AS June 2024 Paper 1 Q15

AQACurrent spec7 marksDifferentiation & Maclaurin

(a) Use Maclaurin’s series expansion for \(\ln(1 + x)\) to show that the first three terms of the Maclaurin’s series expansion of \(\ln(1 + 3x)\) are\[3x - \frac{9}{2}x^2 + 9x^3\] [1 mark]

(b) Julia attempts to use the series expansion found in part (a) to find an approximation for \(\ln 4\)

Julia’s incorrect working is shown below.

\[\begin{aligned} \text{Let} \quad 1 + 3x &= 4 \\ 3x &= 3 \\ x &= 1 \end{aligned}\]\[\begin{aligned} \text{So} \quad \ln 4 &\approx 3 \times 1 - \frac{9}{2} \times 1^2 + 9 \times 1^3 \\ &\approx 3 - 4.5 + 9 \\ &\approx 7.5 \end{aligned}\]

Explain the error in Julia’s working. [2 marks]

Fully justify your answer. [4 marks]

Mark scheme (a)

Scheme	Marks	AO
Substitutes \(3x\) for \(x\) in the \(\ln(1 + x)\) series. and Simplifies to \(3x - \dfrac{9}{2}x^2 + 9x^3\) Condone missing LHS.	R1	2.1
	(1)

Typical solution

\[\ln(1 + 3x) = 3x - \frac{(3x)^2}{2} + \frac{(3x)^3}{3} - \ldots\]\[\ln(1 + 3x) \approx 3x - \frac{9x^2}{2} + \frac{27x^3}{3}\]\[\ln(1 + 3x) \approx 3x - \frac{9}{2}x^2 + 9x^3\]

Mark scheme (b)

Scheme	Marks	AO
States \((-1 \lt)\ 3x \leqslant 1\) Condone \(3x \lt 1\)	M1	3.1a
Explains that \(\boldsymbol{x = 1}\) is not a valid value.	E1	2.3
	(2)

Typical solution

The expansion is only valid for

\[-1 \lt 3x \leqslant 1\]\[\Rightarrow -\frac{1}{3} \lt x \leqslant \frac{1}{3}\]

\(x = 1\) is not in this valid range

So Julia should not have substituted \(x = 1\)

Mark scheme (c)

Scheme	Marks	AO
Substitutes \(x = -\dfrac{1}{6}\) into \(3x - \dfrac{9}{2}x^2 + 9x^3\) and \(\ln(1 + 3x)\) PI by \(-2 \times \left(3\left(-\dfrac{1}{6}\right) - \dfrac{9}{2}\left(-\dfrac{1}{6}\right)^2 + 9\left(-\dfrac{1}{6}\right)^3\right)\)	M1	1.1a
Obtains \(\ln\dfrac{1}{2} \approx -\dfrac{2}{3}\) Accept AWRT \(-0.667\) PI by \(\ln 4 \approx -2 \times \left(3\left(-\dfrac{1}{6}\right) - \dfrac{9}{2}\left(-\dfrac{1}{6}\right)^2 + 9\left(-\dfrac{1}{6}\right)^3\right)\) or \(\ln 4 \approx -2 \times \left(-\dfrac{2}{3}\right)\)	A1	1.1b
Uses \(\ln 4 = -2\ln\dfrac{1}{2}\)	M1	1.1a
Deduces \(\ln 4 \approx \dfrac{4}{3}\) Accept AWRT 1.33 Condone an equals sign instead of an approximation sign for all four marks.	R1	2.2a
	(4)
	(7 marks)

Typical solution

\[\ln\left(1 + 3\left(-\frac{1}{6}\right)\right)\]\[\approx 3\left(-\frac{1}{6}\right) - \frac{9}{2}\left(-\frac{1}{6}\right)^2 + 9\left(-\frac{1}{6}\right)^3\]\[\ln\frac{1}{2} \approx -\frac{1}{2} - \frac{1}{8} - \frac{1}{24}\]\[\ln\frac{1}{2} \approx -\frac{2}{3}\]\[\text{But} \quad \ln 4 = 2\ln 2 = -2\ln\frac{1}{2}\]\[\therefore \quad \ln 4 \approx -2 \times -\frac{2}{3}\]\[\ln 4 \approx \frac{4}{3}\]