| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Verify composite identity |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard C3 topics (exponentials, logarithms, integration, inverse functions) with clear guidance at each step. Parts (i) and (iii) are simple algebraic verification ('show that'), part (ii) is routine integration, and part (iv) applies the standard inverse function derivative relationship. The geometrical interpretation is a textbook result. Slightly easier than average due to the scaffolding and routine nature of all components. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.06a Exponential function: a^x and e^x graphs and properties1.06c Logarithm definition: log_a(x) as inverse of a^x1.07b Gradient as rate of change: dy/dx notation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \frac{1}{2}(e^x - 1)\), swap \(x\) and \(y\): \(x = \frac{1}{2}(e^y - 1)\) | ||
| \(2x = e^y - 1\) | M1 | Attempt to invert – one valid step; merely swapping \(x\) and \(y\) is not 'one step' |
| \(2x + 1 = e^y\) | A1 | |
| \(\ln(2x+1) = y\) | A1 | \(y = \ln(2x+1)\) or \(g(x) = \ln(2x+1)\) AG |
| \(g(x) = \ln(2x+1)\) | apply similar scheme if they start with \(g(x)\) and invert to get \(f(x)\); or \(gf(x) = g\left(\frac{e^x-1}{2}\right) = \ln(1+e^x-1) = \ln(e^x) = x\) M1A1 | |
| Sketch: recognisable attempt to reflect in \(y = x\) | M1 | through O and \((a,a)\) |
| Good shape | A1 | no obvious inflection or TP, extends to third quadrant, without gradient becoming too negative; similar scheme for \(fg\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = \frac{1}{2}e^x\) | B1 | |
| \(g'(x) = \frac{2}{2x+1}\) | M1 | \(\frac{1}{2x+1}\) (or \(\frac{1}{u}\) with \(u = 2x+1\))… |
| A1 | …\(\times 2\) to get \(\frac{2}{2x+1}\) | |
| \(g'(a) = \frac{2}{2a+1}\), \(f'(a) = \frac{1}{2}e^a\) | B1 | either \(g'(a)\) or \(f'(a)\) correct soi |
| so \(g'(a) = \frac{2}{e^a}\) or \(f'(a) = \frac{1}{2}(2a+1)\) | M1 | substituting \(e^a = 1+2a\) |
| \(= \frac{1}{\frac{1}{2}e^a}\) \(\quad = \frac{a+1}{2}\) | A1 | establishing \(f'(a) = 1/g'(a)\); either way round |
| \([= 1/f'(a)]\) \(\quad [= 1/g'(a)]\) | ||
| tangents are reflections in \(y = x\) | B1 | must mention tangents |
## Question 2:
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{1}{2}(e^x - 1)$, swap $x$ and $y$: $x = \frac{1}{2}(e^y - 1)$ | | |
| $2x = e^y - 1$ | M1 | Attempt to invert – one valid step; merely swapping $x$ and $y$ is not 'one step' |
| $2x + 1 = e^y$ | A1 | |
| $\ln(2x+1) = y$ | A1 | $y = \ln(2x+1)$ or $g(x) = \ln(2x+1)$ **AG** |
| $g(x) = \ln(2x+1)$ | | apply similar scheme if they start with $g(x)$ and invert to get $f(x)$; or $gf(x) = g\left(\frac{e^x-1}{2}\right) = \ln(1+e^x-1) = \ln(e^x) = x$ M1A1 |
| Sketch: recognisable attempt to reflect in $y = x$ | M1 | through O and $(a,a)$ |
| Good shape | A1 | no obvious inflection or TP, extends to third quadrant, without gradient becoming too negative; similar scheme for $fg$ |
### Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = \frac{1}{2}e^x$ | B1 | |
| $g'(x) = \frac{2}{2x+1}$ | M1 | $\frac{1}{2x+1}$ (or $\frac{1}{u}$ with $u = 2x+1$)… |
| | A1 | …$\times 2$ to get $\frac{2}{2x+1}$ |
| $g'(a) = \frac{2}{2a+1}$, $f'(a) = \frac{1}{2}e^a$ | B1 | either $g'(a)$ or $f'(a)$ correct soi |
| so $g'(a) = \frac{2}{e^a}$ or $f'(a) = \frac{1}{2}(2a+1)$ | M1 | substituting $e^a = 1+2a$ |
| $= \frac{1}{\frac{1}{2}e^a}$ $\quad = \frac{a+1}{2}$ | A1 | establishing $f'(a) = 1/g'(a)$; either way round |
| $[= 1/f'(a)]$ $\quad [= 1/g'(a)]$ | | |
| tangents are reflections in $y = x$ | B1 | must mention tangents |
---2 Fig. 9 shows the line $y = x$ and the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } - 1 \right)$. The line and the curve intersect at the origin and at the point $\mathrm { P } ( a , a )$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7825ba53-67eb-4050-a671-85e37a30150a-2_681_880_461_606}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
(i) Show that $\mathrm { e } ^ { a } = 1 + 2 a$.\\
(ii) Show that the area of the region enclosed by the curve, the $x$-axis and the line $x = a$ is $\frac { 1 } { 2 } a$. Hence find, in terms of $a$, the area enclosed by the curve and the line $y = x$.\\
(iii) Show that the inverse function of $\mathrm { f } ( x )$ is $\mathrm { g } ( x )$, where $\mathrm { g } ( x ) = \ln ( 1 + 2 x )$. Add a sketch of $y = \mathrm { g } ( x )$ to the copy of Fig. 9.\\
(iv) Find the derivatives of $\mathrm { f } ( x )$ and $\mathrm { g } ( x )$. Hence verify that $\mathrm { g } ^ { \prime } ( a ) = \frac { 1 } { \mathrm { f } ^ { \prime } ( a ) }$.
Give a geometrical interpretation of this result.
\hfill \mbox{\textit{OCR MEI C3 Q2 [19]}}