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: 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

Note: Condone absence of reference to the value of \(\dfrac{x}{y}\) being undefined when \(x = y = 0\)

M1: For the key step in attempting to take logarithms with the same base of both sides and apply the power law to both sides, e.g., \(x = y\log_3 7\) or \(y = x\log_7 3\) or \(x\ln 3 = y\ln 7\)
Condone e.g. \(x = \log_3 7y\) but not \(x = \log_3(7y)\)
Alternatively, takes a root of both sides (either \(x\) or \(y\)), takes logs with the same base of both sides and applies the power law e.g., \(3^x = 7^y \rightarrow 3^{\frac{x}{y}} = 7 \rightarrow \dfrac{x}{y}\log 3 = \log 7\)
May be implied by a correct answer.

A1: \(\left(\dfrac{x}{y} =\right) \dfrac{\log 7}{\log 3}\) or equivalent, e.g, \(\dfrac{\ln 7}{\ln 3}\) or \(\log_3 7\) or \(\dfrac{1}{\log_7 3}\) Condone e.g. \(\dfrac{\ln|7|}{\ln|3|}\)
Correct answer only scores both marks provided there is no incorrect log work.
Any base \(k\) may be used for e.g. \(\dfrac{\log_k 7}{\log_k 3}\) provided it is the same base in both logs, although watch out for e.g. \(\left(\dfrac{\log_3 7}{\log_7 3}\right)^{\frac{1}{2}}\) or \(\dfrac{\log_3 k}{\log_7 k}\) for constant \(k \gt 0\) which are exceptions and correct.
There is no need to see \(\dfrac{x}{y} =\) but it should be clear what their answer is.
Do not ISW if they incorrectly apply log laws e.g. \(\dfrac{\log 7}{\log 3} = \log\dfrac{7}{3}\) or \(= \log 7 - \log 3\) or \(= \log(7 - 3)\) all of which score M1A0.
You may ISW after a correct answer if they go on to provide a decimal approximation.
A decimal approximation with no correct log work seen (1.7712…) scores no marks.

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

Special cases:

SC 1: For those that verify rather than prove a SC 100 is awarded for substituting \(k = \dfrac{5}{2}\) into all three terms to correctly obtain 243, 81 and 27 with a statement that this is geometric with \(r = \dfrac{1}{3}\) (or equivalent reason). All statements must be correct.

SC 2: Be aware that e.g. \(\dfrac{9^{7-2k}}{3^{4k-5}} = \dfrac{3^{2(k-1)}}{9^{7-2k}} \Rightarrow 81^{2(7-2k)} = 9^{6k-7}\) is an incorrect process (without some indication that they have intentionally squared both sides) that fortuitously leads to the correct answer and may score maximum SC 010.

M1: Uses the 3 terms to set up an equation in \(k\) and

• either reaches a common base by replacing 9 with \(3^2\) or by replacing 3 with \(9^{0.5}\) and uses the power law of indices correctly
• or uses the laws of indices correctly to reach \(9^{14-4k} = 3^{6k-7}\) condoning slips in e.g. expanding brackets.

Writing down e.g. \(2(7-2k) - (4k-5) = 2(k-1) - 2(7-2k)\) is sufficient to imply the M1.

dM1: Correct processing leading to a value for \(k\).

A1*: Correct value following correct working. Allow \(k = 2.5\) in place of \(k = \dfrac{5}{2}\)
Condone missing/invisible brackets provided they are recovered correctly.

Alt 1 Using Base 9:

Score dM1 when a value of \(k\) is achieved using a correct process and A1* if fully correct.

Alt 2 Finding \(r\) in terms of \(k\) and using e.g. \(u_3 = ar^2\):

Score dM1 when a value of \(k\) is achieved using a correct process and A1* if fully correct.

Alt 3 Using Logs Way 1:

Score dM1 when a value of \(k\) is achieved using a correct process and A1* if fully correct.

Alt 4 Using Logs Way 2:

Score dM1 when a value of \(k\) is achieved using a correct process and A1* if fully correct.

