Notes

M1: Substitutes the equation of \(C\) into \(x = r\cos\theta\) and differentiates to the required form. The \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta}\) need not be seen – if it is a clear attempt at differentiation allow the marks.

A1: Fully correct differentiation.

M1: Uses correct trig work to solve \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = 0\) to achieve a value for \(\sin\theta\) via a 3 TQ in sin (oe)

A1: Correct values for \(\sin\theta\). The \(-1\) need not be stated but any other values found will be A0.

M1: Uses \(r = a(1 + \sin\theta)\) and their \(\theta = \sin^{-1}(\ldots)\) (or \(\cos^{-1}\) if by error they ended up with a value for \(\cos\theta\) – and may have come from \(\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = 0\)) to find a polar coordinate. May be implied by a correct coordinate following \(\sin\theta = \dfrac{1}{2}\) having been found.

A1: Deduces the correct polar coordinates from correct work for \(A\) and \(B\) with no incorrect extras. May be listed, or may given as \((\theta,\ r)\). You may ignore values from \(\sin\theta = -1\) that are not used.

NB: Allow the final two marks in (a) if it is initially forgotten but the correct work is done as part of answering part (b).

Notes

M1: Uses \(2x = 10\) where \(x = r\cos\theta\) to find a value for \(a\) by any valid method. Several are shown in the scheme, there may be variations on these.

A1: Correct value achieved with no errors cso. Must see a correct unsimplified equation in \(a\) with trig terms evaluated before the final answer.

Notes

M1: Attempts to use the model and area \(= \left(\dfrac{1}{2}\right)\displaystyle\int r^2\,\mathrm{d}\theta\), multiplies out, uses the identity \(\sin^2\theta = \dfrac{1}{2}(\pm 1 \pm \cos 2\theta)\) to get into an integrable form and integrates. Limits are not required and the \(\dfrac{1}{2}\) may be missing for this mark. Condone the \(a\) not being squared.
NB: There may be less direct methods attempted (e.g. integration by parts). Such attempts must be complete attempts reaching an integral of the correct form. If you are unsure in any cases then use the review system.

A1: Correct integration of \(\left(\dfrac{1}{2}\right)\displaystyle\int r^2\,\mathrm{d}\theta\) condoning the \(\dfrac{1}{2}\) not being present at this stage. Allow with \(a^2\) used instead of the value.

dM1: Dependent on the first method mark. Full method for the area, with correct \(\dfrac{1}{2}\displaystyle\int_\alpha^\beta r^2\,\mathrm{d}\theta\) and applies correct limits over \(2\pi\) (\(\theta = 0\) and \(\theta = 2\pi\) or \(\theta = -\pi\) and \(\theta = \pi\)), or applies \(\displaystyle\int_\alpha^\beta r^2\,\mathrm{d}\theta\) with limits \(-\dfrac{\pi}{2}\) to \(\dfrac{\pi}{2}\). Accept equivalent combinations of limits split as two integrals and summed, but must be a full method. If no evidence of the substitution of limits is seen accept answers following the integral as long as suitable limits were linked to the integral. Condone if just \(-0\) is shown for a lower limit of 0 substituted.

A1: Correct area. Must have substituted for the \(a\).

Notes

M1: Uses all the information given for the model and realises the maximum value of \(r\) is \((4 + a)\) and uses the radius of the circle to find a value for \(a\).

A1: Deduces the correct value of \(a\). Note \(a = -1.5\) can potentially gain M1A0.

Notes

Note accept with their \(a\), a made up \(a\) or even \(a\) itself for the first 5 marks. Note use of \(a = -1.5\) can score full marks.

M1: Adopts a correct strategy for the area of the pool. This requires the correct use of the polar area formula including the \(\frac{1}{2}\).
Note the \(\frac{1}{2}\) may be implied by choice of limits (e.g. any span of \(\pi\) radians without the half implies an attempt at doubling so the \(\frac{1}{2}\) may not appear (even if the symmetry is incorrect).)

M1: Squares the bracket, achieving three terms, and applies \(\sin^2 3\theta = \dfrac{\pm 1 \pm \cos 6\theta}{2}\) in order to reach an integrable form. Condone numerical slips when expanding.

