| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 18 |
| 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, inverse functions, integration by parts). Parts (i)-(ii) involve routine verification and algebraic manipulation. Parts (iii)-(iv) require standard integration techniques with some guidance provided. Part (v) requires combining results but follows a predictable pattern. While comprehensive, each component is a textbook exercise requiring no novel insight, making it slightly easier than average. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.06a Exponential function: a^x and e^x graphs and properties1.06d Natural logarithm: ln(x) function and properties1.08e Area between curve and x-axis: using definite integrals1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 + \ln x = 0 \Rightarrow \ln x = -1\) so A is \((e^{-1}, 0)\) \(\Rightarrow\) \(x = e^{-1}\) | M1, A1, B1 | SCI if obtained using symmetry condone use of symmetry Penalise A = \(e^{-1}\), B = \(e^{-1}\), or co-ords wrong way round, but condone labelling errors. |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 0, y = e^{0-1} = e^{-1}\) so B is \((0, e^{-1})\) | E1, E1 | [5] |
| Answer | Marks |
|---|---|
| \(f(1) = 1 \cdot e^{1} = e^1\) \(g(1) = 1 + \ln 1 = 1\) | E1, E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Either by inversion: e.g. \(y = e^x \Rightarrow x = e^y\) | M1 | Taking lns or exps |
| \(\Rightarrow\) \(\ln x = y - 1\) | E1 | |
| \(\Rightarrow\) \(1 + \ln x = y\) | [Dotted line] | |
| or by composing e.g. \(f(g(x)) = f(1 + \ln x) = e^{1+\ln x-1} = e^{\ln x} = x\) | M1, E1 | \(e^{1+\ln x-1}\) or \(1 + (\ln(e^x - 1))\) |
| E1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_c^{e^t} e^{x-1} dx = [e^{x-1}]_c\) | M1 | \([e^{x-1}]\) o.e or \(u = x - 1 \Rightarrow [e^u]\) |
| \(= e^c - e^{-1}\) | M1, A1cao | Substituting correct limits for \(x\) or \(u\) o.e. not \(e^c\), must be exact. [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \ln x dx = \int \ln x \cdot \frac{d}{dx}(x) dx = x \ln x - \int x \cdot \frac{1}{x} dx = x \ln x - x + c\) | M1, A1 | parts: \(u = \ln x\), \(du/dx = 1/x\), \(v = x\), \(dv/dx = 1\) |
| \(\Rightarrow \int_c g(x) dx = \int_c (1 + \ln x) dx = [x + x \ln x - x]_c\) | B1ft | ft their '\(x \ln x - x^?\)' (provided 'algebraic') |
| \(= [\ln x)]_c\) | DM1 | Substituting limits dep B1 |
| \(= \ln 1 - 1 - e^{-1} \ln(e^{-1}) = e^{-1}\) * | E1 | www [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Area = \(\int_c^1 f(x) dx - \int_c g(x) dx = (1 - e^{-1}) - e^{-1} = 1 - 2e\) | M1, A1cao | Must have correct limits 0.264 or better. |
| or | ||
| Area OCB = area under curve – triangle \(= 1 - e^{-1} - \frac{1}{2} \times 1 \times 1 = \frac{1}{2} - e^{-1}\) | M1 | OCA or OCB = \(\frac{1}{2} - e^{-1}\) |
| or | ||
| Area OAC = triangle – area under curve \(= \frac{1}{2} \times 1 \times 1 - e^{-1} = \frac{1}{2} - e^{-1}\) | ||
| Total area = \(2(\frac{1}{2} - e^{-1}) = 1 - 2e\) | A1cao | 0.264 or better [2] |
