Notes

Must be seen in (a) to score

M1: Examples to verify include:

• substitutes \(x = 6\) into both \((y =)\ \dfrac{15x}{(2x+3)(x-3)}\) and \((y =)\ 2x - 10\) and finds the \(y\) values for both
• substitutes \(x = 6\) into one of the two equations to find \(y = 2\) and then substitutes this into the other equation and solves to find \(x\)
• sets \(\dfrac{15x}{(2x+3)(x-3)} = 2x - 10 \Rightarrow\) cubic equation (=0 implied) and either substitutes \(x = 6\) into the expression, attempts \(\mathrm{f}(6)\) or else attempts to divide the cubic = 0 by \((x - 6)\).
  Condone without calculations for this mark only. i.e. \(\mathrm{f}(6) = 0\)

Condone slips for this mark in any of the outlined approaches. There may be variations of these as well.

A1*: Correct calculations must be seen with a minimal conclusion that they intersect
e.g. \(\dfrac{15 \times 6}{(2 \times 6 + 3)(6 - 3)} = 2\) and \(12 - 10 = 2\) so they intersect (allow eg meet/cross/intercept)
e.g. \(\dfrac{15 \times 6}{(2 \times 6 + 3)(6 - 3)} = 2 \Rightarrow 2 = 2x - 10 \Rightarrow x = 6\) so they intersect
Acceptable alternatives are:
\(\mathrm{f}(x) = 4x^3 - 26x^2 - 3x + 90,\ \mathrm{f}(6) = 4 \times 6^3 - 26 \times 6^2 - 3 \times 6 + 90 = 0 \Rightarrow\) so they intersect
\(\mathrm{f}(x) = 4x^3 - 26x^2 - 3x + 90 \Rightarrow (x - 6)(4x^2 - 2x - 15)\) so \(x = 6\) is a root so they intersect
OR \((x - 6)\) is a factor hence they intersect
Only accept verified, QED etc if there is a preamble mentioning intersection about how it will be shown.

Special case: (\(\mathrm{f}(x) = 4x^3 - 26x^2 - 3x + 90 = 0\) o.e.) \(\Rightarrow\) awrt \(-1.7\), 6, awrt 2.2 Scores M1A0 with or without a conclusion. Requires all 3 roots.

Notes

Work seen in (a) must be used in (b) to score

M1: Sets \(\dfrac{15x}{(2x+3)(x-3)} = 2x - 10\) and attempts to cross multiply to form a cubic equation.
Score for \(15x = (2x - 10)(2x + 3)(x - 3)\) or may be implied by an attempt to multiply two of the three brackets out e.g. \(15x = (2x - 10)(2x^2 - 3x - 9)\) or
\(15x = (x - 3)(4x^2 - 14x - 30)\) or \(15x = (2x + 3)(2x^2 - 16x + 30)\) (may be unsimplified)
Condone slips provided the intention is clear.
Condone invisible brackets to be implied by further work which is not the given answer.
Do not allow this mark to be scored for proceeding straight from \(\dfrac{15x}{(2x+3)(x-3)} = 2x - 10\) to a simplified expression of \(15x = 4x^3 - 26x^2 + 12x + 90\)

A1*: Achieves the given answer with sufficient intermediate steps seen and no errors including invisible brackets in the main body of their solution (ignore “side workings”)
Expect to see on the right hand side the product of a linear and a quadratic expression and an expression with all of the brackets multiplied out before simplifying to the given answer.
e.g.

\(\Rightarrow 4x^3 - 26x^2 - 3x + 90 = 0\) * which is A1*

Notes

Work seen in (a) or (b) must be used in (c) to score

M1: For the key step in dividing the cubic by \((x - 6)\). There are various methods or ways to present their solution so typically methods

• by inspection look for first and last terms e.g. \((x - 6)(4x^2 \pm \ldots x \pm 15)\)
• by division look for a three term quadratic quotient of \(4x^2 - 2x \pm \ldots\)

A1: \(4x^3 - 26x^2 - 3x + 90 = (x - 6)\left(4x^2 - 2x - 15\right)\) This may be implied by sight of \(\left(4x^2 - 2x - 15\right)\) in their working.

