Notes

M1: Attempts the gradient of the straight line either way round.
Look for \(\dfrac{300 - 1500}{50 - 30} = (-60)\) o.e. or \(\dfrac{50 - 30}{300 - 1500} = \left(-\dfrac{1}{60}\right)\) o.e.
Condone 1 sign/copying slip only if a correct formula is seen or implied e.g. \(\dfrac{y_2 - y_1}{x_2 - x_1}\)
Alternatively, sets up simultaneous equations \(1500 = 30a + b\) and \(300 = 50a + b\) (or \(30 = 1500a + b\) and \(50 = 300a + b\)) and attempts to solve, leading to a value for \(a\) and/or \(b\).
If using simultaneous equations, condone slips.

dM1: Attempts the equation relating \(x\) and \(y\) with either coordinate (50, 300) or (30, 1500) used correctly. Look for \(y - 1500 = \text{``}{-}60\text{''}(x - 30)\) or \(x - 30 = \text{``}{-}\dfrac{1}{60}\text{''}(y - 1500)\) or using (50, 300). Note that e.g. \(y - 30 = \text{``}{-}60\text{''}(x - 1500)\) scores dM0 as the values have been used incorrectly. If using \(y = mx + c\) they must proceed to \(c = \ldots\)
If using simultaneous equations, they must proceed to values for \(a\) and \(b\)
(Note for reference \(x\) in terms of \(y\) is \(x = -\dfrac{1}{60}y + 55\))
If they use e.g. \(\dfrac{y - 300}{1500 - 300} = \dfrac{x - 50}{30 - 50}\) both M’s can be scored together.
Condone 1 sign/copying slip only if a correct formula is seen or implied.

A1: cao Either \(y = 3300 - 60x\) or \(y = -60x + 3300\) or equivalent with \(y\) in terms of \(x\)
The correct equation with no working scores full marks.

Notes

M1: Attempts either \(\pm x\left(\text{``}3300 - 60x\text{''}\right)\) or \(\pm 10\left(\text{``}3300 - 60x\text{''}\right)\) with or without dividing by 1000.
May see both in one step, i.e., \(\pm(x \pm 10)\left(\text{``}3300 - 60x\text{''}\right)\)
Alternatively, allow \((P =)\ \pm xy \pm 10y \pm 8000\) with or without dividing by 1000.
Allow their model for \(y\) to be any expression in \(x\) (i.e. it does not need to be linear) for M1.

dM1: Full attempt to find \(P\) in terms of \(x\) only. There is no need to see a LHS for this mark.
\((P =)\ \dfrac{x\left(\text{``}3300 - 60x\text{''}\right) - 10\left(\text{``}3300 - 60x\text{''}\right) - 8000}{1000}\) or \((P =)\ \dfrac{(x - 10)\left(\text{``}3300 - 60x\text{''}\right) - 8000}{1000}\)
Condone the consistent absence of division by 1000 e.g.
\((P =)\ x\left(\text{``}3300 - 60x\text{''}\right) - 10\left(\text{``}3300 - 60x\text{''}\right) - 8000\) or \((P =)\ (x - 10)\left(\text{``}3300 - 60x\text{''}\right) - 8000\)
Condone invisible brackets if recovered. Dependent on the previous method mark.
Their model for \(y\) must now be a linear expression in \(x\) for dM1.

A1*: Achieves the given answer through rigorous argument with all elements pulled together clearly. Requires \(P =\) (not just “profit =”) at some point which may be on a previous line.
Any brackets seen must have been expanded in an intermediate line before the given answer.
Allow otherwise correct work leading to \(P = -60x^2 + 3900x - 41000\) followed by \(P = -0.06x^2 + 3.9x - 41\) (without justifying the division by 1000).
Allow recovery if they write e.g. 800 but it is recovered before the given answer.

Note: Attempts by verification are unlikely to score any marks. If unsure, send to review.
e.g. \(x = 10 \rightarrow P = -0.06(10)^2 + 3.9(10) - 41 = -8\) without an accompanying argument involving the \((x, y)\) values (50, 300) and (30, 1500) scores M0dM0A0*.

Notes

M1: Attempts to find either end point of the interval which may come from a calculator.
The exact values are \(\dfrac{195 \pm 5\sqrt{537}}{6}\) but may just be seen in the quadratic formula, e.g., \(\dfrac{-3.9 \pm \sqrt{3.9^2 - 4(-0.06)(-41)}}{2 \times -0.06}\) or using completing the square – see general marking principles.
Accept awrt 13.2 or awrt 51.8 for this mark.

A1: Either \(13.18 \lt x \lt 51.82\) or \(13.19 \leqslant x \leqslant 51.81\) No units are required but 2d.p. are required.
Correct answer scores both marks. Allow e.g. \(13.18 \lt x \leqslant 51.81\) or \(13.19 \leqslant x \lt 51.82\)
Allow e.g. “\(x \gt 13.18\) and \(x \lt 51.82\)” or “\(x \gt 13.18 \cap x \lt 51.82\)” but not use of exact values and not “\(x \gt 13.18\), \(x \lt 51.82\)” or “\(x \gt 13.18 \cup x \lt 51.82\)” or “\(x \gt 13.18\) or \(x \lt 51.82\)”
Must be \(x\) or e.g. “selling price” for the range, not \(y\) or \(P\).

Notes

B1ft: £32.50. Units and 2dp required. Not £32.50p. Ignore attempts to find maximum profit.
Allow FT on the mean of their end points from (c) (their end points must be > 0, their answer requires £ and 2dp).
May come directly from the given answer to (b) using \(x = -\dfrac{b}{2a} = -\dfrac{3.9}{2 \times -0.06} = \text{£}32.50\) or from \(\dfrac{\mathrm{d}P}{\mathrm{d}x} = -0.12x + 3.9 = 0\) or directly from a calculator. No working required.

Note: Following use of 13.2 and/or 51.8 (not e.g. 51.80) with either type of inequality in part (c), £32.5 will score B1ft so that the missing decimal place(s) is only penalised once.