A1ft: Correct integration in any form (\(\frac{1}{2}\) not needed here) (follow through their \(a\)). ie as shown in scheme or if gathered it is \(\left(32 + a^2\right)\dfrac{\theta}{2} + \dfrac{8a}{3}\cos 3\theta - a^2\dfrac{\sin 6\theta}{12}\)

dM1: Depends on previous M. Uses appropriate limits for their integrated function. Allow limits other than 0 and \(2\pi\) by using symmetry provided the correct multiple is used, e.g. \(\dfrac{\pi}{2}\) and \(\dfrac{3\pi}{2}\) followed by doubling (which may be cancelled with the \(\frac{1}{2}\)) or \(-\dfrac{\pi}{6}\) and \(\dfrac{\pi}{2}\) and multiplying by 3

DM1: Depends on first M. Fully correct strategy for obtaining the area of \(T\). Must have a correct attempt at the circle area (maybe be via integration) and area inside the curve. Symmetry may have been used.

A1: Correct area from fully correct work (all previous marks scored). Units not required. The decimal answer \(18.875\pi\) is acceptable, but answer must be exact, not a rounded decimal.

Note: \(\sin^2 3\theta = \dfrac{\pm 1 \pm \cos 2\theta}{2}\) use will score a maximum of M1M0A0dM0DM1A0

Note: if \(\dfrac{1}{2}\displaystyle\int_0^{2\pi} 6^2 - (4 - a\sin 3\theta)^2\,\mathrm{d}\theta\) is used the scheme will follow as above with A1ft for \(10\theta - \dfrac{8a}{6}\cos 3\theta - \dfrac{a^2}{4}\left(\theta - \dfrac{\sin 6\theta}{6}\right)\) and dM1DM1 gained together.

Note: Integrating over 0 to \(2\pi\) means the trig terms will disappear so watch out for incorrect trig terms in the integration, which will lead to the correct answer but will lose both A marks.

Notes

B1: Recalls the correct cardioid shape for this type of polar curve.

B1: Correctly placed with the pole, initial line and point where curve crosses the initial line all indicated in some way.

Notes

M1: Uses \(y = r\sin\theta\) with the curve and attempts to differentiate. Accept any correct form but may have slips in coefficients, so e.g. as shown or \(A\cos\theta + B\cos^2\theta + C\sin^2\theta\) can score M1.

A1: Correct differentiation. Accept equivalents, e.g. \(3\cos\theta + \sqrt{5}\cos^2\theta - \sqrt{5}\sin^2\theta\)

M1: Sets their derivative equal to zero (may be implied), using trig identities to form a quadratics for \(\cos\theta\)
Allow this mark form the use of \(r\cos\theta\) leads to \(-3\sin\theta - 2\sqrt{5}\sin\theta\cos\theta = 0\), score M1 for factorising out \(\sin\theta\) and finds a value for \(\cos\theta\)

A1: Solves their quadratic and selects the correct value for \(\cos\theta\). If the other value is given it is A0 unless clearly rejected.

Notes

B1: Correct value for \(r\)

Notes

B1: Correct area formula applied, including the \(\dfrac{1}{2}\) and correct limits, may be seen later, \(\mathrm{d}\theta\) may be implied.

M1: Attempts the integration \(\sinh\theta \rightarrow \pm\cosh\theta\) and \(\cosh\theta \rightarrow \pm\sinh\theta\) or in terms of exponentials \(\displaystyle\int \mathrm{e}^{\lambda\theta}\,\mathrm{d}\theta = \frac{1}{\lambda}\mathrm{e}^{\lambda\theta}\)

M1: Applies their limits to the integral, subtracts (there must be an attempt to integrate) and uses exponential definitions to achieve answer in suitable form. Condone the inclusion of i or a missing ½ from the definitions. This can be implied.

A1: Correct exact answer, no i

Notes

M1: Substitutes the equation of \(C\) into \(x = r\cos\theta\) and differentiates to the required form

A1: Fully correct differentiation

dM1: Dependent on previous method mark. Sets their \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\) and uses correct trig identities to find a value for \(\theta\). Alternatively substitutes \(\theta = \dfrac{\pi}{4}\) into their \(\frac{\mathrm{d}x}{\mathrm{d}\theta}\) and shows equals 0.

A1*: Shows that \(r = 2\) and hence the polar coordinates \(\left(2, \dfrac{\pi}{4}\right)\) from correct working

Notes

M1: Applies area \(= \dfrac{1}{2}\int r^2\ \mathrm{d}\theta\), multiplies out, uses the identity \(\pm 1 \pm \tan^2\theta = \sec^2\theta\) to get into an integrable form and integrates. Condone missing \(\mathrm{d}\theta\), limits are not required for this mark

A1: Correct integration. Note may include \(\theta - \theta\) if the one’s were not cancelled earlier.

dM1: Dependent on the first method mark. Applies the limits of \(\theta = 0\) and \(\theta = \dfrac{\pi}{4}\) and subtracts the correct way round. Since substitution of the limit \(\theta = 0\) is 0 so may be implied

M1: Correct method to find the area of triangle seen. This may be minimal but area = 1 only is M0, they need to show some method.

