Notes

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

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

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

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

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

Notes

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

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

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

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

Notes

Mark (b) and (c) together.

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

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

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

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

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

Notes

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

May use \(t\) or another variable which is acceptable.

M1: Uses the model with \(v = 0\) and proceeds to \(\mathrm{e}^{0.2T} = 15\) Do not be concerned by the use of an inequality sign instead of an equals.
May be implied by awrt 13.5

A1: \((T =)\) awrt 13.5 units are not required but if given they must be seconds (or e.g. secs or s)

Notes

If no attempt is seen for (b) then allow differentiation seen in (a) to score in (b)

M1: Attempts to use the product rule to differentiate \(t\,\mathrm{e}^{0.2t}\) achieving the form \(A\mathrm{e}^{0.2t} + Bt\mathrm{e}^{0.2t}\) (\(A\) and \(B\) both non zero but may be 1) which may be unsimplified. It is likely to be part of an expression.

A1: \(\left(\dfrac{\mathrm{d}v}{\mathrm{d}t} =\right) 15 - \left(0.2t\,\mathrm{e}^{0.2t} + \mathrm{e}^{0.2t}\right)\) o.e. which may be unsimplified. (Condone a missing trailing bracket)
Do not allow recovery of signs to score this mark if they initially write
e.g. \(15 - 0.2t\,\mathrm{e}^{0.2t} + \mathrm{e}^{0.2t} = 0\) and on a later line correct this. e.g. \(15 - 0.2t\,\mathrm{e}^{0.2t} - \mathrm{e}^{0.2t} = 0\)

dM1: Sets \(15 \pm A\mathrm{e}^{0.2t} \pm Bt\mathrm{e}^{0.2t} = 0\) (the =0 may be implied), attempts to make \(\mathrm{e}^{\pm 0.2t}\) (or \(C\mathrm{e}^{\pm 0.2t}\)) the subject and proceeds to the form
\(C\mathrm{e}^{0.2t} = \dfrac{D}{E + Ft}\) or \(C\mathrm{e}^{-0.2t} = \dfrac{E + Ft}{D}\) (where \(C\) can be 1 and \(A, B, C, D, E, F \neq 0\))
It is dependent on the previous method mark.
May see \(15 - \text{``}0.2t\,\mathrm{e}^{0.2t} - \mathrm{e}^{0.2t}\text{''} = 0 \Rightarrow \mathrm{e}^{0.2t} = \dfrac{15}{0.2t + 1}\) which scores dM1
They must take out a factor of \(\mathrm{e}^{\pm 0.2t}\) (or \(C\mathrm{e}^{\pm 0.2t}\)) and divide by their bracket. Condone sign slips in their rearrangement, however, if they take logs of both sides first, the rearrangement must be correct (with no sign slips).
Allow invisible brackets to be implied by further work which is not the given answer.

A1*: Achieves the given answer with no errors seen including use of invisible brackets (but condone a missing trailing bracket) All previous marks in (b) must have been scored. The = 0 must have been seen somewhere in their solution. Do not allow this mark to be scored for proceeding directly from
\(\mathrm{e}^{0.2t}(0.2t + 1) = 15 \Rightarrow t = 5\ln\left(\dfrac{75}{t + 5}\right)\) which is A0*
We must see either \(\mathrm{e}^{0.2t} = \dfrac{15}{0.2t + 1}\) o.e. or an unsimplified expression for \(t\)
e.g. \(t = 5\ln\left(\dfrac{15}{0.2t + 1}\right)\) before achieving the given answer. Condone \(t = 5\ln\dfrac{75}{t + 5}\)

Notes

(c)(i) Check by the question. If there is a contradiction between answers, the answer in the main body of the script takes precedence.

M1: Attempts to use the iteration formula at least once. \(t_2 = 5\ln\left(\dfrac{75}{8 + 5}\right)\). May be implied by awrt 8.76 or awrt 8.48 or awrt 8.58. It is not implied by awrt 8.55

A1: awrt 8.478 (on its own can score M1A1)

(c)(ii) This mark can only be scored provided in (c)(i) M1 has been scored so M0A0A1 is not a possible mark profile.

A1: awrt 8.55 seconds (e.g. s or secs) Requires units.
If the candidate lists their iterations but does not select an answer then take the final value, which still requires units to be stated (which in many cases is likely to be omitted)
Note that awrt 8.55 does not imply M1.

Notes

The first two marks are the same for all approaches.

M1: Reduces the power by one for any term in \(\mathrm{f}(x)\). Look for \(x^n \rightarrow x^{n-1}\) and allow for \(ax \rightarrow a\) including \(ax^1 \rightarrow ax^0\) or \(-4x \rightarrow -4\)

A1: Correct differentiation \(\left(\mathrm{f}^{\prime}(x) =\right)\ 4x^3 + x^2 - 16x + a\) or \(4x^3 + x^2 - 16x - 4\)
Ignore the absence of the LHS. May be in terms of \(a\) or have \(-4\) substituted at this point as above.

Main Scheme

dM1: Substitutes \(x = -\dfrac{1}{4}\) into their \(\mathrm{f}^{\prime}(x)\), sets = 0 (= 0 may be implied) and solves for \(a\).
Condone \(4 + a = 0\) as evidence of substitution for the dM1 only.
This mark is not available if their \(\mathrm{f}^{\prime}(x)\) does not have a term with \(a\) in it using this approach.

A1*: cso Requires:

• a correct derivative
• correct substitution of \(x = -\dfrac{1}{4}\) into \(\mathrm{f}^{\prime}(x)\) and set = 0 (now the = 0 must be seen somewhere for this mark). Each term does not need to be evaluated.
• achieves \(a = -4\)

For the A1*, evidence of correct substitution might be \(-\dfrac{1}{16} + \dfrac{1}{16} + 4 + a = 0\) but not \(4 + a = 0\)
Condone invisible brackets around \(-\dfrac{1}{4}\) provided this is recovered before the given answer e.g. \(-\dfrac{1}{4}^2\) recovered to \(+\dfrac{1}{16}\)

Alt 1: Verification

dM1: Substitutes \(x = -\dfrac{1}{4}\) and \(a = -4\) into their \(\mathrm{f}^{\prime}(x)\), and finds a value.
The \(a = -4\) may already have been substituted into \(\mathrm{f}(x)\) as above.

A1*: cso Requires:

• a correct derivative
• correct substitution of \(x = -\dfrac{1}{4}\) into \(\mathrm{f}^{\prime}(x)\) and use of \(a = -4\) (at some stage) to achieve \(\mathrm{f}^{\prime}\left(-\dfrac{1}{4}\right) = \ldots = 0\). The substitution must be embedded in \(\mathrm{f}^{\prime}(x)\) or evaluated in an intermediate step e.g. \(-\dfrac{1}{16} + \dfrac{1}{16} + 4 - 4\) before the = 0
• a minimal conclusion e.g. “hence proven” or “\(\therefore a = -4\)” or e.g. a tick.
  but not e.g. “max = \(-\dfrac{1}{4}\)” or just “0 = 0”

Alt 2: Algebraic division

dM1: Attempts to divide their \(\mathrm{f}^{\prime}(x)\) by \((4x + 1)\) \(\left(\text{or } \left(x + \dfrac{1}{4}\right)\right)\), achieves a linear remainder in \(a\) only, sets = 0 and achieves a value for \(a\). Condone slips in their calculations.
As a minimum, expect to see \(x^2 + \lambda x + \mu\) \(\left(\text{or } 4x^2 + \lambda x + \mu\right)\) as their quotient (with \(\mu \neq 0\)) leading to a linear remainder in \(a\) only set = 0 leading to a value for \(a\).

A1*: cso Requires:

• a correct derivative
• correct quotient \(x^2 - 4\) \(\left(\text{or } \left(4x^2 - 16\right)\right)\) and correct remainder \(a + 4\) set = 0
• achieves \(a = -4\)

Notes

B1: \(\dfrac{4735}{768}\) cao Allow exact equivalents e.g. \(6\dfrac{127}{768}\) or the recurring decimal \(6.16536458\dot{3}\)
There is no need to see a calculation. Decimal approximations e.g. \(6.1653645833\ldots\) score B0.

Notes

Note: Candidates who do not score the first M mark may score maximum M0A0M1A1ft via the special case at the end of the mark scheme for part (c).

M1: For the key step in using the factor theorem to take \((4x + 1)\) or \(\left(x + \dfrac{1}{4}\right)\) out as a factor of their \(\mathrm{f}^{\prime}(x)\), not \(\mathrm{f}(x)\). Look for \((4x + 1)\left(x^2 + \lambda x - 4\right)\) or \((4x + 1)\left(x^2 \pm \mu\right)\) or \(\left(x + \dfrac{1}{4}\right)\left(4x^2 \pm \mu\right)\) or \(\left(x + \dfrac{1}{4}\right)\left(4x^2 \pm \lambda x - 16\right)\) (where the \(\lambda\) or \(\mu\) is a constant) either by inspection or division.
If they attempt division in (a) then they must use their quotient in (c) to score the marks.
They may “spot” that one of \(\mathrm{f}^{\prime}(\pm 2) = 0\) and factor out \((x \pm 2)\left(4x^2 + \lambda x \pm 2\right)\) without evidence that \(\mathrm{f}^{\prime}(\pm 2) = 0\)
There must be some factorisation or algebraic division present – not just roots stated from a calculator.