Notes

M1: Attempts to substitute 35 (and not any other value) into the given equation for \(P\) to find a value for \(P\).
May be implied by 22 or 22000. Condone slips.
Alternatively, substitutes \(P = 21.75\) into the model, and solves via any acceptable method, including by calculator, to find a value for \(x\). If using 21750 it must be substituted into their \((P =)\ -60x^2 + 3900x - 41000\). Note \(x = 35.73\) or 29.27

A1: Requires:

• A correct value. Usually (£)22 000 but accept, e.g., 22 thousand (pound). If using 22 they must compare with 21.75. The alternative requires \(x =\) awrt 35.7 to be compared to 35.
• A correct comparison. Some examples below:
  • “Close to (£)21 750” o.e. e.g. “the model predicts 22000 and they make close to this amount”
  • “Agree to 2sf”. Do not condone “it rounds to 22 000” without mention of the degree of accuracy.
  • 1.1% error (accept approx. 1%) Do not be concerned about the mechanics of any percentage error calculation seen.
  • “The difference is small”
  • Condone e.g. \(22 \approx 21.75\) o.e.
  • “Only £250 off”. However, just stating “£250 off” without suggesting this is a (relatively) small difference is not sufficient for this component. Similarly, “£350 off is a small difference” is incorrect (requires correct £250).
• A conclusion. e.g. Concludes it (the model) is suitable / reliable / good / fairly accurate / accurate.

Notes

M1: Forms one correct equation either \(17.6 = A - 8B\) or \(11.9 = A - 27B\)
May be unsimplified e.g. \(17.6 = A - B \times 4^{1.5}\) or \(11.9 = A - B \times 9^{1.5}\)

A1: \(17.6 = A - 8B\) and \(11.9 = A - 27B\) which may be unsimplified

dM1: Solves simultaneously to find values for \(A\) and \(B\). It is dependent on the previous method mark so at least one equation must be correct. Do not be concerned with the process as calculators may be used. Score if values for \(A\) and \(B\) are reached from a pair of simultaneous equations.

A1: \(h = 20 - 0.3\,t^{1.5}\) o.e. e.g. \(t = \left(\dfrac{10}{3}(20 - h)\right)^{\frac{2}{3}}\) isw once a correct equation is found.
Requires the complete equation including \(h = \ldots\)
Just stating the values for \(A\) and \(B\) is A0 but allow A1 to be scored if the correct equation is seen in (b) or (c)

Notes

B1ft: 20 metres but ft on their \(A\) metres (provided \(A \gt 0\)). Condone if they have a value for \(A\) which is given to a greater degree of accuracy than 3sf but they round this to 3sf or better in (b).
Requires units as well (allow m for metres)

Notes

Do not be concerned with the use of t or T. If awrt 16.4 is seen in (a) it must be seen or used in (c) to score.

M1: Sets \(\text{``}20\text{''} - \text{``}0.3\text{''}\,T^{1.5} = 0\) and proceeds to a value or expression for \(T\) e.g. \(\left(\dfrac{\text{``}20\text{''}}{\text{``}0.3\text{''}}\right)^{\frac{2}{3}}\)
Do not be concerned by the use of any inequalities instead of “=”
You do not need to be concerned by the mechanics of the rearrangement, they just need to achieve a value or expression.

If no equation is seen then may be implied by a correct value for \(T\) (to the nearest integer e.g. awrt “16”) or a correct expression for \(T\) for their \(A\) and \(B\). You may need to check \(\left(\dfrac{\text{``}20\text{''}}{\text{``}0.3\text{''}}\right)^{\frac{2}{3}}\) on your calculator. They may also achieve this by trial and improvement.

A1: \((T =)\) awrt 16.4 including \(\left(\dfrac{200}{3}\right)^{\frac{2}{3}}\) or exact equivalent e.g. \(\sqrt[\frac{3}{2}]{\dfrac{200}{3}}\) ignore any units for time. If an exact value is given condone the radical symbol not fully covering the fraction provided it is not clearly e.g. \(\dfrac{\sqrt[\frac{3}{2}]{200}}{3}\)
Condone \((0 \leqslant)\ T \leqslant\) awrt 16.4 o.e. but not \((0 \leqslant)\ T \lt\) awrt 16.4
Do not accept any greater than (or equal) inequalities e.g. \(T \geqslant\) awrt 16.4
isw if there is a written response to (c) once a correct value or valid expression for \(T\) is seen.

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

Note: Using e.g. \(x\) throughout is acceptable for both marks.

M1: Starts the process of showing the given expression is positive.
Completing the square requires \((k-2)^2 \pm \ldots\)
Differentiation requires them to differentiate to a linear expression in \(k\), set = 0 (which may be implied), solve for \(k\) and substitute their \(k\) into \(k^2 - 4k + 5\) to reach a value. Ignore e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
Discriminant requires a calculation of \(b^2 - 4ac = (\pm 4)^2 - 4 \times 1 \times 5\ \{= -4\}\) and might be seen embedded in the quadratic formula.
Sketches on their own are insufficient without working to find the minimum point algebraically.
Stating the minimum is (2, 1) without any evidence is M0.

A1*: Completes the proof with no errors and correct reasoning.
Accept e.g. “hence proved” in place of “so \(k^2 - 4k + 5\) is always positive” as long as there is sufficient justification for it being “proved”.
As a minimum expect to see e.g.:

• \(k^2 - 4k + 5 = (k-2)^2 + 1\) which is always positive as \(\underline{(k-2)^2 \geqslant 0}\) (or as squares are always positive or zero). Must be a correct statement. Do not condone e.g. \((k-2)^2 \gt 0\)
• \(k^2 - 4k + 5 = (k-2)^2 + 1\) so (2, 1) is the minimum (point) so \(k^2 - 4k + 5\) is always positive.
• \(k^2 - 4k + 5 = (k-2)^2 + 1\) so \(\underline{k = 2 \pm \sqrt{-1}}\) hence \(k^2 - 4k + 5\) has no real roots and as k² has a positive coefficient (condone e.g. positive \(k^2\) or positive quadratic or \(a \gt 0\)), hence proved.
• \(2k - 4 = 0 \Rightarrow k = 2 \Rightarrow k^2 - 4k + 5 = 1\) is the minimum value so \(k^2 - 4k + 5\) is always positive.
• \(b^2 - 4ac = 16 - 20 \underline{\lt 0}\) so \(k^2 - 4k + 5 = 0\) has no real roots and as k² has a positive coefficient (condone e.g. positive \(k^2\) or positive quadratic or \(a \gt 0\)), hence proved.

