| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Integrate using hyperbolic substitution |
| Difficulty | Challenging +1.2 This is a structured multi-part question on surface of revolution with hyperbolic substitution. Part (a) is standard formula application, part (b) requires careful substitution work but is guided, part (c) is routine integration using double angle formulas, and part (d) is evaluation. While it involves Further Maths content (hyperbolic functions), the question provides substantial scaffolding through its parts, making it moderately above average difficulty but not requiring significant novel insight. |
| Spec | 1.08h Integration by substitution4.07d Differentiate/integrate: hyperbolic functions4.07f Inverse hyperbolic: logarithmic forms4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = e^{-x} \Rightarrow \frac{dy}{dx} = -e^{-x}\) | B1 | Correct derivative |
| \(S = 2\pi\int y\sqrt{1+(y')^2}\,dx = 2\pi\int e^{-x}\sqrt{1+e^{-2x}}\,dx\) | M1A1 | M1: Use of correct formula; A1: Correct proof with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(e^{-x} = \sinh u \Rightarrow -e^{-x} = \cosh u\,\frac{du}{dx}\) | B1 | Correct differentiation |
| \(S = 2\pi\int e^{-x}\sqrt{1+e^{-2x}}\,dx = 2\pi\int \sinh u\sqrt{1+\sinh^2 u}\cdot\frac{\cosh u}{-\sinh u}\,du\) | M1 | A complete substitution |
| \(= -2\pi\int \cosh^2 u\,du\) | A1 | |
| \(x=0 \Rightarrow u = \text{arcsinh}(1)(= \ln(1+\sqrt{2}))\) and \(x=\ln 3 \Rightarrow u = \text{arcsinh}\!\left(\tfrac{1}{3}\right)\!\left(=\ln\!\left(\tfrac{1}{3}+\sqrt{1+\tfrac{1}{9}}\right)\right)\) | B1 | Both limits correct |
| \(S = 2\pi\int_{\text{arcsinh}(\frac{1}{3})}^{\text{arcsinh}(1)} \cosh^2 u\,du\) | A1 | Correct completion with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2\int\cosh^2 u\,du = \int(\cosh 2u + 1)\,du\) | M1 | Uses \(2\cosh^2 u = \pm\cosh 2u \pm 1\) |
| \(= \frac{1}{2}\sinh 2u + u\,(+k)\) | A1* | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(S = \pi\!\left(\tfrac{1}{2}\sinh 2(\text{arcsinh}\,\beta) + \text{arcsinh}\,\beta - \tfrac{1}{2}\sinh 2(\text{arcsinh}\,\alpha) - \text{arcsinh}\,\alpha\right)\) | M1 | Attempt to use their limits (subtracting either way round); allow omission of \(\pi\); allow \(2\pi\) instead of \(\pi\); must show evidence of use of limits (e.g. answer of 5.08 with no working loses this mark) |
| \(= 5.079\) | A1 | cao (Allow recovery from \(-5.079\)) |
# Question 7:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = e^{-x} \Rightarrow \frac{dy}{dx} = -e^{-x}$ | B1 | Correct derivative |
| $S = 2\pi\int y\sqrt{1+(y')^2}\,dx = 2\pi\int e^{-x}\sqrt{1+e^{-2x}}\,dx$ | M1A1 | M1: Use of correct formula; A1: Correct proof with no errors |
**(3 marks)**
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-x} = \sinh u \Rightarrow -e^{-x} = \cosh u\,\frac{du}{dx}$ | B1 | Correct differentiation |
| $S = 2\pi\int e^{-x}\sqrt{1+e^{-2x}}\,dx = 2\pi\int \sinh u\sqrt{1+\sinh^2 u}\cdot\frac{\cosh u}{-\sinh u}\,du$ | M1 | A complete substitution |
| $= -2\pi\int \cosh^2 u\,du$ | A1 | |
| $x=0 \Rightarrow u = \text{arcsinh}(1)(= \ln(1+\sqrt{2}))$ and $x=\ln 3 \Rightarrow u = \text{arcsinh}\!\left(\tfrac{1}{3}\right)\!\left(=\ln\!\left(\tfrac{1}{3}+\sqrt{1+\tfrac{1}{9}}\right)\right)$ | B1 | Both limits correct |
| $S = 2\pi\int_{\text{arcsinh}(\frac{1}{3})}^{\text{arcsinh}(1)} \cosh^2 u\,du$ | A1 | Correct completion with no errors |
**(5 marks)**
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $2\int\cosh^2 u\,du = \int(\cosh 2u + 1)\,du$ | M1 | Uses $2\cosh^2 u = \pm\cosh 2u \pm 1$ |
| $= \frac{1}{2}\sinh 2u + u\,(+k)$ | A1* | cso |
**(2 marks)**
## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $S = \pi\!\left(\tfrac{1}{2}\sinh 2(\text{arcsinh}\,\beta) + \text{arcsinh}\,\beta - \tfrac{1}{2}\sinh 2(\text{arcsinh}\,\alpha) - \text{arcsinh}\,\alpha\right)$ | M1 | Attempt to use their limits (subtracting either way round); allow omission of $\pi$; allow $2\pi$ instead of $\pi$; must show evidence of use of limits (e.g. answer of 5.08 with no working loses this mark) |
| $= 5.079$ | A1 | cao (Allow recovery from $-5.079$) |
> **NB:** $S = \pi(\sqrt{2} + \ln(1+\sqrt{2}) - \frac{1}{3}\frac{\sqrt{10}}{3} - \ln(\frac{1}{3}+\frac{\sqrt{10}}{3})) = 5.079241597$
**(2 marks) — Total: 12**
---7. The curve $C$ has equation
$$y = \mathrm { e } ^ { - x } , \quad x \in \mathbb { R }$$
The part of the curve $C$ between $x = 0$ and $x = \ln 3$ is rotated through $2 \pi$ radians about the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\item Show that the area $S$ of the curved surface generated is given by
$$S = 2 \pi \int _ { 0 } ^ { \ln 3 } \mathrm { e } ^ { - x } \sqrt { 1 + \mathrm { e } ^ { - 2 x } } \mathrm {~d} x$$
\item Use the substitution $\mathrm { e } ^ { - x } = \sinh u$ to show that
$$S = 2 \pi \int _ { \operatorname { arsinh } \alpha } ^ { \operatorname { arsinh } \beta } \cosh ^ { 2 } u \mathrm {~d} u$$
where $\alpha$ and $\beta$ are constants to be determined.
\item Show that
$$2 \int \cosh ^ { 2 } u \mathrm {~d} u = \frac { 1 } { 2 } \sinh 2 u + u + k$$
where $k$ is an arbitrary constant.
\item Hence find the value of $S$, giving your answer to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2014 Q7 [12]}}