| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Surface area of revolution with hyperbolics |
| Difficulty | Challenging +1.2 This is a standard Further Maths question on hyperbolic functions (cosh x) requiring three routine calculations: mean value (straightforward integration), arc length (using the standard identity 1 + sinh²x = cosh²x), and surface area of revolution (applying the formula with the arc length element). While it involves multiple parts and Further Maths content, each step follows directly from standard techniques without requiring novel insight or complex problem-solving. |
| Spec | 4.08e Mean value of function: using integral8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{M.V.} = \frac{\int_0^{\ln 5}\frac{1}{2}(e^x + e^{-x})\,dx}{\ln 5 - 0} = \frac{\left[\frac{1}{2}(e^x - e^{-x})\right]_0^{\ln 5}}{\ln 5}\) | M1A1 | Uses mean value formula and integrates |
| \(= \frac{\frac{1}{2}\left(5 - \frac{1}{5}\right)}{\ln 5} = \frac{12}{5\ln 5}\) (= 1.49) | M1A1 | Substitutes limits and evaluates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y' = \frac{1}{2}(e^x - e^{-x}) \Rightarrow 1 + (y')^2 = \left\{\frac{1}{2}(e^x + e^{-x})\right\}^2\) | M1A1 | Differentiates and finds \(1+(y')^2\) |
| \(s = \frac{1}{2}\int_0^{\ln 5}(e^x + e^{-x})\,dx = \frac{1}{2}\left[e^x - e^{-x}\right]_0^{\ln 5}\) | M1A1 | Integrates and obtains result |
| \(= \frac{1}{2}\left[5 - \frac{1}{5}\right] = \frac{12}{5}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(S = 2\pi\int_0^{\ln 5}\frac{1}{2}(e^x+e^{-x})\cdot\frac{1}{2}(e^x+e^{-x})\,dx\) | M1 | Uses surface area formula |
| \(= \frac{\pi}{2}\int_0^{\ln 5}(e^{2x} + 2 + e^{-2x})\,dx\) | M1 | Integrates |
| \(= \frac{\pi}{2}\left[\frac{e^{2x}}{2} + 2x - \frac{e^{-2x}}{2}\right]_0^{\ln 5}\) | ||
| \(= \frac{\pi}{2}\left\{\left[\frac{25}{2} + 2\ln 5 - \frac{1}{50}\right] - \left[\frac{1}{2} + 0 - \frac{1}{2}\right]\right\}\) | A1 | Substitutes limits |
| \(= \pi\left(\frac{156}{25} + \ln 5\right)\) (= 24.7) | A1 | Evaluates |
# Question 9:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{M.V.} = \frac{\int_0^{\ln 5}\frac{1}{2}(e^x + e^{-x})\,dx}{\ln 5 - 0} = \frac{\left[\frac{1}{2}(e^x - e^{-x})\right]_0^{\ln 5}}{\ln 5}$ | M1A1 | Uses mean value formula and integrates |
| $= \frac{\frac{1}{2}\left(5 - \frac{1}{5}\right)}{\ln 5} = \frac{12}{5\ln 5}$ (= 1.49) | M1A1 | Substitutes limits and evaluates | **Part total: 4** |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y' = \frac{1}{2}(e^x - e^{-x}) \Rightarrow 1 + (y')^2 = \left\{\frac{1}{2}(e^x + e^{-x})\right\}^2$ | M1A1 | Differentiates and finds $1+(y')^2$ |
| $s = \frac{1}{2}\int_0^{\ln 5}(e^x + e^{-x})\,dx = \frac{1}{2}\left[e^x - e^{-x}\right]_0^{\ln 5}$ | M1A1 | Integrates and obtains result |
| $= \frac{1}{2}\left[5 - \frac{1}{5}\right] = \frac{12}{5}$ | | | **Part total: 4** |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S = 2\pi\int_0^{\ln 5}\frac{1}{2}(e^x+e^{-x})\cdot\frac{1}{2}(e^x+e^{-x})\,dx$ | M1 | Uses surface area formula |
| $= \frac{\pi}{2}\int_0^{\ln 5}(e^{2x} + 2 + e^{-2x})\,dx$ | M1 | Integrates |
| $= \frac{\pi}{2}\left[\frac{e^{2x}}{2} + 2x - \frac{e^{-2x}}{2}\right]_0^{\ln 5}$ | | |
| $= \frac{\pi}{2}\left\{\left[\frac{25}{2} + 2\ln 5 - \frac{1}{50}\right] - \left[\frac{1}{2} + 0 - \frac{1}{2}\right]\right\}$ | A1 | Substitutes limits |
| $= \pi\left(\frac{156}{25} + \ln 5\right)$ (= 24.7) | A1 | Evaluates | **Part total: 4** |
---9 The curve $C$ has equation $y = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } \right)$ for $0 \leqslant x \leqslant \ln 5$. Find\\
(i) the mean value of $y$ with respect to $x$ over the interval $0 \leqslant x \leqslant \ln 5$,\\
(ii) the arc length of $C$,\\
(iii) the surface area generated when $C$ is rotated through $2 \pi$ radians about the $x$-axis.
\hfill \mbox{\textit{CAIE FP1 2011 Q9 [12]}}