Notes

M1: Attempts \(4\left|1 - 3\right| - 5 = \ldots\) or \(4\left|-1 - 3\right| - 5 = \ldots\) proceeding to a value.
Either correct value (3 or 11) that is clearly their answer can imply the M mark.
Alternatively, attempts any of \(-4(-1 - 3) - 5\) or \(-4(1 - 3) - 5\) or \(-4(-1) + 7\) or \(-4(1) + 7\) i.e. without the modulus signs using the ‘left’ branch of the function.

A1: Both 3 and 11 found and no other values that are clearly meant to be their answers (i.e. ignore reference to \(a = 1\) or \(a = -1\)). Accept “3, 11” or “3 or 11” or “3 and 11” or {3, 11} but not (3, 11). Condone \(y = 3, 11\) but not \(x = 3, 11\).

Notes

M1: Attempts \(2(-5) + 17\) which may be implied by sight of 7 used in their range, including in an incorrect range e.g. \(\mathrm{f}(x) \gt 7\)
Alternatively, attempts \(\mathrm{gf}(x) = 2\left(4\left|x - 3\right| - 5\right) + 17\ \left(= 8\left|x - 3\right| + \text{``}7\text{''}\right)\) and replaces \(\left|x - 3\right|\) with 0 or \(x\) with 3 or simply deduces the minimum value of their “7” in this case.
Must be an attempt at gf and not fg.

A1: Deduces the range \(\mathrm{gf}(x) \geqslant 7\) using correct notation.
Condone \(y \geqslant 7\) but not e.g. \(x \geqslant 7\) or \(\mathrm{g}(x) \geqslant 7\) or \(\mathrm{f}(x) \geqslant 7\) or \(\mathrm{fg}(x) \geqslant 7\)
Other acceptable notation includes: \(\mathrm{gf}(x) \in [7, \infty)\)
Do not accept \(\mathrm{gf}(x) \in [7, \infty]\) or \(\mathrm{gf}(x) \in (7, \infty)\)

Notes

B1: Deduces \(k \ldots -4\) allow any equality or inequality here. Condone \(k \ldots \pm 4\) (but not just \(k \ldots 4\)) and condone e.g. “gradient … – 4”
May be seen coming from e.g. \(kx = -4x + 7 \rightarrow (k + 4)x = 7\) or \(x = \dfrac{7}{k + 4}\) which is acceptable but see the note below. Should not be \(x\) or \(y\) for this mark but may be \(m\).
Note: Finding the ‘discriminant’ of a linear equation e.g. \((k + 4)x - 7 = 0\) in order to obtain \(k \ldots -4\) is invalid and cannot earn either the B1 or the final A1 mark unless there is an alternative valid reason given.

M1: Attempts to use the vertex to find the (upper) limit for \(k\). Look for \(\dfrac{\pm 5}{\pm 3}\) but allow this mark to be scored for \(\dfrac{\pm B}{\pm A}\) if there is clear evidence that they think the vertex is at \((A, B)\)
Allow use of \(m\), \(x\), \(y\) or another variable for this mark.

Alt 1 via squaring and the discriminant

\[\begin{aligned}kx = 4\left|x - 3\right| - 5 &\Rightarrow kx + 5 = 4\left|x - 3\right|\\&\Rightarrow k^2x^2 + 10kx + 25 = 16\left(x^2 - 6x + 9\right)\\&\Rightarrow \left(k^2 - 16\right)x^2 + (10k + 96)x - 119 = 0\\&\Rightarrow (10k + 96)^2 - 4\left(k^2 - 16\right)(-119) = 0\\&\Rightarrow 576k^2 + 1920k + 1600 = 0 \Rightarrow k = -\frac{5}{3}\end{aligned}\]

Scores M1 for setting \(kx = 4\left|x - 3\right| - 5\), isolating \(\left|x - 3\right|\) (or \(4\left|x - 3\right|\)), squaring both sides, using \(b^2 - 4ac \ldots 0\) where … is any equality or inequality, and solving the resulting 3TQ using the usual rules and may be by calculator, leading to a value for \(k\).
Condone slips in expanding the brackets.

