Notes

Mark (a) and (b) together

M1: Uses the equation of the model with \(t = 0, V = 20\,000 \Rightarrow 20000 = 1500 + A\,\mathrm{e}^0\) o.e.
May be implied by 18500

A1: \(A = 18\,500\) (18500 with no working seen scores M1A1) Ignore £ if present.

dM1: Attempts to use the equation of the model \(t = 2.5, V = 12\,000\)
\(\Rightarrow 12000 = 1500 + \text{``}18\,500\text{''}\mathrm{e}^{-2.5k}\) and proceeds to \(C\mathrm{e}^{\pm k \times 2.5} = D\) (where \(C \times D \gt 0\) and allow \(C = 1\)) before proceeding to find a value for \(k\). Allow them to have a non-numerical \(C\) for this mark.
Note it cannot be implied by their awrt \(\pm 0.227\) so
\(12000 = 1500 + \text{``}18\,500\text{''}\mathrm{e}^{-2.5k} \Rightarrow k = 0.227\) scores dM0A0 as we need to see the intermediate stage \(C\mathrm{e}^{\pm k \times 2.5} = D\). (typically look for \(18\,500\mathrm{e}^{-k \times 2.5} = 10\,500 \Rightarrow k = \ldots\) or condone to be implied by a correct expression involving logarithms for their \(A\))
It is dependent on the previous method mark.

A1: \(V = 1500 + 18500\,\mathrm{e}^{\text{awrt} -0.227t}\) o.e. e.g. \(t = \dfrac{\ln\left(\dfrac{V - 1500}{18500}\right)}{\text{awrt } -0.227}\) Allow \(k\) to be exact.

Notes

Mark (a) and (b) together

B1: Differentiates to a form \(\left(\dfrac{\mathrm{d}V}{\mathrm{d}t} =\right) -kA\mathrm{e}^{-kt}\) where \(k\) and \(A\) may be their values from (a)
e.g. \(\text{``}{-}0.227\text{''} \times \text{``}18500\text{''}\mathrm{e}^{\text{``}{-}0.227\text{''}t}\) May just see e.g. (using an exact \(k\)) \(\text{``}{-}4191.3\ldots\text{''}\mathrm{e}^{\text{``}{-}0.227\text{''}t}\) or e.g. (using \(k\) to 3sf) \(\text{``}{-}4199.5\text{''}\mathrm{e}^{\text{``}{-}0.227\text{''}t}\) but do not be too concerned by over rounding provided the intention is clear that it is their \(-kA\)
Do not be too concerned by the left hand side / poor labelling of the derivative.

M1: Substitutes \(A\,\mathrm{e}^{-kt} = V - 1500\) into their \(\left(\dfrac{\mathrm{d}V}{\mathrm{d}t} =\right) \pm kA\mathrm{e}^{-kt}\) to form an expression for \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\) in terms of \(V\). May see e.g. \(\text{``}{-}0.227\text{''} \times \text{``}18500\text{''}\mathrm{e}^{\text{``}{-}0.227\text{''}t} \Rightarrow \text{``}{-}0.227\text{''}(V - 1500)\).
Condone recovery of a sign slip on the index on \(\mathrm{e}^{-kt} \rightarrow \mathrm{e}^{kt}\) provided it is before they substitute \(A\,\mathrm{e}^{-kt} = V - 1500\) for this mark.
Using the given answer: substitutes \(V - 1500 = A\,\mathrm{e}^{-kt}\) and shows that \(-k(V - 1500) = -kA\mathrm{e}^{-kt}\) (may be in terms of their \(A\) and \(k\))

A1*: Full and complete proof with sight of \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\) oe seen somewhere in their solution and no errors seen. Must see \(-kA\mathrm{e}^{-kt}\) (or using their values) before proceeding to the given answer which may be written using their numerical value for \(k\)
Using the given answer they must conclude that \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = -k(V - 1500)\) which may be written using their numerical value for \(k\)

Alt (b) Separating the variables – may be in terms of their numerical value for \(k\)

B1ft: Separates the variables correctly and integrates to \(\ln(V - 1500) = -kt\) with or without \(+c\)
\(\displaystyle\int \dfrac{1}{(V - 1500)}\,\mathrm{d}V = \int -k\,\mathrm{d}t \Rightarrow \ln(V - 1500) = -kt\ \ (+c)\) oe

M1: Proceeds from \(\ldots\ln(V - 1500) = \ldots kt + c\) o.e. and rearranges to make \(V\) the subject. Condone slips.
e.g. \(\ln(V - 1500) = -kt + c \Rightarrow V - 1500 = A\mathrm{e}^{-kt} \Rightarrow V = 1500 + A\mathrm{e}^{-kt}\) or \(V = 1500 + \mathrm{e}^{-kt + c}\)
(just look for proceeding to \(V = \ldots\) for this mark though.) May be in terms of their numerical values for \(k\) and \(A\)

A1*: Achieves \(V = 1500 + A\mathrm{e}^{-kt}\) with no errors seen and concludes that \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = -k(V - 1500)\)
May be in terms of their numerical values for \(k\) and \(A\)

Alt (b) Rearranging to make \(t\) the subject

B1ft: For a correct rearrangement to \(t = -\dfrac{1}{k}\ln\left(\dfrac{V - 1500}{18500}\right)\) ft on their \(k\)

M1: Differentiates \(\ldots\ln(V - 1500)\) to \(\dfrac{\ldots}{V - 1500}\) and then finding \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = \dfrac{1}{\left(\dfrac{\mathrm{d}t}{\mathrm{d}V}\right)}\)
Typically look for \(t = -\dfrac{1}{k}\ln\left(\dfrac{V - 1500}{18500}\right) \Rightarrow \dfrac{\mathrm{d}t}{\mathrm{d}V} = -\dfrac{1}{k}\left(\dfrac{1}{V - 1500}\right)\) o.e. so they may have an unsimplified version of this e.g.

May be in terms of their numerical \(k\). Condone slips.

A1*: Achieves the given answer with no errors and \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\) seen somewhere in their solution

Notes

Note the question asks for a limitation of the model. If there is ambiguity over whether the response is referring to the model then try putting “the model suggests” or “the model” in front of their comment to see if this is a valid limitation.
Ignore comments which do not contradict a valid limitation.
If values are given then they must be correct and if it is the value of the car then it must have units (£ or pounds)

B1: Suitable limitations in context referring to the limitation of the model which score B1
e.g. (the model suggests) the value/price of the car will never go below £1500
e.g. (the model suggests) the rate of decrease of the value of the car is proportionally the same each year
e.g. (the model suggests) after a certain period of time the car will no longer lose value (condone this property of the model that it will tend to a limit)
e.g. (the model) only takes account of age
e.g. (the model) does not take into account damage / alteration to the car/ mileage
e.g. (the model) predicts that the car’s value will always go down
e.g. after many years the car may become worthless whereas the model does not allow for this (valid limitation comparing the car to the model)
e.g. the value of the car may go up where as it only decreases according to the model

Do not accept vague/incorrect/irrelevant or non-contextual comments which score B0
e.g. after many years the car may become worthless
e.g. the value of the car may increase
e.g. damage or alterations to the car may impact the value
e.g. the car will still have value when it is very old (should refer to £1500)
e.g. (the model suggests) the minimum value is 1500 (no units for money)
e.g. (the model suggests) the (value of the) car cannot be negative
e.g. car value will fluctuate which the model will not show (a model is not for this purpose)
e.g. the rate of decrease is proportionally the same each year (no context)

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

Marks for part (a) may not be scored in part (b)

M1: Either \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 0.45\) or \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = \pm 0.3V\) o.e. e.g. \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = \dfrac{9}{20}\) or \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = \pm\dfrac{3}{10}V\) seen or implied by e.g. \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 0.45 - \dfrac{3}{10}V\) (but not implied by just stating the given answer). Condone use of \(\dot{V}\)
It may be seen as part of their \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\) e.g. \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 0.45 + V + 0.3V\) scores M1A0*
Condone e.g. change in volume = (inflow – outflow =) \(0.45 - 0.3V\) for this mark.

A1*: Achieves \(20\dfrac{\mathrm{d}V}{\mathrm{d}t} = 9 - 6V\) with no errors, following \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 0.45 - \dfrac{3}{10}V\) o.e. (including the \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\) or \(\dot{V}\) but note that it must be \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\) in the final line and not \(\dot{V}\)).
change in volume \(= 0.45 - 0.3V \rightarrow 20\dfrac{\mathrm{d}V}{\mathrm{d}t} = 9 - 6V\) scores M1A0*.
Ignore any units used in their working for both marks.

