A2 June 2024 Q1
1.
| \(\boldsymbol{x}\) | 2 | 2.5 | 3 | 3.5 | 4 | 4.5 | 5 |
|---|---|---|---|---|---|---|---|
| \(\boldsymbol{y}\) | 1.946 | 2.225 | 2.725 | 2.944 | 3.146 | 3.332 |
In part (b) you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
| Scheme | Marks | AO |
|---|---|---|
| Awrt 2.485 | B1 | 1.1b |
| (1) |
Notes
B1: Correct value, seen in table or in the work. Accept awrt.
| Scheme | Marks | AO |
|---|---|---|
| \(h = 0.5\) | B1 | 1.1b |
| \(\dfrac{1}{3} \times \text{“}0.5\text{”}\left[1.946 + 3.332 + 2(2.485 + 2.944) + 4(2.225 + 2.725 + 3.146)\right]\) \(= \dfrac{1}{6} \times 48.52\) | M1 | 1.1b |
| 8.09 cao | A1 | 1.1b |
| (3) |
Notes
B1: Correct step length used, stated or clearly implied in working.
M1: Correct structure for Simpson’s rule \(\dfrac{1}{3}\text{“}h\text{”}\left[\text{ends} + 2\text{evens} + 4\text{odds}\right]\) using their value of \(h\). If no value of \(h\) is stated nor is implied and the \(\dfrac{1}{3}h\) is not clearly seen, award bod for any multiple in front of the bracket that does not imply a clearly incorrect method – but the insides must be correct. Condone a missing closing bracket at the end, but otherwise bracketing must be correct or implied correct by the answer. Condone minor miscopies of the ordinates as long as they are in correct positions.
A1: For 8.09, correct answer only, must be to 3.s.f. Must have scored the M.
Note calculator gives 8.086 …
| Scheme | Marks | AO |
|---|---|---|
| \(0.5 \times \text{“}8.09\text{”} = 4.045\) or 4.04 or 4.05 (awrt either) | B1ft | 2.2a |
| (1) | ||
| (5 marks) |
Notes
B1ft: Deduces the value by finding \(0.5 \times\) their answer to (b). Allow for awrt 3.s.f. answers giving tolerance with regards to prior rounding. Accept as a fraction, e.g. \(\dfrac{1213}{300}\) as long as it is half their answer to (b), or rounds correctly to it.