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

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

M1: Either

• \(x \leqslant -1\) o.e. e.g. \(-1 \geqslant x\)
• \(2 \leqslant x \leqslant 5\) o.e.

but condone use of strict inequalities anywhere for this mark.
e.g. \(2 \lt x \lt 5\) or \(2 \lt x \leqslant 5\) or \(2 \leqslant x \lt 5\) May also write e.g. \(x \lt 5\) and \(x \gt 2\) which scores M1 but not "\(x \lt 5\) or \(x \gt 2\)"

Allow interval notation such as e.g. \([2, 5]\) or \((-\infty, -1]\) or condone e.g. \((2, 5)\)

Ignore incorrect inequality statements not related to the one which is valid.
e.g. “\(2 \leqslant x \lt 5\) and \(x \gt -1\)” which scores M1 for the first inequality.

A1: Requires \(\{\ \}\) and \(\cup\)

\(\{x : x \leqslant -1\} \cup \{x : 2 \leqslant x \leqslant 5\}\) or \(\{x \mid x \leqslant -1\} \cup \{x \mid 2 \leqslant x \leqslant 5\}\) either way round

but condone \(\{x \leqslant -1\} \cup \{2 \leqslant x \leqslant 5\}\), \(\{x \leqslant -1 \cup 2 \leqslant x \leqslant 5\}\).

Allow e.g. \(\{x : x \leqslant -1\} \cup \{x : 2 \leqslant x \cap x \leqslant 5\}\)

Use of \(\cap\) to join the two separate regions is A0
It is acceptable (but not required) to mention \(\mathbb{R}\)
e.g. \(\{x : x \in \mathbb{R}, x \leqslant -1\} \cup \{x : x \in \mathbb{R}, 2 \leqslant x \leqslant 5\}\)

Condone use of a lower limit written as e.g. \(\{x : -\infty \leqslant x \leqslant -1\} \cup \{x : 2 \leqslant x \leqslant 5\}\)

Notes

Note a correct equation written down scores all 3 marks.
A correct expression but missing e.g. \(y = \ldots\) or \(\mathrm{f}(x) = \ldots\) scores M1dM1A0

M1: Forms the equation of the form \((y =)\ \alpha(x+1)^2(x-5)^2\). Condone \(\alpha = 1\)
Award for sight of \(\alpha(x+1)^2(x-5)^2\) even with \(\alpha = 1\) i.e. \((x+1)^2(x-5)^2\)

dM1: Substitutes \((0,\ -75)\) into the form \(y = \alpha(x+1)^2(x-5)^2\) and attempts to find the value for \(\alpha\). It is dependent on the previous method mark.

A1: \(y = -3(x+1)^2(x-5)^2\) o.e. (e.g. \(y = -3x^4 + 24x^3 - 18x^2 - 120x - 75\))
isw after a correct answer. Condone \(\mathrm{f}(x) = -3(x+1)^2(x-5)^2\) but not \(C = -3(x+1)^2(x-5)^2\)
A correct equation scores all 3 marks. Allow if seen in (c)
isw if they attempt to multiply out.

Alternative I part (b):

Using the form \(y = ax^4 + bx^3 + cx^2 + dx + e\), then setting up and solving simultaneous equations.
There are various versions of this but can be marked similarly.

M1: Sets \(e\) equal to \(-75\) (may just be seen in their equation) and forms three correct different equations in \(a\), \(b\), \(c\) and \(d\) which may be unsimplified.
Note that the form \(y = ax^4 + bx^3 + cx^2 + dx + e\) is M0 until e is set equal to \(-75\)
There are 5 equations that can be formed, only 3 are necessary for this mark.
Do not condone slips.

dM1: Forms four correct different equations and solves to find values for \(a\), \(b\), \(c\) and \(d\). You do not need to be concerned by the process of solving. A calculator can be used to solve the equations.

A1: \(y = -3x^4 + 24x^3 - 18x^2 - 120x - 75\) o.e. isw if they attempt to factorise but withhold this mark if they e.g. divide all terms by 3.
Condone \(\mathrm{f}(x) = \ldots\) but not \(C = \ldots\)
A correct equation scores all 3 marks. Allow if seen in (c)

Alternative II part (b): Uses the form \(y = (x+1)(x-5)(ax^2 + bx + c)\)

M1: Substitutes \(x = 0,\ y = -75\)   \(-75 = -5c \Rightarrow c = 15\), multiplies out, differentiates
\(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = (2x - 4)(ax^2 + bx + 15) + (x^2 - 4x - 5)(2ax + b)\)
and forms a correct equation in \(a\) and \(b\) which may be unsimplified.

dM1: Forms two correct different equations and solves to find values for \(a\) and \(b\). You do not need to be concerned by the process of solving. A calculator can be used to solve the equations.

