June 2018 Paper 2 Q16
16 \(A = 2 - 5x \qquad B = 3x - 1 \qquad C = x^2\)
Show that \(\qquad (2A + 3B)^2 \equiv A + B + C\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(2(2 - 5x) + 3(3x - 1)\) or \(4 - 10x\) or \(9x - 3\) | M1 | |
| \(4 - 10x + 9x - 3 = 1 - x\) | M1dep | |
| \((1 - x)^2 = 1 - 2x + x^2\) | A1 | must see working for M2 |
| \(2 - 5x + 3x - 1 + x^2 = 1 - 2x + x^2\) | B1 | |
| Alternative method 2 | ||
| \(4(2 - 5x)^2 + 6(2 - 5x)(3x - 1)\) \(+\ 6(2 - 5x)(3x - 1) + 9(3x - 1)^2\) | M1 | oe allow \(+\ 12(2 - 5x)(3x - 1)\) for \(+\ 6(2 - 5x)(3x - 1) + 6(2 - 5x)(3x - 1)\) |
| \(4(4 - 10x - 10x + 25x^2)\) \(+\ 6(6x - 2 - 15x^2 + 5x)\) \(+\ 6(6x - 2 - 15x^2 + 5x)\) \(+\ 9(9x^2 - 3x - 3x + 1)\) \(= 16 - 40x - 40x + 100x^2 + 36x - 12\) \(-\ 90x^2 + 30x + 36x - 12 - 90x^2\) \(+\ 30x + 81x^2 - 27x - 27x + 9\) | M1dep | oe must see expansions must see working for 1st M1 allow \(+\ 12(6x - 2 - 15x^2 + 5x)\) for \(+\ 6(6x - 2 - 15x^2 + 5x)\) \(+\ 6(6x - 2 - 15x^2 + 5x)\) |
| \(1 - 2x + x^2\) | A1 | must see working for M2 |
| \(2 - 5x + 3x - 1 + x^2 = 1 - 2x + x^2\) | B1 | |
| Alternative method 3 | ||
| \(2(2 - 5x) + 3(3x - 1)\) or \(4 - 10x\) or \(9x - 3\) | M1 | oe |
| \((4 - 10x + 9x - 3)^2\) \(= 16 - 40x + 36x - 12 - 40x + 100x^2\) \(-\ 90x^2 + 30x + 36x - 90x^2 + 81x^2\) \(-\ 27x - 12 + 30x - 27x + 9\) | M1dep | oe must see expansions |
| \(1 - 2x + x^2\) | A1 | must see working for M2 |
| \(2 - 5x + 3x - 1 + x^2 = 1 - 2x + x^2\) | B1 | |
Additional guidance
| Allow working down both sides of an equation/identity | |
| M2A1 is for working on \((2A + 3B)^2\) | |
| B1 is for working on \(A + B + C\) | |
| \(1 - 2x + x^2\) with working for M2 seen and \(2 - 5x + 3x - 1 + x^2 = x^2 - 2x + 1\) | 4 marks |
| \(1 - x^2 = 1 - 2x + x^2\) (do not allow missing brackets even if recovered) |