A2 June 2023 Q6

EdexcelCurrent spec12 marksTaylor Series

\[y = \ln\left(\mathrm{e}^{2x}\cos 3x\right) \qquad -\frac{1}{2} \lt x \lt \frac{1}{2}\]

(a) Show that\[\frac{\mathrm{d}y}{\mathrm{d}x} = 2 - 3\tan 3x\] (2)

(b) Determine \(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}}\) (3)

(c) Hence determine the first 3 non-zero terms in ascending powers of \(x\) of the Maclaurin series expansion of \(\ln\left(\mathrm{e}^{2x}\cos 3x\right)\), giving each coefficient in simplest form. (3)

(d) Use the Maclaurin series expansion for \(\ln(1 + x)\) to write down the first 4 non-zero terms in ascending powers of \(x\) of the Maclaurin series expansion of \(\ln(1 + kx)\), where \(k\) is a constant. (1)

(e) Hence determine the value of \(k\) for which\[\lim_{x \to 0}\left(\frac{1}{x^2}\ln\frac{\mathrm{e}^{2x}\cos 3x}{1 + kx}\right)\]exists. (3)

Mark scheme (a)

Scheme	Marks	AO
\(y = \ln\left(\mathrm{e}^{2x}\cos 3x\right) \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2\mathrm{e}^{2x}\cos 3x - 3\mathrm{e}^{2x}\sin 3x}{\mathrm{e}^{2x}\cos 3x}\) or \(y = \ln\mathrm{e}^{2x} + \ln\cos 3x = 2x + \ln\cos 3x \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2 + \dfrac{-3\sin 3x}{\cos 3x}\)	M1	2.1
\(= 2 - 3\tan 3x\ *\)	A1*	1.1b
	(2)

Notes

M1: Attempts differentiation of the given function by applying both the chain rule and the product rule. Look for the correct form but they may make slips with signs or coefficients. Alternatively, applies the sum law for logs and differentiates using the chain rule.

A1*: Correct proof.

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = -9\sec^2 3x\)	B1	1.1b
\(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = -54\sec^2 3x\tan 3x\) \(\Rightarrow \dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = -162\sec^4 3x - 324\sec^2 3x\tan^2 3x\) (oe)	M1 A1	2.1 1.1b
	(3)

Notes

B1: Correct second derivative.

M1: Continues the differentiation using the chain rule and product rule to reach the 4^th derivative. Look for correct forms, allowing for slips in signs or coefficients only when differentiating. May be given in terms of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}, \dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\) etc for this mark.

A1: Correct 4^th derivative in terms of \(x\). Accept alternatives. Like terms should be gathered and coefficients simplified but isw after a correct suitable answer.

Mark scheme (c)

Scheme	Marks	AO
\((y)_0 = 0,\ \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_0 = 2,\ \left(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\right)_0 = -9,\ \left(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\right)_0 = 0,\ \left(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}}\right)_0 = -162\)	M1	1.1b
\(y = \mathrm{f}(0) + x\mathrm{f}^{\prime}(0) + \dfrac{x^2}{2!}\mathrm{f}^{\prime\prime}(0) + \dfrac{x^3}{3!}\mathrm{f}^{\prime\prime\prime}(0) + \ldots\)	M1	2.1
\((y =)\,0 + 2x - \dfrac{9x^2}{2} - \dfrac{162x^4}{24} + \ldots = 2x - \dfrac{9x^2}{2} - \dfrac{27x^4}{4}\)	A1cso	1.1b
	(3)

Notes

M1: Attempts the values of the derivatives at \(x = 0\) up to at least the third derivative. If substitution not seen at least two values should be correct for their derivatives, with values for others.

M1: Substitutes the values of their derivatives into the correct Maclaurin expansion formula up to at least the third non-zero value for their derivatives.

A1cso: Must have obtained a correct 4^th derivative in (b) (though need not have been in terms of \(x\)). Correct expansion with simplified coefficients. May be missing the “\(y =\)” and allow “\(\mathrm{f}(x) =\)”. Note the correct series will be obtained from various incorrect 4^th derivatives due to some terms evaluating to 0, but use of a clearly incorrect 4^th derivative score A0 here. However, an initially correct 4^th derivative, with incorrect simplification, can score A1 bod if no incorrect work is shown (ie correct values stated with no substitution shown, as they may have used the initially correct answer).

Scheme	Marks	AO
\(\ln(1 + kx) = kx - \dfrac{k^2x^2}{2} + \dfrac{k^3x^3}{3} - \dfrac{k^4x^4}{4} + \ldots\)	B1	2.2a
	(1)

Mark scheme (e)

Scheme	Marks	AO
\(\ln\dfrac{\mathrm{e}^{2x}\cos 3x}{1 + kx} = \ln\left(\mathrm{e}^{2x}\cos 3x\right) - \ln(1 + kx)\) \(= \ln\left(\mathrm{e}^{2x}\cos 3x\right) - \ln(1 + kx)\) \(= 2x - \dfrac{9x^2}{2} - \dfrac{27x^4}{4} + \ldots - \left(kx - \dfrac{k^2x^2}{2} + \ldots\right)\)	M1	3.1a
\(= \dfrac{1}{x^2}\left((2 - k)x - \dfrac{\left(9 - k^2\right)x^2}{2} - \dfrac{k^3x^3}{3} - \dfrac{\left(27 - k^4\right)x^4}{4} + \ldots\right)\) For the limit to exist \(2 - k = 0 \Rightarrow k = \ldots\)	M1	3.1a
\(k = 2\)	A1	2.2a
	(3)
	(12 marks)

Notes

M1: Applies the subtraction law of logarithms and then substitutes their expansions.

M1: Realises that the \(x\) terms will determine the required value of \(k\) and so collects the \(x\) terms and sets the coefficient to zero and solves for \(k\) (may be implied).

A1: For \(k = 2\)