M1: Finds the required area = area of triangle – area bounded by the curve

A1*: Correct answer, with no errors or omissions. cso

Alternative

M1: Applies area \(= \dfrac{1}{2}\int r^2\ \mathrm{d}\theta\), multiplies out, uses the substitution \(u = \tan\theta\) to get into an integrable form and integrates. Limits are not required for this mark

A1: Correct integration

dM1: Dependent on the first method mark. Applies the limits of \(u = 0\) and \(u = 1\) or substitutes back using \(u = \tan\theta\) and uses the limits \(\theta = 0\) and \(\theta = \dfrac{\pi}{4}\) and subtracts the correct way round. Since substitution of the limit \(\theta = 0\) is 0 so may be implied

M1: Correct method to find the area of triangle

M1: Finds the required area = area of triangle – area bounded by the curve

A1*: Correct answer, with no errors or omissions. cso

(Corrected from the printed mark scheme: both M1 notes print the area formula as \(\dfrac{1}{2}\int r^2\theta\ \mathrm{d}\theta\).)

Notes

(a)
M1: Complete method to find the correct form for \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta}\)
A1: Correct \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta}\)
M1: Sets \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = 0\) and factorises to find values for either \(\sin\theta\) or \(\cos\theta\).
B1: Deduces that as \(\sin\theta = 0\) this provides two tangents. This can be implied by 2 values for \(\theta\)
A1*: Concludes that as \(\cos\theta = -\dfrac{p}{4} \gt -1 \Rightarrow p \lt 4\) and \(p\) is a positive constant \(\therefore 2 \lt p \lt 4\)

(corrected from the printed mark scheme: the conclusion in this note is printed as \(0 \lt p \lt 4\); the scheme and the question give \(2 \lt p \lt 4\))

Notes

(b)
B1: Correct shape and position.

Notes

(c)
M1: Uses the model to find the area of the cross section \(2 \times \dfrac{1}{2}\displaystyle\int_0^{\pi}\left[20(3 + 2\cos\theta)\right]^2\,\mathrm{d}\theta\) or \(\dfrac{1}{2}\displaystyle\int_0^{2\pi}\left[20(3 + 2\cos\theta)\right]^2\,\mathrm{d}\theta\)
M1: Uses the identity \(\cos 2\theta = 2\cos^2\theta - 1\) to integrate to the required form.
A1: Correct integration.
M1: Uses limits \(\theta = 0\) and \(\theta = \pi\) or \(\theta = 0\) and \(\theta = 2\pi\) as appropriate and subtracts the correct way around provided there is an attempt at integration.
Note if first M1 is not awarded for incorrect limits then award this mark for their limits used.
M1: Multiplies their area by 90 (cm).
M1: Divides their volume by 50000
A1: 25 (minutes)

Notes

(d)
B1: See scheme for examples. Any reference to the flow of water is B0

Notes

M1: Realises that the angles at the intersection are required and solves \(C_1 = C_2\) to obtain a value for \(\theta\)
A1: Correct value for \(\theta\). Must be in radians – if given in degrees you may need to check later to see if they convert to radians before substitution.
M1: Evidence selecting the correct polar area formula on either curve
M1: Fully expands both expressions for \(r^2\) either as parts of separate integrals or as one complete integral. (Can be scored from incorrect polar area formula, e.g. missing the ½)
A1: Correct expansions for both curves (may be unsimplified)
M1: Selects the correct strategy by applying the correct double angle identity in order to reach an integrable form and attempting the integration of at least one of the curves.
A1: Correct integration (of both integrals if done separately),
FYI: If done separately the correct integrals are
\(\displaystyle\int (1 + \sin\theta)^2\,\mathrm{d}\theta = \theta - 2\cos\theta + \frac{1}{2}\left(\theta - \frac{1}{2}\sin 2\theta\right) = \frac{3}{2}\theta - 2\cos\theta - \frac{1}{4}\sin 2\theta\) and
\(\displaystyle\int 9(1 - \sin\theta)^2\,\mathrm{d}\theta = 9\theta + 18\cos\theta + \frac{9}{2}\left(\theta - \frac{1}{2}\sin 2\theta\right) = \frac{27}{2}\theta + 18\cos\theta - \frac{9}{4}\sin 2\theta\)
DM1: Depends on all previous M’s. For a fully correct strategy with appropriate limits correctly applied to their integral or integrals and terms combined if necessary. Make sure that if limits of \(\dfrac{\pi}{6}\) and \(\dfrac{\pi}{2}\) are used that the area is doubled as part of the strategy.
A1: Correct area

Notes

M1: Interprets the information from the model and realises that the maximum value of \(r\) gives half the length of the table top (or equivalent) and solves to find a value for \(a\). Use \(\theta = 0\) and \(r = 0.6\) or \(\theta = \pi\) and \(r = -0.6\) to find a value for \(a\).