Notes

B1: Separates the variables correctly, e.g., \(\displaystyle\int \dfrac{1}{9 - 6V}\,\mathrm{d}V = \int \dfrac{1}{20}\,\mathrm{d}t\) or \(\displaystyle\int \dfrac{20}{9 - 6V}\,\mathrm{d}V = \int \{1\}\,\mathrm{d}t\) o.e.
The integral symbol and/or \(\mathrm{d}V\) and/or \(\mathrm{d}t\) may be implied if they go on to integrate both sides to the correct form \(\ldots\ln\left|\alpha(9 - 6V)\right| = \ldots t\ \ (+c)\) with or without the modulus brackets.

M1: Attempts to integrate the reciprocal term \(\dfrac{\beta}{9 - 6V} \rightarrow \ldots\ln\left|9 - 6V\right|\) or \(\rightarrow \ldots\ln\left|\alpha(6V - 9)\right|\) for some constant \(\beta\) (and \(\alpha\) if used). Condone e.g. \(\dfrac{20}{9 - 6V} \rightarrow \ldots\ln 9 - 6V\) or \(\rightarrow \ldots\ln 6V - 9\)

A1: Correct integration for both sides. They do not need the + \(c\) for this mark.
Note scoring this mark implies the earlier B1 (unless it is a verification attempt – see SC).
Note that e.g. \(-\dfrac{1}{6}\ln\left|3 - 2V\right| = \dfrac{t}{20}\ \ (+c)\) or \(-\dfrac{10}{3}\ln\left|2V - 3\right| = t\ \ (+c)\) are also correct.
\(-\dfrac{1}{6}\ln(9 - 6V) = \dfrac{t}{20}\ \ (+c)\) is also correct.
Condone log being used in place of ln.

dM1: Requires constant of integration now. Substitutes (or states) \(t = 0\) and \(V = 0.25\) and finds a value for \(c\), which may be “\(A\)” \(= \mathrm{e}^c\) if they rearrange first to eliminate ln terms.
Dependent on the previous method mark.
Do not be concerned about their processing to find \(c\) or “\(A\)” \(= \mathrm{e}^c\) and does not need to be exact.

A1: Achieves the required form e.g. \(V = \dfrac{3}{2} - \dfrac{5}{4}\mathrm{e}^{-\frac{3t}{10}}\) with no errors and clear working.
Allow equivalent fractions or decimals e.g. \(V = 1.5 - 1.25\mathrm{e}^{-\frac{6}{20}t}\)

SC: Attempts by verification may score maximum B0M1A1dM1A0 – see below.

Alt: Use of an integrating factor – see below.

Special Case: Attempts by verification may score maximum B0M1A1dM1A0

B0: This mark may not be scored via this approach.

M1: Differentiates \(V = P - Q\mathrm{e}^{-kt}\) to the form \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = \alpha\mathrm{e}^{-kt}\) where \(\alpha\) is a constant (note it should be \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = Qk\mathrm{e}^{-kt}\)) and substitutes both this and \(V = P - Q\mathrm{e}^{-kt}\) into \(20\dfrac{\mathrm{d}V}{\mathrm{d}t} = 9 - 6V\) and deduces a value for \(P\) or \(k\) by comparing coefficients.

A1: Correct values for both \(P\) and \(k\).

dM1: Substitutes (or states) \(t = 0\) and \(V = 0.25\) and finds a value for \(Q\).
Requires a value for \(P\) to have been found using the above approach.

A0: This mark may not be scored via this approach.

Alternative: Using Integrating Factor (Further Maths)

B1: Deduces the correct integrating factor for the equation, \(\mathrm{e}^{0.3t}\)
This should come from \(\dfrac{\mathrm{d}V}{\mathrm{d}t} + 0.3V = 0.45 \Rightarrow \text{I.F.} = \mathrm{e}^{\int 0.3\,\mathrm{d}t} = \mathrm{e}^{0.3t}\)
May be implied by sight of \(\dfrac{\mathrm{d}\left(V\mathrm{e}^{0.3t}\right)}{\mathrm{d}t} = \ldots\)

M1: Fully multiplies through by their integrating factor and integrates both sides.
Score for \(V\mathrm{e}^{kt} = \displaystyle\int \ldots\mathrm{e}^{kt}\,\mathrm{d}t = \ldots\mathrm{e}^{kt}\) Condone missing \(\mathrm{d}t\)

A1: Correct integration \(V\mathrm{e}^{0.3t} = \displaystyle\int 0.45\mathrm{e}^{0.3t}\,\mathrm{d}t = \dfrac{3}{2}\mathrm{e}^{0.3t}\ \ (+c)\)

dM1: As main scheme.

A1: As main scheme.

Notes

M1: See main scheme. If using the answer to part (b) it must be of the form \(V = P - Q\mathrm{e}^{-kt}\) but there is no limitation on the values of their \(P\), \(Q\) or \(k\).
Substitution of a large value for \(t\) may score this mark but it is unlikely to be recovered to score the A1 unless they reference e.g. \(V_{\max}\) being “1.5”.
Reference to an (upper) limit of “1.5” or their \(P\) can imply the method mark.
If setting \(V = 2\) in their equation they must reach either ln(–ve) or solve the equation to reach a value for \(t\) to score this mark.

A1ft: Must conclude “no” or equivalent e.g. “the container will not become full”.
Makes a correct interpretation for their method (see bullets 1-8) with a clear conclusion e.g. “no”.
To score this mark through ft, their \(V\) must be of the form \(V = P - Q\mathrm{e}^{-kt}\) with \(k \gt 0\), \(Q \gt 0\) and \(0 \lt P \lt 2\) if used (but note that they can still use the answer to part (a) to score both marks via bullets 1, 3, 5 or 6). Allow “it” in place of the “container”/“tank”.
Just stating “the equation cannot be solved when \(V = 2\)” without any evidence is M0A0.
There must be no incorrect working if solving their equation or contradictory statements such as “\(t\) cannot be negative” but condone notational errors provided the intention is clear.

Notes

M1: Substitutes \(t = 3\) into the given model and proceeds to a value.
Implied by awrt 1007, 1008 or 1010
Condone copying slips e.g. \(-0.75\) or \(-0.0075\) in place of \(-0.075\) or 500 in place of 5000 – in such cases you may need to check their calculation if the substitution isn’t shown.

A1: 1007 but allow 1008 or 1010 (3sf). Must be a whole number.
Correct answer only scores both marks. A0 for e.g. 1007.4
Do not ISW if they go on to sum their values of \(N\) from \(t = 1\) to 3

Notes

M1: Sets \(5000 - 5000\mathrm{e}^{-0.075T}\) equal to 3000 and proceeds to either \(A\mathrm{e}^{-0.075T} = B\) or \(\mathrm{e}^{-0.075T} = k\) where \(A\), \(B\), \(k\) are constants with no restrictions for this mark.
May use \(t\) instead of \(T\). Condone slips.
Condone copying slips e.g. \(-0.75\) or \(-0.0075\) in place of \(-0.075\) or 500 in place of 5000.

dM1: Uses the correct order of operations and correct log work from an equation of the form \(A\mathrm{e}^{-0.075T} = B\) with \(AB \gt 0\) or \(\mathrm{e}^{-0.075T} = k\) with \(k \gt 0\) to proceed to a value for \(T\) or \(t\) which may be a numerical expression.
e.g. \(A\mathrm{e}^{-0.075T} = B \Rightarrow \ln A - 0.075T = \ln B \Rightarrow T = \dfrac{\ln A - \ln B}{0.075}\) or \(= \dfrac{\ln\left(A/B\right)}{0.075}\)
Condone log being used in place of ln unless there is clear evidence that they have used an inconsistent base.
May not be scored if “recovered” from e.g. \(\mathrm{e}^{-0.075T} = -\dfrac{2}{5}\)
Condone copying slips e.g. \(-0.75\) or \(-0.0075\) in place of \(-0.075\) or 500 in place of 5000.

A1: 12.22 cao but must see a correct equation following use of logs, e.g., \(-0.075T = \ln\dfrac{2}{5}\) or \(T = \dfrac{1}{-0.075} \times -0.916\) which may have intermediate rounding. Ignore any units given.
This mark is available following the occasional slip in writing \(-0.075\) as \(-0.75\) or \(-0.0075\) provided it is not consistently used.

Notes

M1: Differentiates once to the form \(\lambda\mathrm{e}^{-0.075t}\), \(\lambda\) a constant (even \(\pm 5000\)), and substitutes in \(t = 3\)
Award for an expression of the form \(k\mathrm{e}^{-0.075 \times 3}\) with constant \(k\) if no incorrect working is seen.
The substitution of \(t = 3\) may be implied by a correct value for their derivative of the required form, provided the derivative is seen. Do not be concerned with what they call their derivative.

