Higher January 2020 Paper 2R Q25
25 \(N\) is a multiple of 5
\(A = N + 1\)
\(B = N - 1\)
Prove, using algebra, that \(A^2 - B^2\) is always a multiple of 20
(3)
| Scheme | Marks |
|---|---|
| \((N + 1)^2 = (N^2 + 2N + 1)\) and \((N - 1)^2 = (N^2 - 2N + 1)\) | M1 |
| \((N^2 + 2N + 1) - (N^2 - 2N + 1) = 4N\) | M1 |
| Working required Answer: \(N = 5x\) oe Therefore \(4N = 20x\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: Must reach \(4N\) correctly
A1: Dep. on M2. A correct conclusion (i.e. 20 “\(x\)”) following fully correct working
Alternative method
| Scheme | Marks |
|---|---|
| \(N = 5x\) oe in both \(A\) and \(B\) | M1 |
| \((5x + 1)^2 = (25x^2 + 10x + 1)\) and \((5x - 1)^2 = (25x^2 - 10x + 1)\) | M1 |
| Working required Answer: \((25x^2 + 10x + 1) - (25x^2 - 10x + 1) = 20x\) | A1 |
Notes
A1: Dep. on M2. Subtraction of two correct brackets to reach 20 “\(x\)”
Alternative method
| Scheme | Marks |
|---|---|
| \(A^2 - B^2 = (A + B)(A - B)\) \(A + B = 2N\) and \(A - B = 2\) | M1 |
| \(A^2 - B^2 = 2N \times 2 = 4N\) | M1 |
| Working required Answer: \(N = 5x\) oe Therefore \(4N = 20x\) | A1 |
Notes
A1: Dep. on M2. A correct conclusion (i.e. 20 “\(x\)”) following fully correct working