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Date November 2021 Marks available 2 Reference code 21N.2.AHL.TZ0.6
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Prove Question number 6 Adapted from N/A

Question

Prove the identity (p+q)3-3pq(p+q)p3+q3.

[2]
a.

The equation 2x2-5x+1=0 has two real roots, α and β.

Consider the equation x2+mx+n=0, where m, n and which has roots 1α3 and 1β3.
Without solving 2x2-5x+1=0, determine the values of m and n.

[6]
b.

Markscheme

METHOD 1

(p+q)3-3pq(p+q)p3+q3

attempts to expand (p+q)3                 M1

p3+3p2q+3pq2+q3

(p+q)3-3pq(p+q)p3+3p2q+3pq2+q3-3pq(p+q)

p3+3p2q+3pq2+q3-3p2q-3pq2                 A1

p3+q3                 AG


Note: Condone the use of equals signs throughout.

 

METHOD 2

(p+q)3-3pq(p+q)p3+q3

attempts to factorise (p+q)3-3pq(p+q)                 M1

(p+q)((p+q)2-3pq) ((p+q)(p2-pq+q2))

p3-p2q+pq2+p2q-pq2+q3                 A1

p3+q3                 AG


Note: Condone the use of equals signs throughout.

 

METHOD 3

p3+q3(p+q)3-3pq(p+q)

attempts to factorise p3+q3                 M1

(p+q)(p2-pq+q2)

(p+q)((p+q)2-3pq)                 A1

(p+q)3-3pq(p+q)                 AG


Note: 
Condone the use of equals signs throughout.


[2 marks]

a.

Note: Award a maximum of A1M0A0A1M0A0 for m=-95 and n=8 found by using α,β=5±174 (α,β=0.219, 2.28).
Condone, as appropriate, solutions that state but clearly do not use the values of α and β.
Special case: Award a maximum of A1M1A0A1M0A0 for m=-95 and n=8 obtained by solving simultaneously for α and β from product of roots and sum of roots equations.


product of roots of x2-52x+12=0

αβ=12 (seen anywhere)                      A1

considers (1α3)(1β3) by stating 1(αβ)3(=n)                      M1


Note: Award M1 for attempting to substitute their value of αβ into 1(αβ)3.

1(αβ)3=1(12)3

n=8                      A1

sum of roots of x2-52x+12=0

α+β=52 (seen anywhere)                A1

considers 1α3 and 1β3 by stating (α+β)3-3αβ(α+β)(αβ)3 ((α+βαβ)3-3(α+β)(αβ)2)(=-m)                      M1


Note: Award M1 for attempting to substitute their values of α+b and αβ into their expression. Award M0 for use of (α+β)3-3αβ(α+β) only.


=(52)3-(32)(52)18 (=125-30=95)

m=-95                A1

(x2-95x+8=0)


[6 marks]

b.

Examiners report

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a.
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b.

Syllabus sections

Topic 1—Number and algebra » SL 1.6—Simple proof
Show 41 related questions
Topic 2—Functions » AHL 2.12—Factor and remainder theorems, sum and product of roots
Topic 1—Number and algebra
Topic 2—Functions

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