A2 June 2025 Q1
1.
| \(x\) | 1.5 | 1.75 | 2 | 2.25 | 2.5 |
|---|---|---|---|---|---|
| \(y\) | 2.73 | 2.81 | 5.22 | 16.31 |
| Scheme | Marks | AO |
|---|---|---|
| 3.35 cao | B1 | 1.1b |
| (1) |
Notes
B1: Correct value 3.35 only, this may be seen in the table or in the script
| Scheme | Marks | AO |
|---|---|---|
| Step length = 0.25 | B1 | 1.1b |
| \(\dfrac{1}{3} \times \text{“}0.25\text{”}\left[2.73 + 16.31 + 2(\text{“}3.35\text{”}) + 4(2.81 + 5.22)\right]\) | M1 | 1.1b |
| \(= 4.8\) | A1 | 1.1b |
| (3) |
Notes
B1: Correct step length, may be seen in the application of Simpson’s rule formula. Could be seen as \(1.75 - 1.5\) or \(\dfrac{2.5 - 1.5}{4}\)
M1: Correct application of Simpson’s rule using their step length. If no step length stated and uses \(\dfrac{1}{3}\left[\text{ends} + 2(\text{evens}) + 4(\text{odd})\right]\) this is M0
A1: Awrt 4.8 must be using 3.35 from (a)
Note calculator answer is 4.67 is no marks
Note Answer only with no working is no marks,
Note: If they write \(\dfrac{1}{3} \times h\left[\text{ends} + 2(\text{evens}) + 4(\text{odd})\right]\) which leads to awrt 4.8 this gets B1M1A1
If their answer is incorrect this would be B0 (if no \(h\) stated) M0A0
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int_{1.5}^{2.5} \mathrm{e}^{\cot^2 x}\,\mathrm{d}x = \int_{1.5}^{2.5} \mathrm{e}^{\operatorname{cosec}^2 x - 1}\,\mathrm{d}x = \frac{1}{\mathrm{e}}\int_{1.5}^{2.5} \mathrm{e}^{\operatorname{cosec}^2 x}\,\mathrm{d}x = \frac{1}{\mathrm{e}} \times \text{“}4.8\text{”}\) | M1 | 3.1a |
| \(=\) awrt 1.8 | A1ft | 1.1b |
| (2) | ||
| (6 marks) |
Notes
M1: Uses a correct identity to express \(\displaystyle\int \mathrm{e}^{\cot^2 x}\,\mathrm{d}x\) in terms of \(\displaystyle\int \mathrm{e}^{\operatorname{cosec}^2 x}\,\mathrm{d}x\) and then uses correct index work to allow \(\displaystyle\int_{1.5}^{2.5} \mathrm{e}^{\cot^2 x}\,\mathrm{d}x\) to be evaluated using part (b) \(\dfrac{1}{\mathrm{e}} \times \text{“}4.8\text{”}\)
A1ft: Awrt 1.8 or follow through \(\dfrac{1}{\mathrm{e}} \times \text{their (b)}\) which must be evaluated and correct for their (b) rounded to 1 d.p.
If uses Simpson’s rule again this is M0 A0