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

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

M1: Correct method, e.g. attempts to find g(2) \(\left(= \dfrac{5}{4-9}\right)\) and substitutes its value into f to achieve a value.
Alternatively, attempts \(\mathrm{fg}(x) = 4 - 3\left(\dfrac{5}{2x-9}\right)^2\), condoning slips, and substitutes \(x = 2\) to achieve a value.

A1: Correct answer only. If gf(2) is also attempted, then mark the final attempt which is the most complete.

Notes

M1: Eliminates the fraction and puts the \(xy\) term (or \(x\) term) onto one side of the equation.
Alternatively swaps \(x\)’s and \(y\)’ s, eliminates the fraction and puts the \(xy\) term (or \(y\) term) onto one side of the equation. Condone minor slips in rearranging e.g. \(-9y\) instead of \(+9y\)

A1: Correct expression for the inverse, \(x\) in terms of \(y\) or \(y\) in terms of \(x\). Need not be simplified.

Note that \(y = \dfrac{5}{2x-9} \Rightarrow 2x - 9 = \dfrac{5}{y} \Rightarrow 2x = \dfrac{5}{y} + 9\) is M1 and \(\Rightarrow x = \dfrac{\frac{5}{y} + 9}{2}\) is A1

A1: Fully correct notation for the inverse including its domain and including the e.g. \(\mathrm{g}^{-1} =\).
Condone \(x \neq 0\) without \(x \in \mathbb{R}\)  Need not be simplified.
Do not be too worried about \(\mathrm{g}^{-1}\) looking a bit like \(y^{-1}\) due to poor handwriting but if it is clearly \(y^{-1}\) then withhold this mark.

Accept e.g. \(\mathrm{g}^{-1}(x) = \frac{5+9x}{2x}\ x \in \mathbb{R}, x \neq 0\) or \(\mathrm{g}^{-1}(x) = \dfrac{5}{2x} + \dfrac{9}{2}\ \ x \neq 0\) or \(\mathrm{g}^{-1} = \dfrac{\frac{5}{x} + 9}{2}\ \ x \neq 0\)

Ignore any reference to the range of g.

Notes

(c)(i)

M1: Correct method. Attempts to substitute f into g, condoning slips, e.g. missing the 3.

A1: Correct simplified fraction. Ignore any reference to domains. Do not isw.
There is no need to include the \(\mathrm{gf}(x) =\)
If \(\mathrm{fg}(x)\) is also attempted, then mark the final attempt which is the most complete.

(ii)

B1: Deduces the correct range. May be scored even if \(\mathrm{gf}(x)\) is incorrect (but not a follow through).
Allow e.g. \(-5 \leqslant y \lt 0,\ y \in [-5, 0)\), \([-5, 0)\)
Do not allow e.g. \(-5 \leqslant x \lt 0,\ y \in (-5, 0)\), \(-5 \leqslant f(x) \lt 0\), \(-5 \leqslant g(x) \lt 0\)

Notes

M1: Sets \(\mathrm{f}(x) = \mathrm{h}(x)\) and attempts to collect terms to obtain a 3TQ = 0
The = 0 may be implied by use of the discriminant. Condone copying slips in \(\mathrm{f}(x)\) and \(\mathrm{h}(x)\).

dM1: Recognises the need to use "\(b^2 - 4ac \ldots 0\)" on their 3TQ and uses this to establish a value or range of values for \(k\). Allow for an attempt to solve \(b^2 - 4ac \ldots 0\) or \(b^2 \ldots 4ac\), which must be in terms of \(k\) only, where … is an equality or any inequality.

