| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Session | Specimen |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Express in form R cosh(x±α) or R sinh(x±α) |
| Difficulty | Challenging +1.3 This is a structured Further Maths question on hyperbolic functions requiring proof of the inverse tanh formula, derivation of R-formula parameters using hyperbolic identities, and solving an equation. While it involves multiple steps and Further Maths content (inherently harder), the techniques are standard: part (a) uses definitions and logarithm laws, part (b) expands and equates coefficients, part (c) applies the derived result. The 17 marks reflect length rather than exceptional insight—this is a thorough but methodical question testing core hyperbolic function manipulation. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07f Inverse hyperbolic: logarithmic forms |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(y = \tanh^{-1}x\) so \(x = \tanh y\) | M1 | AO2 |
| \(= \frac{e^y - e^{-y}}{e^y + e^{-y}}\) | A1 | AO2 |
| \(xe^y + xe^{-y} = e^y - e^{-y}\) | A1 | AO2 |
| \(e^{2y} = \frac{1+x}{1-x}\) | A1 | AO2 |
| \(y = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\) | A1 | AO2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(a\cosh x + b\sinh x \equiv r\cosh(x+\alpha)\) | M1 | AO2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(r\cosh\alpha = a\) | A1 | AO2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tanh\alpha = \frac{b}{a}\) | M1 | AO2 |
| \(\alpha = \tanh^{-1}\left(\frac{b}{a}\right)\) | A1 | AO2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(r^2(\cosh^2\alpha - \sinh^2\alpha) = a^2 - b^2\) | M1 | AO1 |
| \(r = \sqrt{a^2 - b^2}\) | A1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| Here \(r = 3\) | B1 | AO1 |
| \(\alpha = \frac{1}{2}\ln 9 = \ln 3\) | B1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(3\cosh(x + \ln 3) = 10\) | B1 | AO1 |
| \(x + \ln 3 = (\pm)\cosh^{-1}\left(\frac{10}{3}\right)\) | M1 | AO1 |
| \(x = 0.775\) or \(x = -2.97\) | A1, A1 | AO1 |
## Part (a)
Let $y = \tanh^{-1}x$ so $x = \tanh y$ | M1 | AO2
$= \frac{e^y - e^{-y}}{e^y + e^{-y}}$ | A1 | AO2
$xe^y + xe^{-y} = e^y - e^{-y}$ | A1 | AO2
$e^{2y} = \frac{1+x}{1-x}$ | A1 | AO2
$y = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ | A1 | AO2
## Part (b)
$a\cosh x + b\sinh x \equiv r\cosh(x+\alpha)$ | M1 | AO2
$= r\cosh x \cosh\alpha + r\sinh x \sinh\alpha$
Equating like terms:
$r\cosh\alpha = a$ | A1 | AO2
$r\sinh\alpha = b$
Dividing:
$\tanh\alpha = \frac{b}{a}$ | M1 | AO2
$\alpha = \tanh^{-1}\left(\frac{b}{a}\right)$ | A1 | AO2
$= \frac{1}{2}\ln\left(\frac{1+b/a}{1-b/a}\right) = \frac{1}{2}\ln\left(\frac{a+b}{a-b}\right)$
Squaring and subtracting the above equations:
$r^2(\cosh^2\alpha - \sinh^2\alpha) = a^2 - b^2$ | M1 | AO1
$r = \sqrt{a^2 - b^2}$ | A1 | AO1
## Part (c)
Here $r = 3$ | B1 | AO1
$\alpha = \frac{1}{2}\ln 9 = \ln 3$ | B1 | AO1
The equation simplifies to:
$3\cosh(x + \ln 3) = 10$ | B1 | AO1
$x + \ln 3 = (\pm)\cosh^{-1}\left(\frac{10}{3}\right)$ | M1 | AO1
$x = 0.775$ or $x = -2.97$ | A1, A1 | AO1
**Total: [17]**
---\begin{enumerate}[label=(\alph*)]
\item Show that
$$\tanh^{-1} x = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right), \quad \text{where } -1 < x < 1.$$ [4]
\item Given that
$$a \cosh x + b \sinh x \equiv r \cosh(x + \alpha), \quad \text{where } a > b > 0,$$
show that
$$\alpha = \frac{1}{2} \ln \left(\frac{a+b}{a-b}\right)$$
and find an expression for $r$ in terms of $a$ and $b$. [7]
\item Hence solve the equation
$$5 \cosh x + 4 \sinh x = 10,$$
giving your answers correct to three significant figures. [6]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 Q11 [17]}}