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Date May 2022 Marks available 5 Reference code 22M.1.SL.TZ2.6
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 2
Command term Find Question number 6 Adapted from N/A

Question

Consider the binomial expansion (x+1)7=x7+ax6+bx5+35x4++1 where x0 and a, b+.

Show that b=21.

[2]
a.

The third term in the expansion is the mean of the second term and the fourth term in the expansion.

Find the possible values of x.

[5]
b.

Markscheme

EITHER

recognises the required term (or coefficient) in the expansion           (M1)

bx5=C27x512   OR   b=C27  OR  C57

b=7!2!5! =7!2!7-2!

correct working           A1

7×6×5×4×3×2×12×1×5×4×3×2×1   OR   7×62!   OR   422


OR

lists terms from row 7 of Pascal’s triangle           (M1)

1, 7, 21,           A1


THEN

b=21           AG

 

[2 marks]

a.

a=7            (A1)

correct equation            A1

21x5=ax6+35x42   OR   21x5=7x6+35x42

correct quadratic equation            A1

7x2-42x+35=0  OR  x2-6x+5=0  (or equivalent)

valid attempt to solve their quadratic            (M1)

x-1x-5=0   OR   x=6±-62-41521

x=1, x=5            A1

 

Note: Award final A0 for obtaining x=0, x=1, x=5.

 

[5 marks]

b.

Examiners report

The majority of candidates answered part (a) correctly, either by using the Crn formula or Pascal's Triangle. In part (b) of the question, most candidates were able to correctly find the value of a=7 and set up a correct equation showing the mean of the second and fourth terms. While some struggled to complete the required algebra to solve the equation, the majority of candidates who found a correct quadratic equation were able to solve it correctly. A few candidates included x=0 in their final answer, thus not earning the final mark.

a.
[N/A]
b.

Syllabus sections

Topic 2—Functions » SL 2.7—Solutions of quadratic equations and inequalities, discriminant and nature of roots
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Topic 1—Number and algebra » SL 1.9—Binomial theorem where n is an integer
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