Date | May 2015 | Marks available | 5 | Reference code | 15M.2.sl.TZ2.4 |
Level | SL only | Paper | 2 | Time zone | TZ2 |
Command term | Find | Question number | 4 | Adapted from | N/A |
Question
The third term in the expansion of (x+k)8 is 63x6. Find the possible values of k.
Markscheme
valid approach to find the required term (M1)
eg(8r)x8−rkr, Pascal’s triangle to 8th row, x8+8x7k+28x6k2+…
identifying correct term (may be indicated in expansion) (A1)
eg(82)x6k2, (86)x6k2, r=2
setting up equation in k with their coefficient/term (M1)
eg28k2x6=63x6, (86)k2=63
k=±1.5 (exact) A1A1 N3
[5 marks]
Examiners report
Candidates who recognized that the third term is required usually completed the question successfully, although some candidates only gave a single value for k. A few candidates attempted to fully expand algebraically, which proved to be a fruitless enterprise.