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Date November 2016 Marks available 6 Reference code 16N.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

By successive differentiation find the first four non-zero terms in the Maclaurin series for f(x)=(x+1)ln(1+x)x.

[11]
a.

Deduce that, for n2, the coefficient of xn in this series is (1)n1n(n1).

[1]
b.

By applying the ratio test, find the radius of convergence for this Maclaurin series.

[6]
c.

Markscheme

f(x)=(x+1)ln(1+x)x     f(0)=0    A1

f(x)=ln(1+x)+x+11+x1 (=ln(1+x))     f(0)=0    M1A1A1

f(x)=(1+x)1     f(0)=1    A1A1

f(x)=(1+x)2     f(0)=1    A1

f(4)(x)=2(1+x)3     f(4)(0)=2    A1

f(5)(x)=3×2(1+x)4     f(5)(0)=3×2    A1

f(x)=x22!1x33!+2x44!6x55!    M1A1

f(x)=x21×2x32×3+x43×4x54×5

f(x)=x22x36+x412x520

 

Note: Allow follow through from the first error in a derivative (provided future derivatives also include the chain rule), no follow through after a second error in a derivative.

 

[11 marks]

a.

f(n)(0)=(1)n(n2)! So coefficient of xn=(1)n(n2)!n!     A1

coefficient of xn is (1)n1n(n1)     AG

[1 mark]

b.

applying the ratio test to the series of absolute terms

lim    M1A1

= \mathop {\lim }\limits_{n \to \infty } \left| x \right|\frac{{(n - 1)}}{{(n + 1)}}    A1

= \left| x \right|    A1

so for convergence \left| x \right| < 1, giving radius of convergence as 1     (M1)A1

[6 marks]

c.

Examiners report

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Syllabus sections

Topic 9 - Option: Calculus » 9.6 » Mean value theorem.

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