| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Express hyperbolic in exponential form |
| Difficulty | Standard +0.3 Part (a) is direct substitution of standard hyperbolic definitions into exponential form (routine manipulation). Part (b) is a straightforward quadratic in e^x after substitution. Part (c) requires recognizing the integral form and evaluating with given limits, but follows a standard method once the exponential form is established. This is a typical Further Maths question testing standard techniques without requiring novel insight. |
| Spec | 1.08h Integration by substitution4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(f(x) = 5\cosh x - 4\sinh x = 5 \times \frac{1}{2}(e^x + e^{-x}) - 4 \times \frac{1}{2}(e^x - e^{-x})\) | M1 | Replacing both \(\cosh x\) and \(\sinh x\) by terms in \(e^x\) and \(e^{-x}\), condone sign errors |
| \(= \frac{1}{2}(e^x + 9e^{-x})\) ✱ | A1cso | Answer given |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{1}{2}(e^x + 9e^{-x}) = 5 \Rightarrow e^{2x} - 10e^x + 9 = 0\) | M1 A1 | Getting a three term quadratic in \(e^x\); cao |
| So \(e^x = 9\) or \(1\) and \(x = \ln 9\) or \(0\) | M1 A1 | Solving to \(x=\); cao need \(\ln 9\) (o.e.) and \(0\) (not \(\ln 1\)) |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Integral may be written \(\int \frac{2e^x}{e^{2x}+9}\,dx\) | B1 | cao, getting into suitable form, may substitute first |
| This is \(\frac{2}{3}\arctan\left(\frac{e^x}{3}\right)\) | M1 A1 | Integrating to give term in arctan; cao |
| Uses limits to give \(\left[\frac{2}{3}\arctan 1 - \frac{2}{3}\arctan\!\left(\frac{1}{\sqrt{3}}\right)\right] = \left[\frac{2}{3}\times\frac{\pi}{4} - \frac{2}{3}\times\frac{\pi}{6}\right] = \frac{\pi}{18}\) ✱ | DM1 A1cso | Depends on previous M mark. Correct use of \(\ln 3\) and \(\frac{1}{2}\ln 3\) as limits; cso must see them subtracting two terms in \(\pi\) |
| (5) |
# Question 7:
## Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $f(x) = 5\cosh x - 4\sinh x = 5 \times \frac{1}{2}(e^x + e^{-x}) - 4 \times \frac{1}{2}(e^x - e^{-x})$ | M1 | Replacing both $\cosh x$ and $\sinh x$ by terms in $e^x$ and $e^{-x}$, condone sign errors |
| $= \frac{1}{2}(e^x + 9e^{-x})$ ✱ | A1cso | Answer given |
| | **(2)** | |
## Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{1}{2}(e^x + 9e^{-x}) = 5 \Rightarrow e^{2x} - 10e^x + 9 = 0$ | M1 A1 | Getting a three term quadratic in $e^x$; cao |
| So $e^x = 9$ or $1$ and $x = \ln 9$ or $0$ | M1 A1 | Solving to $x=$; cao need $\ln 9$ (o.e.) and $0$ (not $\ln 1$) |
| | **(4)** | |
## Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| Integral may be written $\int \frac{2e^x}{e^{2x}+9}\,dx$ | B1 | cao, getting into suitable form, may substitute first |
| This is $\frac{2}{3}\arctan\left(\frac{e^x}{3}\right)$ | M1 A1 | Integrating to give term in arctan; cao |
| Uses limits to give $\left[\frac{2}{3}\arctan 1 - \frac{2}{3}\arctan\!\left(\frac{1}{\sqrt{3}}\right)\right] = \left[\frac{2}{3}\times\frac{\pi}{4} - \frac{2}{3}\times\frac{\pi}{6}\right] = \frac{\pi}{18}$ ✱ | DM1 A1cso | Depends on previous M mark. Correct use of $\ln 3$ and $\frac{1}{2}\ln 3$ as limits; cso must see them subtracting two terms in $\pi$ |
| | **(5)** | |
---7.
$$\mathrm { f } ( x ) = 5 \cosh x - 4 \sinh x , \quad x \in \mathbb { R }$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { f } ( x ) = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } + 9 \mathrm { e } ^ { - x } \right)$
Hence
\item solve $\mathrm { f } ( x ) = 5$
\item show that $\int _ { \frac { 1 } { 2 } \ln 3 } ^ { \ln 3 } \frac { 1 } { 5 \cosh x - 4 \sinh x } \mathrm {~d} x = \frac { \pi } { 18 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2012 Q7 [11]}}