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Date	May 2021	Marks available	1	Reference code	21M.2.AHL.TZ2.9
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Write down	Question number	9	Adapted from	N/A

Question

Write down the first three terms of the binomial expansion of $(1 + t)^{- 1}$ $(1 + t)^{- 1}$ in ascending powers of $t$ $t$ .

[1]

By using the Maclaurin series for $cos x$ $\cos x$ and the result from part (a), show that the Maclaurin series for $sec x$ $sec x$ up to and including the term in $x^{4}$ $x^{4}$ is $1 + \frac{x^{2}}{2} + \frac{5 x^{4}}{24}$ $1 + \frac{x^{2}}{2} + \frac{5 x^{4}}{24}$ .

[4]

By using the Maclaurin series for $arctan x$ $arctan x$ and the result from part (b), find $lim x \to 0 (\frac{x arctan 2 x}{sec x - 1})$ $\lim_{x \to 0} (\frac{x arctan 2 x}{sec x - 1})$ .

[3]

Markscheme

$1 - t + t^{2}$ $1 - t + t^{2}$ A1

Note: Accept $1, - t$ $1, - t$ and $t^{2}$ $t^{2}$ .

[1 mark]

$sec x = \frac{1}{1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} (- \dots)} (= {(1 - \frac{x^{2}}{2!} + (\frac{x^{4}}{4!} (- \dots)))}^{- 1})$ $sec x = \frac{1}{1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} (- \dots)} (= {(1 - \frac{x^{2}}{2!} + (\frac{x^{4}}{4!} (- \dots)))}^{- 1})$ (M1)

$t = cos x - 1$ $t = \cos x - 1$ or $sec x = 1 - (cos x - 1) + {(cos x - 1)}^{2}$ $sec x = 1 - (\cos x - 1) + {(\cos x - 1)}^{2}$ (M1)

$= 1 - (- \frac{x^{2}}{2!} + \frac{x^{4}}{4!} (- \dots)) + {(- \frac{x^{2}}{2!} + \frac{x^{4}}{4!} (- \dots))}^{2}$ $= 1 - (- \frac{x^{2}}{2!} + \frac{x^{4}}{4!} (- \dots)) + {(- \frac{x^{2}}{2!} + \frac{x^{4}}{4!} (- \dots))}^{2}$ A1

$= 1 + \frac{x^{2}}{2} - \frac{x^{4}}{24} + \frac{x^{4}}{4}$ $= 1 + \frac{x^{2}}{2} - \frac{x^{4}}{24} + \frac{x^{4}}{4}$ A1

so the Maclaurin series for $sec x$ $sec x$ up to and including the term in $x^{4}$ $x^{4}$ is $1 + \frac{x^{2}}{2} + \frac{5 x^{4}}{24}$ $1 + \frac{x^{2}}{2} + \frac{5 x^{4}}{24}$ AG

Note: Condone the absence of ‘…’

[4 marks]

$arctan 2 x = 2 x - \frac{{(2 x)}^{3}}{3} + \dots$ $arctan 2 x = 2 x - \frac{{(2 x)}^{3}}{3} + \dots$

$lim x \to 0 (\frac{x arctan 2 x}{sec x - 1}) = lim x \to 0 ⎛ ⎜ ⎝ \frac{x (2 x - \frac{{(2 x)}^{3}}{3} + \dots)}{(1 + \frac{x^{2}}{2} + \frac{5 x^{4}}{24}) - 1} ⎞ ⎟ ⎠$ $\lim_{x \to 0} (\frac{x arctan 2 x}{sec x - 1}) = \lim_{x \to 0} (\frac{x (2 x - \frac{{(2 x)}^{3}}{3} + \dots)}{(1 + \frac{x^{2}}{2} + \frac{5 x^{4}}{24}) - 1})$ M1

$= lim x \to 0 ⎛ ⎝ \frac{2 x^{2} - \frac{8 x^{4}}{3} + \dots}{\frac{x^{2}}{2} + \frac{5 x^{4}}{24}} ⎞ ⎠$ $= \lim_{x \to 0} (\frac{2 x^{2} - \frac{8 x^{4}}{3} + \dots}{\frac{x^{2}}{2} + \frac{5 x^{4}}{24}})$ A1

$= lim x \to 0 ⎛ ⎜ ⎝ \frac{2 x^{2} (1 - \frac{4 x^{2}}{3})}{\frac{x^{2}}{2} (1 + \frac{5 x^{2}}{12})} ⎞ ⎟ ⎠$ $= \lim_{x \to 0} (\frac{2 x^{2} (1 - \frac{4 x^{2}}{3})}{\frac{x^{2}}{2} (1 + \frac{5 x^{2}}{12})})$

$= 4$ $= 4$ A1

Note: Condone missing ‘lim’ and errors in higher derivatives.
Do not award M1 unless $x$ $x$ is replaced by $2 x$ $2 x$ in $arctan$ $arctan$ .

[3 marks]

Examiners report

[N/A]

Syllabus sections

Topic 1—Number and algebra » AHL 1.10—Perms and combs, binomial with negative and fractional indices

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