## Question 10(i):

### Algebraic Proof:

| Answer/Working | Mark | Guidance |
|---|---|---|
| For $n = 2m$: $n^2 + 2 = 4m^2 + 2$ | M1 | Setting up proof for even numbers |
| Concludes number is not divisible by 4 (explanation is trivial) | A1 | Correctly concludes with reason for $n$ even |
| For $n = 2m+1$: $n^2 + 2 = (2m+1)^2 + 2 = \ldots$ FYI $(4m^2 + 4m + 3)$ | dM1 | Setting up proof for odd numbers |
| Correct working; concludes this is a number in the 4 times table add 3, writes $4(m^2 + m) + 3$ **AND** states hence true for all | A1* | Fully correct proof with valid explanation and conclusion for all $n$ |

# Question 10 (i) - Algebraic Proof that $n^2 + 2$ is not divisible by 4:

**Very Similar Algebraic Proof:**

For $n = 2m$: $\frac{4m^2+2}{4} = m^2 + \frac{1}{2}$ | M1 | 2.1

Not divisible by 4 due to the $\frac{1}{2}$ (a suitable reason is required) | A1 | 1.1b

For $n = 2m+1$: $\frac{n^2+2}{4} = \frac{4m^2+4m+3}{4} = m^2+m+\frac{3}{4}$ | dM1 | 2.1

Not divisible by 4 due to $\frac{3}{4}$ AND states hence for all $n$, $n^2+2$ is not divisible by 4 | A1* | 2.4

**Total: (4)**

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**Proof via Logic:**

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 true for all $n$ | A1* | 2.4

**Total: (4)**

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**Proof via Contradiction:**

Sets up contradiction: Assume $n^2+2$ is divisible by $4 \Rightarrow n^2+2 = 4k$ | M1 | 2.1

$\Rightarrow n^2 = 4k-2 = 2(2k-1)$ and concludes even. Note M mark for setting up contradiction must have been awarded | A1 | 1.1b

States $n^2$ is even, then $n$ is even and hence $n^2$ is a multiple of 4 | dM1 | 2.1

Explains if $n^2$ is a multiple of 4 then $n^2+2$ cannot be a multiple of 4, hence contradiction, hence true for all $n$ | A1* | 2.4

**Total: (4)**

*Alternative contradiction: A1: $n^2 = 2(2k-1) \Rightarrow n = \sqrt{2} \times \sqrt{2k-1}$; dM1: States $2k-1$ is odd so does not have a factor of 2, meaning $n$ is irrational*

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# Question 10 (ii):

**Proof using numerical values:**

SOMETIMES TRUE; chooses $x: 9.25 < x < 9.5$ and shows false. E.g. $x=9.4$, $|3x-28|=0.2$ and $x-9=0.4$ ✗ | M1 | 2.3

Then chooses a number where it is true. E.g. $x=12$, $|3x-28|=8$, $x-9=3$ ✓ | A1 | 2.4

**Total: (2)**

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**Graphical Proof:**

States or implies "sometimes true"; sketches both graphs on same axes; V shape on $+$ve $x$-axis; linear graph with $+$ve gradient intersecting twice | M1 | 2.3

Graphs accurate and explains that there are points where $|3x-28| < x-9$ and points where $|3x-28| > x-9$ | A1 | 2.4

**Total: (2)**

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**Proof via Algebra:**

States sometimes true and attempts to solve both $3x-28 < x-9$ and $-3x+28 < x-9$ or one of these with bound $9.\dot{3}$ | M1 | 2.3

States it is false when $9.25 < x < 9.5$ or $9.25 < x < 9.\dot{3}$ or $9.\dot{3} < x < 9.5$ | A1 | 2.4

**Total: (2)**

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