M1: Whilst not dependent on the previous method mark, it does require an attempt at dividing by \((x - 6)\) however poor to achieve a three term quadratic factor
Solves their \(4x^2 - 2x - 15 = 0\) using either the quadratic formula or completing the square. Usual rules apply for solving a quadratic. It cannot be scored for stating the roots directly via use of a calculator which is M0A0

A1: \((x =)\ \dfrac{1 - \sqrt{61}}{4}\) or exact equivalent e.g. \(\dfrac{1}{4} - \sqrt{\dfrac{61}{16}}\)
The root \((x =)\ \dfrac{1 + \sqrt{61}}{4}\) must be rejected if found. Isw if they put their exact value in decimal form.

Notes

Note: Condone different uses of ‘\(A\)’, ‘\(B\)’ etc for the M marks provided the intention is clear but values must be correctly attributed to each letter or implied (if embedded in a final expression) for the A mark.

B1: Any one of \(A = 3\), \(B = -2\), \(C = 4\), \(D = -5\) correct (usually \(A = 3\) is deduced) from correct work.
May be implied by \(3x\) seen. Condone \(A = 3x\) for this mark.

M1: This mark is for finding the value of \(B\), \(C\) or \(D\), using a valid method, which may be in one of the following ways:

• Attempts algebraic division of \(\left(3x^3 - 8x^2 - 6x - 11\right)\) by their expanded \((x + 1)(x - 3) = \left(\text{``}x^2 - 2x - 3\text{''}\right)\) or separately by both \((x + 1)\) and \((x - 3)\) as far as \(\boldsymbol{3x \pm 2}\)
  Ignore any remainder for this mark. For reference, the correct remainder is \(-x - 17\)
• Algebraic division might be seen as
  \(3x\left(x^2 - 2x - 3\right) \pm 2\left(x^2 - 2x - 3\right) + \text{``}9x - 4x - 6 - 6x - 11\text{''}\) o.e. and scores the B1 for the \(3x\) in the correct place and M1 for both the \(3x\) and the \(\pm 2\) in the correct places.
  
  In the next two bullets, they should achieve
  \(3x^3 - 8x^2 - 6x - 11 = (Ax + B)(x + 1)(x - 3) + C(x - 3) + D(x + 1)\) o.e. but condone small slips e.g. sign errors or a slip on a constant. The bracketed expressions should be in the correct places and \(Ax\) and \(B\) should both be multiplied by \((x + 1)(x - 3)\).
  Note that \(3x^3 - 8x^2 - 6x - 11 = C(x - 3) + D(x + 1)\) without sight of the identity above (condoning the previously mentioned slips) is an incorrect method and scores M0.
• Attempts to multiply up by \((x + 1)(x - 3)\) and substitutes one of \(x = -1\) or \(x = 3\) leading to \(C = \ldots\) or \(D = \ldots\) (you do not need to check the accuracy of their substitution)
• or attempts to multiply up by \((x + 1)(x - 3)\), expands and compares coefficients leading to a value for \(B\), \(C\) or \(D\). Do not be concerned about how they expand the brackets.
• Attempts at the cover up method may be seen. If unsure then send to review.

M1: For finding values of all four constants via a valid method.
Not dependent on the previous method mark so B1M0M1A0 is possible.

• If attempting algebraic division, they must have a quotient of \(3x \pm k\) for constant \(k\) (\(3x \pm 2\) is not required for this mark) and they must go on to deal with their remainder using a correct method, i.e., \(\text{``}{-}x - 17\text{''} = C(x - 3) + D(x + 1)\) followed by either comparing coefficients or substitution of appropriate values. You do not need to check the accuracy of their substitution.
• For those attempting repeated algebraic division, the values of \(C\) and \(D\) must come from a correct method and so cannot be equal to e.g. the remainders from successive division by \((x + 1)\) and \((x - 3)\) or vice versa. If, e.g., they divide by \((x + 1)\) (to get remainder \(R_1\)) and then \((x - 3)\) (to get remainder \(R_2\)) they should get \(3x \pm k + \dfrac{R_1}{(x + 1)(x - 3)} + \dfrac{R_2}{(x - 3)}\). They can then either combine the two fraction terms before using partial fractions or they can use partial fractions on \(\dfrac{R_1}{(x + 1)(x - 3)}\) before combining their two terms in \(\dfrac{1}{(x - 3)}\).
• If using \(3x^3 - 8x^2 - 6x - 11 = (Ax + B)(x + 1)(x - 3) + C(x - 3) + D(x + 1)\) o.e. they may use a combination of substitution and comparing coefficients, but the method must be valid. Do not be concerned about how they expand the brackets, and you do not need to check the accuracy of their substitution.
• If using substitution only they are likely to use \(x = -1\), 3 and 0 but the final value for \(x\) could be anything (and in fact they could use three/four different values for \(x\)). You do not need to check the accuracy of their substitution.
• If attempting to compare coefficients, do not be concerned about how they expand the brackets.

