| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| 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, and differentiation). Part (i) is straightforward substitution. Part (ii) requires routine integration and area calculation. Part (iii) involves finding an inverse function (standard procedure) and sketching. Part (iv) applies the inverse function derivative rule. While it has multiple parts (19 marks total), each step follows predictable techniques with clear guidance, making it slightly easier than the average A-level question. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.06a Exponential function: a^x and e^x graphs and properties1.07a Derivative as gradient: of tangent to curve1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \frac{1}{2}(e^t - 1)\) swap \(x\) and \(y\) | M1 [5] | Attempt to invert – one valid step; merely swapping \(x\) and \(y\) is not 'one step' |
| \(x = \frac{1}{2}(e^y - 1)\) | ||
| \(\Rightarrow 2x = e^y - 1\) | A1 | |
| \(\Rightarrow 2x + 1 = e^y\) | A1 | |
| \(\Rightarrow \ln(2x + 1) = y\) * | A1 | \(y = \ln(2x + 1)\) or \(g(x) = \ln(2x + 1)\) AG; apply a similar scheme if they start with \(g(x)\) and invert to get \(f(x)\). or \(g \, f(x) = g((e^x - 1)/2)\) M1 \(= \ln(1 + e^x - 1) = \ln(e^x)\) A1 \(= x\) A1 |
| \(\Rightarrow g(x) = \ln(2x + 1)\) | ||
| Sketch: recognisable attempt to reflect in \(y = x\) | M1 | through \(O\) and \((a, a)\); no obvious inflexion or TP, extends to third quadrant, without gradient becoming too negative; similar scheme for fg; See appendix for examples |
| Good shape | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f'(x) = \frac{1}{2}e^x\) | B1 [7] | |
| \(g'(x) = \frac{2}{(2x + 1)}\) | M1, A1 | \(\frac{1}{(2x + 1)}\) (or \(\frac{1}{u}\) with \(u = 2x + 1\)) \(\ldots \times 2\) to get \(\frac{2}{(2x + 1)}\); either \(g'(a)\) or \(f'(a)\) correct soi; substituting \(e^a = 1 + 2a\); establishing \(f'(a) = 1 / g'(a)\); either way round |
| \(g'(a) = \frac{2}{(2a+1)}, f'(a) = \frac{1}{2}e^a\) | B1, M1 | |
| so \(g'(a) = \frac{2}{e^a} \quad\) or \(\quad f'(a) = \frac{(2a+1)}{2}\) | ||
| \(= \frac{1}{(\frac{1}{2}e^a)} = \frac{1}{f'(a)}\) | A1 | |
| [= \(1/f'(a)\)] | ||
| tangents are reflections in \(y = x\) | B1 | must mention tangents |
### (iii)
$y = \frac{1}{2}(e^t - 1)$ swap $x$ and $y$ | M1 [5] | Attempt to invert – one valid step; merely swapping $x$ and $y$ is not 'one step'
$x = \frac{1}{2}(e^y - 1)$ | |
$\Rightarrow 2x = e^y - 1$ | A1 |
$\Rightarrow 2x + 1 = e^y$ | A1 |
$\Rightarrow \ln(2x + 1) = y$ * | A1 | $y = \ln(2x + 1)$ or $g(x) = \ln(2x + 1)$ AG; apply a similar scheme if they start with $g(x)$ and invert to get $f(x)$. or $g \, f(x) = g((e^x - 1)/2)$ M1 $= \ln(1 + e^x - 1) = \ln(e^x)$ A1 $= x$ A1
$\Rightarrow g(x) = \ln(2x + 1)$ | |
Sketch: recognisable attempt to reflect in $y = x$ | M1 | through $O$ and $(a, a)$; no obvious inflexion or TP, extends to third quadrant, without gradient becoming too negative; similar scheme for fg; See appendix for examples
Good shape | A1 |
### (iv)
$f'(x) = \frac{1}{2}e^x$ | B1 [7] |
$g'(x) = \frac{2}{(2x + 1)}$ | M1, A1 | $\frac{1}{(2x + 1)}$ (or $\frac{1}{u}$ with $u = 2x + 1$) $\ldots \times 2$ to get $\frac{2}{(2x + 1)}$; either $g'(a)$ or $f'(a)$ correct soi; substituting $e^a = 1 + 2a$; establishing $f'(a) = 1 / g'(a)$; either way round
$g'(a) = \frac{2}{(2a+1)}, f'(a) = \frac{1}{2}e^a$ | B1, M1 |
so $g'(a) = \frac{2}{e^a} \quad$ or $\quad f'(a) = \frac{(2a+1)}{2}$ | |
$= \frac{1}{(\frac{1}{2}e^a)} = \frac{1}{f'(a)}$ | A1 |
[= $1/f'(a)$] | |
tangents are reflections in $y = x$ | B1 | must mention tangents
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**End of Mark Scheme Extract**Fig. 9 shows the line $y = x$ and the curve $y = f(x)$, where $f(x) = \frac{1}{2}(e^x - 1)$. The line and the curve intersect at the origin and at the point P$(a, a)$.
\includegraphics{figure_9}
\begin{enumerate}[label=(\roman*)]
\item Show that $e^a = 1 + 2a$. [1]
\item 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 $y = x$. [6]
\item Show that the inverse function of f$(x)$ is g$(x)$, where g$(x) = \ln(1 + 2x)$. Add a sketch of $y = g(x)$ to the copy of Fig. 9. [5]
\item Find the derivatives of f$(x)$ and g$(x)$. Hence verify that $g'(a) = \frac{1}{f'(a)}$.
Give a geometrical interpretation of this result. [7]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 2013 Q9 [19]}}