Edexcel A-Level Mathematics A2 P4.1 Proof Questions

Assume that there exists a positive number k such that $k+\frac9k<6$.
$$k+\frac9k<6\Rightarrow k^2+9<6k\Rightarrow k^2-6k+9<0$$$$(k-3)^2<0$$
But numbers squared are $\ge 0$, hence $k+\frac9k\ge 6$.
A1*

Assume the curves meet:
$$8x-\frac{15x^2}{2}+8=16-8x+\frac{x^2}{2}+\frac{x^3}{24}$$$$0=8-16x+8x^2+\frac{x^3}{24} =8(x-1)^2+\frac{x^3}{24}$$
For $0<x<\frac83$,
$$8(x-1)^2\ge0,\qquad \frac{x^3}{24}>0$$
so
$$8(x-1)^2+\frac{x^3}{24}>0$$
This contradicts the equation $0=8(x-1)^2+\frac{x^3}{24}$, so the curves do not meet.
(4)

B1: For setting up the contradiction.
Eg Assume that there exists a number n that isn't a multiple of 3, yet $n^{2}$ is a multiple of 3 As a minimum accept something like " define a number n such that n is not a multiple of 3 but ifi js"
M1: States that m=3 p+1 or m=3 p+2 and attempts to square.
Alternatives exist such as m=3 p+1 or m=3 p-1
Using modulo 3 arithmetic it would be $1 \rightarrow 1$ and $2 \rightarrow 4=1$
M1: States that m=3 p+1 AND m=3 p+2 and attempts to square o.e.
A1: Achieves forms that can be argued as to why they are NOT a multiple of 3
E.g. $m^{2}=(3 p+1)^{2}=3\left(3 p^{2}+2 p\right)+1$ or even $9 p^{2}+6 p+1$
and $m^{2}=(3 p+2)^{2}=3\left(3 p^{2}+4 p+1\right)+1$ or even $9 p^{2}+12 p+4$
A1: Correct proof which requires
- Correct calculations
- Correct reasons. E.g. $9 p^{2}+12 p+4$ is not a multiple of 3 as 4 is not a multiple of 3 There are many ways to argue these. E.g $m^{2}=(3 p+1)^{2}=3\left(3 p^{2}+2 p\right)+1$ is sufficient as long as followed (or preceded by) "not a multiple of 3 "
- Minimal conclusion such as. Note that B0 M1 M1 M1 A1 is possible