A1: awrt 299 (car sales per month). Answer only scores no marks. Condone 300 following 299.4
Requires a correct derivative to be seen, e.g., \(375\mathrm{e}^{-0.075t}\) o.e., with or without the substitution of \(t = 3\) present.
Units may be omitted, but score A0 if an incorrect time frame is given e.g. cars per week

Notes

B1: Acceptable refinement. Some examples:

• Change the 5000 to 6500 (condone ambiguity about which 5000)
• Change the first 5000 to 6500
• Change both 5000s to 6500s
• Multiply the model by 1.3 (or e.g. \(\dfrac{65}{50}\) or \(\dfrac{13}{10}\))
• Add 1500 to the model
• Change the constant to 6500

The statement of an acceptable refined model e.g. \((N =)\ 6500 - 6500\mathrm{e}^{-0.075t}\) or e.g. \((N =)\ 6500 - 5000\mathrm{e}^{-0.075t}\) scores B1.
Ignore any extra unnecessary refinements such as increase/decrease the 0.075.

The following score B0:

• Change 5000 in the model (not specific enough – they haven’t said to 6500)
• Change the second 5000 to 6500 (incorrect)

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: Correctly sets up the model. \(\dfrac{\mathrm{d}r}{\mathrm{d}t} \propto \dfrac{1}{\sqrt{r}}\) scores B0 unless e.g. \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \pm\dfrac{k}{\sqrt{r}}\) is seen but condone \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \pm\dfrac{k}{\sqrt{r}}\) being seen at the start of (b). They may use any letter except \(t\) or \(r\) in place of \(k\).
You may see \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \pm\dfrac{1}{k\sqrt{r}}\) which is acceptable provided it is clear that it is not the \(k^{\text{th}}\) root.

Notes

Note: candidates using \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \pm\dfrac{1}{\sqrt{r}}\) in (b) can score maximum B0M1A1dM0A0

B1: \(k = 3.6\) coming from \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \dfrac{k}{\sqrt{r}}\) (or \(k = \dfrac{5}{18}\) coming from \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \dfrac{1}{k\sqrt{r}}\)) and from use of \(r = 16\) and \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = 0.9\) but note that this may occur later in their working, which is perfectly fine provided it is from acceptable work. Note e.g. \(k = -3.6\) coming from \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = -\dfrac{k}{\sqrt{r}}\) is correct.

Note, however, that an attempt to find \(k\) from comparing coefficients between \(r^{\frac{3}{2}} = 1.5kt + c\) and \(r^{\frac{3}{2}} = 5.4t + 10\) is not acceptable (see special case).
They can also find \(k\) by differentiating their \(r = \mathrm{f}(t)\) and substituting \(t = 10\), \(r = 16\) and \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = 0.9\) if they have also used \(t = 10\), \(r = 16\) in their \(r = \mathrm{f}(t)\). This sets up simultaneous equations where \(c\) can be eliminated. Use Review if you are unsure if their approach is acceptable.

M1: Separates the variables for their differential equation correctly and attempts to integrate both sides.
Must be a differential equation of the form \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \mathrm{f}(r)\) for some function \(\mathrm{f}(r)\) independent of \(t\).
Evidence of \(r^n \rightarrow r^{n+1}\), or e.g. \(\dfrac{1}{r} \rightarrow \ln r\) is sufficient for their attempt to integrate in \(r\), but \(k\) must be integrated to \(kt\) o.e. e.g. \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \dfrac{k}{r^2} \Rightarrow r^2\dfrac{\mathrm{d}r}{\mathrm{d}t} = k \Rightarrow \lambda r^3 = kt\,\{+c\}\) would score this mark.
Note that they may divide by \(k\) (or 3.6) prior to integrating. Here, 1 must be integrated to \(t\).

A1: Correct integration for their \(k\). Allow this mark if they have not found \(k\), so allow e.g. \(\dfrac{2}{3}r^{\frac{3}{2}} = kt\,\{+c\}\) with/without the constant of integration but the \(\dfrac{2}{3}\) must be evident in some way.
Condone spurious notation for this intermediate mark e.g. integral signs left in after integrating.

dM1: Uses \(t = 10\), \(r = 16\) in their equation to find the constant of integration.
This mark is dependent on the first method mark.
Must have already found a value for \(k\) using a valid strategy and the constant of integration must be present.
Those that found \(k\) from comparing coefficients between \(r^{\frac{3}{2}} = \dfrac{3}{2}kt + c\) and \(r^{\frac{3}{2}} = 5.4t + 10\) may not score this mark.

A1*: Correct equation from correct working. May be seen at the start of (c).
Must follow A1 earlier so do check if this has been obtained fortuitously.

SC: It is possible to compare coefficients (following integration) between \(r^{\frac{3}{2}} = \dfrac{3}{2}kt + c\) and \(r^{\frac{3}{2}} = 5.4t + 10\) to deduce the value of \(k\) as 3.6 (or just write their coefficient as 5.4).
The maximum that can be scored this way or by a similar invalid approach is B0M1A1dM0A0

Notes

M1: Substitutes \(t = 20\) into the given equation and uses correct processing to find the value of \(r\)
e.g. substitutes \(t = 20\) into \(\dfrac{2}{3}r^{\frac{3}{2}} = 72 + \dfrac{20}{3} \Rightarrow r = \left(\dfrac{3}{2}\left(72 + \dfrac{20}{3}\right)\right)^{\frac{2}{3}} = \ldots\)
Their work should lead to the correct answer so the index work must be correct e.g. \(\sqrt{118^3}\) is M0.
\(\sqrt[3]{118^2}\) or \(118^{\frac{2}{3}}\) are acceptable as values, i.e., the bracket must be evaluated.
May be implied by awrt 24 (cm) following \(r^{\frac{3}{2}} = 118\) or by awrt 24.1 (cm). Ignore units for M1.

A1: cao 241mm or 241 or 24.1cm but do not accept 24.1 or e.g. \(\sqrt[3]{118^2}\) cm
Correct answer with units implies both marks.

Notes

B1: Examples of acceptable answers (which must relate to the model in context):

• (The model will not hold indefinitely as) the balloon may burst/pop
• The balloon is unlikely to be (perfectly) spherical (condone circular)
• The model predicts the balloon will increase in size without limit (which is unrealistic)

Note \(t \rightarrow \infty \Rightarrow r \rightarrow \infty\) is unrealistic / impossible scores B0 unless they reference e.g. the radius.
Condone the presence of additional remarks such as “the balloon may not inflate at the same rate” or “the radius of the balloon might not start at 0” that have already been addressed in the model, but these answers alone score B0.

Notes

(a) Must be using base 10 in (a). Ignore any units associated with \(a\) and \(b\) in part (a).

M1: Correct strategy to get a numerical expression or value for \(a\) or \(b\) e.g. \(a = 10^{0.81}\) or \(b = 10^{0.0054}\). This may be implied by \(a =\) awrt 6.46 or \(b =\) awrt 1.01 if no incorrect work is seen.

A1: Correct value for \(a\) or \(b\). Allow 3 sf for this mark so allow \(a =\) awrt 6.46 or \(b =\) awrt 1.01.
May be seen embedded in their formula.

M1: Correct strategy to get a numerical expression or value for \(a\) and \(b\) e.g. \(a = 10^{0.81}\) and \(b = 10^{0.0054}\). This may be implied by \(a =\) awrt 6.46 and \(b =\) awrt 1.01 if no incorrect work is seen.

A1: Correct values. This requires \(a =\) awrt 6.457 and \(b =\) awrt 1.013 for this mark.
May be seen embedded in their formula.
Isw once correct answers are seen.

Special case: Constants the wrong way round:
\(a = 1.013\) and \(b = 6.457\) with or without working scores M1A1M1A0 unless the equation is formed correctly in which case the final A mark can be recovered.

Note that having found the value of \(a\), it is possible to find \(b\) by substituting e.g. \(t = 1\) as follows:

Note that a misread of 0.0054 as 0.054 is quite common and may score 1110 as it does not simplify the question.

Notes

(b)(i) Follow through their \(a\).

B1ft: Correct interpretation for \(a\) but must reference “billions”.
Allow equivalent alternatives e.g.

• The original/initial population in billions
• The population in 2004 was “6.46” billion

(b)(ii) Follow through their \(b\).

B1ft: Correct interpretation for \(b\) but must reference “each year” or e.g. “yearly” oe
Allow equivalent alternatives e.g.

• The proportional increase/change in each year.
• The population will rise by “1.3%” each year. Must follow their value for \(b\).
• The rate/factor at which the population is rising/increasing/changing per annum.
• “1.013” is the multiplier representing the year on year increase.

Do not accept

• The amount it is rising
• How much it is rising
• The rate the population increases
• The percentage increase each year
• The rate of increase in billions annually

If they are not labelled (b)(i) and (b)(ii) mark in the order given but accept any way round as long as clearly labelled "\(a\) is..........." and "\(b\) is ..............."

Notes

M1: Substitutes \(t = 25\) or 26 or 27 into their model to find a value for \(P\)
Must be using their \(a\) and \(b\) correctly in \(P = ab^t\)
May be implied by sight of “9” or 9 billion if no incorrect working is seen.

