A2 June 2023 Paper 2 Q3
3
| Scheme | Marks | AO |
|---|---|---|
| \(y = \sinh^{-1}u \Rightarrow \sinh y = u\) \(\Rightarrow \cosh y\dfrac{\mathrm{d}y}{\mathrm{d}u} = 1 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}u} = \dfrac{1}{\cosh y}\) \(\therefore \dfrac{\mathrm{d}y}{\mathrm{d}u} = \dfrac{1}{\pm\sqrt{1 + \sinh^2 y}} = \dfrac{1}{\pm\sqrt{u^2 + 1}}\) | M1 | 2.1 |
| But gradient of \(y = \sinh^{-1}u\) is never negative so \(\dfrac{\mathrm{d}y}{\mathrm{d}u} = \dfrac{1}{\sqrt{u^2 + 1}}\) | A1 | 2.4 |
| [2] |
Notes
M1: AG. Taking sinh of both sides, differentiating and using \(\cosh^2 y - \sinh^2 y = 1\)
Condone missing \(\pm\) for M1. Accept \(\frac{\mathrm{d}u}{\mathrm{d}y}\) at this stage
A1: AG so reason required (accept “always positive”); poor notation can be recovered
Alternative method
| Scheme | Marks |
|---|---|
| \(y = \sinh^{-1}u = \ln\left(u + \sqrt{u^2 + 1}\right)\) \(\therefore \dfrac{\mathrm{d}y}{\mathrm{d}u} = \dfrac{1 + \frac{1}{2} \times 2u\left(u^2 + 1\right)^{-\frac{1}{2}}}{u + \sqrt{u^2 + 1}}\) | M1 |
| \(= \dfrac{\left(u^2 + 1\right)^{\frac{1}{2}} + u}{\left(u^2 + 1\right)^{\frac{1}{2}}\left(u + \left(u^2 + 1\right)^{\frac{1}{2}}\right)} = \dfrac{1}{\left(u^2 + 1\right)^{\frac{1}{2}}} = \dfrac{1}{\sqrt{u^2 + 1}}\) | A1 |
M1: AG. Using the definition of \(\sinh^{-1}\) in logarithmic form and attempting to differentiate using the chain rule on ln function.
A1: AG so some intermediate working must be seen. www
Alternative method 2
| Scheme | Marks |
|---|---|
| \(I = \displaystyle\int\frac{1}{\sqrt{u^2 + 1}}\,\mathrm{d}u\) \(u = \sinh v \Rightarrow \mathrm{d}u = \cosh v\,\mathrm{d}v\) \(\Rightarrow I = \displaystyle\int\frac{1}{\sqrt{\sinh^2 v + 1}}\cosh v\,\mathrm{d}v = \int\frac{\cosh v}{\sqrt{\cosh^2 v}}\,\mathrm{d}v = \int 1\,\mathrm{d}v = v\ (+c)\) \(= \sinh^{-1}u\ (+c)\) | M1 |
| Differentiating both sides wrt \(u\) gives \(\dfrac{1}{\sqrt{u^2 + 1}} = \dfrac{\mathrm{d}}{\mathrm{d}u}\left(\sinh^{-1}u\right)\) | A1 |
M1: Correctly integrates RHS. Condone omission of \(c\)
| Scheme | Marks | AO |
|---|---|---|
| \(y = \sinh^{-1}2x\) \(\therefore \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2}{\sqrt{(2x)^2 + 1}} = \dfrac{2}{\sqrt{4x^2 + 1}}\) | M1 | 1.1 |
| \(x = \sqrt{6} \Rightarrow y = \sinh^{-1}2\sqrt{6}\ \left(= \ln\left(5 + 2\sqrt{6}\right)\right)\) | M1 | 1.1 |
| \(x = \sqrt{6} \Rightarrow \left.\dfrac{\mathrm{d}y}{\mathrm{d}x}\right|_{x = \sqrt{6}} = \dfrac{2}{5}\) \(\therefore m = -\dfrac{5}{2}\) | M1 | 1.1 |
| \(y - \ln\left(5 + 2\sqrt{6}\right) = -\dfrac{5}{2}\left(x - \sqrt{6}\right)\) \(\therefore y = -\dfrac{5}{2}x + \ln\left(5 + 2\sqrt{6}\right) + \dfrac{5\sqrt{6}}{2}\) | A1 | 1.1 |
| [4] |
Notes
M1: (1st) Differentiating using chain rule or the formula booklet, giving \(\dfrac{1}{\sqrt{\frac{1}{4} + x^2}}\)
M1: (2nd) Substituting the \(x\)-value into the equation to find the \(y\) coordinate of the point
M1: (3rd) Substituting the \(x\)-value into their gradient and taking negative reciprocal
A1: oe in correct form