Alt 2 via solving simultaneous equations

\[\begin{aligned}&\text{e.g. } kx = 4(x - 3) - 5 \Rightarrow x = \frac{-17}{k - 4}\\&kx = 4(3 - x) - 5 \Rightarrow (k + 4)x = 7\\&\Rightarrow (k + 4)\left(\frac{-17}{k - 4}\right) = 7 \Rightarrow k = -\frac{5}{3}\end{aligned}\]

Scores M1 for setting \(kx = 4(x - 3) - 5\) and \(kx = 4(3 - x) - 5\), eliminating \(x\), and solving for \(k\)

A1ft: Uses their value of \(k\), found using one of the above methods, as the upper end of the range for \(k\), i.e., \(k \lt \text{``}{-}\dfrac{5}{3}\text{''}\) which may be seen as part of their range. Condone \(k \leqslant \text{``}{-}\dfrac{5}{3}\text{''}\) for this mark.
Score once seen as an upper limit and do not withhold if they then incorrectly combine as e.g. \(-\dfrac{5}{3} \lt k \lt -4\)
Condone the use of \(m\) for this mark, but not \(x\) or \(y\).

A1: \(-4 \leqslant k \lt -\dfrac{5}{3}\) o.e. and no other solutions seen. Their range must use \(k\) and not e.g. \(x\), \(y\) or \(m\).
Other acceptable notation includes: \(k \in \left[-4, -\dfrac{5}{3}\right)\), “\(k \lt -\dfrac{5}{3}\) and \(k \geqslant -4\)”, \(k \lt -\dfrac{5}{3} \cap k \geqslant -4\)
Do not accept e.g. “\(k \lt -\dfrac{5}{3}\), \(k \geqslant -4\)” or “\(k \lt -\dfrac{5}{3}\) or \(k \geqslant -4\)” or “\(k \lt -\dfrac{5}{3} \cup k \geqslant -4\)”
Allow \(-1.\dot{6}\) but not \(-1.6\) or \(-1.6\ldots\) for \(-\dfrac{5}{3}\)

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

General guidelines for all parts:

Remember to check answers written against the questions.
If there is any contradiction, mark the answers given in the body of the scripts.
If there is no labelling, mark the responses in the order given.
The coordinates need to be values not just a calculation e.g. not \(6 - 2\) for 4
Points can be written as a coordinate pair or separately as \(x = \ldots,\ y = \ldots\)
Do not allow coordinates written the wrong way round but isw if necessary e.g. \(x = 4, y = -4 \rightarrow (-4,\ 4)\) scores B1 and isw
Condone missing brackets (one or both) e.g. \(x = 4, y = -4\) or \((4,\ -4\) or \(4,\ -4\) for \((4,\ -4)\)
Condone a missing comma e.g. \((4\ \ -4)\) for \((4,\ -4)\)
Condone use of a semi-colon e.g. \((4\ ;\ -4)\) for \((4,\ -4)\)
Condone vector notation e.g. \(\begin{pmatrix}4\\-4\end{pmatrix}\) for \((4,\ -4)\) and condone \(\left(\dfrac{4}{-4}\right)\)

B1: \((4,\ -4)\) o.e. see above

Notes

General guidelines for all parts:

Remember to check answers written against the questions.
If there is any contradiction, mark the answers given in the body of the scripts.
If there is no labelling, mark the responses in the order given.
The coordinates need to be values not just a calculation e.g. not \(6 - 2\) for 4
Points can be written as a coordinate pair or separately as \(x = \ldots,\ y = \ldots\)
Do not allow coordinates written the wrong way round but isw if necessary e.g. \(x = 4, y = -4 \rightarrow (-4,\ 4)\) scores B1 and isw
Condone missing brackets (one or both) e.g. \(x = 4, y = -4\) or \((4,\ -4\) or \(4,\ -4\) for \((4,\ -4)\)
Condone a missing comma e.g. \((4\ \ -4)\) for \((4,\ -4)\)
Condone use of a semi-colon e.g. \((4\ ;\ -4)\) for \((4,\ -4)\)
Condone vector notation e.g. \(\begin{pmatrix}4\\-4\end{pmatrix}\) for \((4,\ -4)\) and condone \(\left(\dfrac{4}{-4}\right)\)

B1: \((-4,\ 6)\) o.e. see above

Notes

General guidelines for all parts:

Remember to check answers written against the questions.
If there is any contradiction, mark the answers given in the body of the scripts.
If there is no labelling, mark the responses in the order given.
The coordinates need to be values not just a calculation e.g. not \(6 - 2\) for 4
Points can be written as a coordinate pair or separately as \(x = \ldots,\ y = \ldots\)
Do not allow coordinates written the wrong way round but isw if necessary e.g. \(x = 4, y = -4 \rightarrow (-4,\ 4)\) scores B1 and isw
Condone missing brackets (one or both) e.g. \(x = 4, y = -4\) or \((4,\ -4\) or \(4,\ -4\) for \((4,\ -4)\)
Condone a missing comma e.g. \((4\ \ -4)\) for \((4,\ -4)\)
Condone use of a semi-colon e.g. \((4\ ;\ -4)\) for \((4,\ -4)\)
Condone vector notation e.g. \(\begin{pmatrix}4\\-4\end{pmatrix}\) for \((4,\ -4)\) and condone \(\left(\dfrac{4}{-4}\right)\)

M1: One correct coordinate. See above.

A1: Both coordinates correct. See above.
Note that M0A1 is not a possible mark profile.
Note that in part (c), some candidates show their thinking by transforming the point piecewise e.g. \((6,\ -4) \rightarrow (6,\ 4) \rightarrow (6,\ 8) \rightarrow (6,\ 5)\)
In such cases, mark their final pair of coordinates

Notes

B1: One correct coordinate. Either \(x = 2\) or \(y = 5\) or (2, …) or (…,5) seen.

B1: Deduces \((2,\ 5)\) Accept written separately e.g. \(x = 2\), \(y = 5\) isw after a correct answer.
Condone 2, 5 without the brackets.

Notes

M1: Attempts to solve the correct equation without modulus signs \(16 - 4x = 3(x-2) + 5 \Rightarrow x = \ldots\)
Must reach a value for \(x\). Ignore attempts at e.g. \(16 - 4x = 3(2-x) + 5\)

A1: \(x = \dfrac{17}{7}\) o.e. exact answer and no other values. If other values have been found they must be rejected or the \(x = \dfrac{17}{7}\) clearly selected. Answer only implies both marks.

Note: \(x =\) “2.75” coming from \(5 = 16 - 4x\) may be found as part of their working to establish which branch of the modulus graph the line \(y = 16 - 4x\) intersects. If this is the case it need not be “rejected” provided it is not clearly stated as one of their solutions.

Those that achieve \(|x| = \dfrac{17}{7}\) can score BOD M1A0.

Alternative by squaring:

M1: Isolates the \(\lvert x-2 \rvert\) (or \(3\lvert x-2 \rvert\)), squares both sides and solves the resulting 3TQ using the usual rules and may be by calculator, leading to a value for \(x\).

A1: Selects the \(\dfrac{17}{7}\) or rejects any other values as in main scheme.

Notes

M1: Correct method to find either critical value (following through on their \(P\)).

Either \(k\) {=} 3 or \(k\) {=} \(\dfrac{\text{``}5\text{''} - 4}{\text{``}2\text{''}}\) scores M1 without evidence of an incorrect method.

Note that \(k = 3\) occasionally appears from use of the discriminant on \(x(k-3) + 5 = 0,\) and scores M0 unless there is an alternative valid reason given.
Allow the use of e.g. \(m =\) in place of \(k =\) here but do not allow \(x =\) or \(y =\)

A1: Correct range in terms of \(k\) in acceptable notation with no incorrect method seen.
Use of e.g. \(x\) is A0.  Allow “and” or “\(\cap\)” to join the regions but not “or” or “,” or “\(\cup\)”

Accept e.g. \((0.5, 3)\) ; \(k \in \left(\dfrac{1}{2}, 3\right)\) ; \(k \lt 3\) and \(k \gt \dfrac{1}{2}\) ; \(k \gt \dfrac{1}{2} \cap k \lt 3\)

but not \(\dfrac{1}{2} \lt x \lt 3\) ; \(\dfrac{1}{2} \leqslant k \leqslant 3\) ; \(\left[\dfrac{1}{2}, 3\right]\) ; \(k \gt \dfrac{1}{2} \cup k \lt 3\) ; \(k \gt \dfrac{1}{2}, k \lt 3\) ; \(k \gt \dfrac{1}{2}\) or \(k \lt 3\)

Alt 1 via solving simultaneous equations:

M1: Sets \(kx + 4 = 3(x-2) + 5\) and \(kx + 4 = 3(2-x) + 5\), eliminates \(x\), and solves for \(k\)

A1: As main scheme.