A1: Requires:

• all of \(A = 3\), \(B = -2\), \(C = 4\) and \(D = -5\) which may be stated separately in their work (\(A = 3x\) that is not recovered would score final A0).
• or \(3x - 2 + \dfrac{4}{x + 1} - \dfrac{5}{x - 3}\) seen. Condone \(3x + -2 + \dfrac{4}{x + 1} + \dfrac{-5}{x - 3}\)

There is no requirement to write out the form given in the question once correct values for \(A\), \(B\), \(C\) and \(D\) have been seen.

ISW after the four correct values (e.g. if they swap \(C\) and \(D\) in the fractions so \(A = 3\), \(B = -2\), \(C = 4\), \(D = -5\) followed by \(3x - 2 - \dfrac{5}{x + 1} + \dfrac{4}{x - 3}\) scores full marks).

Repeated Algebraic Division

An example for repeated algebraic division. Note that the repeated division could be carried out in the other order – i.e. dividing by \((x - 3)\) first followed by \((x + 1)\).

Dividing \(3x^3 - 8x^2 - 6x - 11\) by \(x + 1\) gives quotient \(3x^2 - 11x + 5\) and remainder \(-16\)
(working: \(3x^3 + 3x^2\); \(-11x^2 - 6x\); \(-11x^2 - 11x\); \(5x - 11\); \(5x + 5\); \(-16\))

Followed by dividing \(3x^2 - 11x + 5\) by \(x - 3\), giving quotient \(3x - 2\) and remainder \(-1\)
(working: \(3x^2 - 9x\); \(-2x + 5\); \(-2x + 6\); \(-1\))
B1 Scored for either ‘3’ or ‘\(-2\)’ embedded in the quotient.
1st M1 Scored for achieving \(3x \pm 2\)

\(\dfrac{3x^3 - 8x^2 - 6x - 11}{(x + 1)(x - 3)} = 3x - 2 + \dfrac{-16}{(x + 1)(x - 3)} + \dfrac{-1}{(x - 3)}\)
\(\dfrac{3x^3 - 8x^2 - 6x - 11}{(x + 1)(x - 3)} = 3x - 2 + \dfrac{-16}{(x + 1)(x - 3)} + \dfrac{-(x + 1)}{(x + 1)(x - 3)}\)
Correctly deals with both remainders from the repeated division.

\(\dfrac{3x^3 - 8x^2 - 6x - 11}{(x + 1)(x - 3)} = 3x - 2 + \dfrac{-x - 17}{(x + 1)(x - 3)}\)
\(-x - 17 = C(x - 3) + D(x + 1)\)   Sets up an identity in \(x\) and \(C\) and \(D\) using partial fractions.
\(x = 3 \Rightarrow D = -5\) and \(x = -1 \Rightarrow C = 4\)   2nd M1 Finds the values of all four constants via a correct method (in this case by substitution of appropriate values).
\(A = 3\), \(B = -2\), \(C = 4\), \(D = -5\)   A1 \(A = 3\), \(B = -2\), \(C = 4\) and \(D = -5\)

