AS June 2023 Q2
2.
| Scheme | Marks | AO |
|---|---|---|
| \(3\cos x - 2\sin x \Rightarrow 3\left(\dfrac{1 - t^2}{1 + t^2}\right) - 2\left(\dfrac{2t}{1 + t^2}\right)\) | M1 | 1.1b |
| \(3\left(\dfrac{1 - t^2}{1 + t^2}\right) - 2\left(\dfrac{2t}{1 + t^2}\right) = 1 \Rightarrow 3(1 - t^2) - 4t = 1 + t^2\) | M1 | 1.1b |
| \(2t^2 + 2t - 1 = 0\ *\) | A1* | 2.1 |
| (3) |
Notes
M1: Uses at least one of \(\sin x = \dfrac{2t}{1 + t^2}\) or \(\cos x = \dfrac{1 - t^2}{1 + t^2}\) to express \(3\cos x - 2\sin x\) in terms of \(t\) only.
M1: Uses both correct formulae for \(\sin x = \dfrac{2t}{1 + t^2}\) and \(\cos x = \dfrac{1 - t^2}{1 + t^2}\) in \(3\cos x - 2\sin x\) equates their expression to 1 and multiplies through by \(1 + t^2\) to achieve a quadratic expression in \(t\) (there may be slips in coefficients). Alternatively, gather all terms to a single fraction over denominator \(1 + t^2\).
A1*: Collects terms to one side and simplifies to achieve the printed answer, with no errors seen.
(Brackets that are missing from the printed mark scheme because of a font fault have been restored.)
| Scheme | Marks | AO |
|---|---|---|
| \(t = \dfrac{-2 \pm \sqrt{(2)^2 - 4(2)(-1)}}{2(2)}\ \left\{= \dfrac{-1 \pm \sqrt{3}}{2} = -1.366\ldots, 0.366\ldots\right\}\) | M1 | 1.1b |
| \(\dfrac{x}{2} = \arctan(-1.366\ldots)\) or \(\dfrac{x}{2} = \arctan(0.366\ldots)\) and \(\Rightarrow x = \ldots\) | dM1 | 3.1a |
| \(x = \text{awrt } -107.6^\circ\) or awrt \(40.2^\circ\) | A1 | 1.1b |
| \(x = -107.6^\circ,\ 40.2^\circ\) cao | A1 | 1.1b |
| (4) | ||
| (7 marks) |
Notes
Note: The question says “hence solve”, so there needs to be evidence of use of part (a). Answers only scores no marks. Minimal requirement would be to see the solutions for \(t\) before stating the final answers.
M1: Selects a correct process to solve \(2t^2 + 2t - 1 = 0\) (calculator, quadratic formula, completing the square) to obtain at least one value for \(t\). Attempts by factorisation are M0.
dM1: Adopts a correct strategy of taking arctan(their value for \(t\)) and multiplies the result by 2 to obtain at least one value for \(x\) within the range \(-180^\circ \lt x \lt 180^\circ\). Allow slips on, or miscopies of their roots if the process is correct.
A1: One correct answer awrt 1 d.p.
A1: Both correct answers to 1 d.p. and no incorrect answers in the range \(-180^\circ \lt x \lt 180^\circ\)
(Brackets that are missing from the printed mark scheme because of a font fault have been restored.)