| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Area with Exponential or Logarithmic Curves |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard C3 techniques (finding intersections, integration with exponentials, inverse functions, and derivatives). Each part is guided with 'show that' prompts, requiring routine algebraic manipulation rather than problem-solving insight. The geometrical interpretation at the end is a standard result about inverse function gradients. Slightly easier than average due to the scaffolding. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.06a Exponential function: a^x and e^x graphs and properties1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| At \(P(a,a)\), \(g(a) = a\) so \(\frac{1}{2}(e^a - 1) = a \Rightarrow e^a = 1 + 2a\) | B1 | NB AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = \int_0^a \frac{1}{2}(e^x - 1)\,dx\) | M1 | Correct integral and limits; limits can be implied from subsequent work |
| \(= \frac{1}{2}\left[e^x - x\right]_0^a\) | B1 | Integral of \(e^x - 1\) is \(e^x - x\) |
| \(= \frac{1}{2}(e^a - a - e^0)\) | A1 | |
| \(= \frac{1}{2}(1 + 2a - a - 1) = \frac{1}{2}a\) | A1 | NB AG |
| Area of triangle \(= \frac{1}{2}a^2\) | B1 | |
| Area between curve and line \(= \frac{1}{2}a^2 - \frac{1}{2}a\) | B1cao | Mark final answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \frac{1}{2}(e^x - 1)\), swap \(x\) and \(y\): \(x = \frac{1}{2}(e^y - 1)\) | ||
| \(\Rightarrow 2x = e^y - 1\) | M1 | Attempt to invert – one valid step; merely swapping \(x\) and \(y\) is not 'one step' |
| \(\Rightarrow 2x + 1 = e^y\) | A1 | |
| \(\Rightarrow \ln(2x+1) = y\) | A1 | \(y = \ln(2x+1)\) or \(g(x) = \ln(2x+1)\) AG |
| \(\Rightarrow g(x) = \ln(2x+1)\) | ||
| Sketch: recognisable attempt to reflect in \(y = x\) | M1 | Through \(O\) and \((a,a)\) |
| Good shape | A1 | No obvious inflexion or TP, extends to third quadrant, without gradient becoming too negative |
| Answer | Marks | Guidance |
|---|---|---|
| \(f'(x) = \frac{1}{2}e^x\) | B1 | |
| \(g'(x) = \frac{2}{2x+1}\) | M1, A1 | \(\frac{1}{(2x+1)}\) (or \(\frac{1}{u}\) with \(u = 2x+1\)) \(\ldots \times 2\) to get \(\frac{2}{2x+1}\) |
| \(g'(a) = \frac{2}{2a+1}\), \(f'(a) = \frac{1}{2}e^a\) | B1 | Either \(g'(a)\) or \(f'(a)\) correct |
| So \(g'(a) = \frac{2}{e^a} = \frac{1}{\frac{1}{2}e^a}\), or \(f'(a) = \frac{1}{2}(2a+1) = \frac{a+1}{2}\) | M1 | Substituting \(e^a = 1 + 2a\), establishing \(f'(a) = 1/g'(a)\) |
| \([= 1/f'(a)]\) \(\quad [= 1\ g'(a)]\) | A1 | Either way round |
| Tangents are reflections in \(y = x\) | B1 | Must mention tangents |
## Question 7:
### Part (i)
| At $P(a,a)$, $g(a) = a$ so $\frac{1}{2}(e^a - 1) = a \Rightarrow e^a = 1 + 2a$ | B1 | **NB AG** |
|---|---|---|
### Part (ii)
| $A = \int_0^a \frac{1}{2}(e^x - 1)\,dx$ | M1 | Correct integral and limits; limits can be implied from subsequent work |
|---|---|---|
| $= \frac{1}{2}\left[e^x - x\right]_0^a$ | B1 | Integral of $e^x - 1$ is $e^x - x$ |
| $= \frac{1}{2}(e^a - a - e^0)$ | A1 | |
| $= \frac{1}{2}(1 + 2a - a - 1) = \frac{1}{2}a$ | A1 | **NB AG** |
| Area of triangle $= \frac{1}{2}a^2$ | B1 | |
| Area between curve and line $= \frac{1}{2}a^2 - \frac{1}{2}a$ | B1cao | Mark final answer |
### Part (iii)
| $y = \frac{1}{2}(e^x - 1)$, swap $x$ and $y$: $x = \frac{1}{2}(e^y - 1)$ | | |
|---|---|---|
| $\Rightarrow 2x = e^y - 1$ | M1 | Attempt to invert – one valid step; merely swapping $x$ and $y$ is not 'one step' |
| $\Rightarrow 2x + 1 = e^y$ | A1 | |
| $\Rightarrow \ln(2x+1) = y$ | A1 | $y = \ln(2x+1)$ or $g(x) = \ln(2x+1)$ **AG** |
| $\Rightarrow g(x) = \ln(2x+1)$ | | |
| Sketch: recognisable attempt to reflect in $y = x$ | M1 | Through $O$ and $(a,a)$ |
| Good shape | A1 | No obvious inflexion or TP, extends to third quadrant, without gradient becoming too negative |
### Part (iv)
| $f'(x) = \frac{1}{2}e^x$ | B1 | |
|---|---|---|
| $g'(x) = \frac{2}{2x+1}$ | M1, A1 | $\frac{1}{(2x+1)}$ (or $\frac{1}{u}$ with $u = 2x+1$) $\ldots \times 2$ to get $\frac{2}{2x+1}$ |
| $g'(a) = \frac{2}{2a+1}$, $f'(a) = \frac{1}{2}e^a$ | B1 | Either $g'(a)$ or $f'(a)$ correct |
| So $g'(a) = \frac{2}{e^a} = \frac{1}{\frac{1}{2}e^a}$, or $f'(a) = \frac{1}{2}(2a+1) = \frac{a+1}{2}$ | M1 | Substituting $e^a = 1 + 2a$, establishing $f'(a) = 1/g'(a)$ |
| $[= 1/f'(a)]$ $\quad [= 1\ g'(a)]$ | A1 | Either way round |
| Tangents are reflections in $y = x$ | B1 | Must mention tangents |
---7 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]{d4aa92fb-d21b-4387-b711-b0a6b0d57baa-3_694_886_430_604}
\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 Q7 [19]}}