A1: Correct value including units (allow awrt 9 billion) from a correct model but condone incorrect/premature rounding or truncating in an otherwise correct model that leads to the correct value of awrt 9 billion.
Allow e.g. awrt 9 000 000 000 or e.g. awrt \(9 \times 10^9\)
Just awrt 9 without the “billions” is A0

Notes

B1: The response must refer to the fact that the answer is unreliable together with a reference to the fact that the data used for the model is a long way from 2030
Examples:

• Not good as 2030 is a long way from 2004 – 2007
• Unreliable as based on old data
• Questionable as it has been extrapolated over a long time
• Not reliable due to how far out we have extrapolated
• By the time 2030 arrives it will be unreliable

But not e.g.

• Unreliable, extrapolation
• Not good as outside the range
• Not good as the population rises 101.3% each year
• Disease may happen
• Reliable as based on old data

Notes

Note: Candidates working in radians are able to score all the M and B marks in this question.
Condone the absence of the degrees symbol throughout the whole question.

B1: Correct value for \(K\). Condone \(R = 500\)

M1: Award for \(\tan\alpha = \pm\dfrac{480}{140} \Rightarrow \alpha = \ldots\), \(\tan\alpha = \pm\dfrac{140}{480} \Rightarrow \alpha = \ldots\), \(\sin\alpha = \pm\dfrac{480}{\text{``}500\text{''}} \Rightarrow \alpha = \ldots\) or \(\cos\alpha = \pm\dfrac{140}{\text{``}500\text{''}} \Rightarrow \alpha = \ldots\)
Note \(\alpha =\) awrt 1.3 (rad) implies this mark.

A1: \(\alpha =\) awrt \(73.74\{^\circ\}\) or correct expression \(500\cos(\theta + 73.74)\{^\circ\}\)

Notes

(b)(i) Note: mark parts (b)(i) and (b)(ii) together.

B1ft: Correct equation of the model in either form including the \(R\) = following through on their numerical \(K\) (\(0 \lt K \leqslant 750\)) and their numerical \(\alpha\).
Allow for e.g. \(R = 1500 - \text{``}500\text{''} + \text{``}500\text{''}\cos(30t + \text{``}73.74\text{''})\{^\circ\}\) or for \(R = 1500 - \text{``}500\text{''} + 140\cos(30t)^\circ - 480\sin(30t)^\circ\) but not e.g. \(R = 1500 - K + K\cos(30t + \alpha)^\circ\)
\(R = 1000 + 140\cos 30t - 480\sin 30t\) (without the brackets) is correct.
Allow this mark if they have truncated or rounded an otherwise correct \(\alpha\) (to 3s.f.)

(b)(ii)

B1ft: 500 or follow through on (their \(A\) – their \(K\)) or (1500 – 2 × their \(K\)) provided it is non-negative and less than 1500. It must be clear this is their answer to (b)(ii) so expect to see e.g. (b) or \(R_{\min} =\) or an indication it is the minimum.

Notes

Note: if \(\theta\) is used in place of \(30t\) then they must revert back to \(30t\) correctly to access the marks.

M1: Substitutes \(t = 3.5\) into their model for the number of rabbits (you may need to check if no method is shown)
or substitutes \(t = 3.5\) into their \(\cos(30t + \alpha)^\circ\)
Condone substitution of a value of \(t\) in the range \(3 \leqslant t \leqslant 4.5\) for this mark.

A1: \(R =\) awrt 500.1… or 500 (not awrt) following substitution of \(t = 3.5\), suggesting that the model is valid/reliable/appropriate/good.
or \(\cos(30(3.5) + 73.74)\{^\circ\} \approx -1\) suggesting that the model is valid/reliable/appropriate/good.
Allow this mark if they have truncated or rounded an otherwise correct \(\alpha\) (to 3s.f.)

Alt:

M1: Minimum occurs when \(A + K\cos(30t + \alpha)^\circ = R_{\min} \Rightarrow \cos(30t + \alpha)^\circ = \lambda\) with \(|\lambda| \leqslant 1\) leading to \(t = \ldots\)
May just see \(\cos(30t + 73.74)^\circ = -1 \Rightarrow t = \ldots\) (or their \(\cos(30t + \alpha)^\circ = -1 \Rightarrow t = \ldots\))
\(30t + \alpha = 180 \Rightarrow t = \ldots\) implies this mark. Condone \(30t + \alpha = \pi \Rightarrow t = \ldots\) for this mark.

A1: \(t = 3.54\ldots\) (i.e. the middle of April) so the model is valid/reliable/appropriate/good.
Do not condone incorrect statements, e.g., \(t = 3.54\ldots\) i.e. the middle of March so close to middle of April. If using \(A + K\cos(30t + \alpha)^\circ = R_{\min}\) then \(R_{\min}\) must be = their \(A -\) their \(K\)
Allow this mark if they have truncated or rounded an otherwise correct \(\alpha\) (to 3s.f.)

Alt 2 using differentiation (Condoned)

M1: Condone finding the minimum using \(\ldots\sin(30t + 73.74)^\circ = 0 \Rightarrow t = \ldots\) (or their \(\sin(30t + \alpha)^\circ = 0 \Rightarrow t = \ldots\))
\(30t + \alpha = 180 \Rightarrow t = \ldots\) implies this mark. Condone \(30t + \alpha = \pi \Rightarrow t = \ldots\) for this mark.

A1: \(t = 3.54\ldots\) (i.e. the middle of April) suggesting that the model is valid/reliable/appropriate.
Do not condone incorrect statements, e.g., \(t = 3.54\ldots\) i.e. the middle of March so close to middle of April.
The complete derivative for \(\dfrac{\mathrm{d}R}{\mathrm{d}t}\) does not need to be seen.
Allow this mark if they have truncated or rounded an otherwise correct \(\alpha\) (to 3s.f.)

Notes

Note: if \(\theta\) is used in place of \(30t\) then they must revert back to \(30t\) correctly to access the marks.

M1: Realises that \(\sin(30t + 70)^\circ = -1\), reaches \(30t + 70 = 270\) or \(-90\) and attempts to find \(t\) (or \(30t\))
Condone attempts using differentiation. The minimum occurs when \(\cos(30t + 70)^\circ = 0 \Rightarrow 30t + 70 = 270 \Rightarrow 30t = \ldots\) (or \(t = \ldots\)). They must use 270 or \(-90\) and not 90 to achieve the minimum. Condone \(30t + \alpha = \dfrac{3\pi}{2} \Rightarrow t = \ldots\) for this mark but not \(30t + \alpha = \dfrac{\pi}{2} \Rightarrow t = \ldots\).

A1: Correct value for \(30t\) (or \(t\)) Accept rounded or truncated values to at least 3s.f. e.g. 6.66 or 6.67

dM1: Substitutes their value of \(t \gt 0\) (or \(30t \gt 0\)) coming from \(30t + 70 = \mathbf{270}\) into their model for \(R\)

A1: Correct number of rabbits. Allow 1032 or 1033 but must be whole numbers and not just 1030.
Allow this mark if they have truncated or rounded an otherwise correct \(\alpha\) (to 3s.f.)

Notes

(a) If parametric differentiation is not used in part (a) (e.g. uses Cartesian form) then only the B mark is available but see alternative below.

B1: Uses the given Cartesian coordinates to deduce the correct value for \(t\).
If more than one value for \(t\) e.g. \(t = -5\) is given and \(t = -1\) is not “selected” score B0 but if just \(t = -1\) is used subsequently allow recovery and score B1

M1: Attempts to use \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t} \times \dfrac{\mathrm{d}t}{\mathrm{d}x}\) or equivalent with their differentiated equations.
There must be an attempt to differentiate both parameters, however poor, and divide or multiply correctly so using \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{y}{x}\) scores M0. Both parameters must be “changed”.
Condone confusion with the variables e.g. referring to \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) as \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) if the intention is clear.
This may be implied by e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = -3t^2,\ \dfrac{\mathrm{d}x}{\mathrm{d}t} = 2(t+3), t = -1,\ \dfrac{\mathrm{d}y}{\mathrm{d}t} = -3,\ \dfrac{\mathrm{d}x}{\mathrm{d}t} = 4 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{3}{4}\)

M1: Uses their numerical value of \(t\) (not 4) in their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) to obtain a value.
Condone attempts with different values of \(t\) e.g. \(t = -1\) and \(t = -5\)

ddM1: Applies a correct straight line method with their value of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) which has come from an attempt to use parametric differentiation with their value of \(t\) (not 4) and with \(x = 4\) and \(y = 2\) correctly placed. An attempt at the equation of the normal is M0.
If using \(y = mx + c\) they must reach as far as \(c = \ldots\)
Depends on both previous M marks.

A1*: Correct equation as printed with no errors but condone \(4y + 3x = 20\ *\)
Allow equivalents e.g. \(20 = 4y + 3x\ *\) or \(3x + 4y = 20\ *\)
This is a printed answer so there must be at least one intermediate step as shown in the main scheme.

