Notes

M1: Proceeds to \(\dfrac{1}{A\sin^2 u\sqrt{\pm C \pm C\sin^2 u}} \times B\cos u\) o.e. or may be implied by \(\dfrac{\cos u}{D\sin^2 u\cos u}\) o.e in terms of \(u\) only \(A, B, C, D \neq 0\)
Requires \(\mathrm{d}x \rightarrow \ldots\cos u\ (\mathrm{d}u)\) o.e. Condone the omission of the integral sign and d\(u\) for this mark. Condone the \(\pm B\cos u\) appearing after the d\(u\) if present
It cannot be implied by \(\dfrac{1}{D\sin^2 u}\) or \(\dfrac{\operatorname{cosec}^2 u}{D}\) where \(D \neq 0\)

dM1: Attempts to use \(\pm 1 \pm \sin^2 u = \pm\cos^2 u\) to change \(\sqrt{\pm C \pm C\sin^2 u}\) to \(p\cos u\) o.e. (Must be seen somewhere in their solution which may be part of side-workings)
May be seen as e.g. \(\dfrac{1}{A\sin^2 u\sqrt{C\cos^2 u}} \times B\cos u\) or \(\dfrac{1}{A\sin^2 u \times F\cos u} \times B\cos u\)
It is dependent on the previous method mark.

A1: \(\displaystyle\int \dfrac{1}{4}\operatorname{cosec}^2 u\,\mathrm{d}u\) Ignore any limits or the absence of them for this mark.
Requires the integral sign and d\(u\). Do not isw. Must be seen in (a)

B1: Deduces \(a = \dfrac{\pi}{6}\) and \(b = \dfrac{\pi}{3}\) (in radians) May be with their final integral in the correct positions or stated separately which value is \(a\) and which value is \(b\). Must be seen in (a).

Alternative part (a) (Further Maths)

M1: \(u = \arcsin\left(\dfrac{x}{2}\right) \Rightarrow \dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{M}{\sqrt{1 - Nx^2}}\) It cannot be implied by \(\dfrac{1}{D\sin^2 u}\) or \(\dfrac{\operatorname{cosec}^2 u}{D},\ D \neq 0\)

dM1: Uses \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{1}{\sqrt{4 - x^2}}\) o.e, substitutes \(x = 2\sin u\) and proceeds to \(\dfrac{1}{4\sin^2 u}\) It is dependent on the previous method mark.

A1: \(\displaystyle\int \dfrac{1}{4}\operatorname{cosec}^2 u\,\mathrm{d}u\) Ignore any limits or the absence of them for this mark.
Requires the integral sign and d\(u\). Do not isw. Must be seen in (a)

B1: Deduces \(a = \dfrac{\pi}{6}\) and \(b = \dfrac{\pi}{3}\) (must be in radians)
May be with their final integral in the correct positions or stated separately which value is \(a\) and which value is \(b\). Must be seen in (a).

Notes

Condone x instead of u provided the appropriate limits are substituted into the function

B1ft: \(k\operatorname{cosec}^2 u \rightarrow -k\cot u\) (may be left in terms of \(k \neq 0\))
e.g. \(-2\operatorname{cosec}^2 u \rightarrow 2\cot u\) is B1 (look for the sign to change as well)
Note some candidates may prefer to change

M1: Uses the limits \(\dfrac{\pi}{6}\) and \(\dfrac{\pi}{3}\) either way round (or condone use of \(30^\circ\) and \(60^\circ\)) in an expression of the form \(\pm q\cot u\) and subtracts (either way round).
Allow \(q = 1\). May be implied by their final answer provided B1ft has been scored.
May write as e.g. \(-\dfrac{1}{4}\dfrac{\cos\frac{\pi}{3}}{\sin\frac{\pi}{3}} + \dfrac{1}{4}\dfrac{\cos\frac{\pi}{6}}{\sin\frac{\pi}{6}}\)
Do not condone a sign slip here e.g. \(-\dfrac{1}{4}\cot\dfrac{\pi}{3} - \dfrac{1}{4}\cot\dfrac{\pi}{6}\) is M0.
The expression is sufficient but if just a value is stated following integration to the required form then you may need to check this on your calculator.
If no algebraic integration is seen then M0
If only decimal values are seen then M0

A1: \(\dfrac{1}{2\sqrt{3}}\) or \(\dfrac{\sqrt{3}}{6}\) provided the previous two marks have been scored.

Note that incorrect integration e.g. \(\displaystyle\int k\ \operatorname{cosec}^2 u\,\mathrm{d}u = k\cot u \rightarrow \dfrac{1}{2\sqrt{3}}\) scores B0ftM1A0

Notes

Condone a complete proof entirely in \(\theta\) (or another variable) instead of x.
Do not be concerned with the omission of degrees.

M1: Attempts to use both compound angle expansions to set up an expression in \(\sin x\) and \(\cos x\)
i.e. \(\pm\sin x\cos 30^\circ \pm \cos x\sin 30^\circ \pm \sqrt{3}\left(\pm\cos x\cos 30^\circ \pm \sin x\sin 30^\circ\right)\)
The terms must be correct but condone sign errors and a slip on the multiplication of the \(\sqrt{3}\) if they attempt to multiply out the brackets. (The \(\sqrt{3}\) may be omitted entirely)
This mark may be implied by further work
e.g. \(\pm\dfrac{\sqrt{3}}{2}\sin x \pm \dfrac{1}{2}\cos x \pm \sqrt{3}\left(\pm\dfrac{\sqrt{3}}{2}\cos x \pm \dfrac{1}{2}\sin x\right)\)

A1: Correct expression \(\sin x\cos 30^\circ + \cos x\sin 30^\circ + \sqrt{3}\left(\cos x\cos 30^\circ - \sin x\sin 30^\circ\right)\) o.e.
May be implied by
\(\dfrac{\sqrt{3}}{2}\sin x + \dfrac{1}{2}\cos x + \sqrt{3}\left(\dfrac{\sqrt{3}}{2}\cos x - \dfrac{1}{2}\sin x\right)\) (or implied if multiplied out)

A1*: Proceeds to the given answer with no errors seen including invisible brackets (condone a missing trailing bracket). We must see the exact numerical values used for \(\sin 30^\circ\) and \(\cos 30^\circ\) before proceeding to the given answer.
Minimum required
\(\dfrac{\sqrt{3}}{2}\sin x + \dfrac{1}{2}\cos x + \sqrt{3}\left(\dfrac{\sqrt{3}}{2}\cos x - \dfrac{1}{2}\sin x\right) = 2\cos x\) which scores M1A1A1*
\(\sin x\cos 30^\circ + \cos x\sin 30^\circ + \sqrt{3}\left(\cos x\cos 30^\circ - \sin x\sin 30^\circ\right) = 2\cos x\) scores M1A1A0*
There should not be any notational or bracketing errors and no mixed or missing variables.

Alternative for part (a)

M1: Writes the left hand side of the equation as
\(\dfrac{1}{2}\sin(x + 30^\circ) + \dfrac{\sqrt{3}}{2}\cos(x + 30^\circ) = \sin 30^\circ\sin(x + 30^\circ) + \cos 30^\circ\cos(x + 30^\circ)\)

A1: Correct expression for \(\sin 30^\circ\sin(x + 30^\circ) + \cos 30^\circ\cos(x + 30^\circ) = \cos x\)

A1*: Proceeds to the given answer \(\sin(x + 30^\circ) + \sqrt{3}\cos(x + 30^\circ) = 2\cos(x + 30^\circ - 30^\circ) = 2\cos x\) with no errors seen including invisible brackets (condone a missing trailing bracket).
There should not be any notational or bracketing errors and no mixed or missing variables.

Alternative part (a) – using the R-alpha method

M1: States e.g. \(R\cos(x + 30 \pm \alpha) = \sqrt{3}\cos(x + 30) \mp \sin(x + 30)\)
and attempts to find either \(R\) or \(\alpha\) correctly. (may be implied)

A1: Achieves \(2\cos(x + 30 - 30)\)

A1*: Achieves \(2\cos x\) with no errors seen and both stages of working shown e.g.

• States \(\left(R\cos(x + 30 \pm \alpha) =\right) \sqrt{3}\cos(x + 30)\cos\alpha \mp \sin(x + 30)\sin\alpha\) oe
• Shows \(R = \sqrt{1 + 3} = 2\) and \(\tan\alpha = \left(\dfrac{1}{\sqrt{3}}\right)\) o.e. e.g. using cosine or sine

Notes

Condone the use of x for \(\theta\) and mixed variables

M1: Sets \(2\cos\theta = 3\sin 2\theta\) and proceeds to \(A\cos\theta = B\sin\theta\cos\theta\). (allow \(A = B\))
May be implied by \(\sin\theta = k\ \ (k \neq 0, 1)\)

A1: \(\sin\theta = \dfrac{1}{3}\)

dM1: Finds at least one of their values of \(\theta\) for their \(\sin\theta = k\ (k \neq 0, 1)\) It is dependent on the previous method mark. You may need to check their value(s) (in degrees or radians) but may be implied by e.g. awrt \(19^\circ\) (or awrt \(20^\circ\)) or awrt \(161^\circ\) (or awrt \(160^\circ\))
(awrt 0.34 or awrt 2.8 in radians)

A1: Deduces that \((\theta =)\ 90^\circ\) as well as giving \((\theta =)\) awrt \(19.5^\circ\), awrt \(160.5^\circ\) with no other values in the given range (ignore any found outside of the range)
The degree symbol is not required. (Note the angles are 19.4712206…and 160.528779…)
Answers in radians score A0.

Notes

B1: cao \((n =)\ 3\)

Notes

B1: Either:

• Deduces 30 solutions
• or deduces the correct total number of solutions for \(\sin(nx) = k\) and \(\sin(nx) = -k\) (or \(\sin^2(nx) = k^2\)) in any relevant interval (0 to \(\pi\), \(2\pi\), \(4\pi\) or \(5\pi\))
• or deduces the correct number of solutions in 0 to \(5\pi\) for either \(\sin(nx) = k\) or \(\sin(nx) = -k\)
• or deduces the correct total number of solutions for \(\sin(x) = k\) and \(\sin(x) = -k\) in any relevant interval (0 to \(\pi\), \(2\pi\), \(4\pi\) or \(5\pi\))

dB1: Requires 30 (solutions) and correct justification comprising all the elements in one of the bullet points below. See the tables below for the correct values.

• Deduces the correct total number of solutions for \(\sin(nx) = k\) and \(\sin(nx) = -k\) (or \(\sin^2(nx) = k^2\)) in any relevant interval (0 to \(\pi\), \(2\pi\) or \(4\pi\)) and scales the interval to 0 to \(5\pi\)
• or deduces the correct number of solutions in 0 to \(5\pi\) for \(\sin(nx) = k\) and for \(\sin(nx) = -k\) and then adds
• or finds the total correct number of solutions for \(\sin(x) = k\) and \(\sin(x) = -k\) in any relevant interval (0 to \(\pi\), \(2\pi\), \(4\pi\) or \(5\pi\)) and then scales the interval to 0 to \(5\pi\) and multiplies by 3

Just stating e.g. \(2 \times 6 \times 2.5\) is insufficient for the dB1 mark without further justification.
There are many acceptable variations, and some examples are below.