Note that stating \(k^2 - 4k + 5 = (k-2)^2 + 1 \gt 0\) alone is not sufficient for the A1*
Similarly, stating \(k^2 - 4k + 5 = (k-2)^2 + 1 \gt 0\) because \((k-2)^2 \gt 0\) (or \((k-2)^2\) is positive) is incorrect and scores A0.

SC: Attempts at completing the square for \(k = 2n\) and/or \(k = 2n + 1\) (or \(k = 2n - 1\)) can score M1 provided at least one is attempted as far as \((\ldots)^2 + \ldots\) e.g. to \((2n-2)^2 + \ldots\) or e.g. \(4(n-1)^2 + \ldots\)
Candidates are unlikely to complete the argument to score A1 (it is not sufficient to just consider odd/even) but if you think they might deserve the A1 then send to Review.

Notes

General Note: Throughout this question we are condoning if candidates do not reference the following two points which are deemed fairly trivial and acceptable to be omitted at A-level:

• As \(x\) and \(y\) are integers then both \(3x + 2y\) and \(2x - 5y\) are integers.
• As \(x\) and \(y\) are positive then \(3x + 2y \gt 0\)

As such, we only require candidates to prove that there are no positive integer solutions \(x\) and \(y\) to simultaneous equations that have a positive RHS.

Note: Any attempt to solve the given simultaneous equations does not score any marks on its own.
Any attempts that use substitutions such as \(x = 2n\) etc. are unlikely to score any marks.
There are other methods to eliminate some pairs of simultaneous equations e.g. by showing that the only solution to \(3x + 2y = 7\) is (1, 2) which is not on \(2x - 5y = 4\). Use Review in such cases.

M1: Attempt to solve any one of the other 11 possible cases (labelled A-K) below to find a value for \(x\) or a value for \(y\). The attempt may be implied by a value for \(x\) or \(y\) which need not be correct.

dM1: Attempts to solve any two of the other 11 possible cases below to find a value for \(x\) or a value for \(y\) with at least one correct and correctly rejected. It is not necessary to find both values of \(x\) and \(y\) unless the correct value found does not cause a contradiction (see cases D and G).

ddM1: Attempts to solve all 5 pairs (A-E) of the relevant simultaneous equations with positive RHS or attempts to solve cases A and B and justifies that these are the only cases that need checking using either:

• \(3x + 2y \geqslant 5\) (because \(x\) and \(y\) are positive) or
• \(3x + 2y \gt 2x - 5y\) (because \(x\) and \(y\) are positive) [cases F, G and H do not need checking using this approach because of the general note]

Their values for \(x\) and \(y\) do not need to be correct for this mark as long as the dM1 is scored.

A1: cso Shows that all necessary cases are impossible with correct values, correct reasons, and a minimal conclusion e.g. “hence proved”. There is no need to say “contradiction” or restate the objective (you can also ignore any inaccurate attempt to restate the objective).
All 5 pairs of simultaneous may be rejected in one go if the rejection is sufficiently clear.
The mechanics of solving the simultaneous equations does not need to be shown.

Notes

(a) Way 1 Notes

Condone use of \(y\) for \(H\) for the method marks.
A model of the form \(H = x^2 + ax + b\) or \(H = -x^2 + ax + b\) will score no marks.
Note that it is possible to identify (by symmetry) that the points \((-2, 0.8)\) and \((18, 2)\) also lie on the parabola so you may see valid use of these points.

M1: Uses the equation \(H = \pm ax^2 \pm bx \pm c\) to model the path and uses \(x = 0\) and \(H = 2\) correctly placed to establish the value of the constant term and uses \(x = 20\) and \(H = 0.8\) or \(x = 9, \dfrac{\mathrm{d}H}{\mathrm{d}x} = 0\) to give an equation in ‘\(a\)’ and ‘\(b\)’ with \(\dfrac{\mathrm{d}H}{\mathrm{d}x}\) of the form \(\ldots\alpha x + \beta\)
An alternative is to recognise that the maximum occurs when \(x = -\dfrac{b}{2a} = 9\) or equivalent
e.g. maximum when \(x = 9 \Rightarrow H = a(x-9)^2 + \ldots = ax^2 - 18ax + \ldots \Rightarrow b = -18a\)
Award for \(\pm\dfrac{b}{2a} = 9\) or equivalent.
They may also use e.g. \((-2, 0.8)\) or \((18, 2)\) to give an equation in \(a\) and \(b\).

dM1: This mark requires:

• uses the equation \(H = \pm ax^2 \pm bx \pm c\) to model the path and uses \(x = 0\) and \(H = 2\) correctly placed to establish the value of the constant term
• uses \(x = 20\) and \(H = 0.8\) correctly placed and \(x = 9, \dfrac{\mathrm{d}H}{\mathrm{d}x} = 0\) to give 2 equations in ‘\(a\)’ and ‘\(b\)’ with \(\dfrac{\mathrm{d}H}{\mathrm{d}x}\) of the form \(\ldots ax + b\) or as above using \(\pm\dfrac{b}{2a} = 9\)
  They may also use e.g. \((-2, 0.8)\) or \((18, 2)\) to give an equation in \(a\) and \(b\).

ddM1: Solves their 2 equations in “\(a\)” and “\(b\)” to find their ‘\(a\)’ and ‘\(b\)’.
This may be done on a calculator. You do not need to check their method for solving.

A1: Correct equation. Must be \(H = \mathrm{f}(x)\).

(a) Way 2

(a) Way 2 Notes

Condone use of \(y\) for \(H\) for the method marks.
A model of the form \(H = A \pm (x-9)^2\) will score no marks.

