Date | May 2022 | Marks available | 5 | Reference code | 22M.1.AHL.TZ2.9 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 2 |
Command term | Prove | Question number | 9 | Adapted from | N/A |
Question
Prove by contradiction that the equation 2x3+6x+1=0 has no integer roots.
Markscheme
METHOD 1 (rearranging the equation)
assume there exists some α∈ℤ such that 2α3+6α+1=0 M1
Note: Award M1 for equivalent statements such as ‘assume that α is an integer root of 2α3+6α+1=0’. Condone the use of x throughout the proof.
Award M1 for an assumption involving α3+3α+12=0.
Note: Award M0 for statements such as “let’s consider the equation has integer roots…” ,“let α∈ℤ be a root of 2α3+6α+1=0…”
Note: Subsequent marks after this M1 are independent of this M1 and can be awarded.
attempts to rearrange their equation into a suitable form M1
EITHER
2α3+6α=-1 A1
α∈ℤ⇒2α3+6α is even R1
2α3+6α=-1 which is not even and so α cannot be an integer R1
Note: Accept ‘2α3+6α=-1 which gives a contradiction’.
OR
1=2(-α3-3α) A1
α∈ℤ⇒(-α3-3α)∈ℤ R1
⇒1 is even which is not true and so α cannot be an integer R1
Note: Accept ‘⇒1 is even which gives a contradiction’.
OR
12=-α3-3α A1
α∈ℤ⇒(-α3-3α)∈ℤ R1
-α3-3α is is not an integer (=12) and so α cannot be an integer R1
Note: Accept ‘ -α3-3α is not an integer (=12) which gives a contradiction’.
OR
α=-12(α2+3) A1
α∈ℤ⇒-12(α2+3)∈ℤ R1
-12(α2+3) is not an integer and so α cannot be an integer R1
Note: Accept -12(α2+3) is not an integer which gives a contradiction’.
THEN
so the equation 2x3+6x+1=0 has no integer roots AG
METHOD 2
assume there exists some α∈ℤ such that 2α3+6α+1=0 M1
Note: Award M1 for equivalent statements such as ‘assume that α is an integer root of 2α3+6α+1=0’. Condone the use of x throughout the proof. Award M1 for an assumption involving α3+3α+12=0 and award subsequent marks based on this.
Note: Award M0 for statements such as “let’s consider the equation has integer roots…” ,“let α∈ℤ be a root of 2α3+6α+1=0…”
Note: Subsequent marks after this M1 are independent of this M1 and can be awarded.
let f(x)=2x3+6x+1 (and f(α)=0)
f'(x)=6x2+6>0 for all x∈ℝ is a (strictly) increasing function M1A1
and R1
thus has only one real root between and , which gives a contradiction
(or therefore, contradicting the assumption that for some ), R1
so the equation has no integer roots AG
[5 marks]