(Alt 1) Attempts to complete the square for their 3TQ (usual rules) and uses its minimum value set … 0 to establish a value or range of values for \(k\). Their expression for the minimum value must be in terms of \(k\) only. Condone any equality or inequality when comparing their minimum value to 0.
e.g. \(5x^2 - 6x + k - 4 \rightarrow 5\left(x - \dfrac{3}{5}\right)^2 - \dfrac{29}{5} + k \rightarrow \text{``}-\dfrac{29}{5} + k\text{''} \gt 0 \Rightarrow k \gt \ldots\) scores dM1

(Alt 2) Differentiates their 3TQ with respect to \(x\) to achieve a linear expression, sets = 0 (which may be implied), solves for \(x\) and substitutes \(x\) back into their 3TQ set … 0 to establish a value or range of values for \(k\). Here … can be any equality or inequality.
e.g. \(5x^2 - 6x + k - 4 \rightarrow 10x - 6 \rightarrow x = 0.6 \Rightarrow 5(0.6)^2 - 6(0.6) + k - 4 \gt 0 \Rightarrow k \gt \ldots\) scores dM1

A1: Deduces the correct range for \(k\), e.g. \(k \gt \dfrac{29}{5}\) o.e. Must be in terms of \(k\) and not e.g. \(x\)
Accept e.g. \(k \in (5.8, \infty)\) or just \(\left(\dfrac{29}{5}, \infty\right)\) but not e.g. \(k \geqslant \dfrac{29}{5}\) or \(x \gt \dfrac{29}{5}\) or \([5.8, \infty)\)

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.

Typical solution

Range \(\{y : y \geqslant 5\}\)

Typical solution

\(\mathrm{f}(-1) = 6 = \mathrm{f}(1)\)

\(\therefore\) f is many to one, so it does not have an inverse

Typical solution

Typical solution

(d)(i) \(\sqrt{x^2 + 5}\)

(d)(ii) \(\mathrm{h}(x) \geqslant \sqrt{5}\)

Typical solution

(12, 2)

Typical solution

\((12, -8)\)

Typical solution

Typical solution

(d)(i)

(d)(ii)

Typical solution

Typical solution

Typical solution

(i)

(ii)

The function \(\mathrm{h}\) does not have an inverse as \(\mathrm{h}\) is not one-to-one.

For example

and

Typical solution

\(\mathrm{f}^{-1}(x) = \ln(x-1)\)

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

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

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

Typical solution

(i)

(ii)

Typical solution

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

Typical solution

Typical solution

Notes

B1: For \(2x + k\)

M1: Correct order of operations for the composite function \(\mathrm{f}^{-1}\mathrm{g}\)

Alternative for first two marks in part (a)

B1: soi

M1: Condone sign errors only

M1*: Expanding and simplifying to obtain a three-term quadratic expression in \(x\) (allow sign errors only in the simplification and collection of terms) – all terms must be on the same side and the constant term must be a two-term linear expression in \(k\)

Must come from considering \(a(x^2 + kx + 5) + bk = 4 - 2kx\) where \(a, b \neq 0\)
Condone \(2x^2 + 4kx + (6 + k) \gt 0\) for M1 but cannot score the final A1 mark (as www)

M1dep*: Calculating correct discriminant for their 3TQ – allow \(4k^2\) recovered to \(16k^2\) for M1 only – do not award this mark if only embedded in quadratic formula

Condone \(= 0\), \(\lt 0\) etc. for this M1 mark

A1: AG www – sufficient working must be seen. Discriminant must have been set \(\gt 0\) before dividing by 8

A0 for setting discriminant \(= 0\) or \(\lt 0\) (oe wrong inequality) and later replacing with the correct inequality

Notes

M1: Correct method for solving their 3 term quadratic which would lead to the critical values of \(k\)

See appendix for solving 3TQ

B1: Correct set of values for \(k\) so must be strict inequalities – allow in terms of \(x\) and/or \(k\). Ignore any working seen

Condone poor set notation if intention is clear

B1FT: Correct union set notation for their values of \(k\) so must be in terms of \(k\) (and not \(x\)). Must be of the form \(\{k : k \lt k_1\} \cup \{k : k \gt k_2\}\) where \(k_1 \lt k_2\)
B0 for interval notation

Note that, for example, \(\left\{k \lt -\frac{3}{2}\right\} \cup \{k \gt 2\}\) and \(\left\{k : k \lt -\frac{3}{2} \cup k : k \gt 2\right\}\) etc. are B0