Using \(\theta = 2\pi\) is M0

A1*: Correct value for \(a\).

Alternative

M1: Uses \(a = 0.2\) and \(\theta = 0\) to find a value for \(r\)

A1: Finds \(r = 0.6\) and concludes that \(a = 0.2\)

Notes

B1: \(1.2 \times 0.6\) or 0.72

M1: A correct strategy identified for finding an area enclosed by the polar curve using a correct formula with \(r\) substituted. Attempt at area \(= \dfrac{1}{2}\displaystyle\int (0.4 + 0.2\cos 2\theta)^2\,\mathrm{d}\theta = \ldots\)

Look for \(= \lambda \times \dfrac{1}{2}\displaystyle\int (0.4 + 0.2\cos 2\theta)^2\,\mathrm{d}\theta = \ldots\)

If the \(\dfrac{1}{2}\) is not explicitly seen then look at the limits and it must be either

Condone missing \(\mathrm{d}\theta\)

M1: Squares to achieve three terms and uses \(\cos^2 2\theta = \dfrac{\pm 1 \pm \cos 4\theta}{2}\) to obtain an expression in an integrable form.

A1ft: Correct follow through integration as long as the previous two method marks have been awarded.

dM1: Dependent of first method mark. Finds the required area enclosed by the curve using the correct limits.

There are only three cases either \(\dfrac{1}{2}\displaystyle\int_0^{2\pi} (0.4 + 0.2\cos 2\theta)^2\,\mathrm{d}\theta\) or \(\displaystyle\int_0^{\pi} (0.4 + 0.2\cos 2\theta)^2\,\mathrm{d}\theta\) or \(2\displaystyle\int_0^{\frac{\pi}{2}} (0.4 + 0.2\cos 2\theta)^2\,\mathrm{d}\theta\)

The use of the limit 0 can be implied if it gives 0 but the use of 0 must been seen or implied if it does not result in 0 (just writing 0 is insufficient)

A1: Correct area of the glass following fully correct working. Do not award for the correct answer following incorrect working.

M1: Subtracts their area of the glass from their area of the rectangle, as long as it does not give a negative area

A1: awrt 0.155 or awrt 0.155 m² (If the units are stated they must be correct)

Note: Using a calculator to find the area scores a maximum of B1M0M0A0M0A0M1A1

Typical solution

Typical solution

When \(\theta = 0\) then

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

At \(A\), \(r = 2 + \tan\dfrac{\pi}{4} = 3\)

Area of triangle \(OAB\)

Typical solution

Area enclosed by curve

Shaded area

Typical solution

Typical solution

Typical solution

\(r\) is minimum when \(\cos\theta = 1\), and maximum when \(\cos\theta = -1\)

Typical solution

Typical solution

Starting from \(y^2 + \dfrac{16x^2}{25} = 1\):

The required transformation is a translation by \(\begin{bmatrix} -\frac{3}{4} \\ 0 \end{bmatrix}\)

Typical solution

Typical solution

Typical solution

(i)

(ii)

\(y\) is replaced with \(x\), and vice versa

Reflection in \(y = x\)

Typical solution

Typical solution

Roberto has used incorrect limits of integration.

He should only have included values of \(\theta\) which make \(\sin 2\theta\) positive, because \(r^2\) must be positive

Typical solution

by symmetry

Typical solution

Max. value of \(r = 3\)

For \(r\) maximum, \(\sin 2\theta = 1\)

\(P\left(3, \dfrac{\pi}{4}\right)\) and \(Q\left(3, -\dfrac{3\pi}{4}\right)\)

Typical solution

(i)

Cartesian coordinates of \(P\) are \(\left(\dfrac{3\sqrt{2}}{2}, \dfrac{3\sqrt{2}}{2}\right)\)

For \(P^{\prime}\), \(r^2 = \dfrac{81}{2} + \dfrac{9}{2} = 45 \Rightarrow r = 3\sqrt{5}\)

\(P^{\prime}\left(3\sqrt{5}, \tan^{-1}\left(\dfrac{1}{3}\right)\right)\)

(ii)

