Date | May 2017 | Marks available | 2 | Reference code | 17M.3ca.hl.TZ0.2 |
Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ0 |
Command term | Find | Question number | 2 | Adapted from | N/A |
Question
Let the Maclaurin series for tanxtanx be
tanx=a1x+a3x3+a5x5+…tanx=a1x+a3x3+a5x5+…
where a1a1, a3a3 and a5a5 are constants.
Find series for sec2xsec2x, in terms of a1a1, a3a3 and a5a5, up to and including the x4x4 term
by differentiating the above series for tanxtanx;
Find series for sec2xsec2x, in terms of a1a1, a3a3 and a5a5, up to and including the x4x4 term
by using the relationship sec2x=1+tan2xsec2x=1+tan2x.
Hence, by comparing your two series, determine the values of a1a1, a3a3 and a5a5.
Markscheme
(sec2x=) a1+3a3x2+5a5x4+…(sec2x=) a1+3a3x2+5a5x4+… A1
[1 mark]
sec2x=1+(a1x+a3x3+a5x5+…)2sec2x=1+(a1x+a3x3+a5x5+…)2
=1+a21x2+2a1a3x4+…=1+a21x2+2a1a3x4+… M1A1
Note: Condone the presence of terms with powers greater than four.
[2 marks]
equating constant terms: a1=1a1=1 A1
equating x2x2 terms: 3a3=a21=1⇒a3=133a3=a21=1⇒a3=13 A1
equating x4x4 terms: 5a5=2a1a3=23⇒a5=2155a5=2a1a3=23⇒a5=215 A1
[3 marks]