Notes

Note that just finding a correct equation for the area and/or a correct equation for the perimeter (before any substitution) is insufficient to score any marks.

M1: Uses area formulae to form an equation of the form \(\alpha r^2\theta + \beta r^2 = 240\) o.e. (\(\alpha, \beta \neq 0\)) and rearranges to make \(r\theta\), \(\theta\) or \(r\theta + \dfrac{1}{5}r\) the subject. Look for:
\(r\theta = \dfrac{M \pm Nr^2}{r}\left(= \dfrac{M}{r} \pm Nr\right)\) o.e. or \(\theta = \dfrac{M \pm Nr^2}{r^2}\left(= \dfrac{M}{r^2} \pm N\right)\) o.e. where \(M, N \neq 0\)
or \(r\theta + \dfrac{1}{5}r = \dfrac{L}{r}\ \ L \neq 0\) o.e. May work in degrees.

A1: A correct rearrangement for \(\theta\) or \(r\theta\) or \(r\theta + \dfrac{1}{5}r\) which may be unsimplified (may be in degrees)
\(r\theta = \dfrac{240 - \frac{1}{10}r^2}{\frac{1}{2}r}\) o.e. e.g. \(r\theta = \dfrac{2400 - r^2}{5r}\) or \(r\theta = \dfrac{480 - 0.2r^2}{r}\)
or \(r\theta + \dfrac{1}{5}r = \dfrac{480}{r}\) o.e.
or \(\theta = \dfrac{240 - \frac{1}{10}r^2}{\frac{1}{2}r^2}\) o.e. e.g. \(\theta = \dfrac{2400 - r^2}{5r^2}\) or \(\theta = \dfrac{480}{r^2} - \dfrac{1}{5}\) or \(\theta = 480r^{-2} - 0.2\)

dM1: Substitutes their \(r\theta = \dfrac{M \pm Nr^2}{r}\) o.e. or \(\theta = \dfrac{M \pm Nr^2}{r^2}\) o.e. or \(r\theta + \dfrac{1}{5}r = \dfrac{L}{r}\) into an expression of the form \((P =)\ r\theta + Qr,\ Q \neq 0\) (typically \(P = r\theta + \dfrac{11}{5}r\)) which may be unsimplified or in degrees. It is dependent on the previous method mark. It is acceptable for their valid expression for \(\theta\), \(r\theta\) or \(r\theta + \dfrac{1}{5}r\) to be substituted into the perimeter expression directly (without first seeing them in the perimeter expression).

A1*: \(P = 2r + \dfrac{480}{r}\) following a correct method (condone slips to be recovered) and all previous marks scored. Condone invisible brackets to be recovered.
\(P =\), Perimeter \(=\) must be seen at least once in their solution in the correct place.

Notes

Mark (b) and (c) together. There is no requirement to see the notation \(\dfrac{\mathrm{d}P}{\mathrm{d}r}\) in part (b). It may even be called \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). Allow use of e.g. \(P^{\prime}\) or e.g. \(y^{\prime}\)

M1: \(\left(\dfrac{\mathrm{d}P}{\mathrm{d}r} =\right) p \pm \dfrac{q}{r^2}\) where \(p\) and \(q\) are non-zero constants

dM1: Sets or implies that their \(\dfrac{\mathrm{d}P}{\mathrm{d}r} = 0\) and proceeds to \(mr^{\pm 2} = n,\ \ m \times n \gt 0\). It is dependent on the previous method mark. Do not be concerned by the mechanics of the rearrangement.
This mark may be implied by a correct answer to their \(p - \dfrac{q}{r^2} = 0\). You may need to check this on your calculator.

A1: \(r =\) awrt 15.5 or \(\sqrt{240}\ \left(= 4\sqrt{15}\right)\) Do not accept \(\pm\) (ignore any units if given)

Notes

Mark (b) and (c) together.

Condone other letters used instead of P and r for \(\dfrac{\mathrm{d}^2P}{\mathrm{d}r^2}\) e.g. \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) for M1 only.
Just using \(\dfrac{\mathrm{d}P}{\mathrm{d}r}\) and considering a sign change is M0A0

M1: Differentiates and finds \(\left(\dfrac{\mathrm{d}^2P}{\mathrm{d}r^2} =\right) \pm\dfrac{f}{r^3}\) (do not be concerned about the sign)

A1: Note if they score A0 in (b) then this mark cannot be scored.
Requires

• a correct a correct expression for \(\dfrac{\mathrm{d}^2P}{\mathrm{d}r^2}\)
• a correct value for \(\left(\dfrac{\mathrm{d}^2P}{\mathrm{d}r^2} =\right) \dfrac{960}{r^3} =\) awrt 0.26 using awrt 15.5 (but allow 0.23(43..) if using 16)
• a correct comparison with 0 and a conclusion e.g. minimum

The expression for the second derivative does not need to be labelled but if it is then it must be \(\dfrac{\mathrm{d}^2P}{\mathrm{d}r^2}\) o.e. or accept e.g. \(P^{\prime\prime}\) BUT \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) used in their conclusion is A0

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

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

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

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Area of sectors =

Area of rhombus =

Typical solution

(i)

Max area occurs when \(\dfrac{\mathrm{d}A}{\mathrm{d}\theta} = 0\)

When \(\theta = \dfrac{\pi}{3}\)

(ii)

Typical solution

The angle would be the same.