Alt 5 Recognising that taking \(\log_{3}\) forms an Arithmetic Sequence:

Score dM1 when a value of \(k\) is achieved using a correct process and A1* if fully correct.
There is no need to see any mention of log in this approach.

Notes

M1: Deduces expressions for the first term and the common ratio using \(k = \dfrac{5}{2}\) in the correct formulae and finds at least one of \(a = 243\) or \(r = \dfrac{1}{3}\). Allow if seen in (a). May be implied by correct values for \(a\) and \(r\).  For reference, \(a = 3^{4(2.5)-5}\) and \(r = \dfrac{9^{7-2(2.5)}}{3^{4(2.5)-5}}\ \left\{\text{or } r = \dfrac{3^{2(2.5-1)}}{9^{7-2(2.5)}}\right\}\)

M1: Recalls the sum to infinity formula and substitutes their values for \(a\) and \(r\) provided \(|r| \lt 1\)
Dependent on a correct attempt to find both \(a\) and \(r\) using \(k = 2.5\) but allow if neither value is correct or if they are unprocessed e.g. \(\dfrac{3^{4(2.5)-5}}{1 - \dfrac{9^{7-2(2.5)}}{3^{4(2.5)-5}}}\) scores this mark.

A1: cao. Correct sum to infinity. Answer only (with no working) scores full marks. Apply isw.

Typical solution

Typical solution

Typical solution

Typical solution

Since \(x \gt 0,\ x^{k-1} \neq 0\)

Typical solution

1

Typical solution

When \(k = 1\)

If there is a point of inflection \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0\)

Therefore, \(C\) has no point of inflection.

Typical solution

Typical solution

(b)(i)

(b)(ii)

(b)(iii)

Typical solution

Time = 4 hrs 30 mins

Typical solution

Typical solution

Typical solution

2

Typical solution

Typical solution

Increasing and convex

Typical solution

\(\log_{3} 2\)

Typical solution

\(3\mathrm{e}^{x}\)

Typical solution

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

Typical solution

Let \(\mathrm{f}(x) = x^3 - \mathrm{e}^{6 - 2x}\)

Hence \(\alpha\) lies between 0 and 4

Typical solution

Typical solution

(i)

(ii)

(iii)

It is not possible to evaluate \(x_2\) as ln 0 has no value

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

The 3rd graph: a decreasing curve lying below the \(x\)-axis, crossing the \(y\)-axis below \(O\)

Typical solution

\(-4\) is not a solution as \(\log_{10} -4\) has no real value. Therefore, the equation has exactly one solution.

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

Typical solution

Typical solution

(i)

(ii)

Typical solution

Average distance is approximately 4500 million kilometres

Typical solution

Notes

(a)(i)

B1: State correct scale factor

(a)(ii)

B1: Identify direction – correct language needed. Allow ‘horizontal’
B0 for ‘in’, ‘on’ or ‘along’ the \(x\)-axis, or for right / left
Could be implied by \(\begin{pmatrix} k \\ 0 \end{pmatrix}\)

B1: Correct magnitude of translation. Not dep on previous B1, but must have indicated horizontal translation in some way, including informal language

Do not penalise mentioning the stretch from (a)(i) again in (a)(ii)

\(\begin{pmatrix} \frac{1}{2} \\ 0 \end{pmatrix}\) is B2

Notes

B1: State correct scale factor. Any exact equiv

Notes

M1: State either correct derivative or \(\frac{1}{2}\mathrm{e}^{2x-1}\)

A1: Obtain correct gradient, following correct derivative. ‘Show that’ so correct derivative must be seen first

B1: State, or use, correct \(y\)-value

A1: Use correct numerical \(m\), \(x\) and \(y\) values to obtain given equation of tangent. Could also use \(y = mx + c\) to obtain \(c = 0\) and hence given equation

Notes

B1: State correct integral. Any equiv exact form. Could be in terms of \(u\) if using substitution

M1*: Attempt use of correct limits. Correct order and subtraction, working with exact values. Could be using \(u\) limits (must be \(-1\) and 0)

M1dep*: Attempt area of region ie their definite integral minus attempt at area of triangle. Could be implied by their definite integral minus 0.25 www. Must be using exact values
Area of a triangle must either be \(\frac{1}{2} \times \frac{1}{2} \times\) their \(y\) or from \(\displaystyle\int_0^{0.5} 2x\,\mathrm{d}x\)

