Notes

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

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

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

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

Alternative part (a) (Further Maths)

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

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

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

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

Notes

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

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

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

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

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

Notes

There may be other valid ways to differentiate or to solve the resulting equation.

M1: Attempts the quotient rule to achieve \(\dfrac{\pm\lambda x\left(1 + x^2\right)^2 \pm \mu x\left(1 - x^2\right)\left(1 + x^2\right)}{\left(1 + x^2\right)^4}\)
but condone \(\dfrac{\pm\lambda x\left(1 + x^2\right)^2 \pm \mu x\left(1 - x^2\right)\left(1 + x^2\right)}{\left(1 + x^2\right)^2}\) provided an incorrect quotient rule is not seen.
Alternatively, attempts the product rule on \(\mathrm{f}(x) = \left(1 - x^2\right)\left(1 + x^2\right)^{-2}\) to achieve \(\pm\lambda x\left(1 + x^2\right)^{-2} \pm \mu x\left(1 - x^2\right)\left(1 + x^2\right)^{-3}\)
There may be attempts at splitting the numerator to \(\mathrm{f}(x) = \left(1 + x^2\right)^{-2} - x^2\left(1 + x^2\right)^{-2}\) which should differentiate to \(\pm\lambda x\left(1 + x^2\right)^{-3} \pm \mu x\left(1 + x^2\right)^{-2} \pm \gamma x^3\left(1 + x^2\right)^{-3}\)
In all cases, do not penalise signs of \(\lambda\), \(\mu\) and/or \(\gamma\) unless an incorrect quotient rule or an incorrect product rule is stated.
Any occurrences of \(\left(1 + x^2\right)^2\) may be replaced with \(1 + 2x^2 + x^4\)
There is no need for a LHS e.g. \(\mathrm{f}^{\prime}(x) =\) to be present so ignore an incorrect LHS.
Invisible brackets may be recovered/implied by later work.

A1: Correct differentiation. May be unsimplified. Ignore absence of (or incorrect) LHS.
The correct derivative using the quotient rule is \(= \dfrac{-2x\left(1 + x^2\right)^2 - 4x\left(1 - x^2\right)\left(1 + x^2\right)}{\left(1 + x^2\right)^4}\) o.e.
using the product rule is \(-2x\left(1 + x^2\right)^{-2} - 4x\left(1 - x^2\right)\left(1 + x^2\right)^{-3}\) o.e.
and splitting the numerator is \(-4x\left(1 + x^2\right)^{-3} - 2x\left(1 + x^2\right)^{-2} + 4x^3\left(1 + x^2\right)^{-3}\) o.e.
You may need to check carefully for equivalent derivatives and ISW after a correct expression is seen. Note that some candidates may set = 0 at the start and may omit the denominator as a result – send to review if you are unsure in such cases.

dM1: Attempts to reduce the expression to a suitable form so that the roots can be found, either by cancelling a factor of \(\left(1 + x^2\right)\) and simplifying the other brackets
e.g. \(\dfrac{-2x\left(1 + x^2\right) - 4x\left(1 - x^2\right)}{\left(1 + x^2\right)^3} \rightarrow\) e.g. \(\dfrac{-6x + 2x^3}{\left(1 + x^2\right)^3}\) or \(\dfrac{2x\left(x^2 - 3\right)}{\left(1 + x^2\right)^3}\) or \(2x\left(x^2 - 3\right)\ (= 0)\)
or by attempting to take \(x\), \(\left(1 + x^2\right)\) or \(\pm 2x\left(1 + x^2\right)\) out as a factor and simplify the other brackets. If they only take out a factor of \(x\) then they must have \(x\left(\pm Ax^4 \pm Bx^2 \pm C\right)\)
e.g. \(\dfrac{-2x\left(1 + x^2\right)\left[\left(1 + x^2\right) + 2\left(1 - x^2\right)\right]}{\left(1 + x^2\right)^4} \rightarrow\) e.g. \(\dfrac{-2x\left(1 + x^2\right)\left(3 - x^2\right)}{\left(1 + x^2\right)^4}\) or \(\dfrac{x\left(2x^4 - 4x^2 - 6\right)}{\left(1 + x^2\right)^4}\)
Depends on the first M mark.
Note: they might attempt to expand the brackets first, or set = 0 and multiply through by \(\left(1 + x^2\right)^k\) and/or divide through by \(x\) or \(2x\) at any stage.
Do not be concerned if any factors of \(\left(1 + x^2\right)\), \(x\) or 2 go “missing” during their work.
The = 0 does not explicitly need to be seen.
Look for \(\pm\dfrac{2x\left(1 + x^2\right)\left(Ax^2 + B\right)}{\left(1 + x^2\right)^k}\) or \(\pm\dfrac{\left(1 + x^2\right)\left(Ax^3 + Bx\right)}{\left(1 + x^2\right)^k}\) or \(\pm\dfrac{x\left(Ax^4 + Bx^2 + C\right)}{\left(1 + x^2\right)^k}\) where

• \(k\) may be 0, 1, 2, 3 or 4
• \(A\) and \(B\) (and \(C\) if present) are non-zero
• any of \(2x\) or \(\left(1 + x^2\right)\) in the numerator and/or \(\left(1 + x^2\right)^k\) in the denominator may be absent

ddM1: Attempts to solve their numerator set = 0 using a valid non-calculator method and uses a solution for \(x\) to find a corresponding value for \(y\).
The substitution may be implied by their value of \(y\) but, if not, the substitution must be seen.

• For either \(Ax^2 + B\) or \(Ax^3 + Bx\), we require \(A \neq B\) and \(A \times B \lt 0\) From either expression they can write down their \(x = \pm\sqrt{\dfrac{-B}{A}}\) without working.
• For \(Ax^4 + Bx^2 + C\) they must show a valid non-calculator method for solving the quartic, treating it as a quadratic in \(x^2\) and using the usual rules for solving a quadratic algebraically, e.g. \(a = x^2 \rightarrow 2a^2 - 4a - 6\ (= 0) \rightarrow (a + 1)(2a - 6)\ (= 0) \rightarrow x = \pm\sqrt{3}\). They must reach a value for \(x\) (and not just \(x^2\)) as well as finding a value for \(y\).

Dependent on both previous method marks. They must use a value for \(x\) that is not \(-1\), 0 or 1.

A1: Deduces the correct exact coordinates for \(P\) \(\left(-\sqrt{3}, -\dfrac{1}{8}\right)\) o.e. \(\left(-\sqrt{3}, -0.125\right)\)
Requires all the previous marks to have been scored.
Condone \(x = -\sqrt{3}\), \(y = -0.125\) or e.g. \(-\sqrt{3}, -\dfrac{1}{8}\)
If there is more than one pair of coordinates given, then the correct coordinates must be clearly selected or any others clearly rejected.
Ignore any mistakes that occur if they multiply out the denominator \(\left(1 + x^2\right)^4\) after differentiation.

Notes

Mark parts (b) and (c) together.

B1: Deduces the correct limits for the integral in \(\theta\) which may be seen separately as side working or within their integral work. This mark cannot be scored working in degrees.
Allow \(\dfrac{3\pi}{4}\) instead of \(-\dfrac{\pi}{4}\) but not decimal approximations.

B1: \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = \sec^2\theta\) or equivalent e.g. \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = 1 + x^2\) or \(\dfrac{\cos^2\theta - -\sin^2\theta}{\cos^2\theta}\) (coming from \(x = \dfrac{\sin\theta}{\cos\theta}\)) but must be correct, so not e.g. \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = \sec^2 x\) unless recovered.

M1: Makes a complete attempt at using the substitution \(x = \tan\theta\)
Requires:

• An attempt at using \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta}\) to replace \(\mathrm{d}x\) with \(\mathrm{d}\theta\) either way round, so if \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = \mathrm{g}(\theta)\) allow either \(\mathrm{d}x = \mathrm{g}(\theta)\{\mathrm{d}\theta\}\) or \(\mathrm{d}x = \dfrac{1}{\mathrm{g}(\theta)}\{\mathrm{d}\theta\}\). \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta}\) must be a function of \(\theta\) and not a constant.
• All terms in \(x\) replaced with \(\tan\theta\) (or \(1 + x^2\) with \(\sec^2\theta\))

Condone the absence of \(\mathrm{d}\theta\) but \(\mathrm{d}x\) must no longer be present.
Condone if they fail to square the denominator or e.g. a slip in missing a \(\theta\)
Use of notation such as \(\mathrm{d}(\tan\theta)\) is correct but does not score the M1 until replaced with \(\sec^2\theta\{\mathrm{d}\theta\}\) or their derivative of \(\tan\theta\) (which must not be a constant multiple of \(\tan\theta\)).

dM1: Uses trigonometric identities e.g. \(\pm 1 \pm \tan^2\theta = \pm\sec^2\theta\), \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\), \(\sec\theta = \dfrac{1}{\cos\theta}\), \(\pm\cos^2\theta \pm \sin^2\theta = \pm 1\) to simplify the integral to an expression of the form \(\pm a\sin^2\theta \pm b\cos^2\theta \pm c\) where one of \(a\), \(b\) or \(c\) may be 0.
The algebra should essentially be correct but condone e.g. sign slips or errors collecting terms.
Alternatively, e.g. \(\displaystyle\int \dfrac{1 - \tan^2\theta}{1 + \tan^2\theta}\,\mathrm{d}\theta = \int \dfrac{\cos^2\theta - \sin^2\theta}{\cos^2\theta + \sin^2\theta}\,\mathrm{d}\theta = \int \dfrac{\cos^2\theta - \sin^2\theta}{1}\,\mathrm{d}\theta\)
They must have replaced \(\mathrm{d}x\) with \(\mathrm{d}\theta\) the correct way round for this mark, so if \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = \mathrm{g}(\theta)\) then they must have \(\mathrm{d}x = \mathrm{g}(\theta)\{\mathrm{d}\theta\}\) and not \(\mathrm{d}x = \dfrac{1}{\mathrm{g}(\theta)}\{\mathrm{d}\theta\}\)
It is acceptable to replace e.g. \(\tan^2\theta\cos^2\theta\) with \(\sin^2\theta\)
Dependent on the previous method mark.

A1*: cso Arrives at \(\displaystyle\int \cos 2\theta\,\mathrm{d}\theta\) with sufficient working shown and no incorrect work ignoring limits. The final line must be fully correct, including the integral sign and \(\mathrm{d}\theta\) (ignoring limits). Note that this may be seen in part (c) and may score the mark.
All trigonometric identities must be fully correct.
Condone one or two slips in a missing \(\theta\) but not frequent omissions.
Condone missing integral signs in their intermediate work, but it must be present on the final line.
\(\mathrm{d}x\) must be replaced with \(\mathrm{d}\theta\) at some stage before the final line but does not need to be present in every intermediate line of working.
Condone notational errors throughout e.g. \(\sin\theta^2\) provided they are recovered.
This mark is independent from any work to do with limits, i.e., B0B1M1dM1A1* is possible.

Notes

Mark parts (b) and (c) together.

M1: \(\displaystyle\int \cos 2\theta\,\mathrm{d}\theta \rightarrow \pm\dfrac{1}{2}\sin 2\theta\ \{+c\}\)

dM1: Substitutes their changed limits (not 1 and \(-1\)) into \(\pm\dfrac{1}{2}\sin 2\theta\) and subtracts either way round.
May be implied e.g. \(\dfrac{1}{2} + \dfrac{1}{2}\) provided they have the correct limits.
Dependent on the previous method mark.

A1: Area of 1 found following correct work. This mark requires clear substitution of the correct limits into \(\dfrac{1}{2}\sin 2\theta\) the correct way round. If they have the limits the wrong way round and achieve an answer of \(-1\) they cannot just make the answer positive for this mark.
They may use limits from 0 to \(\dfrac{\pi}{4}\) and multiply their result by 2.
\(\dfrac{3\pi}{4}\) may be used in place of \(-\dfrac{\pi}{4}\) as the lower limit which is acceptable.
Condone any spurious integral symbol accompanying \(\dfrac{1}{2}\sin 2\theta\)
The substitution may be implied by e.g. \(\dfrac{1}{2} - -\dfrac{1}{2}\) or \(\dfrac{1}{2}\sin\left(\dfrac{\pi}{2}\right) - \dfrac{1}{2}\sin\left(-\dfrac{\pi}{2}\right)\) but must follow sight of \(\dfrac{1}{2}\sin 2\theta\)
Condone use of limits \(-45\) and 45 (degrees) the correct way round for full marks in part (c).
Note that use of a calculator on the original integral will give the correct answer. An answer of 1 scores no marks without evidence of scoring both method marks above including the substitution.

Notes

13 Alt I

M1: Obtains \(\alpha x(2x+1)^{-2} \pm \beta\displaystyle\int (2x+1)^{-2}\,(\mathrm{d}x)\) o.e. where \(\alpha, \beta \neq 0\) but may be equal to each other (you do not need to be concerned about how they arrive at this)

dM1: Uses a correct method to integrate an expression of the form \(\pm\beta\displaystyle\int (2x+1)^{-2}\,(\mathrm{d}x) \rightarrow \pm\gamma(2x+1)^{-1},\ \ \beta, \gamma \neq 0\)
It is dependent on the previous method mark.

