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.
The equation 2x2-5x+1=0 has two real roots, α and β.
Consider the equation x2+mx+n=0, where m, n∈ℤ and which has roots 1α3 and 1β3.
Without solving 2x2-5x+1=0, determine the values of m and n.
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]
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]