Alt 2 via squaring and the discriminant:

M1: Sets \(kx + 4 = 3\lvert x-2 \rvert + 5\), isolates \(\lvert x-2 \rvert\) (or \(3\lvert x-2 \rvert\)), squares both sides, uses \(b^2 - 4ac \ldots 0\) where … is any equality or inequality, and solves the resulting 3TQ using the usual rules and may be by calculator, leading to a value for \(k\).
Condone slips in expanding the brackets.

A1: As main scheme.

Alt 3 via Domain for the right hand branch of the modulus graph:

M1: Sets \(kx + 4 = 3(x-2) + 5\), makes \(x\) the subject, sets > 2 and deduces a critical value.

A1: As main scheme.

Notes

General guidelines for all parts:
Remember to check answers written against the questions.
If there is any contradiction, mark the answers given in the body of the script.
If there is no labelling, mark the responses in the order given.
The coordinates need to be values not just a calculation e.g. not \(-2 \times 3 + 5\) for \(-1\)
Points can be written as a coordinate pair or separately as \(x = \ldots,\ y = \ldots\)
Do not allow coordinates written the wrong way round but isw if necessary e.g. \(x = 5, y = -2 \rightarrow (-2, 5)\) scores B1 and isw
Condone missing brackets (one or both) e.g. 5, \(-2\) or (5, \(-2\) or 5, \(-2\)) for (5, \(-2\))
Condone a missing comma e.g. \((5\ \ -2)\) for \((5, -2)\)
Condone use of a semi-colon e.g. \((5\ ; -2)\) for \((5, -2)\)
Condone vector notation e.g. \(\begin{pmatrix}5\\-2\end{pmatrix}\) for \((5, -2)\) and condone \(\left(\dfrac{5}{-2}\right)\)

(i)

B1: \((5, -2)\) o.e. see above

(ii)

B1: \((1.5, -2)\) o.e. see above

(iii)

B1: One correct coordinate. See above.

B1: Both coordinates correct. See above.
Note that B0B1 is not a possible mark profile.
Note that in part (iii), some candidates show their thinking by transforming the point piecewise e.g. \((3, -2) \rightarrow (-3, -2) \rightarrow (-3, -6) \rightarrow (-3, -1)\)
In such cases, mark their final pair of coordinates.

Typical solution

Stretch in the \(y\)-direction by scale factor 3 and

followed by translation \(\begin{bmatrix}4\\0\end{bmatrix}\)

Typical solution

Typical solution

There could be more than one solution between the two points.

The two points are either side of the asymptote.

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Stretch, scale factor 2, parallel to the \(y\)-axis

Typical solution

Typical solution

Typical solution

Typical solution

(i)

(ii)

(iii)

The translated graph will only have one real solution.

Typical 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

Typical solution

Typical solution

The curve can be stretched in the \(y\)-direction by a scale factor of 9

The curve can be stretched in the \(x\)-direction by a scale factor of 3

Beth and Paul are both correct.

Typical solution

Typical solution

Typical solution

Typical solution

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

Typical solution

Typical solution

The top right graph.

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

M1: M1 for \(\dfrac{1}{(x \pm 4)^3}\) only (so no misreads) - need not be an equation for this mark

oe e.g. \((x \pm 4)^{-3}\)

A1: Must be an equation of the form \(y = \ldots\) but A0 for only \(\mathrm{f}(x) = \ldots\)

ISW if expanded either correctly or incorrectly

Notes

B1: B1 for correct explicit \(x\)-coordinate (so not embedded in a calculation for \(y\))

B1: B1 for correct explicit simplified \(y\)-coordinate (so must be 180)

B1 for \(x = 3\)
B1 for \(y = 180\)
Allow missing bracket round coordinates

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 least one equation, involving a constant of proportionality.
Allow BOD if the constants of proportionality are the same in two equations.
Allow \(\propto\) to be used.

A1: Correct simplified equation seen.
Equation must be simplified, so A0 for eg \(z = k\sqrt[3]{\dfrac{1}{x^6}}\)
Must involve just a single constant of proportionality ie \(k\)
A0 if the same constant of proportionality was used in both initial equations, or if \(k\) was used in either of the initial equations.

(a)(ii)

B1: Not dependent on correct equation in (i).
B0 if more than one Fig. identified.