A1: \(-\dfrac{x}{4(2x+1)^2} - \dfrac{1}{8(2x+1)}\) o.e. Allow this to be unsimplified
Watch out for the DI method

(arrows in the table go diagonally from \(x\) to \(-\dfrac{(2x+1)^{-2}}{4}\) and from 1 to \(\dfrac{(2x+1)^{-1}}{8}\))

Giving correct integration e.g. \(\displaystyle\int \dfrac{x}{(2x+1)^3}\,\mathrm{d}x = -\dfrac{x(2x+1)^{-2}}{4} - \dfrac{(2x+1)^{-1}}{8}\)
Score M1dM1 for obtaining \(\pm px(2x+1)^{-2} \pm q(2x+1)^{-1},\ \ p, q \neq 0\) and A1 for both correct.

ddM1: Substitutes 0 and 2 into an expression of the form \(\pm px(2x+1)^{-2} \pm q(2x+1)^{-1},\ \ p, q \neq 0\) or equivalent and subtracts either way round. Evidence of limits used cannot be the given answer. Condone slips but not directly evaluating the expression as 0 when \(x\) is 0. It is dependent on both of the previous method marks.

A1*: Shows evidence of evaluating after or when the limits are substituted in before proceeding to the given answer with no errors seen including invisible brackets and all previous marks scored. Condone missing/poor notation e.g. missing the d\(x\) throughout

Alternative method I – substitution using \(u = 2x + 1\)

M1: Uses a suitable substitution e.g. \(u = 2x + 1\) and proceeds to \(A\displaystyle\int \dfrac{u - 1}{u^3}\,(\mathrm{d}u)\) o.e.

dM1: Splits into separate fractions and attempts to integrate \(A\displaystyle\int u^{-2} - u^{-3}\,(\mathrm{d}u)\) Look for at least one correct index for one of the two terms i.e. \(u^{-2} \rightarrow u^{-1}\) or \(u^{-3} \rightarrow u^{-2}\). It is dependent on the previous method mark.

A1: \(-\dfrac{1}{4}u^{-1} + \dfrac{1}{8}u^{-2}\) o.e.

ddM1: Substitutes correct limits (1 and 5 if in terms of \(u\)) into an expression of the correct form \(\ldots u^{-1} + \ldots u^{-2}\) (or may have substituted back in terms of \(x\) and substitutes in 0 and 2 into an expression of the correct form \(\pm p(2x+1)^{-2} \pm q(2x+1)^{-1},\ \ p, q \neq 0\) o.e and subtracts either way round – see above for ddM1). Evidence of limits used cannot be the given answer. Condone slips. It is dependent on both of the previous method marks.

A1*: Shows evidence of evaluating after or when the limits are substituted in before proceeding to the given answer with no errors seen including invisible brackets and all previous marks scored. Condone missing/poor notation e.g. missing the d\(u\) throughout

Alternative method II – partial fractions

M1: Writes \(\dfrac{x}{(2x+1)^3}\) as \(\dfrac{0.5}{(2x+1)^2} - \dfrac{0.5}{(2x+1)^3}\) o.e. Allow \(\pm\dfrac{M}{(2x+1)^2} \pm \dfrac{N}{(2x+1)^3}\) (where \(M\) and \(N\) are constants)

dM1: \(\displaystyle\int \dfrac{x}{(2x+1)^3}\,(\mathrm{d}x) = \int \dfrac{\text{``}0.5\text{''}}{(2x+1)^2}\,(\mathrm{d}x) - \int \dfrac{\text{``}0.5\text{''}}{(2x+1)^3}\,(\mathrm{d}x) = \pm\ldots(2x+1)^{-1} \pm \ldots(2x+1)^{-2}\)

A1: \(-\dfrac{1}{4(2x+1)^1} + \dfrac{1}{8(2x+1)^2}\) o.e.

ddM1: Substitutes 0 and 2 into an expression of the correct form \(\pm\ldots(2x+1)^{-1} \pm \ldots(2x+1)^{-2}\) and subtracts either way round. Evidence of limits used cannot be the given answer. Condone slips but not directly evaluating the expression as 0 when \(x\) is 0. It is dependent on both of the previous method marks.

A1*: Shows evidence of evaluating after or when the limits are substituted in before proceeding to the given answer with no errors seen including invisible brackets and all previous marks scored. Condone missing/poor notation e.g. missing the d\(x\) throughout.

Notes

Marks for part (a) may not be scored in part (b)

M1: Either \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 0.45\) or \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = \pm 0.3V\) o.e. e.g. \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = \dfrac{9}{20}\) or \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = \pm\dfrac{3}{10}V\) seen or implied by e.g. \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 0.45 - \dfrac{3}{10}V\) (but not implied by just stating the given answer). Condone use of \(\dot{V}\)
It may be seen as part of their \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\) e.g. \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 0.45 + V + 0.3V\) scores M1A0*
Condone e.g. change in volume = (inflow – outflow =) \(0.45 - 0.3V\) for this mark.

A1*: Achieves \(20\dfrac{\mathrm{d}V}{\mathrm{d}t} = 9 - 6V\) with no errors, following \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 0.45 - \dfrac{3}{10}V\) o.e. (including the \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\) or \(\dot{V}\) but note that it must be \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\) in the final line and not \(\dot{V}\)).
change in volume \(= 0.45 - 0.3V \rightarrow 20\dfrac{\mathrm{d}V}{\mathrm{d}t} = 9 - 6V\) scores M1A0*.
Ignore any units used in their working for both marks.

Notes

B1: Separates the variables correctly, e.g., \(\displaystyle\int \dfrac{1}{9 - 6V}\,\mathrm{d}V = \int \dfrac{1}{20}\,\mathrm{d}t\) or \(\displaystyle\int \dfrac{20}{9 - 6V}\,\mathrm{d}V = \int \{1\}\,\mathrm{d}t\) o.e.
The integral symbol and/or \(\mathrm{d}V\) and/or \(\mathrm{d}t\) may be implied if they go on to integrate both sides to the correct form \(\ldots\ln\left|\alpha(9 - 6V)\right| = \ldots t\ \ (+c)\) with or without the modulus brackets.

M1: Attempts to integrate the reciprocal term \(\dfrac{\beta}{9 - 6V} \rightarrow \ldots\ln\left|9 - 6V\right|\) or \(\rightarrow \ldots\ln\left|\alpha(6V - 9)\right|\) for some constant \(\beta\) (and \(\alpha\) if used). Condone e.g. \(\dfrac{20}{9 - 6V} \rightarrow \ldots\ln 9 - 6V\) or \(\rightarrow \ldots\ln 6V - 9\)

A1: Correct integration for both sides. They do not need the + \(c\) for this mark.
Note scoring this mark implies the earlier B1 (unless it is a verification attempt – see SC).
Note that e.g. \(-\dfrac{1}{6}\ln\left|3 - 2V\right| = \dfrac{t}{20}\ \ (+c)\) or \(-\dfrac{10}{3}\ln\left|2V - 3\right| = t\ \ (+c)\) are also correct.
\(-\dfrac{1}{6}\ln(9 - 6V) = \dfrac{t}{20}\ \ (+c)\) is also correct.
Condone log being used in place of ln.

dM1: Requires constant of integration now. Substitutes (or states) \(t = 0\) and \(V = 0.25\) and finds a value for \(c\), which may be “\(A\)” \(= \mathrm{e}^c\) if they rearrange first to eliminate ln terms.
Dependent on the previous method mark.
Do not be concerned about their processing to find \(c\) or “\(A\)” \(= \mathrm{e}^c\) and does not need to be exact.

A1: Achieves the required form e.g. \(V = \dfrac{3}{2} - \dfrac{5}{4}\mathrm{e}^{-\frac{3t}{10}}\) with no errors and clear working.
Allow equivalent fractions or decimals e.g. \(V = 1.5 - 1.25\mathrm{e}^{-\frac{6}{20}t}\)

SC: Attempts by verification may score maximum B0M1A1dM1A0 – see below.

Alt: Use of an integrating factor – see below.

Special Case: Attempts by verification may score maximum B0M1A1dM1A0

B0: This mark may not be scored via this approach.

M1: Differentiates \(V = P - Q\mathrm{e}^{-kt}\) to the form \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = \alpha\mathrm{e}^{-kt}\) where \(\alpha\) is a constant (note it should be \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = Qk\mathrm{e}^{-kt}\)) and substitutes both this and \(V = P - Q\mathrm{e}^{-kt}\) into \(20\dfrac{\mathrm{d}V}{\mathrm{d}t} = 9 - 6V\) and deduces a value for \(P\) or \(k\) by comparing coefficients.

A1: Correct values for both \(P\) and \(k\).

dM1: Substitutes (or states) \(t = 0\) and \(V = 0.25\) and finds a value for \(Q\).
Requires a value for \(P\) to have been found using the above approach.

A0: This mark may not be scored via this approach.

Alternative: Using Integrating Factor (Further Maths)

B1: Deduces the correct integrating factor for the equation, \(\mathrm{e}^{0.3t}\)
This should come from \(\dfrac{\mathrm{d}V}{\mathrm{d}t} + 0.3V = 0.45 \Rightarrow \text{I.F.} = \mathrm{e}^{\int 0.3\,\mathrm{d}t} = \mathrm{e}^{0.3t}\)
May be implied by sight of \(\dfrac{\mathrm{d}\left(V\mathrm{e}^{0.3t}\right)}{\mathrm{d}t} = \ldots\)

M1: Fully multiplies through by their integrating factor and integrates both sides.
Score for \(V\mathrm{e}^{kt} = \displaystyle\int \ldots\mathrm{e}^{kt}\,\mathrm{d}t = \ldots\mathrm{e}^{kt}\) Condone missing \(\mathrm{d}t\)

A1: Correct integration \(V\mathrm{e}^{0.3t} = \displaystyle\int 0.45\mathrm{e}^{0.3t}\,\mathrm{d}t = \dfrac{3}{2}\mathrm{e}^{0.3t}\ \ (+c)\)

dM1: As main scheme.

A1: As main scheme.

Notes

M1: See main scheme. If using the answer to part (b) it must be of the form \(V = P - Q\mathrm{e}^{-kt}\) but there is no limitation on the values of their \(P\), \(Q\) or \(k\).
Substitution of a large value for \(t\) may score this mark but it is unlikely to be recovered to score the A1 unless they reference e.g. \(V_{\max}\) being “1.5”.
Reference to an (upper) limit of “1.5” or their \(P\) can imply the method mark.
If setting \(V = 2\) in their equation they must reach either ln(–ve) or solve the equation to reach a value for \(t\) to score this mark.

A1ft: Must conclude “no” or equivalent e.g. “the container will not become full”.
Makes a correct interpretation for their method (see bullets 1-8) with a clear conclusion e.g. “no”.
To score this mark through ft, their \(V\) must be of the form \(V = P - Q\mathrm{e}^{-kt}\) with \(k \gt 0\), \(Q \gt 0\) and \(0 \lt P \lt 2\) if used (but note that they can still use the answer to part (a) to score both marks via bullets 1, 3, 5 or 6). Allow “it” in place of the “container”/“tank”.
Just stating “the equation cannot be solved when \(V = 2\)” without any evidence is M0A0.
There must be no incorrect working if solving their equation or contradictory statements such as “\(t\) cannot be negative” but condone notational errors provided the intention is clear.

Notes

Mark (a) and (b) together

B1: States that \(\displaystyle\int_{1.44}^{2.89} \dfrac{2}{\sqrt{x}}\,\mathrm{d}x\) or equivalent such as \(\displaystyle 2\int_{1.44}^{2.89} x^{-\frac{1}{2}}\,\mathrm{d}x\) or \(\displaystyle 2\int_{1.44}^{2.89} x^{-0.5}\,\mathrm{d}x\) but must include the limits and the d\(x\). Condone \(\mathrm{d}x \leftrightarrow \delta x\) as it is very difficult to tell one from another sometimes.

Notes

Mark (a) and (b) together

M1: Uses \(\displaystyle\int \dfrac{1}{\sqrt{x}}\,\mathrm{d}x = a\sqrt{x}\) or \(ax^{\frac{1}{2}}\) (allow \(a\) to be 1) and applies the given limits to their \(ax^{\frac{1}{2}}\) subtracting either way round. (Condone with the constant of integration included)
You do not need to be concerned by fractions within fractions as this is still of the required form e.g. \(\dfrac{2x^{\frac{1}{2}}}{\frac{1}{2}}\). Only condone transcription errors of 2.89 or 1.44 when substituting the limits into the expression.
This mark can be scored for

e.g. \(\left[4\sqrt{x}\right]_{1.44}^{2.89} = 4 \times \sqrt{2.89} - 4 \times \sqrt{1.44}\) or e.g. \(\dfrac{2x^{\frac{1}{2}}}{\frac{1}{2}} \rightarrow \dfrac{2(2.89)^{\frac{1}{2}} - 2(1.44)^{\frac{1}{2}}}{\frac{1}{2}}\)

May already be partially evaluated so allow e.g. \(\dfrac{34}{5} - \dfrac{24}{5}\) o.e. provided it is not just 2.

A1: 2 cso
The method mark must have been awarded. Do not withhold this mark for poor notation or e.g. a missing d\(x\) in their solution.