The area would be quadrupled.

Typical solution

Typical solution

Area of sector = \(\dfrac{1}{2}r^2\theta\)

Area of triangle = \(\dfrac{1}{2}ab\sin C\)

Hence

Typical solution

Let \(\mathrm{f}(\theta) = \theta - \sin 2\theta\)

Hence solution lies between \(\dfrac{\pi}{5}\) and \(\dfrac{2\pi}{5}\)

Typical solution

(i)

(ii)

Use more iterations

(iii)

The value is on a stationary point

Typical solution

Typical solution

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

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

For reference only: \(\frac{1}{2}(3x+1)^2(2) \lt 44x - 7\)

B1*: Correct use of \(A = \frac{1}{2}r^2\theta\)

B1dep*: Expand and re-arrange to correct 3TQ expression in \(x\)
Allow any inequality sign or equals

M1dep*: SEE APPENDIX for awarding this mark (solving 3TQ expressions) - dependent on first B mark only (this mark is for solving their 3TQ but not for solving \((3x+1)^2 = 0\))
Correct quadratic followed immediately by correct critical values (with no working) is M0

B1: Correct critical values of \(x\) (if factorisation shown then it must imply these two c.v.)
Must be \(\frac{2}{9}\) or \(0.\dot{2}\) but B0 for 0.222…

B1FT dep*: FT their two positive critical values \(x_1, x_2\) e.g. \(\{x : x_1 \lt x \lt x_2\}\) where \(x_2 \gt x_1\); allow \(\left\{x : x \gt \frac{2}{9}\right\} \cap \{x : x \lt 4\}\) but ‘union’ is B0
B0 for interval notation e.g. \(\left(\frac{2}{9}, 4\right)\)
Answer must be in set notation for this mark – dependent on first B mark only

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

B1: Correct (possibly unsimplified) area of \(OMB\)
Could use other than \(r\) for the radius
Could set their variable equal to \(OM\), giving a radius that is double this
eg \(OM = x\) so area \(= x^2\sin\theta\)

M1: Attempt to use ratio on two correct areas
Using two of \(OMB\) \(\left(\tfrac{1}{4}r^2\sin\theta\right)\), \(MAB\) \(\left(\tfrac{1}{2}r^2\theta - \tfrac{1}{4}r^2\sin\theta\right)\) and \(OAB\) \(\left(\tfrac{1}{2}r^2\theta\right)\) oe with their variable
Must be two correct areas
Must be using the correct ratio for their two areas ie 2:3 if using \(OMB\) and \(MAB\), 2:5 if using \(OMB\) and \(OAB\) or 3:5 if using \(MAB\) and \(OAB\)
Allow ratio to be used the wrong way around eg \(2OMB = 3MAB\)

A1: Any correct statement linking the two areas
Could use other than \(r\) for the radius
Or \(2x^2\theta - x^2\sin\theta\)

A1: Simplify to given answer
At least one line of working once ratio used

Notes

B1: Obtain correct first iterate
3sf or better – more accurate answer is 0.599281923...
Condone truncating if more sig fig given

M1: Attempt correct iterative process to find at least 2 more values
M1 is for the correct process for finding \(\theta_3\) and \(\theta_4\), but these may be incorrect
M0 if working in degrees

A1: Obtain 1.13
Possibly following B0 if first iterate is wrong but process then self corrects
Must follow M1 ie a clear attempt to use the correct iterative process
Must be 3sf
Once M1 is awarded, allow A1 for 1.13 even if an incorrect iterate seen, as process will recover

Notes

B1*: Draw straight line, starting at the origin which intersects the graph
Allow point of intersection to be greater than \(\theta = \frac{1}{2}\pi\)
Ignore incorrect labels, such as \(y = x\)

B1dep*: Vertically into the curve, then horizontally into the straight line, as far as the root
Initial value should be before root
Needs point of intersection to be before \(\theta = \frac{1}{2}\pi\)

B1: Mark independently from other parts of question, including an incorrect diagram, as staircase can be deduced from the iterates in (b)

Notes

B1*: Just need correct shape for \(y = \sin^{-1}k\theta\) graph – a one to one function that starts at the origin (ignore any \(\theta \lt 0\)) and has increasing gradient for all \(\theta\)

B1dep*: Straight line from the origin to intersect their graph
Diagram is sufficient for B1 – no comment or explanation required

Notes

Allow \(h\) or other letter for \(\delta x\) throughout

M1: or \(\displaystyle\lim_{\delta x \to 0} \frac{\cos(x + \delta x) - \cos x}{\delta x}\) or may be seen later. Must include \(\displaystyle\lim_{\delta x \to 0}\).

