Notes

B1: See scheme. This can appear anywhere in the proof.
Accept \(2^4\sin^4\theta\) for \(16\sin^4\theta\), but not \((2\sin\theta)^4\)
Alternatively, they may instead substitute \(\sin\theta = \dfrac{1}{2\mathrm{i}}\left(z - \dfrac{1}{z}\right)\) into the expression \(8\sin^4\theta\). This can be implied but must come from correct work. This can appear anywhere in their proof.
e.g. \(8\sin^4\theta = 8\left[\dfrac{1}{2\mathrm{i}}\left(z - \dfrac{1}{z}\right)\right]^4\) or \(8\sin^4\theta = 8\left[\left(z - \dfrac{1}{z}\right)\right]^4\dfrac{1}{2^4}\)
and allow \(8\sin^4\theta = \dfrac{1}{2}\left[\left(z - \dfrac{1}{z}\right)\right]^4\)

M1: Finds the expansion of \(\left(z - \dfrac{1}{z}\right)^4\) which may be unsimplified. All five terms must be present.
Condone sign slips only

A1: Correct expansion, with terms grouped.

M1: Uses \(z^n + \dfrac{1}{z^n} = 2\cos n\theta\) to write in terms of \(\cos 4\theta\) and \(\cos 2\theta\)

A1*: Achieves the printed answer with no errors or omissions. Cso Follows A0.

Notes

B1: Correct formula \(\pi\displaystyle\int \left(\sin^2\left(\dfrac{1}{2}y\right)\right)^2\,\mathrm{d}y\) (but not \(\text{vol} = \pi\displaystyle\int x^2\,\mathrm{d}y\)) used to find a volume, stated or implied, ignore limits. If there is a missing \(\pi\) or d\(y\) in their integral, then withhold this mark only. However, this mark may be awarded if \(\pi\) and d\(y\) are seen together later in their integral.

Do not award the following marks if algebraic integration is not used. For example finding an answer of 7.3 without algebraic integration will obtain M0A0dM0A0

M1: Uses the result in part (a) to express the volume in an integrable form and attempts to integrate.
Award for an integral of the form \(\displaystyle\int \dfrac{1}{8}\big(A\cos(2y) + B\cos(y) + C\big)\ (\mathrm{d}y)\) with at least one term integrated correctly. Do not be concerned if they use a different variable such as \(\theta\) for \(y\).

Special Case:
If they have not used part (a) and instead use the double angle formulae:
\(\sin^2\alpha = \dfrac{1}{2} - \dfrac{1}{2}\cos 2\alpha\) and \(\sin^4\alpha = \left(\dfrac{1}{2} - \dfrac{1}{2}\cos 2\alpha\right)^2\) with \(\alpha = \dfrac{1}{2}y\)
In this case they must obtain an exact integral equivalent to \(\displaystyle\int \dfrac{1}{8}\big(\cos(2y) - 4\cos(y) + 3\big)\ (\mathrm{d}y)\) and proceed to integrate at least one term correctly.

A1: Correct integration. May be in terms of another variable such as \(\theta\). Ignore \(\pi\)

dM1: Dependent on previous method mark. Finds the required volume using \(\pi\displaystyle\int_0^{\frac{8\pi}{5}} x^2\,\mathrm{d}y\) and applies their limits to their integral and subtracts the correct way round.
If there are no limits seen substituted, then a correct final answer implies the correct use of limits and the inclusion of \(\pi\). This is provided they have already achieved an integrated expression of \(\dfrac{1}{8}\left(\dfrac{1}{2}\sin(2y) - 4\sin(y) + 3y\right)\) oe with correct limits seen, possibly on their integral.
If their integration is incorrect, then there must be evidence of substituting both limits in each of their terms and subtracting. Allow the omission of subtracting zero provided their integration would produce zero for the lower limit.

A1: awrt 7.3

Notes

M1: Finds the mass of the ornament by multiplying their volume by 0.85

A1ft: Draws an appropriate conclusion about the suitability of the model, comparing their two masses.
If the masses differ by 10% then they must conclude it is a good model.
If the masses differ between 10% and 20% then they may conclude it is either a good model or poor model.
If the masses differ by greater than 20% then they must conclude it is a poor model.

Alternative 1

M1: Finds the volume of the ornament by dividing the mass of 6 grams by 0.85 g/cm\(^3\)

A1ft: Draws an appropriate conclusion about the suitability of the model, comparing their two volumes.
If the volumes differ by 10% then they must conclude it is a good model.
If the volumes differ between 10% and 20% then they may conclude it is either a good model or poor model.
If the volumes differ by greater than 20% then they must conclude it is a poor model.

Alternative 2

M1: Finds the density of the ornament by dividing the mass of 6 grams by 7.3 cm\(^3\) (corrected from the printed mark scheme: printed as 7.3 g/cm\(^3\))

A1ft: Draws an appropriate conclusion about the suitability of the model, comparing their two densities.
If the densities differ by 10% then they must conclude it is a good model.
If the densities differ between 10% and 20% then they may conclude it is either a good model or poor model.
If the densities differ by greater than 20% then they must conclude it is a poor model.

Notes

B1: Recalls the exponential definitions for \(\sinh x\) and at least one of \(\cosh x\) or \(\tanh x\), which may be embedded in their workings.

M1: Substitutes the correct exponential definitions to form an equation leading to a quartic equation for \(\mathrm{e}^x\). Any identities used such as \(\sinh 2x = 2\sinh x\cosh x\) must be correct

A1*: Correct equation following at least one intermediate stage, with no errors seen. They cannot go directly from their substitution to the given answer. cso

Notes

B1: Substitutes \(\mathrm{e}^x = 3\) into each term of the equation, shows \(= 0\) and states therefore a solution or writes \(\left(\mathrm{e}^x - 3\right)\) is a factor. Allow a tick, box, QED or appropriate conclusion.

OR
Substitutes \(x = \ln 3\) into each term of the equation, shows \(= 0\) and states therefore a solution or writes \(\left(\mathrm{e}^x - 3\right)\) is a factor. Allow a tick, box, QED or appropriate conclusion.

OR
Factorises their equation:
\(15\mathrm{e}^{4x} - 48\mathrm{e}^{3x} + 32\mathrm{e}^x - 15 = 0\)
\(\Rightarrow \left(\mathrm{e}^x - 3\right)\left(15\mathrm{e}^{3x} - 3\mathrm{e}^{2x} - 9\mathrm{e}^x + 5\right) = 0\)
and states hence \(\mathrm{e}^x = 3\) is a solution

Do not award this mark for simply solving a quartic such as \(15y^4 - 48y^3 + 32y - 15 = 0\) on the calculator and stating \(y = 3\), hence \(\mathrm{e}^x = 3\) is a solution.

Notes

B1: States the correct exact coordinates. Allow \(x = \ldots,\ y = \ldots\) Do not accept decimals.

Notes

M1: Use the substitution \(u = \pm\dfrac{1}{x}\) to obtain an integral of the form \(\displaystyle\int \lambda\mathrm{e}^u\,\mathrm{d}u\) oe and integrates to \(\lambda\mathrm{e}^u\)
Award for a sight of \(\pm\lambda\mathrm{e}^{\frac{1}{x}}\) as the answer to their integral.
Alternatively applies the reverse of the chain rule \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(e^{\frac{1}{x}}\right) = \pm\dfrac{1}{x^2}e^{\frac{1}{x}}\) and integrates to reach \(\pm\lambda\mathrm{e}^{\frac{1}{x}}\)
Answers using integration by parts are unlikely to lead to a correct solution.

