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Date May 2012 Marks available 7 Reference code 12M.2.hl.TZ1.9
Level HL only Paper 2 Time zone TZ1
Command term Find Question number 9 Adapted from N/A

Question

Find the constant term in the expansion of (x2x)4(x2+2x)3(x2x)4(x2+2x)3.

Markscheme

(x2x)4=x48x2+2432x2+16x4(x2x)4=x48x2+2432x2+16x4     (M1)(A1)

(x2+2x)3=x6+6x3+12+8x3(x2+2x)3=x6+6x3+12+8x3     (M1)(A1)

 Note: Accept unsimplified or uncalculated coefficients in the constant term

 

=24×12=24×12     (M1)(A1)

=288=288     A1

[7 marks]

Examiners report

Many correct answers were seen, although most candidates used rather inefficient methods (e.g. expanding the brackets in multiple steps). In a very few cases candidates used the binomial theorem to obtain the answer quickly.

Syllabus sections

Topic 1 - Core: Algebra » 1.3 » The binomial theorem: expansion of (a+b)n(a+b)n, nNnN .
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