| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using substitution u = cosh x or u = sinh x |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing understanding of substitution and hyperbolic functions. Part (a) requires recognizing that the sum of roots formula applies to sinh θ values, not θ values directly (conceptual understanding). Part (b) involves solving a quadratic in sinh θ, then using arsinh and log laws to combine the answers—all standard techniques for this topic with no novel insight required. Easier than average A-level due to being routine application of learned methods. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The roots of the quadratic are \(\sinh\theta_1\) and \(\sinh\theta_2\), so \(\sinh\theta_1 + \sinh\theta_2 = 1\) | E1 | The roots of the quadratic are not the solutions of the equation; the roots of the equation are not \(\theta\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((\sinh\theta - 2)(\sinh\theta + 1) = 0\) so \(\sinh\theta = 2\) or \(\sinh\theta = -1\) | M1 | Finds the roots of the quadratic. PI by two of \(1.44\) or \(-0.88\) or \(0.56\) or better |
| \(\theta_1 = \ln(2 + \sqrt{5})\) and \(\theta_2 = \ln(-1 + \sqrt{2})\) | A1 | Selects a method to find a value of \(\theta_1\) or \(\theta_2\). PI by \(1.44\) or \(-0.88\) or \(0.56\) or better. Condone an incorrect base. Obtains the correct exact values of \(\theta_1\) and \(\theta_2\) in log form. PI by correct sum in log form. May be unsimplified. |
| \(\theta_1 + \theta_2 = \ln(2+\sqrt{5}) + \ln(-1+\sqrt{2})\) | M1 | Correctly changes the sum of two log expressions into one log. Condone an incorrect base. |
| \(= \ln\!\Big((2+\sqrt{5})(-1+\sqrt{2})\Big)\) | A1 | Obtains the correct sum as a single log. ACF |
## Question 15(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| The roots of the quadratic are $\sinh\theta_1$ and $\sinh\theta_2$, so $\sinh\theta_1 + \sinh\theta_2 = 1$ | E1 | The roots of the quadratic are not the solutions of the equation; the roots of the equation are not $\theta$ |
**Subtotal: 1 mark**
---
## Question 15(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(\sinh\theta - 2)(\sinh\theta + 1) = 0$ so $\sinh\theta = 2$ or $\sinh\theta = -1$ | M1 | Finds the roots of the quadratic. PI by two of $1.44$ or $-0.88$ or $0.56$ or better |
| $\theta_1 = \ln(2 + \sqrt{5})$ and $\theta_2 = \ln(-1 + \sqrt{2})$ | A1 | Selects a method to find a value of $\theta_1$ **or** $\theta_2$. PI by $1.44$ or $-0.88$ or $0.56$ or better. Condone an incorrect base. Obtains the correct exact values of $\theta_1$ and $\theta_2$ in log form. PI by correct sum in log form. May be unsimplified. |
| $\theta_1 + \theta_2 = \ln(2+\sqrt{5}) + \ln(-1+\sqrt{2})$ | M1 | Correctly changes the sum of two log expressions into one log. Condone an incorrect base. |
| $= \ln\!\Big((2+\sqrt{5})(-1+\sqrt{2})\Big)$ | A1 | Obtains the correct sum as a single log. ACF |
**Subtotal: 5 marks**
**Question total: 6 marks**
**Question Paper total: 80 marks**15 The two values of $\theta$ that satisfy the equation
$$\sinh ^ { 2 } \theta - \sinh \theta - 2 = 0$$
are $\theta _ { 1 }$ and $\theta _ { 2 }$\\
15
\begin{enumerate}[label=(\alph*)]
\item Hamzah is asked to find the value of $\theta _ { 1 } + \theta _ { 2 }$\\
He writes his answer as follows:\\
The quadratic coefficients are $a = 1 , b = - 1 , c = - 2$\\
The sum of the roots is $- \frac { b } { a }$\\
So $\theta _ { 1 } + \theta _ { 2 } = - \frac { - 1 } { 1 } = 1$\\
Explain Hamzah's error.\\[0pt]
[1 mark]
15
\item Find the correct value of $\theta _ { 1 } + \theta _ { 2 }$
Give your answer as a single logarithm.\\
\includegraphics[max width=\textwidth, alt={}, center]{fd9715c4-9ce1-4608-aed6-f3d4f71208b5-28_2492_1721_217_150}
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2022 Q15 [6]}}