Notes

M1: For increasing any power by one. Score for \(x^n \rightarrow x^{n+1}\) in any term, including, \(-3 \rightarrow \ldots x\) where … is a constant, but not for + \(c\). Allow the indices to be unprocessed, e.g., \(x^{4+1}\)

A1: One correct term which may be unsimplified and indices may be unprocessed.
Condone e.g. \(-3x^1\) or \(-\dfrac{6}{\left(\frac{1}{2} + 1\right)}x^{\frac{1}{2} + 1}\) for this mark. Not scored for + \(c\)

A1: Two correct terms which may be unsimplified but indices must be processed.
Condone \(-3x^1\) for this mark. Not scored for + \(c\)

A1: cao Requires all terms simplified and + \(c\)
Ignore the LHS i.e. ignore what they call their integral.
Allow \(0.2x^5\) for \(\dfrac{1}{5}x^5\) and \(-4\sqrt{x^3}\) or \(-4\sqrt{x}^3\) or \(-4x\sqrt{x}\) or \(-4x^{1.5}\) for \(-4x^{\frac{3}{2}}\)
Do not allow \(-3x^1\) for this mark.
Condone spurious integral signs e.g. \(\displaystyle\int \dfrac{1}{5}x^5 - 4x^{\frac{3}{2}} - 3x + c\) or d\(x\) left in their answer
ISW after a correct expression seen e.g. if they multiply through by 5 or e.g. try to solve = 0
Do not allow e.g. \(\dfrac{1}{5}x^5 + -4x^{\frac{3}{2}} + -3x + c\)

Notes

B1: Correctly sets up the model. \(\dfrac{\mathrm{d}r}{\mathrm{d}t} \propto \dfrac{1}{\sqrt{r}}\) scores B0 unless e.g. \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \pm\dfrac{k}{\sqrt{r}}\) is seen but condone \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \pm\dfrac{k}{\sqrt{r}}\) being seen at the start of (b). They may use any letter except \(t\) or \(r\) in place of \(k\).
You may see \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \pm\dfrac{1}{k\sqrt{r}}\) which is acceptable provided it is clear that it is not the \(k^{\text{th}}\) root.

Notes

Note: candidates using \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \pm\dfrac{1}{\sqrt{r}}\) in (b) can score maximum B0M1A1dM0A0

B1: \(k = 3.6\) coming from \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \dfrac{k}{\sqrt{r}}\) (or \(k = \dfrac{5}{18}\) coming from \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \dfrac{1}{k\sqrt{r}}\)) and from use of \(r = 16\) and \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = 0.9\) but note that this may occur later in their working, which is perfectly fine provided it is from acceptable work. Note e.g. \(k = -3.6\) coming from \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = -\dfrac{k}{\sqrt{r}}\) is correct.

Note, however, that an attempt to find \(k\) from comparing coefficients between \(r^{\frac{3}{2}} = 1.5kt + c\) and \(r^{\frac{3}{2}} = 5.4t + 10\) is not acceptable (see special case).
They can also find \(k\) by differentiating their \(r = \mathrm{f}(t)\) and substituting \(t = 10\), \(r = 16\) and \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = 0.9\) if they have also used \(t = 10\), \(r = 16\) in their \(r = \mathrm{f}(t)\). This sets up simultaneous equations where \(c\) can be eliminated. Use Review if you are unsure if their approach is acceptable.

M1: Separates the variables for their differential equation correctly and attempts to integrate both sides.
Must be a differential equation of the form \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \mathrm{f}(r)\) for some function \(\mathrm{f}(r)\) independent of \(t\).
Evidence of \(r^n \rightarrow r^{n+1}\), or e.g. \(\dfrac{1}{r} \rightarrow \ln r\) is sufficient for their attempt to integrate in \(r\), but \(k\) must be integrated to \(kt\) o.e. e.g. \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \dfrac{k}{r^2} \Rightarrow r^2\dfrac{\mathrm{d}r}{\mathrm{d}t} = k \Rightarrow \lambda r^3 = kt\,\{+c\}\) would score this mark.
Note that they may divide by \(k\) (or 3.6) prior to integrating. Here, 1 must be integrated to \(t\).

A1: Correct integration for their \(k\). Allow this mark if they have not found \(k\), so allow e.g. \(\dfrac{2}{3}r^{\frac{3}{2}} = kt\,\{+c\}\) with/without the constant of integration but the \(\dfrac{2}{3}\) must be evident in some way.
Condone spurious notation for this intermediate mark e.g. integral signs left in after integrating.

dM1: Uses \(t = 10\), \(r = 16\) in their equation to find the constant of integration.
This mark is dependent on the first method mark.
Must have already found a value for \(k\) using a valid strategy and the constant of integration must be present.
Those that found \(k\) from comparing coefficients between \(r^{\frac{3}{2}} = \dfrac{3}{2}kt + c\) and \(r^{\frac{3}{2}} = 5.4t + 10\) may not score this mark.

A1*: Correct equation from correct working. May be seen at the start of (c).
Must follow A1 earlier so do check if this has been obtained fortuitously.

SC: It is possible to compare coefficients (following integration) between \(r^{\frac{3}{2}} = \dfrac{3}{2}kt + c\) and \(r^{\frac{3}{2}} = 5.4t + 10\) to deduce the value of \(k\) as 3.6 (or just write their coefficient as 5.4).
The maximum that can be scored this way or by a similar invalid approach is B0M1A1dM0A0

Notes

M1: Substitutes \(t = 20\) into the given equation and uses correct processing to find the value of \(r\)
e.g. substitutes \(t = 20\) into \(\dfrac{2}{3}r^{\frac{3}{2}} = 72 + \dfrac{20}{3} \Rightarrow r = \left(\dfrac{3}{2}\left(72 + \dfrac{20}{3}\right)\right)^{\frac{2}{3}} = \ldots\)
Their work should lead to the correct answer so the index work must be correct e.g. \(\sqrt{118^3}\) is M0.
\(\sqrt[3]{118^2}\) or \(118^{\frac{2}{3}}\) are acceptable as values, i.e., the bracket must be evaluated.
May be implied by awrt 24 (cm) following \(r^{\frac{3}{2}} = 118\) or by awrt 24.1 (cm). Ignore units for M1.

A1: cao 241mm or 241 or 24.1cm but do not accept 24.1 or e.g. \(\sqrt[3]{118^2}\) cm
Correct answer with units implies both marks.

Notes

B1: Examples of acceptable answers (which must relate to the model in context):

• (The model will not hold indefinitely as) the balloon may burst/pop
• The balloon is unlikely to be (perfectly) spherical (condone circular)
• The model predicts the balloon will increase in size without limit (which is unrealistic)

Note \(t \rightarrow \infty \Rightarrow r \rightarrow \infty\) is unrealistic / impossible scores B0 unless they reference e.g. the radius.
Condone the presence of additional remarks such as “the balloon may not inflate at the same rate” or “the radius of the balloon might not start at 0” that have already been addressed in the model, but these answers alone score B0.

Notes

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

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

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

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

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

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

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

Alternative 13(a) working backwards:

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

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

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

Notes

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

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

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

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

Alternative 13(b) via IBP Way 1:

Alternative 13(b) via IBP Way 2:

Notes

M1: Sets \(\dfrac{1}{V(25-V)} = \dfrac{P}{V} + \dfrac{Q}{25-V}\) and uses a correct method to identify the value of at least one constant.
Do not condone incorrect work e.g. \(\dfrac{1}{V(25-V)} = \dfrac{P}{V} + \dfrac{Q}{25-V} \Rightarrow 1 = PV + Q(25-V)\) etc. this scores M0

A1: Correct partial fractions in any form e.g.
\(\dfrac{1}{25V} + \dfrac{1}{25(25-V)},\ \dfrac{1}{25V} + \dfrac{1}{625-25V},\ \dfrac{\frac{1}{25}}{V} + \dfrac{\frac{1}{25}}{(25-V)},\ \dfrac{1}{25V} - \dfrac{1}{25(V-25)},\ \dfrac{1}{25}\left(\dfrac{1}{V} + \dfrac{1}{25-V}\right)\) etc.
Note that this mark is not just for the correct constants, it is for the correctly written fractions either seen in part (a) or used in part (b). Allow 0.04 for \(\dfrac{1}{25}\).
Correct partial fractions only scores both marks.
If the correct fractions are obtained following incorrect work score M0A0 but allow full recovery in the rest of the question.

Notes

Alternative for the final 3 marks:

Mark (b) and (c) together

M1: Realises that \(\displaystyle\int \frac{\ldots}{V(25-V)}\,(\mathrm{d}V)\) is required and reaches the form \(p\ln\alpha V \pm q\ln\beta(25-V)\) (or e.g. \(p\ln\alpha V \pm q\ln\beta(V-25)\)) or equivalent for this integration e.g. \(p\ln 25V - q\ln(625-25V)\) with \(p\) and \(q\) non-zero.
But note that \(\displaystyle\int \frac{\ldots}{V(25-V)}\,\mathrm{d}V = \ln V(25-V)\) does not score this mark unless we see an attempt to integrate the partial fractions first.
Condone missing brackets e.g. around the \(V - 25\) for this mark
Note that the rhs may be incorrect or missing for this mark.

A1ft: Fully correct equation following through their \(P\) and \(Q\). The “\(+\,c\)” is not required here.
You may need to check carefully when awarding this mark as there will be various alternative correct (or correct ft) forms e.g. these are correct (for correct PF’s):
\(\dfrac{1}{25}\ln 25V - \dfrac{1}{25}\ln(625-25V) = \dfrac{1}{10}t\,(+c),\ \ln V - \ln(25-V) = 2.5t\,(+c),\)
\(\dfrac{2}{5}\ln 25V - \dfrac{2}{5}\ln(25-V) = t\,(+c),\ \dfrac{2}{5}\ln 5V - \dfrac{2}{5}\ln(125-5V) = t\,(+c)\)
In general look for an equation of the form \(P\ln\alpha V - Q\ln\beta(25-V) = \dfrac{1}{10}t\,(+c)\) or a multiple of this equation. Do not condone missing brackets unless they are implied by later work e.g. \(\ln 25 - V\) for \(\ln(25-V)\)
Allow ( ) or | | around the arguments of the ln’s and condone “log” for ln.

M1: States or uses \(t = 0\) and \(V = 20\) consistently leading to a constant of integration which may be simplified or unsimplified. May be implied by their constant so may need to be checked.
This mark is not formally dependent but depends on having made some attempt to integrate both sides, however poor.

dM1: States or uses \(V = 24\) and proceeds to find a value for \(t\) (even if \(t \lt 0\)). You do not need to check the processing provided they reach a value for \(t\). Depends on the previous method mark and depends on an attempt to integrate both sides however poor.
May be implied by their value for \(t\) so may need to be checked.

A1: Correct value of 43 or awrt 43.0 or exact value of \(24\ln 6\).
Units are not required but if any are given it must be minutes or condone “m”.
Note that in hours the time is 0.7167037877… and scores A0

Alternative for final 3 marks:

M1: \(\left[\dfrac{1}{25}\ln V - \dfrac{1}{25}\ln(25-V)\right]_{20}^{24} = \left[\dfrac{1}{10}t\right]_0^T \Rightarrow \dfrac{1}{25}\ln 24 - \dfrac{1}{25}\ln 4 = \dfrac{1}{10}T\)
Applies the limits 20 and 24 to lhs and 0 to “\(T\)” or e.g. “\(t\)” on rhs
This mark is not formally dependent but depends on having made some attempt to integrate both sides, however poor.

dM1: \(T = \dfrac{10}{25}\ln 24 - \dfrac{10}{25}\ln 4 = \ldots\) Solves to find “\(T\)”. You do not need to check the processing provided they reach a value for \(t/T\). Depends on the previous method mark and depends on having made some attempt to integrate both sides however poor.

A1: Correct value of 43 or awrt 43.0 or exact value of \(24\ln 6\).
Units are not required but if any are given it must be minutes or condone “m”.
Note that in hours the time is 0.7167037877… and scores A0

Note

is also correct integration and scores M1A1
The subsequent marks are also available as described above and could lead to the correct answer if the limits and constant of integration are dealt with correctly.
Use review for any examples like these if you are unsure but generally apply the MS as above.

Notes

(c) The marks in (c) depend on having integrated their partial fractions to obtain an equation of the form \(\pm\ldots\ln\ldots V \pm \ldots\ln\ldots(25-V) = \pm kt \pm c,\ \ k, c \neq 0\) and … are non-zero constants or equivalent if they have already attempted to eliminate the ln’s in (b)
e.g. \(\dfrac{\ldots V}{\ldots(25-V)} = \mathrm{e}^{\pm kt \pm c}\) oe

M1: Uses fully correct log work, having obtained a constant of integration, to eliminate all the ln’s including from e.g. \(\mathrm{e}^{\ln 4}\). We condone sign or coefficient slips only.

M1: Proceeds from an equation of the form \(\dfrac{\ldots V}{\ldots(25-V)} = \ldots\mathrm{e}^{\ldots t}\) oe using correct algebra to \(V = \ldots\) e.g.
\(\dfrac{\ldots V}{\ldots(25-V)} = \ldots\mathrm{e}^{\ldots t} \Rightarrow \ldots V = \ldots(25-V)\ldots\mathrm{e}^{\ldots t} \Rightarrow (\ldots \pm \ldots)V = \ldots\mathrm{e}^{\ldots t} \Rightarrow V = \ldots\)
Condone sign/coefficient slips only.

