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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 x22x412x22x412.

(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)

=2sec4x4sec2xtan2x     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

=x22x412     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

ln2π292!2π4814!     A1

ln2π218+π4972=π29(12+π2108)     A1

OR

Taking x=π4     A1

ln12(π4)22!2(π4)44!     A1

12ln2π2162!2π42564!     A1

ln2π216+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.

Syllabus sections

Topic 9 - Option: Calculus » 9.6 » Maclaurin series for ex , sinx , cosx , ln(1+x) , (1+x)p , PQ .

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