A2 October 2021 Q2
2.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\sin x - \cos x + 1}{\sin x + \cos x - 1} = \dfrac{\frac{2t}{1 + t^2} - \frac{1 - t^2}{1 + t^2} + 1}{\frac{2t}{1 + t^2} + \frac{1 - t^2}{1 + t^2} - 1} = \ldots\) | M1 | 1.1b |
| \(= \dfrac{2t - \left(1 - t^2\right) + 1 + t^2}{2t + 1 - t^2 - \left(1 + t^2\right)}\) or \(= \dfrac{2t - 1 + t^2 + 1 + t^2}{2t + 1 - t^2 - 1 - t^2}\) numerator \(= \dfrac{2t - \left(1 - t^2\right) + 1 + t^2}{1 + t^2}\) denominator \(= \dfrac{2t + 1 - t^2 - \left(1 + t^2\right)}{1 + t^2}\) and divides | M1 | 2.1 |
| \(= \dfrac{2t^2 + 2t}{2t - 2t^2}\left(= \dfrac{1 + t}{1 - t}\right)\) | A1 | 1.1b |
| \(= \dfrac{t + 1}{1 - t} \times \dfrac{1 + t}{1 + t} = \dfrac{t^2 + 2t + 1}{1 - t^2} = \dfrac{1 + t^2}{1 - t^2} + \dfrac{2t}{1 - t^2}\) Alt: \(\sec x + \tan x = \dfrac{1 + t^2}{1 - t^2} + \dfrac{2t}{1 - t^2} = \dfrac{1 + 2t + t^2}{1 - t^2}\) | M1 | 3.1a |
| \(= \dfrac{1}{\cos x} + \tan x = \sec x + \tan x\ *\) Alt: \(= \dfrac{(t + 1)^2}{(1 - t)(1 + t)} = \dfrac{1 + t}{1 - t} = LHS\) hence result proved.* | A1* | 2.1 |
| (5) |
Notes
M1: Applies the \(t\)-formulae to the left-hand side of expression. Allow slips in signs of the terms.
M1: Multiplies numerator and denominator through by \(1 + t^2\) (allow if they forget to multiply the 1’s). Alternative works separately on the numerator and denominator to combine terms and then divides.
A1: Correct \(\dfrac{\text{quadratic}}{\text{quadratic}}\) with terms gathered, award where first seen, need not have cancelled \(2t\) for this mark.
M1: Cancels \(2t\), multiplies numerator and denominator by \(1 + t\) and splits to sum of two terms. If working from both sides, this mark is for substituting the \(t\)-formulae into the right-hand side and combining to single fraction.
A1*: Correct completion to given result. No errors in proof. If working from both sides, a suitable conclusion is needed, e.g “hence proven”.
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int_{(0)}^{\left(\frac{\pi}{2}\right)} \frac{5}{4 + 2\cos\theta}\,\mathrm{d}\theta = \int_{(0)}^{(1)} \frac{5}{4 + 2\frac{1 - t^2}{1 + t^2}} \times \frac{2}{1 + t^2}\,\mathrm{d}t\) Alternatively \(\dfrac{\mathrm{d}t}{\mathrm{d}\theta} = \dfrac{1}{2}\sec^2\left(\dfrac{\theta}{2}\right) = \dfrac{1}{\cos\theta + 1}\) leading to \(\displaystyle\int_{(0)}^{\left(\frac{\pi}{2}\right)} \frac{5}{4 + 2\cos\theta}\,\mathrm{d}\theta = \int_{(0)}^{(1)} \frac{5\cos\theta + 5}{4 + 2\cos\theta}\,\mathrm{d}t = \int_{(0)}^{(1)} \frac{5\left(\frac{1 - t^2}{1 + t^2}\right) + 5}{4 + 2\left(\frac{1 - t^2}{1 + t^2}\right)}\,\mathrm{d}t\) | M1 | 2.1 |
| \(\displaystyle = \int_{(0)}^{(1)} \frac{10}{4\left(1 + t^2\right) + 2\left(1 - t^2\right)}\,\mathrm{d}t = \int_{(0)}^{(1)} \frac{5}{3 + t^2}\,\mathrm{d}t\) o.e. | A1 | 1.1b |
| \(= \left[5 \times \dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{t}{\sqrt{3}}\right)\right]_{(0)}^{(1)}\) | M1 | 1.1b |
| \(= \dfrac{5}{\sqrt{3}}\left(\arctan\left(\dfrac{1}{\sqrt{3}}\right) - 0\right)\) or \(\dfrac{5}{\sqrt{3}}\left(\arctan\left(\dfrac{\tan\left(\frac{\pi}{4}\right)}{\sqrt{3}}\right) - \arctan\left(\dfrac{\tan(0)}{\sqrt{3}}\right)\right)\) | M1 | 2.2a |
| \(= \dfrac{5\pi\sqrt{3}}{18}\) oe in a surd form e.g. \(\dfrac{5\pi}{6\sqrt{3}}\) | A1 | 1.1b |
| (5) | ||
| (10 marks) |
Notes
M1: Applies the substitution including the use of \(\mathrm{d}\theta = \dfrac{2}{1 + t^2}\,\mathrm{d}t\) (Limits not needed for first three marks).
A1: Simplifies correctly to a recognisable integrable form.
M1: Integrates to the form \(K\arctan\left(\dfrac{t}{a}\right)\) where \(a^2\) is their constant term.
M1: Deduces correct limits and applies them the correct way round OR deduces integral in terms of \(\theta\) from their integration and applies original limits the correct way.
A1: \(\dfrac{5\pi\sqrt{3}}{18}\) or equivalent in surd form.
Note use of calculator does not lead to the exact value required in the question \(= 1.51149947\)
This can score M1 A1 M0 M1 A0