Date | May 2021 | Marks available | 5 | Reference code | 21M.1.SL.TZ2.4 |
Level | Standard Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 2 |
Command term | Find | Question number | 4 | Adapted from | N/A |
Question
In the expansion of (x+k)7, where k∈ℝ, the coefficient of the term in x5 is 63.
Find the possible values of k.
Markscheme
EITHER
attempt to use the binomial expansion of (x+k)7 (M1)
C07x7k0+C17x6k1+C27x5k2+… (or C07k7x0+C17k5x1+C27k5x2+…)
identifying the correct term C27x5k2 (or C57k2x5) (A1)
OR
attempt to use the general term Cr7xrk7-r (or Cr7krx7-r) (M1)
r=2 (or r=5) (A1)
THEN
C27=21 (or C57=21 (seen anywhere) (A1)
21x5k2=63x5 (21k2=63 , k2=3) A1
k=±√3 A1
Note: If working shown, award M1A1A1A1A0 for k=√3.
[5 marks]