A1: Correct area, in any exact form with like terms combined. Must have worked exactly throughout

Alternative method (integrating wrt \(y\))

M1: Change subject and attempt full integration by parts(using correct parts)

M1*: Attempt use of correct limits. Correct order and subtraction, working with exact values. Must use correct \(y\) limits

M1dep*: Attempt area of region ie attempt at area of triangle minus their definite integral. Could be implied by 0.25 minus their definite integral www. Must be using exact values
Area of a triangle must either be \(\frac{1}{2} \times \frac{1}{2} \times\) their \(y\) or from \(\displaystyle\int_0^{1} \frac{1}{2}y\,\mathrm{d}y\)

A1: Correct area, in any exact form with like terms combined. Must have worked exactly throughout

Notes

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

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

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

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

Notes

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

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

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

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

Notes

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

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

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

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

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

See Appendix for those using T and I

Appendix: Trial and Improvement in 3(c)

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

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

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

Notes

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

Notes

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

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

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

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

Notes

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

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

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

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

Notes

M1: Attempt to show reduction to linear form
Introduce logs on both sides, and correctly split to the sum of two terms

A1: Obtain correct equation
Condone no base; any bases seen must be 10
A0 for \(\log_{10}bt\) unless previously seen as \(t\log_{10}b\)

A1: Link to equation of straight line
Could instead refer to linear relationship

If M0 then allow SC B1 for statement such as \(S\) against \(t\) is an exponential function so \(\log S\) against \(t\) will give a straight line

Alternative method

M1: Attempt equation of straight line, and attempt expression for \(S\)
Must be using \(\log_{10}S\) against \(t\)
Must use base of 10

A1: Correctly split into two terms

A1: Link to exponential model

Notes

B1: Link gradient of line of best fit to linear form and confirm \(b \approx 1.15\)
Allow \(m\) in range [0.055, 0.065]
Or \(\log_{10}1.15 = 0.06\) and compare to gradient

B1: Link intercept of line of best fit to linear form and confirm \(a \approx 120\)
Allow \(c\) in range [2.075, 2.085]
Or \(\log_{10}120 = 2.08\) and compare to intercept

Plotted points are linear so may not see line of best fit drawn
If substituting into formula (either given model or linear reduction) then

• B1 for finding and verifying any point that would be on the line of best fit
• B1 for finding and verifying a second point

Notes

M1: Substitute \(t = 7\) into given model
soi

A1: Conclude with integer value
Accept 320 items

Notes

(d)(i)

B1: State or imply sum of GP with \(a\) as 120 or 138, and \(r\) as 1.15
Could be implied by attempt to use GP sum formula (but not just \(n\)^th term) – allow slip as long as clearly sum being considered

M1*: Attempt sum of GP, with \(a = 120\) or 138 and \(r = 1.15\), related to 70000
Must be correct sum formula
May have \(n\) not \(t\) throughout
Allow \(r = 1.15^{t+1}\) with \(a = 120\) but this is B1 M1 only, as not a valid method)

M1dep*: Attempt to rearrange equation as far as \(1.15^t =\)
Must now have \(a = 138\) (or equiv)
Allow sign errors only
Allow T&I as not DR

A1: Obtain 32 (‘months not required’)
www
If 32 given as answer only then allow full marks; if any method shown then mark using the main scheme
Allow BOD with any inequality signs

(d)(ii)

B1: State or imply that the model is unlikely to be valid, with a sensible reason why – could refer to reason why pattern may not continue or extrapolation not being reliable

• Decrease in demand
• Increase in competition
• No values beyond \(t = 6\) so pattern unknown
• Reason why sales are likely to level off / plateau or unlikely to continue to increase (‘other factors’ not enough)
• Item sales may vary according to season

