Higher November 2024 Paper 2 Q23
23 Show that \(\dfrac{16x^2 - 36}{x - 7} \div \dfrac{2x^2 + 7x + 6}{x^2 - 5x - 14} - (7 + 8x) = n\)
where \(n\) is an integer to be found.
Show clear algebraic working.
(4)
| Scheme | Marks |
|---|---|
| \(16x^2 - 36 = (4x - 6)(4x + 6)\;[= 4(2x + 3)(2x - 3)]\) oe or \(16x^2 - 36 = (8x - 12)(2x + 3)\;[= 4(2x + 3)(2x - 3)]\) oe | M1indep |
| \(2x^2 + 7x + 6 = (2x + 3)(x + 2)\) and \(x^2 - 5x - 14 = (x - 7)(x + 2)\) [We will make an exception of \(2x^2 + 7x + 6 = (4x + 6)(0.5x + 1)\) as this then cancels with \((4x + 6)\)] | M1indep |
\(4(2x - 3) - (7 + 8x)\;(= n)\) [allow invisible brackets ie \(8x - 12 - 7 + 8x\)] or an expression that clearly shows the numerator is – 19 times the denominator eg \(\dfrac{-38x - 57}{2x + 3}\;(= n)\) | M1 |
| Working required Answer: –19 | A1 |
| (4) | |
| (4 marks) |
Notes
M1indep: [NB: the two fractions when divided and cancelled give an answer of \(4(2x - 3)\) or \(2(4x - 6)\) or \(8x - 12\) (any one of these gain 2 marks) ]
M2 for any fraction with completely simplified non-linear numerator and non-linear denominator that will cancel to –19
M1: a linear expression that should give the correct value for \(n\)
(this mark implies previous M marks as not all factorising is necessary)
A1: dep on M2