A1: Correct integration. Award for correct answer with minimal workings as this can be done by inspection using the reverse of the chain rule. However, withhold this mark for incorrect workings.

M1: Uses correct notation to write the integral as the limit as \(t \to 0\), with the limits of \(t\) (or any such variable) and \(-4\). Applies correctly the limits of \(-4\) and \(t\) with correct limit notation, with limits seen substituted the right way round.
Withhold this mark if there is no evidence of using the limiting process. We must see as a minimum \(\displaystyle\lim_{t \to 0}\) oe at some stage in their work.

A1*: Produces an argument that includes an upper limit that approaches 0.
e.g. States that \(t \to 0 \Rightarrow \mathrm{e}^{\frac{1}{t}} \to 0\) and reaches a value of \(\mathrm{e}^{-\frac{1}{4}}\)

Alternative method: changing limits using \(u = \dfrac{1}{x}\)

Notes

B1: Uses \(x = 0\) and \(y = 12\) in the full equation to show that \(A = 144\)

Note Uses \(x = 0\) and \(y = -12\) is B0

Notes

(Corrected from the printed mark scheme: the value at the lower limit is printed as \(-74659.6\), here and in the notes; the correct value is \(-74649.6\).)

B1: Uses the model to set up the volume for the balloon, with limits, the d\(y\) may be implied

M1: Multiplies out the brackets and integrates \(\int x^n\,\mathrm{d}x \rightarrow x^{n+1}\). Alternatively uses integration by parts the correct way. Condone a slip when multiplying out

A1: Correct integration

M1: Uses the limits of \(-12\) and 12, subtracts the correct way round.
If the integration is correct this can be evidenced by for example \((323481.6) - (-74649.6)\) or 2903.56 − (− 670.05) if including \(\dfrac{\pi}{350}\)
If the integration is incorrect we must see the substitution of 12 and \(-12\) into their integrated function
Candidates may use limits of \(-12\) to 0 and then 0 to 12 and add which is fine.

A1: Correct volume 3600 m³, unit required and 2 s.f.

Note: No Evidence of integration maximum B1 M0A0 M0A0 if a correct answer stated

Notes

B1: See scheme for equation

Notes

B1: Correct limitation, see scheme, must be about the balloon part not the basket. Balloon might not be smooth is B0

Notes

M1: Selects the correct form for partial fractions.
Must be of the form \(A + \dfrac{B}{x + 2} + \dfrac{Cx + D}{x^2 + 3}\) so do not award if their \(A\) is not present.
Allow if they set up an expression of the form \(A + \dfrac{\mu x + B}{x + 2} + \dfrac{Cx + D}{x^2 + 3}\) or \(A + \mu x + \dfrac{B}{x + 2} + \dfrac{Cx + D}{x^2 + 3}\) provided they then go on to show their \(\mu = 0\)

dM1: Dependent on having the correct form for the partial fractions.
Complete method for finding the value of at least 3 constants in \(A + \dfrac{B}{x + 2} + \dfrac{Cx + D}{x^2 + 3}\)
Allow slips, provided their intention is clear.

B1: Correct value for \(A\). This is independent of any method and should be awarded regardless of an incorrect use of partial fractions.

A1: Correct solution written as partial fractions, but not for listing values for constants.

Alternative: long division

Scheme	Marks	AO
or states \(2 + \dfrac{\ldots}{2x^3 + 10x^2 + 9x + 22}\) \(\dfrac{6x^2 + 3x + 10}{(x + 2)\left(x^2 + 3\right)} = \dfrac{P}{x + 2} + \dfrac{Qx + R}{x^2 + 3}\)	M1	1.1a
\(6x^2 + 3x + 10 = P\left(x^2 + 3\right) + (Qx + R)(x + 2)\) Correct method to find at least two of the values \(P\), \(Q\) and \(R\) for example \(x = -2 \Rightarrow P = \ldots\{28 = 7P\}\) coeff \(x^2 \Rightarrow Q = \ldots\{6 = P + Q\}\) \(x = 0 \Rightarrow R = \ldots\{10 = 3P + 2R\}\) coeff \(x \Rightarrow R = \ldots\{3 = 2Q + R\}\)	dM1	3.1a
or states \(2 + \dfrac{\ldots}{2x^3 + 10x^2 + 9x + 22}\) or states \(A = 2\)	B1	1.1b
\(2 + \dfrac{4}{x + 2} + \dfrac{2x - 1}{x^2 + 3}\)	A1	1.1b
	(4)

M1: Selects the correct form for partial fractions, (following an attempt at algebraic long division which led to an integer quotient, which may or may not be 2, and their remainder must be a 3TQ)
Their partial fractions must be of the form \(\dfrac{P}{x + 2} + \dfrac{Qx + R}{x^2 + 3}\)

dM1: Complete method for finding the value of at least 2 constants. Dependent on having the correct form for the partial fractions. Allow slips, provided their intention is clear.

B1: Correct constant 2, may be seen in their long division and is likely to appear at the beginning of their workings. This is independent of any method and should be awarded regardless of any incorrect use of partial fractions.

A1: Correct solution written as partial fractions, but not for listing values for constants.

Notes

M1: Rewrites the integral (without the constant term) into an integrable form and integrates to the correct form \(\alpha\ln(x + 2) + \beta\ln\left(x^2 + 3\right) + \lambda\arctan\left(\dfrac{x}{\sqrt{3}}\right)\), where \(\alpha\), \(\beta\) and \(\lambda\) are non-zero constants.

A1: Fully correct integration for the whole expression.

dM1: Uses the limits of 0 and 1, subtracts the correct way round and combines their ln terms correctly.

A1*: Correct answer cso (but can be written in any order)

Notes

B1: Uses the model to obtain a correct value for \(a\) or \(b\)

B1: Uses the model to obtain correct values for \(a\) and \(b\)

Notes

M1: Uses the parametric curve for the model and applies \((\pi)\displaystyle\int x^2\frac{\mathrm{d}y}{\mathrm{d}t}\,\mathrm{d}t\)
The \(\pi\) symbol may be missing here. Also do not be concerned about a missing \(\mathrm{d}t\) at the end of their integral.
Must see an attempt at squaring \(x\) and finding \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) where \(x = (a + 3\sin 2t)\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = k\sin t\) with their \(a\) and \(b\) or the letters \(a\) and \(b\) in their integral.

M1: Expands and makes progress to an integrable form by applying \(\sin 2t = 2\sin t\cos t\) at least once
Integrals in terms of \(a\) and \(b\) will be equivalent to:

Integrals with correct values of \(a\) and \(b\) will be equivalent to:

A1ft: (dependent on both M marks) At least 2 terms integrated correctly; follow through their \(a\) and \(b\), however their \(a\) and \(b\) must now be numerical.

A1: All correct from correct values for \(a\) and \(b\)
Their integral will be equivalent to:

B1: Uses the model to deduce the correct volume of the cylinder; need not be simplified.

ddM1: Dependent on both previous M marks. Fully correct strategy for the volume.
Applies the correct limits (either way round) and adds this to the volume of the cylinder which must have been obtained from \(\pi \times 2^2 \times 4.5\) but condone \(\pi \times 4^2 \times 4.5\)
Both volumes must be positive when combined.

