Date | May 2008 | Marks available | 14 | Reference code | 08M.3ca.hl.TZ2.4 |
Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ2 |
Command term | Find and Show that | Question number | 4 | Adapted from | N/A |
Question
(a) Given that y=lncosxy=lncosx , show that the first two non-zero terms of the Maclaurin series for y are −x22−x412−x22−x412.
(b) Use this series to find an approximation in terms of π for ln2π for ln2 .
Markscheme
(a) f(x)=lncosxf(x)=lncosx
f′(x)=−sinxcosx=−tanx M1A1
f″(x)=−sec2x M1
f‴(x)=−2secxsecxtanx A1
fiv(x)=−2sec2x(sec2x)−2tanx(2sec2xtanx)
=−2sec4x−4sec2xtan2x A1
f(x)=f(0)+xf′(0)+x22!f″(0)+x33!f‴(0)+x44!fiv(0)+…
f(0)=0, M1
f′(0)=0,
f″(0)=−1,
f‴(0)=0,
fiv(0)=−2, A1
Notes: Award the A1 if all the substitutions are correct.
Allow FT from their derivatives.
ln(cosx)≈−x22!−2x44! A1
=−x22−x412 AG
[8 marks]
(b) Some consideration of the manipulation of ln 2 (M1)
Attempt to find an angle (M1)
EITHER
Taking x=π3 A1
ln12≈−(π3)22!−2(π3)44! A1
−ln2≈−π292!−2π4814! A1
ln2≈π218+π4972=π29(12+π2108) A1
OR
Taking x=π4 A1
ln1√2≈−(π4)22!−2(π4)44! A1
−12ln2≈−π2162!−2π42564! A1
ln2≈π216+π41536=π28(12+π2192) A1
[6 marks]
Total [14 marks]
Examiners report
Some candidates had difficulty organizing the derivatives but most were successful in getting the series. Using the series to find the approximation for ln2 in terms of π was another story and it was rare to see a good solution.