Notes

M1: Attempt to differentiate using the chain rule
Obtain derivative of form \(\mathrm{f}(x)\mathrm{e}^{x^2 + 3x}\)
Could also split into two terms and use product rule to obtain derivative of form \(\mathrm{f}(x)\mathrm{e}^{x^2}\mathrm{e}^{3x} + k\mathrm{e}^{x^2}\mathrm{e}^{3x}\ (k \neq 0)\)
M0 if attempt to split results in sum not product

A1: Obtain correct derivative
Brackets must be seen, or implied by later work
aef eg \(\left(2x\mathrm{e}^{x^2}\right)\left(\mathrm{e}^{3x}\right) + \left(\mathrm{e}^{x^2}\right)\left(3\mathrm{e}^{3x}\right)\) from splitting into two terms first
Could be in terms of \(u\), as long as \(u\) clearly defined

A1: Equate correct derivative to 0 soi and solve to obtain \(x = -\frac{3}{2}\)
ISW any \(y\)-coordinates if given
A0 if any additional solutions for \(x\)
Must see differentiation, so \(x = -\frac{3}{2}\) with no supporting method gets no credit (as question is ‘determine’)

B1FT: Indicate no solutions from the exponential term
FT their derivative as long as of form \(\mathrm{f}(x)\mathrm{e}^{x^2 + 3x}\) or \(\mathrm{f}(x)\mathrm{e}^u\)
Allow BOD for explanations such as \(\mathrm{e}^x \gt 0\) for all \(x\)
Must have some reason, eg ‘\(\mathrm{e}^{x^2 + 3x}\) is always positive’, ‘\(\mathrm{e}^{x^2 + 3x}\) cannot be negative’, ‘cannot take ln of a negative number’, ‘not defined’, ‘not real’, ‘no solutions’
A0 for ‘math error’ or ‘doesn’t work’

Alternative method

M1: Take ln and attempt implicit differentiation
Must deal correctly with \(\ln y\)

A1: Obtain correct derivative
May still have \(\dfrac{1}{y}\) on LHS

A1: Equate correct derivative to 0 soi and solve to obtain \(x = -\frac{3}{2}\)

B1: Indicate no solutions from the exponential term
See main MS for guidance
Could also explain why no solutions from \(\dfrac{1}{y}\)

Notes

M1: Attempt to differentiate again using the product rule correctly
Obtain derivative of form \(\left(ax^2 + bx + c\right)\mathrm{e}^{x^2 + 3x}\) aef

A1: Obtain correct derivative
aef eg (depending on method) \(2\mathrm{e}^{x^2 + 3x} + 2x(2x + 3)\mathrm{e}^{x^2 + 3x} + 3(2x + 3)\mathrm{e}^{x^2 + 3x}\)
or \(\left(\left(2\mathrm{e}^{x^2} + 4x^2\mathrm{e}^{x^2}\right)\mathrm{e}^{3x} + \left(2x\mathrm{e}^{x^2}\right)3\mathrm{e}^{3x}\right) + \left(6x\mathrm{e}^{x^2}\mathrm{e}^{3x} + 9\mathrm{e}^{x^2}\mathrm{e}^{3x}\right)\)
Could be in terms of \(u\), as long as \(u\) clearly defined

B1: State, or clearly imply, correct condition at any point in proof
Must be general statement, and not just \(\gt 0\) from testing the stationary point

M1: Explain why correct quadratic is always positive
Could note minimum value of 2 as completed square form
Could use expanded quadratic, which should be \(4x^2 + 12x + 11\)
If showing no real roots then must also say that it is a positive quadratic
Condone \(\gt\) / \(\geqslant\) muddles for M1 only
Could show that there are no points of inflection and \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} \gt 0\) for at least one point

A1: Full and convincing proof to show that curve is convex for all \(x\)
www

Alt method for first 2 marks

M1: Attempt second derivative, using implicit differentiation and the product rule
If still \(\dfrac{1}{y}\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x + 3\) then must be a correct attempt to differentiate the LHS

A1: Obtain correct derivative
aef
Will need to use \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = y(2x + 3)\) to make further progress

Notes

M1: Integrate and obtain expression involving ln. This may be combined with the next step. Ignore limits. Brackets \((x + 2)\) soi
If a substitution is used then only award this mark when \(\ln(x + 2)\) reached or an equivalent integral with appropriate limits (typically \(\ln x\) with limits 2 and 4.5).

