A2 October 2020 Q8
8.
\[\mathrm{f}(x) = \frac{3}{13 + 6\sin x - 5\cos x}\]Using the substitution \(t = \tan\left(\dfrac{x}{2}\right)\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{f}(x) = \dfrac{3}{13 + 6 \times \dfrac{2t}{1 + t^2} - 5 \times \dfrac{1 - t^2}{1 + t^2}}\) | M1 | 1.1b |
| \(= \dfrac{3(1 + t^2)}{13(1 + t^2) + 12t - 5(1 - t^2)}\) | M1 | 1.1b |
| \(= \dfrac{3(1 + t^2)}{18t^2 + 12t + 8} \Rightarrow\) for example \(\dfrac{3(1 + t^2)}{2(9t^2 + 6t + 1) + 6}\) or \(\dfrac{3(1 + t^2)}{2\left[(3t + 1)^2 - 1\right] + 8}\) \(\Rightarrow \dfrac{3(1 + t^2)}{2(3t + 1)^2 + 6}\ *\) | A1* | 2.1 |
| (3) |
Notes
M1: Uses one correct substitution
M1: Both substitutions correct and attempts to multiply through numerator and denominator by \(1 + t^2\).
A1*: Completes to the correct expression with no errors seen. Must see an intermediate step simplifying the denominator – most likely one of the ones seen in the scheme.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{f}(x) = \dfrac{3}{7} \Rightarrow \dfrac{3(1 + t^2)}{2(3t + 1)^2 + 6} = \dfrac{3}{7} \Rightarrow 21 + 21t^2 = 54t^2 + 36t + 24\) \(\Rightarrow 11t^2 + 12t + 1 = 0\) | M1 | 1.1b |
| \(\Rightarrow (11t + 1)(t + 1) = 0 \Rightarrow t = \ldots\) | M1 | 1.1b |
| \(t = -1,\ t = -\dfrac{1}{11}\) | A1 | 1.1b |
| \(\Rightarrow x = 2\arctan\left(\text{“}\textit{their } t\text{”}\right) + 2\pi\) for a negative \(t\) | dM1 | 3.1a |
| \(x = \dfrac{3\pi}{2}\) or awrt \(4.71\) and awrt \(x = 6.10\) | A1 | 1.1b |
| (5) |
Notes
M1: Equates the result in (a) to \(\dfrac{3}{7}\) and simplifies to a 3TQ
M1: Solves their equation by any valid means.
A1: Correct values for \(t\)
dM1: Dependent on first method mark. Applies the correct process to find at least one value for \(x\) from a negative value for \(t\). (If two positive values are found in error, this mark cannot be scored.)
A1: Both answers correct and no others in range.
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \mathrm{f}(x) = \int \dfrac{3(1 + t^2)}{2(3t + 1)^2 + 6} \times \dfrac{2}{1 + t^2}\,\mathrm{d}t = \int \dfrac{3}{(3t + 1)^2 + 3}\,\mathrm{d}t\) | B1 | 2.1 |
| \(= K\arctan\left(M(3t + 1)\right)\) or \(u = (3t + 1) \Rightarrow K\arctan\left(Mu\right)\) | M1 | 1.1b |
| \(= \dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{3t + 1}{\sqrt{3}}\right)\) or \(= \dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{u}{\sqrt{3}}\right)\) | A1 | 1.1b |
| \(\displaystyle\int_{\frac{\pi}{3}}^{\frac{4\pi}{3}}\mathrm{f}(x)\,\mathrm{d}x = \int_{\frac{\pi}{3}}^{\pi}\mathrm{f}(x)\,\mathrm{d}x + \int_{\pi}^{\frac{4\pi}{3}}\mathrm{f}(x)\,\mathrm{d}x\) \(\displaystyle = \int_{\frac{\sqrt{3}}{3}}^{\infty}\ldots\,\mathrm{d}t + \int_{-\infty}^{-\sqrt{3}}\ldots\,\mathrm{d}t\) or \(\displaystyle\int_{\sqrt{3} + 1}^{\infty}\ldots\,\mathrm{d}u + \int_{-\infty}^{1 - 3\sqrt{3}}\ldots\,\mathrm{d}u\) | B1 | 3.1a |
| \(= \dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{3\left(-\sqrt{3}\right) + 1}{\sqrt{3}}\right) - \dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{3\left(\frac{\sqrt{3}}{3}\right) + 1}{\sqrt{3}}\right) + \ldots\) | M1 | 1.1b |
| \(= \dfrac{\sqrt{3}}{3}\left(\arctan\left(\dfrac{\sqrt{3} - 9}{3}\right) - \arctan\left(\dfrac{\sqrt{3} + 3}{3}\right)\right) + \ldots\) | A1 | 1.1b |
| \(= \ldots + \lim\limits_{t \to \infty}\dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{3t + 1}{\sqrt{3}}\right) - \lim\limits_{t \to -\infty}\dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{3t + 1}{\sqrt{3}}\right)\) \(= \ldots + \dfrac{\pi}{2\sqrt{3}} - \left(-\dfrac{\pi}{2\sqrt{3}}\right)\) | M1 | 3.1a |
| \(= \dfrac{\sqrt{3}}{3}\left(\arctan\left(\dfrac{\sqrt{3} - 9}{3}\right) - \arctan\left(\dfrac{\sqrt{3} + 3}{3}\right) + \pi\right)\) | A1 | 2.1 |
| (8) | ||
| (16 marks) |
Notes
B1: Applies the substitution including the use of \(\mathrm{d}x = \dfrac{2}{1 + t^2}\,\mathrm{d}t\)
M1: Attempts the integration to achieve \(K\arctan\left(M(1 + 3t)\right)\) or \(K\arctan\left(Mu\right)\) if using a substitution of \(u = (3t + 1)\).
May use substitution \(3t + 1 = \sqrt{3}\tan\theta \Rightarrow \dfrac{\mathrm{d}t}{\mathrm{d}\theta} = \dfrac{\sqrt{3}}{3}\sec^2\theta \Rightarrow \displaystyle\int\dfrac{\sqrt{3}}{3}\,\mathrm{d}\theta = \dfrac{\sqrt{3}}{3}\theta = \dfrac{\sqrt{3}}{3}\arctan\left(\dfrac{3t + 1}{\sqrt{3}}\right)\)
A1: Correct integral.
B1: Changes the limits and splits the integral around \(\pi\)
M1: Applies their limits ‘\(\frac{1}{\sqrt{3}}\)’ and ‘\(-\sqrt{3}\)’ to their integrand.
A1: Correct “arctan” expressions.
M1: Correct work to evaluate the \(\pm\infty\) limits
A1: Fully correct solution.