A2 June 2023 Paper 2 Q6
6. Given that
\[y = \mathrm{e}^{2x}\sinh x\]prove by induction that for \(n \in \mathbb{N}\)
\[\frac{\mathrm{d}^n y}{\mathrm{d}x^n} = \mathrm{e}^{2x}\left(\frac{3^n + 1}{2}\sinh x + \frac{3^n - 1}{2}\cosh x\right)\](6)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\mathrm{e}^{2x}\sinh x + \mathrm{e}^{2x}\cosh x = a\mathrm{e}^{2x}\sinh x + b\mathrm{e}^{2x}\cosh x\) or \(\mathrm{e}^{2x}(a\sinh x + b\cosh x)\) | M1 | 2.2a |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{2x}(2\sinh x + \cosh x)\) \(n = 1\) then \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{2x}\left(\frac{3 + 1}{2}\sinh x + \frac{3 - 1}{2}\cosh x\right)\) {so the result is true for \(n = 1\)} | A1 | 2.4 |
| (Assume the result is true for \(n = k\), then) Must be an attempt at the product rule, with \(k\)’s in all terms \(\displaystyle \frac{\mathrm{d}^{k+1}y}{\mathrm{d}x^{k+1}} = A\mathrm{e}^{2x}\left(\mathrm{f}(k)\sinh x + \mathrm{g}(k)\cosh x\right) + \mathrm{e}^{2x}\left(\mathrm{f}(k)\cosh x + \mathrm{g}(k)\sinh x\right)\) \(\displaystyle \frac{\mathrm{d}^{k+1}y}{\mathrm{d}x^{k+1}} = 2\mathrm{e}^{2x}\left(\frac{3^k + 1}{2}\sinh x + \frac{3^k - 1}{2}\cosh x\right) + \mathrm{e}^{2x}\left(\frac{3^k + 1}{2}\cosh x + \frac{3^k - 1}{2}\sinh x\right)\) | M1 | 2.1 |
| \(\displaystyle = \mathrm{e}^{2x}\left(\left(3^k + 1 + \frac{3^k - 1}{2}\right)\sinh x + \left(3^k - 1 + \frac{3^k + 1}{2}\right)\cosh x\right)\) or \(\displaystyle = \mathrm{e}^{2x}\left(\frac{3 \times 3^k + 1}{2}\sinh x + \frac{3 \times 3^k - 1}{2}\cosh x\right)\) | dM1 | 1.1b |
| \(\displaystyle = \mathrm{e}^{2x}\left(\frac{3^{k+1} + 1}{2}\sinh x + \frac{3^{k+1} - 1}{2}\cosh x\right)\) | A1 | 2.1 |
| If true for \(n = k\) then true for \(n = k + 1\), and as also true for \(n = 1\), so the result is true for all positive integers or true \(n \in \mathbb{N}\) | A1 | 2.4 |
| (6) | ||
| (6 marks) |
Notes
Alternative using exponential definitions
| Scheme | Marks | AO |
|---|---|---|
| \(y = \mathrm{e}^{2x}\sinh x \Rightarrow y = \mathrm{e}^{2x}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right) = \dfrac{1}{2}\left(\mathrm{e}^{3x} - \mathrm{e}^{x}\right)\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}\left(3\mathrm{e}^{3x} - \mathrm{e}^{x}\right)\) | M1 | 2.2a |
| \(n = 1\) then \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{2x}\left(\frac{3 + 1}{2}\sinh x + \frac{3 - 1}{2}\cosh x\right) = \mathrm{e}^{2x}\left(2\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} + \dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)\) \(= \dfrac{1}{2}\left(3\mathrm{e}^{3x} - \mathrm{e}^{x}\right)\) | A1 | 2.4 |
| (Assume the result is true for \(n = k\), then) \(\displaystyle \frac{\mathrm{d}^k y}{\mathrm{d}x^k} = \mathrm{e}^{2x}\left(\frac{3^k + 1}{2}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right) + \frac{3^k - 1}{2}\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)\right)\) \(\displaystyle = \mathrm{e}^{2x}\left(\frac{3^k}{2}\mathrm{e}^x - \frac{1}{2}\mathrm{e}^{-x}\right) = \frac{3^k}{2}\mathrm{e}^{3x} - \frac{1}{2}\mathrm{e}^x\) Then differentiates \(\displaystyle \frac{\mathrm{d}^{k+1}y}{\mathrm{d}x^{k+1}} = \mathrm{e}^{2x}\left(\frac{3^{k+1} + 1}{2}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right) + \frac{3^{k+1} - 1}{2}\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)\right)\) | M1 | 2.1 |
