Date | May 2008 | Marks available | 5 | Reference code | 08M.2.hl.TZ1.1 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Determine | Question number | 1 | Adapted from | N/A |
Question
Determine the first three terms in the expansion of (1−2x)5(1+x)7 in ascending powers of x.
Markscheme
METHOD 1
constant term: (50)(−2x)0(70)x0=1 A1
term in x: (71)x+(51)(−2x)=−3x (M1)A1
term in x2 : (72)x2+(52)(−2x)2+(71)x(51)(−2x)=−9x2 M1A1 N3
[5 marks]
METHOD 2
(1−2x)5(1+x)7=(1+5(−2x)+5×4(−2x)22!+...)(1+7x+7×62x2+...) M1M1
=(1−10x+40x2+...)(1+7x+21x2+…)
=1+7x+21x2−10x−70x2+40x2+…
=1−3x−9x2+… A1A1A1 N3
[5 marks]
Examiners report
Although the majority of the candidates understood the question and attempted it, excessive time was spent on actually expanding the expression without consideration of the binomial theorem. A fair amount of students confused “ascending order”, giving the last three instead of the first three terms.