Det M = 1

Area enclosed by \(C_2\) = (Area enclosed by \(C_1\)) × det M

= 9 × 1 = 9

Typical solution

\(\therefore\) \(\left(3, \dfrac{\pi}{3}\right)\) lies on \(C\)

Typical solution

furthest from \(O\) is \((6, \pi)\)

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

And \(x = r\cos\theta\)

So in Cartesian coordinates: \(x = \frac{7}{4}\) which is perpendicular to the \(x\)-axis

\(\therefore L\) is perpendicular to the initial line

Typical solution

At points of intersection

\(\cos\theta = -\frac{7}{2}\) (reject as \(-1 \leqslant \cos\theta \leqslant 1\))
or \(\cos\theta = \frac{1}{2}\)

Points are \(\left(\frac{7}{2}, \frac{\pi}{3}\right)\) and \(\left(\frac{7}{2}, -\frac{\pi}{3}\right)\)

Typical solution

Area of triangle \(OPQ\)

Area of sector \(OPQ = 2\left(\int_0^{\frac{\pi}{3}} \frac{1}{2}r^2\,\mathrm{d}\theta\right)\) (by symmetry)

Typical solution

Typical solution

Typical solution

Typical solution

Point of intersection:

\(\theta\) is acute so \(\quad \theta = \sin^{-1}\left(\dfrac{\sqrt{5} - 1}{2}\right) = \alpha\) as defined in the question.

Area enclosed \(= A_1 + A_2\)

Typical solution

Typical solution

\(x = r\cos\theta\) and \(y = r\sin\theta\)

Typical solution

\(\therefore\) angle \(STP = \theta + b\)

\(\therefore\) angle \(OPT = b\) (exterior angle of a triangle)

So the angle between the line \(OP\) and the tangent \(TPQ\) does not depend on \(\theta\).

Typical solution

Typical solution

Typical solution

Typical solution

Notes

M1: Can be implied by correct value in final answer
0 does not have to be explicitly rejected.

A1: Notation must be correct, accept \((4, 2)\) but not \([2, 4]\)

Notes

B1: Constant and limits must be correct (BOD for missing \(\mathrm{d}\theta\))

M1: Condone integrals switched or missing \(\frac{1}{2}\) (ignore limits)
Must have \((2\theta)^2\) and \((\theta^2)^2\) and a subtraction for this mark.

A1FT: Integrating their expression correctly (ignore limits)
FT their integrals.

A1: Answer must be positive but condone recovery by removing negative sign at end

Notes

M1*: Using \(\frac{1}{2}\int r^2\,\mathrm{d}\theta\) with correct expression for \(r\) - condone lack of (or incorrect) limits. Condone missing \(\frac{1}{2}\) only.

M1: Expanding and using correct \(\cos^2\theta = \dfrac{1}{2}(1 + \cos 2\theta)\) - need not be in an integral.

M1dep*: Integrating their \(\int k_1\cos 2\theta + k_2\cos\theta + k_3\,\mathrm{d}\theta\) correctly to \(\frac{1}{2}k_1\sin 2\theta + k_2\sin\theta + k_3\theta\) with \(k_1, k_2, k_3 \neq 0\)

M1: Substitute correct limits 0 and \(2\pi\) - dependent on previous M mark. Need not see zeros from 0 limit or other terms that would be zero when evaluated. (Corrected from the printed mark scheme: the last term of this line is printed as \(2\pi\left(\frac{k^2}{2} + \left(k + \frac{10}{k}\right)^2\right)\); after the factor \(\frac{1}{2}\) it is \(\pi\left(\ldots\right)\), as the next line shows.)

A1: cao from correct integral and correct integration. Condone \(\pi\left(3k^2 + 40 + 200k^{-2}\right)\) from missing \(\frac{1}{2}\).

M1: Differentiating their \(A_k\) correctly which must be of the form \(ak^2 + b + ck^{-2}\) with \(a, b, c \neq 0\). Dependent on all previous M marks.

A1: Dependent on all previous marks and no errors. A0 if correct \(k\) obtained but \(\frac{1}{2}\) missing from area formula (unless justified). Accept any exact equivalent. Condone omission of \(\pi\) (or other constant factor) for final two marks.

Notes

B1: Condone use of “origin” rather than “pole” in explanations.

B1: Must link domain of \(\theta\) with quadrant.

B1: Convincingly explaining why it is a single loop. \(\theta\) values/\(r = 0\) must be linked to loop.
or “There are no other solutions to \(r = 0\) (in the domain)”.