A1: Correct volume, allow awrt 426 (cm³), units are not required but if any are given, they should be correct.

Notes

B1: See scheme. Award for a correct statement. If there is more than one statement ignore any incorrect statements as long as there are no statements that contradict their correct statement.

Comments relating to facts about the shape that are not worthy of marks include:

• the cross section is not a circle
• The vase is not solid
• the cylinder might not be vertical
• it depends on the material used to make the vase
• the vase may not have the same density throughout

Notes

M1: Substitutes (0.4, 4) into the equation modelling the curve in an attempt to find the value of \(k\)

A1: Infers from the data in the model, the value of \(k\)

Notes

B1: Uses the information given in the model to establish the correct volume of the cylinder

M1: Uses the model and applies \(\pi\displaystyle\int x^2\{\mathrm{d}y\}\), d\(y\) not required and \(\pi\) may appear later in their solution. If they find an expression for \(x^2\) first and then substitutes into the formula score M1 even if an incorrect expansion.

A1ft: Correct expression for the volume generated by the curve with the bracket expanded (follow through their \(k\) value), d\(y\) not required and \(\pi\) may appear later in their solution. Indices need to be processed for this mark, may be seen later in the solution.

M1: Attempts to integrate with at least one power raised by 1

A1ft: Correct integration (follow through on their expression for \(x^2\) as long as there are 3 terms). Need not be simplified.

M1: Uses the correct limits and finds the sum of the 2 volumes. Must come from an attempt at \(\pi\displaystyle\int_0^4 x^2\{\mathrm{d}y\}\) and an attempt at the volume of the cylinder, condone incorrect formula used as long as it is 3 dimensional not an area.

A1: \(\dfrac{2462\pi}{875}\)

Use of calculator scores a maximum of B1M1A0M0A0M1A0 volume = \(\boldsymbol{\pi}\)2.7337…

Notes

B1: States an acceptable limitation of the model, which is the curve but accept flaws/bubbles in the glass. Measurements may not be accurate, or anything related to thickness is B0

Notes

B1ft: Compares the actual volume to their answer to part (b) and makes an assessment of the model with a reason. If using a percentage error then they must use 9 as the true volume.

Notes

B1: Suitable explanation stating that one of the bounds/limits is infinite oe isw
e.g. “one of the limits is unbounded” or “the integral is unbounded”

Do not allow this mark if they only say the limit or integral is undefined, unless they go on to say it is undefined at infinity.
Commenting that the function is undefined at infinity is not enough to award this mark on its own, this question concerns the limits.
If the candidates state any extra incorrect comments about the limits withhold this mark e.g. not defined at \(\dfrac{4}{3}\)

Notes

M1: Integrates to obtain \(\alpha\arctan(\beta x)\) where \(\beta \neq 1\)

A1: Correct integration, unsimplified or simplified.

dM1: Applies correct limits, "\(t\)" and \(\dfrac{4}{3}\) with evidence of applying the infinite limit to obtain a non-zero value.
Allow with \(\infty\) used as the limit (which may be implied by \(\dfrac{\pi}{2}\))

A1: Correct value obtained with evidence of use of limiting process on the upper bound. Withhold this mark if there is no evidence of using the limiting process. We must see as a minimum \(\displaystyle\lim_{t \to \infty}\) oe at some stage in their work.

e.g. \(\left[\dfrac{1}{12}\arctan\left(\dfrac{3x}{4}\right)\right]_{\frac{4}{3}}^{\infty} = \dfrac{1}{12}\left(\dfrac{\pi}{2} - \dfrac{\pi}{4}\right) = \dfrac{\pi}{48}\) would score dM1A0

Notes

B1: Selects the correct volume of revolution formula to use, with correct limits in evidence, could appear later in their working

M1: Attempts to integrate with the correct form for the constant and term in \(y^2\)

M1: Applies integration by parts fully on the \(y\cos\left(\frac{5}{2}y\right)\) term in the correct direction. Allow slips in the coefficients, but the form must be correct.

A1: Correct integration of the \(x^2\) equation.

M1: Applies the correct limits to their integral provided there was some attempt at integration. No need for the \(\pi\) for this mark. Condone working in degrees (5.74 …) - (-5.38…), award this mark following correct integration. If their integration is incorrect you will need to check the use of limits in radians only if they do not show the substitution.

A1: Correct volume including units. Accept awrt 2.26 cm³. Allow \(0.719\pi\) cm³. Correct solution only
Note: All previous marks must have been scored to award this final accuracy mark
Special case: Use of calculator for all the integration can score a maximum of B1M0M0A0M1A0

Notes

(corrected from the printed mark scheme: the subtotal for (b) is printed as (1); part (b) carries 2 marks, B1ft B1ft)

B1ft: Attempts to estimate the volume of juice produced by 100 berries - look for their (a) multiplied by 100.

B1ft: Draws a suitable conclusion with reason given, see scheme

Notes

M1: Sets up the correct equation for the model using the information in the question.

dM1: Uses the initial conditions to find the constant of proportionality for flow out.
Condone use of \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = +3\). Depends on the first mark.

A1*: Correct equation shown from correct work proceeding via \(10k = 4\) to find \(k\).

Attempts in (a) using verification score no marks:
E.g. \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 3 - \dfrac{4}{1 + \mathrm{e}^{0.8t}} - \dfrac{2}{5}V \Rightarrow -3 = 3 - \dfrac{4}{2} - 0.4V \Rightarrow V = \dfrac{4}{0.4} = 10\)

Notes

Alternative to part (b):

Notes

M1: Differentiates to achieve the form shown. Allow \(k = 1\)

A1: Correct derivative in any form. Need not be simplified.

Alternative:
M1: Takes tan of both sides and differentiates implicitly and reaches \(\dfrac{1}{1 + \left(\mathrm{e}^{0.4t}\right)^2} \times k\mathrm{e}^{0.4t}\). Allow \(k = 1\).
A1: Correct derivative in any form. Need not be simplified.

Notes

B1: Deduces the correct integrating factor for the equation. May be implied by sight of \(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\mathrm{e}^{0.4t}V\right) = \ldots\) or equivalent work.

M1: Fully multiplies through by their integrating factor and integrates the LHS (look for \(I.F. \times V = \displaystyle\int I.F. \times \left(3 - \frac{4}{1 + \mathrm{e}^{0.8t}}\right)\mathrm{d}t\) though condone missing \(\mathrm{d}t\).

M1: Attempts the integral of the RHS.
Award for \(\displaystyle\int \alpha\mathrm{e}^{0.4t}\,\mathrm{d}t = \beta\mathrm{e}^{0.4t}\ \ \alpha \neq \beta\) or \(\displaystyle\int \frac{\alpha\mathrm{e}^{0.4t}}{1 + \mathrm{e}^{0.8t}}\,\mathrm{d}t = \beta\arctan\mathrm{e}^{0.4t},\ \ \beta \neq 0\)

A1: Correct integration, need not be simplified. Allow if the \(+ c\) is missing for this mark.

M1: Attempts to find their constant – which must have been treated correctly from point of integration.
Note that this is not formally dependent but there must have been an attempt to integrate.