A1: Correct factorisation of the cubic to a linear and quadratic product, so
\((4x + 1)\left(x^2 - 4\right)\) or \(\left(x + \dfrac{1}{4}\right)\left(4x^2 - 16\right)\) or \((x - 2)\left(4x^2 + 9x + 2\right)\) or \((x + 2)\left(4x^2 - 7x - 2\right)\)
or for the correct quadratic seen e.g. \(x^2 - 4\) or \(4x^2 - 16\) (which may be the quotient in their algebraic division and may come from their work in (a) provided it is used in (c)).
There is no need to complete the long division if the appropriate quotient is found.
Note that proceeding directly to \((4x + 1)(x - 2)(x + 2)\) does not score either mark without sight of the e.g. \((4x + 1)\left(x^2 - 4\right)\) but they can score the final 2 marks via the special case.
Similarly, proceeding directly to \(\left(x + \dfrac{1}{4}\right)(2x - 4)(2x + 4)\) or \(4\left(x + \dfrac{1}{4}\right)(x - 2)(x + 2)\) does not score either mark without sight of \(\left(x + \dfrac{1}{4}\right)\left(4x^2 - 16\right)\) but they can score marks via the SC.

M1: Solves their quadratic factor = 0, substitutes one of these solutions (not \(x = -\dfrac{1}{4}\)) into \(\mathrm{f}(x)\) to find a value and selects the inside region between this and their answer to (b).
If their quadratic is of the form \(Ax^2 - B\) then they can write down their solution for \(x\) but if it is a 3TQ then they must find their value of \(x\) using the usual non-calculator rules.
Condone “incorrect” factorisation following the correct quadratic leading to \(x = \pm 2\)
e.g. \(\left(x + \dfrac{1}{4}\right)\left(4x^2 - 16\right) \rightarrow \left(x + \dfrac{1}{4}\right)(x - 2)(x + 2) = 0 \rightarrow x = \pm 2\)
\(y\) (or \(k\)) \(= -5\) (or \(-\dfrac{47}{3}\)) with nothing else can imply substitution of \(x = -2\) (or 2)
Note that if they “spot” that \(\mathrm{f}^{\prime}(\pm 2) = 0\) earlier then they can go straight to substitution.
Condone the use of \(y\) but not \(x\) in their region. Condone e.g. \(k \gt -5, k \lt \text{``}\dfrac{4735}{768}\text{''}\) for this mark.
Condone the use of \(\leqslant\) in place of \(\lt\) for this mark. Note that \(\left(2, -\dfrac{47}{3}\right)\) is the other minimum.

A1ft: \(\left\{k : -5 \lt k \lt \text{``}\dfrac{4735}{768}\text{''}\right\}\) Requires correct use of set notation but condone absence of \(\in \mathbb{R}\)
Follow through on their upper limit from (b) which may be a decimal but must be positive.
Must be strict inequalities i.e. not include the end points and must now be in terms of \(k\).
Other acceptable notation includes: \(k \in \left(-5, \text{``}\dfrac{4735}{768}\text{''}\right)\), \(\{k, k \gt -5\} \cap \left\{k, k \lt \text{``}\dfrac{4735}{768}\text{''}\right\}\)
Condone \(\left\{k : k \gt -5 \cap k \lt \text{``}\dfrac{4735}{768}\text{''}\right\}\) or \(\left\{-5 \lt k \lt \text{``}\dfrac{4735}{768}\text{''}\right\}\)
Do not allow e.g. \(\{k : k \gt -5\} \cup \left\{k : k \lt \text{``}\dfrac{4735}{768}\text{''}\right\}\) or just \(-5 \lt k \lt \text{``}\dfrac{4735}{768}\text{''}\)

SC: Candidates that have not scored the first M1 may go on to score the final two marks by finding an acceptable range between \(-5\) and \(\text{``}\dfrac{4735}{768}\text{''}\).
Use of their “\(-5\)” or “\(-\dfrac{47}{3}\)”, coming from \(x = \pm 2\), (or \(-5\) or \(-\dfrac{47}{3}\) directly from a calculator) with \(\text{``}\dfrac{4735}{768}\text{''}\) will score M1 if the inside region is selected. A correct \(\left\{k : -5 \lt k \lt \text{``}\dfrac{4735}{768}\text{''}\right\}\) will score A1ft
There is no need to see any working to achieve the \(-5\) (or \(-\dfrac{47}{3}\)).
All the relevant guidance regarding the region in the M1 and A1ft notes above continues to apply. e.g. for the M1 we will condone the use of \(y\) but not \(x\) in their region while for the A1ft we require use of \(k\).

For example:

• \(\left\{k : -5 \lt k \lt \text{``}\dfrac{4735}{768}\text{''}\right\}\) without scoring the first M scores M0A0M1A1ft
• \(\text{``}{-}5\text{''} \lt k \lt \text{``}\dfrac{4735}{768}\text{''}\) without scoring the first M scores M0A0M1A0ft
• \(\left\{k : \text{``}{-}\dfrac{47}{3}\text{''} \lt k \lt \text{``}\dfrac{4735}{768}\text{''}\right\}\) without scoring the first M scores M0A0M1A0ft

Notes

Mark parts (a) and (b) together.

M1: Substitutes \(x = -3\) into \(x = \dfrac{t - 1}{2}\), attempts to find \(t\), and substitutes their \(t\) into \(y = 5(t + 2)^4\)
Condone slips. May be implied by substitution of \(t = -5\) into \(y\) or by 405 or by \(y = 5(-3)^4\)
Alternatively, they may substitute \(x = -3\) into their \(y = \mathrm{f}(x)\)

A1: cao Answer only scores full marks and ISW after seeing 405 e.g. \((-5,\ 405)\)
Accept \((-3,\ 405)\) or e.g. \(P = 405\)

Notes

Mark parts (a) and (b) together.

M1: Attempts to make \(t\) the subject of \(x = \dfrac{t - 1}{2}\) using the correct order of operations and substitutes into \(y = 5(t + 2)^4\) Condone slips e.g. dividing by 2 first: \(t = \dfrac{x}{2} + 1\) or missing the +2 in \(y = 5(t + 2)^4\)
Alt 1: Writes \(t = \left(\dfrac{y}{5}\right)^{\frac{1}{4}} - 2\), substitutes into \(x = \dfrac{t - 1}{2}\) and attempts to make \(y\) the subject using the correct order of operations, i.e., \(2x = \left(\dfrac{y}{5}\right)^{\frac{1}{4}} - 3 \Rightarrow (2x + 3)^4 = \dfrac{y}{5} \Rightarrow y = 5(2x + 3)^4\)
Alt 2: Makes \(t\) the subject and then substitutes into an expanded \(5(t + 2)^4\)
Do not be too concerned about their expansion, but it must contain \(t^4\) and a constant term.

A1: cao and ISW after a correct answer seen. May be scored for \(y = 5(2x + 1 + 2)^4\)
Allow e.g. \(y = 80x^4 + 480x^3 + 1080x^2 + 1080x + 405\) or e.g.
\(y = 5(2x + 1)^4 + 40(2x + 1)^3 + 120(2x + 1)^2 + 160(2x + 1) + 80\)
Their RHS may be unsimplified but do not allow if e.g. binomial coefficients are still present.
Do not accept \(\mathrm{f}(x) = \ldots\) It must be \(y = \ldots\)

Notes

Note: Differentiation seen in (a) or (b) can score marks if used in (c)

M1: Reduces the power of their \((\text{``}2x + 3\text{''})^n\) by one to \(Q(\text{``}2x + 3\text{''})^{n-1}\) where \(Q\) is a constant and could be 1. There should be no other terms using this method.
Alternatively, attempts to expand their \((\text{``}2x + 3\text{''})^n\) (may have been expanded in (b)) and reduces the power of \(x\) by one in at least one term.
They may use parametric differentiation, i.e., \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = a(t + 2)^3\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = b\) where \(a\) and \(b\) are constants, leading to \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{a(t + 2)^3}{b}\) i.e. they must divide the correct way round.
Condone attempts at the chain rule that reach e.g. \(y^{\prime} = \ldots u^3\) but make a slip when substituting back in for \(x\) or \(t\), e.g., \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = (5x + 3)^3\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = (t + 2)^2\), provided the intention is clear.

dM1: Substitutes \(x = -3\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\) which may be implied by their answer.
Using parametric differentiation, substitutes their \(t\) (found from an attempt at substituting \(x = -3\) into \(x = \dfrac{t - 1}{2}\) which may have been seen in (a)) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\).
Note for reference, if correct, the parametric differentiation is \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 20(t + 2)^3\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{1}{2}\) leading to \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 40(t + 2)^3\)
They may substitute their value of \(t\) into \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) first before using the chain rule to reach \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) which is acceptable and implies the first M mark.

A1: \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) -1080\)
Correct answer only scores full marks.
May be seen labelled as \(m\) or something else.