| Simplifies \(\displaystyle \frac{\mathrm{d}^{k+1}y}{\mathrm{d}x^{k+1}} = \mathrm{e}^{2x}\left(\frac{3^{k+1}}{2}\mathrm{e}^x - \frac{1}{2}\mathrm{e}^{-x}\right) = \frac{3^{k+1}}{2}\mathrm{e}^{3x} - \frac{1}{2}\mathrm{e}^x\) | dM1 | 1.1b |
| Using the given result \(\displaystyle \frac{\mathrm{d}^{k+1}y}{\mathrm{d}x^{k+1}} = \mathrm{e}^{2x}\left(\frac{3^{k+1} + 1}{2}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right) + \frac{3^{k+1} - 1}{2}\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)\right)\) \(\displaystyle = 3 \times \frac{3^k}{2}\mathrm{e}^{3x} - \frac{1}{2}\mathrm{e}^x = \frac{3^{k+1}}{2}\mathrm{e}^{3x} - \frac{1}{2}\mathrm{e}^x\) | A1 | 2.1 |
| If true for \(n = k\) then true for \(n = k + 1\), and as also true for \(n = 1\), so the result is true for all positive integers or true \(n \in \mathbb{N}\) | A1 | 2.4 |
| (6) |
Notes
M1: Differentiates to a correct form
A1: Correct derivative and reaches appropriate form to deduce the result is true for \(n = 1\), minimum \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{2x}\left(\dfrac{4}{2}\sinh x + \dfrac{2}{2}\cosh x\right)\)
M1: (Makes the inductive assumption and) attempts the \((k + 1)\) th derivative from the \(k\) th derivative. Allow slips in coefficients but must be evidence of the use of the product rule. Must be of the form
\[\frac{\mathrm{d}^{k+1}y}{\mathrm{d}x^{k+1}} = A\mathrm{e}^{2x}\left(\mathrm{f}(k)\sinh x + \mathrm{g}(k)\cosh x\right) + \mathrm{e}^{2x}\left(\mathrm{f}(k)\cosh x + \mathrm{g}(k)\sinh x\right)\]dM1: Dependent on the previous method mark. Factors out the exponential and gathers the \(\sinh x\) and \(\cosh x\) terms. Accept either form shown or equivalent.
A1: Reaches the correct form from correct work. Must have the “\(k\)+1” showing. Depends on the previous two method marks.
A1: Makes appropriate concluding sentence covering the points indicated in scheme. Depends on all method marks having been scored. Must have reached at least the second line shown in the dM mark, and made an attempt at checking \(n = 1\) (though the first A mark need not have been scored if insufficient detail shown).
Maybe seen as a narrative throughout their solution.
This mark requires all necessary brackets throughout.
Alternative: using exponentials
M1: Uses the exponential definition of \(\sinh x\) and differentiates to a correct form
A1: Correct derivative and uses the exponential definitions of \(\sinh x\) and \(\cosh x\) to deduce the result is true for \(n = 1\)
M1: (Makes the inductive assumption and) attempts the \((k + 1)\) th derivative from the \(k\) th derivative. Uses the exponential definition of \(\sinh x\) first and then differentiates to a correct form
dM1: Dependent on the previous method mark. Collects exponential terms and simplifies
A1: Reaches the \(k + 1\) th derivative. Uses the exponential definitions of \(\sinh x\) and \(\cosh x\) in the result, simplifies and achieves the correct result
A1: Makes appropriate concluding sentence covering the points indicated in scheme. Depends on all method marks having been scored. Must have reached at least the second line shown in the dM mark, and made an attempt at checking \(n = 1\) (though the first A mark need not have been scored if insufficient detail shown).
Maybe seen as a narrative throughout their solution.
This mark requires all necessary brackets throughout.