Higher June 2021 Paper 1 Q11
11 \(\sqrt{2} \times 16 = 2^x\)
(a) Find the value of \(x\).
Show your working clearly. (2)
Show your working clearly. (2)
\(\dfrac{(11^{-6})^5}{11^4} = 11^n\)
(b) Find the value of \(n\).
Show your working clearly. (2)
Show your working clearly. (2)
| Scheme | Marks |
|---|---|
| \(2^{\frac{1}{2}} \times 2^4\) or eg \(2 \times (2^4)^2 = (2^x)^2\) or \(2^9 = 2^{2x}\) | M1 |
Working required Answer: \(\dfrac{9}{2}\) | A1 |
| (2) |
Notes
M1: for a correct expression in powers of 2 that is equivalent to \(2^x\) eg \(2^{\frac{1}{2}} \times 2^4\)
or for showing \(\sqrt{2} = 2^{\frac{1}{2}}\) and \(16 = 2^4\)
or for writing the equation in powers of 2 eg \(2 \times (2^4)^2 = (2^x)^2\) or \(2^9 = 2^{2x}\)
A1: or 4.5 or 4½ dependent on M1
| Scheme | Marks |
|---|---|
\(\dfrac{11^{-30}}{11^4}\) or \(-30 - 4 = n\) or \(-30 = n + 4\) oe | M1 |
| Working required Answer: −34 | A1 |
| (2) | |
| (4 marks) |
Notes
M1: For \((11^{-6})^5\) written as \(11^{-30}\) in the equation or \((11^{-6})^5 = 11^{-30}\) shown in working
or a correct equation with indices only
(no marks for \(3.914\ldots \times 10^{-36}\))
A1: dep on M1 (as we have asked for working)