Date | November 2019 | Marks available | 4 | Reference code | 19N.1.SL.TZ0.S_4 |
Level | Standard Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Find and Hence or otherwise | Question number | S_4 | Adapted from | N/A |
Question
Consider (11a)=11!a!9!.
Find the value of a.
Hence or otherwise find the coefficient of the term in x9 in the expansion of (x+3)11.
Markscheme
valid approach (M1)
eg 11−a=9, 11!9!(11−9)!
a=2 A1 N2
[2 marks]
valid approach for expansion using n=11 (M1)
eg (11r)x11−r3r, a11b0+(111)a10b1+(112)a9b2+…
evidence of choosing correct term A1
eg (112)32, (112)x932, (119)32
correct working for binomial coefficient (seen anywhere, do not accept factorials) A1
eg 55, (112)=55, 55×32, (55×9)x9, 11×102×9
495 A1 N2
Note: If there is clear evidence of adding instead of multiplying, award A1 for the correct working for binomial coefficient, but no other marks. For example, 55x9×32 would earn M0A0A1A0.
Do not award final A1 for a final answer of 495x9, even if 495 is seen previously. If no working shown, award N1 for 495x9.
[4 marks]