Note that for full marks to be awarded no incorrect working seen

Ignore inclusion of \(\in \mathbb{R}\) inside the \(\{\ \}\) brackets

Appendix: Rules for solving quadratics in question 4(b) ONLY

In question 4(b) candidates are required to solve a 3 term quadratic (3TQ) using DR – therefore we must see a correct, complete method for solving this quadratic – the correct answers do not imply the corresponding M mark, for example \(2x^2 - x - 6 = 0 \Rightarrow x = 2\) or \(x = -\frac{3}{2}\) is M0

Rules for factorising:
\(at^2 + bt + c \Rightarrow (mt + n)(pt + q)\) where \(a = mp\) and one of \(mq + np = b\) or \(c = nq\) (so when expanding their factorised expression it must give the correct quadratic term and one other term correct of the preceding 3TQ expression/equation)

e.g. in question 4b:
\(2x^2 - x - 6 = \left(x + \frac{3}{2}\right)(x - 2)\) is M0 (but the following B1 for the correct \(x \lt -\frac{3}{2}\) and \(x \gt 2\) can still be awarded)
\(2x^2 - x - 6 = (2x - 3)(x + 2)\) is M1 (when expanded the \(x^2\) and constant terms are correct)

Allow correct part factorisation for their 3TQ expression e.g. \(2x(x - 2) + 3(x - 2)\) scores M1

Rules for the formula:
Must apply the correct formula for their three-term quadratic (no errors even if correct formula is stated) – note that stating the formula (in terms of \(a\), \(b\) and \(c\)) followed immediately by the corresponding roots is M0 – we must see the formula being applied e.g. \(2x^2 - x - 6 = 0 \Rightarrow x = \dfrac{1 \pm \sqrt{1^2 - 4(2)(-6)}}{2(2)}\). Minimal acceptable working would be \(x = \dfrac{1 \pm \sqrt{49}}{4}\) (so must explicitly see the value of the discriminant) for M1

Rules for completing the square
The M1 is not awarded until correctly getting to the stage of \(x - \dfrac{1}{4} = \pm\sqrt{\dfrac{49}{16}}\) (must include \(\pm\) so implying two roots) with no errors (so consistent with applying the formula correctly)

Notes

M1: V-shape with the cusp on the positive \(x\)-axis – need not be symmetrical – this mark can be awarded if V-shape appears in the first quadrant only (so stops at \(y\)-axis)

M0 if V-shape appears in third or fourth quadrants but condone if RH branch appears in third and fourth quadrants (BOD that this is a continuation of the line with equation \(y = 2x - 5\))

A1: Fully correct (so appearing in first and second quadrants) with \(x\)-intercept labelled as \(\frac{5}{2}\) and \(y\)-intercept labelled as 5 – need not be symmetrical– condone if appearing in third and fourth quadrants only as a dashed line. Ignore horizontal line if drawn at \(y = 1\) (assume working for part (b))

Condone \((5, 0)\) labelled on the \(y\)-axis, and \(\left(0, \frac{5}{2}\right)\) labelled on the \(x\)-axis

Notes

M1: Re-write as \(-1 \lt 2x - 5 \lt 1\) followed by \(a \lt 2x \lt b\) with either \(a\) or \(b\) correct or \(c \lt x - \frac{5}{2} \lt d\) with either \(c\) or \(d\) correct or for either \(x \gt 2\) or \(x \lt 3\) (so dealing with one inequality correctly) or for both critical values (so 2 and 3 www)

\(a = 4\), \(b = 6\)
\(c = -\frac{1}{2}\), \(d = \frac{1}{2}\)
M1 for \(|2x - 5| \lt 1 \Rightarrow (2x - 4)(2x - 6) \lt 0\)

A1: cao – allow as two separate inequalities and allow e.g. ‘\(x \gt 2\) or \(x \lt 3\)’. ‘\(x \gt 2\), \(x \lt 3\)’, ‘\(x \gt 2\) and \(x \lt 3\)’ etc.

