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Date November 2019 Marks available 2 Reference code 19N.1.SL.TZ0.S_4
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Find Question number S_4 Adapted from N/A

Question

Consider  ( 11 a ) = 11 ! a ! 9 ! .

Find the value of a .

[2]
a.

Hence or otherwise find the coefficient of the term in  x 9 in the expansion of ( x + 3 ) 11 .

[4]
b.

Markscheme

valid approach    (M1)

eg    11 a = 9 11 ! 9 ! ( 11 9 ) !

a = 2         A1  N2

[2 marks]

a.

valid approach for expansion using n = 11     (M1)

eg    ( 11 r ) x 11 r 3 r a 11 b 0 + ( 11 1 ) a 10 b 1 + ( 11 2 ) a 9 b 2 +

evidence of choosing correct term         A1

eg     ( 11 2 ) 3 2 ,   ( 11 2 ) x 9 3 2 ,   ( 11 9 ) 3 2

correct working for binomial coefficient (seen anywhere, do not accept factorials)         A1

eg     55 ,   ( 11 2 ) = 55 ,   55 × 3 2 ,   ( 55 × 9 ) x 9 ,   11 × 10 2 × 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,  55 x 9 × 3 2  would earn M0A0A1A0.

Do not award final A1 for a final answer of  495 x 9 , even if  495  is seen previously. If no working shown, award N1 for  495 x 9 .

[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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