^{\ln(4/3)}=\frac{2}{3}\) | B1 | \(2-e^0=1\) and \(2-e^{\ln(4/3)}=\frac{2}{3}\) seen |
| \(\Rightarrow \int_0^{\ln(4/3)}g(x)\,dx = \int_1^{2/3}-\frac{1}{u}\,du\) | M1 | \(\int -1/u\,du\); condone \(\int 1/u\,du\) |
| \(=[-\ln(u)]_1^{2/3}=-\ln(2/3)+\ln 1=\ln(3/2)\) | A1, A1cao | \([-\ln(u)]\); NB AG |
| Shaded region \(=\) rectangle \(-\) integral | M1 | |
| \(=2\ln(4/3)-\ln(3/2)\) | B1 | rectangle area \(=2\ln(4/3)\); NB AG must show at least one step from \(2\ln(4/3)-\ln(3/2)\) |
| \(=\ln(16/9\times 2/3)=\ln(32/27)\) | A1cao |
# Question 3:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| When $x=1$, $f(1)=\ln(2/2)=\ln 1=0$ so $P$ is $(1,0)$ | B1 | or $\ln(2x/(1+x))=0 \Rightarrow 2x/(1+x)=1 \Rightarrow 2x=1+x \Rightarrow x=1$ |
| $f(2)=\ln(4/3)$ | B1 | if approximated, can isw after $\ln(4/3)$ |
**[2]**
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $y=\ln(2x)-\ln(1+x)$ | | condone lack of brackets |
| $\Rightarrow \frac{dy}{dx}=\frac{2}{2x}-\frac{1}{1+x}$ | M1 | one term correct |
| | A1cao | mark final ans; $2/2x$ or $-1/(1+x)$ |
| **OR** $\frac{d}{dx}\left(\frac{2x}{1+x}\right)=\frac{(1+x)2-2x\cdot1}{(1+x)^2}=\frac{2}{(1+x)^2}$ | B1 | correct quotient or product rule; need not be simplified |
| $\frac{dy}{dx}=\frac{2}{(1+x)^2}\cdot\frac{2x/(1+x)}{1}=\frac{1}{x(1+x)}$ | M1 | chain rule attempted |
| | A1 | o.e., but mark final ans; need not be simplified |
| At $P$, $\frac{dy}{dx}=1-\frac{1}{2}=\frac{1}{2}$ | A1cao | |
**[4]**
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x=\ln\!\left[\frac{2y}{1+y}\right]$ or $e^x=\frac{2y}{1+y}$ | B1 | ($x\leftrightarrow y$ here or at end to complete) |
| $e^x(1+y)=2y$ | B1 | |
| $e^x=2y-e^xy=y(2-e^x)$ | B1 | |
| $y=\frac{e^x}{2-e^x}\ [=g(x)]$ | B1 | completion |
| **OR** $gf(x)=g(2x/(1+x))=e^{\ln[2x/(1+x)]}/\{2-e^{\ln[2x/(1+x)]}\}$ | M1 | forming $gf$ or $fg$ |
| $=\frac{2x/(1+x)}{2-2x/(1+x)}=\frac{2x/(1+x)}{(2+2x-2x)/(1+x)}=\frac{2x}{2}=x$ | A1, M1A1 | |
| gradient at $R = 1/\frac{1}{2}=2$ | B1ft | $1/\text{their ans in (ii)}$ unless $\pm1$ or $0$ |
**[5]**
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| Let $u=2-e^x \Rightarrow du/dx=-e^x$ | | |
| $x=0,\ u=1$;\ $x=\ln(4/3),\ u=2-e^{\ln(4/3)}=\frac{2}{3}$ | B1 | $2-e^0=1$ and $2-e^{\ln(4/3)}=\frac{2}{3}$ seen |
| $\Rightarrow \int_0^{\ln(4/3)}g(x)\,dx = \int_1^{2/3}-\frac{1}{u}\,du$ | M1 | $\int -1/u\,du$; condone $\int 1/u\,du$ |
| $=[-\ln(u)]_1^{2/3}=-\ln(2/3)+\ln 1=\ln(3/2)$ | A1, A1cao | $[-\ln(u)]$; NB AG |
| Shaded region $=$ rectangle $-$ integral | M1 | |
| $=2\ln(4/3)-\ln(3/2)$ | B1 | rectangle area $=2\ln(4/3)$; NB AG must show at least one step from $2\ln(4/3)-\ln(3/2)$ |
| $=\ln(16/9\times 2/3)=\ln(32/27)$ | A1cao | |
**[7]**3 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]{75eebbfb-7bfa-4382-a6d7-1c5a7f3f419a-3_559_644_622_745}
\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 $\ln \frac { 32 } { 27 }$. [Hint: consider its reflection in $y = x$.]
\hfill \mbox{\textit{OCR MEI C3 Q3 [18]}}