Ignore incorrect set notation provided the intention of the correct inequalities is clear

Notes

M1*: Use their answer to part (b) to re-write in terms of \(\mathrm{e}^{0.1N}\). Must imply either \(\mathrm{e}^{0.1N} \gt k_1\) and \(\mathrm{e}^{0.1N} \lt k_2\) where \(k_1 \lt k_2\) with both positive or \(\mathrm{e}^{0.1N} = k_1\) and \(\mathrm{e}^{0.1N} = k_2\) where \(k_1 \neq k_2\) with both positive
Allow \(n\), \(x\) etc. for \(N\) throughout

\(k_1\) and \(k_2\) are their critical values from part (b)
M0 for \(\mathrm{e}^{0.1N} \gt k_2\) and \(\mathrm{e}^{0.1N} \lt k_1\) where \(k_1 \lt k_2\)

M1dep*: Correctly take logs of both sides and make \(N\) the subject of both inequalities/equations e.g. \(\dfrac{\ln 2}{0.1} \lt N \lt \dfrac{\ln 3}{0.1}\), \(10 \times 0.693.. \lt N \lt 10 \times 1.098\ldots\) (at least to 1 dp), \(N = 10\ln 2\) and \(N = 10\ln 3\)

Must be correct inequalities/equations for \(N\) following through their values of \(k_1\) and \(k_2\)
If M0 M0 then SC B1 for 6.93… and 10.98… or \(10\ln 2\) and \(10\ln 3\)

A1: cao www – must see at least \(6.9\ldots \lt N \lt 10.9\ldots\) followed by all four integer values (and no others) stated explicitly or correct four integer values stated from correct critical values of 6.9… and 10.9…
A0 for \(7 \lt N \lt 10\) oe e.g. 7 – 10

See Appendix for those using T and I

Appendix: Trial and Improvement in 3(c)

Trial and Improvement in 3(c) – if only the first M mark awarded by main scheme, then for B2 (so for possibly full marks) allow one of the two following cases (the working below is the minimum required)

Case 1:
\(\mathrm{e}^{0.1N} \lt 3 \Rightarrow N \lt 10.98\ldots\) and \(\mathrm{e}^{0.1N} \gt 2\)
When \(N = 7, \mathrm{e}^{0.1 \times 7} = 2.0137\ldots \gt 2\) and when \(N = 6, \mathrm{e}^{0.1 \times 6} = 1.822 \lt 2\) therefore \(N = 7, 8, 9\) and \(10\)

Case 2:
\(\mathrm{e}^{0.1N} \lt 3, \mathrm{e}^{0.1N} \gt 2 \Rightarrow N \gt 6.93\ldots\)
When \(N = 10, \mathrm{e}^{0.1 \times 10} = 2.7182\ldots \lt 3\) and when \(N = 11, \mathrm{e}^{0.1 \times 11} = 3.0041\ldots \gt 3\) therefore \(N = 7, 8, 9\) and \(10\)

Notes

B1: Correct inequality
B0 for \(x \leqslant 0\)
Could use interval notation ie \((-\infty, 0)\)
Condone an incorrect attempt at set notation, as long as intention is clear

Notes

(b)(i) B1: State correct value, and no other
Value of \(n\) not required, but B0 if 19 comes from a clearly incorrect \(n\)
B0 if additional solution
B0 for \(n \leqslant 19\)

(b)(ii) B1: State correct value, and no other
Value of \(n\) not required, but B0 if 1 comes from a clearly incorrect \(n\)
B0 if additional solution
B0 for \(n \geqslant 1\)

Notes

(c)(i)

B1: Obtain \(x = \tfrac{4}{3}\) oe

M1: Attempt to solve equation with all signs reversed on one side of the equation, or square both sides and attempt to solve
M0 for eg \(\tfrac{1}{2}x - 1 = -2x - 3\)