**Part (i)**
**Part A**
| $1 + \ln x = 0 \Rightarrow \ln x = -1$ so A is $(e^{-1}, 0)$ $\Rightarrow$ $x = e^{-1}$ | M1, A1, B1 | SCI if obtained using symmetry condone use of symmetry Penalise A = $e^{-1}$, B = $e^{-1}$, or co-ords wrong way round, but condone labelling errors. |
**Part B**
| $x = 0, y = e^{0-1} = e^{-1}$ so B is $(0, e^{-1})$ | E1, E1 | [5] |
**Part C**
| $f(1) = 1 \cdot e^{1} = e^1$ $g(1) = 1 + \ln 1 = 1$ | E1, E1 | |
**Part (ii)**
| Either by inversion: e.g. $y = e^x \Rightarrow x = e^y$ | M1 | Taking lns or exps |
| $\Rightarrow$ $\ln x = y - 1$ | E1 | |
| $\Rightarrow$ $1 + \ln x = y$ | | [Dotted line] |
| or by composing e.g. $f(g(x)) = f(1 + \ln x) = e^{1+\ln x-1} = e^{\ln x} = x$ | M1, E1 | $e^{1+\ln x-1}$ or $1 + (\ln(e^x - 1))$ |
| | E1 | [2] |
**Part (iii)**
| $\int_c^{e^t} e^{x-1} dx = [e^{x-1}]_c$ | M1 | $[e^{x-1}]$ o.e or $u = x - 1 \Rightarrow [e^u]$ |
| $= e^c - e^{-1}$ | M1, A1cao | Substituting correct limits for $x$ or $u$ o.e. not $e^c$, must be exact. [3] |
**Part (iv)**
| $\int \ln x dx = \int \ln x \cdot \frac{d}{dx}(x) dx = x \ln x - \int x \cdot \frac{1}{x} dx = x \ln x - x + c$ | M1, A1 | parts: $u = \ln x$, $du/dx = 1/x$, $v = x$, $dv/dx = 1$ |
| $\Rightarrow \int_c g(x) dx = \int_c (1 + \ln x) dx = [x + x \ln x - x]_c$ | B1ft | ft their '$x \ln x - x^?$' (provided 'algebraic') |
| $= [\ln x)]_c$ | DM1 | Substituting limits dep B1 |
| $= \ln 1 - 1 - e^{-1} \ln(e^{-1}) = e^{-1}$ * | E1 | www [6] |
**Part (v)**
| Area = $\int_c^1 f(x) dx - \int_c g(x) dx = (1 - e^{-1}) - e^{-1} = 1 - 2e$ | M1, A1cao | Must have correct limits 0.264 or better. |
| | | |
| or | | |
| Area OCB = area under curve – triangle $= 1 - e^{-1} - \frac{1}{2} \times 1 \times 1 = \frac{1}{2} - e^{-1}$ | M1 | OCA or OCB = $\frac{1}{2} - e^{-1}$ |
| or | | |
| Area OAC = triangle – area under curve $= \frac{1}{2} \times 1 \times 1 - e^{-1} = \frac{1}{2} - e^{-1}$ | | |
| Total area = $2(\frac{1}{2} - e^{-1}) = 1 - 2e$ | A1cao | 0.264 or better [2] |
---8 Fig. 8 shows the line $y = x$ and parts of the curves $y = \mathrm { f } ( x )$ and $y = \mathrm { g } ( x )$, where
$$\mathrm { f } ( x ) = \mathrm { e } ^ { x - 1 } , \quad \mathrm {~g} ( x ) = 1 + \ln x$$
The curves intersect the axes at the points A and B , as shown. The curves and the line $y = x$ meet at the point C .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1167a0e5-48c8-48e0-b2d1-76a50bad03ad-3_807_897_1016_625}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Find the exact coordinates of A and B . Verify that the coordinates of C are $( 1,1 )$.\\
(ii) Prove algebraically that $\mathrm { g } ( x )$ is the inverse of $\mathrm { f } ( x )$.\\
(iii) Evaluate $\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x$, giving your answer in terms of e.\\
(iv) Use integration by parts to find $\int \ln x \mathrm {~d} x$.
Hence show that $\int _ { \mathrm { e } ^ { - 1 } } ^ { 1 } \mathrm {~g} ( x ) \mathrm { d } x = \frac { 1 } { \mathrm { e } }$.\\
(v) Find the area of the region enclosed by the lines OA and OB , and the arcs AC and BC .
\hfill \mbox{\textit{OCR MEI C3 2009 Q8 [18]}}