M1: Substitute and use limits 0 & 2.5 (or appropriately changed limits in the case of a substitution) in their ln integral. Must see this step.

A1: www, any equivalent exact form, but must be a single term expression. Do not accept \(\ln\frac{4.5}{2}\) for this mark.

Correct answer with no working SC B1 [1/3]
Ignore use of modulus signs throughout.

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

M1*: Take logs of both sides (any base) correctly and use power law correctly at least once. Common correct answers that score M1 are: \(2x + 1 = \log_4 5^x\), \(\log_5 4^{2x + 1} = x\), \((2x + 1)\ln 4 = \ln(5^x)\)
Condone lack of bracket on the \(2x + 1\) term

M1dep*: Re-arrange to get an equation with a single term in \(x\) – condone sign errors only e.g. the following score the first two M marks: \(\dfrac{1}{x} + 2 = \dfrac{\log 5}{\log 4}\), \(x\left(2 - \dfrac{\log 5}{\log 4}\right) = -1\), \(x(\log_4 5 - 2) = 1\), \(x(1 - 2\log_5 4) = \log_5 4\), \(x(2\ln 4 - \ln 5) + \ln 4 = 0\)
This mark can be implied by a correct answer provided the first M mark was awarded (that is, we must see logs being taken and the power law used at least once)

A1: awrt \(-1.19\)
\(-1.19184404\ldots\)
Correct answer with no working scores no marks

Alternative method

B1: Correctly re-writes the given equation in the form \(a^x = b\) with \(a\) and \(b\) correct (check carefully for other equivalent correct equations)

M1: Taking logs correctly of their \(a^x = b\), where \(a\) and \(b\) are both positive, to obtain either \(x\log a = \log b\) (any base) or \(x = \log_a b\) (for their \(a\) and \(b\))
Not dependent on the B mark

A1: awrt \(-1.19\)
\(-1.19184404\ldots\)
Correct answer with no working scores no marks

Notes

M1: Separate variables and attempt integration of at least one side
Allow \(k\mathrm{e}^{3y}\), with \(k \ne 1\), as ‘attempt’ at integration of LHS
‘Attempt’ at RHS may not be use of integration by parts
Allow BOD on missing integral sign / missing \(\mathrm{d}y\) / missing \(\mathrm{d}x\) as long as intention clear

B1: Correct LHS
B0 if still part of an expression also involving \(x\)

M1: Attempt integration by parts on RHS – must have correct parts
As far as attempt at \(x^3\ln x - \displaystyle\int x^2\,\mathrm{d}x\), possibly with \(\displaystyle\int \frac{1}{x}x^3\,\mathrm{d}x\) not yet simplified

A1: Correct RHS (condone no \(+ c\))
Condone no modulus sign on \(\ln x\)

M1: Attempt use of \((\mathrm{e}, 1)\) to find \(c\)
Used in an equation involving \(x\), \(y\) and \(c\), following some integration attempt of both sides
As far as finding \(c\), either exact or as a decimal
M1 can be implied by sight of \(-\frac{1}{3}\mathrm{e}^3\) or \(-6.695\ldots\) following a correct equation

A1: Obtain correct equation, in required form
Any equivalent form on the RHS, but must be \(\mathrm{e}^{3y} = \ldots\)
A0 if decimal approximation for \(\mathrm{e}^3\)

Notes

M1*: Substitute \(x = \mathrm{e}^2\), into their integral involving \(\ln x\), and attempt to simplify
\(\ln x\) may be \(\ln x^p\) if any coefficient has been taken into the ln term
As far as correctly simplifying the ln term to remove ln
Must be working exactly, so M0 if decimals seen before ln dealt with

M1dep*: Introduce logs correctly, and attempt to rearrange to given form
Their equation must have two terms, or possibly more, with the terms having a common factor of \(\mathrm{e}^k\)
Attempt must go as far as splitting into the sum of two terms, with \(\ln\mathrm{e}^k\) simplified to \(k\)

A1: Obtain \(y = 1 + \frac{1}{3}\ln(5\mathrm{e}^3 - 1)\)
No need to state \(a\), \(b\) and \(c\) explicitly

