A2 June 2022 Paper 1 Q6
6 Let \(y = x\cosh x\).
Prove by induction that, for all integers \(n \geqslant 1\), \(\dfrac{\mathrm{d}^{2n-1}y}{\mathrm{d}x^{2n-1}} = x\sinh x + (2n - 1)\cosh x\). [6]
| Scheme | Marks | AO |
|---|---|---|
| Base case: \(y = x\cosh x \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \cosh x + x\sinh x\) \(= (2 \times 1 - 1)\cosh x + x\sinh x\) So true for \(n = 1\) | B1 | 2.5 |
| Assume result holds for \(n = k\) \(\dfrac{\mathrm{d}^{2k-1}y}{\mathrm{d}x^{2k-1}} = x\sinh x + (2k - 1)\cosh x\) | M1* | 2.1 |
| \(\Rightarrow \dfrac{\mathrm{d}^{2k}y}{\mathrm{d}x^{2k}} = \sinh x + x\cosh x + (2k - 1)\sinh x\) | M1dep | 1.1 |
| \(\Rightarrow \dfrac{\mathrm{d}^{2k+1}y}{\mathrm{d}x^{2k+1}} = \cosh x + \cosh x + x\sinh x + (2k - 1)\cosh x\) \(= x\sinh x + (2k + 1)\cosh x\) | M1dep | 3.1a |
| i.e. \(\dfrac{\mathrm{d}^{2(k+1)-1}y}{\mathrm{d}x^{2(k+1)-1}} = x\sinh x + \left(2(k + 1) - 1\right)\cosh x\) | A1* | 2.2a |
| So if true for \(n = k\) then also true for \(n = k + 1\) But it is true for \(n = 1\) and so is true generally | A1dep | 2.4 |
| [6] |
Notes
B1: Evidence of correct differentiation using product rule and a substitution \(n = 1\) into RHS
M1dep: Differentiate using the product rule
M1dep: Differentiate second time using the product rule
A1*: Must be equated to correct derivative form
A1dep: Dependent on previous A1 and the B1