| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find stationary points - polynomial/exponential products |
| Difficulty | Standard +0.3 This is a standard C3 question requiring routine differentiation of exponential functions, solving equations, integration, and finding an inverse function. While multi-part with several techniques, each step follows textbook procedures without requiring novel insight—slightly easier than average due to the straightforward algebraic manipulations involved. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.06a Exponential function: a^x and e^x graphs and properties1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| At P, \((e^x - 2)^2 - 1 = 0\) | M1 | square rooting – condone no ± |
| \(\Rightarrow e^x - 2 = [\pm]1,\) | ||
| \(e^x = [1 \text{ or }] 3\) | ||
| \(\Rightarrow x = [0 \text{ or}] \ln 3\) | A1 | x-coordinate of P is ln 3; must be exact; condone P in 3, but not y = ln 3 |
| or \((e^x)^2 - 4e^x + 3 = 0\) | M1 | expanding to correct quadratic and solve by factorising or using quadratic formula |
| \(\Rightarrow (e^x - 1)(e^x - 3) = 0\), \(e^x = 1\) or \(3\) | ||
| \(\Rightarrow x = [0 \text{ or}] \ln 3\) | A1 | condone \(e^x \not\times 2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f'(x) = 2(e^x - 2)e^x\) | M1 | chain rule; e.g. \(2u \times\) their deriv of \(e^x\) |
| \(= 0\) when \(e^x = 2\), \(x = \ln 2 \,*\) | A1 | not from wrong working |
| \(= 0 \text{ when } e^x = 2, x = \ln 2 \,*\) | A1 | or verified by substitution |
| or \(f(x) = e^{2x} - 4e^x + 3\) | M1 | expanding to 3 term quadratic with \((e^x)^2\) or \(e^{2x}\); condone \(e^x \not\times 2\) |
| \(\Rightarrow f'(x) = 2e^{2x} - 4e^x\) | A1 | correct derivative, not from wrong working |
| \(= 0\) when \(2e^{2x} = 4e^x\), \(e^x = 2\), \(x = \ln 2 \,*\) | A1 | or \(2e^{2x}(e^x - 2) = 0 \Rightarrow e^x = 2, x = \ln 2\); or verified by substitution |
| \(y = f(\ln(2)) = -1\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^{\ln 3} [(e^x - 2)^2 - 1] dx = \int_0^{\ln 3} [(e^x)^2 - 4e^x + 4 - 1] dx\) | M1 | expanding brackets; must have 3 terms: \((e^x)^2 - 4e^x + 4\) is M0, condone \(e^x \not\times 2\) |
| \(= \int_0^{\ln 3} [e^{2x} - 4e^x + 3] dx\) | A1 | |
| \(= \left[\frac{1}{2}e^{2x} - 4e^x + 3x\right]_0^{\ln 3}\) | B1 | \([\frac{1}{2}e^{2x} - 4e^x + 3x]\) |
| \(= (4.5 - 12 + 3\ln 3) - (0.5 - 4)\) | A1 | condone 3ln3 – 4 as final ans; mark final ans |
| \(= 3\ln 3 - 4\) [so area \(= 4 - 3\ln 3\)] | A1 | |
| Total: [5] | or if \(u = e^x\), \(\int [u^2 - 4u + 4 - 1]/u \, du = \int u - 4 + 3/u \, du = [\frac{1}{2}u^2 - 4u + 3\ln u]\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = (e^x - 2)^2 - 1 \, x \leftrightarrow y\) | ||
| \(x = (e^y - 2)^2 - 1\) | M1 | attempt to solve for y (might be indicated by expanding and then taking lns) |
| \(\Rightarrow x + 1 = (e^y - 2)^2\) | A1 | condone no ± |
| \(\Rightarrow \pm\sqrt{x+1} = e^y - 2 \,(+ \text{ for } y \geq \ln 2)\) | A1 | |
| \(\Rightarrow 2 + \sqrt{x+1}) = e^y\) | A1 | must have interchanged \(x\) and \(y\) in final ans |
| \(\Rightarrow y = \ln(2 + \sqrt{(x+1)}) = f^{-1}(x)\) | A1 | must be \(\geq -1\) and (not y) |
| Domain is \(x \geq -1\) | B1 | must be \(\geq -1\); if not specified, assume first ans is domain and second range |
| Range is \(y \geq \ln 2\) | B1 | or \(f^{-1}(x) \geq \ln 2\), must be \(\geq\) (not \(x\) or \(f(x)\)); if \(x > -1\) and \(y > \ln 2\) SCB1 |
| M1 | recognisable attempt to reflect curve, or any part of curve, in \(y = x\) | |
| A1 | good shape, cross on \(y = x\) (if shown), correct domain and range indicated; [see extra sheet for examples] | |
| Total: [7] | y = x shown indicative but not essential; e.g. –1 and ln 2 marked on axes |