Some examples:

• 6 solutions in each interval of length \(\pi\) (B1) and 5 intervals of length \(\pi\) so 30 (dB1)
• 12 solutions in each interval of length \(2\pi\) (B1) and 2.5 intervals of length \(2\pi\) so 30 (dB1)
• 2 solutions for \(\sin x = k\) up to \(2\pi\), so for \(\sin^2 x\) there are 4 per \(2\pi\) (B1) and since \(0 \leqslant x \lt 5\pi\) then \(0 \leqslant 3x \lt 15\pi\) so \(\dfrac{15}{2} \times 4 = 30\) (dB1)
• Double the 6 solutions since \(\sin(nx) = \pm k\) (or \(\sin^2(nx) = k^2\)) so 12 (B1) and so 24 in \(4\pi\), hence 30 in \(5\pi\) (dB1)
• 16 solutions for \(\sin(nx) = k\) (B1) and 14 (solutions) for \(\sin(nx) = -k \rightarrow 30\) (dB1)
• \(\sin^2(3x) = \dfrac{1 - \cos(6x)}{2}\) and as \(\cos x = k\) has 5 solutions in 0 to \(5\pi\) (B1) and so \(\cos 6x\) has \(5 \times 6 = 30\) (dB1)

Graphical approaches may be used but these must be convincing, clearly showing the correct number of solutions in one of the relevant intervals. Be lenient with the shape of the curve.

e.g. 30 solutions in \(0 \leqslant x \lt 5\pi\) [with 6 solutions on, for example, either sketch below]
left: \(y = \sin^2(3x)\ \ 0 \leqslant x \lt \pi\); right: \(y = \sin(3x)\ \ 0 \leqslant x \lt \pi\)

Note that a suitable graph of \(\sin x\) showing, e.g. 10 solutions up to \(5\pi\), followed by 30 (calculation of \(\times 3\) clearly implied) would also be eligible to score the dB1.

The following are examples of an incorrect justification but scores B1dB0 for reaching 30:

• \(\sin(nx) = k\) has \(3 \times 6\) solutions in the interval 0 to \(5\pi\), \(\sin(nx) = -k\) has \(3 \times 4\) solutions (in the interval 0 to \(5\pi\)) so \(18 + 12 = 30\) solutions in total.
• \(\sin(nx) = k\) has 15 solutions in the interval 0 to \(5\pi\), so for \(\sin^2(nx)\) has \(15 \times 2 = 30\) solutions in total.

For reference, the tables below show the total number of solutions for each branch of \(\sin^2(nx) = k^2\) with \(n = 3\) and the bold values score the first B1 provided they are in the correct interval.

and \(n = 1\) for those that deal with \(nx\) (\(3x\)) last.

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.

Notes

It is acceptable in this question to work in degrees and convert if necessary.
NB Angle ABC in degrees is 22.591…°

M1: Attempts to use the cosine rule with values correctly placed. May be implied by, e.g., \(\cos\theta = \dfrac{8^2 + 13^2 - 6.4^2}{2 \times 8 \times 13}\) or \(\cos^{-1}\left(\dfrac{6.4^2 - 8^2 - 13^2}{-2 \times 8 \times 13}\right)\)
Condone slips in substitution if a correct cosine formula is seen. The placement of the values should be correct. So, condone e.g. a missing 2 once a correct cosine formula is seen.
Alt 1: attempts the cosine rule and finds angle \(ADB\) (= awrt 0.5 or \(29^\circ\)) or angle \(BAD\) (= awrt 2.2 or 2.3 or \(129^\circ\)) and uses the sine rule correctly with angle \(ABC\) involved.
Alt 2: attempts the cosine rule and finds angle \(ADB\) (= awrt 0.5 or \(29^\circ\)) and angle \(BAD\) (= awrt 2.2 or 2.3 or \(129^\circ\)) and sums angles \(ADB\), \(BAD\), and \(ABC\) to \(\pi\) (or \(180^\circ\))
Alt 3: using Pythagoras and simultaneous equations to find the length of \(AN\) or \(BN\) (see diagram below) and uses e.g. \(\cos ABC = \dfrac{BN}{8}\) or \(\sin ABC = \dfrac{AN}{8}\)

A1*: cso Correct proof.
If starting with \(6.4^2 = 8^2 + 13^2 - 2(8)(13)\cos ABC\) then there must be an intermediate line such as

• \(\cos ABC = \dfrac{8^2 + 13^2 - 6.4^2}{2 \times 8 \times 13}\) or angle \(ABC = \cos^{-1}\left(\dfrac{8^2 + 13^2 - 6.4^2}{2 \times 8 \times 13}\right)\)
• \(\cos ABC =\) awrt 0.92 or \(\dfrac{4801}{5200}\)
• angle \(ABC =\) awrt 0.3943

The minimum required is e.g. \(\cos^{-1}\left(\dfrac{8^2 + 13^2 - 6.4^2}{2 \times 8 \times 13}\right)\) or \(\cos\theta = \dfrac{8^2 + 13^2 - 6.4^2}{2 \times 8 \times 13} \rightarrow \theta = 0.394\) both of which score M1A1*.
The final value must be 0.394 and not e.g. 0.3943. There should be no obvious incorrect statements in the proof e.g. \(\cos(0.92) = 0.394\) and no obvious incorrect work.
Allow the use of, e.g., \(\theta\), \(x\), or even e.g. \(A\) throughout. Mention of radians is not required.
Those working in degrees and achieve 22.6 can convert to 0.394 without calculation for M1A1
Attempts via verification e.g. \(AD^2 = 8^2 + 13^2 - 2(8)(13)\cos 0.394 \rightarrow AD =\) awrt 6.4 score maximum M1A0*.

Helpful Diagram:

Notes

It is acceptable in this question to work in degrees and convert if necessary.
NB Angle ABC in degrees is 22.591…°

Note: you may need to check the diagram for working.

M1: Attempts the area of the sector or the area of the triangle \(ABD\) via a correct method.
May be implied by a full attempt at the area of \(R\).
The sector area may be implied by \(\dfrac{1576}{125}\) or by \(\dfrac{0.394}{2\pi} \times \pi \times 8^2\) or \(\dfrac{22.6}{360} \times \pi \times 8^2\)
The angle \(ABC\) is given in the question as 0.394 and should be used (or a more accurate value). Do not allow use of a different value for angle \(ABC\) (other than the angle in degrees).

There are many acceptable alternative approaches to find the area of triangle \(ABD\), e.g. using their angle \(ADB\) or their angle \(BAD\) or using right-angled triangles
e.g. \(\dfrac{1}{2}(13)(\text{``}3.07\text{''})\) or \(\dfrac{1}{2}(\text{``}7.39\text{''})(\text{``}3.07\text{''}) + \dfrac{1}{2}(\text{``}5.614\text{''})(\text{``}3.07\text{''})\) and could be implied by e.g. 11.3 + 8.6

Alternatively, attempts the area of the triangle \(ACD\) \(\left(\text{e.g. } \dfrac{1}{2}(5)(6.4)\sin\text{``}0.501\text{''} = \text{awrt } 7.68\right)\)
or the area of the segment (in sector \(ABC\)) \(\left(\dfrac{1}{2}\left(8^2\right)(0.394) - \dfrac{1}{2}\left(8^2\right)\sin 0.394 = \text{awrt } 0.324\right)\)
via a correct method. The area of the triangle \(ABC\) alone is insufficient for this mark.
Note that \(\sin 0.394\) might be seen as \(\sqrt{1 - \left(\dfrac{4801}{5200}\right)^2}\)
For values, see the diagram below.

dM1: Complete and correct method for the area of \(R\). Requires correct attempts at both the area of the sector and the area of the triangle \(ABD\) and subtraction of the two values (or expressions). Allow sector – triangle \(ABD\) if this is then recovered by making the area positive.

Alternatively, correct attempts at both the area of the triangle \(ACD\) and the area of the segment (in sector \(ABC\)) and subtraction of the two values (or expressions).

Condone slips in substitution provided a correct formula is seen.
May be implied by a correct answer.

A1: awrt 7.34 to 7.37 no units required but penalise incorrect units.
ISW after an acceptable answer is seen.

Helpful Diagram:

Notes

B1: \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = 2a\sin\theta\cos\theta\) or \(\mathrm{d}x = 2a\sin\theta\cos\theta\,\mathrm{d}\theta\) o.e. seen or implied by their substitution.
Note that writing \(x = a\sin^2\theta = \dfrac{a}{2}(1 - \cos 2\theta) \rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}\theta} = a\sin 2\theta\) is correct. Condone use of \(\dfrac{\mathrm{d}x}{\mathrm{d}a}\)

M1: Attempts to substitute, fully replacing \(x^{\frac{1}{2}}\) and \(\sqrt{a-x}\) with \(\theta\)’s and d\(x\) with their d\(x\) = …

Look for \(x^{\frac{1}{2}}\sqrt{a-x}\,\mathrm{d}x \rightarrow \mathrm{f}(\theta)\mathrm{g}(\theta)\mathrm{h}(\theta)\) where

• \(\mathrm{f}(\theta)\) is an attempt at \(\sqrt{a\sin^2\theta}\) e.g. allow \(a\sin\theta\) but just \(a\sin^2\theta\) is not condoned
• \(\mathrm{g}(\theta)\) is an attempt at \(\sqrt{a - a\sin^2\theta}\) but not \(\sqrt{a} - \sqrt{a\sin^2\theta}\) unless \(\sqrt{a - a\sin^2\theta}\) is attempted first
• \(\mathrm{h}(\theta) =\) their d\(x\) or their \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta}\) or \(\dfrac{1}{\text{their } \dfrac{\mathrm{d}x}{\mathrm{d}\theta}}\) (in terms of \(\theta\) only but condone d\(a\) seen)

Condone slips provided the intention is clear, e.g. \(x^{\frac{1}{2}} \rightarrow \sqrt{a}\sin^2\theta\) but \(x\) must be eliminated.
There is no need to replace the limits and condone the absence of \(\mathrm{d}\theta\) and/or the integral sign for this mark.

dM1: Attempts to use \(\sin 2\theta = 2\sin\theta\cos\theta\) to convert an integral of the form \(\displaystyle\int \sin^2\theta\cos^2\theta\,\mathrm{d}\theta\) or e.g. the form \(\displaystyle\int \sin\theta\cos\theta\sin 2\theta\,\mathrm{d}\theta\) to \(\displaystyle\int \ldots\sin^2 2\theta\,\mathrm{d}\theta\)
There is no need to replace the limits and condone the absence of \(\mathrm{d}\theta\) and/or the integral sign for this mark.

A1*: Replaces or considers limits \(\{x = 0 \Rightarrow\}\ \theta = 0,\ \{x = a \Rightarrow\}\ \theta = \dfrac{\pi}{2}\) at some stage before the given answer and proceeds with no errors to the given answer. The replaced limits may appear with their integral symbol and do not have to be justified and do not have to appear on every line.
Condone infrequent slips in notation, e.g. \(\sin\theta^2\) in a line as long as it is not consistently poor.
You must see the integral sign with the correct limits and the \(\mathrm{d}\theta\) together in the given answer.

Alternative 13(a) working backwards:

Score M1 here for using the double angle identity and replacing \(\ldots\sin\theta\cos\theta\,\mathrm{d}\theta\) with d\(x\)

Score dM1 here for a full attempt to replace all trig leading to everything in terms of \(x\) only
Must come from the form \(\displaystyle\int \ldots\sin\theta\cos\theta\,\mathrm{d}x\)

A1 fully correct with limits replaced / considered before the final line and the final line fully correct with limits, integral sign and d\(x\) as per the main scheme.