A1: \(y = (x+1)(x-5)(-3x^2 + 12x + 15)\) o.e. isw if they attempt to multiply out or factorise
Condone \(\mathrm{f}(x) = \ldots\) but not \(C = \ldots\) but withhold this mark if they e.g. divide all terms by 3. A correct equation scores all 3 marks. Allow if seen in (c)

Alternative III part (b): Uses \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \beta(x+1)(x-2)(x-5)\) (\(\beta\) may be 1) and integrates.

M1: Integrates \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) \beta(x+1)(x-2)(x-5)\) to \((y =)\ \beta\left(\dfrac{1}{4}x^4 - 2x^3 + \dfrac{3}{2}x^2 + 10x + k\right)\) and forms one correct equation using either \((0,\ -75)\): \(-75 = \beta k\) (allow \(-75 = k\))
\((-1,\ 0)\): \(0 = \beta\left(\dfrac{1}{4} + 2 + \dfrac{3}{2} - 10 + k\right)\)   \((5,\ 0)\): \(0 = \beta\left(\dfrac{625}{4} - 250 + \dfrac{75}{2} + 50 + k\right)\)

dM1: Forms a different equation using one of \((0,\ -75),\ (-1,\ 0),\ (5,\ 0)\) and solves to find values for \(\beta\) and \(k\). You do not need to be concerned by the process of solving. A calculator can be used to solve the equations.

A1: \(y = -12\left(\dfrac{1}{4}x^4 - 2x^3 + \dfrac{3}{2}x^2 + 10x + \dfrac{25}{4}\right)\) o.e. isw if they attempt to multiply out or factorise
Condone \(\mathrm{f}(x) = \ldots\) but not \(C = \ldots\) but withhold this mark if they e.g. divide all terms by 3. A correct equation scores all 3 marks. Allow if seen in (c)

Notes

M1: Substitutes \(x = 2\) into their \(y = -3(x+1)^2(x-5)^2\) (must be a quartic in any form) and proceeds to find a value for \(y\). Sight of their \(\pm y\) (or \(\pm 243\)) scores M1.
You may need to check this on your calculator if only a value is seen.

A1ft: \(0 \lt k \lt 243\) o.e. ft on their negative \(y\) value at \(x = 2\).
Allow use of set notation, interval notation and allow e.g. \(k \lt 243,\ k \gt 0\) but do not allow OR or \(\cup\). Do not accept \(0 \leqslant k \leqslant 243\) o.e.
If there are multiple attempts at describing the region, mark what appears to be their final answer.
This mark can only be scored if they have a negative quartic graph function
i.e. \(\alpha \lt 0\) for their \(y = \alpha(x+1)^2(x-5)^2\) or \(a \lt 0\) for their \(y = ax^4 + bx^3 + cx^2 + dx + e\)

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

B1: States or uses \(a = 4\) e.g. may be seen in their attempt at dividing algebraically by \(x + 4\)

M1: Attempts to divide \(\mathrm{f}(x)\) by \((x + 4)\) to find a three-term quadratic \(Q(x)\). There are various methods or ways to present their solution so typically methods

• by inspection look for \(2x^3 + 3x^2 - 16x + 16 = (x + 4)(2x^2 + \ldots x \pm 4)\)
• by division look for a quadratic quotient of \(2x^2 - 5x \pm \ldots\)

A1: \((x + 4)(2x^2 - 5x + 4)\) condone the missing trailing bracket i.e. \((x + 4)(2x^2 - 5x + 4\) isw if they attempt to factorise their quadratic factor.
Allow to be scored if seen in (b).

