June 2025 Paper 2 Q6

Notes

M1: Attempts to use \(\cos\theta \approx 1 - \dfrac{\theta^2}{2}\) with \(\theta\), \(2\theta\), \(x\) or \(2x\) and squares the resulting expression.
Condone poor squaring, e.g., \(\left(1 - 2x^2\right)^2 = 1 - 4x^4\) or missing brackets \(\left(1 - \dfrac{2x^2}{2}\right)^2 = \left(1 - x^2\right)^2 = \ldots\)
but not e.g. \(\cos 2x \approx 2\left(1 - \dfrac{x^2}{2}\right)\)
May be implied by e.g. \(1 - \left(1 - 2x^2\right)^2 = 2x^2\left(2 - 2x^2\right)\) from difference of two squares.

A1*: Correct proof with an intermediate line such as \(1 - \left(1 - 4x^2 + 4x^4\right)\)
Do not be concerned if the LHS does not appear and ignore any spurious = 0 in their work.
There should be no obvious incorrect statements in the proof e.g. \(1 - \left(1 + 4x^2 - 4x^4\right)\)
Do not condone incorrect work including invisible brackets such as \(1 - 1 - 4x^2 + 4x^4\) or \(\left(1 - \dfrac{2x^2}{2}\right)\) but condone a missing trailing bracket e.g. \(1 - (1 - 4x^2 + 4x^4\)
Condone an attempt that starts in stages, which may not deal with the full expression e.g., \(1 - \left(1 - \dfrac{4x^2}{2}\right) \rightarrow 1 - \left(1 - 2x^2\right)^2 = \ldots\) has a missing square on the first bracket but is recovered in the next stage/step.
The final line should be in terms of \(x\) but condone a slip to e.g. \(\theta\) in the workings of the proof.
If they complete the proof using \(\theta\) they must revert to \(x\) to score this mark.

Special Cases: In each of these cases below, please send to review.
You may see attempts that use e.g.

• \(\cos 2x = \pm 2\cos^2 x \pm 1\) or \(\cos^2 2x = \pm\dfrac{1}{2} \pm \dfrac{1}{2}\cos 4x\)
• \(\cos 2x = \pm 1 \pm 2\sin^2 x\) followed by the small angle approximation for \(\sin x\)
• Maclaurin expansions

Notes

M1: Uses the given answer to part (a) and writes \(\sin\left(\dfrac{x}{3}\right)\) and \(\tan\left(\dfrac{x}{2}\right)\) using correct small angle approximations. Only allow misreads that are clearly misreads e.g. they cannot replace \(\tan\left(\dfrac{x}{2}\right)\) with \(\dfrac{x}{3}\) unless \(\tan\left(\dfrac{x}{3}\right)\) is seen first. In such cases they will lose the A mark in (b) but both of the marks in (c) are available. Condone mixed variables for this mark.

A1: \(24 - 24x^2\) but condone e.g. \(24 + -24x^2\) or e.g. \(a = 24\) and \(b = -24\)
Condone recovery of missing/invisible brackets but the work must otherwise be correct.
Do not condone mixed variables being recovered unless explicitly replaced with \(x\) before cancelling \(x^2\).
No marks are scored in (b) for using the Maclaurin expansions for \(\sin\left(\dfrac{x}{3}\right)\) and/or \(\tan\left(\dfrac{x}{2}\right)\)

Notes

B1ft: 24 but this must follow from the non-zero constant term in their answer to (b).
Do not allow e.g. \(x = 24\)
Allow follow through on their non-zero constant term from a polynomial in \(x\).

dB1ft: Suitable reason given but it should refer to their \(x^2\) term in some way (and any additional terms if they have any). Ignore spurious remarks e.g. “if \(x \lt 1\)” unless contradictory.
Dependent on the previous B1ft mark.
Some acceptable examples:

• \(\text{``}24\text{''}x^2 \rightarrow 0\) or \(x^2 \rightarrow 0\)
• “We can ignore the \(x^2\) term”
• “\(x^2\) is much smaller than 24”
• As \(x \rightarrow 0\) their \(a + bx^2 \rightarrow a\) (condone as \(x \rightarrow 0\), \(bx^2\) “becomes” 0)
• \(\displaystyle\lim_{x \to 0} a + bx^2 = \left(a + b(0)^2\right) = a\)
• Since \(x\) is very small, \(24 - 24(0)^2 = 24\) (and condone if the squared is missing).

They cannot just substitute in 0 or a very small value for \(x\) to score the mark for the reason.
A reason such as “the answer rounds to 24” is not acceptable.
There must be some justification, either “as \(x \rightarrow 0\)” or “since \(x\) is very small”
so e.g. “\(24 - 24(0)^2 = 24\) on its own scores B1ft dB0ft
This mark may be scored from any polynomial in \(x\) that includes at least a constant term and an \(x^2\) term.
They cannot go back to e.g. \(4x^4\) or \(\dfrac{4x^4}{\frac{1}{6}x^2}\) and say this is negligible or \(\rightarrow 0\) as they haven’t dealt with the \(x\) in the denominator.