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: M1 for \(k(2x + 9)^{\frac{3}{2}}\) with non-zero \(k\), \(k \ne 1\)

A1: cao (allow unsimplified)

A1: Uses correct limits (or implies correct limits) to get 9. Condone limits the wrong way round leading to \(-9\) but must be changed to \(+9\)

M1*: Attempt to solve \(4\mathrm{e}^{-2x} - 1 = 0\) by correctly taking logs of both sides leading to \(\pm 2x = \pm\ln\alpha\) where \(\alpha \gt 0\)
Allow sign errors and other minor slips only

A1: Or equivalent exact value (soi possibly by correct exact value used later) e.g. \(\frac{1}{2}\ln 4\) or \(\ln 2\)

M1*: Integrate \(4\mathrm{e}^{-2x} - 1\) to obtain \(c\mathrm{e}^{-2x} \pm x\)
Where \(c\) is non-zero and \(c \ne 4\)

M1dep*: Uses limits correctly \(F\left(\frac{1}{2}\ln 4\right) - F(0)\) (with their \(\frac{1}{2}\ln 4\)) – dependent on the previous two M marks (allow non-exact top limit). Condone limits the wrong way round only if the sign of their answer is subsequently changed
If zero limit is assumed to give 0 (with no working) then M0

A1: If the values of the integral(s) are changed from negative to positive (e.g. from limits the wrong way round) with no justification given then A0
\(p\) and \(q\) need not be explicitly stated. \(p = \dfrac{21}{2}\) (oe) and \(q = -1\)

Notes

(a)(i)

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

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

(a)(ii)

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

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

Notes

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

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

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

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

DR so no credit for answer only

Notes

M1: M1 for \(x(x + 2) - (x + 1)(x - 1)\) oe

M1: Multiply out brackets. Allow one error
Ignore denominator even if “= 0”

A1: NB correct with no working: SC B1

Alternative method

M1: M1 for attempt “cross-multiply”.

M1: Multiply out brackets. Allow one error

Notes

M1: or cubic in \(x\). Condone quadratic in \(x\) with \(x = \frac{1}{x^3}\) or \(x = x^3\)
Must see attempt at correct method for this mark
Allow arithmetical errors

B1: Can be scored without M1. Condone \(x = 1, -\frac{1}{8}\) or \(x = 1, -8\)
Ignore \(x^3 = 0\), if seen, for this mark

B1f: ft their \(x^3\) or \(\dfrac{1}{x^3}\). If also \(x = 0\), B0
NB correct with no working: M0B0B1

Notes

M1: Condone incorrect or omitted brackets
Any correct step after log(both sides)
or ANY correct step using indices

A1: NB correct with no working or T & I: SC B1

Notes

B1: Or show that \(2\mathrm{e}^{-3t} = 0\) has no solutions
Need to see \(y \neq 0\) or \(y \gt 0\) and reason relating to exponential (or logarithmic) function
Must clearly be referring to \(y\) or \(2\mathrm{e}^{-3t}\)

Notes

M1: Equate to 0 and attempt to solve disguised quadratic
‘determine’ so some evidence of method needed

A1: Obtain both correct values
A0 for \(\ln 1\) and not 0

Notes

B1: Correct \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\)
Mark derivative and condone no/wrong label

B1: Correct \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\)
Mark derivative and condone no/wrong label

M1: Attempt correct method to combine derivatives
Combine their derivatives correctly

A1: Show manipulation to given answer
Need to see some evidence of how \(\mathrm{e}^{-3t}\) is dealt with
AG so method must be fully correct

Notes

M1: Equate denominator to 0
Or \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = 0\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0\)

A1: Solve for \(t\) to obtain \(t = \ln 2\)
No need to see \(\mathrm{e}^{4t} = 0\) discounted

A1: Obtain correct coordinate
Or \(x = -1\), \(y = \frac{1}{4}\)