A1: Correct answer. The question says “simplest form” but allow equivalent expressions e.g. \(V = 7.5 - \dfrac{10\arctan\left(\mathrm{e}^{0.4t}\right)}{\mathrm{e}^{0.4t}} + \dfrac{5\pi}{2\mathrm{e}^{0.4t}} + \dfrac{5}{2\mathrm{e}^{0.4t}}\) but do not allow inexact values for the constants.

Notes

B1ft: Evaluates \(V\) where \(V \gt 0\) at \(t = 10\) and makes an appropriate comment.
For the evaluation, allow if a value of \(V\) is obtained even if there is no evidence of substitution provided that it is clear that \(t = 10\) has not been substituted into something that is not \(V\). So you do not need to check their value.
For the tolerance you may need to use your own judgement but a general guide is:

Notes

B1: Sets up the model to find a correct expression for the volume, including limits, d\(y\) may be implied. The limits may be seen later.

M1: Integrates to the form \(\alpha y \pm \beta y^2\)

A1: Correct integration
M1: Substitutes their \(y\) limits the correct way round and subtracts, must be a changed expression
M1: Multiplies their volume by 900 to find the mass
A1: 5700 cao

Note incorrect upper limit of \(\sqrt{2}\) leads to 5200kg Scores B0 M1 A1 M1 M1 A0

Note: Finding the volume around the \(x\)-axis can score B0 M0 A0 M0 M1 A0 only

Note If they use their calculator to find the value of the definite integration and achieve the correct answer the maximum they can score is B1M0A0M1M1A1

Notes

B1: See scheme, must be referring to the model and not the value of the density etc

Notes

B1ft: See scheme, follow through on their answer to (a). If using a calculation, it must be correct. Ignore any contradictory comments e.g. 3.6% out so it’s fairly close so it’s a good model but it’s an overestimate which isn’t good. You may need to use your own judgement but any sensible comment comparing their value to 5500 is acceptable.

Notes

B1: Correct completed square form. Allow \(3^2\) for 9.

Notes

M1: Achieves a correct form for the integration for their \(p\) and \(q\) from part (a):

\(\operatorname{arcosh}\left(\dfrac{x + p}{\sqrt{q}}\right)(+c)\) or \(\ln\left(x + p + \sqrt{(x + p)^2 - q}\right)(+c)\) or e.g. \(\ln\left(\dfrac{x + p}{\sqrt{q}} + \sqrt{\left(\dfrac{x + p}{\sqrt{q}}\right)^2 - 1}\right)(+c)\)
where \(p \neq 0,\ q \neq 1\)

Allow \(\cosh^{-1}\) for arcosh
Allow attempts that use substitution following an attempt to complete the square but must be an appropriate substitution e.g. \(x + p = \sqrt{q}\cosh u\) leading to a correct form as above.

A1: Correct integration. The “\(+ c\)” is not required. Apply isw once a correct expression is seen.
Note that \(\ln\left(\dfrac{x + 2}{3} + \sqrt{\left(\dfrac{x + 2}{3}\right)^2 - 1}\right)(+c)\) is also correct

Notes

B1: Recalls the definition of a mean function accurately. \(\dfrac{1}{13 - 3}\displaystyle\int_3^{13} \frac{1}{\sqrt{x^2 + 4x - 5}}\,\mathrm{d}x\) seen or implied.
Note that the \(\dfrac{1}{13 - 3}\) may appear at the end. \(\dfrac{1}{13 - 3}\displaystyle\int_3^{13} \mathrm{f}(x)\,\mathrm{d}x\) is sufficient as f(\(x\)) is defined in the question. Also allow it to be implied by e.g. \(\dfrac{1}{10}\left[\mathrm{g}(x)\right]_3^{13}\) where g(\(x\)) is their integrated function.

M1: Applies the correct limits the right way round to whatever they think the answer to part (b) is.
This can be awarded if the \(\dfrac{1}{10}\) is present or not.

A1: Correct answer in correct form. Allow equivalents e.g. \(\dfrac{1}{10}\ln\left(\dfrac{5}{3} + \dfrac{2\sqrt{6}}{3}\right)\), \(\dfrac{1}{20}\ln\left(\dfrac{49}{9} + \dfrac{20\sqrt{6}}{9}\right)\)
And allow if the surd is not simplified e.g. \(\dfrac{1}{10}\ln\left(\dfrac{5 + \sqrt{24}}{3}\right)\), \(\dfrac{1}{20}\ln\left(\dfrac{49 + \sqrt{2400}}{9}\right)\)
Apply isw once a correct answer is seen.
The brackets must be present in forms such as \(\dfrac{1}{10}\ln\left(\dfrac{5}{3} + \dfrac{2\sqrt{6}}{3}\right)\), \(\dfrac{1}{20}\ln\left(\dfrac{49}{9} + \dfrac{20\sqrt{6}}{9}\right)\) but not in e.g. \(\dfrac{1}{10}\ln\dfrac{5 + \sqrt{24}}{3}\)
If extra values are offered then score A0

Notes

B1: For a suitable explanation. Technically this should refer to the interval being unbounded, but this is unlikely to be seen. Accept “Because the upper limit is infinity”, but not “because it is infinity” without reference to what “it” is. Do not accept “the upper limit tends to infinity” or “the integral is unbounded”.

Notes

B1: Writes the integral in terms of a limit as \(t \to \infty\) (or other variable) with limits 0 and “\(t\)”, or implies the integral is a limit by subsequent working by correct language.

M1: Integrates \(\cosh x\) correctly either as \(\sinh x\) or in terms of exponentials and applies correctly the limits of 0 and “\(t\)”. The bottom limit zero may be implied. No need for the \(\lim\limits_{t \to \infty}\) for this mark but substitution of \(\infty\) is M0.

A1: cso States that (as \(t \to \infty\)) \(\sinh t \to \infty\) or \(\mathrm{e}^t \to \infty\) and \(\mathrm{e}^{-t} \to 0\) therefore divergent (or not convergent), or equivalent working. Accept \(\sinh t\) is undefined as \(t \to \infty\)

Notes

M1: Divides through by \(\cosh x\) to find an expression involving \(\tanh x\)
Alternative: uses the correct exponential definitions and finds an expression for \(\mathrm{e}^{2x}\) or solves a quadratic in \(\mathrm{e}^{2x}\)

A1: Deduces the correct inequality for \(p\). Note \(|p| \lt 4\) is a correct inequality for \(p\).

Notes

Units not required in this question

B1: For \(a = 4\), ignore any reference to units.

Notes

B1: Correct distance from base for Model A is 4

B1ft: Correct width at widest point. Follow through their ‘\(a\)’, so \(2 \times (\text{‘}a\text{’} + 2)\).

M1: Attempts the derivative for Model B’s equation, reduce any power by 1

A1: Sets \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) and finds correct \(x\) coordinate of the stationary point (accept \(\pm\))

A1: For \(8 - \dfrac{8}{\sqrt{3}}\) or awrt 3.38 cso

dM1: Dependent on previous M mark. Uses their value of \(x\) to find the value of \(y\). If no working shown the value of \(y\) must come from their \(x\) value.
Note using \(x = 4.62\) give \(y = 2.029\ldots\)

A1: Correct diameter, awrt 11.9 follow through their ‘\(a\)’, so \([2a + 3.94\ldots\text{ ft}]\)

Note: Correct answers with no working send to review

Trial and error approach
Candidates could score B1 B1 for model A however if working in integers it is unlikely that they will find the correct value for \(x\) (they are using \(x = -5\)) not a valid method M0A0A0dM0A0

Notes

B1ft: They must have answers for all parts in (b). Accept any well-reasoned comment that follows their answers to (b) If the answers are correct, they must conclude that model B is more appropriate.

