| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using sech/tanh identities |
| Difficulty | Challenging +1.3 This is a multi-part Further Maths question requiring manipulation of hyperbolic functions and their inverses. Part (a) is algebraic manipulation with surds (routine), part (b) requires deriving the inverse hyperbolic secant formula using exponential definitions (standard FP3 technique), and part (c) involves solving a hyperbolic equation using identities and substitution. While it requires multiple techniques and careful algebra across 13 marks, these are all standard FP3 procedures without requiring novel insight—moderately harder than average A-level but typical for Further Maths Pure. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07e Inverse hyperbolic: definitions, domains, ranges |
\begin{enumerate}[label=(\alph*)]
\item Show that, for $0 < x \leq 1$,
$$\ln \left(\frac{1 - \sqrt{1-x^2}}{x}\right) = -\ln \left(\frac{1 + \sqrt{1-x^2}}{x}\right).$$ [3]
\item Using the definition of $\cosh x$ or $\operatorname{sech} x$ in terms of exponentials, show that, for $0 < x \leq 1$,
$$\operatorname{arsech} x = \ln \left(\frac{1 + \sqrt{1-x^2}}{x}\right).$$ [5]
\item Solve the equation
$$3 \tanh^2 x - 4 \operatorname{sech} x + 1 = 0,$$
giving exact answers in terms of natural logarithms. [5]
\end{enumerate}
(Total 13 marks)
\hfill \mbox{\textit{Edexcel FP3 Q34 [13]}}