A2 June 2024 Q5

EdexcelCurrent spec9 marksMethods in Calculus

\[y = \mathrm{e}^{3x}\sin x\]

(a) Use Leibnitz’s theorem to show that\[\frac{\mathrm{d}^4 y}{\mathrm{d}x^4} = 28\mathrm{e}^{3x}\sin x + 96\mathrm{e}^{3x}\cos x\] (6)

(b) Hence express \(\dfrac{\mathrm{d}^4 y}{\mathrm{d}x^4}\) in the form\[R\mathrm{e}^{3x}\sin(x + \alpha)\]where \(R\) and \(\alpha\) are constants to be determined, \(R \gt 0\) and \(0 \lt \alpha \lt \dfrac{\pi}{2}\) (3)

Mark scheme (a)

Scheme	Marks	AO
\(y = \mathrm{e}^{3x}\sin x \qquad u = \mathrm{e}^{3x} \qquad v = \sin x\)
\(u^{\prime} = 3\mathrm{e}^{3x},\ u^{\prime\prime} = 9\mathrm{e}^{3x},\ u^{\prime\prime\prime} = 27\mathrm{e}^{3x},\ u^{(4)} = 81\mathrm{e}^{3x}\)	M1 A1	1.1b 1.1b
\(v^{\prime} = \cos x,\ v^{\prime\prime} = -\sin x,\ v^{\prime\prime\prime} = -\cos x,\ v^{(4)} = \sin x\)	M1 A1	1.1b 1.1b
Thus \(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = \left(\mathrm{e}^{3x} \times \sin x\right) + \left(4 \times 3\mathrm{e}^{3x} \times -\cos x\right) + \left(6 \times 9\mathrm{e}^{3x} \times -\sin x\right)\) \(\qquad + \left(4 \times 27\mathrm{e}^{3x} \times \cos x\right) + \left(81\mathrm{e}^{3x} \times \sin x\right)\)	M1	2.1
\(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = 28\mathrm{e}^{3x}\sin x + 96\mathrm{e}^{3x}\cos x\ *\)	A1*	1.1b
	(6)

Notes

M1: Finds the correct form of the first four derivatives of \(\mathrm{e}^{3x}\). Must all be of the form \(\alpha\mathrm{e}^{3x}\).
SC: M1 may also be awarded for at least four relevant derivatives found if they do not find all four for the \(\mathrm{e}^{3x}\) or the \(\sin x\).

A1: Correct derivatives. Allow if the general form is identified rather than individual ones given.
SC allow for at least four correct derivatives if not all are found.

M1: Finds the correct form of the first four derivatives of \(\sin x\), condone sign slips only.

A1: Correct derivatives.

M1: Applies Leibnitz’s theorem to get the 4^th derivative with their expressions. Binomial coefficients must be present and correct numerical expressions (not notational forms).

A1*: Correct simplified 4^th derivative from fully correct working, including the \(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} =\) seen at some stage.

Note: If do not use Leibnitz’s theorem then up to the first 4 marks can be awarded for using the product rule to differentiate. Accept correct forms for the derivatives.

M1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \pm\mathrm{e}^{3x}\cos x + A\mathrm{e}^{3x}\sin x\) and \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = C\mathrm{e}^{3x}\cos x + D\mathrm{e}^{3x}\sin x\)

A1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{3x}\cos x + 3\mathrm{e}^{3x}\sin x\) and \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 6\mathrm{e}^{3x}\cos x + 8\mathrm{e}^{3x}\sin x\)

dM1: Forms correct next two. A1: \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = 26\mathrm{e}^{3x}\cos x + 18\mathrm{e}^{3x}\sin x\) and \(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = 28\mathrm{e}^{3x}\sin x + 96\mathrm{e}^{3x}\cos x\)

Mark scheme (b)

Scheme	Marks	AO
\(R\left(= \sqrt{28^2 + 96^2}\right) = 100\)	B1	1.1b
\(\tan\alpha = \pm\dfrac{96}{28} \Rightarrow \alpha = \ldots\ (\alpha = 1.287\ldots)\)	M1	1.1b
\(\left(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} =\right) 100\mathrm{e}^{3x}\sin(x + 1.29)\)	A1	2.1
	(3)
	(9 marks)

Notes

(b) Note this appears on epen as MAA but is being marked as BMA

B1cao: Deduces the correct value of \(R\), simplified, \(R = 100\)

M1: Attempts the value of \(\alpha\) via \(\tan\alpha = \pm\dfrac{B}{A}\) or \(\tan\alpha = \pm\dfrac{A}{B}\) or \(\sin\alpha = \pm\dfrac{B}{R}\) or \(\cos\alpha = \pm\dfrac{A}{R}\)
Note that \(\alpha =\) awrt 1.29 or \(\alpha =\) awrt \(74^\circ\) imply this mark.

A1: Correct answer, the \(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}}\) may be missing but must be given as the correct expression with \(\alpha =\) awrt 1.29 not just separate values for \(R\) and \(\alpha\)