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Date November 2011 Marks available 4 Reference code 11N.3ca.hl.TZ0.4
Level HL only Paper Paper 3 Calculus Time zone TZ0
Command term Show Question number 4 Adapted from N/A

Question

Using the integral test, show that n=114n2+1 is convergent.

[4]
a.

(i)     Show, by means of a diagram, that n=114n2+1<14×12+1+114x2+1dx.

(ii)     Hence find an upper bound for n=114n2+1

[4]
b.

Markscheme

14x2+1dx=12arctan2x+k     (M1)(A1)

Note: Do not penalize the absence of “+k”.

 

114x2+1dx=12lima[arctan2x]a1     (M1)

Note: Accept 12[arctan2x]1.

 

=12(π2arctan2)(=0.232)     A1

hence the series converges     AG

[4 marks]

a.

(i)

     A2

The shaded rectangles lie within the area below the graph so that n=214n2+1<114x2+1dx. Adding the first term in the series, 14×12+1, gives n=114n2+1<14×12+1+114x2+1dx     R1AG

 

(ii)     upper bound =15+12(π2arctan2)(=0.432)     A1

[4 marks]

b.

Examiners report

This proved to be a hard question for most candidates. A number of fully correct answers to (a) were seen, but a significant number were unable to integrate 14x2+1 successfully. Part (b) was found the hardest by candidates with most candidates unable to draw a relevant diagram, without which the proof of the inequality was virtually impossible.

a.

This proved to be a hard question for most candidates. A number of fully correct answers to (a) were seen, but a significant number were unable to integrate 14x2+1 successfully. Part (b) was found the hardest by candidates with most candidates unable to draw a relevant diagram, without which the proof of the inequality was virtually impossible.

b.

Syllabus sections

Topic 9 - Option: Calculus » 9.2 » Convergence of infinite series.

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