Notes

M1: Adopts an appropriate strategy by using the double angle identity to obtain an integrable form \(\displaystyle\ldots\int \sin^2 2\theta\,\mathrm{d}\theta \rightarrow \ldots\int \frac{\pm 1 \pm \cos 4\theta}{2}\,\mathrm{d}\theta\) which may be seen as \(\displaystyle\lambda\int \sin^2 2\theta\,\mathrm{d}\theta \rightarrow \frac{\lambda}{2}\int \pm 1 \pm \cos 4\theta\,\mathrm{d}\theta\) with the \(\dfrac{1}{2}\) absorbed into their coefficient of the integral.

dM1: Integrates into the form \(\pm p\theta \pm q\sin 4\theta\)

A1: Correct integration of \(\displaystyle\mu\int \sin^2 2\theta\,\mathrm{d}\theta\) to \(\dfrac{\mu}{2}\left(\theta - \dfrac{1}{4}\sin 4\theta\right)\). Here \(\mu\) may be 1.
Condone lack of limits here.

A1: Applies limits to the correct integral and proceeds to \(\dfrac{1}{8}\pi a^2\) following correct work.
There is no need to see 0 substituted in and condone any omission of integral signs and/or \(\mathrm{d}\theta\)
Note that \(\dfrac{a^2}{4}\left[\theta + \dfrac{1}{4}\sin 4\theta\right]_0^{\frac{\pi}{2}} = \dfrac{1}{8}\pi a^2\) is incorrect and scores M1dM1A0A0
Use of \(\sin^2 2\theta = \dfrac{\pm 1 \pm \cos k\theta}{2}\) with \(k \neq 4\) scores M0dM0A0A0 but may lead to \(\dfrac{1}{8}\pi a^2\)
Condone use of \(x\) in place of \(\theta\) e.g. \(\dfrac{a^2}{4}\left[x - \dfrac{1}{4}\sin 4x\right]_0^{\frac{\pi}{2}} = \dfrac{1}{8}\pi a^2\)

Alternative 13(b) via IBP Way 1:

Alternative 13(b) via IBP Way 2:

Notes

Note: Candidates working in radians are able to score all the M and B marks in this question.
Condone the absence of the degrees symbol throughout the whole question.

B1: Correct value for \(K\). Condone \(R = 500\)

M1: Award for \(\tan\alpha = \pm\dfrac{480}{140} \Rightarrow \alpha = \ldots\), \(\tan\alpha = \pm\dfrac{140}{480} \Rightarrow \alpha = \ldots\), \(\sin\alpha = \pm\dfrac{480}{\text{``}500\text{''}} \Rightarrow \alpha = \ldots\) or \(\cos\alpha = \pm\dfrac{140}{\text{``}500\text{''}} \Rightarrow \alpha = \ldots\)
Note \(\alpha =\) awrt 1.3 (rad) implies this mark.

A1: \(\alpha =\) awrt \(73.74\{^\circ\}\) or correct expression \(500\cos(\theta + 73.74)\{^\circ\}\)

Notes

(b)(i) Note: mark parts (b)(i) and (b)(ii) together.

B1ft: Correct equation of the model in either form including the \(R\) = following through on their numerical \(K\) (\(0 \lt K \leqslant 750\)) and their numerical \(\alpha\).
Allow for e.g. \(R = 1500 - \text{``}500\text{''} + \text{``}500\text{''}\cos(30t + \text{``}73.74\text{''})\{^\circ\}\) or for \(R = 1500 - \text{``}500\text{''} + 140\cos(30t)^\circ - 480\sin(30t)^\circ\) but not e.g. \(R = 1500 - K + K\cos(30t + \alpha)^\circ\)
\(R = 1000 + 140\cos 30t - 480\sin 30t\) (without the brackets) is correct.
Allow this mark if they have truncated or rounded an otherwise correct \(\alpha\) (to 3s.f.)

(b)(ii)

B1ft: 500 or follow through on (their \(A\) – their \(K\)) or (1500 – 2 × their \(K\)) provided it is non-negative and less than 1500. It must be clear this is their answer to (b)(ii) so expect to see e.g. (b) or \(R_{\min} =\) or an indication it is the minimum.

Notes

Note: if \(\theta\) is used in place of \(30t\) then they must revert back to \(30t\) correctly to access the marks.

M1: Substitutes \(t = 3.5\) into their model for the number of rabbits (you may need to check if no method is shown)
or substitutes \(t = 3.5\) into their \(\cos(30t + \alpha)^\circ\)
Condone substitution of a value of \(t\) in the range \(3 \leqslant t \leqslant 4.5\) for this mark.

A1: \(R =\) awrt 500.1… or 500 (not awrt) following substitution of \(t = 3.5\), suggesting that the model is valid/reliable/appropriate/good.
or \(\cos(30(3.5) + 73.74)\{^\circ\} \approx -1\) suggesting that the model is valid/reliable/appropriate/good.
Allow this mark if they have truncated or rounded an otherwise correct \(\alpha\) (to 3s.f.)

Alt:

M1: Minimum occurs when \(A + K\cos(30t + \alpha)^\circ = R_{\min} \Rightarrow \cos(30t + \alpha)^\circ = \lambda\) with \(|\lambda| \leqslant 1\) leading to \(t = \ldots\)
May just see \(\cos(30t + 73.74)^\circ = -1 \Rightarrow t = \ldots\) (or their \(\cos(30t + \alpha)^\circ = -1 \Rightarrow t = \ldots\))
\(30t + \alpha = 180 \Rightarrow t = \ldots\) implies this mark. Condone \(30t + \alpha = \pi \Rightarrow t = \ldots\) for this mark.

A1: \(t = 3.54\ldots\) (i.e. the middle of April) so the model is valid/reliable/appropriate/good.
Do not condone incorrect statements, e.g., \(t = 3.54\ldots\) i.e. the middle of March so close to middle of April. If using \(A + K\cos(30t + \alpha)^\circ = R_{\min}\) then \(R_{\min}\) must be = their \(A -\) their \(K\)
Allow this mark if they have truncated or rounded an otherwise correct \(\alpha\) (to 3s.f.)

Alt 2 using differentiation (Condoned)

M1: Condone finding the minimum using \(\ldots\sin(30t + 73.74)^\circ = 0 \Rightarrow t = \ldots\) (or their \(\sin(30t + \alpha)^\circ = 0 \Rightarrow t = \ldots\))
\(30t + \alpha = 180 \Rightarrow t = \ldots\) implies this mark. Condone \(30t + \alpha = \pi \Rightarrow t = \ldots\) for this mark.

A1: \(t = 3.54\ldots\) (i.e. the middle of April) suggesting that the model is valid/reliable/appropriate.
Do not condone incorrect statements, e.g., \(t = 3.54\ldots\) i.e. the middle of March so close to middle of April.
The complete derivative for \(\dfrac{\mathrm{d}R}{\mathrm{d}t}\) does not need to be seen.
Allow this mark if they have truncated or rounded an otherwise correct \(\alpha\) (to 3s.f.)

Notes

Note: if \(\theta\) is used in place of \(30t\) then they must revert back to \(30t\) correctly to access the marks.

M1: Realises that \(\sin(30t + 70)^\circ = -1\), reaches \(30t + 70 = 270\) or \(-90\) and attempts to find \(t\) (or \(30t\))
Condone attempts using differentiation. The minimum occurs when \(\cos(30t + 70)^\circ = 0 \Rightarrow 30t + 70 = 270 \Rightarrow 30t = \ldots\) (or \(t = \ldots\)). They must use 270 or \(-90\) and not 90 to achieve the minimum. Condone \(30t + \alpha = \dfrac{3\pi}{2} \Rightarrow t = \ldots\) for this mark but not \(30t + \alpha = \dfrac{\pi}{2} \Rightarrow t = \ldots\).

A1: Correct value for \(30t\) (or \(t\)) Accept rounded or truncated values to at least 3s.f. e.g. 6.66 or 6.67

dM1: Substitutes their value of \(t \gt 0\) (or \(30t \gt 0\)) coming from \(30t + 70 = \mathbf{270}\) into their model for \(R\)

A1: Correct number of rabbits. Allow 1032 or 1033 but must be whole numbers and not just 1030.
Allow this mark if they have truncated or rounded an otherwise correct \(\alpha\) (to 3s.f.)

Notes

Note: Use of degrees, if used in formulae in degrees, can score full marks.

B1: Deduces angle \(BAO\) or angle \(BOA\) is \(\dfrac{\pi}{3}\) radians or \(60^\circ\) or deduces that angle \(BOC\) is \(\dfrac{2\pi}{3}\) radians or \(120^\circ\). May be seen on the diagram or their own sketches or embedded in a formula.
May be implied if they find e.g. \(\dfrac{1}{6} \times \pi \times 5^2\) for the area of the minor sector.

M1: Uses a correct process and an angle of \(\dfrac{\pi}{3}\) radians or \(60^\circ\) to find

• the area of the segment bounded by the arc \(OB\) and straight line \(OB\)
• or the area of the segment bounded by the arc \(AB\) and straight line \(AB\)
• or the area of unshaded region \(AOB\).

Allow decimal values to imply the method (require 3sf rounded or truncated).
Use of \(60^\circ\) must be in a correct formula in degrees. \(0.5 \times 5^2 \times 60\) scores M0.
For reference these are the areas of the regions:

• segment \(= \dfrac{1}{2} \times 5^2 \times \dfrac{\pi}{3} - \dfrac{1}{2} \times 5^2 \times \sin \dfrac{\pi}{3}\left\{= \dfrac{25\pi}{6} - \dfrac{25\sqrt{3}}{4} = 2.26\right\}\) (3 s.f.) scores M1
• \(AOB = \dfrac{1}{2} \times 5^2 \times \dfrac{\pi}{3} + \dfrac{1}{2} \times 5^2 \times \dfrac{\pi}{3} - \dfrac{1}{2} \times 5^2 \times \sin \dfrac{\pi}{3}\left\{= \dfrac{25\pi}{3} - \dfrac{25\sqrt{3}}{4} = 15.35\right\}\) (2 d.p.) scores M1
• Minor sector \(= \dfrac{1}{2} \times 5^2 \times \dfrac{\pi}{3}\left\{= \dfrac{25\pi}{6} = 13.09\right\}\) (2 d.p.) scores M0 on its own
• Major sector \(= \dfrac{1}{2} \times 5^2 \times \dfrac{2\pi}{3}\left\{= \dfrac{25\pi}{3} = 26.18\right\}\) (2 d.p.) scores M0 on its own
• Triangle \(AOB = \dfrac{1}{2} \times 5^2 \times \sin \dfrac{\pi}{3}\left\{= \dfrac{25\sqrt{3}}{4} = 10.8\right\}\) (3 s.f.) scores M0 on its own.

This mark may be implied if seen as part of a correct strategy for the area of \(R\).
Note that the area of triangle \(AOB\) may also be found using Pythagoras and \(\dfrac{1}{2}bh\) but their method must be correct.

dM1: A fully correct strategy for finding the area of \(R\).
Follow through on incorrectly simplified areas but the method must be correct.
Allow decimal values to imply the method (require 3sf rounded or truncated).
For reference, the area is approximately 23.92 (2d.p.) and is likely to imply B1M1dM1 but their work should be checked.
Either:

• major sector – segment \(= \dfrac{1}{2} \times 5^2 \times \dfrac{2\pi}{3} - \left(\dfrac{1}{2} \times 5^2 \times \dfrac{\pi}{3} - \dfrac{1}{2} \times 5^2 \sin \dfrac{\pi}{3}\right)\)

or

• semicircle – \(AOB\) \(= \dfrac{1}{2} \times \pi \times 5^2 - \left(\dfrac{1}{2} \times 5^2 \times \dfrac{\pi}{3} - \dfrac{1}{2} \times 5^2 \sin \dfrac{\pi}{3} + \dfrac{1}{2} \times 5^2 \times \dfrac{\pi}{3}\right)\)

or

• semicircle – 2 × sector \(AOB\) + triangle \(AOB\) \(= \dfrac{1}{2} \times \pi \times 5^2 - 2 \times \dfrac{1}{2} \times 5^2 \times \dfrac{\pi}{3} + \dfrac{1}{2} \times 5^2 \sin \dfrac{\pi}{3}\)

or

• sector \(AOB\) + triangle \(AOB\) \(= \dfrac{1}{2} \times 5^2 \times \dfrac{\pi}{3} + \dfrac{1}{2} \times 5^2 \sin \dfrac{\pi}{3}\) (this also implies the first M1)

If they go straight to one of these expressions then we would imply the first M1 (and the B1).