A1: Correct expression not just values for the constants.

Notes

B1: Correct value of 25 seen
Allow e.g. \(\lt 25\) or \(\leqslant 25\)
Condone “\(\gt 25\)” but the following mark is then not available

B1: Depends on a correct final equation in any form in (c) e.g. \(V = \dfrac{100\mathrm{e}^{2.5t}}{1 + 4\mathrm{e}^{2.5t}}\) oe and one of:

• Considers the behaviour as \(t \rightarrow \infty\) e.g. states that as \(t \rightarrow \infty,\ \mathrm{e}^{-2.5t} \rightarrow 0\) (condone “= 0”) oe
• \(V \lt 25\) as \(\ln(25 - V)\) is not possible when \(V \geqslant 25\)
• Verifies the 25 using a value of \(t,\ t \geqslant 9\)

Using the differential equation:

B1: Correct value of 25 seen
Allow e.g. “\(\lt 25\)” or \(\leqslant 25\)
Condone “\(\gt 25\)” but the following mark is then not available

B1: E.g. when \(V = 25,\ \dfrac{\mathrm{d}V}{\mathrm{d}t} = 0\) or \(\dfrac{\mathrm{d}V}{\mathrm{d}t} \lt 0\) if \(V \gt 25\)

Notes

Mark positively in this question and do not penalise poor notation such as a missing “\(\mathrm{d}x\)” or spurious integral signs, “\(+\,c\)” etc. as long as the intention is clear.

M1: Obtains \(\displaystyle\pm\alpha x^2\mathrm{e}^{-3x} \pm \beta\int x\,\mathrm{e}^{-3x}\,\mathrm{d}x\)
(you do not need to be concerned about how they arrive at this)

A1: Correct expression simplified or unsimplified. E.g. allow \(\displaystyle-\frac{8x^2}{3}\mathrm{e}^{-3x} - \int -\frac{16x}{3}\mathrm{e}^{-3x}\,\mathrm{d}x\)
Note that we condone the “8” missing for this mark so allow e.g. \(\displaystyle-\frac{x^2}{3}\mathrm{e}^{-3x} - \int -\frac{2x}{3}\mathrm{e}^{-3x}\,\mathrm{d}x\)
Note that notation may be poor here but the intention clear e.g. if they obtain \(-\dfrac{8x^2}{3}\mathrm{e}^{-3x} + \left[\dfrac{16x}{3}\mathrm{e}^{-3x}\right]\) and then attempt to integrate \(\dfrac{16x}{3}\mathrm{e}^{-3x}\) both marks can be implied.

dM1: Attempts parts again on \(\displaystyle\pm\beta\int x\,\mathrm{e}^{-3x}\,\mathrm{d}x\) to obtain \(\displaystyle\pm Ax\mathrm{e}^{-3x} \pm B\int \mathrm{e}^{-3x}\,\mathrm{d}x\)
This may be seen in isolation and does not need to be seen as part of the complete integration. Depends on the first method mark.
Watch for the DI method (with or without the 8):

(In the mark scheme, arrows join each entry in the D column to the entry in the I column one row below.)

Giving the correct integration e.g. \(\displaystyle\int 8x^2\mathrm{e}^{-3x}\,\mathrm{d}x = -\frac{8x^2}{3}\mathrm{e}^{-3x} - \frac{16x}{9}\mathrm{e}^{-3x} - \frac{16}{27}\mathrm{e}^{-3x}\)
In such cases score M1dM1 for obtaining \(\pm px^2\mathrm{e}^{-3x} \pm qx\mathrm{e}^{-3x} \pm r\mathrm{e}^{-3x},\ \ p, q, r \neq 0\) and then A1 for the correct first 2 terms, with or without the factor of 8.
Note that for this approach M1A1dM0 is not possible.

M1: Substitutes the limits 1 and 0 into an expression of the form \(\pm\alpha x^2\mathrm{e}^{-3x} \pm \beta x\mathrm{e}^{-3x} \pm \gamma\mathrm{e}^{-3x},\ \ \alpha, \beta, \gamma \neq 0\) and subtracts the right way round.
Must see evidence of the use of both limits and subtraction and use of \(\mathrm{e}^0 = 1\).
Note that some candidates apply the limits as they go e.g. to the \(\left[-\dfrac{8x^2}{3}\mathrm{e}^{-3x}\right]\) which is acceptable but you will need to check carefully that overall they are satisfying the conditions above.
Condone not realising that the first 2 terms evaluate to 0 when substituting \(x = 0\) e.g.
condone \(-\dfrac{8}{3}\mathrm{e}^{-3} - \dfrac{16}{9}\mathrm{e}^{-3} - \dfrac{16}{27}\mathrm{e}^{-3} - \left(-\dfrac{8}{3} - \dfrac{16}{9} - \dfrac{16}{27}\right)\) as we have evidence of \(\mathrm{e}^0 = 1\)
Note that e.g. \(-\dfrac{8}{3}\mathrm{e}^{-3} - \dfrac{16}{9}\mathrm{e}^{-3} - \dfrac{16}{27}\mathrm{e}^{-3} - \left(-\dfrac{16}{27}\mathrm{e}^0\right) = -\dfrac{136}{27}\mathrm{e}^{-3} - \dfrac{16}{27}\) scores M0 as it suggests that \(\mathrm{e}^0 = -1\) not \(+1\).

A1: Correct answer of \(\dfrac{16}{27} - \dfrac{136}{27}\mathrm{e}^{-3}\) but allow equivalent exact fractions and condone \(\dfrac{16}{27} - \dfrac{136}{27\mathrm{e}^3}\). Isw once the correct answer is seen.

Candidates who consistently misread \(8x^2\mathrm{e}^{-3x}\) as \(8x^2\mathrm{e}^{3x}\):

Scores a maximum of M1A0dM1M1A0
The main scheme can be applied similarly e.g.

M1: Attempts parts to obtain \(\displaystyle\alpha x^2\mathrm{e}^{3x} - \beta\int x\,\mathrm{e}^{3x}\,\mathrm{d}x,\ \ \alpha, \beta \gt 0\)

A0: Not available

dM1: Attempts parts again on \(\displaystyle\beta\int x\,\mathrm{e}^{3x}\,\mathrm{d}x\) to obtain \(\displaystyle Cx\mathrm{e}^{3x} - D\int \mathrm{e}^{3x}\,\mathrm{d}x,\ \ C, D \gt 0\)

M1: Substitutes the limits 1 and 0 into an expression of the form \(\pm\lambda x^2\mathrm{e}^{3x} \pm \mu x\mathrm{e}^{3x} \pm \gamma\mathrm{e}^{3x},\ \ \lambda, \mu, \gamma \neq 0\) and subtracts the right way round.
Must see evidence of the use of both limits and subtraction and use of \(\mathrm{e}^0 = 1\).

A0: Not available

But note, do not allow mixing of \(3x\)’s and \(-3x\)’s. If there are a mixture, apply the main scheme.

Notes

B1: Substitutes \(x = 4\) into the equation of the curve and verifies that \(y = 0\). Accept “\(8(4) - 4^{\frac{5}{2}} = 0\)”
Alternatively, sets \(8x - x^{\frac{5}{2}} = 0\) and solves with correct processing to achieve \(x = 4\).
As a minimum accept e.g. \(8x - x^{\frac{5}{2}} = 0 \Rightarrow x^{\frac{3}{2}} = 8 \Rightarrow \{x=\}\ 4\) which may follow factorisation.

Notes

B1: Correct differentiation. The \(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\) need not be present.

M1: Correct method for finding the equation of the tangent at \(A(4, 0)\).
Requires substitution of \(x = 4\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and an attempt at the equation of the line using this gradient.  If using \(y - y_1 = m(x - x_1)\) then condone the omission of the \(-\,0\).
If \(y = mx + c\) is used they must proceed as far as \(c = \ldots\)
Accept \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -12\) or \(m = -12\) without explicit substitution of \(x = 4\) provided \(8 - \dfrac{5}{2}x^{\frac{3}{2}}\) is seen.

A1*: Correct work leading to the given equation having scored B1M1.
Condone \(y + 12x = 48\) and apply isw once seen.
Do not condone \(12x + y - 48 = 0\) (unless a correct equation = 48 is seen).

Notes

Note: Condone poor notation such as missing d\(x\) or spurious \(\displaystyle\int\) symbols throughout.

M1: Attempts to find either the \(x\) or \(y\) coordinate of the intersection of line \(l_1\) and line \(l_2\)
You may need to check the diagram or limits to their integrals.

dM1: Correct method for the area of the triangle. e.g. Triangle area is \(\dfrac{1}{2} \times 4 \times \text{``}19.2\text{''}\left(= 38.4 \text{ or } \dfrac{192}{5}\right)\)
If integration is attempted then condone slips in their rearrangement of \(12x + y = 48\) to \(y = 48 - 12x\) and note that their integrals do need not to be evaluated, so for example look for \(\displaystyle\int_0^{\text{``}2.4\text{''}} 8x\,\mathrm{d}x + \int_{\text{``}2.4\text{''}}^{4} \text{``}(48 - 12x)\text{''}\,\mathrm{d}x\quad \left\{= \dfrac{576}{25} + \dfrac{384}{25} = 23.04 + 15.36\right\}\)

B1: Correct integration of curve ignoring limits, i.e. \(4x^2 - \dfrac{2}{7}x^{\frac{7}{2}}\) but condone e.g. \(\dfrac{8x^{1+1}}{2} - \dfrac{x^{\frac{5}{2}+1}}{\frac{7}{2}}\)

ddM1: Fully correct strategy including substitution which would lead to an exact area.
Does not need to reach a value. Dependent on both previous M marks.
Implied by \(38.4 - \dfrac{192}{7}\) or a correct final answer \(\dfrac{384}{35}\)
Note that the decimal approximation that might be seen is 10.97142857 and implies ddM0 unless there is evidence of substitution (which need not be evaluated).

A1: Correct exact value. Either \(\dfrac{384}{35}\) or \(10\dfrac{34}{35}\)

Alternative using lines – curve:

M1: Attempts to find either the \(x\) or \(y\) coordinate of the intersection of line \(l_1\) and line \(l_2\)
You may need to check the diagram or limits to their integrals.

dM1: Correct method for at least one part (0 to “2.4” or “2.4” to 4) of the area of \(R\) including limits.
Condone slips in their rearrangement of \(12x + y = 48\) to \(y = 48 - 12x\) and note that their integrals do need not to be evaluated, so for example look for \(\displaystyle\int_0^{\text{``}2.4\text{''}} 8x - \left(8x - x^{\frac{5}{2}}\right)\mathrm{d}x\) or \(\displaystyle\int_{\text{``}2.4\text{''}}^{4} \text{``}(48 - 12x)\text{''} - \left(8x - x^{\frac{5}{2}}\right)\mathrm{d}x\) (or a sum of both)

B1: Correct integration of both regions ignoring limits. May be completed as a sum or separately.
Condone e.g. \(\dfrac{x^{\frac{5}{2}+1}}{\frac{7}{2}}\) in place of \(\dfrac{2}{7}x^{\frac{7}{2}}\)  Note that each integral may have been simplified.

ddM1: Fully correct strategy including substitution which would lead to an exact area.
Does not need to reach a value. Dependent on both previous M marks.
This approach requires:

• substitution of 0, their 2.4 and 4 in the correct places
• the \(\dfrac{2}{7}(2.4)^{\frac{7}{2}} - \dfrac{2}{7}(2.4)^{\frac{7}{2}}\) to be cancelled (may be implied by a correct final answer \(\dfrac{384}{35}\))

Note that the decimal approximation that might be seen is 10.97142857 and implies ddM0 unless there is evidence that the \(\dfrac{2}{7}(2.4)^{\frac{7}{2}} - \dfrac{2}{7}(2.4)^{\frac{7}{2}}\) has been cancelled e.g. \(\cancel{6.118\ldots} - \cancel{6.118\ldots}\)

A1: Correct exact value. Either \(\dfrac{384}{35}\) or \(10\dfrac{34}{35}\)

Notes

M1: Attempts integration to achieve \(\{H=\}\,k\,\mathrm{e}^{-0.2t}\,\{+c\}\) with \(k\) a numerical constant \(\neq -0.12\)
Note that we will condone \(k = (-0.2)(-0.12)\ \{= 0.024\}\)
If they divide by \(-0.12\) first before integrating they need \(aH = b\mathrm{e}^{-0.2t}\,\{+c\}\) with \(a\) and \(b\) numerical and \(b \neq 1\). Condone a spurious integral symbol remaining after integration.

dM1: Uses \(t = 0, H = 1.5\) and a model of the form \(H = k\,\mathrm{e}^{-0.2t} + c\) (or \(aH = b\mathrm{e}^{-0.2t} + c\)) to find the value of the constant \(c\). They cannot just “make up” a value for \(k\).
Do not be concerned with their processing to find \(c\) but they cannot just state \(B\) (or \(c\)) is 0.
For reference if they divide by \(-0.12\) first they should reach \(-\dfrac{25}{3}H = -5\mathrm{e}^{-0.2t} - 7.5\) o.e.