Notes

M1: Attempts to show that their three-term quadratic “\(2x^2 - 5x + 4\)” does not have any roots:

• Attempts the discriminant
  e.g. \(b^2 - 4ac = 25 - 4 \times 2 \times 4\ (= -7)\) (may be embedded in the quadratic formula)
• Attempts to use the quadratic formula
  e.g. \((x =)\ \dfrac{5 \pm \sqrt{(-5)^2 - 4 \times 2 \times 4}}{4}\) but do not allow directly from a calculator \(\dfrac{5 \pm \sqrt{7}i}{4}\)
• Attempts to complete the square
  e.g. \(2x^2 - 5x + 4 = 2\left(x^2 - \dfrac{5}{2}x\right) + 4 = 2\left(x - \dfrac{5}{4}\right)^2 + \ldots\ \left(= 2\left(\left(x - \dfrac{5}{4}\right)^2 - \dfrac{25}{8} + 4\right)\right)\)
• Uses calculus to find the turning point
  e.g. \(\dfrac{\mathrm{d}(2x^2 - 5x + 4)}{\mathrm{d}x} = 4x - 5 = 0 \Rightarrow x = \dfrac{5}{4} \Rightarrow y = \ldots\)

Note that any attempts using the discriminant or quadratic formula must have the values embedded in the correct places (may be partially evaluated) to score M1

A1: Dependent on a correct \(Q(x) = 2x^2 - 5x + 4\)

Fully correct argument that requires:

• Fully correct work
• A justification depending on strategy and no incorrect reasoning seen
• A conclusion

Examples below – Note we must see working before they proceed to a correct root or minimum value – see M1 for guidance

Note that it is possible to justify and conclude in one step by using phrases e.g. “no more (real) roots” or “no other (real) roots”

e.g. \(2x^2 - 5x + 4 \Rightarrow b^2 - 4ac = 25 - 32 = -7\) so no more roots which scores M1A1

Condone \(25 - 32 \lt 0\) as a justification that the quadratic has no real roots

Condone the conclusion \(-4\) is the only (real) solution (instead of (real) root).

Note if \((x + 4)\) is described as a root or \(-4\) is described as a factor this scores A0

Notes

B1: Takes a factor of \(2x\) or \(x\) out of the given cubic correctly.
Dividing by 2 and then achieving \(x\left(x^2 - 12x + 20\right)\) does not score this mark unless recovered.

M1: Attempts to factorise their quadratic. Invisible brackets may be implied by later work.
Score for \((x \pm b)(x \pm d)\) where \(|bd| = 20\) coming from \(x^2 \pm 12x \pm 20\)
or for \((ax \pm b)(cx \pm d)\) where \(|ac| = 2\) and \(|bd| = 40\) coming from \(2x^2 \pm 24x \pm 40\)
May be scored if they have divided by \(x\) but not from e.g. \(2x^3 - 24x^2 + 40x \rightarrow (2x - 4)(x - 10)\) without clear indication that they have divided by or taken out a factor of \(x\) or \(2x\).
There may be incorrect intermediate steps such as \((2x - 20)(2x - 4)\) which do not score the mark on their own but may be ignored if they return to e.g. \((2x - 20)(x - 2)\).

A1: \(2x(x - 2)(x - 10)\) or e.g. \(2(x - 10)(x - 2)x\). Do not accept e.g. \(x(2x - 4)(x - 10)\). Ignore = 0
ISW after a fully correct factorisation e.g. \(2x(x - 2)(x - 10)\) that becomes \(x(x - 2)(x - 10)\) and ignore any attempt to find roots (before or after factorisation). Allow \((2x)(x - 2)(x - 10)\)

Alternative:

B1: Takes a factor of \((x - 2)\) or \((x - 10)\) out correctly i.e. \((x - 2)\left(2x^2 - 20x\right)\) or \((x - 10)\left(2x^2 - 4x\right)\). Must be seen as a product of factors and not just in a division attempt but may be implied by later work.

M1: Attempts to factorise their quadratic. Invisible brackets may be implied by later work.
Score for \(ax(bx \pm c)\) where \(|ab| = \text{``}2\text{''}\) and \(|ac| =\) their \(\text{``}20\text{''}\) or \(\text{``}4\text{''}\) coming from \(Ax^2 + Bx\)

A1: As main scheme.

Note: Solutions that solve the cubic = 0 to achieve \(x = 0\), 2 and 10 and then arrive at e.g. \(x(x - 2)(x - 10)\) score no marks without a prior line of working such as \(x\left(x^2 - 12x + 20\right)\) (would score M1) or \(x\left(2x^2 - 24x + 40\right)\) (would score B1).

