A2 June 2025 Paper 2 Q10
10
- \(u = 2\)
- \(u = 5\)
- \(u = 2\)
- \(u = 5\)
A function f is defined for all integers \(n\) by \(\mathrm{f}(n) = \sinh(0.01n) - 5\cosh(0.005n) - 9\tanh n\).
With the help of suitable approximations, use an algebraic method to determine the smallest value of \(n\) for which \(\mathrm{f}(n) \gt 100\). You should verify your answer, once found, by direct calculation. You may assume that the required value of \(n\) is large. [6]
| Scheme | Marks | AO |
|---|---|---|
| \(\left(\dfrac{\frac{1}{2}\mathrm{e}^{-2}}{\sinh 2} \times 100\% =\right)\) awrt 1.87% \(\left(\dfrac{\frac{1}{2}\mathrm{e}^{-5}}{\sinh 5} \times 100\% =\right)\) awrt 0.00454% | B1 | 1.1 |
| [1] |
Notes
B1: For both. % symbol not necessary.
\(1.865736036\ldots\)
\(4.540199101\ldots \times 10^{-3}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\left(\dfrac{\cosh 2 - \sinh 2}{\sinh 2} \times 100\% =\right)\) awrt \(\pm\)3.73% \(\left(\dfrac{\cosh 5 - \sinh 5}{\sinh 5} \times 100\% =\right)\) awrt \(\pm\)0.00908% | B1 | 1.1 |
| [1] |
Notes
B1: For both. % symbol not necessary.
\(3.731472073\ldots\)
\(9.08039820 \times 10^{-3}\)
| Scheme | Marks | AO |
|---|---|---|
| DR (If \(n\) large then) \(\sinh(0.01n) \sim 0.5\mathrm{e}^{0.01n}\) or \(\tanh n \sim 1\) or \(\cosh(0.005n) \sim 0.5\mathrm{e}^{0.005n}\) | B1 | 3.1a |
| \(0.5\mathrm{e}^{0.01n} - 2.5\mathrm{e}^{0.005n} - 9 \gt 100\) | M1 | 2.2a |
| \(y = \mathrm{e}^{0.005n} \Rightarrow 0.5y^2 - 2.5y - 109 \gt 0\) \(y^2 - 5y - 218 \gt 0\) | M1* | 3.1a |
| \(\therefore \mathrm{e}^{0.005n} \gt 17.47\ldots\) | M1dep* | 2.2a |
| \(\therefore n \gt 200\ln 17.47\ldots = 572.15\ldots\) But \(n\) is an integer so \(n_{\min} = 573\) | A1 | 3.2a |
| \(\mathrm{f}(572) = 99.6\ldots \lt 100\) and \(\mathrm{f}(573) = 100.96\ldots \gt 100\) | A1 | 1.1 |
| [6] |
Notes
B1: One correct approximation soi.
M1: Correct use of approximations to derive an inequality in \(n\).
Accept equations in place of inequalities for all M marks.
M1*: Rearranging to a 3 term quadratic inequality in \(\mathrm{e}^{0.005n}\)
May see quadratic in \(\mathrm{e}^{0.005n}\).
M1dep*: Correctly solves their quadratic equation to find expression for \(\mathrm{e}^{0.005n}\)
\(17.47497913\ldots\) Or \(\left(\frac{5 + \sqrt{897}}{2}\right)\). Other root does not need to be explicitly rejected.
A1: Must see values, not just an assertion.
NB no marks for \(\mathrm{f}(572) = 99.6\ldots \lt 100\) and \(\mathrm{f}(573) = 100.96\ldots \gt 100\) without justification.
Alternative method
| Scheme | Marks |
|---|---|
| DR (If \(n\) large then) \(\sinh x \sim \cosh x\) or \(\tanh n \sim 1\) | B1 |
| \(\sinh 0.01n - 5\sinh 0.005n - 9 \gt 100\) | M1 |
| \(u = 0.005n \Rightarrow \sinh 2u - 5\sinh u - 9 \gt 100\) \(\Rightarrow 2\sinh u\cosh u - 5\sinh u - 9 \gt 100\) \(\Rightarrow 2\sinh^2 u - 5\sinh u - 9 \gt 100\) \(\Rightarrow 2\sinh^2 u - 5\sinh u - 109 \gt 0\) | M1* |
| \(\therefore \sinh u \gt 8.737\ldots\) \([\Rightarrow u \gt 2.864\ldots]\) | M1dep* |
| \(\therefore n \gt 200 \times 2.864\ldots = 572.8057\ldots\) But \(n\) is an integer so \(n_{\min} = 573\) | A1 |
| \(\mathrm{f}(572) = 99.6\ldots \lt 100\) and \(\mathrm{f}(573) = 100.96\ldots \gt 100\) | A1 |
B1: One correct approximation soi.
M1: Correct use of both approximations to derive an inequality in sinh or cosh only. May see \(\cosh 0.01n - 5\cosh 0.005n - 9\) and use of \(\cosh x\) throughout.
Accept equations in place of inequalities for all M marks.
M1*: Using an appropriate identity (e.g. \(\sinh 2x \equiv 2\sinh x\cosh x\)) and using approximation to rearrange to a 3 term quadratic inequality in sinh or cosh only.
May see quadratic in e.g. \(y = \sinh u = \sinh 0.005n\) i.e. \(2y^2 - 5y - 109 \gt 0\).
M1dep*: Correctly solves their quadratic equation to find expression for \(\cosh u\) or \(\sinh u\)
\(y \gt 8.737489\ldots\) or \(\frac{5 + \sqrt{897}}{4}\). Other root does not need to be explicitly rejected.
A1: Must see values, not just an assertion.
NB no marks for \(\mathrm{f}(572) = 99.6\ldots \lt 100\) and \(\mathrm{f}(573) = 100.96\ldots \gt 100\) without justification.