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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 limx0(1cosx6x12).

Markscheme

METHOD 1

f(0)=00, hence using l’Hôpital’s Rule,     (M1)

g(x)=1cos(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 limx01cosx6x12=limx0sinx62x6     A1

=12limx0sinx62x6     A1

=12 since limx0sinx62x6=1     A1 (R1)

METHOD 2

substituting x6 for x in the expansion cosx=1x22+x424     (M1)

1cosx6x12=1(1x122+x2424)x12     M1A1

=12x1224+...     A1A1

limx01cosx6x12=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.

Syllabus sections

Topic 9 - Option: Calculus » 9.7 » The evaluation of limits of the form limxaf(x)g(x) and limxf(x)g(x) .

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