A1: Correct expression \(\dfrac{25}{4}\sqrt{3} + \dfrac{25}{6}\pi\) o.e. in the correct form e.g. \(\dfrac{50}{8}\sqrt{3} + \dfrac{75}{18}\pi\)
Ignore any reference to (or absence of) units.
Do not apply isw if they go on to add or subtract additional areas.
However, we can condone poor simplification e.g. \(\dfrac{25}{12}\left(\sqrt{3} + \pi\right)\) following a correct answer.

Note: Attempts via integration for the area underneath a circle or polar coordinates are unlikely but if seen then use Review. The schemes for these approaches follow the main scheme but the first M1 may also be scored for finding half of the area of \(AOB\).

You might find the following diagram helpful:

Notes

(a) Way 2

(a) Condone a complete proof entirely in \(x\) (or another variable) instead of \(\theta\)
Condone “=” for “\(\equiv\)”
Note that we are marking this as B1M1A1 not M1M1A1

B1: Adds the fractions to obtain a correct single fraction (not fractions over fractions) in any form. Condone missing brackets when they combine their fractions as long as they are recovered to give a correct fraction.
This can be done in a variety of ways but when combined, the fraction must be correct e.g.
\(\dfrac{\operatorname{cosec}\theta + 1 + \operatorname{cosec}\theta - 1}{(\operatorname{cosec}\theta - 1)(\operatorname{cosec}\theta + 1)}\) or \(\dfrac{2\operatorname{cosec}\theta}{(\operatorname{cosec}\theta - 1)(\operatorname{cosec}\theta + 1)}\) or \(\dfrac{2\operatorname{cosec}\theta}{\operatorname{cosec}^2\theta - 1}\)
or e.g. \(\left(\dfrac{1}{\frac{1}{\sin\theta} - 1} + \dfrac{1}{\frac{1}{\sin\theta} + 1} = \dfrac{\sin\theta}{1 - \sin\theta} + \dfrac{\sin\theta}{1 + \sin\theta}\right) = \underline{\dfrac{2\sin\theta}{1 - \sin^2\theta}}\) etc.

M1: Uses a correct Pythagorean identity anywhere in their attempt e.g.
\(\operatorname{cosec}^2\theta - 1 = \cot^2\theta,\ \sin^2\theta + \cos^2\theta = 1\) etc. or equivalent

A1*: Correct work with all necessary steps shown leading to the given answer. See scheme for the necessary steps. They need to proceed via sine and cosine to the given answer. There should be no notational or bracketing errors and no mixed or missing variables. E.g. we would consider \(\cos^2\theta\) written as \(\cos\theta^2\) a notational error.
Condone reaching \(2\sec\theta\tan\theta\ *\)

Way 2 (Meet in the middle)

B1: See Way 1

M1: See Way 1

A1*: Correct work on the RHS with all necessary steps shown leading to showing the equivalence with the LHS. See scheme for the necessary steps. There should be no notational or bracketing errors and no mixed or missing variables.
For this approach there must be a (minimal) conclusion e.g. “= LHS”, “QED”, “Hence proven” etc.

It is possible to start with the rhs e.g.:

B1: Correctly reaches \(2\tan\theta\sec\theta = \dfrac{2\operatorname{cosec}\theta}{(\operatorname{cosec}\theta - 1)(\operatorname{cosec}\theta + 1)}\)

M1: Uses a correct Pythagorean identity anywhere in their attempt e.g.
\(\operatorname{cosec}^2\theta - 1 = \cot^2\theta,\ \sin^2\theta + \cos^2\theta = 1\) etc. or equivalent

A1*: Correct work with all necessary steps shown leading to the lhs. See scheme for the necessary steps. There should be no notational or bracketing errors and no mixed or missing variables.

Notes

(b) Note that attempts solve an equation of the form:
\(2\tan x\sec x = \cot 2x\sec 2x\) or e.g. \(2\tan\theta\sec\theta = \cot 2\theta\sec 2\theta\) or e.g. \(2\tan\theta\sec\theta = \cot 2x\sec 2x\)
Will generally score no marks in part (b)

Condone the use of \(\theta\) instead of \(x\) here.

B1: Deduces the correct equation using the result from part (a)

M1: Factors out or cancels the \(\sec 2x\) to obtain \(\ldots\tan 2x \pm \ldots\cot 2x = 0\) oe e.g. \(\ldots\tan 2x = \pm\ldots\cot 2x\) leading to an equation of the form:
\(\tan^2 2x = \alpha\) oe e.g. \(\cot^2 2x = \dfrac{1}{\alpha}\) where \(\alpha \gt 0\)
Or uses \(\tan 2x = \dfrac{\sin 2x}{\cos 2x}\) and \(\cot 2x = \dfrac{\cos 2x}{\sin 2x}\) and then \(\sin^2 2x = \pm 1 \pm \cos^2 2x\) or \(\cos^2 2x = \pm 1 \pm \sin^2 2x\) to obtain an equation of the form \(\sin^2 2x = \beta\) or \(\cos^2 2x = \gamma\) oe e.g. \(\operatorname{cosec}^2 2x = \dfrac{1}{\beta}\) or \(\sec^2 2x = \dfrac{1}{\gamma}\) where \(0 \lt \beta \lt 1\) or \(0 \lt \gamma \lt 1\)
Or uses \(\tan 2x = \dfrac{2\tan x}{1 - \tan^2 x}\) and \(\cot 2x = \dfrac{1 - \tan^2 x}{2\tan x}\) to obtain a 3TQ in \(\tan^2 x\) (or possibly in \(\sec^2 x\))
Or uses \(\tan 2x = \dfrac{\sin 2x}{\cos 2x}\) and \(\cot 2x = \dfrac{\cos 2x}{\sin 2x}\) and then \(2\sin^2 2x = \pm 1 \pm \cos 4x\) and \(2\cos^2 2x = \pm 1 \pm \cos 4x\) to obtain an equation of the form \(\cos 4x = k,\ \ 0 \lt k \lt 1\)

M1: Correct order of operations from \(\tan^2 2x = \alpha\) or \(\sin^2 2x = \beta\) or \(\cos^2 2x = \gamma\) or \(\cos 4x = k\) or equivalents e.g. \(\operatorname{cosec}^2 2x = \dfrac{1}{\beta}\) where \(\alpha \gt 0\) or \(0 \lt \beta \lt 1\) or \(0 \lt \gamma \lt 1\) or \(0 \lt k \lt 1\) leading to at least one value for \(x\) e.g. square roots, finds inverse tan/sin/cos/cosec and divides by 2 or inverse cos and divides by 4
or from \(\tan^2 x = k,\ k \gt 0\) which follows their equation (may need to check) and then finds \(x = \tan^{-1}\sqrt{k}\)
You may need to check their value(s) (in degrees or radians) to see if the correct order of operations has been used. May be implied by e.g. \(17.6^\circ\) or \(17.7^\circ\) provided no incorrect work is seen.

A1: Correct values. Allow awrt \(17.6^\circ\) and awrt \(72.4^\circ\). The degrees symbol is not required. Ignore any values outside the range (correct or incorrect) but if there are extra angles in range score A0. Answers in radians score A0.

Note that some candidates may convert to \(\sin x\) or \(\cos x\) and then solve:

E.g.

These can be marked in a similar way.

Alternative not using part (a):

Score as:

M1: For correct work leading to one of the forms in the main scheme e.g.
\(\tan^2 2x = \alpha\) oe e.g. \(\cot^2 2x = \dfrac{1}{\alpha}\) where \(\alpha \gt 0\)
or
\(\sin^2 2x = \beta\) or \(\cos^2 2x = \gamma\) oe e.g. \(\operatorname{cosec}^2 2x = \dfrac{1}{\beta}\) or \(\sec^2 2x = \dfrac{1}{\gamma}\) where \(0 \lt \beta \lt 1\) or \(0 \lt \gamma \lt 1\)

B1: Any correct equation e.g. \(\tan^2 2x = \dfrac{1}{2},\ \cot^2 2x = 2,\ \sin^2 2x = \dfrac{1}{3}\) etc.

Then M1A1 as main scheme

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:

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:

Scores B1M1A1

Scores B1M1A1

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):

Scores B1M1A0

Scores B1M1A0

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

Typical solution

The angles in a triangle add up to \(\pi\) radians.

Angle \(BAC = \pi - \dfrac{\pi}{3} - x = \dfrac{2\pi}{3} - x\)

Typical solution

(b)(i)

(b)(ii) \(\dfrac{\pi}{4}\)

Typical solution

area above

area below

Typical solution

Hence \(\sqrt{29}\sin(x + 68.2)\)

Typical solution

(b)(i)

(b)(ii)

(b)(iii)

Typical solution

1

Typical solution

Typical solution

\(x = 90,\ x = 270\)

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

(i)

\(\cos\theta \neq 1\) as \(\operatorname{cosec}\theta\) is undefined

(ii)

Typical solution

Typical solution

(−4, 6)

Typical solution

Typical solution

Typical solution

since \(\theta\) is obtuse \(\sin\theta = \dfrac{\sqrt{6}}{3}\)

Typical solution

Typical solution

The bottom-right graph: a decreasing curve through \(\left(0, \dfrac{\pi}{2}\right)\) ending at \((1, 0)\) on the \(x\)-axis.

Typical solution

1

Typical solution

(i)

(ii)

All three graphs intersect at the same point.

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

(i)

–180-30 = -210

(ii)

Typical solution

(i)

(ii)

Typical solution

Typical solution

Typical solution

Typical solution

(i)

(ii)

(iii)

Typical solution

(i)

(ii)

Typical solution

(i)

(ii)

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Notes

M1: Attempt correct process to find \(R\)

M1: Attempt correct process to find \(\tan\alpha\) (or equiv with \(\sin\alpha\) or \(\cos\alpha\))
Condone \(\cos\alpha = 3\), \(\sin\alpha = 4\) seen in method, leading to \(\tan\alpha = \frac{4}{3}\)
M0 for \(\tan\alpha = \frac{3}{4}\)

A1: Obtain \(5\cos(2x - 53.1)\). Could be implied by \(R = 5\) and \(\alpha = 53.1\) (awrt 53.1)
Condone not being written out in required form, and ISW any error when attempting to do so
A0 for \(\alpha = 0.927\) radians

Notes

M1: Clear fraction, attempt to rearrange and gather like terms

A1FT: Obtain correct harmonic equation, with like terms gathered. FT their \(R\cos(2x - \alpha)\)
Could still be in terms of \(\cos 2x\) and \(\sin 2x\) eg \(9\cos 2x + 12\sin 2x - 10\ (= 0)\)

M1*: Correct process to find at least one value for \(x\). Must be solving equation of the form \(\cos(2x \pm \alpha) = k\)
Allow substitution soi eg \(n\) = their \(5\cos(2x - 53.1)\) to obtain \(3n = 10\)

A1: Obtain at least one correct value. Likely to be using 48.2
Allow answers in range [50.6, 50.7] – see later A1 detail for range for other solutions

M1dep*: Correct process to find (at least) one further value for \(x\)

