A2 June 2022 Paper 1 Q9
9 The function \(\mathrm{f}(x)\) is defined by \(\mathrm{f}(x) = \ln(1 + \sinh x)\).
(a) Given that \(k\) lies in the domain of this function, explain why \(k\) must be greater than \(\ln\left(\sqrt{2} - 1\right)\). [2]
(b)
(i) Find \(\mathrm{f}'(x)\). [2]
(ii) Show that \(\mathrm{f}''(x) = \dfrac{a\sinh x + b}{(1 + \sinh x)^2}\), where \(a\) and \(b\) are integers to be determined. [3]
(c) Hence find a quadratic approximation to \(\mathrm{f}(x)\) for small values of \(x\). [3]
(d) Find the percentage error in this approximation when \(x = 0.1\). [2]
| Scheme | Marks | AO |
|---|---|---|
| \(\sinh k \gt -1\) \(\Rightarrow k \gt \sinh^{-1}(-1) = \ln\left(-1 + \sqrt{2}\right)\) | M1 A1 | 2.1 2.2a |
| [2] |
Notes
M1: May be in exponential form
A1: AG
| Scheme | Marks | AO |
|---|---|---|
| (i) \(f'(x) = \dfrac{\cosh x}{1 + \sinh x}\) | M1 A1 | 1.1 1.1 |
| [2] |
Notes
M1: chain rule
| Scheme | Marks | AO |
|---|---|---|
| (ii) \(f''(x) = \dfrac{(1 + \sinh x)\sinh x - \cosh^2 x}{(1 + \sinh x)^2}\) | M1 | 1.1 |
| \(f''(x) = \dfrac{\sinh x - 1}{(1 + \sinh x)^2}\) | M1 A1 | 3.1a 1.1 |
| [3] |
Notes
M1: quotient or product rule
M1: \(\cosh^2 x - \sinh^2 x = 1\) used
| Scheme | Marks | AO |
|---|---|---|
| \(f(0) = 0,\ f'(0) = 1,\ f''(0) = -1\) | B1ft | 1.1 |
| \(f(x) = x - \dfrac{1}{2}x^2\) | M1 A1cao | 1.1 1.1 |
| [3] |
Notes
B1ft: soi
M1: Maclaurin expansion attempted, must see their values substituted in
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\ln(1 + \sinh(0.1)) - 0.095}{\ln(1 + \sinh(0.1))} \times 100\) | M1 | 1.1 |
| \(= 0.48\%\) | A1 | 1.1 |
| [2] |
Notes
A1: Allow \(-0.48\%\)