A1: Correct complete equation in the required form: \(H = 0.6\mathrm{e}^{-0.2t} + 0.9\) with the \(H\) = present.
May be awarded if seen at the start of (b) but not in (c). Condone \(-\dfrac{1}{5}\) in place of \(-0.2\)
Finding correct values for \(A\) and \(B\) is insufficient for this mark.
Allow exact equivalents but they must be in the required form, e.g. \(H = \dfrac{6}{10}\mathrm{e}^{-0.2t} + \dfrac{9}{10}\)
A minimally acceptable answer is \(\{H=\}\,0.6\mathrm{e}^{-0.2t} + c \rightarrow H = 0.6\mathrm{e}^{-0.2t} + 0.9\) score M1dM1A1.
Note: sight of differentiating the given form to e.g. \(\dfrac{\mathrm{d}H}{\mathrm{d}t} = -0.2A\mathrm{e}^{-0.2t}\) in their working without clear evidence of integration of the original differential equation should be marked using the special case below.

SC: For candidates starting with the given answer \(H = A\mathrm{e}^{-0.2t} + B\) it is possible to use \(\dfrac{\mathrm{d}H}{\mathrm{d}t} = -0.2A\mathrm{e}^{-0.2t} = -0.12\mathrm{e}^{-0.2t}\) to deduce that \(A = 0.6\). This can be awarded SC M1dM0A0
If they go on to find \(B\) as in the main scheme then this can be awarded SC M1dM1A0
Answer with no working scores 110.

Note: If the special case is applied they may go on to achieve the rest of the marks in (b) and (c).

Notes

Note: \(A\) and \(B\) must be numbers but may be “made up” if they did not have an answer to (a).

M1: Uses \(H = 1.2\) in a model of the form \(H = A\mathrm{e}^{-0.2t} + B,\ B \neq 0\) and rearranges to make \(A\mathrm{e}^{\pm 0.2t}\) or \(\mathrm{e}^{\pm 0.2t}\) the subject. Condone slips in rearranging, e.g. dividing the LHS by 0.9 instead of subtracting 0.9. Rearranging first before substituting is acceptable but they must get to \(A\mathrm{e}^{\pm 0.2t}\) or \(\mathrm{e}^{\pm 0.2t}\) as the subject.

dM1: Correct use of ln to make \(t\) the subject.  Requires \(A \gt 0,\ 0 \lt B \lt 1.2\) and \(\mathrm{e}^{\pm 0.2t} = \lambda \gt 0\)
If they had a negative value for \(A\) in part (a) they cannot just make it positive at this stage.
Any of \(5\ln 2\) or \(-5\ln\dfrac{1}{2}\) or awrt 3.46 or awrt 3.47 following a correct equation will imply M1dM1.
If they do not show their method for an incorrect \(H = A\mathrm{e}^{-0.2t} + B\) with \(A \gt 0,\ 0 \lt B \lt 1.2\) you may need to check their value for \(t \gt 0\) as it may imply M1dM1.

A1: Correct time in hours and minutes \(\{t=\}\) 3 hours 28 minutes, but condone e.g. 3h 28m
Must come from correct values of \(A\) and \(B\) in (a).

Note: If their \(B = 0\) then they should end up with \(t = -3.46\ldots\) however, they did not score the first M1. They cannot “recover” this by making it positive and finding \(t =\) 3 hours 28 minutes.

Notes

Note that 0.9 or 0.9m must come from a correct value of \(B\) in (a) to score any marks.

M1: Identifies the requirement to establish the limit as \(t\) tends to infinity.
It can be implied by stating that \(H = A\mathrm{e}^{-0.2t} + B \rightarrow B\) or \(\left(\lim\limits_{t \to \infty}\left[0.6\mathrm{e}^{-0.2t} + 0.9\right]\right) = 0.9\)
Stating “\(B\)” on its own will score this mark.
Substituting a large value is M0 unless it leads to their value for \(B\) at which point the A1 is available as well.

A1ft: Correct height including units. Follow through on their value of \(B\) where \(0 \lt B \lt 1.2\)
Correct ft height including units implies M1A1, while e.g. 0.9 (no units) would imply M1A0.
Evidence of an incorrect method such as \(1.5 - 0.6\mathrm{e}^{-0.2(0)} = 0.9\) m scores M0A0.

Misreading as \(\dfrac{\mathrm{d}H}{\mathrm{d}t} = -0.12\mathrm{e}^{0.2t}\) can score a maximum (a) 110 (b) 100 (c) 10.

Typical solution

Typical solution

Typical solution

area above

area below

Typical solution

Typical solution

Typical solution

Typical solution

when \(h = 130\), \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = -1.5\)

Typical solution

Typical solution

61 minutes

Typical solution

Let \(u = 2x + 1\)

Typical solution

Typical solution

Since the area of the rectangles is an underestimate, increasing their number will give an increased total \(\gt A\)

No matter how many rectangles are used their total will be less than the exact area so \(\lt \dfrac{1322}{15}\)

Typical solution

Typical solution

Typical solution

Typical solution

125

Typical solution

Typical solution

(i)

(ii)

(iii)

Typical solution

(i)

(ii)

Typical solution

(i)

(ii)

(iii)

Typical solution

Typical solution

Typical solution

30

Typical solution

(i)

(ii)

(iii)

Typical solution

(i)

(ii)

Typical solution

Area \(\approx\) 5.28

Typical solution

Shaded region is below \(x\)-axis

area = \(32\ln 2 - \dfrac{33}{2}\)

Typical solution

(i)

(ii)

The value predicted by the exponential model will grow without limit.

This can’t be true as there are only 900 tomato plants in the greenhouse.

Typical solution

(i)

(ii)

(iii)

It takes 11 days from when the damage is first noticed until half of the plants are damaged by insects

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Notes

B1: Identify correct partial fractions

M1: Clear fractions and attempt (at least) one coefficient. From using their 3 distinct linear factors

A1: Obtain one correct value www

A1: Obtain fully correct partial fractions. Possibly implied by \(A = 2\) etc

Notes

M1*: Attempt to separate variables to obtain \(\int \mathrm{f}(y)\,\mathrm{d}y = \int \mathrm{g}(x)\,\mathrm{d}x\)
Condone no integral signs
BOD if no d\(x\) and/or d\(y\) or still present as a single operator

B1FT: Correct integration of their 3 linear partial fractions from part (a). B0 if any \(y\) term still on RHS
Condone no constant of integration
Condone brackets not modulus signs
Brackets / modulus could be implied by further working, otherwise B0

B1: Correct integration of \(\tan y\). B0 if any \(x\) term on LHS, or if \(\tan y\) is still in term on RHS
\(\ln|\sec y|\) or \(-\ln|\cos y|\)

M1dep*: Correct process to combine all terms and remove logs. Must deal correctly with indices and correctly use rules for combining logs
Must include constant of integration (could be \(\mathrm{e}^c\) rather than \(k\)). but could be numerical already (possibly incorrect)

A1FT: Obtain a correct equation no longer involving logs. FT on an incorrect numerical constant of integration only
NB \((x - 1)^3\) not \((1 - x)^3\) is A0
All algebraic terms must be fully correct (no FT on these)

M1dep*: Use \(\left(\frac{1}{2}, \frac{1}{4}\pi\right)\) in a correct equation to attempt to find \(k\). Could be awarded before combining terms

A1: Obtain fully correct equation, in required form. Surd may not be rationalised, but no fractions within fractions

Notes

(a)(i)

B1: State correct scale factor

(a)(ii)

B1: Identify direction – correct language needed. Allow ‘horizontal’
B0 for ‘in’, ‘on’ or ‘along’ the \(x\)-axis, or for right / left
Could be implied by \(\begin{pmatrix} k \\ 0 \end{pmatrix}\)

B1: Correct magnitude of translation. Not dep on previous B1, but must have indicated horizontal translation in some way, including informal language

Do not penalise mentioning the stretch from (a)(i) again in (a)(ii)

\(\begin{pmatrix} \frac{1}{2} \\ 0 \end{pmatrix}\) is B2

Notes

B1: State correct scale factor. Any exact equiv

Notes

M1: State either correct derivative or \(\frac{1}{2}\mathrm{e}^{2x-1}\)

A1: Obtain correct gradient, following correct derivative. ‘Show that’ so correct derivative must be seen first

B1: State, or use, correct \(y\)-value

A1: Use correct numerical \(m\), \(x\) and \(y\) values to obtain given equation of tangent. Could also use \(y = mx + c\) to obtain \(c = 0\) and hence given equation

Notes

B1: State correct integral. Any equiv exact form. Could be in terms of \(u\) if using substitution

M1*: Attempt use of correct limits. Correct order and subtraction, working with exact values. Could be using \(u\) limits (must be \(-1\) and 0)

M1dep*: Attempt area of region ie their definite integral minus attempt at area of triangle. Could be implied by their definite integral minus 0.25 www. Must be using exact values
Area of a triangle must either be \(\frac{1}{2} \times \frac{1}{2} \times\) their \(y\) or from \(\displaystyle\int_0^{0.5} 2x\,\mathrm{d}x\)

A1: Correct area, in any exact form with like terms combined. Must have worked exactly throughout

Alternative method (integrating wrt \(y\))

M1: Change subject and attempt full integration by parts(using correct parts)

M1*: Attempt use of correct limits. Correct order and subtraction, working with exact values. Must use correct \(y\) limits

M1dep*: Attempt area of region ie attempt at area of triangle minus their definite integral. Could be implied by 0.25 minus their definite integral www. Must be using exact values
Area of a triangle must either be \(\frac{1}{2} \times \frac{1}{2} \times\) their \(y\) or from \(\displaystyle\int_0^{1} \frac{1}{2}y\,\mathrm{d}y\)

A1: Correct area, in any exact form with like terms combined. Must have worked exactly throughout

Notes

Condone missing ‘\(\mathrm{d}x\)’ or ‘\(\mathrm{d}u\)’ throughout all methods.

M1*: Attempt integration by parts – must have correct parts, possibly with \(\int x \times (-2) \times \frac{1}{3 - 2x}\,\mathrm{d}x\) not yet simplified. Allow a sign error or missing (−)2. No limits needed for M1M1A1 (first 3 marks)

M1** dep*: Attempt partial fractions; allow one sign error

M1 dep**: Attempt substitute correct limits \(0, \frac{1}{2}\) and subtract (allow one slip). FT their integral. Dependent on both previous M marks.

A1: For any correct exact form, including \(\left(\frac{1}{2}\ln 2 - \frac{1}{2} - \frac{3}{2}\ln 2\right) - \left(-\frac{3}{2}\ln 3\right)\), need not be simplified. ISW.
Answer \(0.45477\ldots\) scores A0 unless exact answer also seen

Alternative method using substitution following correct setup of integration by parts

If candidates apply integration by parts and then use substitution, apply this method. (NB candidates can switch from this back to the main method if they revert correctly from \(u\) to \(x\) so look for limits used correctly in either \(x\) or \(u\) as appropriate).

M1*: Attempt integration by parts – must have correct parts, possibly with \(\int x \times (-2) \times \frac{1}{3 - 2x}\,\mathrm{d}x\) not yet simplified. Allow a sign error or missing ‘2’. No limits needed for M1M1A1 (first 3 marks)

M1** dep*: Using e.g. \(u = 3 - 2x\), allow one sign error.

A1: Ignore any additional constant term e.g. \(+c\) or \(+k\) for any \(k\).

M1 dep**: Attempt to substitute transformed limits 3, 2 and subtract (condone swapped limits). FT their integral. Dep on both previous M marks.

A1: For any correct exact form, including \(\left(\frac{1}{2}\ln 2 - \frac{1}{2} - \frac{3}{2}\ln 2\right) - \left(-\frac{3}{2}\ln 3\right)\), need not be simplified. ISW.
Answer \(0.45477\ldots\) scores A0 unless exact answer also seen

Alternative method for Q8 applying substitution first

If candidates start by applying a substitution, then apply this method.

M1*: Attempting to integrate by substitution using e.g. \(u = 3 - 2x\) (no limits required for this mark). Allow a sign error or missing \(-\frac{1}{2}\).

A1: Fully correct integral with transformed limits (may be seen later, condone swapped limits).

M1 dep*: Use of integration by parts. This mark is not implied by later correct working – must see this form or e.g. \(\ln u\), \(\frac{1}{u}\), \(u\) and 1 as setup for integration by parts. Ignore limits for this mark. Condone a sign error or missing \(-\frac{1}{2}\).

A1: oe but must be correct

A1: For any correct form (as above). ISW
NB candidates who do not show evidence of integration by parts used anywhere can score max. 2/5 under this method.
Answer \(0.45477\ldots\) scores A0 unless exact answer also seen

Alternative method for Q8 assuming formula for \(\int \ln x\,\mathrm{d}x\) (without substitution) (Only 2 marks, since this Q is DR and required integration by parts)

If no evidence of integration by parts, or of substitution, then apply this method.