A1: Obtain one further correct value. Correct process to find \(x\) from a second value for their \(\cos^{-1}\left(\frac{2}{3}\right)\) Or their first solution \(+\ 180^\circ\) oe
Answers in range [2.46, 2.48], [182, 183], [230, 231]

A1: Obtain all 4 correct values, and no others in given range. Answers in range [2.46, 2.48], [182, 183], [230, 231]

Alt method if not harmonic form

M1: Clear fraction, attempt to rearrange and gather like terms. Could have terms in \(\cos 2x\) and \(\sin 2x\) or even \(\tan x\), \(\cos x\), \(\sin x\), \(\cos x\sin x\) (if double angle formulae used)

A1FT: Obtain correct equation in a single trig ratio, with like terms gathered. Could be in terms of \(\sin x\) or \(\cos x\), if double angle formulae used but equation never seen in terms of just \(\cos 2x\) or \(\sin 2x\)

M1*: Correct process to find at least one value for \(x\). Solving a 3 term quadratic (or possibly quartic) in a single trig ratio (condone BC) and also attempt \(x\)

A1: Obtain at least one correct value. Allow answers in range [50.6, 50.7] – see later A1 detail for range for other solutions

M1dep*: Correct process to find one further value for \(x\). Must be finding a secondary value from their principal value. M0 if just finding principal value from other quadratic root

A1: Obtain one further correct value. Answers in range [2.46, 2.48], [182, 183], [230, 231]

A1: Obtain all 4 correct values, and no others in given range. Answers in range [2.46, 2.48], [182, 183], [230, 231]

Notes

B1: Use the correct formula for the area of a sector to obtain 0.8 www. Must be 0.8 and not equivalent fractions

Condone verification
Must work exactly throughout so B0 if via decimals

Notes

M1: Attempt use of correct formula for area of a triangle or any equiv complete method. Using 0.8 radians or \(45.8^\circ\)

A1: Obtain correct \(AC\). 3sf or better (awrt 20.1)

Notes

B1: Correct arc length (possibly unsimplified) soi. Condone awrt 4.8, if working in degrees

M1: Attempt use of correct cosine rule, or any equiv complete method

A1: Obtain correct \(BC\) (awrt 16.5) soi

M1: Attempt full method for perimeter. Their arc \(BD\) + (their \(AC - 6\)) + their \(BC\)
NB using \(BD = 4.7\) (or better) is likely to be chord \(BD\) so M0 unless evidence that arc has actually been attempted

A1: Obtain correct perimeter (awrt 35.3). Accept without units

Notes

B1: AG Must see as a minimum the following three components:

[1]: the correct expressions for \(A\) and \(B\) stated explicitly (not just in a final answer or as part of trig. expressions) or \(2\cos\left(\dfrac{A + B + A - B}{2}\right)\sin\left(\dfrac{A + B - (A - B)}{2}\right) = 2\cos A\sin B\) (if working right to left)

[2]: \(\sin C - \sin D = 2\cos A\sin B\)

[3]: the given answer or both \(\sin C - \sin D = 2\cos A\sin B\) and \(2\cos A\sin B = 2\cos\left(\dfrac{C + D}{2}\right)\sin\left(\dfrac{C - D}{2}\right)\) (so linking the two parts of the given answer explicitly via the expression \(2\cos A\sin B\))

Notes

B1*: Can be implied by the correct expression for the area of \(\Delta PQR\)

B1 for either expression soi in the area of \(\Delta PQR\)

B1*: One correct area expression

B1dep*: Correct sandwich inequality in terms of \(r\) and \(\theta\) - must be seen at some point in this form and not just as separate inequalities

This result with no working scores B3 but \(\sin\theta \lt \theta \lt \tan\theta\) with no working is no marks

B1: AG – www sufficient working seen therefore tan must be replaced with \(\dfrac{\sin}{\cos}\) and therefore \(\sin\theta \lt \theta \lt \tan\theta \Rightarrow 1 \lt \dfrac{\theta}{\sin\theta} \lt \dfrac{1}{\cos\theta}\) is B0
Dependent on all previous B marks

Notes

(c)(i)

B1: or \(1 \gt \dfrac{\sin\theta}{\theta} \gt \cos\theta\) but not as two separate inequalities (unless correctly brought together)

condone \(\dfrac{\cos\theta}{1} \lt \dfrac{\sin\theta}{\theta} \lt 1\)

(c)(ii)

B1: Ignore any working seen but B0 if implying \(\lt 1\), \(\gt 1\) etc.

Notes

B1*: cao

Denominator must be simplified to \(h\)

M1dep*: Using the given expression for \(\sin C - \sin D\) from part (a) with \(C = x + h\) and \(D = x\) – allow un-simplified

If not using result from part (a) then M0

A1: Using the result that \(\displaystyle\lim_{h \to 0}\cos\left(\frac{2x + h}{2}\right) = \cos x\) to obtain \(\displaystyle\cos x \times \lim_{h \to 0}\left[\frac{2}{h}\sin\left(\frac{h}{2}\right)\right]\)

Must see limit present for this A mark

A1: Re-writing \(\dfrac{2}{h}\sin\left(\dfrac{h}{2}\right)\) so that the result from part (c)(ii) can be applied and obtaining the correct derivative

Or equivalent limit argument e.g. \(\displaystyle\lim_{h \to 0}\left(\frac{\sin\left(\frac{1}{2}h\right)}{h}\right) = \frac{1}{2}\)

Notes

B1: Clearly identify the error made e.g. stating that it should be \(-\cos\left(\frac{\pi}{2} - x\right)\) or ‘the student forgot to multiply \(\cos\left(\frac{\pi}{2} - x\right)\) by \(-1\)’ or ‘\(\sin\left(\frac{\pi}{2} - x\right)\) does not differentiate to \(\cos\left(\frac{\pi}{2} - x\right)\)’ oe
B0 for ‘the student forgot to multiply by \(-1\)’ only
B0 for ‘the student forgot to divide \(\cos\left(\frac{\pi}{2} - x\right)\) by \(-1\)’

Just stating that the derivative should be \(-\sin x\) is B0

If mentioning that ‘the student forgot to multiply/include the \(-1\)’ then it must be clear this is on line 3

Notes

Allow “=” throughout Q4

M1: M1 for expanding correctly

A1: Must see final statement www

Alternative method

A1: Must see final statement www

Notes

M1: For writing \(\tan 2\theta\) as \(\dfrac{\sin 2\theta}{\cos 2\theta}\) or \(\dfrac{2\tan\theta}{1 - \tan^2\theta}\). May be seen embedded or implied by next A1 (Allow this mark if \(\sec 2\theta\) still used)

A1: May see \(1 + 2\cos\theta\sin\theta\) for the numerator here

A1: Allow A1 for correct numerator or correct denominator seen in terms of \(\theta\), not \(2\theta\). Allow this mark even if subsequently \(\surd\) incorrectly.

A1: correctly obtained www, ie must see previous line

Alternative method

M1: or \(\dfrac{(\cos\theta + \sin\theta)^2}{\cos^2\theta - \sin^2\theta}\) seen

A1: correct numerator seen

A1: correct denominator seen in terms of \(2\theta\), not \(\theta\)

A1: correctly obtained www, ie must see previous line

Alternative method using \(t\)-formulae

A1: www

Notes

M1: Set \(\cos 2x = 1\) soi
\(1 - \cos 2x = 0\), then \(x = 0\), would imply M1
Allow \(\cos 2x \geqslant 1\), but not \(\cos 2x \leqslant 1\) (unless recovered by final answer)
Allow \(\cos 2x \neq 1\) if considering the values that \(x\) cannot take

A1: Identify all multiples of \(\pi\), including negatives
Allow any clear notation, but must include negative integers as well
eg \(x = 0, \pm\pi, \pm 2\pi \ldots\) (allow ‘etc’ for ‘…’)
Condone working in degrees, as long as final answer is in radians

Notes

M1: Attempt to differentiate
Obtain \(\dfrac{k\sin 2x}{1 - \cos 2x}\), or unsimplified equiv
Other derivatives may be seen if trig identities, or log laws, used before differentiation; allow coefficient errors only
Could use implicit differentiation on \(1 - \cos 2x = \mathrm{e}^{\frac{1}{5}y}\) oe to obtain \(a\sin 2x = b\mathrm{e}^{\frac{1}{5}y}\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

M1: Equate their derivative to 0 and attempt to solve for a non-zero \(x\) value
Allow M1 as long as their numerator involves a trig term
Allow M1 if working in degrees

A1: Obtain \(x = \tfrac{1}{2}\pi\) only
Must be exact
A0 if \(x = 0\) also given in final answer

A1: Obtain \(y = 5\ln 2\) (or \(\ln 32\))
Must be exact, simplified, and from \(x = \tfrac{1}{2}\pi\)
Allow A1 if \(5\ln 2\) comes from \(90^\circ\)

Notes

(c)(i)

M1*: Attempt differentiation using an appropriate method on their first derivative
Starting with \(\dfrac{k\sin 2x}{1 - \cos 2x}\), or a multiple of any other correct first derivative, including eg \(k\cot x\)
Must be correct structure for the differentiation method being attempted, allowing coefficient errors only
Could use implicit differentiation on \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 10\sin 2x \times \mathrm{e}^{-\frac{1}{5}y}\)

A1: If using implicit differentiation then A1 can be awarded if \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) is still present

M1d*: Attempt to simplify their second derivative using at least one trigonometric identity correctly
Only award M1 for trig identities used after differentiation

M1: Correctly replace \(1 - \cos 2x\) with \(\mathrm{e}^{\frac{1}{5}y}\) or vice versa
Used either in their second derivative or in the given answer
If using implicit differentiation then the M1 will be awarded before the differentiation attempt

A1: Obtain / confirm given answer www
Penalise any clearly incorrect equations, but allow BOD if denominator disappears (eg when using trig identities) but then reappears when relevant

(c)(ii)

B1: Correct conclusion, with justification
Refer to the exponential term being positive, hence second derivative must be negative, hence maxima
Could refer to \(\mathrm{e}^k\) not \(\mathrm{e}^{-\frac{1}{5}y}\)
Could refer to the correct second derivative of \(\dfrac{-20}{1 - \cos 2x}\) and explain why this is always negative, hence maxima (so no need to refer to exponential term with this approach)

Notes

(a)(i) B1: State 0.8 m, units required
Units may be given as m or metres
Could be \(\tfrac{4}{5}\) m
Could be 80 cm

(a)(ii)

M1: Identify that minimum occurs when \(\cos(30t - 60)\) is \(-1\), so need \(30t - 60 = 180\)
M1 does not require attempt at solution for \(t\)
No FT on an incorrect \(d\) being used from the previous part eg \(d = 2.45\) from using \(t = 0\)

A1: No units needed, as value of \(t\) is requested
Condone 0800 or 8am
Ignore additional values of \(t\) that are greater than 8, but A0 for a smaller positive value of \(t\) also given

Notes

M1: Equate model to 1, rearrange and use \(\cos^{-1}\)
As far as \(30t - 60 = k\), using correct order of operations
Allow M1 if working in radians (gives \(30t - 60 = 2.529\))

A1: Obtain correct first value of \(t\)
Implied by first time of 0650 with no errors seen
3sf or better
Ignore inequality signs if used

M1: Attempt all further values of \(t\) within \(0 \lt t \lt 24\)
Using a valid method
M0 if using radians
Values of \(t\) could also be found using the symmetry of the curve eg \(8 + (8 - 6.83) = 9.17\)

A1: Allow answers to 3sf
Ignore inequality signs if used
Correct time periods would imply \(t\) values

A1: Correct two periods, given as time intervals
Could also be given as 6:50am to 9:10am, and 6:50pm to 9:10pm
Must be given as intervals and not just times eg A0 for ‘0650 and 0910’ etc
BOD if correct intervals given following any incorrect inequality signs
Accept 0649 to 0911 and 1849 to 2111 www (from checking times and realising that rounded answers give depths of less than 1 metre)
A0 if giving answers in minutes, or hours and minutes, after midnight and not times (eg ‘410 minutes to 550 minutes’ or ‘6 hours 50 minutes to 9 hours 10 minutes’)
Condone attempt at interval notation / inequalities as long as intention is clear, and allow BOD if written as a strict inequality such as \(0650 \lt t \lt 0910\)

Special Case
If M1A1M0 awarded, then allow SC B1 for giving a correct time period eg 0650 to 0910

Notes

B1: Any sensible suggestion that suggests that amplitudes of the tides will be reduced due to the exponential term
Must refer to the effect of the exponential term in context in some way
Condone reference to the exponential term having a ‘damping’ effect on the tides
Cannot just restate the question eg ‘river gets shallower’

Notes

(a)(i)

Condone a small slip in early irrelevant working before substituting, but must be fully correct after this.

