| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Differentiate inverse hyperbolic functions |
| Difficulty | Standard +0.8 This is a Further Maths question requiring differentiation of inverse hyperbolic functions using the product rule, then applying integration by parts (recognizing the 'hence' connection). While the techniques are standard for FP3, inverse hyperbolic functions are inherently more advanced than core A-level content, and the integration requires careful algebraic manipulation to reach the exact form requested. |
| Spec | 1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{arsinh}\, 2x + x \cdot \dfrac{2}{\sqrt{1+4x^2}}\) | M1 A1, A1 | Differentiating getting an arsinh term and a term of form \(\frac{px}{\sqrt{1\pm qx^2}}\); cao \(\text{arsinh}\, 2x\); cao \(+\frac{2x}{\sqrt{1+4x^2}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^{\sqrt{2}}\text{arsinh}\, 2x\, dx = \left[x\,\text{arsinh}\, 2x\right]_0^{\sqrt{2}} - \int_0^{\sqrt{2}}\frac{2x}{\sqrt{1+4x^2}}dx\) | 1M1 1A1ft | Rearranging answer to (a) OR setting up parts; ft from (a) OR setting up parts correctly |
| \(= \left[x\,\text{arsinh}\, 2x\right]_0^{\sqrt{2}} - \left[\frac{1}{2}(1+4x^2)^{\frac{1}{2}}\right]_0^{\sqrt{2}}\) | 2M1 2A1 | Integrating getting an arsinh or arcosh term and a \((1\pm ax^2)^{\frac{1}{2}}\) term; cao |
| \(= \sqrt{2}\,\text{arsinh}\,2\sqrt{2} - \left[\frac{3}{2} - \frac{1}{2}\right]\) | 3DM1 | Depends on previous M, correct use of \(\sqrt{2}\) and \(0\) as limits |
| \(= \sqrt{2}\ln(3 + 2\sqrt{2}) - 1\) | 4M1 3A1 | Converting to log form; cao depends on all previous M marks |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{arsinh}\, 2x + x \cdot \dfrac{2}{\sqrt{1+4x^2}}$ | M1 A1, A1 | Differentiating getting an arsinh term and a term of form $\frac{px}{\sqrt{1\pm qx^2}}$; cao $\text{arsinh}\, 2x$; cao $+\frac{2x}{\sqrt{1+4x^2}}$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^{\sqrt{2}}\text{arsinh}\, 2x\, dx = \left[x\,\text{arsinh}\, 2x\right]_0^{\sqrt{2}} - \int_0^{\sqrt{2}}\frac{2x}{\sqrt{1+4x^2}}dx$ | 1M1 1A1ft | Rearranging answer to (a) OR setting up parts; ft from (a) OR setting up parts correctly |
| $= \left[x\,\text{arsinh}\, 2x\right]_0^{\sqrt{2}} - \left[\frac{1}{2}(1+4x^2)^{\frac{1}{2}}\right]_0^{\sqrt{2}}$ | 2M1 2A1 | Integrating getting an arsinh or arcosh term and a $(1\pm ax^2)^{\frac{1}{2}}$ term; cao |
| $= \sqrt{2}\,\text{arsinh}\,2\sqrt{2} - \left[\frac{3}{2} - \frac{1}{2}\right]$ | 3DM1 | Depends on previous M, correct use of $\sqrt{2}$ and $0$ as limits |
| $= \sqrt{2}\ln(3 + 2\sqrt{2}) - 1$ | 4M1 3A1 | Converting to log form; cao depends on all previous M marks |
---\begin{enumerate}
\item (a) Differentiate $x \operatorname { arsinh } 2 x$ with respect to $x$.\\
(b) Hence, or otherwise, find the exact value of
\end{enumerate}
$$\int _ { 0 } ^ { \sqrt { 2 } } \operatorname { arsinh } 2 x \mathrm {~d} x$$
giving your answer in the form $A \ln B + C$, where $A , B$ and $C$ are real.
\hfill \mbox{\textit{Edexcel FP3 2012 Q5 [10]}}