A1: Obtain \(x = \tfrac{8}{5}\) oe
Maximum of 2 marks if additional solutions

Alternative method

M1: Square both sides to obtain two 3 term quadratics, and attempt to solve
Possibly BC

A1: Obtain \(x = \tfrac{4}{3}\)

A1: Obtain \(x = \tfrac{8}{5}\)
Maximum of 2 marks if additional solutions

(c)(ii)

B1: State \(x = \tfrac{8}{5}\), with reason as to why other solution is not valid
Both values of \(x\) must be correct ie no FT

For the sketch method: the gradient of \(y = 2x - 3\) must be clearly greater than the gradient of \(y = \tfrac{1}{2}x - 1\)
Correct sketch, but no scale needed (ISW any incorrect intercepts), but intercept of the lines must be to the left of the minimum point

Notes

(a)(i)

B1: (\(a = 2\)) Either stated or embedded in equation
eg \(|2x - b|\) seen
ignore any other values seen
B0 for \(a = -2\), unless subsequently corrected

B1: (\(b = 6\)) Either stated or embedded in equation
eg \(|ax - 6|\) seen
ignore any other values seen

B1: (\(c = 1\)) Either stated or embedded in equation
eg \(|ax - b| + 1\) seen
ignore any other values seen

(a)(ii)

B1: Any correct reason
Condone no explicit example
Could also say ‘because f is not one to one’
B1 BOD for ‘it is not one to one’
If referring to ‘one to many’ or ‘many to one’ it must be clear whether this is f or \(\mathrm{f}^{-1}\) (just ‘it’ or ‘the function’ is not enough)
Allow implication of function eg ‘as it is a many to one function there is no inverse function’
May also refer to the ‘horizontal line test’, but need to state outcome eg horizontal line would cross graph of \(y = \mathrm{f}(x)\) twice

Notes

(b)(i)

M1: Complete attempt to find inverse function of \(\mathrm{f}(x) = px - q\)
Correct order of operations, allow sign error only
Could use coordinate geometry and reflection in \(y = x\)
Allow M1 BOD if more than one function is being considered

A1: Obtain correct inverse, in terms of \(x\)
Could be single term ie \(\mathrm{g}^{-1}(x) = \frac{x + q}{p}\)
A1 for just \(\frac{1}{p}x + \frac{q}{p}\), ie \(\mathrm{g}^{-1}(x)\) can be omitted
If LHS seen, it must be \(\mathrm{g}^{-1}(x)\) or \(y\) (allow BOD for \(\mathrm{g}^{-1}\), or using f not g)
BOD if modulus sign included
A0 if additional equations given

B1: Correct domain
B0 for \(x \gt 0\)
Independent of the first two marks
If in words then must be correct, so B1 for ‘any non-negative \(x\)’ but B0 for ‘any positive \(x\)’
\(\mathrm{g}^{-1}(x) \geqslant 0\) is B0
Condone incorrect set notation as long as intention is clear

(b)(ii)

B1: Correct set of values, any notation
No need for \(0 \lt p\) as specified in question, so B1 for \(p \leqslant 1\)
B0 for \(p \lt 1\)
B0 for any additional incorrect values
B0 if just single example and not set of values
Condone incorrect set notation as long as intention is clear

Notes

B4 for all 5 rows correct
B3 for 3 or 4 rows correct
B2 for 2 rows correct
B1 for 1 row correct

Notes

B1: \(\geqslant \frac{1}{2}\) soi, no top limit (except \(\infty\)). Allow \(\gt \frac{1}{2}\)
Allow \(\mathrm{f}(x)\) or \(\dfrac{1}{x}\) or any letter or none for 1^st B1

B1: Correct range in set notation. Any letter (not \(x\)) or \(\dfrac{1}{x}\) or \(\mathrm{f}(x)\)

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

B1: B0 for \(|x| = 6\) unless replaced with \(x = 6\)

M1: oe e.g. \((2x - 3)^2 = 81\) or \(2x - 3 = \pm 9\)
M0 for \((2x - 3)^2 = 9\)

