Higher January 2021 Paper 1R Q12
12
Given that \(4^{k + 3} = 16 \times 2^k\)
Show your working clearly. (4)
| Scheme | Marks |
|---|---|
| \(4e^5f^3\) | B2 |
| (2) |
Notes
B2: (B1 for 2 out of 3 terms correct in a 3 term product)
| Scheme | Marks |
|---|---|
E.g. \(\dfrac{3(2x + 1) + 4(x - 2)}{12}\) or \(\dfrac{3(2x + 1)}{12} + \dfrac{4(x - 2)}{12}\) | M1 |
E.g. \(\dfrac{6x + 3 + 4x - 8}{12}\) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{10x - 5}{12}\) | A1 |
| (3) |
Notes
M1: for expressing both fractions correctly with a common denominator.
Allow as two separate fractions.
M1: for removing brackets correctly in a correct single fraction
A1: accept \(\dfrac{5(2x - 1)}{12}\)
| Scheme | Marks |
|---|---|
\((4^{k+3} =)\;(2^2)^{k+3}\) oe or \((16 =)\;2^4\) \((16 =)\;4^2\) or \((2^k =)\;\left(4^{\frac{1}{2}}\right)^k\) oe \((4^{k+3} =)\;\left(16^{\frac{1}{2}}\right)^{k+3}\) oe or \((2^k =)\;\left(16^{\frac{1}{4}}\right)^k\) oe | M1 |
\((4^{k+3} =)\;(2^2)^{k+3}\) oe and \((16 =)\;2^4\) \((16 =)\;4^2\) and \((2^k =)\;\left(4^{\frac{1}{2}}\right)^k\) oe \((4^{k+3} =)\;\left(16^{\frac{1}{2}}\right)^{k+3}\) oe and \((2^k =)\;\left(16^{\frac{1}{4}}\right)^k\) oe | M1 |
E.g. \(2k + 6 = 4 + k\) or \(k + 3 = 2 + \dfrac{1}{2}k\) or \(\dfrac{1}{2}(k + 3) = 1 + \dfrac{1}{4}k\) | M1 |
| Working required Answer: \(-2\) | A1 |
| (4) | |
| (9 marks) |
Notes
M1: for \((2^2)^{k+3}\) oe or \(2^4\) or \(4^2\) or \(\left(4^{\frac{1}{2}}\right)^k\) oe or \(\left(16^{\frac{1}{2}}\right)^{k+3}\) oe or \(\left(16^{\frac{1}{4}}\right)^k\) oe
M1: for \((2^2)^{k+3}\) oe and \(2^4\) or \(4^2\) and \(\left(4^{\frac{1}{2}}\right)^k\) oe or \(\left(16^{\frac{1}{2}}\right)^{k+3}\) oe and \(\left(16^{\frac{1}{4}}\right)^k\) oe
M1: for a correct linear equation in \(k\)
A1: dep on at least M2
In the first two rows, \(4^{k+3}\) is written as \(\left(16^{\frac{1}{2}}\right)^{k+3}\) (corrected from the printed mark scheme: “\(\left(16^{\frac{1}{4}}\right)^{k+3}\)”).