Higher June 2018 Paper 1 Q15
15
(a) Simplify fully \(\left(\dfrac{256x^{20}}{y^8}\right)^{-\frac{1}{4}}\) (2)
(b) Express \(\dfrac{1}{9x^2 - 25} - \dfrac{1}{6x + 10}\) as a single fraction in its simplest form. (3)
| Scheme | Marks |
|---|---|
E.g. \(\left(\dfrac{y^8}{256x^{20}}\right)^{\frac{1}{4}}\) or \(\left(\dfrac{4x^5}{y^2}\right)^{-1}\) or \(\dfrac{x^{-5}}{4y^{-2}}\) or \(\dfrac{\frac{1}{4}x^{-5}}{y^{-2}}\) or \(k\dfrac{y^a}{x^b}\) or \(\dfrac{ky^a}{x^b}\) with 2 of \(k = \dfrac{1}{4}\) oe, \(a = 2\), \(b = 5\) or \(\dfrac{y^a}{mx^b}\) with 2 of \(m = 4\), \(a = 2\), \(b = 5\) | M1 |
| \(\dfrac{y^2}{4x^5}\) | A1 |
| (2) |
Notes
M1: for a correct first step leading to a correct partially simplified expression
A1: for \(\dfrac{y^2}{4x^5}\) or \(\dfrac{\frac{1}{4}y^2}{x^5}\) or \(0.25\dfrac{y^2}{x^5}\) or \(0.25y^2x^{-5}\)
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{(3x - 5)(3x + 5)} - \dfrac{1}{2(3x + 5)}\) | M1 |
E.g. \(\dfrac{2}{2(3x - 5)(3x + 5)} - \dfrac{1(3x - 5)}{2(3x - 5)(3x + 5)}\) or \(\dfrac{6x + 10}{(9x^2 - 25)(6x + 10)} - \dfrac{9x^2 - 25}{(9x^2 - 25)(6x + 10)}\) | M1 |
| \(\dfrac{7 - 3x}{2(3x - 5)(3x + 5)}\) | A1 |
| (3) | |
| (5 marks) |
Notes
M1: indep for \((3x + 5)(3x - 5)\)
M1: for two correct fractions with a common denominator
if there is any expansion at this stage then it must be correct
A1: accept equivalents eg. \(\dfrac{7 - 3x}{18x^2 - 50}\)
| Scheme | Marks |
|---|---|
| \(\dfrac{6x + 10}{(9x^2 - 25)(6x + 10)} - \dfrac{9x^2 - 25}{(9x^2 - 25)(6x + 10)}\) | M1 |
| \(\dfrac{(7 - 3x)(3x + 5)}{(9x^2 - 25)(6x + 10)}\) | M1 |
| \(\dfrac{7 - 3x}{2(3x - 5)(3x + 5)}\) | A1 |
Notes
M1: for two correct fractions with a common denominator
M1: Numerator expanded and then factorised correctly
A1: accept equivalents