June 2022 Paper 1 Q2
2
| Scheme | Marks | AO |
|---|---|---|
| eg \(1 \gt -2\), but \(1^2 \lt (-2)^2\) as \(1 \lt 4\) | B1 | 2.1 |
| [1] |
Notes
B1: Any correct counterexample, and contradiction identified
Initial inequality soi and then contradiction eg \(-3 \gt -4\) but \(9 \lt 16\) (or \(9 \ngtr 16\))
| Scheme | Marks | AO |
|---|---|---|
| (i) eg \(\sin 150^\circ = 0.5\) as well | B1 | 2.3 |
| [1] | ||
| (ii) \(\sin x^\circ = 0.5 \Leftarrow x^\circ = 30^\circ\) | B1 | 2.5 |
| [1] |
Notes
(b)(i)
B1: Any correct statement
Identifies that \(\sin x = 0.5\) could give values of \(x\) other than \(30^\circ\)
Either specific example or general statement eg ‘many to one’ function
(b)(ii)
B1: Any correct relationship
If attempting to write general solution then must be fully correct eg \(x = 30^\circ + 360n^\circ\), \(x = 150^\circ + 360n^\circ\)
Condone \(\leftarrow\) instead of \(\Leftarrow\)
| Scheme | Marks | AO |
|---|---|---|
| \((4n) + (4n + 4) + (4n + 8) + (4n + 12)\), where \(n\) is an integer | B1* | 2.1 |
| \(= 16n + 24\) \(= 8(2n + 3)\) | M1dep* | 2.1 |
| \(2n + 3\) is an integer, so \(8(2n + 3)\) is a multiple of 8 | A1 | 2.4 |
| [3] |
Notes
B1*: Four consecutive multiples of 4 written correctly in terms of \(n\), or any other variable
Allow BOD if \(n\) not explicitly stated to be an integer
Sufficient to just list the 4 terms, rather than as a sum
Not necessarily starting on \(4n\)
Could also define \(k\) as a multiple of 4 and then have \(k\), \(k + 4\) etc
M1dep*: Correctly sum terms, and correctly take out common factor of 8
Or sum and then consider each term separately
Could be a different factor if using \(k\)
A1: Conclude appropriately
Allow BOD if \(2n + 3\) not explicitly stated to be an integer
If using \(k\)… expect \(8(0.5k + 3)\) then justify \(0.5k\) as an integer, or \(4(k + 6)\) then justify \(k + 6\) is a multiple of 2