M1: Uses the equation \(H = A \pm B(x-9)^2\) or \(H = A \pm B(9-x)^2\) to model the path and uses one of the ‘end points’ correctly placed to give an equation in ‘\(A\)’ and ‘\(B\)’
They may also use e.g. \((-2, 0.8)\) or \((18, 2)\) to give an equation in \(A\) and \(B\).

dM1: Uses the equation \(H = A + B(x-9)^2\) or \(H = A + B(9-x)^2\) to model the path and uses both ‘end points’ correctly placed to give 2 equations in ‘\(A\)’ and ‘\(B\)’
They may also use e.g. \((-2, 0.8)\) or \((18, 2)\) to give an equation in \(A\) and \(B\).

ddM1: Solves their 2 equations in “\(A\)” and “\(B\)” to find their ‘\(A\)’ and ‘\(B\)’.
This may be done on a calculator. You do not need to check their method for solving.

A1: Correct equation. Must be \(H = \mathrm{f}(x)\).

Note that using \(H = A + B(x-9)^2\) followed by the incorrect assumption that \(A = 2\) is unlikely to score any marks as they will subsequently not be able to produce 2 equations in “\(A\)” and “\(B\)”

Possible alternative 3:

M1: Uses the equation \(H = A\left((x-9)^2 - 81\right) + B\) to model the path and uses \(H = 2\) when \(x = 0\) correctly placed to find “\(B\)”

dM1: Uses the equation \(H = A\left((x-9)^2 - 81\right) + B\) to model the path and uses \(H = 0.8\) when \(x = 20\) correctly placed. May also use e.g. \((-2, 0.8)\) or \((18, 2)\)

ddM1: Substitutes their value for “\(B\)” to find a value for “\(A\)”

A1: Correct equation. Must be \(H = \mathrm{f}(x)\).

Notes

B1: Gives a suitable limitation – see scheme
If more than one limitation is given and one is acceptable then award this mark as long as none of the other statements are contradictory (they may be incorrect/inappropriate)

Notes

M1: Substitutes \(x = 16\) into their equation modelling the path to obtain a value for \(H\).
This may be seen explicitly as above or may be implied by their value (you may need to check). Must have a quadratic function in \(x\).

A1: Depends on

• A correct equation
• \(H = 2.96\)
• Correct conclusion that she cannot catch the ball or equivalent

A minimum for M1A1 could be e.g. \(x = 16 \Rightarrow H = 2.96\) “so no”

(c) Alternative:

So Chandra would not be able to catch the ball

M1: Substitutes \(H = 2.5\) into their quadratic equation modelling the path to obtain a value for \(x\). This may be seen explicitly as above or may be implied by their value (you may need to check). Must have a quadratic function in \(x\).

A1: Depends on

• A correct equation
• \(x =\) awrt 17
• Correct conclusion that she cannot catch the ball or equivalent.

A minimum for M1A1 could be e.g. \(H = 2.5 \Rightarrow x = 17\) “so no”

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 \([5.8, \infty)\)

Typical solution

Typical solution

(b)(i)

(b)(ii)

\(r - r^2\) is at its maximum when \(r = \dfrac{1}{2}\)

Maximum value of \(r - r^2 = \dfrac{1}{4}\)

(b)(iii)

Hence \(T_{\infty} \geqslant 240\)

Typical solution

Typical solution

\(x \in (-\infty, -2) \cup (7, \infty)\)

Typical solution

Typical solution

Typical solution

So \(A\left(\dfrac{2 - \sqrt{6}}{2}, \dfrac{\sqrt{6}}{2}\right)\)

Typical solution

Typical solution

Typical solution

Typical solution

Since 42 and \(25n\)-9 are both integers the expression is rational

Typical solution

Typical solution

(i)

(ii)

(iii)

The translated graph will only have one real solution.

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: Attempt to raise both sides to power \(\frac{2}{3}\) (May see \(\sqrt[\frac{3}{2}]{8}\))
May be implied by \(x^2 - 5 = 4\) or \(x^2 = 9\)

A1: www (e.g. A0 if \(8^{\frac{3}{2}} = 4\) seen).

Notes

M1: Attempt to take logs of both sides.

A1: SCB1 for correct answer without working (max 1/2).

Notes

M1: For recognising the given quartic as a quadratic in \(x^2\) or \(u = x^2\) and attempting to solve (allow a sign error in factorisation).
One of these four lines of working must be seen (oe), but the other may be implied by later correct work. Condone missing \(= 0\).

A1: Or \((x + 2)(x - 2)\) seen (NB this mark not implied by correct answers)

A1: Or \(x = 2\) and \(x = -2\) (and no other solutions e.g. A0 if \(x = 4\) seen)

B1: Accept equivalents but there must be an indication that there are no (real) solutions from \(x^2 = -1\) (e.g. ‘N/A’ or ‘X’ etc.) Or \(x = \pm i\)

If M0 then allow SCB2 for \((x - 2)(x + 2) \Rightarrow x = \pm 2\) (max. 2/4)
Answers only without working scores 0/4

Notes

B1: Correct expression. Or \(2x^{-0.25}\). Could be implied by \(p = 2\), \(q = -0.25\)

Notes

B1: Obtain 125. Could be implied by \(p = 125\)

B1: Obtain \(x^{\frac{9}{2}}\). Allow \(4\tfrac{1}{2}\) or 4.5. Could be implied by \(q = 4.5\)

Allow SC B1 if 0 scored but \(x\sqrt{x}\) is seen as \(x^{\frac{3}{2}}\) oe

Notes

B1: Obtain 4. Could be implied by \(p = 4\)

B1: Obtain \(x^4\). Could be implied by \(q = 4\)

Allow SC B1 if 0 scored but \(\sqrt{16x^8}\) seen

Notes

M1*: Obtain either \(ax^2\) or \(3x^b\)

M1dep*: Correctly combine \(x^5\) and their \(3x^2\). Obtain either \(ax^7\) following their \(a\), or \(3x^{(5+b)}\) following their \(b\)

A1: Obtain correct answer. Could be implied by \(p = 3\), \(q = 7\)

Notes

M1: Rearrange and attempt to solve quadratic inequality or equation
Some evidence of finding critical values (e.g. factorising or formula etc.) must be seen. This mark is not implied by correct answers.