Alternative for (a) using parametric differentiation but avoids the need for a value for \(t\):

B1: Either a correct expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\) and/or \(y\) following a correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\) or for \(t = -1\) seen anywhere.

M1: Attempts to use \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t} \times \dfrac{\mathrm{d}t}{\mathrm{d}x}\) with their differentiated equations.
There must be an attempt to differentiate both parameters, however poor, and divide or multiply correctly so using \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{y}{x}\) scores M0
Condone confusion with the variables e.g. referring to \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) as \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) if the intention is clear.
This may be implied by e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = -3t^2,\ \dfrac{\mathrm{d}x}{\mathrm{d}t} = 2(t+3), t = -1,\ \dfrac{\mathrm{d}y}{\mathrm{d}t} = -3,\ \dfrac{\mathrm{d}x}{\mathrm{d}t} = 4 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{3}{4}\)

M1: Attempts to express their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) which is in terms of \(t\), in terms of \(x\) and/or \(y\) and uses \(x = 4\) and \(y = 2\) correctly placed in an attempt to find the gradient of the tangent.

ddM1: Applies a correct straight line method with their value of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) which has come from an attempt to use parametric differentiation with their gradient and with \(x = 4\) and \(y = 2\) correctly placed.
If using \(y = mx + c\) they must reach as far as \(c = \ldots\)
Depends on both previous M marks.

A1*: Correct equation as printed with no errors.
This is a printed answer so there must be at least one intermediate step as shown in the main scheme.

Notes

B1: 9m or equivalent including correct units. Accept e.g. 9 metres, 900cm etc.

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: Attempts integration to achieve \(\{H=\}\,k\,\mathrm{e}^{-0.2t}\,\{+c\}\) with \(k\) a numerical constant \(\neq -0.12\)
Note that we will condone \(k = (-0.2)(-0.12)\ \{= 0.024\}\)
If they divide by \(-0.12\) first before integrating they need \(aH = b\mathrm{e}^{-0.2t}\,\{+c\}\) with \(a\) and \(b\) numerical and \(b \neq 1\). Condone a spurious integral symbol remaining after integration.

dM1: Uses \(t = 0, H = 1.5\) and a model of the form \(H = k\,\mathrm{e}^{-0.2t} + c\) (or \(aH = b\mathrm{e}^{-0.2t} + c\)) to find the value of the constant \(c\). They cannot just “make up” a value for \(k\).
Do not be concerned with their processing to find \(c\) but they cannot just state \(B\) (or \(c\)) is 0.
For reference if they divide by \(-0.12\) first they should reach \(-\dfrac{25}{3}H = -5\mathrm{e}^{-0.2t} - 7.5\) o.e.

A1: Correct complete equation in the required form: \(H = 0.6\mathrm{e}^{-0.2t} + 0.9\) with the \(H\) = present.
May be awarded if seen at the start of (b) but not in (c). Condone \(-\dfrac{1}{5}\) in place of \(-0.2\)
Finding correct values for \(A\) and \(B\) is insufficient for this mark.
Allow exact equivalents but they must be in the required form, e.g. \(H = \dfrac{6}{10}\mathrm{e}^{-0.2t} + \dfrac{9}{10}\)
A minimally acceptable answer is \(\{H=\}\,0.6\mathrm{e}^{-0.2t} + c \rightarrow H = 0.6\mathrm{e}^{-0.2t} + 0.9\) score M1dM1A1.
Note: sight of differentiating the given form to e.g. \(\dfrac{\mathrm{d}H}{\mathrm{d}t} = -0.2A\mathrm{e}^{-0.2t}\) in their working without clear evidence of integration of the original differential equation should be marked using the special case below.

SC: For candidates starting with the given answer \(H = A\mathrm{e}^{-0.2t} + B\) it is possible to use \(\dfrac{\mathrm{d}H}{\mathrm{d}t} = -0.2A\mathrm{e}^{-0.2t} = -0.12\mathrm{e}^{-0.2t}\) to deduce that \(A = 0.6\). This can be awarded SC M1dM0A0
If they go on to find \(B\) as in the main scheme then this can be awarded SC M1dM1A0
Answer with no working scores 110.

Note: If the special case is applied they may go on to achieve the rest of the marks in (b) and (c).

Notes

Note: \(A\) and \(B\) must be numbers but may be “made up” if they did not have an answer to (a).

M1: Uses \(H = 1.2\) in a model of the form \(H = A\mathrm{e}^{-0.2t} + B,\ B \neq 0\) and rearranges to make \(A\mathrm{e}^{\pm 0.2t}\) or \(\mathrm{e}^{\pm 0.2t}\) the subject. Condone slips in rearranging, e.g. dividing the LHS by 0.9 instead of subtracting 0.9. Rearranging first before substituting is acceptable but they must get to \(A\mathrm{e}^{\pm 0.2t}\) or \(\mathrm{e}^{\pm 0.2t}\) as the subject.

dM1: Correct use of ln to make \(t\) the subject.  Requires \(A \gt 0,\ 0 \lt B \lt 1.2\) and \(\mathrm{e}^{\pm 0.2t} = \lambda \gt 0\)
If they had a negative value for \(A\) in part (a) they cannot just make it positive at this stage.
Any of \(5\ln 2\) or \(-5\ln\dfrac{1}{2}\) or awrt 3.46 or awrt 3.47 following a correct equation will imply M1dM1.
If they do not show their method for an incorrect \(H = A\mathrm{e}^{-0.2t} + B\) with \(A \gt 0,\ 0 \lt B \lt 1.2\) you may need to check their value for \(t \gt 0\) as it may imply M1dM1.

A1: Correct time in hours and minutes \(\{t=\}\) 3 hours 28 minutes, but condone e.g. 3h 28m
Must come from correct values of \(A\) and \(B\) in (a).

Note: If their \(B = 0\) then they should end up with \(t = -3.46\ldots\) however, they did not score the first M1. They cannot “recover” this by making it positive and finding \(t =\) 3 hours 28 minutes.

Notes

Note that 0.9 or 0.9m must come from a correct value of \(B\) in (a) to score any marks.

M1: Identifies the requirement to establish the limit as \(t\) tends to infinity.
It can be implied by stating that \(H = A\mathrm{e}^{-0.2t} + B \rightarrow B\) or \(\left(\lim\limits_{t \to \infty}\left[0.6\mathrm{e}^{-0.2t} + 0.9\right]\right) = 0.9\)
Stating “\(B\)” on its own will score this mark.
Substituting a large value is M0 unless it leads to their value for \(B\) at which point the A1 is available as well.

A1ft: Correct height including units. Follow through on their value of \(B\) where \(0 \lt B \lt 1.2\)
Correct ft height including units implies M1A1, while e.g. 0.9 (no units) would imply M1A0.
Evidence of an incorrect method such as \(1.5 - 0.6\mathrm{e}^{-0.2(0)} = 0.9\) m scores M0A0.

Misreading as \(\dfrac{\mathrm{d}H}{\mathrm{d}t} = -0.12\mathrm{e}^{0.2t}\) can score a maximum (a) 110 (b) 100 (c) 10.

Notes

B1*: Correct working to obtain 290. Must be a correct calculation so do not condone missing brackets unless they are recovered. E.g. \(\left(u_{12} =\right) 400 + 12 - 1 \times -10 = 290\) scores B0 unless followed by \(= 400 + 11 \times -10 = 290\). Condone \(\left(u_{12} =\right) 400 + (12-1) - 10 = 290\)
The “£” symbol is not required but the “290” must appear.

Alternative 1:

B1*: Correct working using the 290 to obtain \(n = 12\).
There must be at least one intermediate line after setting up the equation and must be correct work so do not condone missing brackets unless they are recovered (as above).
A conclusion is not required with this approach as long as 12 is correctly obtained.

Alternative 2:

B1*: Correct working using the 290 and 400 to obtain \(d = -10\).
There must be at least one intermediate line after setting up the equation and must be correct work so do not condone missing brackets unless they are recovered (as above).
A conclusion is not required with this approach as long as \(-10\) is correctly obtained.

Allow candidates to list terms and show the 12^th term is 290 e.g.
400, 390, 380, 370, 360, 350, 340, 330, 320, 310, 300, 290
Must list all 12 terms which must be correct and end with 290
Condone if missing 400, 390, 380 as these are given in the question.

Notes

Mark (b) and (c) together

M1: Uses a correct sum formula in terms of \(N\) or \(n\) with \(a = 400\) and \(d = -10\) or \(+10\) and sets = 8100. Condone e.g. > 8100 and allow A1 if this is recovered to become “=” before the final line.
Condone \(8100 = \dfrac{1}{2}N\left(2 \times 400 + (N-1) - 10\right)\) if recovered or not.

A1*: Fully correct proof with sufficient working shown and no unrecovered errors.
Do not condone e.g. missing brackets or e.g. a missing \(N/n\) unless recovered before the final given answer.
Condone the use of \(n\) instead of \(N\) for both marks.
Condone terms in a different order as long as they are correct.
Condone \(0 = N^2 - 81N + 1620\ *\)
Sufficient working requires all brackets to be removed to obtain an unsimplified expanded quadratic before proceeding to the given answer including the “=0”.

