Higher June 2023 Paper 1R Q23
23 Simplify \(\;\left(x^2 - 4\right) \div \left(\dfrac{4x^2 - 7x - 2}{x}\right) - 2x\)
Give your answer in the form \(\dfrac{ax^2}{bx + c}\;\) where \(a\), \(b\) and \(c\) are integers.
(4)
| Scheme | Marks |
|---|---|
| \((x + 2)(x - 2)\) oe or \((4x + 1)(x - 2)\) oe | M1 |
\((x + 2)(x - 2) \times \dfrac{x}{(4x + 1)(x - 2)}\) or \(\dfrac{x(x + 2)(x - 2)}{(4x + 1)(x - 2)}\) or \(\dfrac{x(x + 2)}{(4x + 1)}\) | M1 |
\(\dfrac{x(x + 2) - 2x(4x + 1)}{(4x + 1)}\) or \(\dfrac{x^2 + 2x - 8x^2 - 2x}{(4x + 1)}\) or \(\dfrac{x(x + 2)}{(4x + 1)} - \dfrac{2x(4x + 1)}{(4x + 1)}\) or \(\dfrac{x^2 + 2x}{(4x + 1)} - \dfrac{8x^2 + 2x}{(4x + 1)}\) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{-7x^2}{4x + 1}\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for complete factorisation of \(x^2 - 4\) or \(4x^2 - 7x - 2\)
Each factor must be in the form \((ax \pm b)\) where \(a\) and \(b\) are integers
M1: for complete factorisation of \(4x^2 - 7x - 2\) and \(x^2 - 4\) and inverting and intention to multiply
M1: for a correct single fraction following correct cancellation
or
for two correct fractions with common denominator following correct cancellation
A1: oe but must be in form \(\dfrac{ax^2}{bx + c}\) where \(a\), \(b\) and \(c\) are integers.
| Scheme | Marks |
|---|---|
| \(\dfrac{-7x^3 + 14x^2}{4x^2 - 7x - 2}\) oe | M1 |
| \(\dfrac{-7x^2(x - 2)}{(4x + 1)(x - 2)}\) oe | M1 |
| \(\dfrac{-7x^2(x - 2)}{(4x + 1)(x - 2)}\) oe | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{-7x^2}{4x + 1}\) | A1 |
Notes
M1: for a correct single fraction
M1: for complete factorisation of \(-7x^3 + 14x^2\) or \(4x^2 - 7x - 2\)
Each factor must be in the form \((ax \pm b)\)
M1: for complete factorisation of \(-7x^3 + 14x^2\) and \(4x^2 - 7x - 2\)
Each factor must be in the form \((ax \pm b)\)
A1: oe but must be in form \(\dfrac{ax^2}{bx + c}\) where \(a\), \(b\) and \(c\) are integers.