• If answers for one model are correct ish but other incorrect, or one value is clearly closer
  For example

  	Distance (3)	Diameter (12)	Distance (3)	Diameter (12)
  A	9.4	9.05	4	6
  B	3.38	12.06	4.62	4.06
  Conclusion	Selects B as distance/diameter closet		Select A as diameter closest

• If distances and diameters are similar selects the model which has the most appropriate value for distance or diameter
  For example

  	Distance (3)	Diameter (12)	Distance (3)	Diameter (12)
  A	0.76	6.8	4	20
  B	1.28	10.5	3.38	19.94
  Conclusion	selects B as the diameter is closet		Selects B as distance is closet

• If all values of the distances and diameters are varied any sensible reason stated for selecting a model.

Notes

B1: Applies \(\pi\displaystyle\int_{-8}^{8} y^2\,\mathrm{d}x\) to the model. Must have \(\pi\) and correct limits, with \(y\) substituted in.
Alternatively attempts to square \(y\) first and then substitute in.

M1: Attempts to expand \(y^2\) this can be a poor attempt but must include at least a constant and \(x^6\) terms as long a clear attempt at \(y^2\) (Limits not required for this mark.)

dM1: Attempts the integration, must first be rearranged to an integrable form then look for power increasing by at least 1 in at least two terms. (Limits not required for this mark.)

M1: Applies correct limits to their integral following an attempt at \(y^2\) with at least a constant and \(x^6\) terms.
If there is no working shown, allow this method mark if the correct answer appears from a calculator as it implies correct limits have been applied the correct way round. (So M0dM0M1 is possible.)

A1: awrt 905 cso note it must come from a fully correct solution

Note: For answers that appear from calculator B1M0dM0M1A0 is possible, the question specifies algebraic integration to be used so the integration needs to be seen to score the other marks.

Notes

B1ft: Compares their volume to 900 or compares their volume + 100 to 1 litre or 1000 and comments appropriately. Correct answer in (d) needs to conclude that it is suitable.

Notes

B1: Correct identity or equivalent rearrangement. This can appear anywhere in the proof.

M1: Attempts the expansion of \(\left(z + \dfrac{1}{z}\right)^6\) must have at least 3 correct terms. Combining the powers when expanding is fine.

A1: Correct expansion with \(z\) terms simplified, need not be rearranged. (So a correct expansion will score M1A1.)

M1: Uses \(z^n + \dfrac{1}{z^n} = 2\cos n\theta\) to write the expression in terms multiple angles of \(\cos 6\theta,\ \cos 4\theta\) and \(\cos 2\theta\). Pairing of terms must be seen.

A1*: Achieves the printed answer with no errors or omissions. Cso

For approaches using De Moivre B0M1A1M0A0 may be scored if the binomial expansions is attempted (and correct for the A).

Notes

B1: See scheme

Notes

Note: The question instructs use of algebraic integration and part (a), so answer only can score at most B1 for implied correct formula.

B1ft: Correct expression for the volume of the paperweight or the solid formed through 360° rotation, stated or implied, ignore limits. No need to expand, but must be applied, not just a formula in \(y\), though allow a correct formula followed by correct integral if the \(\pi\) disappears. Follow through their \(H\)

M1: Uses the result in part part (a) to express the volume in an integrable form and attempts to integrate. Note use of \(\theta\) instead of \(x\) is permissible for this mark. Allow if one term is missing or miscopied.

A1: Correct integration in terms of \(x\). Ignore \(\pi\), their \(H^2\) and the \(\dfrac{1}{32}\). Note if \(\theta\) has been used it is A0 unless a correct substitution method has been implied as the coefficients will be incorrect.

dM1: Dependent on previous method mark and must have reached and integral of the correct form -- in terms of \(x\) with correct arguments allowing for one slip. Finds the required volume using either \(\pi\int_0^4 y^2\,\mathrm{d}x\) or \(\dfrac{1}{2}\pi\int_{-4}^{4} y^2\,\mathrm{d}x\) and applies their limits - accept any value following a valid attempt at the integration as an attempt at applying limits.

A1: cao 24.56

Notes

B1: States an appropriate limitation. See scheme for some examples. The limitation should refer to the paperweight, not to paper. Do not accept “it does not take into account thickness of material” as it is a solid, not a shell, being modelled. Award the mark for a correct reason if two reasons are given and one is incorrect.

Notes

M1: Selects the correct form for partial fractions and multiplies through to form suitable identity or uses a method to find at least one value (e.g. cover up rule).

dM1: Full method for finding values for all three constants. Dependent on first M. Allow slips as long as the intention is clear.

A1: Correct constants or partial fractions.

Notes

M1: Splits the integral into an integrable form and integrates at least two terms to the correct form. They may use a substitution on the arctan term

A1: Fully correct Integration.

dM1: Uses the limits of 0 and 2 (or appropriate for a substitution), subtracts the correct way round and combines the ln terms from separate integrals to a single term with evidence of correct ln laws at least once.

A1: Correct answer

Notes

M1: Uses the substitution \(x = \cosh u\) fully to achieve an integral in terms of \(u\) only. Must have replaced the \(\mathrm{d}x\) but allow if the \(\mathrm{d}u\) is missing.

A1: Correct integral in terms of \(u\). (Allow if the \(\mathrm{d}u\) is missing.)

M1: Uses correct identities \(\cosh^2 u - 1 = \sinh^2 u\) and \(\cosh 2u = 2\cosh^2 u - 1\) to achieve an integrand of the required form

M1: Integrates to achieve the correct form, may be sign errors.

M1: Uses the identities \(\sinh 2u = 2\sinh u\cosh u\) and \(\cosh^2 u - 1 = \sinh^2 u\) to attempt to find \(\sinh 2u\) in terms of \(x\). If using exponentials there must be a full and complete method to attempt the correct form.

A1*: Achieves the printed answer with no errors seen, cso

NB attempts at integration by parts are not likely to make progress – to do so would need to split the integrand as \(x\dfrac{x}{\sqrt{x^2 - 1}}\). If you see any attempts that you feel merit credit, use review.

Notes

M1: Uses integration by parts the correct way around to achieve the required form.

A1: Correct integration by parts

B1ft: Deduces the integral by using the result from part (a). Follow through on their ‘\(uv\)’

dM1: Dependent on previous method mark. Uses the limits \(x = 1\) and \(x = 3\) the correct way around and subtracts

A1*cso: Achieves the printed answer with at least one intermediate step showing the evaluation of the arcosh 3, and no errors seen.