Notes

M1*: Use \(x = 4\) and \(z = 3\) to attempt to find \(k\) from their equation of proportionality.
Their equation must involve \(x\), \(z\) and \(k\)
As far as attempting \(k\)

M1d*: Attempt to find \(x\) using \(z = 12\) and their numerical \(k\). Dependent on previous M1.
Their equation involving \(x\), \(z\) and their \(k\)
Attempt at least one value of \(x\)

A1: Both values required.
Must have had correct final equation in (a)(i), but could follow A0 if constants of proportionality were not dealt with correctly.

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

(a)(i)

M1: Indicate horizontal translation (in either direction) in some way with magnitude of 4 (‘units’ not required)
B1 for \(\begin{pmatrix}4\\0\end{pmatrix}\)
Condone informal language as long as intent is clear eg ‘left’ (or even ‘right’, as either direction allowed)
M0 if ambiguous eg ‘in’ or ‘on’ the \(x\)-axis

A1: B2 for \(\begin{pmatrix}-4\\0\end{pmatrix}\)
Must now be correct language so A0 for eg ‘along’ the \(x\)-axis or ‘left’
Allow ‘parallel to the \(x\)-axis’ or ‘horizontal’

(a)(ii)

B1: Identify direction - correct language needed
Allow ‘\(x\)-axis invariant’, ‘parallel to the \(y\)-axis’ or ‘vertical’
Condone ‘positive’ \(y\)-direction (as given function > 0)

B1: ‘scale factor’ or ‘factor’ needed (condone ‘stretch’ factor)
Not dep on previous B1, but must have indicated vertical stretch in some way, including informal language such as ‘upwards’
Cannot be ambiguous language, such as ‘in’, ‘on’, ‘across’ the \(y\)-axis

Notes

M1: Correctly combine two correct log terms
Or \(\log_2(8x) = \log_2\frac{2}{1-x}\)
Or \(3 + \log_2(x(1 - x)) = 1\)
Or \(\log_2(4x(1 - x)) = 0\)
OR use indices base 2 on both sides (ie \(8x = 2^{1 - \log_2(1-x)}\)) and use rules of indices to split eg \(8x = 2 \times 2^{-\log_2(1-x)}\)

M1: Correct method to remove logs
Correctly used on equation of form \(\log_2\mathrm{f}(x) = \log_2\mathrm{g}(x)\) or \(\log_2\mathrm{f}(x) = k\)
OR correct method to deal with log term – expect \(8x = \frac{2}{1-x}\)

A1: Any correct equation not involving logarithms
Could still contain brackets and / or fractions

A1: Obtain \(x = 0.5\)
A0 if additional solutions

DR so no credit for answer only

Notes

M1: Equate, and rearrange to three term quadratic
Attempt to gather like terms, but not necessarily on same side of equation
Condone no ‘= 0’

M1: Attempt to solve quadratic
If factorising then expansion should give \(x^2\) and one other term correct
Quadratic formula should be correct – allow one slip when substituting as long as general formula already seen as correct
Completing the square needs to go as far as \(x - p = \pm\sqrt{q}\)

A1: Obtain both correct \(x\) values
Or one correct \((x, y)\) coordinate following a correct factorisation oe

A1: Obtain both correct pairs of coordinates
Allow as eg \(x = 1\), \(y = 0\) as long as pairings are clear

SC If no method shown for solving quadratic then allow
M1 for obtaining 3 term quadratic
A1 for \(x = 1\), \(x = 3\)
A1 for \((1, 0)\) and \((3, 4)\)

SC If no method at all shown then allow B1 for both \((1, 0)\) and \((3, 4)\)

Notes

M1: No need for line to actually intersect with negative \(y\)-axis as long as it goes beneath positive \(x\)-axis

A1: Must pass through both points

Notes

B1FT: FT any straight line that splits the overlap area into two finite regions, with the lower region identified
Allow for straight line with negative gradient as well, but not \(x = k\)

Notes

B1: Do not accept shift, move, transformation, etc. for first B1

B1: Correct description e.g. correct vector (not as a coordinate), ‘8 units down’. Do not allow second B1 after incorrect type of transformation e.g. stretch/rotation etc. but allow after shift/move etc.
Condone lack of ‘units’ but do not accept ‘factor \(-8\)’, ‘8 places/spaces/steps down’ etc.
For ‘parallel to the \(y\) axis’ allow ‘vertically’, ‘in the \(y\) direction’. Do not accept across/up/along/to/in/towards the \(y\) axis’