Alternatively, from \(\dfrac{3x^3 - 8x^2 - 6x - 11}{(x + 1)(x - 3)} = 3x - 2 + \dfrac{-16}{(x + 1)(x - 3)} + \dfrac{-1}{(x - 3)}\):
\(\dfrac{3x^3 - 8x^2 - 6x - 11}{(x + 1)(x - 3)} = 3x - 2 + \dfrac{F}{(x + 1)} + \dfrac{G}{(x - 3)} + \dfrac{-1}{(x - 3)}\)
\(-16 = G(x + 1) + F(x - 3)\)   Alternatively, splits the first fraction using partial fractions and sets this equal to their remainder divided by \((x + 1)(x - 3)\) and multiplies up correctly.
\(x = -1 \Rightarrow F = 4\), \(x = 3 \Rightarrow G = -4\)   2nd M1 Finds the values of all four constants via a valid method.
\(\dfrac{3x^3 - 8x^2 - 6x - 11}{(x + 1)(x - 3)} = 3x - 2 + \dfrac{4}{(x + 1)} + \dfrac{-4}{(x - 3)} + \dfrac{-1}{(x - 3)}\)
\(\dfrac{3x^3 - 8x^2 - 6x - 11}{(x + 1)(x - 3)} = 3x - 2 + \dfrac{4}{(x + 1)} - \dfrac{5}{(x - 3)}\)   A1 \(A = 3\), \(B = -2\), \(C = 4\) and \(D = -5\) seen embedded in a final expression.

Notes

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

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

Notes

Alternative for the final 3 marks:

Mark (b) and (c) together

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

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

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

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

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

Alternative for final 3 marks:

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

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

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

Note

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

Notes

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

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

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

A1: Correct expression not just values for the constants.

Notes

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

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

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

Using the differential equation:

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

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

Typical solution

Typical solution

Typical solution

(i)

(ii)

\(\dfrac{1}{2}\), \(\dfrac{3}{4}x\) and \(\dfrac{9}{8}x^2\) form a geometric sequence with common ratio \(\dfrac{3}{2}x\)

Typical solution

Let \(x = -\dfrac{1}{3}\)

Typical solution

(i)

(ii)

Typical solution

\(4\)

Typical solution

Typical solution

(i)

(ii)

(iii)

Notes

B1: Identify correct partial fractions

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

A1: Obtain one correct value www

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

Notes

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

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

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

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

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

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

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

Notes

B2: B1 for 6 or \(a^5\) in their final answer or for \(8a^6 \times \frac{3}{4}a^{-1}\)
\(\frac{24}{4}a^5\) as a final answer is B1 B0

Notes

M1: M1 for either \(4x^2 - 9 = (2x-3)(2x+3)\) or \(2x^2 + 5x - 12 = (2x-3)(x+4)\)
A correct answer www implies M1

A1: oe, for example, \((x+4)^{-1}\)
Do not ISW if further incorrect simplification occurs e.g. \(\frac{1}{x+4} = \frac{1}{x} + \frac{1}{4}\) is A0

Notes

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

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

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

Notes

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

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

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

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

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

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

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

Notes

M1: M1 for \(x(x + 2) - (x + 1)(x - 1)\) oe

M1: Multiply out brackets. Allow one error
Ignore denominator even if “= 0”

A1: NB correct with no working: SC B1

Alternative method

M1: M1 for attempt “cross-multiply”.

M1: Multiply out brackets. Allow one error

Notes

M1: or cubic in \(x\). Condone quadratic in \(x\) with \(x = \frac{1}{x^3}\) or \(x = x^3\)
Must see attempt at correct method for this mark
Allow arithmetical errors

B1: Can be scored without M1. Condone \(x = 1, -\frac{1}{8}\) or \(x = 1, -8\)
Ignore \(x^3 = 0\), if seen, for this mark

B1f: ft their \(x^3\) or \(\dfrac{1}{x^3}\). If also \(x = 0\), B0
NB correct with no working: M0B0B1

Notes

M1: Condone incorrect or omitted brackets
Any correct step after log(both sides)
or ANY correct step using indices

A1: NB correct with no working or T & I: SC B1

Notes

M1: Sets up partial fractions of the correct form

M1: Clears the denominator (for their form with at least two fractions). Not dependent on previous M1

M1: Uses at least one value of \(x\) leading to a value for one of the constants (for their form). Allow for equating coefficients for at least one power of \(x\) leading to a value for one of the constants

