Higher June 2024 Paper 2 Q20
20 Given that \(\;k = x - y\;\) and \(\;x = \dfrac{1}{4y}\)
express \(\;\dfrac{5k}{x + 2}\;\) in the form \(\;\dfrac{a - by^2}{c + dy}\;\) where \(a\), \(b\), \(c\) and \(d\) are integers.
(3)
| Scheme | Marks |
|---|---|
\(\dfrac{5\left(\dfrac{1}{4y} - y\right)}{\dfrac{1}{4y} + 2}\left(= \dfrac{\dfrac{5}{4y} - 5y}{\dfrac{1}{4y} + 2}\right)\) oe or \(\dfrac{4y(5x - 5y)}{8y + 1}\) oe | M1 |
\(\dfrac{\dfrac{5}{4y} \times 4y - 5y \times 4y}{\dfrac{1}{4y} \times 4y + 2 \times 4y}\) or \(\dfrac{\dfrac{5 - 20y^2}{4y}}{\dfrac{1 + 8y}{4y}}\) oe or \(\dfrac{4y(5x - 5y)}{8y + 1} = \dfrac{20xy - 20y^2}{8y + 1}\) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{5 - 20y^2}{1 + 8y}\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: For a correct substitution with only values of \(y\)
or
an expression containing \(xy\) (not just \(x\))
or
a correct denominator of \(1 + 8y\)
M1: multiplying every term by \(4y\) or a multiple of \(4y\) or writing numerator and denominator over 4y or a multiple of \(4y\)
or
correctly expanded with an \(xy\) term (\(xy\) could be replaced with 0.25 oe)
or
3 of \(a\), \(b\), \(c\) or \(d\) correct if written in the form \(\dfrac{a - by^2}{c + dy}\) where \(a\), \(b\), \(c\) and \(d\) are integers
A1: oe eg \(\dfrac{-5 + 20y^2}{-1 - 8y}\) or \(\dfrac{20 - 80y^2}{4 + 32y}\)