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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.

[2]
a.

Hence or otherwise find the coefficient of the term in x9 in the expansion of (x+3)11.

[4]
b.

Markscheme

valid approach    (M1)

eg    11a=911!9!(119)!

a=2        A1  N2

[2 marks]

a.

valid approach for expansion using n=11    (M1)

eg    (11r)x11r3ra11b0+(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]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 1—Number and algebra » SL 1.9—Binomial theorem where n is an integer
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Topic 1—Number and algebra

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