Date | November 2010 | Marks available | 7 | Reference code | 10N.3ca.hl.TZ0.1 |
Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ0 |
Command term | Find | Question number | 1 | Adapted from | N/A |
Question
Find limx→0(1−cosx6x12).
Markscheme
METHOD 1
f(0)=00, hence using l’Hôpital’s Rule, (M1)
g(x)=1−cos(x6), h(x)=x12; g′(x)h′(x)=6x5sin(x6)12x11=sin(x6)2x6 A1A1
EITHER
g′(0)h′(0)=00, using l’Hôpital’s Rule again, (M1)
g″(x)h″(x)=6x5cos(x6)12x5=cos(x6)2 A1A1
g″(0)h″(0)=12, hence the limit is 12 A1
OR
So limx→01−cosx6x12=limx→0sinx62x6 A1
=12limx→0sinx62x6 A1
=12 since limx→0sinx62x6=1 A1 (R1)
METHOD 2
substituting x6 for x in the expansion cosx=1−x22+x424… (M1)
1−cosx6x12=1−(1−x122+x2424)…x12 M1A1
=12−x1224+... A1A1
limx→01−cosx6x12=12 M1A1
Note: Accept solutions using Maclaurin expansions.
[7 marks]
Examiners report
Surprisingly, some weaker candidates were more successful in answering this question than stronger candidates. If candidates failed to simplify the expression after the first application of L’Hôpital’s rule, they generally were not successful in correctly differentiating the expression a 2nd time, hence could not achieve the final three A marks.