Alternative (further maths method): Series summation approach:

M1: Attempt to sum an appropriate series with first term 400. Condone use of \(+10\) as in the main scheme.

A1*: As main scheme.

Notes

M1: Solves the given quadratic equation by any correct method including a calculator to obtain at least one value for \(N\). See general guidance for solving a 3-term quadratic.
If values are just written down and only one value is given it must be 45 or 36.
If both values are just written down they must both be correct.

A1: Realises that the smaller value is required and so selects \((N =)\ 36\).
Ignore any units if given.
The “\(N =\)” is not required, just look for the correct value.
It must be clear that this value has been selected. This may be indicated by e.g. underlining the 36 or the omission of the 45. If the 45 is not rejected score A0.
\(N = 36\) with no working scores M1A1

Typical solution

Typical solution

Typical solution

(c)(i)

(c)(ii)

The height is decreasing at a constant rate of 0.018 cm per minute.

Typical solution

Hence \(\sqrt{29}\sin(x + 68.2)\)

Typical solution

(b)(i)

(b)(ii)

(b)(iii)

Typical solution

Typical solution

(b)(i)

(b)(ii)

(b)(iii)

Typical solution

Time = 4 hrs 30 mins

Typical solution

when \(h = 130\), \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = -1.5\)

Typical solution

Typical solution

61 minutes

Typical solution

Typical solution

Typical solution

Value of top prize = £240

Typical solution

(i)

(ii)

(iii)

Typical solution

When \(x = 0.25\)

The model predicts a negative median mass for a monkey that is one week old, therefore it is unsuitable for use with monkeys 1 week old or less.

Typical solution

Room temperature = 20 °C

Typical solution

For large values of \(t\), \(\mathrm{e}^{-kt} \to 0\)

Hence \(T = 220\)

Typical solution

(i)

(ii)

Typical solution

Typical solution

(i)

Total in account = £ 6 787

(ii)

The interest rate is unlikely to remain fixed for the whole 10 years

Typical solution

Typical solution

Typical solution

Typical solution

When \(t = 0.2\), \(y = 10.2\)

\(t = 3\), \(y = \dfrac{11}{3}\)

The slide is safe.

Typical solution

(i)

(ii)

Length of \(RS\) = 2.83 metres

(iii)

Typical solution

(i)

(ii)

\(b\) = 1.14

Typical solution

14%

Typical solution

(i)

£467 900 000

(ii)

Sales may suddenly fall due to unforeseen circumstances such as a pandemic.

Typical solution

Typical solution

(i)

Ziad will need to train for 29 days

(ii)

Swims 47850 metres

47850 < 50 000

Therefore the coach is not correct

Typical solution

Typical solution

Maximum height is approximately 10.5 metres

Typical solution

The entrance to the cave is unlikely to be perfectly smooth

Typical solution

(i)

(ii)

The value predicted by the exponential model will grow without limit.

This can’t be true as there are only 900 tomato plants in the greenhouse.

Typical solution

(i)

(ii)

(iii)

It takes 11 days from when the damage is first noticed until half of the plants are damaged by insects

Typical solution

(i)

(ii)

Typical solution

Average distance is approximately 4500 million kilometres

Notes

B1: State two correct equations soi

M1: Attempt to solve their two linear simultaneous equations. As far as eliminating one variable. M1 implied by correct final answer, if no method shown

A1: Obtain correct value for \(a\)

A1: Obtain correct value for \(b\)

Alternative method (graphical approach)

M1: Attempt gradient of line. M1 implied by correct \(a\)

A1: Obtain \(a = 7\)

M1: Attempt to find value for \(b\). Intercept of line or algebraic method. M1 implied by correct \(b\)

A1: Obtain \(b = 73\)

Notes

(b)(i)

M1: Substitute \(t = 4\) into their model

A1FT: Condone no units, but A0 if incorrect eg 101m. FT their model. Or 1.01 (m)

(b)(ii)

B1: Any sensible reason as to why the model for ‘average height’ may, or may not, be accurate in predicting Sam’s height
eg height varies a lot between children of the same age
eg could depend on different diets and lifestyle
eg might be accurate as long as Sam is ‘average’

B0 for interpolation, including comments about Sam’s height being between 87cm and 108cm

Exemplar responses for Q5(b)(ii)

Notes

B1: Any sensible comment
eg growth spurt at 12 due to puberty
eg people grow quicker when younger

Extrapolation not reliable
B0 for comments about individual variations as question refers to ‘average height’ model

Exemplar responses for Q5(c)

Notes

(a)(i) B1: State 0.8 m, units required
Units may be given as m or metres
Could be \(\tfrac{4}{5}\) m
Could be 80 cm

(a)(ii)

M1: Identify that minimum occurs when \(\cos(30t - 60)\) is \(-1\), so need \(30t - 60 = 180\)
M1 does not require attempt at solution for \(t\)
No FT on an incorrect \(d\) being used from the previous part eg \(d = 2.45\) from using \(t = 0\)

A1: No units needed, as value of \(t\) is requested
Condone 0800 or 8am
Ignore additional values of \(t\) that are greater than 8, but A0 for a smaller positive value of \(t\) also given

Notes

M1: Equate model to 1, rearrange and use \(\cos^{-1}\)
As far as \(30t - 60 = k\), using correct order of operations
Allow M1 if working in radians (gives \(30t - 60 = 2.529\))

A1: Obtain correct first value of \(t\)
Implied by first time of 0650 with no errors seen
3sf or better
Ignore inequality signs if used

M1: Attempt all further values of \(t\) within \(0 \lt t \lt 24\)
Using a valid method
M0 if using radians
Values of \(t\) could also be found using the symmetry of the curve eg \(8 + (8 - 6.83) = 9.17\)

A1: Allow answers to 3sf
Ignore inequality signs if used
Correct time periods would imply \(t\) values

A1: Correct two periods, given as time intervals
Could also be given as 6:50am to 9:10am, and 6:50pm to 9:10pm
Must be given as intervals and not just times eg A0 for ‘0650 and 0910’ etc
BOD if correct intervals given following any incorrect inequality signs
Accept 0649 to 0911 and 1849 to 2111 www (from checking times and realising that rounded answers give depths of less than 1 metre)
A0 if giving answers in minutes, or hours and minutes, after midnight and not times (eg ‘410 minutes to 550 minutes’ or ‘6 hours 50 minutes to 9 hours 10 minutes’)
Condone attempt at interval notation / inequalities as long as intention is clear, and allow BOD if written as a strict inequality such as \(0650 \lt t \lt 0910\)

Special Case
If M1A1M0 awarded, then allow SC B1 for giving a correct time period eg 0650 to 0910

Notes

B1: Any sensible suggestion that suggests that amplitudes of the tides will be reduced due to the exponential term
Must refer to the effect of the exponential term in context in some way
Condone reference to the exponential term having a ‘damping’ effect on the tides
Cannot just restate the question eg ‘river gets shallower’

Notes

B1: Must see use of decrease/decline linked to the negative sign.
Condone answers that do not refer to \(\frac{\mathrm{d}P}{\mathrm{d}t}\)

Examples:

• “it’s \(-\frac{1}{80}P^2\) because decreasing” B1
• “decreasing” [and nothing further] B0
• “decline implies negative sign” B1

Notes

M1*: Attempt to separate variables, must see \(P\) and d\(P\) on the same side
May see \(\frac{\mathrm{d}t}{\mathrm{d}P} = \frac{-80}{P^2}\) as a first step. In this case M1 is implied by a subsequent attempt to integrate the RHS w.r.t. \(P\) (so do not give M1 for this statement alone).

A1: May see \(\frac{1}{P} = \frac{t}{80}\,(+c)\) etc. Allow without \(+c\)

M1 dep*: Substitute (0, 120) and attempt to find \(c\) (which may not be this value). Must reach a value of \(c\) for this mark.

A1: Must be in terms of \(P\) (\(P\)=…) but need not be simplified.
isw incorrect attempts to simplify following \(P\)=…

Notes

M1: Either substitute \(P = 10\) into their equation from (b) or rearrange to make \(t\) the subject.

A1: Accept 7.33 (3sf), ignore units.

