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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 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 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

α2α3+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-α3-3α          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-α3-3α          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=-12α2+3          A1

α-12α2+3          R1

-12α2+3 is not an integer and so α cannot be an integer          R1


Note:
Accept -12α2+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 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 be a root of 2α3+6α+1=0…”

Note: Subsequent marks after this M1 are independent of this M1 and can be awarded.

 

let fx=2x3+6x+1  (and fα=0)

f'x=6x2+6>0 for all x f is a (strictly) increasing function         M1A1

f0=1 and f-1=-7          R1

thus fx=0 has only one real root between -1 and 0, which gives a contradiction

(or therefore, contradicting the assumption that fα=0 for some α),          R1

so the equation 2x3+6x+1=0 has no integer roots           AG

  

[5 marks]

Examiners report

[N/A]

Syllabus sections

Topic 1—Number and algebra » AHL 1.15—Proof by induction, contradiction, counterexamples
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Topic 1—Number and algebra

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