B1: oe in terms of \(\theta\), not \(2\theta\). May be implied by next line

B1: oe in terms of \(\cos\theta\) only
Must reach AG or give a conclusion (even if just ‘QED’) www

Alternative 1

B1: oe in terms of \(\theta\), not \(2\theta\). May be implied by next line

B1: oe in terms of \(\cos\theta\) only
Must reach AG or give a conclusion (even if just ‘QED’) www

Alternative 2

B1: oe in terms of \(\theta\), not \(2\theta\). May be implied by next line

B1: oe in terms of \(\cos\theta\) only
Must reach AG or give a conclusion (even if just ‘QED’) www

(a)(ii)

M1: oe

A1: Allow A1 for two correct values of \(3\theta\). Ignore other values.
This mark is not implied by correct final answers but accept equivalent correct working e.g. a graph of \(\cos 3\theta\).
Accept radians for this mark only: \(3\theta = \frac{\pi}{4}, \frac{7\pi}{4}, \frac{9\pi}{4}\)

A1: cao. Ignore values outside of the range 0-180°, but do not accept radians for this mark. This mark may be given following M1A0.

Notes

(b)(i) B1: Must see a correct multiplied out form and AG or conclusion.

(b)(ii)

B1: oe, must be exact

B1: May see \(x = \frac{2\sqrt{2} \pm \sqrt{24}}{8}\) etc.

Notes

B1: soi - for ‘spotting’ the connection. This mark can be gained regardless of their answers to (a)(ii) and (b)(ii).
\(x = \cos\theta\) or 3 correctly paired roots are sufficient for this mark
Condone \(x = \cos 15\)

B1: Justification for selecting this root (may say e.g. “cos 15 is the only positive root”)
This may be implied by matching each pair of answers correctly (but all three must be present or mentioned).

Notes

M1: Use of both \(\cot x \equiv \dfrac{\cos x}{\sin x}\) and \(\mathrm{cosec}\,x \equiv \dfrac{1}{\sin x}\)
Condone use of \(s\) and \(c\) throughout but final answer must be in terms of sine – allow e.g. \(\theta\) for \(x\) for M marks but must be \(x\) for the A mark

M1: Correct use of \(\sin^2 x + \cos^2 x \equiv 1\) to obtain an equation in \(\sin x\) only
Not dependent on the first M mark

A1: AG (so must be equal to zero) – sufficient working must be shown – any errors seen is A0
A0 if an angle is missing from any trig. expression used in their working

Alternative method

M1: Correct use of \(1 + \cot^2 x \equiv \mathrm{cosec}^2 x\)

M1: Replacing \(\mathrm{cosec}\,x\) with \(\dfrac{1}{\sin x}\) to obtain an equation in \(\sin x\) only
Not dependent on first M mark
Note: \(2\,\mathrm{cosec}^2 x - 9\,\mathrm{cosec}\,x - 5 = 0 \Rightarrow 2 - 9\sin x - 5\sin^2 x = 0\) with no intermediate working is M0 unless explicit mention is made of multiplying through by \(\sin^2 x\) or explicit mention is made of dividing by \(\mathrm{cosec}^2 x\)

A1: AG (so must be equal to zero) – sufficient working must be shown – any errors seen is A0
A0 if an angle is missing from any trig. expression used in their working

Notes

(b)(i)

M1: SEE APPENDIX for awarding this mark (solving 3TQ expressions)
condone using \(x\) for \(\theta\) or \(2\theta\) for the M mark – condone for M1 only \((5\sin\theta - 1)(\sin\theta + 2)\)

B1: Correctly stating that \(\sin 2\theta = 0.2\) and that \(\sin 2\theta\) cannot equal \(-2\) (must explicitly reject the \(-2\) (but no rationale required) - this mark is not implied by correct values for \(\theta\) (as DR required)
Must be solving \(5\sin^2 2\theta + 9\sin 2\theta - 2 = 0\) for the B marks
condone \(\sin 2x = 0.2\)

B1: awrt 0.101 (0.1006789…) www

B1: awrt 1.470 (1.4701173…) www

Ignore additional solutions outside of the range \(0 \lt \theta \lt \pi\), but if any other solutions inside the range, award at most one of the two final B marks for one correct value
SC B1 for awrt 0.10 and awrt 1.47 only if 3 dp or better) not seen
SC B1 for awrt 5.77 and awrt 84.2 only (working in degrees)

(b)(ii)

M1: Use of the small angle approximation \(\sin 2\theta \approx 2\theta\) twice in the given answer from (a) to obtain a three-term quadratic in \(\theta\) (allow un-simplified)
Award M1 only for \(5\theta^2 + 9\theta - 2\,[= 0]\) (so for using \(\theta\) instead of \(2\theta\)) – allow e.g. \(x\) for \(\theta\)

A1: State 0.10 (or better e.g. 0.100…) as a decimal following a correct quadratic in \(\theta\) seen (no method required for solving the quadratic) and comment that this is accurate to 2 dp (as a minimum must mention ‘2 dp’ with the value of 0.10(000…) appearing in this part and 0.10 or 0.101 or 0.100(6789…) appearing in part (b)(i))
This mark is dependent on an awrt 0.10 seen in part (b)(i) or a correct sign change test (see below)
Ignore any consideration of other root(s)

Alternative for M mark

M1: Re-writing at least one of their equations \(\sin 2\theta = k\) with \(-1 \lt k \lt 1\) (from part (a)) as \(2\theta = k\)

Alternative for A mark

A1: Correct values to at least 1 dp (rot) with explanation (‘change of sign’ either stated or comparing values with zero) and correct conclusion (as a minimum must mention ‘2 dp’)

Alternative for (b)(ii) (Appendix)

M1: Use of small angle approximations \(\sin 2\theta \approx 2\theta\) and \(\tan 2\theta \approx 2\theta\) in the given equation from (a) to obtain what would be equivalent to a three-term quadratic in \(\theta\) (allow un-simplified). Allow M1 only for \(\dfrac{2}{\theta^2} - \dfrac{9}{\theta} - 3\,[= 0]\) (so for using \(\theta\) instead of \(2\theta\)) – allow e.g. \(x\) for \(\theta\)

A1: State 0.10 (or better e.g. 0.1039…) as a decimal following a correct quadratic in \(\theta\) (no method required for solving quadratic) and comment that this is accurate to 2 dp (as a minimum must mention ‘2 dp’ with the value of 0.10(39…) appearing in this part and 0.10 or 0.101 or 0.100(6789…) appearing in part (b)(i))

Appendix: rules for solving quadratics in questions 2 and 4(b)(i) only

In questions 2 and 4(b)(i) candidates are required to solve 3 term quadratics (3TQ) using DR – therefore we must see a correct, complete method for solving these quadratics – the correct answers do not imply the corresponding M mark, for example in question 2, \(9x^2 - 38x + 8 = 0 \Rightarrow x = 4\) or \(x = \frac{2}{9}\) is M0

Rules for factorising:
\(at^2 + bt + c \Rightarrow (mt + n)(pt + q)\) where \(a = mp\) and one of \(mq + np = b\) or \(c = nq\) (so when expanding their factorised expression it must give the correct quadratic term and one other term correct of the preceding 3TQ expression/equation)
e.g. in question 2 (and similarly for question 4(b)(i)):
\(9x^2 - 38x + 8 = \left(x - \frac{2}{9}\right)(x - 4)\) is M0 (but the following B1 for the correct c.v. of \(\frac{2}{9}\) and 4 in qu. 2 can still be awarded as they follow from these two factors)
\(9x^2 - 38x + 8 = (3x + 8)(3x + 1)\) is M1 (when expanded the \(x^2\) and constant terms are correct)
Allow correct part factorisation for their 3TQ expression e.g. if correct 3TQ then in question 2 the expression \(9x(x - 4) - 2(x - 4)\) scores M1

Rules for the formula:
Must apply the correct formula for their three-term quadratic (no errors even if correct formula is stated) – note that stating the formula (in terms of \(a\), \(b\) and \(c\)) followed immediately by the corresponding roots is M0 – we must see the formula being applied e.g. \(9x^2 - 38x + 8 = 0 \Rightarrow x = \dfrac{38 \pm \sqrt{38^2 - 4(9)(8)}}{2(9)}\).
Minimal acceptable working would be \(x = \dfrac{38 \pm \sqrt{1156}}{18}\) (so must explicitly see the discriminant) for M1

Rules for completing the square – using \(9x^2 - 38x + 8 = 0\) as an example:
The M1 is not awarded until correctly getting to the stage of \(x - \frac{19}{9} = \pm\sqrt{\frac{289}{81}}\) (must include \(\pm\) so implying two roots) with no errors (so consistent with applying the formula correctly)

Notes

M1*: Attempt at Pythagoras =36 oe, may see \(\sqrt{a^2 + b^2} = 6\)

M1*: oe or \(\dfrac{b}{a} = \tan 60\) or \(\dfrac{b}{a} = 1.73\), condone \(\dfrac{a}{b}\)
May see equivalent statements in sin or cos.

M1 dep*: Substitute both expressions (dependent on both previous M marks) – must reach an equation in \(a\) or \(b\) only.

A1: A0 for negative answers (if not disregarded)
Accept \(b = \sqrt{27}\)
If no (or insufficient) working then SC B1B1 (max 2/4) for each correct answer (must be exact).

Alternative method

M1: Allow this mark for 6cos or 6sin of \(30, 60, 120^\circ\)
Must see this step oe.

M1: Allow this mark for 6cos or 6sin of \(30, 60, 120^\circ\) provided it is consistent with their other expression (i.e. not the same).
Must see this step oe.

If no (or insufficient) working then SC B1B1 (max 2/4) for each correct answer (must be exact).