Notes

Alternative – expands \((x+y)^3\) before differentiating

(a) Some candidates have a spurious “\(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\)” appearing as their intention to differentiate e.g.

This can be condoned for the first 3 marks in both versions.
Allow equivalent notation for the \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) e.g. \(y^{\prime}\)

M1: Award this mark for one of:

• \((x+y)^3 \rightarrow k(x+y)^2\left(\lambda + \dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\) where \(\lambda\) is 1, \(x\) or 0 but condone missing brackets e.g. \(3(x+y)^2 1 + \dfrac{\mathrm{d}y}{\mathrm{d}x}\)
• \(3x^2 - 3y - 2 \rightarrow \ldots x + \ldots\dfrac{\mathrm{d}y}{\mathrm{d}x}\) but condone \(3x^2 - 3y - 2 \rightarrow \ldots x + \ldots\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2\)

A1: Either \(3(x+y)^2\left(1 + \dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\) or \(6x - 3\dfrac{\mathrm{d}y}{\mathrm{d}x}\) oe
May be implied if e.g. they collect terms to one side initially.
Do not condone missing brackets unless they are implied by subsequent work.

A1: \(3(x+y)^2\left(1 + \dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\) and \(6x - 3\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (seen separately or equated)
If they collect terms to one side initially then the signs must be correct.

M1: A valid attempt to make \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) the subject with exactly 2 different terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), one coming from the differentiation of \((x+y)^3\) and the other coming from the differentiation of “\(-3y\)”
Note that here, 2 different terms means terms such as \(3\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(3(x+y)^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and not e.g. \(3\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(-8\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
Look for \((\ldots \pm \ldots)\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) which may be implied by their working.
Condone slips provided the intention is clear.
For those candidates who had a spurious \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) at the start, they may incorporate this in their rearrangement in which case they will have 3 terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and so score M0.
If they ignore it, then this mark is available for the condition as described above.
Note that from \(3(x+y)^2\left(1 + \dfrac{\mathrm{d}y}{\mathrm{d}x}\right) = 6x - 3\dfrac{\mathrm{d}y}{\mathrm{d}x}\), candidates may expand the brackets before rearranging, in which case they would need 4 different \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) terms coming from the appropriate places.
Note that the different \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) terms do not have to be correct as long as the above conditions are satisfied.

A1: Fully correct expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). Allow any equivalent correct forms.
Apply isw as soon as a correct expression is seen.

(a) alternative by expanding:

M1: Award this mark for one of:

• \(3x^2 - 3y - 2 \rightarrow \ldots x + \ldots\dfrac{\mathrm{d}y}{\mathrm{d}x}\) but condone \(3x^2 - 3y - 2 \rightarrow \ldots x + \ldots\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2\)
• Expanding \((x+y)^3\) to obtain either an \(x^2y\) term or an \(xy^2\) term and then uses the product rule to obtain \(\ldots x^2y \rightarrow \ldots x^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + \ldots xy\) or \(\ldots xy^2 \rightarrow \ldots xy\dfrac{\mathrm{d}y}{\mathrm{d}x} + \ldots y^2\)

A1: Either \(3x^2 + 3x^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 6xy + 6xy\dfrac{\mathrm{d}y}{\mathrm{d}x} + 3y^2 + 3y^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(6x - 3\dfrac{\mathrm{d}y}{\mathrm{d}x}\).
May be implied if e.g. they collect terms to one side initially.

A1: \(3x^2 + 3x^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 6xy + 6xy\dfrac{\mathrm{d}y}{\mathrm{d}x} + 3y^2 + 3y^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(6x - 3\dfrac{\mathrm{d}y}{\mathrm{d}x}\) oe. (seen separately or equated) If they collect terms to one side initially then the signs must be correct.

M1: A valid attempt to make \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) the subject with exactly 4 different terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), 3 coming from the differentiation of \((x+y)^3\) and the other coming from the differentiation of “\(-3y\)”
Note that here, 4 different terms means terms such as \(x^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(6xy\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and not e.g. \(3\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(-8\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
Look for \((\ldots \pm \ldots \pm \ldots \pm \ldots)\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) which may be implied by their working.
Condone slips provided the intention is clear.
For those candidates who had a spurious \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) at the start, they may incorporate this in their rearrangement in which case they will have 5 terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and so score M0.
If they ignore it, then this mark is available for the condition as described above.
Note that the different \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) terms do not have to be correct as long as the above conditions are satisfied. E.g. if they have an incorrect term such as \(6x\dfrac{\mathrm{d}y}{\mathrm{d}x}\), this mark is still available.

A1: Fully correct expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). Allow any equivalent correct forms.
Condone e.g. \(3x2y\) for \(6xy\).
Apply isw as soon as a correct expression is seen.

Alternative making \(y\) the subject in (a):

Score as follows:

M1: Cube roots both sides and makes \(x + y\) or \(y\) the subject then award for

• \(\left(3x^2 - 3y - 2\right)^{\frac{1}{3}} \rightarrow \ldots\left(3x^2 - 3y - 2\right)^{-\frac{2}{3}}\) or
• \(3x^2 - 3y - 2 \rightarrow \ldots x + \ldots\dfrac{\mathrm{d}y}{\mathrm{d}x}\) but condone \(3x^2 - 3y - 2 \rightarrow \ldots x + \ldots\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2\)

A1: For the \(\dfrac{1}{3}\left(3x^2 - 3y - 2\right)^{-\frac{2}{3}}\) or \(6x - 3\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

A1: Fully correct

M1: A valid attempt to make \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) the subject with exactly 2 different terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

A1: Correct expression

Using partial derivatives in (a):

Score as follows:

M1: Correct structure for either partial derivative:
Doesn’t expand: \(\dfrac{\partial \mathrm{f}}{\partial x} = \ldots(x+y)^2 + \ldots x\) or \(\dfrac{\partial \mathrm{f}}{\partial y} = \ldots(x+y)^2 + \ldots\)
or
Expands: \(\dfrac{\partial \mathrm{f}}{\partial x} = \ldots x^2 + \ldots xy + \ldots y^2 + \ldots x\) or \(\dfrac{\partial \mathrm{f}}{\partial y} = \ldots x^2 + \ldots xy + \ldots y^2 + \ldots\)
Where “…” are non-zero constants

A1: Correct \(\dfrac{\partial \mathrm{f}}{\partial x}\) or correct \(\dfrac{\partial \mathrm{f}}{\partial y}\)

A1: Correct \(\dfrac{\partial \mathrm{f}}{\partial x}\) and correct \(\dfrac{\partial \mathrm{f}}{\partial y}\)

M1: Attempts \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{\partial \mathrm{f}}{\partial x} \div \dfrac{\partial \mathrm{f}}{\partial y}\)

A1: Correct expression

Notes

(b) Note that the gradient of \(\frac{1}{2}\) could have been deduced from the given equation so you will need to check their solution carefully.

M1: Substitutes \(x = 1\) and \(y = 0\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) to obtain the tangent gradient and then uses the negative reciprocal and \(x = 1\) and \(y = 0\) in a correct straight line method to obtain the normal equation with \(x = 1\) and \(y = 0\) correctly placed.
Note that when finding the normal gradient, they may find the negative reciprocal of their expression from part (a) and then substitute \(x = 1\) and \(y = 0\) which is fine.
If using \(y = mx + c\) they must proceed as far as finding a value for \(c\).
If no substitution of \(x = 1\) and \(y = 0\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) is seen you will need to check their value. If they just state a value for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) then it must follow their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) with \(x = 1\) and \(y = 0\)

A1*: Correct equation with no errors following a correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) from part (a) (unless they start again which is unlikely)
Be aware that some incorrect expressions for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) from part (a) may fortuitously give \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}\) and would generally score A0
In general A1* must follow the final A1 in (a) or correct differentiation in (a)

Notes

M1: Uses the equation from part (a) and substitutes \(y = \pm 2x \pm 2\) or \(x = \dfrac{\pm 2 \pm y}{2}\) to obtain an equation in one variable (usually \(x\)) (not necessarily a cubic equation). Allow slips in rearranging to obtain \(x\) in terms of \(y\) (or \(y\) in terms of \(x\)) as long as the intention is clear.

A1: Correct cubic equation with terms collected and “\(= 0\)” seen or implied.
Note that both \(-x^3 + 3x^2 - 18x + 16 = 0\) and \(-y^3 - 60y = 0\) are correct equations.

To access any of the following marks, candidates must attempt to use either the factor of \((x - 1)\) with their cubic in \(x\) or the factor of \(y\) in their cubic in \(y\) to obtain a quadratic expression in \(x\) or \(y\).
Attempts that just use a calculator to solve the cubic equation score no more marks in this part.

dM1: Uses the fact that \((x - 1)\) or \(y\) is a factor in an attempt to establish the quadratic factor.
For the cubic in \(x\), it must be of the form \(ax^3 + bx^2 + cx + d = 0\ \ a, b, c, d \neq 0\)
For the cubic in \(y\), it must be of the form \(ay^3 + by = 0\ \ a, b \neq 0\)
For the cubic in \(x\), the attempt at the quadratic factor using \((x - 1)\) may be via inspection or e.g. long division to obtain a 3 term quadratic expression. There may or may not be a remainder but they must obtain 3 terms.
For the cubic in \(y\), they would need to take out a factor of \(y\) (or divide through by \(y\)) to obtain a factor of the form \(k\left(y^2 + \alpha\right)\)

ddM1: This mark requires:

• a correct cubic equation in \(x\) or \(y\)
• the correct quadratic factor or a multiple of it e.g. \(k\left(x^2 - 2x + 16\right)\) or \(k\left(y^2 + 60\right)\)
• an attempt to show that the quadratic factor has no real roots

For the quadratic in \(x\) this could be:

Attempts discriminant: e.g. \(b^2 - 4ac = 4 - 4 \times 1 \times 16\) (may be embedded in the quadratic formula)
Attempts to complete the square: e.g. \(x^2 - 2x + 16 = (x - 1)^2 - 1 + 16\)
Uses calculus to find the turning point: e.g. \(\dfrac{\mathrm{d}\left(x^2 - 2x + 16\right)}{\mathrm{d}x} = 2x - 2 = 0 \Rightarrow x = 1 \Rightarrow y = \ldots\)
Attempts to solve: e.g. \(x^2 - 2x + 16 = 0 \Rightarrow x = \dfrac{2 \pm \sqrt{4 - 4 \times 16}}{2}\) or from a calculator \(x = 1 \pm \sqrt{15}\mathrm{i}\)

For the quadratic in \(y\) this is likely to be:

Attempts to solve: e.g. \(y^2 + 60 = 0 \Rightarrow y^2 = -60 \Rightarrow \ldots\)

A1: Fully correct argument that requires:

• fully correct work
• a justification depending on their strategy
• a conclusion depending on their strategy

Via discriminant: \(4 - 4 \times 1 \times 16 \lt 0\) so no real roots so they do not meet again
Via completing the square: \(\rightarrow (x - 1)^2 + 15\) which has a minimum value of 15 so no real roots so they do not meet again.
Via calculus: \(x = 1 \Rightarrow y = 15\) is the minimum so no real roots so they do not meet again.
Via solving: \(x = 1 \pm \sqrt{15}\mathrm{i}\) or e.g. \(x = 1 \pm 3.87\mathrm{i}\) or e.g. \(x = 1 \pm \sqrt{-15}\) so math error so they do not meet again.
For \(y\) it is likely to be more straightforward e.g. \(y^2 \neq -60\) which cannot be solved so they do not meet again.
Allow equivalent statements for “they do not meet again” e.g. so they only meet once.

(But do not condone incorrect statements such as “therefore \(P\) does not meet \(C\) again”)

A minimum justification could be:

\(x^2 - 2x + 16 = 0 \rightarrow b^2 - 4ac = (-2)^2 - 4 \times 1 \times 16\) ddM1
\(4 - 4 \times 1 \times 16 \lt 0\) so no more roots so no more intersections A1

Do not allow e.g.
“\(x^2 - 2x + 16 = 0\) gives a math error so they do not meet again”
as there has been no attempt to show why the “math error” occurs – this scores M0A0

Alternative to (c) by showing the cubic is strictly increasing (or decreasing):

M1A1: As in the main scheme then

Hence there can only be one intersection (at \(x = 1\)) so the normal and curve do not intersect again.

dM1: Differentiates their cubic of the form \(ax^3 + bx^2 + cx + d = 0\ \ a, b, c, d \neq 0\) to obtain a 3 term quadratic expression with only coefficient errors on the non-constant terms.

ddM1: This mark requires:

• a correct cubic equation in \(x\)
• the correct derivative or a multiple of it
• an attempt to show that the quadratic expression is always positive (or negative)

A1: Fully correct concluding argument e.g. that as the derivative is always positive (or always negative) the function is strictly increasing (or decreasing) and therefore there can only be one intersection (at \(x = 1\)) so the normal and curve do not meet again.

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

(a)(i)

B1: Correct derivative in any form. "\(\mathrm{f}^{\prime}(x) =\)" is not required. Apply isw if necessary.

(ii)

B1: Correct derivative in any form. "\(\mathrm{g}^{\prime}(x) =\)" is not required. Apply isw if necessary.

Notes

M1: Eliminates e by setting their \(\mathrm{f}^{\prime}(x) =\) their \(\mathrm{g}^{\prime}(x)\) where \(\mathrm{f}^{\prime}(x) = Ax\mathrm{e}^{4x^2-1}\) oe and \(\mathrm{g}^{\prime}(x) = \dfrac{B}{x}\) oe with \(A \times B \gt 0\) and proceeds via \(\mathrm{e}^{4x^2-1} = \dfrac{\ldots}{x^2}\) or equivalent work (see below) to obtain \(4x^2 - 1 = \ln\dfrac{\ldots}{x^2}\) oe e.g. \(\ln x + 4x^2 - 1 = \ln\dfrac{1}{x}\)
Allow if they use \(\alpha\) for \(x\).
Note that there are various alternatives for this mark but the derivatives must be of the form defined above and the processing must be correct with coefficient/sign slips only.

Examples of equivalent work:

A1*: Obtains the printed answer with sufficient working and no errors.
Sufficient work would require the “e” eliminated before the given answer.
Must follow correct derivatives in part (a).
Condone \(4x^2 + 2\ln|x| - 1 = 0\) and condone \(4\alpha^2 + 2\ln\alpha - 1 = 0\) or \(4\alpha^2 + 2\ln|\alpha| - 1 = 0\)

Note that if both derivatives in (a) are correct we will allow fully correct work using the equation in (b) to work backwards to verify that \(p\mathrm{f}^{\prime}(x) = q\mathrm{g}^{\prime}(x)\) for M1 then obtains \(\mathrm{f}^{\prime}(x) = \mathrm{g}^{\prime}(x)\) with a minimal conclusion for A1
If either derivative in (a) is incorrect or missing, candidates who work backwards score no marks in (b).

Notes

(c)(i)/(ii)

M1: Attempts to use the iterative formula with \(x_1 = 0.6\)
Award this mark for e.g. \((x_2 =)\sqrt{\dfrac{1 - 2\ln 0.6}{4}}\) or may be implied by awrt 0.71 provided no incorrect working is seen.
Candidates sometimes find \(x_3\) (or possibly subsequent terms) rather than \(x_2\) in which case the M1 can be implied. (See table below for first few iterations)

A1: \((x_2 =)\) awrt 0.7109
Sight of \((x_2 =)\) awrt 0.7109 scores M1A1

B1(A1 on ePEN): \((\alpha =)\) 0.6706 (4dp)
Must be this value and not awrt 0.6706

For reference:

Notes

M1: Attempts the quotient rule to obtain an expression of the form \(\dfrac{P\left(x^2+4\right) - Qx(2x-3)}{\left(x^2+4\right)^2}\) \(P, Q \gt 0\) condoning bracketing errors/omissions or minor slips (e.g. \(2x+3\) or \(x+4\)).

Condone, e.g. \(\{\mathrm{f}^{\prime}(x)=\}\dfrac{2\left(x^2+4\right) - 2x(2x-3)}{\left(x^2+4\right)}\) provided an incorrect formula is not quoted.

May also see the product rule applied to \((2x-3)\left(x^2+4\right)^{-1}\) to obtain an expression of the form \(\{\mathrm{f}^{\prime}(x)=\}P\left(x^2+4\right)^{-1} - Qx(2x-3)\left(x^2+4\right)^{-2}\) \(P, Q \gt 0\) condoning bracketing errors/omissions or minor slips (e.g. \(2x+3\) or \(x+4\))

A1: Fully correct differentiation in any form with correct bracketing which may be implied by subsequent working.

A1: \(\mathrm{f}^{\prime}(x) = \dfrac{-2x^2 + 6x + 8}{\left(x^2+4\right)^2}\) or simplified equivalent, e.g. numerator terms in a different order.

Allow recovery from “invisible” brackets earlier and apply isw once a correct answer is seen.
Note that the complete form of the answer is not given so allow candidates to go from e.g. \(\mathrm{f}^{\prime}(x) = \dfrac{2\left(x^2+4\right) - 2x(2x-3)}{\left(x^2+4\right)^2} \rightarrow \dfrac{-2x^2 + 6x + 8}{\left(x^2+4\right)^2}\) for the final mark. The denominator \(\left(x^2+4\right)^2\) may go “missing” on an intermediate line provided it is present in their initial derivative and recovered in the final answer. Allow recovery from incorrect expansion of the denominator.

The \(\mathrm{f}^{\prime}(x) =\) must appear at some point but allow e.g. "\(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\)"

Note that just e.g. \(\mathrm{f}^{\prime}(x) = \dfrac{-2\left(x^2 - 3x - 4\right)}{\left(x^2+4\right)^2}\) without sight of a correct derivative in the correct form scores A0.

Notes

Note: it is possible to score B0M1A1 in this question due to the demand to “use algebra”.

Note: if their numerator from (a) is not a 3 term quadratic then no marks can be scored in (b).

B1ft: Uses algebra to solve their \(ax^2 + bx + c \ldots 0\) with \(a, b, c \neq 0\) where … is any equality or inequality, finding the correct, real critical values for their 3TQ.
The … 0 may be implied by their method.
They must show their working for this mark, so expect to see factorisation, substitution into the correct quadratic formula or completing the square.
Correct values for their quadratic do not imply this mark.
Approaches via factorisation must have completely correct factorisation, e.g.

\(-2x^2 + 6x + 8\{=0\} \Rightarrow -2(x+1)(x-4)\{=0\} \Rightarrow x = -1,\ 4\) scores B1ft
\(-2x^2 + 6x + 8\{=0\} \Rightarrow (2x+2)(4-x)\{=0\} \Rightarrow x = -1,\ 4\) scores B1ft
\(-2x^2 + 6x + 8\{=0\} \Rightarrow x^2 - 3x - 4\{=0\} \Rightarrow (x+1)(x-4)\{=0\} \Rightarrow x = -1,\ 4\) scores B1ft
\(-2x^2 + 6x + 8\{=0\} \Rightarrow (2x+2)(x-4)\{=0\} \Rightarrow x = -1,\ 4\) scores B0ft
\(-2x^2 + 6x + 8\{=0\} \Rightarrow (x+1)(x-4)\{=0\} \Rightarrow x = -1,\ 4\) scores B0ft

M1: Selects the “correct” region for their critical values and their \(a\) from part (a). Must be \(x\) not \(\mathrm{f}(x)\)
CVs may have been found using a calculator and may be implied if they are correct for their 3TQ. CVs may be incorrect due to errors in their calculations (but not errors in their method).

• For \(a \lt 0\) and roots \(\alpha \lt \beta\) they need e.g. \(x \lt \alpha\), \(x \gt \beta\) (or e.g. \(x \leqslant \alpha\) or \(x \geqslant \beta\))
• For \(a \gt 0\) and roots \(\alpha \lt \beta\) they need e.g. \(\alpha \lt x \lt \beta\) (or e.g. \(x \geqslant \alpha,\ x \leqslant \beta\))

Do not be overly concerned about their use of =, >, < in reference to their \(-2x^2 + 6x + 8 \ldots 0\) for this mark or for the A1.
Indicating the region on a sketch is not sufficient. Allow , / or / and / \(\cup\) / \(\cap\) for the M1.
If they have complex roots (or they use the discriminant to find there are no real roots) then they can score this mark for concluding:

• if \(a \lt 0\), “all values for \(x\) (have f decreasing)” or “f is always decreasing” or \(x \in \mathbb{R}\)
• if \(a \gt 0\), “no values for \(x\) (have f decreasing)” or “f is never decreasing”

A1: Correct solution \(x \lt -1\) or \(x \gt 4\) (allow \(x \leqslant -1\) or \(x \geqslant 4\)) coming from the correct numerator.

Do not isw if they go on to select e.g. \(x \gt 4\) or combine incorrectly to \(4 \lt x \lt -1\)
Allow full marks to be scored in (b) from an incorrect denominator (but it must be positive for all \(x\)), e.g. from \(\mathrm{f}^{\prime}(x) = \dfrac{-2x^2 + 6x + 8}{(x+4)^2}\) or \(\mathrm{f}^{\prime}(x) = \dfrac{-2x^2 + 6x + 8}{4x^2}\) or \(\mathrm{f}^{\prime}(x) = \dfrac{-2x^2 + 6x + 8}{x^2 + 16}\)

Examples: Just “\(x \lt -1\) or \(x \gt 4\)” stated scores B0M1A1
\(-2x^2 + 6x + 8\{=0\} \Rightarrow -2(x+1)(x-4) = 0 \Rightarrow x \lt -1, x \gt 4\) scores B1ftM1A1
\(-2x^2 + 6x + 8\{=0\} \Rightarrow x^2 - 3x - 4\{=0\} \Rightarrow (x+1)(x-4) = 0 \Rightarrow x \leqslant -1,\ x \geqslant 4\) scores B1ftM1A1
\(-2x^2 + 6x + 8\{=0\} \Rightarrow (2x+2)(x-4) \Rightarrow x \lt -1,\ x \gt 4\) scores B0ftM1A1
\(-2x^2 + 6x + 8 \lt 0 \Rightarrow x^2 - 3x - 4 \lt 0 \Rightarrow (x+1)(x-4) \lt 0 \Rightarrow x \lt -1, x \gt 4\) scores B1ftM1A1 (as this has correct factorisation shown, the region follows from \(a \lt 0\) (M1) and we condone reference to \(x^2 - 3x - 4 \lt 0\) as part of their working to find critical values (A1).)

Acceptable notation: allow a “,” , “or” , “and” or “\(\cup\)” to link the two regions, which may also be in set notation. e.g. \(x \lt -1\) or \(x \gt 4\)  ;  \(x \leqslant -1,\ x \geqslant 4\)  ;  \(x \lt -1\) and \(x \gt 4\)
\(x \leqslant -1 \cup x \geqslant 4\)  ;  \(\{x : x \lt -1 \cup x \gt 4\}\)  ;  \(\{x \in \mathbb{R} : x \leqslant -1\} \cup \{x \in \mathbb{R} : x \geqslant 4\}\)  ;
\(x \in (-\infty, -1) \cup (4, \infty)\)  ;  \((-\infty, -1] \cup [4, \infty)\) etc.

Do not accept \(4 \lt x \lt -1\) or use of the \(\cap\) symbol e.g. \((-\infty, -1] \cap [4, \infty)\) for the final mark, but they may be condoned for the M1. Note also that \([-\infty, -1] \cup [4, \infty]\) scores A0.

Notes

Note: Throughout the question allow use of \(\delta x\) for \(h\) or any other letter e.g. \(\alpha\) if used consistently.
If \(\delta x\) is used then you can condone e.g. \(\delta^2 x\) for \(\delta x^2\) as well as condoning e.g. poorly formed \(\delta\)’s

M1: Begins the process by writing down the gradient of the chord and attempts to expand the correct squared bracket – you can condone “poor” squaring e.g. \((x+h)^2 = x^2 + h^2\) but the \(-x^2\) must be present.

A1: Reaches a correct fraction o.e. with the \(x^2\) terms cancelled out and with no algebraic errors, e.g. \(\dfrac{\cancel{x^2} + 2xh + h^2 - \cancel{x^2}}{h}\), \(2x + h\) is correct.

A1*: Completes the process by applying a limiting argument and deduces that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x\) with no errors seen. They must have = \(2x\) and not just \(\lim\limits_{h \to 0} 2x\) to complete the proof.

\(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\) or an equivalent e.g. \(\mathrm{f}^{\prime}(x) =\) or “Gradient =” must be evident somewhere in their working or final line. If \(\mathrm{f}^{\prime}(x)\) is used then there is no requirement to see \(\mathrm{f}(x)\) defined first. Condone e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} \rightarrow 2x\) or \(\mathrm{f}^{\prime}(x) \rightarrow 2x\).

Condone missing brackets to allow e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \lim\limits_{h \to 0} \dfrac{2xh + h^2}{h} = \lim\limits_{h \to 0} 2x + h = 2x\)

Do not allow \(h = 0\) if there is never a reference to \(h \to 0\).

e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \lim\limits_{h \to 0} \dfrac{2xh + h^2}{h} = \lim\limits_{h \to 0} 2x + 0 = 2x\) is acceptable

but e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2xh + h^2}{h} = 2x + 0 = 2x\) is not unless \(h \to 0\) is seen.

The \(h \to 0\) does not need to be present throughout the proof e.g. appear on every line but must appear at least once.

They must reach \(2x + h\) at the end and not \(\dfrac{2xh + h^2}{h}\) (without the \(h\)’s cancelled) to complete the limiting argument.

Notes

M1: Attempts both f(3.6) and f(3.7) or a narrower interval that contains the root 3.672… (which may be implied by sight of f(3.6) = … and f(3.7) = … with at least one correct) and obtains at least one correct to 1 significant figure (rounded or truncated) for their interval and considers their signs. Use of degrees is M0.

Some examples for consideration of sign (which are also sufficient for the change of sign part of reasoning for the A1):

• \(\mathrm{f}(3.6) = -0.6 \lt 0\) and \(\mathrm{f}(3.7) = 0.2 \gt 0\)
• \(\mathrm{f}(3.6) \times \mathrm{f}(3.7) \lt 0\)
• \(\mathrm{f}(3.6) = -0.7,\ \mathrm{f}(3.7) = 0.2\) “change in sign”

For reference \(\mathrm{f}(3.6) = -0.68626\ldots\) and \(\mathrm{f}(3.7) = 0.21194\ldots\)

A1*: This mark requires:

• both f(3.6) and f(3.7) correct to 1 significant figure (rounded or truncated) (or their values correct to 1 significant figure if using a narrower interval)
• a reference to sign change
• a reference to continuity {of \(\mathrm{f}(x)\)}
• a (minimal) conclusion, e.g. “hence root”, “proved”, \(\checkmark\), #, QED, \(3.6 \lt \alpha \lt 3.7\)

Accept as a minimum, "change of sign, continuous, root".
Do not condone “change in sign therefore continuous” or other incorrect statements such as “\(x\) is continuous”, “the interval is continuous” – these score A0.
Condone “the graph is continuous”.
Condone reference to \(x\) in place of \(\alpha\) in their conclusion, e.g. “hence \(x\) lies in the interval”.
Condone statements such as “there is at least one root” in place of their conclusion.

Notes

Note: Their answer to (b) may be seen in part (c) provided that they have not clearly attempted part (b) incorrectly, e.g., an attempt at \(\mathrm{f}^{-1}(x)\) in (b).

M1: For \(\tan\left(\dfrac{1}{2}x\right) \rightarrow \ldots\sec^2\left(\dfrac{1}{2}x\right)\) o.e. The brackets are not required. You may see attempts at the quotient rule but the method should be correct and they should reach something equivalent to \(\ldots\sec^2\left(\dfrac{1}{2}x\right)\).

e.g. \(\tan\left(\dfrac{1}{2}x\right) = \dfrac{\sin\left(\dfrac{1}{2}x\right)}{\cos\left(\dfrac{1}{2}x\right)} \rightarrow \dfrac{k\cos\left(\dfrac{1}{2}x\right)\cos\left(\dfrac{1}{2}x\right) - -k\sin\left(\dfrac{1}{2}x\right)\sin\left(\dfrac{1}{2}x\right)}{\cos^2\left(\dfrac{1}{2}x\right)}\) where \(k\) is a positive constant scores M1. If the formula is seen it must be correct.

A1: \(\{\mathrm{f}^{\prime}(x)=\}\ 1 + \dfrac{1}{2}\sec^2\left(\dfrac{1}{2}x\right)\) o.e. which may be unsimplified and apply isw.

The brackets are not required. There is no need for \(\mathrm{f}^{\prime}(x) =\) just look for the expression.

Note that \(\{\mathrm{f}^{\prime}(x)=\}\ \dfrac{3}{2} + \dfrac{1}{2}\tan^2\left(\dfrac{1}{2}x\right)\) is correct and appears occasionally.

\(\{\mathrm{f}^{\prime}(x)=\}\ 1 + \dfrac{1}{2}\sec\dfrac{1}{2}x^2\) is condoned for M1A0 only but \(1 + \dfrac{1}{2}\left(\sec\dfrac{1}{2}x\right)^2\) scores M1A1.

Notes

M1: Attempts \(3.7 - \dfrac{\mathrm{f}(3.7)}{\mathrm{f}^{\prime}(3.7)}\) and obtains a value following through on their \(\mathrm{f}^{\prime}(x)\) as long as it is a “changed” function in terms of \(x\).

Just stating \(3.7 - \dfrac{\mathrm{f}(3.7)}{\mathrm{f}^{\prime}(3.7)} = \ldots\) without evidence of use of 3.7 in \(\mathrm{f}(x)\) (note that this evidence might come from part (a)) and in their \(\mathrm{f}^{\prime}(x)\) is M0 unless implied by a correct value for both \(\mathrm{f}(x)\) and \(\mathrm{f}^{\prime}(x)\) or by their final answer.
Must be a correct N-R formula used – you may need to check their values – accuracy of at least 3s.f. rounded or truncated required.
Allow if attempted in degrees. For reference in degrees \(\mathrm{f}(3.7) = 3.73\ldots\) and \(\mathrm{f}^{\prime}(3.7) = 1.50\ldots\) and gives \(\alpha = 1.21\ldots\)
Note that the full N-R accuracy is 3.672051617.
For reference, the value of \(\alpha\) is approximately 3.673194406… and scores M0A0 without other valid work.

A1: For awrt 3.672  Ignore any subsequent iterations.

Notes

(a)(i) If “\(+\,c\)” is included with either derivative penalise it only once on the first occurrence.

M1: Award for \(x^3 \rightarrow x^2\) or \(x^2 \rightarrow x\) or \(5x \rightarrow 5\) or \(-10 \rightarrow 0\)
Indices may be unprocessed e.g. \(x^3 \rightarrow x^{3-1}\) or \(x^2 \rightarrow x^{2-1}\) or \(5x \rightarrow 5x^0\)

A1: Correct simplified expression with indices processed \(12x^2 - 14x + 5\).
Do not allow \(x^1\) for \(x\) or \(5x^0\) for 5.
Apply isw if necessary once a correct answer is seen.
The “\(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\)” is not required.

(ii)

A1ft: Correct simplified second derivative \(24x - 14\) or follow through their first derivative.
Must be simplified so do not allow e.g. \(x^1\) for \(x\) or \(x^0\) for 1 as above.
Apply isw if necessary once a correct answer is seen.
The “\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} =\)” is not required.

Notes

M1: Sets their second derivative of the form \(ax + b,\ a, b \neq 0\) equal to 0 and proceeds to a value for \(x\). Condone slips in rearranging as long as a value for \(x\) is obtained.
This may be implied by their value of \(x\) or may be implied by their working e.g.
\(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} =\right) 24x - 14 \rightarrow 24x = 14 \Rightarrow x = \ldots\)
Condone one slip in copying their second derivative.
Also condone if they “cancel” e.g. \(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} =\right) 24x - 14 \rightarrow 12x - 7 = 0 \Rightarrow x = \ldots\)

A1: Correct value from correct work and a correct second derivative but allow recovery if they “cancel” their second derivative to obtain e.g. \(12x - 7\).
Allow exact equivalents e.g. \(\dfrac{14}{24}\) but not rounded decimals e.g. 0.583
Allow recurring decimal if clearly indicated e.g. \(0.58\dot{3}\)
Correct answer only from a correct second derivative (or correctly cancelled second derivative) scores both marks.
Isw after a correct answer is seen.

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

The positive square root is used since \(-\dfrac{1}{4} \leqslant t \leqslant 0\)

Typical solution

Typical solution

Typical solution

(c)(i)

(c)(ii)

The height is decreasing at a constant rate of 0.018 cm per minute.

Typical solution

Typical solution

Typical solution

Typical solution

Since \(x \gt 0,\ x^{k-1} \neq 0\)

Typical solution

1

Typical solution

When \(k = 1\)

If there is a point of inflection \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0\)

Therefore, \(C\) has no point of inflection.

Typical solution

Typical solution

Typical solution

Typical solution

Stationary points occur when \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\)

At \(B\), \(x = -1\)

Therefore, stationary point is \((-1, 5)\)

Typical solution

(a)(i) 7.2155

(a)(ii)

Typical solution

The curve is convex, so the trapezium rule gives an over-estimate.

Therefore, the student is incorrect.

Typical solution

Increasing and convex

Typical solution

\(3\mathrm{e}^{x}\)

Typical solution

Typical solution

\(\mathrm{f}^{\prime\prime}(x)\) does not change sign either side of \(x\) = 0

Therefore, the curve does not have a point of inflection at \(x\) = 0

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

At stationary point \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\)

So

Typical solution

(4, 2)

Typical solution

Area of sectors =

Area of rhombus =

Typical solution

(i)

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

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

(ii)

Typical solution

The angle would be the same.

The area would be quadrupled.

Typical solution

Typical solution

Typical solution

Typical solution

As \(x \gt 0\),

Equation has no solutions so the curve has no stationary points.

Typical solution

Typical solution

Typical solution

Typical solution

Maximum height is approximately 10.5 metres

Typical solution

The entrance to the cave is unlikely to be perfectly smooth

Typical solution

Typical solution

Many-to-one function is when two or more \(x\) values give the same \(y\) value.

This graph is many-to-one because you can draw a horizontal line and it will cross the graph twice.

Typical solution

(i)

(ii)

Typical solution

This is the same equation solved in part c(i) so \(P\) and \(Q\) must be stationary points.

Typical solution

Typical solution

When \(t\) = 3

Typical solution

= 0.501 cm s⁻¹

Typical solution

Since \(A = q(15 - q^2) \gt 0\)

then \(q\)’s upper limit \(c = \sqrt{15}\)

Typical solution

Max occurs at \(\dfrac{\mathrm{d}A}{\mathrm{d}q} = 0\)

Typical solution

Length = 180.5 cm

Typical solution

(i)

(ii)

Width of surfboard = 58 cm

Hence the width is approximately one third of the length

Typical solution

When \(x\) = 0

Typical solution

The 1st (top) graph: a line with negative gradient, positive \(y\)-intercept, crossing the positive \(x\)-axis.

Typical solution

The top right graph.

Typical solution

Notes

M1: Attempt to find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). Correctly combine attempts at algebraic or numerical derivatives

A1: Correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). Algebraic or numerical (ie \(\frac{4}{5}\)) www

B1: Correct \(t\) stated or explicitly used. Ignore any additional \(t\) values

M1: Attempt equation of tangent, using \((2, 3)\) and their numerical gradient (from clearly using their \(t\) value in their attempt at parametric differentiation)
If using \(y = mx + c\) then M1 is awarded when value for \(c\) is obtained
M0 if using \(x = 3\), \(y = 2\)

A1: Obtain given answer www

If trying to start with a Cartesian equation there are a number of different, correct, solutions. To gain any credit it would have to be a valid attempt in obtaining the initial equation, condoning only a slip. Please consult with your TL if seen.

Notes

M1*: Attempt to solve equations of tangent and curve simultaneously. Or valid attempt to obtain an equation in \(x\) or \(y\) only

A1: Obtain correct cubic. Brackets expanded and like terms collected
Some (usually BC) methods may not explicitly state simplified cubic; if M1 awarded and \(t = -\frac{7}{4}\) seen www then allow first A1 BOD, with then possibly SCB1

M1dep*: Attempt to solve cubic to obtain \(t\) value other than \(t = 1\). DR so method needed
Factorise to 3 linear factors and solve oe, eg using division
If using \((t - 1)^2\) as repeated factor, M1 for \((t - 1)^2(4t \pm 7)\)
If using \((t - 1)\) as linear factor then need both attempt at quadratic factor (factorisation or division) and also attempt to solve quadratic – BOD for slips

A1: Obtain correct coordinates, in exact form, www. Ignore \((2, 3)\) if also given, but A0 if other point(s) as well

If M1 A1 awarded for correct cubic, but no method shown to solve either cubic, or quadratic from \((t - 1)\)(quadratic factor), then SC B1 for correct coordinates www ie max of 3 marks

See comment in part (a) with regard to using Cartesian equation

Notes

(a)(i)

M1: Attempt \(\frac{\mathrm{d}y}{\mathrm{d}x}\), \(\geqslant 3\) terms correct

A1: All correct (ISW if candidates go on to divide by 12, but this answer oe must be seen)

(a)(ii)

M1: Setting their \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and attempt to solve (i.e. reach one or more values for \(x\)).
May be implied by correct answers

A1: And no others. May be done BC
Sight of \(x = 1\) and \(x = -1\) (and no other solutions) 2/2
ISW any attempt to find corresponding \(y\) values.

Notes

M1: FT their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) (\(\geqslant 2\) terms correct when differentiating their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) a second time).
May be implied by A1 but if both correct roots not seen must be set \(= 0\) for this mark.
Allow \(x^3 - x^2 - x + 1 \Rightarrow 3x^2 - 2x - 1 = 0\)

A1*: A1 for both correct roots and no others, i.e. points of inflection

A1 dep*: Statement needed. Dependent on previous A1.
Accept e.g.:

• One of these \(\left[x = -\frac{1}{3}\right]\) is not a stationary point A1
• \(x = -\frac{1}{3}\) is not a stationary point A1
• Categorising both points (i.e. \(x = 1\) is stationary and a point of inflection, \(x = -\frac{1}{3}\) is not stationary…) A1

But not “only one of these is a point of inflection” A0

Notes

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

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

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

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

Notes

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

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

B1*: One correct area expression

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

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

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

Notes

(c)(i)

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

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

(c)(ii)

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

Notes

B1*: cao

Denominator must be simplified to \(h\)

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

If not using result from part (a) then M0

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

Must see limit present for this A mark

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

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

Notes

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

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

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

Notes

B1: oe in words eg \(a\) is negative

B0 for \(a \leqslant 0\)
B0 if left as \(2a \lt 0\)

Notes

B1: Mark final answer, so B0 if \(-9a\) is then changed to \(9a\)
Must be given as a single term, not eg \(9a - 18a\)

Notes

B1: Accept \(x \leqslant 3\)

oe in words
Do not penalise errors if set notation is attempted, as long as intention is clear

Notes

B3: B2 for 2 correct \(\{\ \}\) components, B1 for 1 correct \(\{\ \}\) component must be of the form \(\pm a(x \pm b)^2 \pm c\) with terms simplified. Award B1 for explicitly stating \(a = 3\), B1 for explicitly stating \(b = 2\) and B1 for explicitly stating \(c = 5\)

SC B2 for \(3(x - 2) + 5\) but no marks for any other non-squared expression

Notes

B1FT: Correct answer or follow through from their answer to part (a) of the form \(\pm a(x \pm b)^2 \pm c\)

Allow \(x = 2\), \(y = 5\) or 2,5

Notes

B1FT: Correct answer or follow through their consistent \(x\)-value from parts (a) and (b)

Must be an equation

Notes

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

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

Notes

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

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

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

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

Notes

(c)(i)

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

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

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

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

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

(c)(ii)

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

Notes

M1*: Either \(\dfrac{\mathrm{d}}{\mathrm{d}x}(2xy) = 2y + 2x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(y^2\right) = 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
Attempt at implicit differentiation wrt \(x\)

A1: Correct equation (any form) – allow = 0 implied by later working
Setting derivative equal to \(y'\) is A0

M1dep*: Set the denominator of their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) equal to zero (oe)
Denominator must be linear in both \(x\) and \(y\)

M1: Eliminate \(x\) correctly and simplify to a 3TQ quadratic in \(y\) (if correct: \(2y^2 - 10y + k = 0\)) or eliminate \(y\) correctly to form a 3TQ quadratic in \(x\) (if correct: \(x^2 - 10x + 2k = 0\))
Dependent on both previous M marks – condone sign slips only in simplification

M1: Setting up the equation for diff. in \(y\) \(\dfrac{\sqrt{b^2 - 4ac}}{a} = \pm 3\), LHS of equation must be correct following through from their 3TQ (the constant term for their 3TQ in \(y\) must contain \(k\))
Dependent on first two M marks – if considering diff. in \(x\) values = 3 then M0
Correct \(k\) or both \(y\)’s www is M1 A1

A1: www (can be implied by correct \(y\)-values)
\(y_P = 1, y_Q = 4\)

A1: Both correct www – do not penalise \(x_P = 8, x_Q = 2\)
Dependent on all previous marks

Alternative for the first 3 marks (implicit differentiation wrt \(y\))

M1: Either \(\dfrac{\mathrm{d}}{\mathrm{d}x}(2xy) = 2x + 2y\dfrac{\mathrm{d}x}{\mathrm{d}y}\) or \(\dfrac{\mathrm{d}}{\mathrm{d}y}(5x) = 5\dfrac{\mathrm{d}x}{\mathrm{d}y}\)

A1: Correct equation (any form) – allow = 0 implied by later working
Setting derivative equal to \(x'\) is A0

M1: Set the numerator of their \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = 0\) oe

Alternative for the first 4 marks (making \(x\) the subject, diff. wrt \(y\))

M1*: Makes \(x\) the subject – must be of the form \(x = \dfrac{\mathrm{f}(y)}{\mathrm{g}(y)}\) where \(\mathrm{f}(y)\) is a two-term quadratic fn. in \(y\) and \(\mathrm{g}(y)\) is a two-term linear fn. in \(y\)

A1: Correct derivative (allow un-simplified)
Any equivalent correct form

M1dep*: Set the numerator of their \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) equal to zero (oe) where \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{\mathrm{g}(y) \times \mathrm{f}'(y) \pm \mathrm{f}(y) \times \mathrm{g}'(y)}{(\mathrm{g}(y))^2}\)
Where \(\mathrm{f}(y)\) and \(\mathrm{g}(y)\) are as defined in the first M mark and derivatives follow through correctly from their f and g

M1: Simplify to a 3TQ quadratic in \(y\) (if correct: \(2y^2 - 10y + k = 0\))
Dependent on both previous M marks – condone sign slips only in simplification to 3 TQ in \(y\)

Alternative 1 for the fourth M mark (roots of polynomials)

M1: Using \(\sum\alpha = -\dfrac{b}{a}\) correctly for their 3TQ in \(y\) with roots that differ by 3
Could use e.g. \(\alpha, \alpha - 3\) etc.
Dependent on first two M marks

Alternative 2 for the fourth M mark (difference in \(x\)-values)

M1: Setting up the equation for diff. in \(x\) \(\dfrac{\sqrt{b^2 - 4ac}}{a} = \pm\)their 6, with their 6 from their \(4(3) - 2x = 0\), LHS of equation must be correct following through from their 3TQ (the constant term for their 3TQ in \(x\) must contain \(k\))
Dependent on first two M marks

Notes

B1: Substitute (4, 8) into the equation of the curve.
Seen anywhere in solution
Allow for unsimplified equation, even if error then occurs

M1: Attempt differentiation.
To obtain derivative of the form \(px + qx^{-0.5}\)
Can still be awarded if \(p\) and \(q\) now incorrect numerical values

M1: Use \(x = 4\) correctly in their derivative.
Must be an attempt at differentiation, but could still follow M0
Their derivative could now be part of an equation or an attempt at a perpendicular gradient

M1: Attempt to use the relationship between the gradients of perpendicular lines.
Attempt the gradient of the tangent, using attempt at gradient of given normal (condone \(-\tfrac{1}{13}x\) if recovered)
OR
Attempt gradient of the normal using their derivative (either algebraic or in terms of \(a\) and \(b\))
Condone slips with fractions within fractions as long as intent is clear
NB \(\left(8a + \tfrac{1}{4}b\right) \times -\dfrac{1}{13} = -1\) would imply next M1 as well

M1: Equate expressions / values for normals or for tangents, with \(x\) now substituted. Could be using gradients or equations.
Must be comparing like with like ie tangent with tangent, or normal with normal
If using equations then would need to equate expressions for either the gradients or the intercepts

A1: Obtain correct linear equation.
If using eg \(-\dfrac{1}{8a + \frac{1}{4}b} = -\dfrac{1}{13}\) then terms in the denominators must have been dealt with correctly

A1: Obtain \(a = 2,\ b = -12\)
BC so no method needed

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

B2: B1 each term, isw incorrect attempts to simplify

Notes

B2: B1 this term correct
B1 no second term

Notes

B1: Must mention ‘rate of change’ of bees, or equiv, and how this relates to wildflower plants
B0 for ‘change’ not ‘rate of change’
B1 for rate of change of bees with respect to plants
B1 for rate of change of bees as the plants change
B1 BOD for rate of change of bees compared to the number of plants
State or imply that it is bees compared to flowers, so B0 if other way around
Must relate to bees and plants, and not just \(B\) and \(F\)
See appendix for further examples (below)

Appendix: exemplar responses for Q9(a)

Notes

B1: Correct derivative
No need to see \(\dfrac{\mathrm{d}F}{\mathrm{d}t}\) notation
Could be unsimplified

B1: Correct derivative
No need to see \(\dfrac{\mathrm{d}B}{\mathrm{d}t}\) notation
NB watch out for \(2 - 3\sin t\)

M1: Correct method to combine their two derivatives – algebraic or numerical
B0B0M1 is possible
Must have \(\mathrm{e}^{0.1t}\) in denominator, or \(\mathrm{e}^{-0.1t}\) in the numerator, but allow muddles with the placing of the 5

A1: Substitute \(t = 4\) to obtain 0.484, or better

Alternative method

B1: Correct expression for \(B\) as a function of \(F\)

M1: Attempt differentiation
May see ln terms split first (possibly even including use of \(\cos(A - B)\))

A1: Obtain correct derivative aef

A1: Substitute \(F = 50\mathrm{e}^{0.4}\) to obtain 0.484, or better
Could use \(t = 4\) if derivative now in terms of \(t\)

Notes

B1: Summer will be over so pattern may not continue
Summer is not greater than 12 weeks
Fewer bees and/or flowers in autumn/winter
Any reason referring to a change in season having an effect on bees and/or flowers
B0 for just considering long-term behaviour eg flowers will not continue to increase exponentially
Reasons must reference seasons / different time of year (could be implied by ‘weather getting colder’)

Notes

M1: Attempt to differentiate using the chain rule
Obtain derivative of form \(\mathrm{f}(x)\mathrm{e}^{x^2 + 3x}\)
Could also split into two terms and use product rule to obtain derivative of form \(\mathrm{f}(x)\mathrm{e}^{x^2}\mathrm{e}^{3x} + k\mathrm{e}^{x^2}\mathrm{e}^{3x}\ (k \neq 0)\)
M0 if attempt to split results in sum not product

A1: Obtain correct derivative
Brackets must be seen, or implied by later work
aef eg \(\left(2x\mathrm{e}^{x^2}\right)\left(\mathrm{e}^{3x}\right) + \left(\mathrm{e}^{x^2}\right)\left(3\mathrm{e}^{3x}\right)\) from splitting into two terms first
Could be in terms of \(u\), as long as \(u\) clearly defined

A1: Equate correct derivative to 0 soi and solve to obtain \(x = -\frac{3}{2}\)
ISW any \(y\)-coordinates if given
A0 if any additional solutions for \(x\)
Must see differentiation, so \(x = -\frac{3}{2}\) with no supporting method gets no credit (as question is ‘determine’)

B1FT: Indicate no solutions from the exponential term
FT their derivative as long as of form \(\mathrm{f}(x)\mathrm{e}^{x^2 + 3x}\) or \(\mathrm{f}(x)\mathrm{e}^u\)
Allow BOD for explanations such as \(\mathrm{e}^x \gt 0\) for all \(x\)
Must have some reason, eg ‘\(\mathrm{e}^{x^2 + 3x}\) is always positive’, ‘\(\mathrm{e}^{x^2 + 3x}\) cannot be negative’, ‘cannot take ln of a negative number’, ‘not defined’, ‘not real’, ‘no solutions’
A0 for ‘math error’ or ‘doesn’t work’

Alternative method

M1: Take ln and attempt implicit differentiation
Must deal correctly with \(\ln y\)

A1: Obtain correct derivative
May still have \(\dfrac{1}{y}\) on LHS

A1: Equate correct derivative to 0 soi and solve to obtain \(x = -\frac{3}{2}\)

B1: Indicate no solutions from the exponential term
See main MS for guidance
Could also explain why no solutions from \(\dfrac{1}{y}\)

Notes

M1: Attempt to differentiate again using the product rule correctly
Obtain derivative of form \(\left(ax^2 + bx + c\right)\mathrm{e}^{x^2 + 3x}\) aef

A1: Obtain correct derivative
aef eg (depending on method) \(2\mathrm{e}^{x^2 + 3x} + 2x(2x + 3)\mathrm{e}^{x^2 + 3x} + 3(2x + 3)\mathrm{e}^{x^2 + 3x}\)
or \(\left(\left(2\mathrm{e}^{x^2} + 4x^2\mathrm{e}^{x^2}\right)\mathrm{e}^{3x} + \left(2x\mathrm{e}^{x^2}\right)3\mathrm{e}^{3x}\right) + \left(6x\mathrm{e}^{x^2}\mathrm{e}^{3x} + 9\mathrm{e}^{x^2}\mathrm{e}^{3x}\right)\)
Could be in terms of \(u\), as long as \(u\) clearly defined

B1: State, or clearly imply, correct condition at any point in proof
Must be general statement, and not just \(\gt 0\) from testing the stationary point

M1: Explain why correct quadratic is always positive
Could note minimum value of 2 as completed square form
Could use expanded quadratic, which should be \(4x^2 + 12x + 11\)
If showing no real roots then must also say that it is a positive quadratic
Condone \(\gt\) / \(\geqslant\) muddles for M1 only
Could show that there are no points of inflection and \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} \gt 0\) for at least one point

A1: Full and convincing proof to show that curve is convex for all \(x\)
www

Alt method for first 2 marks

M1: Attempt second derivative, using implicit differentiation and the product rule
If still \(\dfrac{1}{y}\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x + 3\) then must be a correct attempt to differentiate the LHS

A1: Obtain correct derivative
aef
Will need to use \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = y(2x + 3)\) to make further progress

Notes

(a)(i)

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

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

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

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

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

(a)(ii)

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

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

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

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

Notes

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

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

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

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

Notes

(c)(i)

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

(c)(ii)

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

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

Notes

A1: For correct gradient of either tangent or normal

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

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

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

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

Notes

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

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

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

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

Alternative method (1)

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

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

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

Alternative method (2)

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

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

A1: Must be exact and come from exact working.

Alternative method (3)

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

M1dep: Correct dot product with exact values

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

Notes

M1: Complete the square (for \(x\) and \(y\)) to obtain \((x \pm 3)^2 + (y \pm 5)^2 + \ldots\)
or for \(\ldots \pm k \pm 3^2 \pm 5^2\)

A1: cao – www
allow equivalent in either set notation e.g. \(\{k : k \lt 34\}\) or interval notation e.g. \((-\infty, 34)\) or \((-\infty, 34]\) but not \([-\infty, 34)\) or \([-\infty, 34]\) unless \(k \lt 34\) already seen
If implying a lower limit then A0
Allow \(k \leqslant 34\)

Notes

M1*: Attempt to differentiate the equation for \(C\) implicitly – must be four terms including a \(2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) term and two other terms correct (condone either \(-46\) or \(k\) appearing in their deriv. as a 5^th term) but if the derivative of \(x^2 + y^2 - 6x + 10y - 46 = 0\) is put equal to \(\frac{1}{2}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (and used subsequently) then M0
or applying the chain rule to an expression of the form \(\pm 5 \pm \sqrt{\lambda \pm (x \pm 3)^2}\) for some non-zero \(\lambda\), so must be of the form \(\frac{1}{2}(\mathrm{f}(x))^{-\frac{1}{2}}\mathrm{g}(x)\) where \(\mathrm{f}(x)\) is quadratic and \(\mathrm{g}(x)\) is linear

M1dep*: Sets derivative equal to \(\frac{1}{2}\)
or substitutes \(\frac{1}{2}\) for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

M1: Substitutes their linear expression into the given equation of \(C\) with \(k = -46\) (must be five terms with two quadratic terms and two linear terms in \(x\) oe if using completing the square form from (a)) to obtain an expression/equation in \(x\) (or \(y\)) only
Dependent on first two M marks – if chain rule used this mark is implied by setting \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) equal to \(\frac{1}{2}\)

M1: Simplify to a 3TQ in \(x\) (or \(y\)) (allow sign errors only when simplifying from their five term equation/expression or completing the square equation/expression)
Dependent on all M marks
For \(y\) if correct: \(y^2 + 10y - 39\ (= 0)\)

A1: BC - do not need to be stated as coordinates
Two values of \(x\) and \(y\) only

Alternative method for first two marks

M1*: Finding an expression for the gradient of the line segment that passes through the centre and any point \((x, y)\) on the circumference of \(C\)
Follow through their centre from (a) – gradient expression must be correct for their centre

M1dep*: Equating gradient with \(-2\) (oe)

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 \(\mathrm{f}'(x)\)