Notes

B1*: All 4 entries correct and non-zero ones correct to at least 3 sf. ie awrt 0.912, 1.59, 1.35.

M1: All points, theirs and given, plotted approximately correctly (ie on the \(\theta\) = constant lines and in the correct half between the appropriate \(r\) = constant lines)
Points can be inferred by curve

A1dep*: Smooth curve plotted between correctly plotted points (see criteria above) joining at the pole.
Single maximum \(r\) at some angle between \(\pi/4\) and \(5\pi/12\) (it might be on the wrong side of \(\theta = \pi/3\).
Curve needs to be: smooth (except at pole); curved (not line segments); closed; single loop; cusp at pole

Notes

M1: Attempting to differentiate \(r(\theta)\) using the product rule.
\(a\sinh\frac{1}{3}\theta\cos 2\theta + b\sin 2\theta\cosh\frac{1}{3}\theta\) where \(a\), \(b\) are non-zero constants

A1: All correct (any form)

M1: Setting their derivative to 0 and rearranging to a form containing tan and tanh only.
If using exponential form, must reach tan and reduce to one exponential term.
eg \(\mathrm{e}^{\frac{2}{3}\phi}(6 + \tan 2\phi) = 6 - \tan 2\phi\)

A1: AG so intermediate step must be shown.
Condone use of \(\theta\) rather than \(\phi\) but final answer must be in terms of \(\phi\) (or explicitly stating “\(\theta = \phi\)”).

Notes

B1: 0.0741 comes from using degrees.

Notes

B1: (1st) Correct condition for intersection of curves leading to a correct expression or value for an angle at a PoI

B1ft: For finding (\(\pm\)) the correct area of the sector (or half of it) either using \(\frac{1}{2}r^2\theta\) or \(\frac{1}{2}\int_{\theta_1}^{\theta_2}(5)^2\,\mathrm{d}\theta\). Accept lower limit of 0. May be embedded in a calculation for total area.
Here \(\theta_1 = -\cosh^{-1}\frac{5}{3}\), \(\theta_2 = \cosh^{-1}\frac{5}{3}\) or \((\theta_1 = -\ln 3), (\theta_2 = \ln 3)\), i.e. may later evaluate integral… for some \(k\)

M1: (1st) Correct use of area formula with \(\pm\)their value for limits. Condone lower limit of 0. Here \(\theta_1 = -\cosh^{-1}\frac{5}{3}\), \(\theta_2 = \cosh^{-1}\frac{5}{3}\). Limits can be seen later. Later doubling may be seen

*M1: Correct conversion of \(\cosh^2\theta\) into a form which can be integrated, or \(\int\cosh^2\theta\,\mathrm{d}\theta = \frac{1}{2}\int 1 + \cosh 2\theta\,\mathrm{d}\theta\)

dep*M1: Correctly integrating \(\cosh^2\theta\) and substituting their limits. Could be embedded.
Or \(\displaystyle\int_{-\ln 3}^{\ln 3}\cosh^2\theta\,\mathrm{d}\theta = \frac{1}{2}\int_{-\ln 3}^{\ln 3}1 + \cosh 2\theta\,\mathrm{d}\theta = \frac{1}{2}\left[\theta + \frac{1}{2}\sinh 2\theta\right]_{-\ln 3}^{\ln 3} = \frac{20}{9} + \ln 3\)
NB \(A_{C_2} = \dfrac{9}{2} \times \dfrac{1}{4}\left(\dfrac{80}{9} + 4\ln 3\right) = 10 + \dfrac{9}{2}\ln 3\)

A1: (1st) Accept unsimplified form. If B1B1M1M0M0 can score ScB1 both here and in the next line for correct answers

Notes

M1: Fully shown (AG). Use of double angle formula and \(y = r\sin\theta,\ x = r\cos\theta\) oe

A1: Any correct form of the equation with \(\theta\) eliminated.

A1: Fully shown

Notes

M1: Interchange of \(x\) and \(y\)

A1: Showing/explaining clearly that interchange of \(x\) and \(y\) leaves the equation unchanged.
(corrected from the printed mark scheme: the printed line reads \(9\left(-1^4\right)(x - y)^4\); the factor is \((-1)^4\))

Alternative method

M1: Showing that \(\sin 2\theta\) is symmetrical about \(\theta = \frac{1}{4}\pi\) (or \(\theta = \frac{5}{4}\pi\)). May be done graphically or using trig identities.