Notes

M1: Replace \(B\) with \(-B\) in given identity

A1: State \(\cos(-B) = \cos B\) and \(\sin(-B) = -\sin B\), and conclude with correct identity
Condone \(-\sin A\sin(-B)\) becoming \(\sin A\sin B\) with no intermediate step
\(\cos(-B) = \cos B,\ \sin(-B) = -\sin B\) must be stated, but no justification needed

Notes

B1: Use correct identities, with exact trig values, to obtain a correct expression
Allow BOD for ambiguous positioning of + and – signs in a product, but penalise explicit errors if a single identity is seen in isolation
If expansion done before exact trig values used, then the expression must still be correct at the point that the B1 is awarded

M1: Expand brackets
May be recognised as difference of two squares so no need to see \(\frac{\sqrt{3}}{4}\cos\theta\sin\theta - \frac{\sqrt{3}}{4}\cos\theta\sin\theta\)
To obtain answer of form \(a\cos^2\theta - b\sin^2\theta\ (a \gt 0,\ b \gt 0)\), with possibly \(c\cos\theta\sin\theta - c\cos\theta\sin\theta\) also present

A1: Use Pythagorean identity and simplify to given answer
www eg if middle terms shown for expansion, then these must be correct

Notes

(c)(i)

B1: Correct max value

B1: Correct angle
B0 if any extra angles given
Must be ‘positive’ so B0 for \(0^\circ\)
Must be in degrees

Marks are independent

(c)(ii)

B1: Correct min value

B1: Correct angle
B0 if any extra angles given
Must be in degrees
SC If angles in both parts are correct, but in radians, then penalise only once (mark as B0 in (i) and B1 in (ii))

Marks are independent

Notes

In this question, if candidates attempt more than one method, review all their working to identify which is their ‘final’ or ‘substantially most complete’ answer, then apply one scheme only.
Candidates cannot gain marks from more than one scheme.

M1: Substitute \(y = x\) into equation of circle (need not be simplified – may see this quadratic in \(y\))

A1: A1 for either correct (condone if not specifically identified as \(A\) or \(B\) – may see \(y =\) these values)

M1: Attempt \(CM\) using their values (method must be correct)
May see \(CM = \frac{3\sqrt{2}}{2}\)

M1: Attempt \(BA\) using their values (method must be correct)
May see just \(\sqrt{2(9 - 2a)}\) oe, e.g. \(\sqrt{18 - 4a}\)
Alternative: \(AM = \sqrt{\frac{9}{2} - a}\) (and then \(\text{Area} = 2 \times \frac{1}{2} \times CM \times AM\))

M1: Attempt at area, in terms of \(a\)

A1: Must see correct expression before answer

Alternative Method

M1: Attempt complete the square for \(x\)

A1: Both soi

B1: Possibly coming from \(\dfrac{|3 - 0|}{\sqrt{1^2 + 1^2}}\)

M1: Their radius² (in terms of \(a\)) – their \(CM^2\)

M1: Attempted in terms of \(a\)

A1FT: FT their \(AM\) or \(AB\) (in terms of \(a\)), and their \(CM\)

A1: Must see one correct intermediate step

Alternative method for last three marks

M1: Attempted in terms of \(a\)

A1FT: FT their \(r\) and \(AM\) or \(AB\) (in terms of \(a\)), and their \(CM\)

A1: Must see one correct intermediate step

Alternative method for last three marks

M1: Attempted in terms of \(a\)

A1FT: FT their \(r\) and \(AM\) or \(AB\) (in terms of \(a\)), and their \(CM\) (using double angle formula for \(\sin BCA\))

A1: Must see one correct intermediate step

Notes

(b)(i)

B1: oe

(b)(ii)

B1: Must see this geometrical answer (accept e.g. the line touches the circle)
Accept a diagram that clearly shows the line is a tangent

Notes

B1: Must see this geometrical answer (condone BOD for ‘the line does not touch the circle’)
Accept a diagram that clearly shows the line does not meet the circle

Notes

B1*: Forming a correct expression for \(d\) (or a correct equation containing \(d\)) e.g. \((d =)\,\frac{1}{2}(-\sqrt{3}\cos\theta - 2\sin\theta)\), \((d =)\,\frac{1}{3}\left(\frac{7}{2}\sin\theta - 2\sin\theta\right)(= 0.5\sin\theta)\)
Can be implied e.g. \(\frac{7}{2}\sin\theta = 2\sin\theta + 3(\ldots)\) seen
Can be implied by a correct equation for \(\theta\)

M1dep*: Obtaining an equation of the form \(\tan\theta = k\) from a trigonometric equation which initially had 3 sine and 1 cosine terms or 2 sine and 2 cosine terms e.g. if correct \(\frac{7}{2}\sin\theta = 2\sin\theta + 3\left(\frac{7}{2}\sin\theta + \sqrt{3}\cos\theta\right)\)

A1: Condone \(-\dfrac{\pi}{6}\) stated too but A0 if any other value given in the interval \(\frac{1}{2}\pi \lt \theta \lt \pi\) (but ignore any values that are given outside this range)
Exact answer must be seen at some stage

Notes

B1ft: Correct formula for the sum of an AP with \(a = 2\sin\theta\) (with either \(\theta\) or their value of \(\theta\) substituted) and either \(d\) or their value of \(d\) substituted or their expression for \(d\)
Follow through their values of \(\theta\) and \(d\) if used provided \(\frac{100}{2}[2(2\sin\theta) + (100 - 1)d]\) implied

B1ft: Correct expression for \(d\) using their \(\theta\) (e.g. \(d = \frac{1}{2}(-\sqrt{3}\cos\theta - 2\sin\theta)\), \(d = \frac{1}{3}\left(\frac{7}{2}\sin\theta - 2\sin\theta\right)\))
Follow through their value of \(\theta\) only

B1: www – must have come from \(\theta = \frac{5}{6}\pi\) correctly derived in (a) oe (not for 1338 or 1340 unless 1337.5 seen so isw once 1337.5 (oe e.g. \(\frac{2675}{2}\)) seen)
Correct answer with no working scores all 3 marks

Notes

(a)(i)

M1: Attempt differentiate \(\mathrm{f}(x)\), allow sign errors but both trig functions must be changed.
oe e.g. \(2\cos\left(x + \frac{\pi}{6}\right)\)

M1: Setting their \(\mathrm{f}'(x) = 0\) and correctly manipulating to reach an equation in a single trig function (e.g. \(4\cos^2 x = 1\) or \(4\sin^2 x = 3\)) do not allow incorrect working e.g. \(a - b = 0 \Rightarrow a^2 - b^2 = 0\)

A1: www, must be in radians (allow decimal 1.05 3sf)

A1: www, must be in radians (allow decimal 4.19 3sf), and for this mark have no other solutions in the given range (ignore any outside)

SC B1 for each correct solution (max [2/4]) if insufficient or no working shown, but not from incorrect working.
If both solutions given correctly in degrees (60°, 240°) then can get M1M1SCB1 (max [3/4])

(a)(ii)

M1: Attempt to differentiate their \(\mathrm{f}'(x)\) (allow sign errors but both trig functions must be changed), setting their \(\mathrm{f}''(x) = 0\) and attempting to manipulate. May see \(\mathrm{f}''(x) = -2\sin\left(x + \frac{\pi}{6}\right)\)

A1: For correctly reaching an equation in a single trig function (oe, may see \(4\sin^2 x = 1\) or \(4\cos^2 x = 3\))
Not from incorrect working e.g. \(a - b = 0 \Rightarrow a^2 - b^2 = 0\)

A1: www, for both correct in radians (allow decimals 2.62, 5.76 3sf)

SC B1 for both correct solutions (max [1/3]) if insufficient or no working shown, but not from incorrect working.
Answers in degrees can get M1A1A0 (max [2/3])

Notes

B1FT: FT their \(x\)-values in radians or degrees from (a)(i), both with \(y = 0\)
Both must be correctly labelled with \(A\) and \(C\), allow decimals 3sf

B1FT: FT their \(x\)-values in radians or degrees from (a)(ii), with both \(y = -2\) and \(y = 2\) correct
Both must be correctly labelled with \(B\) and \(D\), allow decimals 3sf

If neither mark awarded then can get one of (max [1/2]):

• SC B1FT for any two correct pairs \(x\) and \(y\), correctly labelled
• SC B1FT for all four \(x\)-coords, correctly labelled and in ascending order (only if 2 distinct answers given in both (a)(i) and (a)(ii))
• SC B1FT for all four \(y\)-coords, correctly labelled

Notes

(c)(i)

B1: Must reference the graph – do not accept just ‘gradient is positive’ or just ‘\(\mathrm{f}'(x) \gt 0\)’

(c)(ii)

B1: Both intervals correctly identified, ignore set notation for this mark
cao (values must be correct) but condone any clear indication of both correct intervals e.g. \(0 \to \frac{\pi}{3}\) and \(\frac{4\pi}{3} \to 2\pi\) (allow decimals)
Accept any combination of \(\leqslant\) and \(\lt\) or ( ) and [ ]

B1: Writing their answer in correct set notation, values may be incorrect for this mark but there must be two separate intervals.
Allow single-tailed inequalities as long as written in correct set notation, e.g. \(\left\{x : x \lt \frac{\pi}{3}\right\} \cup \left\{x : x \gt \frac{4\pi}{3}\right\}\)
Accept any combination of \(\leqslant\) and \(\lt\) or ( ) and [ ] but do not accept \(\cap\) instead of \(\cup\).

Notes

A1: For correct gradient of either tangent or normal

M1*: Using their gradient of normal and (1,1) to form an equation or \(y = -\frac{1}{2}x + \frac{3}{2}\) (must find a value for c for this mark to be awarded) FT their gradient of normal (not = 2)

M1dep*: Substituting \(y = x^2\) and attempting to solve (may see quadratic in \(x\) or \(y\))

A1: Correct quadratic in \(x\) or \(y\), in any form or \(4y^2 - 13y + 9 = 0\)

A1: Allow (−1.5, 2.25). Accept \(x = \ldots\), \(y = \ldots\) but must see both.

Notes

or \(AB = \dfrac{5\sqrt{5}}{4}\), or \(OB = \dfrac{3\sqrt{13}}{4}\), or \(OA = \sqrt{2}\)

M1: Attempt to find all three, squared or not (may see on diagram)

M1dep: Correct use of cos rule in any form, FT their \(AB\), \(OB\) & \(OA\)

A1: Must be exact and come from exact working for \(\cos\alpha\). Must see an exact intermediate step – either \(\frac{1}{\sqrt{26}}\) or \(\frac{\sqrt{26 - 1}}{1}\).
Decimals used throughout can achieve max [2/3] M1M1A0
Do not penalise candidates who write down \(\alpha = 78.69\ldots^\circ\) but the final answer must come from exact working (not BC).

Alternative method (1)

M1: Attempt to find \(\tan\beta\) & \(\tan\gamma\)
\(\beta\) and \(\gamma\) are the angles made by OB and OA respectively with the horizontal
(Alternatively with angles to the vertical, \(\tan\beta = \frac{2}{3}\))

M1dep: Correct use of \(\tan(180 - \theta) = -\tan\theta\) and \(\tan(A + B)\) formula using their \(\tan\beta\) & \(\tan\gamma\) (or, with angles to the vertical, \(\tan\alpha = \dfrac{1 + \frac{2}{3}}{1 - \frac{2}{3} \times 1}\))

A1: Must be exact and come from exact working.
Do not penalise candidates who write down \(\alpha = 78.69\ldots^\circ\) but the final answer must come from exact working (not BC).

Alternative method (2)

M1: Attempt find \(\tan BOx\) & \(\tan AOx\) and use \(\tan(\theta - \varphi)\) formula

M1dep: Correct use of \(\tan(\theta - \varphi)\) using their \(\tan BOx\) & \(\tan AOx\)

A1: Must be exact and come from exact working.

Alternative method (3)

M1: Attempt \(\overrightarrow{OB} \cdot \overrightarrow{OA}\)

M1dep: Correct dot product with exact values

A1: Must be exact and come from exact working for \(\cos\alpha\). Must see an exact intermediate step – either \(\frac{1}{\sqrt{26}}\) or \(\frac{\sqrt{26 - 1}}{1}\).