Notes

(a)
B1: Recalls the definition for \(\sinh\left(\dfrac{1}{2}\ln 3\right)\) or forms an equation for \(\operatorname{arsinh} x\)
M1: Uses logarithms to find a value for \(\alpha\) or forms and solves a correct equation without log
A1: Deduces the correct exact value for \(\alpha\)

Note using the result
\(\ln\left(\dfrac{1}{\sqrt{3}} + \sqrt{\left(\dfrac{1}{\sqrt{3}}\right)^2 + 1}\right) = \ln\left(\dfrac{1}{\sqrt{3}} + \sqrt{\dfrac{4}{3}}\right) = \ln\sqrt{3} = \dfrac{1}{2}\ln 3\) therefore \(\operatorname{arsinh}\left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{1}{2}\ln 3\)
B1 for substituting in \(\alpha\) into \(\operatorname{arsinh} x\), M1 for rearranging to show \(\dfrac{1}{2}\ln 3\), A1 for conclusion

Notes

(b)
B1: Correct expression for the volume \(\pi\displaystyle\int_0^{\frac{1}{2}\ln 3}\sinh^2 y\,\mathrm{d}y\) requires integration signs, \(\mathrm{d}y\) and correct limits.
M1: Uses the exponential formula for \(\sinh y\) or the identity \(\cosh 2y = \pm 1 \pm 2\sinh^2 y\) to write in a form which can be integrated at least one term
dM1: Dependent of previous method mark, integrates.
A1: Correct integration.
M1: Correct use of the limits \(y = 0\) and \(y = \dfrac{1}{2}\ln 3\)
A1: Correct exact volume.

Notes

B1: Correct integration.

M1: Attempt to integrate to a form \(\lambda\mathrm{e}^{-\frac{1}{2}x}\) where \(\lambda \neq 2\), and applies correct limits with some consideration of the infinite limit given (e.g. with the limit statement). Only allow with \(\infty\) used as the limit if subsequent work shows the term is zero.

A1: Correct value

Notes

B1: Recalls the correct formula for finding the mean value of a function. You may see the division by “24” only at the end. No integration is necessary, just a correct statement with an integral.

M1: Integrates to a form \(\alpha t + \beta\cos\left(\dfrac{\pi}{12}t\right) + \delta\sin\left(\dfrac{\pi}{6}t\right)\) and uses the limits of 0 and 24 (the correct way around). If no explicit substitution is seen, accept any value following the integral as an attempt. Answers from a calculator with no correct integral seen score M0 as the question requires calculus to be used.

A1*cso: Achieves 8 with no errors seen following a full attempt at the substitution. Must have seen some evidence of the limits used, minimum required for substitution is \(\left[\left(8(24) + \dfrac{60}{\pi}\right) - \left(\dfrac{60}{\pi}\right)\right]\).

Notes

B1: Accept any reasonable adaptation to the equation that will increase the mean value. E.g. as in scheme, or introduce another positive term, or decrease the constant 5 etc. It must be clear which constant they are referring to in their reason, not just “increase the constant”.

Notes

M1: Deduces the required series by using the Maclaurin series for \(\cos x\), replacing \(x\) with \(\dfrac{x}{3}\) and squares, or first applying the double angle identity (allow sign error) and then applying the series for \(\cos x\) with \(\dfrac{2x}{3}\). Attempts at finding from differentiation score M0 as the cosine series is required.

A1: Correct series

Notes

M1: Divides their series in part (a) by \(x\) and integrates to the form \(A\ln x + Bx^2 + Cx^4\)

A1ft: Correct integration, follow through on their coefficients and need not be simplified.

A1: Deduces the definite integral awrt 0.98295

Notes

B1: Correct value.

Notes

B1: Makes a quantitative statement about the accuracy, so e.g. how many decimal places or significant figures it is correct to, or calculates a percentage accuracy to deduce it is reasonable. Do not accept just “underestimate” or similar without quantitative evidence. Allow for a reasonable comment as long as (b) is correct to at least 2 s.f. but (c) must be the correct value.

Notes

(a)
M1: Uses the given coordinates correctly in the equation modelling the curve to obtain at least one correct equation and attempts to find the values of \(a\) and \(b\)
A1: Correct values

Notes

(b)
B1ft: Uses the model to obtain \(\displaystyle\pi\int \left(\frac{\text{their } a}{y + \text{their } b}\right)^2\mathrm{d}y\). Note the \(\pi\) can be recovered if appears later.
M1: Chooses limits appropriate to the model i.e. 0.5 and 2.5
M1: Integrates to obtain an expression of the form \(k(y + \text{“}1.5\text{”})^{-1}\)
B1: Deduces the correct equation for the circle
M1: Uses their circle equation and \(\displaystyle\pi\int x^2\,\mathrm{d}y\) to attempt the top volume. Note the \(\pi\) can be recovered if appears later.
M1: Identifies limits appropriate to the model i.e. 2.5 and 3 + their radius
A1: Correct integration
dM1: Uses the model to find the volume of the chess piece including the cylindrical base (dependent on all previous method marks)
A1: Correct volume

Notes

(a)
M1: Makes progress in establishing the derivative by taking the tan of both sides and differentiating with respect to \(y\) or implicitly with respect to \(x\)
M1: Use of the correct identity
A1*: Fully correct proof

Notes

(b)
B1: Correct derivative
M1: Uses integration by parts in the correct direction
A1: Correct expression
M1: Adopts a correct strategy for the integration by splitting into two fractions or using a substitution of \(4x = \tan u\) to get to an integrable form
A1: Correct answer

(corrected from the printed mark scheme: \(\displaystyle\frac{1}{32}\int \left(\sec^2 u - 1\right)\mathrm{d}u\) is printed as \(\displaystyle\frac{1}{32}\int \sec^2 u - u\,\mathrm{d}u\))

Notes

(c)
M1: Correctly applies the method for the mean value for their integration. The limit of zero can be implied if it comes to 0.
A1: Correct exact answer. Allow exact equivalents e.g. \(\dfrac{4\pi\sqrt{3} - 9}{72},\ \dfrac{\pi\sqrt{3}}{18} - \dfrac{1}{8}\)

Notes

B1: Correct equation of the circle, may be implied by correct integral

B1: Correct expression for the volume, including limits, d\(x\) may be implied and if using limits \(r\) and 0 the 2 could appear later with reasoning

M1: Integrates to the form \(\alpha x \pm \beta x^3\). Do not award if \(r^2 \to \lambda r^3\)

dM1: Dependent on previous method mark. Correct use of limits \(-r\) and \(r\) or limits of 0 and \(r\) with twice the volume.

A1*: \(V = \dfrac{4}{3}\pi r^3\) * cso

Note: rotation about the \(y\)-axis all marks are available, however for the final accuracy mark must refer to symmetry

Notes

(a)
B1: For a suitable explanation with no contrary reasoning. Technically this should refer to the interval being unbounded, but this is unlikely to be seen. Accept “Because the upper limit is infinity”. Do not award if there are erroneous statements e.g. referring to as \(x = 0\) the integrand is not defined. Do not accept “because one of the limits is undefined” unless they state they mean \(\infty\). Do not accept “it is undefined when \(x = \infty\)” without reference to “it” being the upper limit.