M1: Allow \(k\left[(3 - 2x)\ln(3 - 2x) - (3 - 2x)\right]_0^{\frac{1}{2}}\)

A1: ISW

Answer \(0.45477\ldots\) scores A0 unless exact answer also seen

Notes

B1: B1 for \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = -\sin x\) oe

M1*: Obtain an integral in terms of \(u\) only – must be of the form \(\displaystyle\pm\int \frac{\pm 1 \pm u}{\pm 1 \pm u^2}\,(\mathrm{d}u)\)

Allow \(\mathrm{d}u\) missing throughout but M0 if implying \(\mathrm{d}u = \mathrm{d}x\)

M1dep*: Explicit use of \(1 - u^2 = (1 + u)(1 - u)\) or \(u^2 - 1 = (u + 1)(u - 1)\) and simplify to an integral of the form \(\displaystyle\pm\int \frac{1}{\pm 1 \pm u}\,(\mathrm{d}u)\)

A1: AG www – condone \(\ln|1 - \cos x| + c\)
Allow A1 for \(\displaystyle-\int \frac{1}{1 - u}\,\mathrm{d}u = \ln(1 - u) = \ln(1 - \cos x) + c\) (must see \(\ln(1 - u)\) as AG)
However, \(\displaystyle\int \frac{1}{u - 1}\,\mathrm{d}u = \ln|\cos x - 1| = \ln(1 - \cos x) + c\) is A0 without further correct working or correct explanation/justification
Note that \(\displaystyle\int \frac{1}{u - 1}\,\mathrm{d}u = \ln(\cos x - 1) = \ln(1 - \cos x) + c\) is A0 regardless of any further explanation/justification

\(+c\) must be present in the final given answer
e.g. as a minimum \(\ln|\cos x - 1| = \ln|(-1)(1 - \cos x)| = \ln(1 - \cos x) + c\)

Notes

B1: Possibly implied by later working – condone (use of) 180 (in degrees) or \(\cos^{-1}(-1)\) (so do not need to see this evaluated to \(\pi\))

e.g. as a limit on an integral

M1*: Correct use of their upper limit and correct lower limit to form an equation in \(a\) only so must be of the form \(\ln k - \ln(1 - \cos a) = a\) following through their value of \(k \gt 0\)

Setting equal to \(a^2\) rather than \(a\) is M0

M1dep*: Correct use of log laws and correctly remove natural log to form an equation in \(\mathrm{e}^a\) from \(\ln k - \ln(1 - \cos a) = a\) following through their \(k \gt 0\) and \(k \neq 1\)

A1: AG – sufficient working must be shown and must be \(= 0\)

allow arccos for \(\cos^{-1}\)

Notes

M1: Obtain correct value for either f(1.1) or f(1.2) to at least 1 sf rot

A1: Both correct values to at least 1 sf rot, ‘change of sign’ either shown or stated + conclusion (minimum e.g. ‘root’)

Allow, for example,’ \(-0.1 \lt 0 \lt 0.03\) therefore \(1.1 \lt a \lt 1.2\)’ for A1

Notes

B1: At least the first three values of \(a_i\) stated correctly to at least 2 decimal places rot e.g. \(a_3 = 1.18\) or \(a_3 = 1.19\)

Ignore values after the first three

B1: cao (must be to exactly 2 decimal places) – final answer must be 1.18 or \(a = 1.18\) and not a term of a sequence e.g. \(a_9 = 1.18\)

Not dependent on the previous B1. A correct answer without the correct first three terms is B0 B1

Notes

B1: Correct explanation that the graph or diagram or curve (in Fig. 2) implies that the gradient (either at \(a\) or implying all points) is negative.

Must mention ‘the graph’ oe for this mark but mention of Fig. 1 is B0

B1: Correct justification that as the iterative formula \(\left(a_{n+1} = \mathrm{F}(a_n)\right)\) converged to \(a \Rightarrow |\mathrm{F}^{\prime}(a)| \lt 1\).
As a minimum must state/mention ‘convergence’ and EITHER that the gradient at \(a\) is between \(-1\) and 1 OR that the gradient at \(a\) is between \(-1\) and 0 but only following a correct explanation that the graph (oe) implies a negative gradient

Not \(\Rightarrow |\mathrm{F}^{\prime}(a_n)| \lt 1\).
For both marks to be awarded the interval of (–1,0) must be mentioned

B2 for ‘graph shows negative gradient, and there is convergence, therefore the gradient at \(a\) lies in the interval (–1,0)’

Implying that \(a\) is between –1 and 1 and not the gradient is B0 for the second B mark

Notes

(a)(i)

M1: Correct formula oe with \(\geqslant 4\) \(y\)-values correct or omitting the \(2 \times\) May see \(\frac{1}{2} \times 0.5 \times \{1.732\ldots + 2.645\ldots + 2(1.803\ldots + 2 + 2.291\ldots)\}\)

A1: All correct. May see \(\sqrt{\frac{13}{4}}\) and \(\sqrt{\frac{21}{4}}\) oe

A1: awrt 4.14
Answer only without working scores 0/3

(a)(ii)

B1: \(0.5 \times (1.7321 + 1.8023 + 2 + 2.2913)\) awrt 3.91

B1: \(0.5 \times (1.8028 + 2 + 2.2913 + 2.6458)\) awrt 4.37

Allow B1B1 for these values seen anywhere (e.g. may see lower/upper switched). Must be evaluated.

Notes

M1: M1 for a partial or incomplete expression, e.g. missing (0,2) \(\displaystyle\lim_{\delta x \to 0} \sum \left(\sqrt{x^2 + 3}\right)\delta x\). Allow any expression involving \(\lim\), \(\Sigma\), \(\sqrt{x^2 + 3}\) (or \(y\)) and \(\delta x\) for this mark.

A1: oe (must be in terms of \(x\), not \(y\)).

Notes

M1: Attempt \(\displaystyle\int x\frac{\mathrm{d}y}{\mathrm{d}t}\,\mathrm{d}t\) in terms of \(t\), detail required
Clear indication that integrand is given by \(\displaystyle\int x\frac{\mathrm{d}y}{\mathrm{d}t}\,\mathrm{d}t\) (condone just \(\displaystyle\int x\frac{\mathrm{d}y}{\mathrm{d}t}\)), along with \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 4t + 3\) and full substitution into integrand
Condone no d\(t\) in integrand in initial statement and/or when substituting
May instead see integrand as \(\displaystyle\int x\,\mathrm{d}y\) with \(\mathrm{d}y = (4t + 3)\,\mathrm{d}t\)

A1: Obtain correct given integrand
d\(t\) required throughout

B1: Determine correct lower limit from solving equation
Evidence for \(a = 0\) required eg \(2t^2 + 3t = 0\)
\(a = 0\) doesn’t need to be seen explicitly, and could be implied by 0 appearing as the lower limit on an integral sign once sufficient evidence seen
Mark independently of any integrand attempted

M1: Equate expression for \(y\) to 2
Could be implied by \(t = \tfrac{1}{2}\) seen as a limit

A1: Obtain \(b = \tfrac{1}{2}\) as upper limit www
No need for \(t = -2\) to be explicitly rejected
\(b = \tfrac{1}{2}\) doesn’t need to be seen, and may be implied by appearing as the upper limit on an integral sign
Mark independently of any integrand attempted

Notes

M1*: Use \(u = 2t + 1\) to attempt to change entire integrand to a function of \(u\)
Attempt to write numerator, denominator and d\(t\) in terms of \(u\)

A1: Obtain correct integrand
Condone no d\(u\)

M1*: Attempt integration to obtain integral of form \(au^{-2} + bu^{-3}\)
M0 if additional terms

M1d*: Attempt use of correct limits: either correct \(t\) limits (ie \(a = 0\) and \(b = \tfrac{1}{2}\)) in a \(t\)-integral or commensurate upper and lower limits in an integral involving a substitution (eg with \(u = 2t + 1\), then upper limit must be 2 and lower limit must be 1)
Dependent on M1 M1
Minimum evidence needed is two terms ie \(\left(-\dfrac{7}{24}\right) - \left(-\dfrac{4}{3}\right)\) or \(\left(\dfrac{3}{4}\right) + \left(\dfrac{7}{24}\right)\)
If these values are not seen then M1 can be awarded for term by term substitution seen (ie 4 terms needed), but allow one error

A1: Obtain correct area, any exact equivalent
Explicit use of limits must be seen in a correct integral for A1

Candidates may mix and match methods eg start with substitution and then try to do the actual integration by parts – the MS allows M1A1 for changing the integrand to useable form; M1A1 for doing the integration; M1A1 for use of limits

Alternative method (integration by parts)

M1*: Attempt integration by parts
Correct parts and correct formula

A1: Obtain correct first step
Allow unsimplified

M1*: Attempt integration to obtain integral of form \(a(8t + 6)(2t + 1)^{-3} + b(2t + 1)^{-2}\)

A1: Obtain fully correct integral
Allow unsimplified

M1d*: Attempt use of correct limits
See guidance in main MS

A1: Obtain correct area, any exact equivalent

Alternative method (separate fractions)

M1*: Attempt to rewrite integrand as separate fractions with constant numerators
Could be informal method, or use of partial fractions (extending expected knowledge)
As far as \(\dfrac{P}{(2t + 1)^3} + \dfrac{Q}{(2t + 1)^4}\), with \(P\) and \(Q\) as constants, and no other fractions

A1: Obtain correct integrand

M1*: Attempt integration to obtain integral of form \(\dfrac{a}{(2t + 1)^2} + \dfrac{b}{(2t + 1)^3}\)

A1: Obtain fully correct integral
Allow unsimplified

M1d*: Attempt use of correct limits
See guidance in main MS

A1: Obtain correct area, any exact equivalent

Alternative method (integrating between curve and \(x\)-axis)

M1*: Attempt integration by substitution / integration by parts on correct expression
Apply the same MS as for integrating between curve and \(y\)-axis

A1: Obtain correct integrand
Using substitution eg \(u = 2t + 1\)
Using integration by parts – first stage required for M1

M1*: Apply the same MS as for integrating between curve and \(y\)-axis

A1: Obtain fully correct integral
Allow unsimplified

M1d*: Attempt use of correct limits, and combine with correct area of rectangle \(\left(= \tfrac{1}{4}\right)\)
Limits must be \(\displaystyle\int_{\frac{1}{2}}^{0}\) or commensurate \(u\)-limits, and used in the correct order
See guidance in main MS, but must also add on the correct area of the rectangle

A1: Obtain correct area, any exact equivalent

Notes

B1: Must see use of decrease/decline linked to the negative sign.
Condone answers that do not refer to \(\frac{\mathrm{d}P}{\mathrm{d}t}\)

Examples:

• “it’s \(-\frac{1}{80}P^2\) because decreasing” B1
• “decreasing” [and nothing further] B0
• “decline implies negative sign” B1

Notes

M1*: Attempt to separate variables, must see \(P\) and d\(P\) on the same side
May see \(\frac{\mathrm{d}t}{\mathrm{d}P} = \frac{-80}{P^2}\) as a first step. In this case M1 is implied by a subsequent attempt to integrate the RHS w.r.t. \(P\) (so do not give M1 for this statement alone).

A1: May see \(\frac{1}{P} = \frac{t}{80}\,(+c)\) etc. Allow without \(+c\)

M1 dep*: Substitute (0, 120) and attempt to find \(c\) (which may not be this value). Must reach a value of \(c\) for this mark.

A1: Must be in terms of \(P\) (\(P\)=…) but need not be simplified.
isw incorrect attempts to simplify following \(P\)=…

Notes

M1: Either substitute \(P = 10\) into their equation from (b) or rearrange to make \(t\) the subject.

A1: Accept 7.33 (3sf), ignore units.

Notes

B1: Allow \(t = 160 \Rightarrow P = 0.5\)
(Use of the value \(t\)=160 not required)
Any correct statement in context relating to the model predicting non-integer values between 0 and 1. (Note that the statement ‘will never reach zero’ is given in the question so gains no credit).
Acceptable examples (for a comment, in combination with the statement that \(P\) is small for large \(t\)):

• “cannot have part of an animal”
• “the actual population has reached 0”
• “not possible to have a population of 0.49(8)”
• “the population will go extinct when less than 1”

Notes

(a)(i)

M1: M1 for attempt to differentiate using the product rule (oe) – answer must be of the form \(\left(x^3 - 2x^2\right) \times \dfrac{k_1}{x} + \left(k_2x^2 + k_3x\right)\ln x\) for non-zero constants \(k_1, k_2, k_3\)
Condone invisible brackets for this mark

A1: A1 for a correct first derivative (allow un-simplified)
Condone invisible brackets but only if correctly recovered at some stage

A1: A1 for a correct second derivative (allow un-simplified)
Condone invisible brackets but only if correctly recovered at some stage

M1: Setting the second derivative (which if simplified would be of the form \(ax + b + (cx + d)\ln x\) with non-zero constants \(a\), \(b\), \(c\) and \(d\)) equal to zero

A1: AG – so sufficient working must be shown – at least one intermediate line of working from second derivative set equal to zero to given answer
Any errors seen (e.g. any missing/invisible brackets) is A0

(a)(ii)

B1: Uses given result and given starting value (of 1.1) to obtain correct \(x_2\) and \(x_3\) (so first two iterations after the initial value of 1.1) to at least 2 dp (rot) – but all stated values in these two terms must be correct

B1: Must be stated to 2 dp only – not dependent on the first B mark – can be awarded if either of \(x_2\) and \(x_3\) are incorrect (assume that the iterative process corrected itself or a slip in the candidate writing down an earlier value)
Must be clear that \(x\) is 1.13 (and not the final term shown in the iterative process e.g. \(x_6 = 1.13\) only is B0) – this mark can be awarded from using alternative iterative methods e.g. Newton-Raphson

Notes

B1*: Correct \(x\)-intercepts (soi) – ignore mention of \(x = 0\)
Could be seen as limits on integral(s)

M1*: M1 for attempt at integration by parts – must be of the form \(\left(ax^4 + bx^3\right)\ln x \pm \int \left(cx^4 + dx^3\right) \times \frac{1}{x}\,(\mathrm{d}x)\) for non-zero constants \(a\), \(b\), \(c\) and \(d\)
Limits not required for this and the next two A marks (so condone incorrect limits too for these 3 marks)

A1: correct first application (allow un-simplified)
\(\mathrm{d}x\) not required and integral sign(s) can be implied

A1: cao (allow un-simplified)

M1dep*: Uses correct limits completely correctly \(\pm(F(2) - F(1))\) in their fully integrated expression – need not be simplified (or equivalent)
Do not condone invisible brackets unless recovered

A1: Must be of this form but allow exact equivalents (for example, \(\frac{1}{3}\ln 16 - \frac{89}{144}\)) but the \(p\) and \(r\) must be positive rational numbers and \(q\) must be a positive integer
For reference – one possibility is: \(p = \frac{4}{3}, q = 2, r = \frac{89}{144}\)
These values do not need to be stated explicitly
Be aware of those who consider \(\int_1^2 \left(2x^2 - x^3\right)\ln x\,\mathrm{d}x\) which is correct

Notes

M1: Values of iterates not needed, i.e. condone \(\lt 0\) and \(\gt 0\) etc. but signs (and values if given) must be correct to 1sf – see table.

