Higher June 2018 Paper 2 Q13
13 Show that, for \(\;x \neq -1\)
\(\dfrac{8x^2 - 8}{4x + 4} \quad\) simplifies to the form \(\quad ax + b \quad\) where \(a\) and \(b\) are integers. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| Any correct factorisation of the numerator or the denominator | M1 | eg \(8(x^2 - 1)\) or \(4(x + 1)\) or \(2(4x^2 - 4)\) or \(2(2x + 2)\) or \(4(2x^2 - 2)\) or \((4x + 4)(2x - 2)\) or \((4x - 4)(2x + 2)\) or \((8x + 8)(x - 1)\) or \((8x - 8)(x + 1)\) or \(-2(-4x^2 + 4)\) does not need to be seen in a fraction may be implied eg \(\dfrac{2x^2 - 2}{x + 1}\) or \(\dfrac{4x^2 - 4}{2x + 2}\) |
| Correct fraction with a common algebraic factor in the numerator and the denominator | A1 | eg \(\dfrac{8(x + 1)(x - 1)}{4(x + 1)}\) or \(\dfrac{2(2x + 2)(2x - 2)}{2(2x + 2)}\) or \(\dfrac{2(x + 1)(x - 1)}{(x + 1)}\) or \(\dfrac{4(x + 1)(2x - 2)}{4(x + 1)}\) or \(\dfrac{(4x + 4)(2x - 2)}{4x + 4}\) |
| \(2x - 2\) or \(a = 2\) and \(b = -2\) with M1A1 scored | A1 | |
| Alternative method 2 | ||
| \(4ax^2 + 4ax + 4bx + 4b\) | M1 | oe expands \((ax + b)(4x + 4)\) to 4 terms with at least 3 terms correct |
| Any 2 of \(4a = 8 \qquad 4b = -8 \qquad 4a + 4b = 0\) | A1 | |
| \(a = 2\) and \(b = -2\) and shows that third equation is satisfied with M1A1 scored | A1 | |
Additional guidance
| M1 is implied by the first A1 eg \(\dfrac{8(x + 1)(x - 1)}{4(x + 1)}\) | M1A1 |
| \(1(8x^2 - 8)\) or \(-1(8 - 8x^2)\) etc | M0 |
| \(2x - 2\) without M1A1 scored | M0A0A0 |
| M1A1 scored and \(2x - 2\) followed by attempt to solve \(2x - 2 = 0\) | M1A1A1 |
| M1A1 scored and \(2x - 2\) followed by \(2(x - 1)\) | M1A1A1 |
| M1A1 scored followed by \(2(x - 1)\) but \(2x - 2\) not seen | M1A1A0 |