Notes

(b)
M1: Selects the correct form for partial fractions and proceeds to find values for \(A\) and \(B\)
A1: Correct constants or partial fractions
A1ft: \(\displaystyle\int \frac{p}{x} + \frac{q}{2x + 5}\,\mathrm{d}x = p\ln x + \frac{q}{2}\ln(2x + 5)\) Note that \(\dfrac{1}{5}\ln 5x - \dfrac{1}{5}\ln(10x + 25)\) is correct.
M1: Combines logs correctly. May see \(-\dfrac{1}{5}\ln\left(\dfrac{2x + 5}{x}\right) = -\dfrac{1}{5}\ln\left(2 + \dfrac{5}{x}\right)\)
B1: Correct upper limit for \(x \to \infty\) by recognising the dominant terms. (Simply replacing \(x\) with \(\infty\) scores B0)
A1: Deduces the correct value for the improper integral in the correct form

(b) Way 2

Note the method marks as MAMABA, and should be entered in this order on ePEN.
M1: Expands the denominator and completes the square.
A1: Correct expression
M1: For \(\dfrac{1}{(x + p)^2 - a^2} \to k\ln\left|\dfrac{x + p - a}{x + p + a}\right|\)
A1ft: \(\dfrac{1}{2}\dfrac{1}{(x + a)^2 - a^2} \to \dfrac{1}{2a}\ln\left|\dfrac{x}{x + 2a}\right|\) with their \(a\) (may be simplified as in scheme).
B1: Correct upper limit for \(x \to \infty\) by recognising the dominant terms. (Simply replacing \(x\) with \(\infty\) scores B0) Note in this method the upper limit evaluates to zero.
A1: Deduces the correct value for the improper integral in the correct form. Accept \(-\dfrac{1}{5}\ln\dfrac{2}{7}\)

Notes

M1: A correct overall strategy, either finding full volume rotated by \(2\pi\) first, then performing some kind of scaling, or using \(\alpha\displaystyle\int y^2\,\mathrm{d}x\) for a variable \(\alpha\) (ideally \(\dfrac{\theta}{2}\), but for the strategy accept with any variable multiple), to form an equation in just the angle.

M1: Attempting to square \(y\) to a three or four term expression. Look for correct powers on first and last term with some term(s) in the middle.

A1: Correct expansion in three or four terms – award when first seen.

M1: Integrates \(y^2\) w.r.t. \(x\). Must have at least two terms in their \(y^2\) with fractional indices. Power to be increased by 1 in at least two terms.

A1ft: Two terms of integral correct. Follow through on their expansion. Need not be simplified.

A1: Fully correct integral. Need not be simplified. May still be four terms

M1: Either: Substitutes limits and subtracts correct way round (must be seen or implied by the answer), and equates to \(\dfrac{461}{2}\) if using \(\frac{1}{2}\theta\displaystyle\int y^2\,\mathrm{d}x\) and proceeds to find \(\theta\).
Or: Substitutes limits and subtracts correct way round (seen or implied) and multiplies by \(\pi\) to get the full volume AND then multiplies the result by \(\dfrac{\theta}{2\pi}\) before equating to \(\dfrac{461}{2}\).
The method must be correct for this mark – so they must be using \(\dfrac{\theta}{2}\displaystyle\int y^2\,\mathrm{d}x\) directly or \(\pi\displaystyle\int y^2\,\mathrm{d}x\) and scale by \(\dfrac{\theta}{2\pi}\) when setting equal to \(\dfrac{461}{2}\)

A1: Correct angle found. Accept \(\dfrac{40}{9}\), awrt 4.44 or awrt 255° (as long as the degrees units are made clear – do not accept just 255) isw once a correct value of \(\theta\) is found.

Special case The question specified that algebraic integration must be used, so use of a calculator to find the integral cannot score the marks for integration but may be allowed the strategy and answer marks. A maximum of M1M0A0M0A0A0M1A1 is available in such cases.
Expanding \(y^2\) first but showing no integration can score the second M and first A (if earned) as well.

Note that \(\displaystyle\int_{1/8}^{8}\left(2x^{\frac{1}{3}} + x^{-\frac{2}{3}}\right)^2\mathrm{d}x = \frac{4149}{40} = 103.725\) but just this alone is worth no marks. There must be an attempt to incorporate this within a strategy to gain access to marks.

Notes

B1: Uses the model to obtain a correct value for \(k\). Must be 2.6 not -2.6

Notes

M1: Substitutes their value of \(k\) and \(x = 1.18\) into the given model to find a value for \(y\)

A1: Infers that the depth of the pool could be awrt 0.5 m

Notes

B1ft: Uses the model to obtain \(x\) correctly in terms of \(y\) (follow through their \(k\))

M1: Uses the model to obtain an expression for the volume of the pool using \(\displaystyle\pi\int (\mathit{their}\ f(y))^2\,\mathrm{d}y\) – must expand in order to reach an integrable form (allow poor squaring e.g. \((a + b)^2 = a^2 + b^2\). Note that the \(\pi\) may be recovered later.

A1: Correct integration

M1: Selects limits appropriate to the model (\(h\) and 0) substitutes and clearly shows the use of both limits (i.e. including zero)

A1: Correct expression (allow unsimplified and isw if necessary)

Notes

Way 1

M1: Recognises that \(\dfrac{\mathrm{d}V}{\mathrm{d}h}\) is required and attempts to find \(\dfrac{\mathrm{d}V}{\mathrm{d}h}\) or \(\dfrac{\mathrm{d}h}{\mathrm{d}V}\) from their integration or using the earlier result (before integrating). Must clearly be identified as \(\dfrac{\mathrm{d}V}{\mathrm{d}h}\) or \(\dfrac{\mathrm{d}h}{\mathrm{d}V}\) unless this implied by subsequent work.

M1: Evidence of the correct use of the chain rule (ignore any confusion with units). Look for an attempt to divide 15 or their converted 15 by their \(\dfrac{\mathrm{d}V}{\mathrm{d}h}\) or to multiply 15 or their converted 15 by \(\dfrac{\mathrm{d}h}{\mathrm{d}V}\) but must reach a value for \(\dfrac{\mathrm{d}h}{\mathrm{d}t}\) but you do not need to check their value.

A1: Interprets their solution correctly to obtain the correct answer (awrt 25.4) with the correct units

Alternative: Way 2

M1: Uses \(y = 0.2\) to find \(x\) and the surface area of the water at that instant

M1: Attempts to divide the rate by their area (ignore any confusion with units)

A1: Interprets their solution correctly to obtain the correct answer (awrt 25.4) with the correct units

Notes

B1: Selects an appropriate substitution leading to an integrable form

M1: Demonstrates a fully correct method for the substitution that includes substituting into the function and dealing with the “\(\mathrm{d}x\)”. The substitution being substituted does not need to be “correct” for this mark but the substitution must be an attempt at \(\displaystyle\int \frac{1}{\sqrt{4\left[\mathrm{f}(u)\right]^2 + 9}} \times \mathrm{f}^{\prime}(u)\,\mathrm{d}u\) with the \(\mathrm{f}^{\prime}(u)\) correct for their substitution. E.g. if \(x = \dfrac{1}{2}u\) is used, must see \(\mathrm{d}x = \dfrac{1}{2}\mathrm{d}u\) not \(2\mathrm{d}u\).

A1: Correct simplified integral in terms of \(u\) from correct work and from a correct substitution

A1: Correct answer including “\(+ c\)”. Allow arcsinh or arsinh for \(\sinh^{-1}\) from correct work and from a correct substitution

Alternative: Way 2

(Corrected from the printed mark scheme: the last line is printed as “\(u = \dfrac{1}{2}\sinh^{-1}\left(\dfrac{2x}{3}\right) + c\)”; it is the integral, not \(u\), that equals this.)