Notes

B1: Allow \(t = 160 \Rightarrow P = 0.5\)
(Use of the value \(t\)=160 not required)
Any correct statement in context relating to the model predicting non-integer values between 0 and 1. (Note that the statement ‘will never reach zero’ is given in the question so gains no credit).
Acceptable examples (for a comment, in combination with the statement that \(P\) is small for large \(t\)):

• “cannot have part of an animal”
• “the actual population has reached 0”
• “not possible to have a population of 0.49(8)”
• “the population will go extinct when less than 1”

Notes

B1: Must mention ‘rate of change’ of bees, or equiv, and how this relates to wildflower plants
B0 for ‘change’ not ‘rate of change’
B1 for rate of change of bees with respect to plants
B1 for rate of change of bees as the plants change
B1 BOD for rate of change of bees compared to the number of plants
State or imply that it is bees compared to flowers, so B0 if other way around
Must relate to bees and plants, and not just \(B\) and \(F\)
See appendix for further examples (below)

Appendix: exemplar responses for Q9(a)

Notes

B1: Correct derivative
No need to see \(\dfrac{\mathrm{d}F}{\mathrm{d}t}\) notation
Could be unsimplified

B1: Correct derivative
No need to see \(\dfrac{\mathrm{d}B}{\mathrm{d}t}\) notation
NB watch out for \(2 - 3\sin t\)

M1: Correct method to combine their two derivatives – algebraic or numerical
B0B0M1 is possible
Must have \(\mathrm{e}^{0.1t}\) in denominator, or \(\mathrm{e}^{-0.1t}\) in the numerator, but allow muddles with the placing of the 5

A1: Substitute \(t = 4\) to obtain 0.484, or better

Alternative method

B1: Correct expression for \(B\) as a function of \(F\)

M1: Attempt differentiation
May see ln terms split first (possibly even including use of \(\cos(A - B)\))

A1: Obtain correct derivative aef

A1: Substitute \(F = 50\mathrm{e}^{0.4}\) to obtain 0.484, or better
Could use \(t = 4\) if derivative now in terms of \(t\)

Notes

B1: Summer will be over so pattern may not continue
Summer is not greater than 12 weeks
Fewer bees and/or flowers in autumn/winter
Any reason referring to a change in season having an effect on bees and/or flowers
B0 for just considering long-term behaviour eg flowers will not continue to increase exponentially
Reasons must reference seasons / different time of year (could be implied by ‘weather getting colder’)

Notes

B1: Obtain \(t = 20\)
Allow [19.9, 20.1] from setting up and using exponential model

B1: Obtain \(m = 97.2\)
Allow [97.1, 97.3] from setting up and using exponential model

Notes

(b)(i)

B1: Equate given model to 80
soi, so could be \(\mathrm{e}^{-0.055t} = 0.5\)

M1: Attempt correct process to find value of \(t\), as far as dealing with exponential term
Rearrange to \(\mathrm{e}^{-0.055t} = k\), and hence obtain \(-0.055t = \ln k\)
If introducing logs straight away then need to get as far as \(\ln 160 - 0.055t = \ln(\text{their } 80)\)

A1: Obtain \(t = 12.6\), or better
If more sig fig given, then allow answers which round to 12.60 (the more accurate answer is 12.602676..)

(b)(ii)

B1: Correct derivative soi
Allow unsimplified
No need to see \(\frac{\mathrm{d}m}{\mathrm{d}t} =\)

M1: Substitute \(t = 15\) into their derivative
Must be of the form \(k\mathrm{e}^{-0.055t}\), with \(k \ne 160\)
Possibly still with \(k\) unsimplified
Substitution sufficient, no need to evaluate for M1

A1: Units required, and positive answer
Must follow correct derivative ie negative coefficient
No need to see \(-3.86\) first, but A0 if clear error
Accept 3.9 grams/minute
Accept g/m for grams/minute

Notes

B1: State \(A\), with clear comparison
Insufficient to just say that \(A\) has a greater initial mass – needs to consider decay factor as well

Allow solutions that identify that \(B\) is decaying faster, with supporting evidence
eg after 10 minutes, \(B\)’s mass is 92.3g which is 58% of initial mass whereas \(A\) is 60% of initial mass so \(B\) decaying faster
eg \(A\)’s half-life is 13.6 so \(B\) is decaying faster
eg change initial mass in model \(B\) to 1250 then when \(t = 10\) \(B\)’s mass would be 721g which is less than 750 hence decaying faster
eg compare coefficients of \(t\) (for \(A\), coeff is \(-0.0511\)); \(B\)’s is of a greater magnitude hence decaying faster

For either solution, the conclusion and the supporting evidence must be consistent
Numerical supporting evidence must be correct, allowing for slight inaccuracies from using different numbers of sig fig (see appendix below)

Appendix: supporting evidence for Q8(c)

When comparing % remaining or percentage lost in \(t\) minutes, substance \(B\) is shown to be decreasing at a faster rate. (Choose \(t = 15\).)

Substance A

Substance B

When comparing RATE of decrease at \(t\) minutes, substance \(A\) is shown to be decreasing at a faster rate.

Substance A

Substance B

Notes

M1: Sum of a GP implied. Allow any \(r\), eg \(r = 0.05\)

A1: Correct equation

A1: Allow 0.487 or 0.488

M1: or \(0.95^{14} = 0.4875\) or 0.487 or 0.488 seen. Can be implied by their answer
ft their equation of form \(a^n = b\) (dep M1 gained and \(b \gt 0\))

A1: cao. Allow \(n = 14\). Allow 14.0. Allow \(\approx 14\)

Alternative method

M1: Attempt add \(\geqslant 10\) terms. Allow any value of \(r\) for this mark

A1: Correct 14 terms added

NB Unsupported correct answer: SC B3

Alternative (incorrect) methods using \(r = 1.05\), or \(\frac{1}{0.95}\) or \(\frac{20}{19}\)

(For info’ only: \(r = \dfrac{1}{0.95}\) or \(\dfrac{20}{19}\) comes from misinterpreting “lowest” to mean “shortest”)

M1: Allow any value of \(r\) for this mark

A1: oe using 1.05. Correct equation

M1: oe, eg \(\dfrac{\ln 1.539}{\ln 1.053}\) or \(\dfrac{\ln 1.51}{\ln 1.05}\)   ft their “\(\frac{117}{76}\)”
Can be implied by their answer

Notes

(a)(i) B1: Allow \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = k\) or \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = -3.5\)
Both sides of differential equation required

(a)(ii)

M1: Obtain equation of the form \(\theta = \pm 3.5t + c\), where \(c\) could already be numerical and possibly incorrect
Not dependent on correct differential equation in (i)

A1: Obtain correct equation

Alt method
For M1, integrate to get \(\theta = kt + c\), then use \((0, 160)\) and \((10, 125)\) to attempt \(c\) and hence \(k\)

(a)(iii) B1: Any sensible comment
Cooling rate unlikely to be linear
Identify that limit (ie room temperature) will be reached

Notes

(b)(i) B1: Allow \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = k(\theta - 20)\)
Both sides of differential equation required
ISW if \(k = -3.5\) used once correct equation seen (but B0 if only ever seen with \(-3.5\))

(b)(ii)

M1: Separate variables (or invert each side) and attempt integration
Allow M1 for integration of a differential equation not of this form eg \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = \dfrac{-k}{(\theta - 20)}\), as long as \(t\) and/or \(\theta\) are involved – must be attempt at correct rearrangement of their diff eqn

A1: Obtain correct integral
Or \(\ln|\theta - 20| = kt + c\)
Condone brackets not modulus

M1: Use \(t = 0\), \(\theta = 160\) in an equation involving both \(k\) and \(c\)
Equation must be from integration attempt, but could follow M0
As far as numerical \(c\) or \(k\)
Using both pairs of values as limits in a definite integral is M2

M1: Use \(t = 10\), \(\theta = 125\) in an equation involving both \(k\) and \(c\) (\(c\) possibly now numerical)
As far as numerical \(c\) and \(k\)

M1: Attempt to make \(\theta\) the subject
As far as correctly removing logs
Equation must now be of the correct form ie \(\ln|a\theta + b| = ct + d\)
Could still be in terms of \(c\) and \(k\) to give eg \(\theta = A\mathrm{e}^{kt} + 20\)

A1: Obtain correct equation
Allow \(-0.0288\) (or better) for \(0.1\ln 0.75\) and/or 4.94 (or better) for \(\ln 140\)
Allow \(\theta = 20 + \mathrm{e}^{(0.1\ln 0.75)t + \ln 140}\)
Could see \(\theta = 140(0.75)^{0.1t} + 20\)

Notes

M1: Use \(\theta = 25\) in both of their equations to find values for \(t\)
As far as two numerical values for \(t\)

A1: Obtain 77 minutes
Accept any answer rounding to 77, with no errors seen

Notes

B1: B0 for \(\sqrt{169}\) or for \(\pm 13\)

M1: M1 for \(\tan\alpha = k\) where \(k = \pm\frac{5}{12}, \pm\frac{12}{5}\)
or for \(\cos\alpha = \pm\frac{12}{R}\), \(\sin\alpha = \pm\frac{5}{R}\)

A1: A0 if in degrees (must be stated to exactly 4 significant figures at some point)
For reference: 0.3947911197…

Notes

M1: Sets \(R\cos(t - \alpha)\) (with their \(R\) and \(\alpha\)) equal to either 3 or \(-3\), and attempt to solve (with correct order of operations)
M1 only if in degrees