A1: Fully correct expression must be seen

SC: If \(\dfrac{7x-25}{(x-1)(x-4)^2} = \dfrac{A}{x-1} + \dfrac{Bx+C}{(x-4)^2}\) used, award M0 and up to M1 M1 SC B1 for \(\dfrac{7x-25}{(x-1)(x-4)^2} = \dfrac{-2}{x-1} + \dfrac{2x-7}{(x-4)^2}\)

Notes

M1: separation of variables; condone omission of integral signs or of \(\mathrm{d}x\) and/or \(\mathrm{d}y\); allow 1 slip such as omission of 9 or sign error in bracket

M1: allow 1 sign error in bracket

A1: one of two terms correct

A1: all correct

M1*: any ln integral correct; FT their \(A\) and \(B\); condone omission of \(+\,c\)

A1: all three terms correct including \(+\,c\);
may see \(\frac{1}{9}\ln y = \frac{1}{3}\ln(x-1) - \frac{1}{3}\ln(x+2) + c\)
NB \(\ln y + c\) is equivalent to \(\ln Ay\) where \(A\) is a constant

M1dep*: substitution of (2, 16) in their expression; may be implied by eg \(\ln 16 = 3\ln 1 - 3\ln 4 + c\); must see substitution for incorrect expressions

A1: may see \(c = \frac{5}{9}\ln 4\) oe
allow exact equivalents only

M1: correctly combines their RHS into a single logarithm; their \(+\,c\) must be correctly incorporated into their \(\ln\mathrm{f}(y)\) or their \(\ln\mathrm{f}(x)\)

A1: all correct

M1: correctly exponentiates their expressions, may be awarded before combination into single logarithm

A1: all correct; must see “\(y =\)” at some stage

Alternatively, for the last 6 marks

M1dep*: correctly combines their RHS into a single logarithm;

A1: all correct

M1: correctly exponentiates their expressions; may be awarded before combining into single logarithm

A1: all correct

M1: substitution of (2, 16) in their expression; may be implied by eg \(16 = \frac{1^3}{4^3} \times D\); must see substitution for incorrect expressions; may be awarded before exponentiating

A1: all correct; must see “\(y =\)” at some stage

Notes

B1: both factors seen

M1: setting up partial fractions using their factors. May be implied by correct expression as final answer.

M1: method for finding either \(A\) or \(B\) soi

A1: both \(A\) and \(B\) correct if clear which denominator they apply to
ISW if an error made only in the transcription to final answer

Notes

M1: Clearing the denominators oe

A1: For one correct coefficient

A1: Correct partial fractions seen
Accept just values for \(A\) and \(B\) if defined

Notes

M1: Method mark is implied by correct answer

A1: Final solution needed for A1.
Can be recovered in 10(b)

Notes

M1: Separation of variables – both sides seen
Integral signs, \(\mathrm{d}y\) and \(\mathrm{d}x\) needed

M1: Use of their (a). RHS only needed
Condone no \(\mathrm{d}x\)

M1: Integration. One correct \(x\) term (ft their partial fractions)
Condone missing modulus signs

B1: Finding constant (FT their (a))

M1: For use of \(a\ln m = \ln m^a\) and \(\ln m - \ln n = \ln\dfrac{m}{n}\)
Not dep on \(c\)
Condone missing modulus signs throughout

A1: Answer with correct values of \(A\) and \(B\)

Notes

B1: Denominator may be multiplied out

Notes

M1: Separates variables (inc \(\mathrm{d}x\) and \(\mathrm{d}t\))
400 either side

M1: Use of result from (a) or method for partial fractions

M1: Attempt to integrate (at least one term correct)

A1: All integration correct
Or \(\ln\left|\dfrac{Ax}{400 - x}\right| = t\)

M1: Finding constant ( or \(A = 3\))

M1: Using log laws to combine the \(c\) value

M1: Attempt to rearrange to find \(x\)

A1: Must be rounded to nearest whole number

Alternative method (OR)

M1: Separates variables

M1 M1: Use result from (a) or method for partial fractions
Attempt to integrate (at least one term correct)

A1: All integration correct

M1: Limits applied (condone one error)

M1: Combining log terms together

M1: Attempt to rearrange to find \(X\)

A1: Must be rounded to nearest whole number