| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Integration by parts with inverse trig |
| Difficulty | Challenging +1.3 This is a structured Further Maths question with clear guidance ('hence or otherwise'). Part (a) is routine differentiation of inverse hyperbolic functions using the chain rule. Part (b) requires recognizing the derivative from part (a) in the integrand and applying limits, which is a standard technique once the connection is spotted. The algebraic manipulation to reach the logarithmic form requires care but follows established methods for inverse hyperbolic functions. While this is Further Maths content (inherently harder), the question structure provides significant scaffolding, making it moderately above average difficulty rather than highly challenging. |
| Spec | 1.08h Integration by substitution4.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} \times \frac{1}{\sqrt{1+(\sqrt{x})^2}}\) | B1, M1 | |
| \(\therefore \frac{dy}{dx} = \frac{\frac{1}{2}x^{-\frac{1}{2}}}{\sqrt{1+x}} \left(= \frac{1}{2\sqrt{x(1+x)}}\right)\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\therefore \int_{\frac{1}{4}}^{4} \frac{1}{\sqrt{x(x+1)}}dx = \left[2\,\text{arsinh}\sqrt{x}\right]_{\frac{1}{4}}^{4}\) | M1 | |
| \(= \left[2\,\text{arsinh}\,2 - 2\,\text{arsinh}\left(\frac{1}{2}\right)\right]\) | M1 | |
| \(= \left[2\ln(2+\sqrt{5})\right] - \left[2\ln\left(\frac{1}{2}+\sqrt{\frac{5}{4}}\right)\right]\) | M1 | |
| \(2\ln\frac{(2+\sqrt{5})}{(\frac{1}{2}+\sqrt{(\frac{5}{4})})} = 2\ln\frac{2(2+\sqrt{5})}{(1+\sqrt{5})} = 2\ln\frac{(\sqrt{5}+2)(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)} = 2\ln\frac{(3+\sqrt{5})}{2}\) | M1 | |
| \(= \ln\frac{(3+\sqrt{5})(3+\sqrt{5})}{4} = \ln\frac{14+6\sqrt{5}}{4} = \ln\left(\frac{7}{2}+\frac{3\sqrt{5}}{2}\right)\) | A1 A1 | (6) [9] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use \(\sinh y = \sqrt{x}\) and state \(\cosh y \frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}}\) | B1 | |
| \(\therefore \frac{dy}{dx} = \frac{\frac{1}{2}x^{-\frac{1}{2}}}{\sqrt{1+\sinh^2 y}} = \frac{\frac{1}{2}x^{-\frac{1}{2}}}{\sqrt{1+(\sqrt{x})^2}}\) | M1 | |
| \(\therefore \frac{dy}{dx} = \frac{\frac{1}{2}x^{-\frac{1}{2}}}{\sqrt{1+x}}\left(=\frac{1}{2\sqrt{x(1+x)}}\right)\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use \(x=\tan^2\theta\), \(\frac{dx}{d\theta}=2\tan\theta\sec^2\theta\) to give \(2\int\sec\theta\,d\theta = [2\ln(\sec\theta+\tan\theta)]\) | M1 | |
| \(= [2\ln(\sec\theta+\tan\theta)]_{\tan\theta=\frac{1}{2}}^{\tan\theta=2}\) i.e. use of limits | M1 | |
| then proceed as before from line 3 of scheme |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use \(\int \frac{1}{\sqrt{[(x+\frac{1}{2})^2-\frac{1}{4}]}}dx = \text{arcosh}\frac{x+\frac{1}{2}}{\frac{1}{2}}\) | M1 | |
| \(= \left[\text{arcosh}\,9 - \text{arcosh}\left(\frac{3}{2}\right)\right]\) | M1 | |
| \(= \left[\ln(9+\sqrt{80})\right] - \left[\ln\left(\frac{3}{2}+\frac{1}{2}\sqrt{5}\right)\right]\) | M1 | |
| \(= \ln\frac{(9+\sqrt{80})}{(\frac{3}{2}+\frac{1}{2}\sqrt{5})} = \ln\frac{2(9+\sqrt{80})(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}\) | M1 | |
| \(= \ln\frac{2(9+4\sqrt{5})(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})} = \ln\left(\frac{7}{2}+\frac{3\sqrt{5}}{2}\right)\) | A1 A1 | (6) [9] |
# Question 4:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} \times \frac{1}{\sqrt{1+(\sqrt{x})^2}}$ | B1, M1 | |
| $\therefore \frac{dy}{dx} = \frac{\frac{1}{2}x^{-\frac{1}{2}}}{\sqrt{1+x}} \left(= \frac{1}{2\sqrt{x(1+x)}}\right)$ | A1 | (3) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\therefore \int_{\frac{1}{4}}^{4} \frac{1}{\sqrt{x(x+1)}}dx = \left[2\,\text{arsinh}\sqrt{x}\right]_{\frac{1}{4}}^{4}$ | M1 | |
| $= \left[2\,\text{arsinh}\,2 - 2\,\text{arsinh}\left(\frac{1}{2}\right)\right]$ | M1 | |
| $= \left[2\ln(2+\sqrt{5})\right] - \left[2\ln\left(\frac{1}{2}+\sqrt{\frac{5}{4}}\right)\right]$ | M1 | |
| $2\ln\frac{(2+\sqrt{5})}{(\frac{1}{2}+\sqrt{(\frac{5}{4})})} = 2\ln\frac{2(2+\sqrt{5})}{(1+\sqrt{5})} = 2\ln\frac{(\sqrt{5}+2)(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)} = 2\ln\frac{(3+\sqrt{5})}{2}$ | M1 | |
| $= \ln\frac{(3+\sqrt{5})(3+\sqrt{5})}{4} = \ln\frac{14+6\sqrt{5}}{4} = \ln\left(\frac{7}{2}+\frac{3\sqrt{5}}{2}\right)$ | A1 A1 | (6) **[9]** |
## Alternative (i) for part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use $\sinh y = \sqrt{x}$ and state $\cosh y \frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}}$ | B1 | |
| $\therefore \frac{dy}{dx} = \frac{\frac{1}{2}x^{-\frac{1}{2}}}{\sqrt{1+\sinh^2 y}} = \frac{\frac{1}{2}x^{-\frac{1}{2}}}{\sqrt{1+(\sqrt{x})^2}}$ | M1 | |
| $\therefore \frac{dy}{dx} = \frac{\frac{1}{2}x^{-\frac{1}{2}}}{\sqrt{1+x}}\left(=\frac{1}{2\sqrt{x(1+x)}}\right)$ | A1 | (3) |
## Alternative (i) for part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use $x=\tan^2\theta$, $\frac{dx}{d\theta}=2\tan\theta\sec^2\theta$ to give $2\int\sec\theta\,d\theta = [2\ln(\sec\theta+\tan\theta)]$ | M1 | |
| $= [2\ln(\sec\theta+\tan\theta)]_{\tan\theta=\frac{1}{2}}^{\tan\theta=2}$ i.e. use of limits | M1 | |
| then proceed as before from line 3 of scheme | | |
## Alternative (ii) for part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use $\int \frac{1}{\sqrt{[(x+\frac{1}{2})^2-\frac{1}{4}]}}dx = \text{arcosh}\frac{x+\frac{1}{2}}{\frac{1}{2}}$ | M1 | |
| $= \left[\text{arcosh}\,9 - \text{arcosh}\left(\frac{3}{2}\right)\right]$ | M1 | |
| $= \left[\ln(9+\sqrt{80})\right] - \left[\ln\left(\frac{3}{2}+\frac{1}{2}\sqrt{5}\right)\right]$ | M1 | |
| $= \ln\frac{(9+\sqrt{80})}{(\frac{3}{2}+\frac{1}{2}\sqrt{5})} = \ln\frac{2(9+\sqrt{80})(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}$ | M1 | |
| $= \ln\frac{2(9+4\sqrt{5})(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})} = \ln\left(\frac{7}{2}+\frac{3\sqrt{5}}{2}\right)$ | A1 A1 | (6) **[9]** |
---\begin{enumerate}
\item Given that $y = \operatorname { arsinh } ( \sqrt { } x ) , x > 0$,\\
(a) find $\frac { \mathrm { d } y } { \mathrm {~d} x }$, giving your answer as a simplified fraction.\\
(b) Hence, or otherwise, find
\end{enumerate}
$$\int _ { \frac { 1 } { 4 } } ^ { 4 } \frac { 1 } { \sqrt { [ x ( x + 1 ) ] } } \mathrm { d } x$$
giving your answer in the form $\ln \left( \frac { a + b \sqrt { } 5 } { 2 } \right)$, where $a$ and $b$ are integers.\\
\hfill \mbox{\textit{Edexcel FP3 2009 Q4 [9]}}