A1: so \(r = 3(1 - \sin 2\theta)\) has lines of symmetry at \(\theta = \frac{1}{4}\pi\) and \(\theta = \frac{5}{4}\pi\) and the line \(y = x\) has polar equation \(\theta = \frac{1}{4}\pi\) and \(\theta = \frac{5}{4}\pi\).
Any mention of \(\theta = \frac{5}{4}\pi\) is unnecessary since it is implied by the \(\theta = \frac{1}{4}\pi\) case.

Notes

B1: For identifying appropriate limits correctly

M1: Use of \(A = \dfrac{1}{2}\displaystyle\int r^2\,\mathrm{d}\theta\)

M1*: Expand correctly and use of \(\sin^2 2\theta = \dfrac{1}{2}(1 - \cos 4\theta)\) Ignore limits

M1dep: Integrate

M1: Substituting correct limits and subtracting

A1: Answer for each loop which must be stated

Alternative method for last 2 marks

M1: Substituting correct limits and subtracting

A1: Answer for each loop which must be stated

Notes

B1: or just \(\theta = 2\pi\). ISW

B1: or just \(r = 4\pi\). ISW

Notes

M1: Correct condition for PoI

A1: or \(r = 2\) and \(\theta = 1\). ISW

Notes

B1: \(C_2\) drawn as a smooth curve spiralling out from \([1, 0]\) outside \(C_1\) until a single point in the 1^st quadrant and then inside \(C_1\). Must stop on initial line.
Start at \([1, 0]\).
Intersection in 1^st quadrant (by eye).
\(r\) increasing (by eye).
Reaches initial line and stops.
Ignore labels.

Notes

B1: Both required
Don’t give if any extras within range.
Ignore values outside range

Notes

B1: For \(r\)

B1: For \(\theta\)

Notes

M1: Correct use of formula – ignore limits

M1*: Attempt to use double angle formula (Could be wrong way round, 2 missing or sign wrong)

DepM1: Integrate their integrand

A1: Use correct limits, must be seen

Notes

M1: Using triple angle formula and \(y = r\sin\theta\)

A1: isw

Notes

M1: Substitution
May see “or \(r = 0\)” but not required.

Notes

M1: Correct substitution into their \(\mathrm{f}(\theta)\)

Notes

B1: Allow \(y = x\).
Must have \(\theta =\)

Notes

B1: BC

Notes

B1: For both, ignore extras.

B1: Conclusion – both statements for \(r\) need to be mentioned

Notes

M1: Correct formula for area with \(r\) correctly substituted and their limits. Must be unambiguous but can be implied by correct answer/later work
Incorrect formula = M0A0
Condone missing \(\mathrm{d}\theta\)

A1: BC

Notes

M1: or radius \(R = 0.3466\) (4 sf); condone correct \(R\) or \(D\) without reasoning
It must be clear that the \(r\) value would be the diameter of the circle; the calculation alone is insufficient for M1.
M1 can be implied by area given as \(\pi\left(\dfrac{\ln 2}{2}\right)^2\)

A1: or \(R = 0.3466\) (4 sf) (or \(D = 0.6931\)) \(\Rightarrow A = 0.3773\) (4 sf) which is not 0.4162 (4 sf)
Explanation must include comparison of \(R\)’s, \(D\)’s or \(A\)’s and conclusion. Allow correct working to 3 sf.

Notes

M1: Correct form, in terms of \(\theta\). M1 can be implied by 1.0757… BC

*A1: Integrand has been squared out. Must include limits (can be seen later)

dep*M1: Might be as result of substitution. Allow coefficient error for M1. eg \(\frac{3}{4}\left[\mathrm{e}^{u}\right]_{-\frac{2}{3}}^{\frac{2}{3}}\) or \(\frac{3}{4}\left[\mathrm{e}^{\frac{2}{3}u}\right]_{-1}^{1}\) oe

A1: isw

Notes

*M1: Attempt to differentiate using product and chain rules. Must be in the form \(uv' + u'v\) with at most one of \(u, v, u'\) or \(v'\) incorrect or omitted

dep*M1: Setting \(r'\) to zero and factorising/cancelling to produce a quadratic equation in cos and/or sin

M1: Use of \(\cos^2 + \sin^2 = 1\) to find 3 term quadratic equation in \(\cos\theta\). Or could be in \(\sin^2\theta\): \(4\sin^4\theta + 9\sin^2\theta - 9 = 0\)

*A1: Solving quadratic correctly. \(\sin^2\theta = \frac{3}{4}, -3\)

dep*A1: Explicitly rejecting root. Rejects \(\sin^2\theta = -3\) and \(\sin\theta = -\frac{\sqrt{3}}{2}\)

A1: AG. At least one intermediate step must be seen. Can be awarded even if rejection of root(s) was implicit.