Alternative: Way 3

Notes

M1: Correctly applies the method for the mean value for their integration which must be of the form specified in part (a) and substitutes the limits 0 and 3 but condone omission of 0

A1: Correct exact answer (follow through their \(A\) and \(B\)). Brackets are required if appropriate.

Notes

M1: Selects the correct form for partial fractions.

dM1: Full method for finding values for all three constants. Dependent on having the correct form for the partial fractions. Allow slips as long as the intention is clear.

A1: Correct constants or partial fractions.

A1ft: Integrates \(\displaystyle\int \frac{px}{2x^2 + 3} - \frac{q}{x + 1}\,\mathrm{d}x = \frac{p}{4}\ln\left(2x^2 + 3\right) - q\ln(x + 1)\) and no extra terms

M1: Combines two algebraic log terms correctly.

B1: Correct upper limit for \(x \to \infty\) by recognising the dominant terms. (Simply replacing \(x\) with \(\infty\) scores B0). This can be implied.

A1: Deduces the correct value for the improper integral in the correct form, cao A0 for \(2\ln\dfrac{2}{3}\)

Correct answer with no working seen is no marks.

Note: Incorrect partial fraction form, \(\dfrac{A}{2x^2 + 3} + \dfrac{B}{x + 1}\) or \(\dfrac{Ax}{2x^2 + 3} + \dfrac{B}{x + 1}\) the maximum it can score is M0M0A0A0M1B1A0

Notes

M1: Chooses (4, 14) and (1, 18) and substitutes into the equation modelling the curve to obtain at least one correct equation and attempts to find the values of \(a\) and \(b\).

A1: Infers from the data in the model, the values of \(a\) and \(b\)

Notes

B1: Correct expressions for the 2 cylindrical parts. May be seen as a sum or as separate cylinders.

B1ft: Uses the model to obtain \(\pi\displaystyle\int\left(\frac{y - \text{their } b}{\text{their } a}\right)\mathrm{d}y\) (Note that the \(\pi\) may be recovered later)

M1: Chooses limits appropriate to the model i.e. 14 and 18

M1: Integrates to obtain an expression of the form \(\alpha y + \beta y^2\)

A1: Uses their model correctly to give \(274y - \dfrac{15y^2}{2}\)

ddM1: Uses the model to find the sum of their cylinders + their integrated volume. Must be a fully correct method here and is dependent on both previous method marks. So must have attempted the volumes of the cylinders “AHBG” and “CFED” and adds these to the magnitude of their integrated volume.

A1: \(268\pi\) or awrt 842

Notes

B1: States an acceptable limitation of the model with no contradictory statements. (This is independent of part (b))

Notes

B1ft: Compares the actual volume to their answer to part (b) and makes an assessment of the model with a reason with no contradictory statements.

Typical solution

Let \(s = \sinh x\)

Solution of inequality:

Typical solution

Typical solution

Area of \(R = A_1 + A_2\), where

Area of \(R = \frac{9}{2} + \ln 2 + 15\ln\frac{4}{3}\)

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

\(0 \leqslant x \leqslant 1\)

Typical solution

Typical solution

Typical solution

Let \(x + 3 = \cosh\theta\)

Then

Let \(I = \displaystyle\int_{-2}^{1} \sqrt{x^2 + 6x + 8}\,\mathrm{d}x\)

When \(x = -2\), \(\cosh\theta = 1\) and \(\theta = 0\)

When \(x = 1\), \(\cosh\theta = 4\) and \(\theta = \cosh^{-1}(4)\)

When \(\cosh\theta = 4\), \(\sinh\theta = \sqrt{4^2 - 1} = \sqrt{15}\)

Typical solution

Typical solution

(i)

(ii)

Typical solution

Arc length

Typical solution

Typical solution

Typical solution

The mean of f

Typical solution

\(\dfrac{4}{3}\)

Typical solution

Typical solution

or

let \(u = \ln t\)

Typical solution

Typical solution

\(\mathrm{f}\) is maximum when the denominator is minimum, that is when \(x = -2\)

So the graph of \(y = \mathrm{f}(x)\) has a stationary point at \(\left(-2, \dfrac{1}{3}\right)\)

Typical solution

Typical solution

Typical solution

\(\dfrac{\pi}{2}\) and \(\dfrac{3\pi}{2}\)

Typical solution

Typical solution

By using Simpson’s rule with 5 ordinates

Typical solution

Typical solution

Typical solution

When a strip is rotated it forms a disc which is approximately cylindrical.

If thickness of disc is \(\delta x\) then volume of the disc is (approximately)

Total volume of discs is

Volume of solid

Typical solution

Compare \(x^2\)-coeff: \(B = -8\)

Typical solution

Typical solution

Typical solution

The area of the trapezium is greater than the integral.

Area of trapezium \(= (b - a)y_{\frac{1}{2}}\)

The area of the trapezium is the same as the area of the rectangle from use of the mid-ordinate rule.

This is equal to Sharon’s estimate.

The trapezium includes the area represented by the integral, so Sharon’s estimate is an over-estimate.

Typical solution

Typical solution

At P, \(y^2 - x^2 = 16\) and \(x^2 + y^2 = 32\)

Let \(x = 4\sinh u\)

Then \((x^2 + 16)^{\frac{1}{2}} = 4\cosh u\)

and \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = 4\cosh u\)

When \(x = 2\sqrt{2}\),

\(\sinh u = \dfrac{\sqrt{2}}{2}\) and \(\cosh u = \dfrac{\sqrt{6}}{2}\)

OP makes an angle of \(\dfrac{\pi}{3}\) with the \(x\)-axis

as required

Typical solution

Typical solution

The required integral is \(S_4\) with \(a = 1\)

Typical solution

As \(\lim_{h \to 0}\{h\ln(h)\} = 0\)

Then \(\displaystyle\int_0^8 \ln(x)\,\mathrm{d}x = 8\ln(8) - 8\)

Typical solution

Typical solution

\(17\)

Typical solution

Typical solution

Typical solution

Typical solution

IF: \(\mathrm{e}^{\int 1\,\mathrm{d}t} = \mathrm{e}^t\)

When \(t = 0\), \(v = 0\) \(\therefore c = 1\)

Speed \(= \mathrm{e}^{2\pi} - \mathrm{e}^{-2\pi}\)

Typical solution

Typical solution

Integrating factor \(= \mathrm{e}^{\int P\,\mathrm{d}x} = x^2\)

or

Notes

(corrected from the printed mark scheme: the final A1 guidance prints \(xy = \ldots\); the left-hand side is \(x^2y\), as in the typical solution)

Typical solution

Typical solution

Typical solution

Typical solution

The upper limit is infinity, so it is an improper integral.

Typical solution

Typical solution

Integrating factor

To find

Typical solution

Typical solution

Typical solution

Typical solution

For equal roots \(\Delta = 0\)

Typical solution

Notes

[This can also be done by translating the curve by \(\begin{pmatrix} -b \\ 0 \end{pmatrix}\) and finding \(\pi\int_0^b 4ax\,\mathrm{d}x\), but this must be clearly explained to gain full marks.]

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution