Higher January 2021 Paper 2R Q19
19
(a) Simplify \(8^2 \times \sqrt[3]{4^6}\)
Give your answer in the form \(2^a\) where \(a\) is an integer.
Show each stage of your working clearly. (3)
Give your answer in the form \(2^a\) where \(a\) is an integer.
Show each stage of your working clearly. (3)
Given that \(n^{\left(-\frac{4}{5}\right)} = \left(\dfrac{1}{2}\right)^4\) where \(n \gt 0\)
(b) find the value of \(n\). (4)
| Scheme | Marks |
|---|---|
eg \((2^3)^2 \times \sqrt[3]{(2^2)^6}\) or \((2^3)^2 \times (4)^{\frac{6}{3}}\) or \(4^3 \times 4^2\) or \(2^6\) or \(2^4\) seen or \(2^6 \times 16\) or \(64 \times 4^2\) or \(8^2 \times 4^2\) or \(8^2 \times 16\) or 64 × 16 | M1 |
| \(2^6 \times (2^{12})^{\frac{1}{3}}\) or 1024 or \(32^2\) or \(4^5\) or \(2^6 \times 2^4\) | M1 |
| Working required Answer: \(2^{10}\) | A1 |
| (3) |
Notes
M1: a correct first stage.
M1: dep on 1st M mark.
A1: dependent on first M1
isw if \(2^{10}\) seen but then 10 given as answer.
| Scheme | Marks |
|---|---|
\(\left(n^{-\frac{4}{5}} =\right) \dfrac{1}{16}\) or 0.0625 oe or eg \(\left(n^{-\frac{1}{5}}\right)^4 = \left(\dfrac{1}{2}\right)^4\) | M1 |
\((n =)\; 16^{\frac{5}{4}}\) or \(0.0625^{-\frac{5}{4}}\) oe \((n =)\; 2^5\) or \(\sqrt[4]{1048576}\) oe or \(\dfrac{1}{0.0625^{\frac{5}{4}}}\) or \(\left(\dfrac{1}{16}\right)^{-\frac{5}{4}}\) or eg \((n =)\; \left(\dfrac{1}{2}\right)^{-5}\) | M2 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 32 | A1 |
| (4) | |
| (7 marks) |
Notes
M1: for sight of \(\dfrac{1}{16}\) oe, even if raised to an incorrect power.
or for algebraic approach, separating out the 4, or 5 or −1 in the power
M2: for a correct expression for \(n\)
(M1 for one correct algebraic stage eg \(n^{-\frac{1}{5}} = \dfrac{1}{2}\))