Higher January 2023 Paper 2 Q26
26 Find the values of \(n\) such that
\(\dfrac{10^{4n} \times 2^{3(n^2 - 5n)} \times 5^{2(1 - 2n)}}{20^2} = 1\)
Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
| For one of: \(10^{4n} = (5 \times 2)^{4n}\) or \(5^{4n} \times 2^{4n}\) or oe \(20^2 = (5 \times 2^2)^2\) or \(5^2 \times 2^4\) or \(5^2 \times 2^2 \times 2^2\) or cancelling \(5^2\) on numerator with \(20^2\) to get 16 or \(2^4\) | M1 |
| for getting numerator to the stage (this scores M1M1) \(2^{4n} \times 2^{3n^2 - 15n} \times 5^2\) oe eg \(2^{3n^2 - 11n} \times 5^{-4n + 4n + 2}\) oe or for two of \(10^{4n} = (5 \times 2)^{4n}\) or \(5^{4n} \times 2^{4n}\) or oe \(20^2 = (5 \times 2^2)^2\) or \(5^2 \times 2^4\) or \(5^2 \times 2^2 \times 2^2\) or cancelling \(5^2\) on numerator with \(20^2\) to get 16 or \(2^4\) or getting the numerator to the stage | M1 |
| \(\dfrac{2^{4n} \times 2^{3n^2} \times 2^{-15n}}{2^4}\;[= 1 \text{ or } 2^0]\) or \(2^{4n} \times 2^{3n^2} \times 2^{-15n} = 2^4\) oe | M1 |
| \(3n^2 - 11n - 4\;[= 0]\) | A1 |
working required Answer: \(-\dfrac{1}{3}\), 4 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for writing \(10^{4n}\) correctly as a product of 5 and 2 to the power \(4n\) oe
or
writing \(20^2\) correctly as a product of 5 and \(2^2\) to the power 2 oe
or
cancelling \(5^2\) on numerator with \(20^2\) to get 16 or \(2^4\)
M1: For writing the equation in powers of 2 only
A1: Correct quadratic equation dep on M1
A1: Both answers required dep on M1