A1: For both values (may be seen later e.g. embedded in an incorrectly formed inequality or implied by the final A1)

A1: Condone equivalent forms, e.g. \(x \leqslant 2\) and \(x \geqslant -5\) but do not accept open interval(s)

Answers only without working scores 0/3

Notes

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

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

Notes

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

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

Notes

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

Must be an equation

Notes

B1: Obtain any correct equation in terms of \(r\) and \(x\)
Could be implied by later work
May use other than \(r\)

M1*: Attempt equation in terms of only \(x\)
This equation would imply the B1
Or correct equation in terms of only \(r\)

A1: Obtain any correct equation not involving fractions or brackets
May still have like terms not yet combined
May result in a cubic depending on method (probably \(6x^3 - 39x^2 - 270x - 297 = 0\))

A1: Solve quadratic BC to obtain both correct \(x\) values
Or solve cubic, to obtain three correct roots (third is likely to be \(x = -1.5\))

A1: Obtain at least \(r = \tfrac{4}{5}\)
If second value of \(r\) given then it must be correct (if third value given then it must be consistent with their correct cubic roots)

M1d*: Attempt sum to infinity, using correct formula, with their \(r\) and their attempt at \(a\)
Must be using their numerical values for \(a\) and \(r\) with \(|r| \lt 1\)
ISW using additional value(s) of \(r\)
M0 if using their \(x\) and not attempt at \(a\)

A1: Obtain 125 only
A0 if additional solution

B1: Clear explanation as to why \(r = -2\) is discarded
Must be considering correct \(r\) value, so B0 if rejecting \(x = -3\) as \(|-3| \gt 1\)
Could generate the terms \(-3, 6, (-12)\) and hence conclude with ‘divergent sequence’
If additional solutions for \(x\) and/or \(r\) from cubic then they must also be correct and explicitly rejected

NB Eliminating \(x\) not \(r\) is a valid method, and could gain full credit. When solving their quadratic there is no need to see \(r = -2\) (and hence \(x = -3\)) to award the A marks

Notes

M1: Equate attempt at gradient of line to 2
Must be attempting to use \(\dfrac{y_2 - y_1}{x_2 - x_1}\), with consistent order in numerator and denominator; allow one sign slip
Could also use informal methods

A1: Obtain \(h = \tfrac{5}{2}\) oe

Notes

B1FT: Obtain correct \(x\) coordinate for \(C\), following their \(h\) ie \(2h - 1\)
Must be clear that their \(2h - 1\) is the \(x\) coordinate
BOD if brackets omitted

B1: Must be clear that 13 is the \(y\) coordinate
BOD if brackets omitted

SC B1 if both values correct but given as a vector not a coordinate

Notes

M1: Attempt equation of line through \(A\)
Allow one sign slip, but M0 if \(x\) and \(y\) coordinates transposed
Could use their \(B\) or \(C\) instead, but must still be using gradient of 2

M1: Equate line and curve

A1: Obtain correct quadratic, with like terms collected
Condone no ‘= 0’

M1*: Attempt discriminant of 3 term quadratic, resulting from equating line and curve
Correct discriminant for their quadratic
Condone any inequality or equality, or no, sign for this mark
M0 if using just \(x^2 - 4x + k\)
If the discriminant is initially embedded in the quadratic formula, then M1 is only awarded when it is considered in isolation

M1d*: Use \(b^2 - 4ac \gt 0\)
Inequality sign could be implied by final answer
M0 if incorrect inequality sign, including \(b^2 - 4ac \geqslant 0\)

A1: Obtain \(k \lt 14\)

Alt method for final 3 marks

M1*: Attempt completed square form
Correct expression for their quadratic
Allow unsimplified

M1d*: Set constant term \(\lt 0\)
M0 if constant term \(\leqslant 0\)
May see more informal method to determine inequality sign

A1: Obtain \(k \lt 14\)

Alt method (using differentiation to find point of intersection)

M1: Differentiate equation of curve, equate to 2 and attempt \(x\)

M1: Attempt equation of line through \(A\)
Could use their \(B\) or \(C\) instead

M1: Attempt \(y\) value from line, and use their (3, 11) in equation of curve to attempt \(k\)

A1: Obtain \(k = 14\)

M1: Clear method to determine inequality sign
Could be algebraic or a sketch
Could be implied by final answer

A1: Obtain \(k \lt 14\)

Notes

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Notes

(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

For reference only: \(\frac{1}{2}(3x+1)^2(2) \lt 44x - 7\)

B1*: Correct use of \(A = \frac{1}{2}r^2\theta\)

B1dep*: Expand and re-arrange to correct 3TQ expression in \(x\)
Allow any inequality sign or equals

M1dep*: SEE APPENDIX for awarding this mark (solving 3TQ expressions) - dependent on first B mark only (this mark is for solving their 3TQ but not for solving \((3x+1)^2 = 0\))
Correct quadratic followed immediately by correct critical values (with no working) is M0

B1: Correct critical values of \(x\) (if factorisation shown then it must imply these two c.v.)
Must be \(\frac{2}{9}\) or \(0.\dot{2}\) but B0 for 0.222…

B1FT dep*: FT their two positive critical values \(x_1, x_2\) e.g. \(\{x : x_1 \lt x \lt x_2\}\) where \(x_2 \gt x_1\); allow \(\left\{x : x \gt \frac{2}{9}\right\} \cap \{x : x \lt 4\}\) but ‘union’ is B0
B0 for interval notation e.g. \(\left(\frac{2}{9}, 4\right)\)
Answer must be in set notation for this mark – dependent on first B mark only

