Date | May 2021 | Marks available | 3 | Reference code | 21M.2.AHL.TZ2.9 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 2 |
Command term | Find | Question number | 9 | Adapted from | N/A |
Question
Write down the first three terms of the binomial expansion of (1+t)-1 in ascending powers of t.
By using the Maclaurin series for cos x and the result from part (a), show that the Maclaurin series for sec x up to and including the term in x4 is 1+x22+5x424.
By using the Maclaurin series for arctan x and the result from part (b), find limx→0(x arctan 2xsec x-1).
Markscheme
1-t+t2 A1
Note: Accept 1, -t and t2.
[1 mark]
sec x=11-x22!+x44!(-…) (=(1-x22!+(x44!(-…)))-1) (M1)
t=cos x-1 or sec x=1-(cos x-1)+(cos x-1)2 (M1)
=1-(-x22!+x44!(-…))+(-x22!+x44!(-…))2 A1
=1+x22-x424+x44 A1
so the Maclaurin series for sec x up to and including the term in x4 is 1+x22+5x424 AG
Note: Condone the absence of ‘…’
[4 marks]
arctan 2x=2x-(2x)33+…
limx→0(x arctan 2xsec x-1)=limx→0(x(2x-(2x)33+…)(1+x22+5x424)-1) M1
=limx→0(2x2-8x43+…x22+5x424) A1
=limx→0(2x2(1-4x23)x22(1+5x212))
=4 A1
Note: Condone missing ‘lim’ and errors in higher derivatives.
Do not award M1 unless x is replaced by 2x in arctan.
[3 marks]