Some examples:

• \(2x^3 - 24x^2 + 40x \rightarrow x(x - 2)(x - 10)\) scores B0M0A0
• \(2x^3 - 24x^2 + 40x \rightarrow (2x - 4)(x - 10)\) scores B0M0A0
• \(x^2 - 12x + 20 \rightarrow (x + 2)(x - 10)\) scores B0M1A0
• \(x^3 - 12x^2 + 20x \rightarrow x(x - 2)(x - 10)\) scores B0M1A0
• \(x\left(2x^2 - 24x + 40\right) \rightarrow x(x - 2)(x - 10)\) scores B1M0A0
• \(x\left(2x^2 - 24x + 40\right) \rightarrow \{x(2x + 2)(2x + 40) \rightarrow\}\ x(2x + 2)(x + 20)\) scores B1M1A0
• \(2x\left(x^2 - 12x + 20\right) \rightarrow 2x(x - 1)(x + 20)\) scores B1M1A0
• \(2x(x - 2)(x - 10)\) on its own scores B1M1A1

Notes

Note: Ignore any use of \(\mathrm{f}(x)\) in place of \(\mathrm{g}(x)\) throughout.

M1: Attempts \(\mathrm{g}(3) = 0\) to set up a linear equation in \(k\). The = 0 may implied by their value of \(k\).

Expect to see 3 substituted for \(x\) at least twice but condone minor slips copying the function. May be scored for e.g. \(81 - 180 + 3(k+17) + k = 0\)
Missing brackets may be recovered.

Attempting \(\mathrm{g}(-3) = 0\) scores M0 but note that the second M1 is available.

If algebraic division is attempted, they need to achieve a linear remainder in \(k\) only and set =0   Condone slips in their calculations.

As a minimum, expect to see \(3x^2 + \lambda x,\ \lambda \neq 0\) as their quotient leading to a linear remainder in k only set = 0 (the = 0 may be implied by their value for \(k\)).

For reference, the correct division is

You may also see variations on the table below.
Here, the M1 is scored when the sum of both coefficients of \(x\) are equated to \((k+17)\)

\(33-\dfrac{k}{3}=k+17\) scores M1

M1: Scored for attempting to solve a linear equation in \(k\) having attempted \(\mathrm{g}(\pm 3) = 0\)
Do not be concerned about the process, e.g. \(-81 + 180 - 3(k+17) + k = 0 \rightarrow k = \ldots\) scores M1.

Via division they must have a linear remainder in \(k\) set = 0
The = 0 may be implied by their value for \(k\) in all approaches.

A1: Obtains \(\{k=\}\,12\) only. Do not accept e.g. \(\dfrac{48}{4}\) Allow slips in working to be recovered.

Condone e.g. \(x = 12\) provided it has come from a linear equation in \(k\).

Note that e.g. \(3(3)^3 - 20(3)^2 + 3(k+17) + k\{=0\} \rightarrow k = 12\) and \(81 - 180 + 3k + 51 + k\{=0\} \rightarrow k = 12\) are sufficient to imply M1M1A1.

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

\(\therefore (2x + 1)\) is a factor of \(\mathrm{P}(x)\)

Typical solution

Typical solution

There are two factors. Both factors are integers not equal to 1 so \(4n^3 + 8n^2 + 11n + 4\) is never prime.

Typical solution

Typical solution

Typical solution

\(-5\)

Typical solution

Typical solution

(i)

(ii)

(iii)

The translated graph will only have one real solution.

Typical solution

Hence \((x + 2)\) is a factor of \(\mathrm{p}(x)\) for all values of \(b\)

Typical solution

(i)

(ii)

\(b = 4\) gives only one point of intersection

Notes

B1: Show \(\mathrm{f}\left(\tfrac{1}{4}\right) = 0\), some detail required.
B0 for just f(0.25) = 0, but condone seeing either just the substitution or just the evaluated terms
Could use division by \((4x - 1)\) or \(\left(x - \tfrac{1}{4}\right)\) but must identify remainder of 0

Notes

B1: Identify factor of \((4x - 1)\)
Allow factor of \(\left(x - \tfrac{1}{4}\right)\)

M1: Must be a complete method ie attempt all 3 terms to obtain \(x^2\) and one other correct term (allow one slip in method)
Could be implied by \(A = 1\) and one other correct if using coefficient matching
Condone division by \(\left(x - \tfrac{1}{4}\right)\), to obtain \(4x^2\) and either \(-24x\) or \(-64\)