Appendix: rules for solving quadratics in questions 2 and 4(b)(i) only

In questions 2 and 4(b)(i) candidates are required to solve 3 term quadratics (3TQ) using DR – therefore we must see a correct, complete method for solving these quadratics – the correct answers do not imply the corresponding M mark, for example in question 2, \(9x^2 - 38x + 8 = 0 \Rightarrow x = 4\) or \(x = \frac{2}{9}\) 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 2 (and similarly for question 4(b)(i)):
\(9x^2 - 38x + 8 = \left(x - \frac{2}{9}\right)(x - 4)\) is M0 (but the following B1 for the correct c.v. of \(\frac{2}{9}\) and 4 in qu. 2 can still be awarded as they follow from these two factors)
\(9x^2 - 38x + 8 = (3x + 8)(3x + 1)\) is M1 (when expanded the \(x^2\) and constant terms are correct)
Allow correct part factorisation for their 3TQ expression e.g. if correct 3TQ then in question 2 the expression \(9x(x - 4) - 2(x - 4)\) 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. \(9x^2 - 38x + 8 = 0 \Rightarrow x = \dfrac{38 \pm \sqrt{38^2 - 4(9)(8)}}{2(9)}\).
Minimal acceptable working would be \(x = \dfrac{38 \pm \sqrt{1156}}{18}\) (so must explicitly see the discriminant) for M1

Rules for completing the square – using \(9x^2 - 38x + 8 = 0\) as an example:
The M1 is not awarded until correctly getting to the stage of \(x - \frac{19}{9} = \pm\sqrt{\frac{289}{81}}\) (must include \(\pm\) so implying two roots) with no errors (so consistent with applying the formula correctly)

Notes

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

Notes

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

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

Notes

(a)(i)

M1: Attempt to rewrite fractions using correct common denominator
Common denominator could just appear as \(9 - 4x\)
Must include correct attempt at numerators as well

A1: Obtain correct simplified fraction
No need to state values for \(a\), \(b\) and \(c\) explicitly
www – if middle terms shown for expansion of denominator, then these must be correct
ISW any further attempt to ‘simplify’
SC B1 for answer only, with no method shown

(a)(ii)

M1: Attempt to solve equation – as far as clearing the fraction and combining constant terms
M1 for using their fraction, as long as of correct form
Correct method to clear fraction, so M0 for eg \(6 = 18 - 4x\), but allow sign error when combining constant terms

A1: Obtain \(x = \frac{3}{2}\)
aef, but fractions must be simplified

Notes

M1: Attempt to solve disguised quadratic in \(2^y\)
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 two correct roots (could still be in terms of eg \(u\) if substitution used)
SC If no method shown then award B1 in place of M1A1 for both correct roots (final two marks can still be awarded)

M1: Attempt to solve \(2^y = k\), where \(k \gt 0\)
May just see \(y = 3\), with no explicit use of \(\log_2\)
Allow BOD if attempt at solving \(2^y = -1\) still present
If \(k \neq 8\) then solution method must be seen, even if \(k\) is a power of 2

A1: Obtain \(y = 3\), having rejected \(2^y = -1\) with some reasoning
Must have some reason, eg ‘\(2^y\) is always positive’, ‘\(2^y\) cannot be negative’, ‘cannot take log of a negative number’, ‘not defined’, ‘not real’, ‘no solutions’
A0 for ‘math error’, ‘does not work’, ‘not possible’

SC If no method at all shown then allow B1 for \(y = 3\), with no other solutions

Notes

(a)(i)

B1 B1: B1 for each constant.
For ‘4’ (Allow \(a = 4\))
For ‘\(-5\)’ (Allow \(b = -5\))
ISW

(a)(ii)

B1FT: FT their \(b\)
Accept \((4, -5)\) as the coordinates of the minimum point but must be correct for their \(a\) and \(b\).

Notes

M1: Write \(b^2 - 4ac = 0\) (accept \(\gt 0\) for this mark only) or equate with completed square form.
Accept a sketch or equivalent reasoning.

A1: Final answer must be given as \(k = -5\)

SC B1: correct answer without working (max [1/2])

Notes

M1: Either of the two \(\frac{\mathrm{d}y}{\mathrm{d}x}\) terms correct, allowing sign errors
Condone \(6x^2\mathrm{d}x + 6y\mathrm{d}x + 6x\mathrm{d}y - 6y\mathrm{d}y\)

B1: Both terms correct
Must now be \(6y + 6x\frac{\mathrm{d}y}{\mathrm{d}x}\), or implied in a correct expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\)

A1: Condone missing or incorrect RHS
Must now have \(\frac{\mathrm{d}y}{\mathrm{d}x}\) and not just \(\mathrm{d}y\) or \(\mathrm{d}x\) in terms

M1: Use \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1\) in their equation
Must now be equation, but RHS could be incorrect (eg ‘= 2’)

B1: Solve correct quadratic in \(x\) to obtain two correct roots (possibly BC)
Quadratic must come from correct implicit differentiation
B0 if \(x\) ‘cancelled’ in quadratic to give \(x = -1\) as only root, but M1A1 still available

B1: Explicitly reject \(x = 0\), with reasoning
\(x = 0\) must come from \(x^2 + x = 0\)
eg negative numbers cannot be square rooted or \(y^2 \ne -\frac{2}{3}\) as \(y\) is real
(just \(y^2 \ne -\frac{2}{3}\) is insufficient)
Must be sensible reason and not just ‘math error’ or ‘not possible’
Could say that there are only imaginary (or not real) roots – condone ‘complex’ roots

M1: Attempt to determine the number of real roots of their 3 term quadratic in \(y\)
From substituting their \(x\) value into the equation of the curve
Consider discriminant, or use quadratic formula, or attempt minimum value of function

A1: Obtain correct discriminant from correct quadratic and conclude appropriately
\(x = -1\) must come from \(x^2 + x = 0\)
If using quadratic formula then it must be fully correct and attention drawn to why there are no real roots

Notes

M1: Or for \(m\sec\theta + 3\cos\theta = 4\sin\theta\) \(\left(\Rightarrow \dfrac{m\sec\theta}{\cos\theta} + 3 = 4\dfrac{\sin\theta}{\cos\theta}\right)\) \(m\sec^2\theta + 3 = 4\tan\theta\)
The first M mark is for a valid method arriving at a three term equation containing \(\sec^2\theta\)
Squaring each individual term of the original equation scores no marks

