| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Topic | Hyperbolic functions |
| Type | Solve using sech/tanh identities |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard hyperbolic function techniques. Part (a) is routine algebraic manipulation from the definition, part (b) is a quadratic in tanh x requiring the inverse formula from (a), and part (c) asks for a simple observation about the range of tanh. While it's Further Maths content, the question follows a predictable structure with no novel insights required, making it slightly easier than average overall. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07e Inverse hyperbolic: definitions, domains, ranges4.07f Inverse hyperbolic: logarithmic forms |
\begin{enumerate}[label=(\alph*)]
\item Given that $u = \tanh x$, use the definition of $\tanh x$ in terms of exponentials to show that
$$x = \frac{1}{2}\ln\left(\frac{1+u}{1-u}\right).$$
[4]
\item Solve the equation $4\tanh^2 x + \tanh x - 3 = 0$, giving the solution in the form $a\ln b$ where $a$ and $b$ are rational numbers to be determined. [4]
\item Explain why the equation in part (b) has only one root. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 1 2021 Q4 [9]}}