Higher November 2018 Paper 2 Q12
12
(a) Write \(\dfrac{4x^2 - 9}{6x + 9} \times \dfrac{2x}{x^2 - 3x}\) in the form \(\dfrac{ax + b}{cx + d}\) where \(a\), \(b\), \(c\) and \(d\) are integers. (3)
(b) Express \(\dfrac{3}{x + 1} + \dfrac{1}{x - 2} - \dfrac{4}{x}\) as a single fraction in its simplest form. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\dfrac{4x - 6}{3x - 9}\) | M1 | factorises numerator of \(4x^2 - 9\) eg \((2x - 3)(2x + 3)\) oe |
| M1 | factorises denominator eg \(x(x - 3)\) or \(3(2x + 3)\) or for \(3x(2x^2 - 3x - 9)\) | |
| A1 | cancels to give \(\dfrac{4x - 6}{3x - 9}\) |
Additional guidance
\(\dfrac{2x(2x - 3)(2x + 3)}{3x(2x + 3)(x - 3)}\)
Accept \(a = 4\), \(b = -6\), \(c = 3\), \(d = -9\)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\dfrac{-x + 8}{x(x + 1)(x - 2)}\) | M1 | method to use a common denominator eg \(x(x + 1)(x - 2)\) by multiplying terms |
| M1 | deduce numerator eg \(3x(x - 2) + x(x + 1) - 4(x + 1)(x - 2)\) | |
| A1 | oe |
Additional guidance
Method must involve finding equivalents for all three separate terms; may be done in several stages.
Equivalents must be algebraically equivalent and must have involved full simplification.