M1: Correctly uses the identity \(1 + \tan^2\theta \equiv \sec^2\theta\) to obtain an equation in \(\tan\theta\) only

A1: AG so sufficient working must be shown
A0 if angle missing from any trigonometric terms

Notes

M1*: Considers discriminant of given quadratic equation in tan (\(c\) must be two terms) to get an expression in \(m\) only. M0 for embedded discriminant in quadratic formula unless explicitly considered
Allow \(4^2 - 4m(m + 3)\)

M1dep*: Sets their discriminant equal to zero and obtains an expanded three-term quadratic in \(m\)
Reasoning for setting the discriminant equal to zero is not required for this mark

A1: State or imply \(m = -4\) only

M1: Uses their negative integer value of \(m\), and solves the equivalent of their three term quadratic equation in tan, to obtain (at least) \(\tan\theta = k\), where \(k\) is non-zero - dependent on both previous M marks
Allow \(-4\tan^2\theta - 4\tan\theta - 1 = 0 \Rightarrow \tan\theta = -0.5\) for this mark
If no method shown for solving their quadratic, then award this mark if the solution is correct for their quadratic

A1: For full marks must explain why the discriminant should be set equal to zero – must say that as there is only one value of \(\theta\) or \(\tan\theta \Rightarrow \Delta = 0\) (as a minimum must see explicit mention of ‘one’ together with ‘\(\theta\)’ or ‘\(\tan\theta\)’ for this mark). Allow awrt 2.68
2.677945045…

Notes

B1: Could be implied by \(2(x + q)^2 + r\)

B1: Could be implied by \(p(x + 1.5)^2 + r\)

B1FT: FT on their \(p\) and \(q\) ie \(7 - pq^2\)

Notes

B1FT: FT on their (a)
Could come from differentiation

B1FT: FT on their (a)
No FT on incorrect \(x\)-value from differentiation

Notes

B1FT: FT on their minimum value
Allow BOD if different answers in (a) and (b)
2.5 must be stated as, or clearly intended to be, the minimum value
Just \((\ldots, 2.5)\) is insufficient

M1: Attempt to solve \(\tan\theta =\) their \((-1.5)\)
To obtain numerical value for \(\theta\)
Allow an angle in radians (expect \(-0.983\) rad)
Allow BOD if different answers in (a) and (b)

A1: Obtain \(124^\circ\), or better
A0 if additional solutions

Condone approaches other than ‘hence’ eg
B1 – attempt to solve \(\tan\theta = -1.5\), from correct derivative (expect \(4\tan\theta\sec^2\theta + 6\sec^2\theta = 0\))
B1 – obtain \(\theta = 124^\circ\)
B1 – obtain min value of 2.5 (no FT)

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

M1: Attempt to find value for \(k\)
From \(x = ky^2\sqrt{z}\) or \(x = kz^2\sqrt{y}\) only
Using sum, not product, is M0 but watch for \(+\sqrt{z}\) being used for positive square root

A1: Correct equation
Ignore modulus sign if used around \(\sqrt{z}\)
Allow BOD if initial equation stated explicitly, \(k\) found correctly but then final equation not seen or seen as now incorrect

Notes

M1: Attempt to find \(x\), from equation in terms of \(y\), \(z\) and numerical \(k\)
Could be from using direct proportion and not their equation from (a)

A1: Obtain 112.5
Or any exact equiv

Notes

B1: Correct equation / expression for \(A\)
Or \(40 + 0.04m\)

B1: Correct equation / expression for \(B\)
Or \(0.06(m - 100)\)

B1B0 if units inconsistent in two equations

SC B1 for both \(44 + 0.04m\) and \(0.06m\) (or \(4400 + 4m\) and \(6m\)) – from using \(m = 0\) at 100 minutes

Notes

M1: Attempt to solve simultaneously, from two linear equations in \(m\)
At least one equation must have constant term
Could be implied by final answer of 38hrs 20 mins

A1: Obtain 2300 (minutes)
isw once 2300 seen

Notes

M1*: Attempt discriminant
Allow \(b^2 + 4ac\) for M1, but nothing else

A1: Obtain correct inequality
Not necessarily expanded

M1dep*: Attempt to solve their inequality or equation for \(k\)

A1: Obtain \(k \lt 1\)

OR (completing the square or differentiating)
M1* – attempt to complete the square, or differentiate, and link minimum point to 0
A1 – obtain \((k + 3) - 4 \lt 0\)
M1d* – solve their inequality or equation
A1 – obtain \(k \lt 1\)

OR (using perfect square)
M1* – link \(k + 3\) to 4
A1 – obtain \(k + 3 \lt 4\)
M1d* – solve their inequality or equation
A1 – obtain \(k \lt 1\)

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

M1: Attempts to solve simultaneously

M1: Collects terms. Must see \(=0\).
May be implied by correct discriminant

M1: Forms discriminant of their equation and equates to zero

A1: cao

Alternative method

M1: Uses gradient of 3 to form an equation linking \(k\) and \(x\)

M1: Forms second equation for the point of intersection oe

M1: (Either) Uses above quadratic equation and an expression for \(x\) obtained from gradient to form a quadratic in \(k\) only

A1: cao

M1: (Or) Uses above quadratic equation and an expression for \(k\) to form a quadratic in \(x\) only, leading to values for \(k\)

A1: cao

Notes

M1: Uses their values for \(k\) and attempts to solve simultaneously

A1: Note – using any values of \(k\) will give the same point of intersection for full credit

Notes

M1*: differentiation of all 4 terms with 3 of the 4 terms differentiated correctly

A1: all correct

M1dep*: discriminant for their \(6, 6m\) and \(-9m\);
may see \((2m)^2 - 4 \times 2 \times (-3m)\)

M1: dependent on obtaining discriminant from their derivative;
M0 for use of their discriminant \(\gt 0\) or \(\geqslant 0\)
NB \(4m^2 + 24m \lt 0\) or \(4m^2 + 24m = 0\)

A1: inequality or interval must be strict

Notes

M1: Rationalising denominators
At least one example of multiplying by e.g. \(\dfrac{\sqrt{2} - 1}{\sqrt{2} - 1}\)

A1: Convincing completion including all steps including the denominators of 1 or better

Alternative method

M1: Putting over a common denominator and simplifying all like terms

A1: Convincing completion including all steps

Additional guidance

This is a DR question. There are two solutions possible. The first is to rationalise the denominators. To score the M1 we want to see at least one fraction multiplied by the correct fraction \(\frac{\sqrt{2}-1}{\sqrt{2}-1}\) or \(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\) or \(\frac{2-\sqrt{3}}{2-\sqrt{3}}\).

