| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2018 |
| Session | December |
| Marks | 9 |
| Topic | Hyperbolic functions |
| Type | Prove inverse hyperbolic logarithmic form |
| Difficulty | Standard +0.8 This is a standard Further Maths question on inverse hyperbolic functions requiring algebraic manipulation of exponentials and solving a quadratic in tanh x. Part (a) is bookwork derivation, part (b) applies it, and part (c) requires understanding of the tanh range. Multi-step but follows established techniques without requiring novel insight. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}} = u\); \(\Rightarrow e^x(1-u) = e^{-x}(1+u)\); \(\Rightarrow e^{2x} = \left(\frac{1+u}{1-u}\right) \Rightarrow x = \frac{1}{2}\ln\left(\frac{1+u}{1-u}\right)\) | M1, A1 | Use of exponentials; Attempt to find \(e^{2x}\) |
| (b) \(4\tanh^2 x + \tanh x - 3 = 0\); \(\Rightarrow (4u-3)(u+1) = 0 \Rightarrow u = \frac{3}{4},(-1)\) | M1, A1 | Solve quadratic |
| \(\Rightarrow x = \frac{1}{2}\ln\left(\frac{1+\frac{3}{4}}{1-\frac{3}{4}}\right) = \frac{1}{2}\ln 7\) oe | M1, A1 | Substitute into the standard form |
| i.e. \(a = \frac{1}{2}, b = 7\) oe | A1 | |
| (c) E.g. Because \(-1 < \tanh x < 1\) | B1 | e.g. One root has been rejected as outside range. |
**(a)** $\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}} = u$; $\Rightarrow e^x(1-u) = e^{-x}(1+u)$; $\Rightarrow e^{2x} = \left(\frac{1+u}{1-u}\right) \Rightarrow x = \frac{1}{2}\ln\left(\frac{1+u}{1-u}\right)$ | M1, A1 | Use of exponentials; Attempt to find $e^{2x}$
**(b)** $4\tanh^2 x + \tanh x - 3 = 0$; $\Rightarrow (4u-3)(u+1) = 0 \Rightarrow u = \frac{3}{4},(-1)$ | M1, A1 | Solve quadratic
$\Rightarrow x = \frac{1}{2}\ln\left(\frac{1+\frac{3}{4}}{1-\frac{3}{4}}\right) = \frac{1}{2}\ln 7$ oe | M1, A1 | Substitute into the standard form
i.e. $a = \frac{1}{2}, b = 7$ oe | A1 |
**(c)** E.g. Because $-1 < \tanh x < 1$ | B1 | e.g. One root has been rejected as outside range.
---8
\begin{enumerate}[label=(\alph*)]
\item Given that $u = \tanh x$, use the definition of $\tanh x$ in terms of exponentials to show that
$$x = \frac { 1 } { 2 } \ln \left( \frac { 1 + u } { 1 - u } \right)$$
\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.
\item Explain why the equation in part (b) has only one root.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q8 [9]}}