June 2024 Paper 2 Q5

Notes

B1: Award this mark for \(\theta\tan 2\theta = \theta \times 2\theta\) or \(1 - \cos 3\theta = 1 - \left(1 - \frac{(3\theta)^2}{2}\right)\) or equivalents.
May be seen when working on numerator or denominator separately or within the fraction.
This is a B mark so if awarding for \(\cos 3\theta\) do not condone missing brackets e.g. \(1 - \dfrac{3\theta^2}{2}\) unless they are recovered or are implied by subsequent work.

M1: Attempts to use both correct small angle approximations in the given expression.
For this mark they must have attempted to use \(\tan 2\theta = 2\theta\) and \(\cos 3\theta = 1 - \dfrac{(3\theta)^2}{2}\) in the given expression but condone poor bracketing e.g. \(\dfrac{\theta \times 2\theta}{1 - \left(1 - \tfrac{3\theta^2}{2}\right)}\) or e.g. \(\dfrac{\theta \times 2\theta}{1 - 1 - \tfrac{3\theta^2}{2}}\)
Do not allow e.g. \(\dfrac{\theta \times 2\theta}{1 - \frac{(3\theta)^2}{2}}\) as this suggests they are approximating \(\dfrac{\theta\tan 2\theta}{\cos 3\theta}\)

A1: Correct value. Do not allow rounded decimals e.g. 0.444 but allow if recurring decimals are clearly indicated e.g. \(0.\dot{4}\) Do not allow e.g. \(\dfrac{2}{4.5}\). Ignore any units if given.
Isw once a correct answer is seen.

Examples:

\(\dfrac{\theta \times 2\theta}{1 - \left(1 - \tfrac{3\theta^2}{2}\right)} = \dfrac{4}{3}\left(\text{or } -\dfrac{4}{3}\right)\)	scores B1M1A0 (Missing brackets not recovered)
\(\dfrac{\theta \times 2\theta}{1 - \frac{(3\theta)^2}{2}}\)	scores B1M0A0 (Missing “1 –“ in the denominator so M0)
\(\dfrac{\theta \times 2\theta}{1 + \left(1 - \frac{(3\theta)^2}{2}\right)}\)	scores B1M0A0 (Has “1 +“ in the denominator so M0)
\(\dfrac{\theta \times \theta}{1 - \left(1 - \tfrac{3\theta^2}{2}\right)} = \ldots\)	scores B0M0A0 (The B mark could be recovered but M0 because of the incorrect numerator)
\(\dfrac{\theta \times 2\theta}{1 - \left(1 - \tfrac{3\theta^2}{2}\right)} = \dfrac{2\theta^2}{\frac{9\theta^2}{2}} = \dfrac{2}{9}\)	scores B1M1A0 (Missing brackets recovered)
\(\dfrac{\theta \times 2\theta}{1 - \left(1 - \left(\frac{3\theta}{2}\right)^2\right)}\)	Scores B1M0A0 (The denominator suggests an incorrect expansion – unless it was recovered.)
\(\dfrac{\theta \times 2\theta}{\frac{9\theta^2}{2}} = \dfrac{2}{18}\)	B1M1A0 (The B1 is awarded for the numerator but can be implied by the denominator. The M1 is implied)
\(\dfrac{\theta \times 2\theta}{1 - \left(1 - \tfrac{3\theta^2}{2}\right)} = \dfrac{4}{9}\)	B1M1A1 (The correct value implies correct recovery of missing brackets.)

Note that other approaches are possible using identities.
In such cases we will allow correct work leading to an expression that if terms in \(\theta^3\) and higher can be ignored will lead to \(\dfrac{4}{9}\)
But to score the M mark they must be using correct identities and correct approximations but condone bracketing errors as in the main scheme.