If more than a single transformation, then no marks (unless two translations equivalent to the correct answer)
Mark vector before description/words

Notes

M1: Attempt to make \(x\) the subject (allow sign errors only)
May use either \(\mathrm{f}(x)\) or \(y\)

A1: Must be in terms of \(x\). Allow the expression only e.g. \((x + 8)^{\frac{1}{3}}\), \(\sqrt[3]{x + 8}\), oe
Ignore what this expression is equated to

Notes

B1: Must include both ‘reflection’ or ‘mirror image’ or ‘mirrored’ and ‘\(y = x\)’
B0 for ‘symmetrical’ unless clearly describing reflective symmetry

Notes

M1: Correct left-hand side (including equals sign) and right-hand side must be of the form \(\dfrac{\mathrm{f}(y)}{\mathrm{g}(x)}\) or \(\dfrac{\mathrm{g}(x)}{\mathrm{f}(y)}\)

A1: cao (oe)

Notes

M1*: Separation of variables – dependent on the M mark in (a)
With an indication of integration

A1ft: Follow through their differential eq. from (a)
Condone no constant

M1dep*: Attempt to find \(c\) and eliminate logs

A1: oe (but must be of the form \(y = \mathrm{f}(x)\))

Notes

B1: B0 if another type of transformation stated or if shift/move etc. used

B1ft: Follow through their \(y = (x - p)^3 + q\) - B0 if this transformation is given as a stretch/rotation/reflection/enlargement etc. (but condone no transformation stated or shift/move etc.) – need not be given as a vector
\(p, q \neq 0\)

Notes

(a)(i) B1: Correct coordinate
And no others

(a)(ii) B1: Correct \(x\)-coordinate
B1: Correct \(y\)-coordinate
If more than one solution given then award B1 if either co-ordinate is consistent in all solutions

(a)(iii) B1: Correct coordinate
And no others

Notes

(b)(i) B1: Both \(x\)-coordinates correct, and no others
Ignore any attempt at \(y\) values

(b)(ii) B1: Correct inequality, and no others
Allow \(\leqslant\)
Could be written in set notation

(b)(iii) B1: Correct \(x\)-coordinate, and no others
Ignore any attempt at \(y\) values

Notes

B1: \(y = |x - 1|\) drawn correctly – must touch (but not intersect) the positive \(x\)-axis
Intercepts with axes need not be labelled

B1: \(y = kx^{-1}\) drawn correctly – must not intersect axes

Notes

B1: Dependent on both marks in (a) – must mention that the solution of the equation \(x|x - 1| = k\) corresponds to where the two graphs in (a) intersect (so just stating that the graphs in (a) intersect at only one point is B0)

Notes

M1: Uses graph and sets up quadratic (oe) – allow if \(x^2 - x + 6 = 0\) stated as well (but M0 if this is the only quadratic (oe) considered)
Or setting up a four-term quartic from \((x - 1)^2 = 36x^{-2}\)

A1: BC \(x = -2\) only (www)

SC If no marks awarded, then B1 for \(x = -2\) only and then B1 for explicitly showing that \(-2\lvert -2 - 1\rvert = -2(3) = -6\)

Notes

B1: \(y = x^2\) drawn correctly – must pass through the origin and (roughly) symmetrically about the \(y\)-axis

B1: \(x + y = 2\) drawn correctly – must have positive \(y\)-intercept and negative gradient
Line need not intersect the \(x\)-axis and intercepts need not be labelled

B1: Correct identification of region (dependent on both previous B marks) – ignore shading if \(R\) correctly labelled

Condone dashed line/curve for full marks

Notes

B1: AG
Or complete argument e.g. \(a > 0\) so \(\frac{3a}{a} \geqslant 0\)
Condone \(3 > 0\) or \(\frac{3a}{a} > 0\)
No wrong working seen

Notes

B1: AG
Clear explanation, must be based on \(y^2\) non-negative (\(\geqslant 0\)) and \(x^2\) positive (\(> 0\))
Condone \(\frac{y^2}{x^2} \geqslant 0\) then \(\frac{(3a-x)}{a+x} \geqslant 0\) from rearranging

