C4 June 2016 Q6

Notes

1^st M1: Writing \(\dfrac{3y - 4}{y(3y + 2)} \equiv \dfrac{A}{y} + \dfrac{B}{(3y + 2)}\) and a complete method for finding the value of at least one of their \(A\) or their \(B\).

Note: M1A1 can be implied for writing down either \(\dfrac{3y - 4}{y(3y + 2)} \equiv \dfrac{-2}{y} + \dfrac{\text{their } B}{(3y + 2)}\) or \(\dfrac{3y - 4}{y(3y + 2)} \equiv \dfrac{\text{their } A}{y} + \dfrac{9}{(3y + 2)}\) with no working.

Note: Correct bracketing is not necessary for the penultimate A1ft, but is required for the final A1 in (i)

Note: Give 2^nd M0 for \(\dfrac{3y - 4}{y(3y + 2)}\) going directly to \(\pm\alpha\ln(3y^2 + 2y)\)

Note: …but allow 2^nd M1 for either \(\dfrac{M(6y + 2)}{3y^2 + 2y} \to \pm\alpha\ln(3y^2 + 2y)\) or \(\dfrac{M(3y + 1)}{3y^2 + 2y} \to \pm\alpha\ln(3y^2 + 2y)\)

Way 2

Way 3

Way 4

Notes

1^st M1: Substitutes \(x = 4\sin^2\theta\) and their \(\mathrm{d}x\) \(\left(\text{from their correctly rearranged } \dfrac{\mathrm{d}x}{\mathrm{d}\theta}\right)\) into \(\sqrt{\left(\dfrac{x}{4 - x}\right)}\,\mathrm{d}x\)

Note: \(\mathrm{d}x \neq \lambda\,\mathrm{d}\theta\). For example \(\mathrm{d}x \neq \mathrm{d}\theta\)

Note: Allow substituting \(\mathrm{d}x = 4\sin 2\theta\) for the 1^st M1 after a correct \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = 4\sin 2\theta\) or \(\mathrm{d}x = 4\sin 2\theta\,\mathrm{d}\theta\)

2^nd M1: Applying \(x = 4\sin^2\theta\) to \(\sqrt{\left(\dfrac{x}{4 - x}\right)}\) to give \(\pm K\tan\theta\) or \(\pm K\left(\dfrac{\sin\theta}{\cos\theta}\right)\)

Note: Integral sign is not needed for this mark.

1^st A1: Simplifies to give \(\displaystyle\int 8\sin^2\theta\,\mathrm{d}\theta\) including \(\mathrm{d}\theta\)

2^nd B1: Writes down a correct equation involving \(x = 3\) leading to \(\theta = \dfrac{\pi}{3}\) and no incorrect work seen regarding limits

Note: Allow 2^nd B1 for \(x = 4\sin^2\left(\dfrac{\pi}{3}\right) = 3\) and \(x = 4\sin^2 0 = 0\)

Note: Allow 2^nd B1 for \(\theta = \sin^{-1}\left(\sqrt{\dfrac{x}{4}}\right)\) followed by \(x = 3, \theta = \dfrac{\pi}{3};\ x = 0, \theta = 0\)

Alternative methods for B1M1M1A1 in (ii)(a)

Way 2

Way 3

Notes

M1: Writes down a correct equation involving \(\cos 2\theta\) and \(\sin^2\theta\)
E.g.: \(\cos 2\theta = 1 - 2\sin^2\theta\) or \(\sin^2\theta = \dfrac{1 - \cos 2\theta}{2}\) or \(K\sin^2\theta = K\left(\dfrac{1 - \cos 2\theta}{2}\right)\)
and applies it to their integral. Note: Allow M1 for a correctly stated formula (via an incorrect rearrangement) being applied to their integral.

M1: Integrates to give an expression of the form \(\pm\alpha\theta \pm \beta\sin 2\theta\) or \(k(\pm\alpha\theta \pm \beta\sin 2\theta)\), \(\alpha \neq 0, \beta \neq 0\) (can be simplified or un-simplified).

1^st A1: Integrating \(\sin^2\theta\) to give \(\dfrac{1}{2}\theta - \dfrac{1}{4}\sin 2\theta\), un-simplified or simplified. Correct solution only.
Can be implied by \(k\sin^2\theta\) giving \(\dfrac{k}{2}\theta - \dfrac{k}{4}\sin 2\theta\) or \(\dfrac{k}{4}(2\theta - \sin 2\theta)\) un-simplified or simplified.

2^nd A1: A correct solution in part (ii) leading to a “two term” exact answer of
e.g. \(\dfrac{4}{3}\pi - \sqrt{3}\) or \(\dfrac{8}{6}\pi - \sqrt{3}\) or \(\dfrac{4}{3}\pi - \dfrac{2\sqrt{3}}{2}\) or \(\dfrac{1}{3}\left(4\pi - 3\sqrt{3}\right)\)

Note: A decimal answer of 2.456739397… (without a correct exact answer) is A0.

Note: Candidates can work in terms of \(\lambda\) (note that \(\lambda\) is not given in (ii)) and gain the 1^st three marks (i.e. M1M1A1) in part (b).

Note: If they incorrectly obtain \(\displaystyle\int_0^{\frac{\pi}{3}} 8\sin^2\theta\,\mathrm{d}\theta\) in part (i)(a) (or correctly guess that \(\lambda = 8\)) then the final A1 is available for a correct solution in part (ii)(b).