**(i)**
At P, $(e^x - 2)^2 - 1 = 0$ | M1 | square rooting – condone no ±
$\Rightarrow e^x - 2 = [\pm]1,$ | |
$e^x = [1 \text{ or }] 3$ | |
$\Rightarrow x = [0 \text{ or}] \ln 3$ | A1 | x-coordinate of P is ln 3; must be exact; condone P in 3, but not y = ln 3
**or** $(e^x)^2 - 4e^x + 3 = 0$ | M1 | expanding to correct quadratic and solve by factorising or using quadratic formula
$\Rightarrow (e^x - 1)(e^x - 3) = 0$, $e^x = 1$ or $3$ | |
$\Rightarrow x = [0 \text{ or}] \ln 3$ | A1 | condone $e^x \not\times 2$
**Total: [2]**
**(ii)**
$f'(x) = 2(e^x - 2)e^x$ | M1 | chain rule; e.g. $2u \times$ their deriv of $e^x$
$= 0$ when $e^x = 2$, $x = \ln 2 \,*$ | A1 | not from wrong working | NB AG
$= 0 \text{ when } e^x = 2, x = \ln 2 \,*$ | A1 | or verified by substitution
**or** $f(x) = e^{2x} - 4e^x + 3$ | M1 | expanding to 3 term quadratic with $(e^x)^2$ or $e^{2x}$; condone $e^x \not\times 2$
$\Rightarrow f'(x) = 2e^{2x} - 4e^x$ | A1 | correct derivative, not from wrong working
$= 0$ when $2e^{2x} = 4e^x$, $e^x = 2$, $x = \ln 2 \,*$ | A1 | or $2e^{2x}(e^x - 2) = 0 \Rightarrow e^x = 2, x = \ln 2$; or verified by substitution
$y = f(\ln(2)) = -1$ | B1 |
**Total: [4]**
**(iii)**
$\int_0^{\ln 3} [(e^x - 2)^2 - 1] dx = \int_0^{\ln 3} [(e^x)^2 - 4e^x + 4 - 1] dx$ | M1 | expanding brackets; must have 3 terms: $(e^x)^2 - 4e^x + 4$ is M0, condone $e^x \not\times 2$
$= \int_0^{\ln 3} [e^{2x} - 4e^x + 3] dx$ | A1 |
$= \left[\frac{1}{2}e^{2x} - 4e^x + 3x\right]_0^{\ln 3}$ | B1 | $[\frac{1}{2}e^{2x} - 4e^x + 3x]$
$= (4.5 - 12 + 3\ln 3) - (0.5 - 4)$ | A1 | condone 3ln3 – 4 as final ans; mark final ans
$= 3\ln 3 - 4$ [so area $= 4 - 3\ln 3$] | A1 |
**Total: [5]** | | or if $u = e^x$, $\int [u^2 - 4u + 4 - 1]/u \, du = \int u - 4 + 3/u \, du = [\frac{1}{2}u^2 - 4u + 3\ln u]$
**(iv)**
$y = (e^x - 2)^2 - 1 \, x \leftrightarrow y$ | |
$x = (e^y - 2)^2 - 1$ | M1 | attempt to solve for y (might be indicated by expanding and then taking lns)
$\Rightarrow x + 1 = (e^y - 2)^2$ | A1 | condone no ±
$\Rightarrow \pm\sqrt{x+1} = e^y - 2 \,(+ \text{ for } y \geq \ln 2)$ | A1 |
$\Rightarrow 2 + \sqrt{x+1}) = e^y$ | A1 | must have interchanged $x$ and $y$ in final ans
$\Rightarrow y = \ln(2 + \sqrt{(x+1)}) = f^{-1}(x)$ | A1 | must be $\geq -1$ and (not y)
Domain is $x \geq -1$ | B1 | must be $\geq -1$; if not specified, assume first ans is domain and second range
Range is $y \geq \ln 2$ | B1 | or $f^{-1}(x) \geq \ln 2$, must be $\geq$ (not $x$ or $f(x)$); if $x > -1$ and $y > \ln 2$ SCB1
| M1 | recognisable attempt to reflect curve, or any part of curve, in $y = x$
| A1 | good shape, cross on $y = x$ (if shown), correct domain and range indicated; [see extra sheet for examples]
**Total: [7]** | | y = x shown indicative but not essential; e.g. –1 and ln 2 marked on axes
---9 Fig. 9 shows the curve $y = \mathrm { f } ( x )$, where
$$\mathrm { f } ( x ) = \left( \mathrm { e } ^ { x } - 2 \right) ^ { 2 } - 1 , x \in \mathbb { R } .$$
The curve crosses the $x$-axis at O and P , and has a turning point at Q .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{955bebfb-04a3-4cd9-a33e-a8ba4b73e2ba-5_867_988_497_525}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
(i) Find the exact $x$-coordinate of P .\\
(ii) Show that the $x$-coordinate of Q is $\ln 2$ and find its $y$-coordinate.\\
(iii) Find the exact area of the region enclosed by the curve and the $x$-axis.
The domain of $\mathrm { f } ( x )$ is now restricted to $x \geqslant \ln 2$.\\
(iv) Find the inverse function $\mathrm { f } ^ { - 1 } ( x )$. Write down its domain and range, and sketch its graph on the copy of Fig. 9.
\hfill \mbox{\textit{OCR MEI C3 2015 Q9 [18]}}