B1: cao (Allow this mark even with no/insufficient working).

A1: By showing two values between 0.785-0.795 with opposite signs
Values of iterates not needed, i.e. condone \(\lt 0\) and \(\gt 0\) etc, but if given must be correct to 1sf
e.g. \(x = 0.79 \Rightarrow y = 0.0007\) and \(x = 0.795 \Rightarrow y = -0.0009\)

(Note that exact root is 0.792…)

Notes

M1: For an attempt at differentiating using the product rule, with at least one term correct (their derivative must have two terms).

B1: May be implied by not giving a corresponding solution (provided another solution for \(x\) is reached). (NB this mark can be gained following M0 provided their derivative has \(\mathrm{e}^{1-3x}\) as a factor)

A1: Not decimal, www (i.e. must come from a correct derivative)

Notes

M1: M1 for attempting integration by parts with at least one term correct, must see limits (condone swapped limits, may appear later). Must have two terms.

A1: A1 for both terms correct after 2^nd integral performed

A1: Need not be simplified, isw any incorrect attempts to simplify

Notes

B1: State the 4 correct non-zero \(y\)-values and no others.
Exact values (including unsimplified) or decimal equivs (0, 0.1516, 0.3679, 0.5020, 0.5413), which could be truncated or rounded.
For the first value, if \(0\mathrm{e}^0 = 1\) is seen then allow credit for the unsimplified value; if however it is only ever seen as 1 then this is B0 but M1M1 could still be awarded.
B0 if other ordinates seen, unless clearly not intended to be used.

M1*: Big brackets need to be seen or implied.
Attempts at \(y\)-values must be correctly placed (but no need to see \(y = 0\) explicitly).
If no earlier evidence of \(y\)-values seen (eg in a table) then allow M1 for the correct structure with 4 of the 5 values being correct.
Condone using more than 4 intervals as long as values equally spaced between \(x = 0\) and \(x = 2\).

M1d*: Dep on previous M1.
Or using \(k = 0.5h\), with \(h\) consistent with their different number of intervals.

A1: Obtain 0.646.
Allow answers > 3sf, as long as they round to 0.646.
A0 if not using 4 strips, even if 0.646 is obtained.
No credit if no evidence of using the trapezium rule shown.

Using separate strips (a triangle and then trapezia) is an acceptable method, and marks should be awarded as per the main MS (ie \(y\)-values / structure / widths / final answer).

Notes

B1: Convincing reason.
Allow just ‘more trapezia’ or ‘narrower trapezia’.
Could refer to strips or intervals.

Notes

B1: Curve is both convex and concave.
Comment about the shape
Referring to increasing and decreasing gradients is correct, but increasing and decreasing curve is not.
Allow BOD if muddles about which part of the curve is convex and which is concave.

B1: The tops of trapezia are both above and below the curve.
Comment about the estimates
If candidates refer to ‘it’ rather than ‘trapezia’ then allow BOD.

B marks are independent.
See appendix for further examples (below).

Appendix: exemplar responses for Q1(c)

Notes

B1: Correct statement linking \(\mathrm{d}u\) and \(\mathrm{d}x\)
or \(\mathrm{d}x = \dfrac{1}{u + 2}\,\mathrm{d}u\)

M1: Use \(\mathrm{e}^x = u + 2\) to attempt integrand in terms of \(u\)
Must see clear evidence of substitution, including how \(\mathrm{e}^x\mathrm{d}x\) is dealt with
M0 for going straight from \(7\mathrm{e}^x - 8\) to \(7u + 6\) with no justification
Must include \(\mathrm{d}u\)

A1: Correct integrand
Including both integral sign and \(\mathrm{d}u\) throughout, as AG

Notes

M1: Attempt correct partial fractions
May have \(\dfrac{Au + B}{u^2} + \dfrac{C}{u + 2}\) but M0 for just \(\dfrac{B}{u^2}\) with no \(\dfrac{A}{u}\)
Correct method to combine correct fractions, and at least one constant attempted
If considering \(\dfrac{7}{u^2} + \dfrac{-8}{u^2(u + 2)}\) then must use partial fractions on the second term to get credit

A1: Correct partial fractions
May have \(\dfrac{2u + 3}{u^2} - \dfrac{2}{u + 2}\)
Possibly implied by their \(A\), \(B\), and \(C\) values ie \(A = 2\), \(B = 3\), \(C = -2\)

M1: Attempt integration of \(\dfrac{B}{u^2}\) and at least one of \(\dfrac{A}{u}\) or \(\dfrac{C}{u + 2}\), and no others
Allow errors in coefficients only
Allow M1 if only two fractions, as long as of required form
If using \(\dfrac{Au + B}{u^2}\) then it must be a correct integration attempt (ie split into two fractions first)

A1FT: FT on their two or three fractions as long as \(ku^{-2}\) and one or two fractions each with a linear denominator
Condone brackets not modulus
Condone no brackets as long as implied by later working, eg when limits are used

M1: Attempt use of correct limits – correct order and subtraction; \(u\) or \(x\) but commensurate with their integral
Allow substitution into any function that is clearly attempt at integration

M1: Attempt to rearrange correct numerical integral to required form
Must be correct numerical expression from correct working
Terms may have been combined before use of limits, but must still be correct expression to gain M1
Correct attempt to combine ln terms ie deal with coefficients and correct product / quotient for the sum / differences
Allow one slip

A1: Obtain \(\dfrac{3}{4} + \ln\dfrac{16}{9}\)
Condone \(\dfrac{3}{4} + 2\ln\dfrac{4}{3}\)
Fractions must be simplified
ISW an incorrect attempt to write this answer in a different form, but A0 if further work done eg multiplying by a constant to clear the fractions

Notes

M1*: Setting equation for \(a\) equal to zero and removing \(t^2\) correctly from the denominator e.g. \(8t - \frac{1}{2}(7 + 4t^2) = 0\) to obtain the equivalent of a 3TQ in \(t\) only
This mark can be implied by a correct 3TQ in \(t\)

M1dep*: Correct method for solving their 3TQ in \(t\)
If factorising: \(at^2 + bt + c \Rightarrow (mt + n)(pt + q)\) where \(a = mp\) and one of \(mq + np = b\) or \(c = nq\) so note that \(4t^2 - 16t + 7 = (t - 0.5)(t - 3.5)\) is M0 but e.g. \((-2t + 7)(2t - 1)\) is M1 bod
If using the formula: must apply the correct formula for their three-term quadratic in \(t\) (no errors)
If completing the square: The M mark is not awarded until correctly getting to the stage \(t - 2 = \pm\sqrt{\dfrac{9}{4}}\) for their 3TQ in \(t\) (must include \(\pm\) so implying two roots) with no errors (so consistent with applying the formula correctly)
Must see the method – the correct answers do not imply this mark therefore \(4t^2 - 16t + 7 = 0 \Rightarrow t = 0.5\) and 3.5 scores M1 M0 B1
As a minimum must see (if correct) \(\dfrac{16 \pm \sqrt{144}}{8}\)

B1: This mark is not dependent on the previous M mark(s)
So M1 M0 B1 is common or M0 M0 B1 if no working seen

Notes

B1*: Stating the correct differential equation
Possibly implied by correct separation of variables

B1dep*: Correct lhs

B1dep*: Correct rhs – not multiplied by \(v\) or 5 (or any other constant)
Condone lack of \(+c\) for both B marks

M1*: Uses correct initial conditions to find \(c\) from an equation of the form \(k_1\ln v = k_2\ln(7 + 4t^2) + k_3t + c\) (note that e.g. \(+\,c\) may appear on the lhs)
With non-zero values of \(k_i\) - accept any equivalent form e.g. \(v = A(7 + 4t^2)\mathrm{e}^{-0.5t}\) and then use initial conditions to find \(A\) (if correct then \(A = 2.5\))

M1dep*: Uses \(t = T\), \(v = 5\) to obtain an equation in \(T\) only – dependent on previous M mark
Condone use of \(t\) for \(T\) throughout the remainder of the question

A1: AG so at least one step of intermediate working from substitution of \(t = T\) and \(v = 5\)
Condone \(T = 2\ln\left|\dfrac{7 + 4T^2}{2}\right|\)

Notes

B1: Uses given result and given starting value to obtain correct \(T_1\) and \(T_2\) (so the first two iterations after the initial value of 11.25) to at least 4 sf (rot) – but all stated values in these two terms must be correct

B1: Must be stated to 4 sf only – not dependent on the first B mark – can be awarded if either of \(T_2\) and/or \(T_3\) incorrect (assume that the iterative process corrected itself or a slip in the candidate writing down an earlier value)
Must be clear that \(T\) is 11.01 (and not the final term shown in the iterative process) – this mark can be awarded from using alternative iterative methods e.g. Newton-Raphson

Notes

B1: awrt 7.51
No follow through from incorrect earlier values

Notes

B1*: B1 for \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0.2 \pm \sin t\)

M1dep*: Uses \(\displaystyle\int y\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t\) and replaces \(\sin^2 t\) with \(1 - \cos^2 t\) to obtain an expression involving \(\cos^2 t\) and \(\sin t\)
No limits required for this mark

A1: AG (so must be checked carefully for any errors e.g. must contain relevant brackets around the \(1 - \cos^2 t\) term(s)). A correct expression e.g. \(\displaystyle\int_0^{\pi} k(1 - \cos^2 t)(0.2 + \sin t)\,\mathrm{d}t\) followed by the correct given answer can score this mark – limits and d\(t\) must be seen at least once (but need not be on the final integral)
Do not need to see \(= 1\) anywhere in their solution (and condone lack of brackets around the integrand)

Notes

B1: Correctly applies \(2\cos^2 t \equiv 1 + \cos 2t\) (so not just stating this identity) - implied by seeing \(-0.1t - 0.05\sin 2t\) after integration
e.g. applies could be for an attempt to integrate \(\frac{1}{2}(1 + \cos 2t)\) or stating this identity in an integral

B1: First two terms integrated correctly (Look out for those that have \(0.1t\) only from combining \(0.2t\) with \(-0.1t\))
This mark should be awarded if 0.2 is combined with another constant term and integrated correctly

M1*: M1 for an answer of the form \(\pm pt \pm q\sin 2t \pm r\cos^3 t\) or an answer of the form \(\pm pt \pm q\sin 2t \pm r\sin t\sin 2t \pm u\cos t\cos 2t\) or an answer of the form \(\pm pt \pm q\cos t \pm r\sin 2t \pm u\sin t\sin 2t \pm v\cos t\cos 2t\) for non-zero \(p\), \(q\), \(r\) (and \(u\), \(v\)) from integrating \(-0.2\cos^2 t - \sin t\cos^2 t\)

A1: A1 for the correct remaining three/four terms
(Alternatives: \(-0.1t - 0.05\sin 2t + \frac{1}{6}\sin t\sin 2t + \frac{1}{3}\cos t\cos 2t\) or \(-0.1t + 0.5\cos t - 0.05\sin 2t - \frac{1}{3}\sin t\sin 2t - \frac{1}{6}\cos t\cos 2t\))

M1dep*: Uses correct limits correctly \(F(\pi) - F(0)\) - condone limits the wrong way round only if the sign of their answer is subsequently changed
Must be using exact values (so must include term(s) in \(\pi\))

A1: www oe e.g. \(k = \dfrac{1}{0.1\pi + \frac{4}{3}}\) - isw once a correct expression for \(k\) seen
Correct exact value must be seen at some stage for the final mark – check first A mark carefully

