4MA0 Higher June 2014 Paper 4R Q21
21
(a) Show that \[(a^2 + 1)(c^2 + 1) = (ac - 1)^2 + (a + c)^2\] (3)
(b) By finding suitable values of \(a\) and \(c\), use part (a) to write 650065 as the sum of two square numbers. (3)
| Scheme | Marks |
|---|---|
| LHS \(a^2c^2 + a^2 + c^2 + 1\) | M1 |
| RHS \(a^2c^2 - 2ac + 1 + a^2 + 2ac + c^2\) | M1 |
| Shown | A1 |
| (3) |
Notes
M1: for correct expansion of LHS
M1: for correct expansion of RHS
A1: for convincing progression and conclusion
Alternative (i)
| Scheme | Marks |
|---|---|
| LHS \(a^2c^2 + a^2 + c^2 + 1\) | M1 |
| \(a^2c^2 - 2ac + 1 + a^2 + 2ac + c^2\) | M1 |
| Shown | A1 |
Notes
M1: for correct expansion of LHS
M1: for introduction of \(+2ac\) and \(-2ac\)
A1: for convincing progression and conclusion
Alternative (ii)
| Scheme | Marks |
|---|---|
| RHS \(a^2c^2 - 2ac + 1 + a^2 + 2ac + c^2\) | M1 |
| \(a^2c^2 - 2ac + 1 + a^2 + 2ac + c^2 = a^2c^2 + a^2 + c^2 + 1\) \(= a^2(c^2 + 1) + (c^2 + 1)\) | M1 |
| Shown | A1 |
Notes
M1: for correct expansion of RHS
M1: for correct simplification and convincing factorisation
A1: for convincing progression and conclusion
| Scheme | Marks |
|---|---|
| 65 × 10001 | M1 |
| \(a = 8\), \(c = 100\) \(ac - 1 = 8 \times 100 - 1\) \(a + c = 8 + 100\) | A1 |
| \(108^2,\; 799^2\) | A1 |
| (3) | |
| (6 marks) |
Notes
M1: for 65 × 10001
A1: \(a = 8\), \(c = 100\) or \(a = 100\) and \(c = 8\) or 108 and 799
A1: \(108^2\), \(799^2\) or 11664 and 638401