Notes

B1: Dismissing zero value for \(x + a\)
Allow e.g. \(x + a > 0\) and \(x + a < 0\)
Condone e.g. “\(x + a > 0\) as denominator cannot be 0”

B1: Condone “both positive or both negative”
Considering negative case
Condone e.g. \(3a - x < 0\) and \(a + x < 0\) for this mark

B1: AG
Dismissing the “both negative” case
Dep on previous two B1 marks
Inequality signs must be correct

Alternative method for 3^rd B1

B1: AG
Dismissing the “both negative” case
Dep on previous two B1 marks
Inequality signs must be correct

Notes

B1: Any equivalent inequality/inequalities with \(x\) as a subject
Both inequality signs must be correct
Condone e.g. comma for ‘and’

Notes

B1: \(-a\) alone is insufficient.
oe e.g. \(x + a = 0\)

Notes

B1: cao

Notes

B1: all correct; ignore equations outside the range, B0 if extra equations in the range
B0 for \(x = 0, \frac{\pi}{2}, \pi\); B0 for \(y = \ldots\)

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

B1: V shaped graph (approximately symmetric for vertical mirror line)
Must be for positive and negative values of \(x\)

B1: Single vertex at (1.5, 0), and (0, 3) seen

Notes

B1: Two correct linear equations equating \(2x-3\) to \(\pm(4-x)\) oe

M1: Attempts to solve at least one equation.

A1: Both values seen

Alternative method

B1: Correct equation seen

M1: Collects terms to form 3 term quadratic and attempt to solve

A1: Both values seen

Notes

M1: Use of result from line 19 of article

A1: oe \(\dfrac{73}{2}\)
Accept a fully correct solution by an alternative method for M1A1.
isw after a correct answer

Additional guidance

They must use the formula from line 19 of the article. The M1 is for substituting a=6 and h=5 and the A1 is for the correct answer. Allow 2 marks for a completely correct valid alternative method.

Notes

B1: Zeroes at 0 and close to -2 and 2 (and no others)

B1: Rotational symmetry about the origin (mark intent)

B1: General shape (actual values of gradient may be wrong)

Additional guidance

The 3 marks are independent.

The first one is for showing the crossing points are 0 and near to +2 and -2.

The second is for rotational symmetry (or close to).

The third is for the general shape of a positive cubic But do not condone incorrect curvature. Some candidates are drawing y=x, which scores B0B1B0.

In all cases use your judgement and mark the intention of the candidate.

List all 3 marks in the order of the MS.

Notes

M1: Correct use of one of the laws of logs – award if \(2^x\) seen or if \((y-k) = 10^{x\log 2}\)

A1: LHS must be \(y =\)
Allow \(y = k + 10^{x\log 2}\)

Notes

B1: General shape correct. Positive and negative values of \(x\) should be seen used. FT their exponential (a)

B1: \(y\)-intercept at \(k+1\) on positive \(y\)-axis. FT their exponential (a) provided in terms of \(k\)

B1: Asymptote at \(k\). Horizontal line need not been seen provided the intention is clear

Notes

M1: eg M1 for \(2\mathrm{e}^x - 3\) or \(e^{\frac{1}{2}x-3}\);

A1: must be an equation; B2 for correct answer with no working

Notes

M1: allow one sign or coefficient error; must be four terms

M1: at least two terms correct

M1: may be implied by \(x = -2\) seen unsupported or \((x + 2)\) identified as factor

M1: by inspection or long division; allow one sign error or one coefficient error in trinomial
may be implied by \(x = \frac{1}{2}\) seen unsupported or \((2x - 1)\) oe identified as factor

A1: may see \(x = \frac{1}{2}\) (repeated)
A0 for \(x = -2\) (repeated)

M1*: allow one sign or one coefficient error, FT their \(\frac{\mathrm{d}y}{\mathrm{d}x}\);
allow M1 for \(12x^2 + 8x - 7\)

A1: NB test indecisive at \(x = \frac{1}{2}\)
A0 for just eg \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \gt 0\) so minimum
award M1A1 for consideration of gradient either side of −2, values must be correct to at least 2sf for A1

M1: or eg

A1: values in table must be correct

A1: ignore calculation of associated \(y\)-value
allow any correct decimals to 3 sf or more

Notes

M1: curve with a minimum in 3^rd quadrant and stationary point of inflection in 4^th quadrant and no other stationary points