A1: Integer coefficients now required
Must be written as a product, so cannot be implied by eg correct quotient appearing following division by \((4x - 1)\) but the two factors never combined
Could be \((1 - 4x)(-x^2 + 6x + 16)\)

If division was used in part (a) then quotient must appear in part (b), but evidence for B1M1 could be in (a)
If \((x - 8)(x + 2)\) seen before the quadratic factor then both roots must be justified (eg factor theorem), otherwise M0 (but could still get B1)

Notes

M1: Attempt to find \(y\) from at least one positive root for \(\mathrm{e}^y\)
Attempt to link \(\mathrm{e}^y\) to the root(s) of the cubic in \(x\), and then solve \(\mathrm{e}^y = k\) to obtain \(y = \ln k\), where \(k\) is one of their positive roots

A1: Obtain at least one correct solution in the required form
\(y = -2\ln 2\) comes from the given root, but \(y = 3\ln 2\) must come from the correct solution of the correct quadratic

A1: Must come from the correct solution of the correct quadratic
Allow BOD if \(\ln(-2)\) also seen

B1: Reject \(\mathrm{e}^y = -2\) with a reason
Must have some reason, eg ‘\(\mathrm{e}^y\) is always positive’, ‘\(\mathrm{e}^y\) cannot be negative’, ‘cannot take log of a negative number’, ‘not defined’, ‘not real’, ‘no solutions’
B0 for ‘math error’, ‘does not work’, ‘not possible’, N/A etc
\(\mathrm{e}^y = -2\) must come from the correct solution of the correct quadratic

Notes

(a)(i)

Condone a small slip in early irrelevant working before substituting, but must be fully correct after this.

B1: oe in terms of \(\theta\), not \(2\theta\). May be implied by next line

B1: oe in terms of \(\cos\theta\) only
Must reach AG or give a conclusion (even if just ‘QED’) www

Alternative 1

B1: oe in terms of \(\theta\), not \(2\theta\). May be implied by next line

B1: oe in terms of \(\cos\theta\) only
Must reach AG or give a conclusion (even if just ‘QED’) www

Alternative 2

B1: oe in terms of \(\theta\), not \(2\theta\). May be implied by next line

B1: oe in terms of \(\cos\theta\) only
Must reach AG or give a conclusion (even if just ‘QED’) www

(a)(ii)

M1: oe

A1: Allow A1 for two correct values of \(3\theta\). Ignore other values.
This mark is not implied by correct final answers but accept equivalent correct working e.g. a graph of \(\cos 3\theta\).
Accept radians for this mark only: \(3\theta = \frac{\pi}{4}, \frac{7\pi}{4}, \frac{9\pi}{4}\)

A1: cao. Ignore values outside of the range 0-180°, but do not accept radians for this mark. This mark may be given following M1A0.

Notes

(b)(i) B1: Must see a correct multiplied out form and AG or conclusion.

(b)(ii)

B1: oe, must be exact

B1: May see \(x = \frac{2\sqrt{2} \pm \sqrt{24}}{8}\) etc.

Notes

B1: soi - for ‘spotting’ the connection. This mark can be gained regardless of their answers to (a)(ii) and (b)(ii).
\(x = \cos\theta\) or 3 correctly paired roots are sufficient for this mark
Condone \(x = \cos 15\)

B1: Justification for selecting this root (may say e.g. “cos 15 is the only positive root”)
This may be implied by matching each pair of answers correctly (but all three must be present or mentioned).

Notes

B1: Must see substitution; or factorise showing factor \((x - 3)\)
No conclusion required beyond =0 (so e.g. accept long division leading to remainder 0).

Notes

M1: Attempt quadratic factor by inspection or division.
Look for \((x - 3)\) and a quadratic or two linear factors.
NB this step may be seen in (a).

A1: or \((x - 3)(x + 1)^2\)

Notes

Fig 1.2: Accept values between 1.9 and 2.1

Fig 1.3: Accept values between 0.9 and 1.1

Notes

(a)(i)

M1: Attempt at differentiating \(\mathrm{f}(x)\) with at least one non-zero term correct
M0 for \(\mathrm{f}'(x) = x^2 + p + \dfrac{q}{x}\)

A1: Correct value for \(p\)

(a)(ii)

M1: Substituting \(x = 2\) into \(\mathrm{f}(x)\) and equating to 0 or for correctly re-writing as \((x - 2)(x^2 + 2x + 4 + p)\) (with \(p\) or their value of \(p\) from (a)(i))
Could be in terms of \(q\) only e.g., \(2^3 + 2 \times \mathit{their}(p) + q = 0\)
Possibly seen in an attempt at long division

