| Exam Board | AQA |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Prove hyperbolic identity from exponentials |
| Difficulty | Challenging +1.2 Part (a) is straightforward substitution into a given identity. Part (b) requires recognizing a telescoping series pattern, which is a standard Further Maths technique but demands more insight than routine manipulation. The 'do not write' restriction suggests avoiding a simpler exponential approach, forcing the telescoping method. Overall, moderately above average difficulty for Further Maths. |
| Spec | 4.06b Method of differences: telescoping series4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\sinh(m+1)x = \sinh mx \cosh x + \cosh mx \sinh x\); \(\sinh(m-1)x = \sinh mx \cosh x - \cosh mx \sinh x\) | B1 (1.1b) | Correct expressions for both \(\sinh(m+1)x\) and \(\sinh(m-1)x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Subtracts \(\sinh(m+1)x\) and \(\sinh(m-1)x\) expressions | M1 (3.1a) | |
| \(\sinh(m+1)x - \sinh(m-1)x = 2\cosh mx \sinh x\); \(\sinh 2x - \sinh 0 = 2\cosh x \sinh x\); \(\sinh 3x - \sinh x = 2\cosh 2x \sinh x\); \(\sinh 4x - \sinh 2x = 2\cosh 3x \sinh x\); ... | M1 (3.1a) | Uses method of differences, at least first two terms shown |
| \(\sinh(n-1)x - \sinh(n-3)x = 2\cosh(n-2)x\sinh x\); \(\sinh nx - \sinh(n-2)x = 2\cosh(n-1)x\sinh x\); \(\sinh(n+1)x - \sinh(n-1)x = 2\cosh nx \sinh x\) | A1 (1.1b) | Correct terms, at least first three and last two |
| Deduces \(\sum 2\cosh mx \sinh x\) is a multiple of \(C_n\); \(\sinh(n+1)x + \sinh nx - \sinh x = 2\sinh x(\cosh x + \cosh 2x + \cdots + \cosh nx)\); \(\therefore C_n = \frac{\sinh(n+1)x + \sinh nx - \sinh x}{2\sinh x}\) | M1 (2.2a) | |
| Completes rigorous argument | R1 (2.1) |
## Question 14(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\sinh(m+1)x = \sinh mx \cosh x + \cosh mx \sinh x$; $\sinh(m-1)x = \sinh mx \cosh x - \cosh mx \sinh x$ | B1 (1.1b) | Correct expressions for both $\sinh(m+1)x$ and $\sinh(m-1)x$ |
## Question 14(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Subtracts $\sinh(m+1)x$ and $\sinh(m-1)x$ expressions | M1 (3.1a) | |
| $\sinh(m+1)x - \sinh(m-1)x = 2\cosh mx \sinh x$; $\sinh 2x - \sinh 0 = 2\cosh x \sinh x$; $\sinh 3x - \sinh x = 2\cosh 2x \sinh x$; $\sinh 4x - \sinh 2x = 2\cosh 3x \sinh x$; ... | M1 (3.1a) | Uses method of differences, at least first two terms shown |
| $\sinh(n-1)x - \sinh(n-3)x = 2\cosh(n-2)x\sinh x$; $\sinh nx - \sinh(n-2)x = 2\cosh(n-1)x\sinh x$; $\sinh(n+1)x - \sinh(n-1)x = 2\cosh nx \sinh x$ | A1 (1.1b) | Correct terms, at least first three and last two |
| Deduces $\sum 2\cosh mx \sinh x$ is a multiple of $C_n$; $\sinh(n+1)x + \sinh nx - \sinh x = 2\sinh x(\cosh x + \cosh 2x + \cdots + \cosh nx)$; $\therefore C_n = \frac{\sinh(n+1)x + \sinh nx - \sinh x}{2\sinh x}$ | M1 (2.2a) | |
| Completes rigorous argument | R1 (2.1) | |
**Total: 6 marks**
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**Paper Total: 100 marks**14
\begin{enumerate}[label=(\alph*)]
\item Given that
$$\sinh ( A + B ) = \sinh A \cosh B + \cosh A \sinh B$$
express $\sinh ( m + 1 ) x$ and $\sinh ( m - 1 ) x$ in terms of $\sinh m x , \cosh m x , \sinh x$ and $\cosh x$\\
14
\item Hence find the sum of the series
$$C _ { n } = \cosh x + \cosh 2 x + \cdots + \cosh n x$$
in terms of $\sinh x , \sinh n x$ and $\sinh ( n + 1 ) x$\\
Do not write\\
\includegraphics[max width=\textwidth, alt={}, center]{44e22a98-6424-4fb1-8a37-c965773cb7b6-30_2491_1736_219_139}
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 1 2020 Q14 [6]}}