M1: Allow \(\cos\delta x = 1\) for small \(\delta x\)   \(\left(\text{or } 1 - \dfrac{(\delta x)^2}{2}\right)\)
Allow \(\sin\delta x = \delta x\) for small \(\delta x\)
Both must be explicitly stated for M1
If not stated but implied, M0, but can still possibly gain final A1

A1: Dep on at least B1M1 gained, and approximations either seen explicitly or seen substituted.
and nothing incorrect seen
NB. \(\cos x - \sin x - \cos x = -\sin x\) is incorrect and scores A0

Notes

B1: May be implied by \(\frac{1}{6}\) or \(\frac{5}{6}\) seen

M1: Uses their AOB to find the angle in the major sector soi

M1: Uses sector formula with their \(\theta\) soi

A1: Accept awrt 65.4 or 65.5

Alternative method for last 3 marks

M1: Uses their AOB to find the area of the minor sector soi

M1: Subtracts from \(25\pi\) soi
Do not allow for the difference between the areas of the circle and the triangle

A1: Accept awrt 65.4 or 65.5

Notes

M1: Uses both given small angle approximations in the expression. Must see \(\sqrt{4x}\) and \(\tfrac{1}{2}(2x)^2\) or \(\tfrac{1}{2} \times 4x^2\) condone missing brackets. Also allow for clear use of \(\cos 2x = 1 - 2\sin^2 x\) and \(\sin x \approx x\) used.

A1: AG Convincing argument to reach given answer

Notes

M1: Attempts to integrate the expression in powers of \(x\)
Must be seen

A1: At least 2 correct terms (no FT from (a) as answer given)

A1: Fully correct indefinite integral. Need not be simplified

A1: Correct value from correct indefinite integral. Allow for 0.24 or better if the method is clear.
(Do not allow for 0.24059… which is obtained by integrating the original function by calculator.)

Alternative solution

M1: Also allow for a single trapezium

A1: Soi For example \(\frac{0.1}{2}\ \frac{0.05}{2}, \frac{0.025}{2}, \frac{0.01}{2}\) used for 1, 2, 4 10 strips

A1: Must be ordinates from the approximating function
eg 2, 2.4372… and 2.5924… seen

A1: Accept awrt 0.24

Table of values

Answers

Notes

M1: sine wave with 2 complete cycles between 0 and \(2\pi\) which must be clear from labelling; condone eg waves of unequal amplitude or different length; allow labelling to override poor sketching; allow labelling in degrees for M1 only

A1: all correct with amplitude 1; ignore graph outside \([0, 2\pi]\)

Notes

M1: NB from use of \(\sin^{-1}\left(-\frac{1}{2}\right)\)
NB \(-0.262\) or \(-1.309\); decimals correct to 2 dp or better
or
3.665 or 5.760; decimals correct to 2 dp or better
M1 may be implied by two correct answers in radians

A1: two correct values
NB awrt 1.833, 2.880, 4.974, 6.021 or correct to 2 dp

A1: all four values correct with no extras for \(0 \leqslant \theta \leqslant 2\pi\)
if M0A0A0 allow SCB1 for two correct answers in degrees or SCB2 for all four correct in degrees 105, 165, 285 and 345

Notes

B1: OR \(r = \dfrac{10}{2 + \theta}\)
Expression for one of \(r\), \(\theta\) in terms of the other

M1: OR \(A = \dfrac{100\theta}{2(2+\theta)^2} = \dfrac{50\theta}{(2+\theta)^2}\)
Area in terms of either their \(r\) or their \(\theta\). Need not expand brackets.

M1: Completing the square

B1: Convincing explanation that there is a max

Alternative method 1

Alternative method 2

M1: Reasonable attempt at quotient rule

For information: \(\dfrac{\mathrm{d}^2A}{\mathrm{d}\theta^2} = -\dfrac{200}{256} = -0.78125\) at \(\theta = 2\)

Notes

B1: Substitution seen or implied by partial working
Condone 0.70588…

B1: Soi maybe in substitution

B1: Substitution seen or implied by partial working
Both answers must be \(\frac{12}{17}\)

Notes

M1: Attempt to use small angle approximations scores M0

E1: Convincing completion not necessarily in words

Notes

M1: may be implied by 1.13(4464…) or \(0.36\ldots \times \pi\)

Notes

B1: ignore units

Notes

M1: may see \(\cos x \approx 1 - \frac{x^2}{2}\)

A1: intermediate step needed from here to earn final mark

(corrected from the printed mark scheme: the last line is printed as \(\ln(0.05) - \ln(0.01) + 0.0012 \approx \ln 5\); with \(\mathrm{F}[x] = \ln x - \frac{1}{2}x^2\) the 0.0012 is subtracted)

Notes

M1: or differentiation of \(y\) using quotient rule and use of small angle approximation

A1: from \(-\frac{1}{x^2}\) or \(-\frac{1}{x^2} - 1\) oe

Notes

E1: allow sign change between 0.947745 and 0.9477475

Notes

B1: ignore units

Notes

M1: or \(\dfrac{\theta^\circ}{360} \times \pi \times 8^2\)
\(\theta\) = 120 – 120.3211…

A1: ignore units