A2 June 2024 Paper 2 Q3
3.
| Scheme | Marks | AO |
|---|---|---|
| Because the upper limit is infinite | B1 | 2.4 |
| (1) |
Notes
B1: Suitable explanation stating that one of the bounds/limits is infinite oe isw
e.g. “one of the limits is unbounded” or “the integral is unbounded”
Do not allow this mark if they only say the limit or integral is undefined, unless they go on to say it is undefined at infinity.
Commenting that the function is undefined at infinity is not enough to award this mark on its own, this question concerns the limits.
If the candidates state any extra incorrect comments about the limits withhold this mark e.g. not defined at \(\dfrac{4}{3}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \frac{1}{9x^2 + 16}\,\mathrm{d}x = \frac{1}{12}\arctan\left(\frac{3x}{4}\right)\) | M1 A1 | 3.1a 1.1b |
| \(\displaystyle\int_{\frac{4}{3}}^{\infty} \frac{1}{9x^2 + 16}\,\mathrm{d}x = \frac{1}{12}\lim_{t \to \infty}\left[\arctan\left(\frac{3x}{4}\right)\right]_{\frac{4}{3}}^{t}\) \(\displaystyle = \frac{1}{12}\left(\lim_{t \to \infty}\arctan\left(\frac{3t}{4}\right) - \arctan(1)\right)\) | dM1 | 1.1b |
| \(= \dfrac{1}{12}\left(\dfrac{\pi}{2} - \dfrac{\pi}{4}\right) = \dfrac{\pi}{48}\) | A1 | 2.1 |
| (4) | ||
| (5 marks) |
Notes
M1: Integrates to obtain \(\alpha\arctan(\beta x)\) where \(\beta \neq 1\)
A1: Correct integration, unsimplified or simplified.
dM1: Applies correct limits, "\(t\)" and \(\dfrac{4}{3}\) with evidence of applying the infinite limit to obtain a non-zero value.
Allow with \(\infty\) used as the limit (which may be implied by \(\dfrac{\pi}{2}\))
A1: Correct value obtained with evidence of use of limiting process on the upper bound. Withhold this mark if there is no evidence of using the limiting process. We must see as a minimum \(\displaystyle\lim_{t \to \infty}\) oe at some stage in their work.
e.g. \(\left[\dfrac{1}{12}\arctan\left(\dfrac{3x}{4}\right)\right]_{\frac{4}{3}}^{\infty} = \dfrac{1}{12}\left(\dfrac{\pi}{2} - \dfrac{\pi}{4}\right) = \dfrac{\pi}{48}\) would score dM1A0