A1: A0 for \(|x| = -3\) unless replaced with \(x = -3\)
The correct answer implies the previous M mark
Two marks maximum if more than two answers given

Notes

(a)(i) B1: Allow alternative notation, or worded equivalent
Allow \(y\), or just f, but not \(x\)
Accept just \(\mathbb{R}\)
Allow \((-\infty, \infty)\)

(a)(ii) B1: Allow alternative notation, or worded equivalent
Allow \(y\), or just g, but not \(x\)
Or \((-\infty, \infty)\) with \((-1, 1)\) clearly excluded

Notes

(b)(i)

M1: Attempt correct composition of functions
At least one interim value required

A1: Conclude with 5.33

SC B1 for stating \(2\tan(1 \div \cos 0.6) = 5.33\)

(b)(ii)

B1: Must refer to inverse of f not existing, with reason
Must be clear that referring to the function f

Notes

M1: Attempt use of identity in their equation to obtain quadratic
Allow \(\tan^2 x = \pm\sec^2 x \pm 1\)
Award M1 when reduced to single trig ratio

A1: Obtain correct equation in \(\sec x\) – possibly still with brackets
Or correct quadratic in \(\cos x\) – possibly still with brackets but with no fractions \((4\cos^2 x - 6\cos x - 4 = 0)\)

M1: Solve 3 term quadratic and attempt to find at least one value for \(x\)
Could solve quadratic BC
Must be using root that would give solution for \(x\)

A1: Obtain at least one correct value
Allow decimals or in degrees
Must be from correct solution method of correct quadratic (condone second root not being seen – but must be correct if seen)

A1: Obtain both correct values, and no others, and explain that \(\sec x = 0.5\) has no solutions as outside range
Now exact and in radians
Or equiv explanation for \(\cos x\)

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: \(3x^2\) or \(k\mathrm{e}^{3x}\) where \(k = 1, 3\) or \(\frac{1}{3}\)
Implied by correct answer

B1: Fully correct derivative

Notes

B1: Or \(x = 0\), \(y = 1\).
1 alone is not sufficient.
Condone omission of brackets

Notes

M1: Consideration of relationship between gradients of the two curves where they cross the relevant axes. May be shown in a diagram with sketch of any increasing function reflected in \(y = x\)
Implied by correct answer
FT their derivative

Alternative method for M1

M1: Clear attempt to use implicit differentiation to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\)
Implied by correct answer

A1: nfww

Notes

B1: OR \(\frac{3-x}{x}\)
B1 BOD for \(3/x - 1\) but B0 if ambiguous e.g. \(3 - x/x\)
Mark answers in answer space provided first. If this is blank then mark answer written elsewhere in space.

B1: Or other forms for the domain e.g. \(x \in \mathbb{R},\ x \neq 0\)
Condone missing brackets
Need not use set notation but must have both \(x \in \mathbb{R}\) and \(x \neq 0\)

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: Starting to work through inverse processes
Allow equivalent processes e.g. swapping \(x\) and \(y\) at other stages

B1: B0 if \(y \geqslant 0\)

Additional guidance

Candidates may swap \(x\) and \(y\) at any stage, so you can allow the M1 for \(x = \sqrt{1+2y}\) or \(x^2 = 1 + 2y\).

The A1 is given for the correct answer or equivalent, so \(\frac{x^2}{2} - \frac{1}{2}\) is alright. They do not need to write \(\mathrm{f}^{-1}(x)\) as it is given in the answer space (even if they do not put their answer there).

The B1 is given for giving the correct domain and may not be in the answer space allocated to it so allow it anywhere. But do not allow \(y \geqslant 0\), \(\mathrm{f}^{-1}(x) \geqslant 0\) or \(x \gt 0\)

Notes

E1: Could be via an example e.g. \(x = 2\) and \(x = -2\) give the same value of \(\mathrm{g}(x)\)
Give BOD for use of ‘it’ provided it does not contradict
e.g. \(\mathrm{g}(x)\) is one-to-many does not score
isw after a correct answer

Additional guidance

We are allowing any of the three forms for the answer if using the word ‘It’. ‘It’ could refer to the function \(\mathrm{g}(x)\) or it’s inverse \(\mathrm{g}^{-1}(x)\), so give B1 BOD if you see ‘It is one-to-many’. But if the candidate specifically says ‘\(\mathrm{g}(x)\) is one-to-many’ then that is incorrect and scores B0.