Notes

M1: Attempt expression for \(\mathrm{f}(x + h) - \mathrm{f}(x)\)
Allow sign error from no bracket around final term, ie \((x + h)^2 + 2(x + h) - x^2 + 2x\) is M1, but no other errors allowed
If considering \(x^2\) and \(2x\) separately then expressions for both must be seen

M1: Expand and simplify \(\mathrm{f}(x + h) - \mathrm{f}(x)\)
Expand and gather like terms (either separately, or single expression)
Condone sign errors only, so M0 if collecting like terms after an incorrect attempt to divide by \(h\)
Allow BOD if \(2x \ldots + 2x\) becomes 0 rather than \(4x\)

M1: Attempt \(\dfrac{\mathrm{f}(x + h) - \mathrm{f}(x)}{h}\)
Divide all terms by \(h\)
Allow BOD if previous error results in a term with a denominator of \(h\)

A1: Complete proof by considering limit as \(h \to 0\)
www, including correct signs throughout
Must divide by \(h\) before \(h \to 0\)
Must see ‘lim’, ‘\(h \to 0\)’, and \(\mathrm{f}'(x)\) at some point in their solution and not just when quoting the generic formula, but allow BOD for \(\mathrm{f}'(x) = \frac{\mathrm{f}(x+h) - \mathrm{f}(x)}{h}\) followed by =…, =…, =… on subsequent lines
A0 if ‘lim’ still in final answer
Condone \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in place of \(\mathrm{f}'(x)\)

Notes

B1: State or imply correct equation, including \(+\,c\)
‘\(y =\)’ could be implied by use of 5
\(c\) may be implied by later work

M1: Attempt \(c\) using \((-1, 5)\)
Allow M1 if equation incorrect, as long as from attempt at integrating \(2x + 2\) ie of form \(y = kx^2 + 2x + c\)
\(c\) may be implied by method eg \(y = x^2 + 2x\), followed by \(5 = 1 - 2\) and then an attempt to ‘balance’ the sides
Must use \(x\) and \(y\) the correct way around
As far as attempting a value for \(c\)

A1: Obtain correct equation, including \(y = \ldots\)
Equation must be stated, and not just implied by \(c = 6\) seen
Allow \(\mathrm{f}(x) = \ldots\)
A0 for ‘equation’ \(= x^2 + 2x + 6\)

Just stating \(y = x^2 + 2x + 6\) or \(y = (x + 1)^2 + 5\) gets full marks (may come from observing that \((-1, 5)\) is the minimum point)

Notes

M1: Integrate and obtain expression involving ln. This may be combined with the next step. Ignore limits. Brackets \((x + 2)\) soi
If a substitution is used then only award this mark when \(\ln(x + 2)\) reached or an equivalent integral with appropriate limits (typically \(\ln x\) with limits 2 and 4.5).

M1: Substitute and use limits 0 & 2.5 (or appropriately changed limits in the case of a substitution) in their ln integral. Must see this step.

A1: www, any equivalent exact form, but must be a single term expression. Do not accept \(\ln\frac{4.5}{2}\) for this mark.

Correct answer with no working SC B1 [1/3]
Ignore use of modulus signs throughout.

Notes

M1: Separate variables and attempt integration of at least one side
Allow \(k\mathrm{e}^{3y}\), with \(k \ne 1\), as ‘attempt’ at integration of LHS
‘Attempt’ at RHS may not be use of integration by parts
Allow BOD on missing integral sign / missing \(\mathrm{d}y\) / missing \(\mathrm{d}x\) as long as intention clear

B1: Correct LHS
B0 if still part of an expression also involving \(x\)

M1: Attempt integration by parts on RHS – must have correct parts
As far as attempt at \(x^3\ln x - \displaystyle\int x^2\,\mathrm{d}x\), possibly with \(\displaystyle\int \frac{1}{x}x^3\,\mathrm{d}x\) not yet simplified

A1: Correct RHS (condone no \(+ c\))
Condone no modulus sign on \(\ln x\)

M1: Attempt use of \((\mathrm{e}, 1)\) to find \(c\)
Used in an equation involving \(x\), \(y\) and \(c\), following some integration attempt of both sides
As far as finding \(c\), either exact or as a decimal
M1 can be implied by sight of \(-\frac{1}{3}\mathrm{e}^3\) or \(-6.695\ldots\) following a correct equation

A1: Obtain correct equation, in required form
Any equivalent form on the RHS, but must be \(\mathrm{e}^{3y} = \ldots\)
A0 if decimal approximation for \(\mathrm{e}^3\)

Notes

M1*: Substitute \(x = \mathrm{e}^2\), into their integral involving \(\ln x\), and attempt to simplify
\(\ln x\) may be \(\ln x^p\) if any coefficient has been taken into the ln term
As far as correctly simplifying the ln term to remove ln
Must be working exactly, so M0 if decimals seen before ln dealt with

M1dep*: Introduce logs correctly, and attempt to rearrange to given form
Their equation must have two terms, or possibly more, with the terms having a common factor of \(\mathrm{e}^k\)
Attempt must go as far as splitting into the sum of two terms, with \(\ln\mathrm{e}^k\) simplified to \(k\)

A1: Obtain \(y = 1 + \frac{1}{3}\ln(5\mathrm{e}^3 - 1)\)
No need to state \(a\), \(b\) and \(c\) explicitly

Notes

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

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

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

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

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

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

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

Notes

B1: Correct substitution

M1: NOT if \(h = 50\) or \(r = 50\tan 30\) used

M1: Resulting equation must involve exactly 2 variables

M1: Their equation must involve exactly 2 variables

A1: Ignore limits

M1: Integrals must be of correct forms (see examples below)

Note 1 Candidates who substitute numerical values for \(h\) or \(V\) or \(r\) may be able to score the 2^nd and/or 3^rd M1 marks, but probably nothing else. See the example of this below.

Note 2. There is a special case for candidates who use \(r = h\sin 30\) (answer \(\frac{625\pi}{4}\) or 491).
These can score all 4 M-marks and the final A1

Note 3. The chain rule may be used to find \(\frac{\mathrm{d}V}{\mathrm{d}t}\) or \(\frac{\mathrm{d}h}{\mathrm{d}t}\) or \(\frac{\mathrm{d}V}{\mathrm{d}h}\) or \(\frac{\mathrm{d}V}{\mathrm{d}r}\) or other derivatives. Two of the example methods below illustrate use of \(\frac{\mathrm{d}V}{\mathrm{d}t}\) and \(\frac{\mathrm{d}V}{\mathrm{d}r}\), but use of other derivatives can also lead to correct methods.

Example method 1

B1: or \(V = \frac{\pi}{9}h^3\) oe

M1: Attempt differentiate their \(V\) in terms of \(h\) only
NOT if \(h = 50\) or \(r = 50\tan 30\) used.

M1: Attempt use chain rule for \(\frac{\mathrm{d}V}{\mathrm{d}t}\) or \(\frac{\mathrm{d}h}{\mathrm{d}t}\) in terms of \(t\) & \(h\) only
(Set their \(\frac{\mathrm{d}V}{\mathrm{d}t} = -2h\))

M1: Attempt separate variables in their equation in terms of \(h\) and \(t\) only (not \(V\) or \(r\)). Integral signs not essential

A1: Correct integrals, any limits or none

M1: Substitute correct limits into integrals of forms \(ah^2\) & \(bt\)
OR substitute \(t = 0\) & \(h = 50\) to find \(c\) and substitute \(h = 0\)

A1: Allow without secs   or 10.9 mins or 10 mins 54 secs
or:
SC. Use of \(r = h\sin 30\) (answer \(\frac{625\pi}{4}\) or 491) can score all 4 M-marks and final A1

Example method 2

B1: Subst \(h = \dfrac{r}{\tan 30^\circ}\) into correct formula for \(V\)

M1: Attempt use chain rule to find \(\frac{\mathrm{d}V}{\mathrm{d}t}\) or \(\frac{\mathrm{d}r}{\mathrm{d}t}\) in terms of \(t\) and \(r\)
(Set their \(\frac{\mathrm{d}V}{\mathrm{d}t} = -2r\sqrt{3}\) oe)

M1: Attempt separate variables in their equation in terms of \(r\) and \(t\) only (not \(V\) or \(h\)). Integral signs not essential

A1: Correct integrals, any limits or none

M1: Substitute correct limits into integrals of the form \(ar^2\) & \(bt\)
OR substitute \(t = 0\) & \(r = \frac{50}{\sqrt{3}}\) to find \(c\) and substitute \(r = 0\)

A1: Allow without secs   or 10.9 mins or 10 mins 54 secs
SC. Use of \(r = h\sin 30\) (answer 491) can score M4A1

Example method 3 (NOT using chain rule)

This method is different from the summary method above

B1: or \(V = \frac{\pi}{9}h^3\) oe

M1: Allow \(h = kV^{1/3}\)

M1: \(\frac{\mathrm{d}V}{\mathrm{d}t} = -2 \times\)(their \(h\) in terms of \(V\))

M1: Attempt separate variables in their equation in terms of \(V\) and \(t\) only (not \(h\) or \(r\)). Integral signs not essential

A1: Correct integrals, any limits or none

M1: Substitute correct limits into integrals of forms \(aV^{2/3}\) & \(bt\)
OR substitute \(t = 0\) & \(V = \frac{\pi 50^3}{9}\) to find \(c\) and substitute \(V = 0\)

A1: Allow without secs   or 10.9 mins or 10 mins 54 secs
or:
SC. Use of \(r = h\sin 30\) (answer \(\frac{625\pi}{4}\) or 491) can score all 4 M-marks and final A1

Example incorrect method

Notes

M1: M1 for \(k(2x + 9)^{\frac{3}{2}}\) with non-zero \(k\), \(k \ne 1\)

A1: cao (allow unsimplified)

A1: Uses correct limits (or implies correct limits) to get 9. Condone limits the wrong way round leading to \(-9\) but must be changed to \(+9\)

M1*: Attempt to solve \(4\mathrm{e}^{-2x} - 1 = 0\) by correctly taking logs of both sides leading to \(\pm 2x = \pm\ln\alpha\) where \(\alpha \gt 0\)
Allow sign errors and other minor slips only

A1: Or equivalent exact value (soi possibly by correct exact value used later) e.g. \(\frac{1}{2}\ln 4\) or \(\ln 2\)

M1*: Integrate \(4\mathrm{e}^{-2x} - 1\) to obtain \(c\mathrm{e}^{-2x} \pm x\)
Where \(c\) is non-zero and \(c \ne 4\)

M1dep*: Uses limits correctly \(F\left(\frac{1}{2}\ln 4\right) - F(0)\) (with their \(\frac{1}{2}\ln 4\)) – dependent on the previous two M marks (allow non-exact top limit). Condone limits the wrong way round only if the sign of their answer is subsequently changed
If zero limit is assumed to give 0 (with no working) then M0

A1: If the values of the integral(s) are changed from negative to positive (e.g. from limits the wrong way round) with no justification given then A0
\(p\) and \(q\) need not be explicitly stated. \(p = \dfrac{21}{2}\) (oe) and \(q = -1\)

Notes

B1: oe, eg They may have different constants of integration
Only the “\(c\)”s are different
Not “Both are correct” or “just different correct methods”

Notes

(b)(i)

M1: Attempt integral, must be of form \(k(1 + x)^{-1}\) or \(ku^{-1}\) or \(ku^{-0.5}\) (if from substitution \(u = (1 + x)^2\))
Ignore limits

M1: Attempt substitute appropriate limits into their integral

A1: cao oe single fraction

(b)(ii)

M1: or their new attempt at \(\displaystyle\int_1^a \frac{1}{(1 + x)^2}\,\mathrm{d}x = \frac{1}{3}\)

A1: cao

Notes

Allow incorrect use of brackets throughout
Allow \(\ln(\ldots)\) instead of \(\ln|\ldots|\)

M1: Allow \(k\ln(\sin 2x + 2)\), \(k\) any constant. Ignore limits

M1: Attempt substitute both correct limits into their log integral.
Allow numerical errors

A1: Allow \(\times\) any \(k\), otherwise any correct form without trig.

A1: Correct one-term exact result. ISW, eg ignore decimal
NB No working, no marks.

Alternative methods

M1: Attempt substitute and integrate and obtain \(k\ln u\) or \(k\ln(u + 2)\), \(k\) any constant; ignore limits
(May not see this step)

M1: Attempt substitute their limits into their log integral.
but not limits for wrong variable, eg not \(\ln\frac{\pi}{12} - \ln 0\)
Allow numerical errors

A1: Correct exact result, any form without trig. Allow \(\times\) any \(k\)

A1: cao, ISW, eg ignore decimal answer

Notes

B1: State the 4 correct non-zero \(y\)-values and no others.
Exact values (including unsimplified) or decimal equivs (0, 1.12, 1.73, 2.29, 2.83) – 3sf or better
B0 if other ordinates seen unless clearly not intended to be used

M1*: Big brackets need to be seen or implied
\(y\)-values must be correctly placed
Must be using attempts for at least 4 \(y\)-values (but no need to see \(y = 0\) explicitly)
Condone using other than 4 intervals as long as values equally spaced between \(x = 1\) and \(x = 3\)

M1d*: Dep on previous M1
Or using \(k = 0.5h\), \(h\) consistent with their different number of intervals

A1: Obtain 3.28, or better
Allow answers to > 3sf, as long as they round to 3.28