To score the A1 it must be all correct with the denominators seen as 2-1, 3-2, 4-3 or 1,1,1. There must be something there. This leads to the final answer of 1.

The second method is where they try to combine all three fractions. The M1 is given attempting to find a numerator of \((\sqrt{3} + \sqrt{2})(2 + \sqrt{3}) + (\sqrt{2} + 1)(2 + \sqrt{3}) + (\sqrt{2} + 1)(\sqrt{3} + \sqrt{2})\) all multiplied out to \(3\sqrt{6}+4\sqrt{3}+5\sqrt{2}+7\) and a denominator of \((\sqrt{2} + 1)(\sqrt{3} + \sqrt{2})(2 + \sqrt{3})\) multiplied out to \(3\sqrt{6}+4\sqrt{3}+3\sqrt{2}+2\sqrt{2}+7\).

The A1 is for getting the correct fraction \(\frac{3\sqrt{6}+4\sqrt{3}+5\sqrt{2}+7}{3\sqrt{6}+4\sqrt{3}+5\sqrt{2}+7}\) before cancelling and getting the answer 1.

Notes

M1: must be sum of two squares;
may be implied by correct answer or by \((\pm 4)^2 + (\pm 2)^2\)

A1: mark the final answer; B2 for correct answer unsupported

Notes

M1: An initial step to getting all terms in \(x\) on one side.
Allow incorrect inequality signs

A1: Correct completion

Additional guidance

This is a straightforward inequality question. The M1 is given for seeing either of the two inequalities in the MS. Allow the M1 even if they use the incorrect signs. The A1 is for \(x \gt 5\) only.

Notes

M1: Attempt to find equation of the line using correct gradient formula

M1: Eliminating one variable

M1: oe Three term quadratic seen or implied by correct \(x\)-values

A1: cao

A1: Both \(y\)-coordinates seen.

M1: Uses distance formula for their points (not given points)

A1: Must be exact (allow \(\sqrt{360}\) oe)

Notes

B1: OR \(r = \dfrac{10}{2 + \theta}\)
Expression for one of \(r\), \(\theta\) in terms of the other

M1: OR \(A = \dfrac{100\theta}{2(2+\theta)^2} = \dfrac{50\theta}{(2+\theta)^2}\)
Area in terms of either their \(r\) or their \(\theta\). Need not expand brackets.

M1: Completing the square

B1: Convincing explanation that there is a max

Alternative method 1

Alternative method 2

M1: Reasonable attempt at quotient rule

For information: \(\dfrac{\mathrm{d}^2A}{\mathrm{d}\theta^2} = -\dfrac{200}{256} = -0.78125\) at \(\theta = 2\)

Notes

operations may be in any order, but 3 distinct numerical steps required for 3 marks

if taking reciprocal and one other step are combined into one step, allow B1B1B0

if B0B0 for cubing and square rooting, allow SC1 for \(\left(\sqrt{\frac{49}{64}}\right)^3 = \frac{343}{512}\) or \(\left(\sqrt{\frac{64}{49}}\right)^3 = \frac{512}{343}\) seen

eg

B1: taking reciprocal; may be awarded after simplification

B1: square roots found; may be seen before taking reciprocal

B1: dependent on award of both preceding marks

eg Alternatively

B1: dependent on award of both preceding marks

Notes

M1: Rationalising denominators.
This is the minimum working needed for M1
Accept 1 for “6 – 5” and –1 for “5 – 6” etc

A1: Finding \(k\) convincingly after M1

Alternative

M1: For dealing appropriately with fractions (working with both sides) e.g. clearing the fractions or making k the subject with the RHS as a single fraction.
\(k = \dfrac{\left(\sqrt{5}+\sqrt{7}+2\sqrt{6}\right)\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{5}+\sqrt{6}\right)\left(\sqrt{6}+\sqrt{7}\right)}\)

A1: Expanding brackets and collecting like surds
\(k = \dfrac{12 + 2\sqrt{5}\sqrt{7} + 2\sqrt{5}\sqrt{6} + 2\sqrt{6}\sqrt{7}}{6 + \sqrt{5}\sqrt{7} + \sqrt{5}\sqrt{6} + \sqrt{6}\sqrt{7}}\)

A1: Finding \(k\) convincingly after M1

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

DR: This question included the instruction: In this question you must show detailed reasoning.

B1: Substituting values

M1: Attempt to rationalise denominator for one term
Either \(\times\dfrac{\sqrt{2}-1}{\sqrt{2}-1}\) or \(\dfrac{\sqrt{2}-1}{2-1}\) at least once for M1

A1: All correct

A1: Convincing completion (AG)

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: line from 2 to 7 only (do not need number line)
if M0 allow SC1 for one end correct

A1: one end correctly shaded

A1: both ends correctly shaded

Notes

M1: Elimination of one of \(x\), \(y\)

A1: May be as part of a pair of coordinates

A1: May be as part of a pair of coordinates
Allow correct unsimplified answers

Notes

\((0, k)\) unsupported earns B1 SC
Need not be as coordinates but must be clear which \(y\) goes with which \(x\)
Substituting for \(x\) gets M0 unless it leads to correct values of \(y\)

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

Notes

M1: Attempt to rationalize the denominator

M1: Attempt to expand the brackets

A1: must be in the form \(p + q\sqrt{2}\)
Mark final answer
Allow for instead stating \(p = \dfrac{3a+2}{7},\ q = \dfrac{a+3}{7}\)

Notes

M1: \(\dfrac{1}{x^6}\) soi

A1: Fully simplified. Allow \(x^{-2}\) oe