Notes

B1: allow because it’s one-to-many (mapping)
allow eg because each value of \(x\) is mapped to two values oe
do not allow
eg because it’s a one-to-many function

Notes

B1: do not allow
eg \(x < -1\) and \(x > 1\)
eg \(-1 > x > 1\)

Notes

B1: Going over the given line above the \(x\)-axis and to the left of the \(y\)-axis and then going up from the \(x\)-axis at the same angle (by eye)
Condone right hand line segment dotted/dashed

Notes

M1: Could be treated as two separate inequalities (at least one correct) in \(x\) not \(|x|\)
OR \((5 - 2x)^2 < 9\)
If only one linear inequality in \(x\) stated scores M0 A0 A0
OR \(4x^2 - 20x + 16 < 0\) or \(x^2 - 5x + 4 < 0\)
OR \((x-1)(x-4) < 0\)
Allow M1 if treated as equations in \(x\) not \(|x|\)

A2: A1 if only one inequality correct
OR for \(1 \leqslant x \leqslant 4\)
OR for \(1 < x\), \(x < 4\)
OR for ‘\(1 < x\) or \(x < 4\)’

Notes

M1: Attempt to re-arrange

A1: Ignore labelling of this expression eg \(\mathrm{fh}(x) = \sqrt[3]{(x+8)}\) scores 2

Notes

B1: Condone lack of brackets if meaning is clear

B1: C: FT their B

B1: D: FT their A

Notes

B1: FT their A and B. May be implied by later work

B1: FT -ve reciprocal of their 4

M1: Or using \(y = mx + c\) and attempting to evaluate \(c\)
Must be using their midpoint and their \(-\dfrac{1}{4}\)

A1: Final answer

Notes

M1: May solve all 4 which also earns 3^rd method mark

Or

M1: Identifying 2 perp bisectors of 2 chords eg BD and AC, or AB and CD

Notes

B1: Isw

Notes

M1: Implied by B2

B1: At least one part of the domain correct, condone missing = for this mark, e.g. \(x < 5\)
Can score M0 B1

B1: Domain completely correct and correctly expressed
Eg allow \(x \geqslant -5\) and \(x \leqslant 5\) but not \(x \geqslant -5\) or \(x \leqslant 5\) or other ambiguous notation eg a comma
Condone \(x \leqslant \pm 5\) seen in working if final answer is correct

Notes

B1: At least one part of the range correct, condone missing = for this mark, e.g. \(y < 5\), allow use of \(\mathrm{f}(x)\) or \(\mathrm{g}(x)\) notation here but not \(x\)

B1: Range completely correct and correctly expressed
See notes in 2(b)(i)

Notes

E1: Convincing explanation of given result
Must relate to triangle. Do not need to mention angle BEA.

Notes

B1: Ignore any values on axes

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: expression
mark the final answer

B1: domain

Notes

B1: allow 4.2426406872… rounded to 2 or more sf

Notes

M1: chain rule used
allow one slip in differentiation, eg sign error

A1: all correct

M1: \(\mathrm{h}(x)\) must be correct for first M1

Alternative

M1: allow one slip eg sign error

rearrangement to find \(\mathrm{h}^{-1}(x)\) explicitly in terms of \(x\) followed by differentiation does not score

Notes

M1: E.g. \(x = -1\) or \(x = 3\) seen or \((1 - x)^2 > 2^2\)

A1: May be in final answer

B2: B1 if equals included in inequalities but otherwise correct
(both inequalities needed for B1 or B2)
Allow , (comma) for ‘or’ but do not allow ‘and’