Higher November 2020 Paper 1R Q21
21 Given that \(\;M = \dfrac{18^{4n} \times 2^{3(n^2 - 6n)} \times 3^{2(1 - 4n)}}{12^2}\)
find the values of \(n\) for which \(M = 2\)
(5)
| Scheme | Marks |
|---|---|
| \(12^2 = 2^4 \times 3^2\) or \(2 \times 12^2 = 2^5 \times 3^2\) oe or \(\dfrac{2 \times 12^2}{3^2}(= 32) = 2^5\) | M1 |
| \(18^{4n} = (2 \times 3^2)^{4n}\) or \(2^{4n} \times 3^{2 \times 4n}\) | M1 |
| \(3n^2 - 14n - 5\;(= 0)\) | A1 |
e.g. \((3n + 1)(n - 5)(= 0)\) \(n = \dfrac{14 \pm \sqrt{(-14)^2 - (4 \times 3 \times -5)}}{2 \times 3}\) | M1 |
| \(-\dfrac{1}{3},\; 5\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for solving their 3 term quadratic equation using any correct method - if factorising, allow brackets which expanded give 2 out of 3 terms correct (if using formula or completing the square allow one sign error and some simplification – allow as far as e.g. \(\dfrac{14 \pm \sqrt{196 + 60}}{6}\) oe)
A1: Allow −0.33 or better for \(-\dfrac{1}{3}\)