Date | November 2020 | Marks available | 6 | Reference code | 20N.2.SL.TZ0.S_5 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Find | Question number | S_5 | Adapted from | N/A |
Question
Consider the expansion of (3x2-kx)9, where k>0.
The coefficient of the term in x6 is 6048. Find the value of k.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach for expansion (must have correct substitution for parameters, but accept an incorrect value for r). (M1)
eg (9r)(3x2)9-r(-kx)r , (3x2)9+(91)(3x2)8(-kx)1+(92)(3x2)7(-kx)2+…
valid attempt to identify correct term (M1)
eg 2(9-r)-r=6 , (x2)r(x-1)9-r=x6
identifying correct term (may be indicated in expansion) (A1)
eg r=4, r=5
correct term or coefficient in binominal expansion (A1)
eg (94)(3x2)5(-kx)4 , 126(243x10)(k4x4), 30618k4
correct equation in k (A1)
eg (94)(243)(k4)x6=6048x6 , 30618k4=6048
k=23 (exact) 0.667 A1 N3
Note: Do not award A1 if additional answers given.
[6 marks]