A1: awrt 1.73
1.7327192…

A1: awrt 2.20 (condone 2.2) – ignore other values that are greater than 2.20
2.1984556…

Notes

ft their (a) \(-\) 4000

Notes

Ignore all else
NOT “Increases for 1^st 6 months then decreases”

Notes

M1: May be implied by answer

A1: Allow \(t = 12\), or \(t = 12\) months, or just 12

Notes

M1: Substitute \(P = 4500\) May be implied by next line

A1: Correct rearrangement

M1: Attempt \(30t = \cos^{-1}(\text{their } 0.5)\), giving \(\alpha\) and \(360 - \alpha\).
Condone \(30\mathrm{t} = \frac{\pi}{3}, \frac{5\pi}{3}\)

A1: or after 10 months. Allow \(t = 10\) months, or just 10
SC. (If not gained 1^st M1A1) Correct answer with no or inadequate working and/or T&I: \(t = 10\) stated: B2; \(t = 10\) embedded: B1B0

Alternative methods for 2nd M1A1

M1: \(30t = 60\ (t = 2)\)
A1: (end of 1^st cycle at \(t = 12\)) 2^nd time \(P = 4500\) is when \(t = 12 - 2 = 10\)

Notes

or other good answers
eg \(P = 5000 - (1000\cos(30t)^\circ)^{1/t}\)
\(P = 5000 - \dfrac{1000}{t}\cos(30t)^\circ.\ (t \gt 0)\)

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

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

B1: separation of variables; condone omission of integral signs or of \(\mathrm{d}t\); allow \(P\) instead of \(\int \mathrm{d}P\) on left hand side

M1*: integration by parts; allow sign errors only

A1: all correct;

A1: allow omission of \(+\,c\); allow unsimplified

M1dep*: may be awarded after rearrangement; must see constant of integration;
NB if expression incorrect, need to see full substitution for M1

A1: allow just \(c = 36\) as long as “\(P\) =” seen following completion of integration; allow eg \(P = 36 + 45\left(\frac{1}{2}e^{-0.8t} - 2te^{-0.8t}\right)\); like terms must be collected and fractions must be in lowest terms

Alternative method

B1: separation of variables; condone omission of integral signs or of \(\mathrm{d}t\); allow \(P\) instead of \(\int \mathrm{d}P\) on left hand side

B1: may be embedded in integration by parts

M1* A1: integration by parts; allow sign errors only
all correct

A1: all correct; allow omission of \(+\,c\); allow unsimplified

M1dep*: may be awarded after rearrangement; must see constant of integration

A1: allow just \(c = 36\) as long as “\(P\) =” seen following completion of integration; allow eg \(P = 36 + 45\left(\frac{1}{2}e^{-0.8t} - 2te^{-0.8t}\right)\); like terms must be collected and fractions must be in lowest terms

Notes

M1: or \(\displaystyle\lim_{t\to\infty} (k)\mathrm{e}^{-0.8t} = 0\); \(k\) is a constant;
allow eg as \(t\) tends to infinity / as \(t\) becomes very large / in the long term, \((k)\mathrm{e}^{-0.8t}\) tends to / converges to / approaches zero oe
do not allow
eg substitution of \(t = \infty\)

A1: condone \(P = 36\)

Notes

B1: Uses laws of logs to split into two terms and then to simplify the second term. This line must be seen
AG

Notes

B1: Allow \(n = 1.5\) or better seen

M1: soi

A1: Allow awrt \(6.1\times 10^{-9}\)

For using 2 data points from the table SC1 awrt \(n = 1.5\), SC1 \(5.159\times 10^{-9} \lt k \lt 7.076\times 10^{-9}\)

Notes

M1: Uses the model to predict

A1FT: Appropriate comment based on similarity of their correct value and the given value.
FT their \(k\)
Do not accept a comment that implies the model should give an exact match

Alternative method

M1: Uses the model to predict \(\log T\)

A1: Appropriate comment based on similarity of their correct values.

SC1 “Student equation would not be suitable as the data would have to be extrapolated” oe

Notes

B1: Allow embedded in equation e.g. \(P_{t+1} = \frac{3}{2}P_t\)

Notes

B1: cao

Notes

(i) M1: Both values – might not round answers
Allow \(106.8 \approx 108\)
Condone incorrect values if \(48\mathrm{e}^{0.4}\) and \(48\mathrm{e}^{0.8}\) seen

(i) A1: Must have correct values
e.g. 71.6 is close to 72 and 106.8 is close to 108
Ignore attempts to compare the two models
Condone e.g. ‘close for \(t = 1\)’ and ‘not close for \(t = 2\)’

(ii) M1: Differentiation – might not simplify e.g. \(\frac{\mathrm{d}Q}{\mathrm{d}t} = 19.2\mathrm{e}^{0.4t}\) or \(\frac{\mathrm{d}Q}{\mathrm{d}t} = \frac{96}{5}\mathrm{e}^{0.4t}\)
Implied by \(\frac{\mathrm{d}Q}{\mathrm{d}t} = ke^{0.4t}\)
Condone poor/no notation

(ii) A1: Showing the rate of change is proportional to population e.g. \(\frac{\mathrm{d}Q}{\mathrm{d}t} = 0.4 \times 48\mathrm{e}^{0.4t}\) and \(Q = 48\mathrm{e}^{0.4t}\) soi or \(\dfrac{\frac{\mathrm{d}Q}{\mathrm{d}t}}{48\mathrm{e}^{0.4t}} = 0.4\)
0.4 must be exact and not rounded e.g. from numerical trials
Must see \(\frac{\mathrm{d}Q}{\mathrm{d}t}\) or e.g. \(Q^\prime\)

(iii) M1: Use of exponential equation one month later – may award for correct use of equation for Q to find populations for at least 2 consecutive months (for \(t = 2\) onwards)
For info: \(t = 2, Q = 106.8\); \(t = 3, Q = 159.4\); \(t = 4, Q = 237.7\) to 3 s.f. or better
This M1 may be implied by finding multiplier of \(\mathrm{e}^{0.4}\)

(iii) B1: Finding multiplier for Q
If ‘multiplier’ omitted, then multiplication must be implied by later work

(iii) A1: Completion with convincing general explanation including consideration of an early value (could be \(P > Q\) at \(t = 1\) or both sequences start at 48 or at \(t = 2\) the values are 107 for \(Q\) and 108 for \(P\))

Alternative 1 for A1 (iii)

A1: Convincing completion, must see ‘48’ or both equal populations at \(t = 0\) (or \(t = 1\))
oe e.g. showing \(\dfrac{1.5^t}{\mathrm{e}^{0.4t}} > 1\) and starting values of 48 soi or writing model for P as \(48\mathrm{e}^{(\ln 1.5)t} = 48\,\mathrm{e}^{0.405\ldots t}\)

Alternative 2 for A1 (iii)

A1: Differentiates may use ln 1.5 or 0.405…
Convincing completion

Alternative method using inequalities (iii)

B1: B1 for finding correct model for P

M1: Forming inequality for P and Q

A1: Convincing completion

Notes

B1: Allow embedded in \(A = 8 + 0.8t\)

Notes

B1: Allow for “after 9 days” oe FT their values in (a)

Notes

B1: Must indicate the mis-match between the linear model and the exponential observed results either in general terms or for a particular day
(Amounts are 8, 8.8 and 9.6 for the model and 8, 8.8 9.68 for the 10% increase)
(see appendix)

Appendix: exemplar responses for Q10(c)

Notes

B1: cao

M1: Forming an equation for \(k\) using \(t = 1\) and \(A = 8.8\) oe
FT their value for \(P\)

A1: Allow for \(\ln 1.1\) or a decimal answer to at least 2sf

Notes

M1: Correct use of logs in an attempt to solve indicial equation
Similarly for \(15 = 8 \times 1.1^t\)

A1: FT their \(P\) and \(k\)

Notes

B1: Must describe what the model predicts and compare with the situation being modelled (see appendix)

Appendix: exemplar responses for Q10(f)

Notes

(i) M1: Forms two equations and attempt to solve simultaneously (BC)
Allow if \(10 = a + 2.303b\) and \(200 = a + 4.443b\) used

E1: AG must be 3 s.f.
If by simultaneous equations solved BC, 88.78…..or better must also be seen

(ii) B1: Accept awrt \(-194\) or \(-195\)

Notes

(i) E1: argument based on negativity

(ii) E1: argument based on contrast between ever increasing height predicted by the model and reality

Notes

M1: Also allow for 84.5 days used for 85

E1: Established using a value of \(h\) between 137.9 and 139 needed.

Alternative method

M1: Equate to 100 and solve for \(t\). Condone \(h = 1\) used

E1: Established using a value of \(t\) between 27 and 28 needed.

Notes

M1: Attempt to differentiate to give expression of the form \(\frac{k}{t}\)

M1: equates their derivative to 3

A1: allow 29 or 30 days