A1: Correct value for \(q\)

Notes

B1: Substituting \((x \pm 2)\) into both \(x\) terms of \(y = \mathrm{f}(x)\)

B1: Subtracting 3 (oe) at some stage
Allow with \(p\) and \(q\) or with incorrect values of \(p\) and/or \(q\) – note that using 2 vertically and/or 3 horizontally cannot be treated as a MR

M1: Correct expansion of \((x \pm 2)^3\). Can be unsimplified for M1. If correct bracketing not seen then must be implied by later working

A1: cao – condone just the expression \(x^3 - 6x^2 + 13x - 23\) (so do not need to see \(y = \ldots\))
For reference: \(a = -6\), \(b = 13\), \(c = -23\)

Notes

or \((n + 1)(n^2 - n + 1)\)

Notes

B1: Stated. Allow \(x\) instead of \(n\)
NOT \(n^3 + 1\) can be expressed as \((n + 1)(n^2 - n + 1)\)

B1: Must see \(n \gt 2\)
Allow omission of this step

A1: Conclusion stated, from correct working seen.
Dep at least B1M1 and correct reasoning given

SC: \((n + 1) \gt 1\) or \(n + 1 \gt 3\) (or \(n^2 - n + 1 \gt 1\))   B1
\((n + 1)\) & \((n^2 - n + 1)\) are factors of \(K\)   B1

Notes

B1: Attempt \(\mathrm{f}(0.5)\) and show equal to 0
Must be using factor theorem so B0 for alternative methods
B0 for just \(\mathrm{f}(0.5) = 0\)
Condone \(2(0.5)^3 - 3(0.5)^2 - 11(0.5) + 6 = 0\)

Notes

M1: Attempt complete division by \((2x - 1)\)
DR so need to see quadratic factor
Allow equivalent complete methods eg coefficient matching / inspection / grid method
Condone slip(s) in otherwise correct method

A1: Obtain correct quadratic factor
Seen in division / correct coeffs eg \(A = 1\) etc / at top of grid

A1: Obtain correct fully factorised \(\mathrm{f}(x)\)
Must be seen as a product of all 3 factors
SC B1 for correct factorisation with no DR

Notes

B1: State or imply that \(x = 2^y\)
Could be implied by equating \(2^y\) to at least one of their roots

M1: Attempt to find at least one value of \(y\)
Exact or decimal

A1: Obtain both correct values, and no others
1.58 or better, or \(\dfrac{\log_n 3}{\log_n 2}\) for \(\log_2 3\)
Must give reason for \(2^y = -2\) having no solution
eg cannot log a negative number
\(2^y\) always greater than 0

Notes

B1: arithmetic must be correct and 4 terms must be seen

Notes

M1: may be embedded in long/synthetic division; allow one coefficient error

A1: allow \((x-3)(2x+1)(2x+1)\); mark the final answer; supporting working must be seen for the award of this mark

Alternative method

M1: for any two equations from equating coefficients; may be embedded in expansion; allow one coefficient error

A1: supporting working must be seen for the award of this mark

Notes

M1: cubic of correct orientation; curve touches \(x\)-axis on correct side of origin for their \(-\frac{b}{a}\) and intersects positive \(x\)-axis;

A1: intercepts at \(\left(-\frac{1}{2},\ 0\right)\), \((3,\ 0)\) and \((0,\ -3)\) labelled or identified on sketch; local minimum in 4^th quadrant

Notes

M1: Finding one root or factor by factor theorem or division to a remainder of 0. A conclusion is needed for this M1.

M1: Factorising to find quadratic factor or division (method seen) or factor theorem again (substitution shown) to get a different root/factor

DM1: Completion, all 3 factors/roots seen, dep on previous M1
Might not be in same place

A1: All 3 points as coordinates or pairs of values
Dep on M3

Notes

M1: Substitution seen. Do not allow for division here

A1: Clear conclusion.

Notes

M1: Attempt to divide or factorise

A1: Correct quadratic factor seen

B1: Product of linear factors seen

A1: Do not allow without full working

Also allow M1 A1 for \(\mathrm{f}(x) = (x-3)\left(x^2+4x+4\right)\) if \((x-3)\) also established as a factor by division or factor theorem.