Notes

B1: B0 for \(R = \pm 5, \sqrt{25}\) etc. unless replaced with 5
No working required for this mark. Ignore working

M1: M1 for \(\tan\alpha = k\) where \(k = \pm\frac{3}{4}, \pm\frac{4}{3}\) or equivalent e.g. \(\cos\alpha = \pm\dfrac{3}{R}\), \(\sin\alpha = \pm\dfrac{4}{R}\) with their value of \(R\) (but not just \(R\) and do not allow reciprocals for this mark). 53.1 (or better) with no working implies M1
SC If \(\cos\alpha = 3, \sin\alpha = 4 \Rightarrow \tan\alpha = \dfrac{4}{3}\) explicitly seen then this scores M1 A0 but do not penalise again in (b) (if correct answer seen)

A1: www awrt 53.13 (at least 4 sf required) so 53.1 (or 53) is A0 (but if an awrt 53.13 seen then isw if replaced with a less accurate value)
\(53.13010235\ldots\) - an answer in radians scores A0
53.13 from \(R\sin(x - \alpha)\) soi

Notes

M1: M1 for \(x = \alpha + \arcsin\left(\dfrac{2}{R}\right)\) or \(x - \alpha = \arcsin\left(\dfrac{2}{R}\right)\) with their \(R\) and \(\alpha\) substituted
SC B1 for 76.7 only (in the given range) from using an alternative method e.g. \(9\sin^2 x = (2 + 4\cos x)^2\)

A1: awrt 76.7 (at least 3 sf required) – ignore any answers given outside the range \(0 \lt x \lt 90\) but do not award this mark if any other values in this range are given – www but see SC in (a)
Correct answer with no working seen scores SC B1
Answer in radians scores A0

Notes

M1: Attempt use of cosine rule
Allow either omission of 2, or + not –, but no other errors
Allow other fully complete methods, such as basic trigonometry, possibly combined with Pythagoras

A1: Obtain 10.3cm, or better
If > 3sf then allow 10.25, or answers that round to 10.25
Condone no units

Notes

M1: Attempt use of sine rule
Correct equation seen, with fractions either way up
Could also be implied by method eg \(\sin^{-1}(0.75)\) is M1, but just 0.75 is M0
Allow other fully complete methods

A1: Obtain \(D = 48.6^\circ\), or better
\(D = 48.590377\ldots\)
Allow \(D = 0.848\) radians

A1FT: Obtain \(D = 131^\circ\), or better
FT their first angle as long as \(\lt 150^\circ\)
\(D = 131.409622\ldots\)
A0 if additional angles given as well
Allow \(D = 2.29\) radians (could be FT on incorrect acute angle in radians, as long as \(D \lt 2.618\))

Notes

M1: Attempt to link \(\mathrm{d}x\) and \(\mathrm{d}\theta\)
Allow sign error only

M1: Attempt to write integrand in terms of \(\theta\)
Must substitute for both function and \(\mathrm{d}x\)
Can follow M0 but do not allow just \(\mathrm{d}x = \mathrm{d}\theta\)

A1: Obtain correct integrand in terms of \(\cos\theta\) only
Condone no \(\mathrm{d}\theta\), as long as previously seen

M1: Attempt use of double angle formula
Using \(\cos 2\theta = \pm 2\cos^2\theta \pm 1\)
Integrand must be of form \(k\cos^2\theta\), which must have come from correct method with coefficient errors only

A1FT: Integrate to obtain \(\sin 2\theta + 2\theta\)
FT on \(a\cos 2\theta + b\) only

M1: Attempt use of limits
Must be correct limits (either \(x\) or \(\theta\), as long as consistent with their integral), correct order and subtraction
Allow M1 for use of limits in any integration attempt in terms of \(\theta\). Allow M1 for either expressions that still involve sin, or exact equivs
M0 for decimal values, even if then stated to be the same as \(\frac{1}{3}\pi\)
Condone eg \(\frac{1}{2}\sqrt{3}\) from \(\sin 120^\circ\), but M0 if degrees used in linear term

A1: Obtain given answer of \(\frac{1}{3}\pi\)
Must see both surd values, or an explanation as to why \(\sin\frac{2}{3}\pi = \sin\frac{2}{6}\pi\)

Notes

M1: Or for \(m\sec\theta + 3\cos\theta = 4\sin\theta\) \(\left(\Rightarrow \dfrac{m\sec\theta}{\cos\theta} + 3 = 4\dfrac{\sin\theta}{\cos\theta}\right)\) \(m\sec^2\theta + 3 = 4\tan\theta\)
The first M mark is for a valid method arriving at a three term equation containing \(\sec^2\theta\)
Squaring each individual term of the original equation scores no marks

M1: Correctly uses the identity \(1 + \tan^2\theta \equiv \sec^2\theta\) to obtain an equation in \(\tan\theta\) only

A1: AG so sufficient working must be shown
A0 if angle missing from any trigonometric terms

Notes

M1*: Considers discriminant of given quadratic equation in tan (\(c\) must be two terms) to get an expression in \(m\) only. M0 for embedded discriminant in quadratic formula unless explicitly considered
Allow \(4^2 - 4m(m + 3)\)

M1dep*: Sets their discriminant equal to zero and obtains an expanded three-term quadratic in \(m\)
Reasoning for setting the discriminant equal to zero is not required for this mark

A1: State or imply \(m = -4\) only

M1: Uses their negative integer value of \(m\), and solves the equivalent of their three term quadratic equation in tan, to obtain (at least) \(\tan\theta = k\), where \(k\) is non-zero - dependent on both previous M marks
Allow \(-4\tan^2\theta - 4\tan\theta - 1 = 0 \Rightarrow \tan\theta = -0.5\) for this mark
If no method shown for solving their quadratic, then award this mark if the solution is correct for their quadratic

A1: For full marks must explain why the discriminant should be set equal to zero – must say that as there is only one value of \(\theta\) or \(\tan\theta \Rightarrow \Delta = 0\) (as a minimum must see explicit mention of ‘one’ together with ‘\(\theta\)’ or ‘\(\tan\theta\)’ for this mark). Allow awrt 2.68
2.677945045…

Notes

B1: Could be implied by \(2(x + q)^2 + r\)

B1: Could be implied by \(p(x + 1.5)^2 + r\)

B1FT: FT on their \(p\) and \(q\) ie \(7 - pq^2\)

Notes

B1FT: FT on their (a)
Could come from differentiation

B1FT: FT on their (a)
No FT on incorrect \(x\)-value from differentiation

Notes

B1FT: FT on their minimum value
Allow BOD if different answers in (a) and (b)
2.5 must be stated as, or clearly intended to be, the minimum value
Just \((\ldots, 2.5)\) is insufficient

M1: Attempt to solve \(\tan\theta =\) their \((-1.5)\)
To obtain numerical value for \(\theta\)
Allow an angle in radians (expect \(-0.983\) rad)
Allow BOD if different answers in (a) and (b)

A1: Obtain \(124^\circ\), or better
A0 if additional solutions

Condone approaches other than ‘hence’ eg
B1 – attempt to solve \(\tan\theta = -1.5\), from correct derivative (expect \(4\tan\theta\sec^2\theta + 6\sec^2\theta = 0\))
B1 – obtain \(\theta = 124^\circ\)
B1 – obtain min value of 2.5 (no FT)

Notes

(a)(i) B1: Justification for \(CD\)
Need to see either \(\cos y = \frac{CD}{a}\) or adj = hyp × \(\cos\theta\) before given answer

(a)(ii) B1: Use area of triangle to show given answer
Could quote general expression for area and then show clear substitution
If not, then sides being used need to be clearly identified through statement or diagram
Could also use right-angled triangle, with base as \(AD\)
Condone not being rearranged to given expression

(a)(iii)

B1: Correct \(CD\) in terms of \(b\) and \(x\)

B1: Correct area of triangle \(BCD\)
B0 B1 if correct area stated with no justification

B1: Correct area of triangle \(ABC\)

B1: Equate area of \(ABC\) to the sum of the areas of the two small triangles and complete proof convincingly
Allow alternative proofs eg using lengths

Notes

B1: Correct use of compound angle formulae
Could be implied if exact values used immediately – allow BOD for RHS
May be seen as two separate expressions, not yet equated

M1: Use exact trig values
In either equation or two expressions
Must see all 4 values, but expansions may not be fully correct

M1: Gather like terms and attempt \(\tan\alpha\)
May still have fractions in the fraction
\(\tan\alpha\) does not yet need to be the subject, but must only appear once for M1

M1: Attempt to rationalise their denominator
Clear intention seen to multiply throughout by the conjugate of their denominator

A1: Obtain given answer
With full detail, including (at least) \(3 - 2\) in denominator

Notes

(a)(i) B1: Allow alternative notation, or worded equivalent
Allow \(y\), or just f, but not \(x\)
Accept just \(\mathbb{R}\)
Allow \((-\infty, \infty)\)

(a)(ii) B1: Allow alternative notation, or worded equivalent
Allow \(y\), or just g, but not \(x\)
Or \((-\infty, \infty)\) with \((-1, 1)\) clearly excluded

Notes

(b)(i)

M1: Attempt correct composition of functions
At least one interim value required

A1: Conclude with 5.33

SC B1 for stating \(2\tan(1 \div \cos 0.6) = 5.33\)

(b)(ii)

B1: Must refer to inverse of f not existing, with reason
Must be clear that referring to the function f

Notes

M1: Attempt use of identity in their equation to obtain quadratic
Allow \(\tan^2 x = \pm\sec^2 x \pm 1\)
Award M1 when reduced to single trig ratio

A1: Obtain correct equation in \(\sec x\) – possibly still with brackets
Or correct quadratic in \(\cos x\) – possibly still with brackets but with no fractions \((4\cos^2 x - 6\cos x - 4 = 0)\)

M1: Solve 3 term quadratic and attempt to find at least one value for \(x\)
Could solve quadratic BC
Must be using root that would give solution for \(x\)

A1: Obtain at least one correct value
Allow decimals or in degrees
Must be from correct solution method of correct quadratic (condone second root not being seen – but must be correct if seen)

A1: Obtain both correct values, and no others, and explain that \(\sec x = 0.5\) has no solutions as outside range
Now exact and in radians
Or equiv explanation for \(\cos x\)

Notes

M1: With at least one correct value – values should be given to at least 2 sf (rot)
Allow degrees for M1 only: \(\mathrm{f}(0.6) = 68.8617\ldots\)

A1: Correct values together with explanation in words (change of sign) and conclusion

Notes

M1: Correct order of operations to get \(2x - 1 = \sin\left(kx^2\right)\)
\(k \neq 0\)

A1: \(p = \frac{1}{2},\ q = \frac{1}{2}\) and \(r = \frac{1}{6}\) (oe)

Notes

M1: Uses their iterative formula with correct starting value to produce terms up to at least \(x_2\) to at least 4 significant figures
Allow degrees for M1 only:
For reference:
\(x_1 = 0.5003636\ldots\)
\(x_2 = 0.5003641\ldots\)
\(x_3 = 0.5003641\ldots\)

A1: Must be stated to exactly 4 significant figures

Notes

B1: B0 for \(\sqrt{169}\) or for \(\pm 13\)

M1: M1 for \(\tan\alpha = k\) where \(k = \pm\frac{5}{12}, \pm\frac{12}{5}\)
or for \(\cos\alpha = \pm\frac{12}{R}\), \(\sin\alpha = \pm\frac{5}{R}\)

A1: A0 if in degrees (must be stated to exactly 4 significant figures at some point)
For reference: 0.3947911197…

Notes

M1: Sets \(R\cos(t - \alpha)\) (with their \(R\) and \(\alpha\)) equal to either 3 or \(-3\), and attempt to solve (with correct order of operations)
M1 only if in degrees

A1: awrt 1.73
1.7327192…

A1: awrt 2.20 (condone 2.2) – ignore other values that are greater than 2.20
2.1984556…