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 (x−2x)4(x2+2x)3(x−2x)4(x2+2x)3.
Markscheme
(x−2x)4=x4−8x2+24−32x2+16x4(x−2x)4=x4−8x2+24−32x2+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.