Examples:

\[\frac{\theta\tan 2\theta}{1 - \cos 3\theta} = \frac{\theta \times \dfrac{\sin 2\theta}{\cos 2\theta}}{1 - \cos 3\theta} = \frac{\theta \times \dfrac{2\theta}{1 - \frac{(2\theta)^2}{2}}}{1 - \left(1 - \dfrac{(3\theta)^2}{2}\right)} = \frac{2\theta^2}{1 - 2\theta^2} \times \frac{2}{9\theta^2} = \frac{4\theta^2}{9\theta^2 - 18\theta^4}\]\[= \frac{4\theta^2}{9\theta^2} = \frac{4}{9}\]

Scores B1M1A1

\[\text{Similarly: } \frac{\theta\tan 2\theta}{1 - \cos 3\theta} = \frac{\theta \times \sin 2\theta}{\cos 2\theta(1 - \cos 3\theta)} = \frac{\theta \times 2\theta}{\left(1 - \dfrac{(2\theta)^2}{2}\right)\left(1 - \left(1 - \dfrac{(3\theta)^2}{2}\right)\right)} = \frac{2\theta^2}{1 - 2\theta^2} \times \frac{2}{9\theta^2} \text{ etc.}\]\[\frac{4\theta^2}{9\theta^2 - 18\theta^4} = \frac{4\theta^2}{9\theta^2} = \frac{4}{9}\]

Scores B1M1A1

\[\frac{\theta\tan 2\theta}{1 - \cos 3\theta} = \frac{\theta \times \sin 2\theta}{\cos 2\theta(1 - \cos 3\theta)} = \frac{\theta \times 2\theta}{\left(1 - \dfrac{(2\theta)^2}{2}\right)\left(1 - \left(1 - \dfrac{(3\theta)^2}{2}\right)\right)} = \frac{2\theta^2}{1 - 2\theta^2} \times \frac{2}{9\theta^2}\]\[= \frac{4\theta^2}{9\theta^2 - 18\theta^4} = \frac{4}{9 - 18\theta^2} = \frac{4}{9}\]

Scores B1M1A0
(They cannot just assume the term in \(\theta^2\) is 0 unless they provide a convincing limiting argument e.g. \(\displaystyle\lim_{\theta \to 0} \frac{4}{9 - 18\theta^2} = \frac{4}{9}\) or equivalent)

Example of a candidate’s response (handwritten in the mark scheme):

\[\begin{aligned}\tan 2\theta &= \frac{2\tan\theta}{1 - \tan^2\theta} \approx \frac{2\theta}{1 - \theta^2}\\\cos 3\theta &= \cos\theta\left(1 - 2\sin^2\theta\right) - 2\cos\theta\sin^2\theta\\&= \cos\theta\left(1 - 4\sin^2\theta\right)\\&\approx \left(1 - \frac{\theta^2}{2}\right)\left(1 - 4\theta^2\right)\\&= 1 - \frac{9}{2}\theta^2 + 2\theta^4\end{aligned}\]\[\text{So } \frac{\theta\tan 2\theta}{1 - \cos 3\theta} = \frac{\dfrac{2\theta^2}{1 - \theta^2}}{1 - \left(1 - \frac{9}{2}\theta^2 + 2\theta^4\right)} = \frac{\dfrac{2\theta^2}{1 - \theta^2}}{\frac{9}{2}\theta^2 - 2\theta^4}\]\[= \frac{\dfrac{4}{1 - \theta^2}}{9 - 4\theta^2} = \frac{4}{\left(9 - 4\theta^2\right)\left(1 - \theta^2\right)} = \frac{4}{9 - 13\theta^2 + 4\theta^4} \approx \frac{4}{9 - 13\theta^2}\]

Scores B1M1A0

\[\frac{\theta\tan 2\theta}{1 - \cos 3\theta} = \frac{\theta \times \dfrac{2\tan\theta}{1 - \tan^2\theta}}{1 - \left(4\cos^3\theta - 3\cos\theta\right)} = \frac{\theta \times \dfrac{2\theta}{1 - \theta^2}}{1 - \left(4\left(1 - \dfrac{\theta^2}{2}\right)^3 - 3\left(1 - \dfrac{\theta^2}{2}\right)\right)}\]

Scores B1M1A0

Note that attempts to use expansions in higher powers of \(\theta\) should be sent to review.