B1: \((0, -6)\) identified as \(y\)-intercept (intercept must be below the \(x\)-axis and above \(-20\))

A1: correct curve with intercepts at \((-a, 0)\) and \((b, 0)\), where \(-3 \lt a \lt -2.6\) and \(0.8 \lt b \lt 1.2\); minimum at \((-2, y)\) where \(-90 \lt y \lt -80\) and inflection for \(0 \lt x \lt 1\) and \(y\) is between the \(x\)-axis and the \(y\)-intercept

Notes

B1: General shape with horizontal asymptotes
Allow if asymptote not drawn provided the intention is clear
Must be a one-to-one function

B1: \(y\)-values \(\pm\dfrac{\pi}{2}\) seen

Notes

M1: soi

M1: Attempt to solve \(\cos x = 0\)

A1: Both values in radians needed

M1: Allow for \(x = \tan^{-1}\frac{1}{3}\)

A1: Both values in radians to at least 2 s.f. needed. Do not award if additional values inside the interval \([-\pi, \pi]\)
Ignore additional values outside the interval \([-\pi, \pi]\).
SC1 award for \(18.4^\circ\) and \(-161.6^\circ\) if \(90^\circ\) already seen

A1: Allow awrt 0.90
Notice 0.9 is exact.

Alternative method

M1: soi

M1: Attempt to solve \(\cos x = 0\)

A1: Both values in radians needed

M1: Complete method for finding at least one value for \(\sin x\)

A1: Both correct roots and no others in the range

A1: Allow awrt 0.90
Notice 0.9 is exact.

Notes

M1: decreasing concave up curve in 1^st and 2^nd quadrants which does not cut the \(x\)-axis; mark intent

B1: decreasing curve with intercept \((0, 3)\); may be in one quadrant only

A1: smooth curve from \((-0.5, a)\) through \((2.5, b)\), where \(4.5 \leqslant a \leqslant 5\) and \(0 \lt b \lt 0.5\)

Notes

M1: taking logarithms in any base

M1: 3^rd law of logs used correctly

A1: if M0M0 allow SC1 for 1.44 unsupported

Alternatively

M1: may see \(x\log 0.4 = \log\left(\frac{0.8}{3}\right)\) oe

Notes

M1: Can be implied later by an equation of the correct form

Notes

M1: product rule & chain rule; allow one error

Alternative

M1: NB from \(x^5 - 6x^4 + 12x^3 - 8x^2\)
expand brackets and differentiate; allow one error

Notes

A1: accept – 1.10592 to 2 sf or better

Notes

or for any of the three points: considers \(y\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) either side of correct stationary point accompanied by suitable commentary; must see numerical values

Notes

M1: shape of curve correct with max, min and inflection

A1: correct intercepts marked on sketch or identified next to graph

Notes

M1: Implied by any of the points

A2: A2 for all three points.
A1 for one correct point
Ignore \((3, 1)\)
If other incorrect points A1 max

Notes

M1: \(y = 2x - 3\) correct in first quadrant; may be implied by reflection or by A1

M1: reflection of their straight line to form V shape for positive \(y\) only

A1: symmetrical V shape through (0,3) with vertex at (1.5, 0) ; points may be identified from numbered scale or written separately
condone branches of different lengths

Notes

B1: Allow interchange of \(>\) and \(\geqslant\) or \(<\) and \(\leqslant\) for one inequality as long as the direction is correct

B1: Both inequalities fully correct. Allow \(x + 5 < y \leqslant 8 - 2x - x^2\) oe

Notes

M1: A correct three term quadratic equation (or inequality) must be seen
Any method including BC is acceptable for solving the quadratic equation clearly seen in the form \(ax^2 + bx + c = 0\).

A1: Correct roots of the equation soi
Must be exact

B1: Shows correct inequality FT their roots
Allow B1 for \(-3.79 < x < 0.79\) www

Notes

M1: \(k = \frac{1}{2}\); condone \(k = 2\); condone omission of brackets

A1: must see “\(y =\)” either here or in first line

Notes

May be seen as \((x + 4)^2 - 14\)

Notes

B1: FT their (a)

Notes

B1: If other transformations as well (eg stretch) then B0
‘Shift’ does not score

B1: FT their (a)
Or B1 for translation 4 to left oe
Or B1 for translation 14 down oe
B0 B1 can be awarded for eg Translation, correct vector & stretch