Notes

M1: substituting for all \(x\) and \(y\) terms correctly. Implied by a correct unsimplified equation for \(r^4\) or \((r^2)^2\)

A1: RHS could be unsimplified; \(r^4\) could be left as \((r^2)^2\)

Notes

B1FT: or \(r^2\) maximum \(= \frac{1}{2}\) or correct differentiation of their \(r^2\) (or \(r\)), equating to zero and getting \(\cos 2\theta = 0\), or \(\theta = \frac{\pi}{4}\). soi by correct value for their \(r\). FT from part (a)

B1FT: o.e. must be exact. www. FT from part (a)

Notes

M1: \(\frac{1}{2}\int r^2\,\mathrm{d}\theta\) used with their \(r^2\) provided it is in the form \(a\sin b\theta\). Must be seen. Condone missing \(\mathrm{d}\theta\) and missing or incorrect limits

A1FT: FT their integral (i.e. \(-\frac{a}{2b}\cos b\theta\)). Must be seen.
Note \(\int \sin\theta\cos\theta\,\mathrm{d}\theta = \frac{1}{2}\sin^2\theta\) or \(-\frac{1}{2}\cos^2\theta\),

M1: substituting correct limits after attempt to integrate, soi by e.g. correct answer or \(\frac{1}{4}\left(\frac{1}{2} + \frac{1}{2}\right)\) if correct integral found. Accept correct alternative limits, e.g. \(\pi\) to \(\frac{3\pi}{2}\)
Requires an initial integral of the form \(k\int a\sin b\theta\,\mathrm{d}\theta\)

A1: cao www

Notes

B1*: One loop in correct position (requires initial line drawn)

B1*: Exactly three loops in correct position

B1dep: Lower loop only shown with a broken line. Any coordinates on the curve must be correct (either polar or cartesian)

Notes

M1: \(A = \int \frac{1}{2}a^2\sin^2 3\theta\,\mathrm{d}\theta\). Condone missing \(\mathrm{d}\theta\).

A1: Limits correct. Accept alternative limits between 0 and \(\pi\) provided correct multiplication or division of integral is seen or implied at some stage.

M1: Double angle formula used correctly in their integral. Condone missing \(\mathrm{d}\theta\).

A1: \(k\left[\theta - \frac{1}{6}\sin 6\theta\right]\). Condone incorrect or missing limits

A1: SC B4 for an otherwise fully correct answer using limits outside the range 0 to \(\pi\).

Notes

B1: (A) or \(\theta = 2\pi\)

Notes

M1: Allow \(\sin\theta = \frac{1}{2}\) (or \(\lt \frac{1}{2}\) or \(\leqslant \frac{1}{2}\) or \(\geqslant \frac{1}{2}\))

A1: both

A1: condone \(\leqslant\) for \(\lt\)

Notes

M1: symmetrical loop about the initial line

A1: correct shape with cusp at O

Notes

M1: correct integral and limits, condone missing \(\mathrm{d}\theta\)
limits can be soi by later work
may see \(\int_0^{\pi} a^2(1 - \cos\theta)^2\,\mathrm{d}\theta\)

M1: Expanding correctly

M1: substituting for \(\cos^2\theta\)

B1: \(k\left[3\theta - 4\sin\theta + \dfrac{1}{2}\sin 2\theta\right]\)

Notes

\(\alpha = 1.107\): or \(\arctan(2)\)

\(\theta = 1.107\) rad: or \(\arctan(2)\)

Alternative solution

B1: 2.24 or better

Notes

(b)(i)

M1: attempt to find cartesian eqn

M1: multiplying by \(r\)

M1: completing the square

Notes

SC: Coordinates reversed \((\theta, r)\) award B1B0

Notes

M1: accept even function

Notes

B1: or \(a^2\displaystyle\int_0^{\pi} (3 + 2\cos\theta)^2\,\mathrm{d}\theta\); limits seen in work

M1: substitute an expression involving \(\cos 2\theta\) for \(\cos^2\theta\)

A1: \(= \left[11\theta + 12\sin\theta + \sin 2\theta\right]\)

Notes

M1: substituting for \(r^2\), \(x\) and \(y\)

A1: NB AG

Notes

B1: one loop shown; allow sep diags

B1*: both shown (no extras), in correct quadrant

B1dep: \(-\)ve \(r\) with broken line; dep B1*

Notes

B1: condone missing \(\mathrm{d}\theta\); limits soi

B1: \(\displaystyle\int \sin 2\theta\,\mathrm{d}\theta = -\frac{1}{2}\cos 2\theta\)