| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using double angle formulas |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on hyperbolic functions requiring multiple techniques: deriving an inverse formula from definitions, proving an identity using double/triple angle formulas, and solving an equation. While it involves several steps and Further Maths content (inherently harder), each part follows standard methods with clear guidance, making it moderately challenging but not requiring novel insight. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sinh x = \frac{e^x - e^{-x}}{2}\), \(\cosh x = \frac{e^x + e^{-x}}{2}\) | B1 | Uses definitions of \(\sinh x\) and \(\cosh x\) to obtain expression for \(\tanh x\) |
| \(\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}\); multiplies by \(e^x\) | M1 | Multiplies by \(e^x\) |
| Obtains \(e^{2x}\) | A1 | |
| \(t = \frac{e^{2x}-1}{e^{2x}+1}\), so \(te^{2x} + t = e^{2x} - 1\), giving \(1+t = e^{2x}(1-t)\), \(e^{2x} = \frac{1+t}{1-t}\), \(2x = \ln\frac{1+t}{1-t}\), hence \(x = \frac{1}{2}\ln\frac{1+t}{1-t}\) | R1 | Completes fully correct argument by taking logs; available only if all previous marks awarded |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| To prove: \(\left(\frac{e^x+e^{-x}}{2}\right)^3 = \frac{1}{4}\left(\frac{e^{3x}+e^{-3x}}{2}\right) + \frac{3}{4}\left(\frac{e^x+e^{-x}}{2}\right)\); expresses \(\cosh 3x\) and \(\cosh x\) in exponential form | B1 | Seen anywhere in solution |
| LHS \(= \left(\frac{e^x+e^{-x}}{2}\right)^3 = \frac{1}{8}(e^{3x} + 3e^{2x}\cdot e^{-x} + 3e^x\cdot e^{-2x} + e^{-3x})\) | M1 | Expands LHS; FT 'their' LHS provided first M1 awarded; allow one slip |
| \(= \frac{1}{4}\left(\frac{e^{3x}+e^{-3x}}{2}\right) + 3\frac{(e^x+e^{-x})}{2}\); simplifies and collects terms | M1 | Allow one slip |
| \(= \frac{1}{4}\cosh 3x + \frac{3}{4}\cosh x =\) RHS | R1 | Completes fully correct proof; available only if all previous marks awarded |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((\cosh x)^3 = \frac{1}{4}\times 13\cosh x + \frac{3}{4}\cosh x\); substitutes for \(\cosh 3x\) from (b)(i) | M1 | Allow one slip |
| \((\cosh x)^3 - 4\cosh x = 0\); \(\cosh x\left[(\cosh x)^2 - 4\right] = 0\) | M1 | Obtains equation in \(\cosh x\) and solves it; allow one slip |
| Solutions are \(\cosh x = 0, -2, 2\); solutions \(0\) and \(-2\) are not possible since range of \(\cosh x \geq 1\) | E1 | Eliminates \(0\) and \(-2\) with reason |
| \(\cosh x = 2 \Rightarrow x = \ln(2+\sqrt{3})\) | A1 | States correct solution in exact log form |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sinh x = \frac{e^x - e^{-x}}{2}$, $\cosh x = \frac{e^x + e^{-x}}{2}$ | B1 | Uses definitions of $\sinh x$ and $\cosh x$ to obtain expression for $\tanh x$ |
| $\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$; multiplies by $e^x$ | M1 | Multiplies by $e^x$ |
| Obtains $e^{2x}$ | A1 | |
| $t = \frac{e^{2x}-1}{e^{2x}+1}$, so $te^{2x} + t = e^{2x} - 1$, giving $1+t = e^{2x}(1-t)$, $e^{2x} = \frac{1+t}{1-t}$, $2x = \ln\frac{1+t}{1-t}$, hence $x = \frac{1}{2}\ln\frac{1+t}{1-t}$ | R1 | Completes fully correct argument by taking logs; available only if all previous marks awarded |
---
## Question 6(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| To prove: $\left(\frac{e^x+e^{-x}}{2}\right)^3 = \frac{1}{4}\left(\frac{e^{3x}+e^{-3x}}{2}\right) + \frac{3}{4}\left(\frac{e^x+e^{-x}}{2}\right)$; expresses $\cosh 3x$ and $\cosh x$ in exponential form | B1 | Seen anywhere in solution |
| LHS $= \left(\frac{e^x+e^{-x}}{2}\right)^3 = \frac{1}{8}(e^{3x} + 3e^{2x}\cdot e^{-x} + 3e^x\cdot e^{-2x} + e^{-3x})$ | M1 | Expands LHS; FT 'their' LHS provided first M1 awarded; allow one slip |
| $= \frac{1}{4}\left(\frac{e^{3x}+e^{-3x}}{2}\right) + 3\frac{(e^x+e^{-x})}{2}$; simplifies and collects terms | M1 | Allow one slip |
| $= \frac{1}{4}\cosh 3x + \frac{3}{4}\cosh x =$ RHS | R1 | Completes fully correct proof; available only if all previous marks awarded |
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## Question 6(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\cosh x)^3 = \frac{1}{4}\times 13\cosh x + \frac{3}{4}\cosh x$; substitutes for $\cosh 3x$ from (b)(i) | M1 | Allow one slip |
| $(\cosh x)^3 - 4\cosh x = 0$; $\cosh x\left[(\cosh x)^2 - 4\right] = 0$ | M1 | Obtains equation in $\cosh x$ and solves it; allow one slip |
| Solutions are $\cosh x = 0, -2, 2$; solutions $0$ and $-2$ are not possible since range of $\cosh x \geq 1$ | E1 | Eliminates $0$ and $-2$ with reason |
| $\cosh x = 2 \Rightarrow x = \ln(2+\sqrt{3})$ | A1 | States correct solution in exact log form |
---6
\begin{enumerate}[label=(\alph*)]
\item Use the definitions of $\sinh x$ and $\cosh x$ in terms of $\mathrm { e } ^ { x }$ and $\mathrm { e } ^ { - x }$ to show that $x = \frac { 1 } { 2 } \ln \left( \frac { 1 + t } { 1 - t } \right)$ where $t = \tanh x$\\[0pt]
[4 marks]\\
6
\item (i) Prove $\cosh ^ { 3 } x = \frac { 1 } { 4 } \cosh 3 x + \frac { 3 } { 4 } \cosh x$\\[0pt]
[4 marks]
6 (b) (ii) Show that the equation $\cosh 3 x = 13 \cosh x$ has only one positive solution.\\
Find this solution in exact logarithmic form.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 Q6 [12]}}