| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Prove inverse hyperbolic logarithmic form |
| Difficulty | Standard +0.8 This is a standard Further Maths FP2 question on hyperbolic functions requiring algebraic manipulation to derive the inverse tanh formula, differentiation using the chain rule or logarithmic differentiation, and integration by parts with careful algebraic simplification. While it involves multiple techniques and careful algebra, these are well-practiced procedures for FP2 students with no novel insights required, placing it moderately above average difficulty. |
| Spec | 4.07f Inverse hyperbolic: logarithmic forms4.08g Derivatives: inverse trig and hyperbolic functions |
\begin{enumerate}[label=(\alph*)]
\item Using the definition $\tanh y = \frac{\text{e}^y - \text{e}^{-y}}{\text{e}^y + \text{e}^{-y}}$, show that, for $|x| < 1$,
$$\tanh^{-1} x = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)$$ [3 marks]
\item Hence, or otherwise, show that $\frac{\text{d}}{\text{d}x}(\tanh^{-1} x) = \frac{1}{1-x^2}$. [3 marks]
\item Use integration by parts to show that
$$\int_{0}^{\frac{1}{4}} \tanh^{-1} x \, \text{d}x = \ln \left(\frac{3^m}{2^n}\right)$$
where $m$ and $n$ are positive integers. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2013 Q5 [11]}}