June 2019 Paper 1 Q10

EdexcelCurrent spec6 marksFunctions (including |mod|)Proof

10.

(i) Prove that for all \(n \in \mathbb{N}\), \(n^2 + 2\) is not divisible by 4 (4)

(ii) “Given \(x \in \mathbb{R}\), the value of \(|3x - 28|\) is greater than or equal to the value of \((x - 9)\).”
State, giving a reason, if the above statement is always true, sometimes true or never true. (2)

Mark scheme (i)

Example of an algebraic proof

Scheme	Marks	AO
For \(n = 2m,\quad n^2 + 2 = 4m^2 + 2\)	M1	2.1
Concludes that this number is not divisible by 4 (as the explanation is trivial)	A1	1.1b
For \(n = 2m + 1,\quad n^2 + 2 = (2m+1)^2 + 2 = \ldots\) FYI \((4m^2 + 4m + 3)\)	dM1	2.1
Correct working and concludes that this is a number in the 4 times table add 3 so cannot be divisible by 4 or writes \(4(m^2 + m) + 3\) ..........AND states ......hence true for all	A1*	2.4
	(4)

Notes

Notes: Note that M0 A0 M1 A1 and M0 A0 M1 A0 are not possible due to the way the scheme is set up

M1: Awarded for setting up the proof for either the even or odd numbers.

A1: Concludes correctly with a reason why \(n^2 + 2\) cannot be divisible by 4 for either \(n\) odd or even.

dM1: Awarded for setting up the proof for both even and odd numbers

A1: Fully correct proof with valid explanation and conclusion for all \(n\)

General points for marking question 10 (i):

Students who just try random numbers in part (i) are not going to score any marks.
Students can mix and match methods. Eg you may see odd numbers via logic and even via algebra
Students who state \(4m^2 + 2\) cannot be divided by (instead of is not divisible by) cannot be awarded credit for the accuracy/explanation marks, unless they state correctly that \(4m^2 + 2\) cannot be divided by 4 to give an integer.
Students who write \(n^2 + 2 = 4k \Rightarrow k = \dfrac{1}{4}n^2 + \dfrac{1}{2}\) which is not a whole number gains no credit unless they then start to look at odd and even numbers for instance
Proofs via induction usually tend to go nowhere unless they proceed as in the main scheme
Watch for unusual methods that are worthy of credit (See below)
If the final conclusion is \(n \in \mathbb{R}\) then the final mark is withheld. \(n \in \mathbb{Z}^+\) is correct

Watch for methods that may not be in the scheme that you feel may deserve credit.
If you are uncertain of a method please refer these up to your team leader.

Eg 1. Solving part (i) by modulo arithmetic.

All \(n \in \mathbb{N}\) mod 4	0	1	2	3
All \(n^2 \in \mathbb{N}\) mod 4	0	1	0	1
All \(n^2 + 2 \in \mathbb{N}\) mod 4	2	3	2	3

Hence for all \(n\), \(n^2 + 2\) is not divisible by 4.

Example of a very similar algebraic proof

Scheme	Marks	AO
For \(n = 2m,\quad \dfrac{4m^2 + 2}{4} = m^2 + \dfrac{1}{2}\)	M1	2.1
Concludes that this is not divisible by 4 due to the \(\dfrac{1}{2}\) (A suitable reason is required)	A1	1.1b
For \(n = 2m + 1,\quad \dfrac{n^2 + 2}{4} = \dfrac{4m^2 + 4m + 3}{4} = m^2 + m + \dfrac{3}{4}\)	dM1	2.1
Concludes that this is not divisible by 4 due to the \(\dfrac{3}{4}\) ...AND states ...... hence for all \(n\), \(n^2 + 2\) is not divisible by 4	A1*	2.4
	(4)

Example of a proof via logic

Scheme	Marks	AO
When \(n\) is odd, "odd \(\times\) odd" = odd	M1	2.1
so \(n^2 + 2\) is odd , so (when \(n\) is odd) \(n^2 + 2\) cannot be divisible by 4	A1	1.1b
When \(n\) is even, it is a multiple of 2, so "even \(\times\) even" is a multiple of 4	dM1	2.1
Concludes that when \(n\) is even \(n^2 + 2\) cannot be divisible by 4 because \(n^2\) is divisible by 4…..AND STATES ……..trues for all \(n\).	A1*	2.4
	(4)

Example of proof via contradiction

Scheme	Marks	AO
Sets up the contradiction ‘Assume that \(n^2 + 2\) is divisible by 4 \(\Rightarrow n^2 + 2 = 4k\)’	M1	2.1
\(\Rightarrow n^2 = 4k - 2 = 2(2k - 1)\) and concludes even Note that the M mark (for setting up the contradiction must have been awarded)	A1	1.1b
States that \(n^2\) is even, then \(n\) is even and hence \(n^2\) is a multiple of 4	dM1	2.1
Explains that if \(n^2\) is a multiple of 4 then \(n^2 + 2\) cannot be a multiple of 4 and hence divisible by 4 Hence there is a contradiction and concludes Hence true for all \(n\).	A1*	2.4
	(4)

A similar proof exists via contradiction where

A1: \(n^2 = 2(2k - 1) \Rightarrow n = \sqrt{2} \times \sqrt{2k - 1}\)

dM1: States that \(2k - 1\) is odd, so does not have a factor of 2, meaning that \(n\) is irrational

sketch of the V-shaped graph of y = |3x − 28| with its vertex on the positive x-axis and the line y = x − 9 crossing it twice — Mark scheme (ii)

June 2019 Paper 1 Q10