Notes

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

Notes

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

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

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

Notes

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

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

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

Notes

B1: 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

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

M1: See appendix
Allow \(1^4 - 2^1\) or 1 – 2 and \(2^4 - 2^2\) or 16 – 4
Ignore ‘= 0’
Condone \(-1\) and 12 only provided clearly linked to \(x = 1\) and \(x = 2\) e.g. as coordinates or in a table of values
Allow for using two values between 1 and 2 that are either side of 1.24 e.g. 1.2 and 1.3

A1: Correct values of function and comment on change of sign, allow e.g. \(-1 < 0\) and \(12 > 0\)
Condone no mention of function being sufficiently well-behaved (or continuous)
e.g. 1.2 and 1.3 leading to \(-0.22\) and 0.39 (2 s.f. is sufficient)

Appendix: exemplar responses for Q5(a)

REMEMBER: It is not possible to have an award of M0A1

Notes

B1: AG
Convincing rearrangement to given result
Must show one step of working and no incorrect work seen
Must not work backwards
e.g. \(4\log_2(x) - x = 0\) is B0 as not convincingly from a correct process unless \(4\log_2(x) = x\) is seen first

Notes

M1: \(x_1 =\) awrt 1.07 (= 1.07177...).
soi by any of these values: 1.204…, 1.232…, 1.237…

A1: cao
Must have shown at least one iteration e.g. \(x_1 = 1.07\)

Notes

B1: Vertical line \(x = 0.4\) drawn to meet the curve, could start line from \(y = x\)

B1: Correct completion of staircase diagram to head towards root with at least two vertical and two horizontal lines seen (lines need not extend beyond region between line and curve)
Condone good freehand lines, mark intent
SC B1B0 for using a different starting value e.g. \(x = 0\) with at least two vertical and two horizontal lines seen. Correct staircase diagram converging to root from their starting value.

Notes

B1: AG
Convincing rearrangement to given result
Must show at least one step of working and no incorrect work seen
e.g. \(\log_2(x^4) - \log_2(2^x) = 0\) is B0 as not convincingly from a correct process unless \(\log_2(x^4) = \log_2(2^x)\) is seen first
Must not work backwards

Notes

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

M1: Or \(4\log_2 1 = 0\)

A1: Condone e.g. math error or \(x_2\) is invalid or \(x_2\) has no solution provided log 0 is mentioned

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

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

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

Notes

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

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

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

Notes

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

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

Notes

B1: 2.45 (need not have working)
May be done BC

B1: Convincing completion

Notes

M1: uses laws of logs

A1: AG \(\log\frac{200}{20}\) or \(\log_{10}10\) must be seen explicitly

Alternative method

M1: Uses laws of logs

A1: Must see \(\log_{10}100 = 2\) or \(\log_{10}10 = 1\) explicitly

Notes

M1: Attempts to establish a common difference of 1

A1: argues from a common difference eg “increases by 1 each time”. Must use exact values to establish the difference between the second and third terms.

Alternative method

M1: rewrites two more terms of the sequence and makes a comparison

A1: argues from a common difference eg “increases by 1 each time” Must use exact values to establish the difference between the second and third terms.

Notes

M1: Uses the formula with first term \(\log_{10}20\) and common difference 1

A1: Allow for fully correct expression not simplified. isw
Correct forms include \(50\log_{10}20 + 1225\), \(25\log 400 + 1225\), \(25(\log 400 + 49)\), \(25(\log 4 + 51)\) etc

Notes

M1: Or better

B1: Integral correctly evaluated
ln 1 or “\(\ln n - 0\)” seen.

A1: AG. Convincing completion

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

Notes

M1: Equate to 3

E1: AG Rearrange to quadratic in \(\mathrm{e}^{x}\) Also allow \(\mathrm{e}^{2x} - 3\mathrm{e}^{x} - 4 = 0\)
Expression \(= 0\) must be seen

Notes

M1: May be BC giving at least one root of the quadratic equation (a)

A1: \(x = \ln 4\) must be seen explicitly

E1: must explain why they reject the value \(-1\) for \(\mathrm{e}^{x}\), or state \(\mathrm{e}^{x} + 1\) is never zero

Notes

B1: Condone missing \(+c\) in their integral

M1: Attempt to evaluate \(c\)

M1: Substituting \(x = 1\) into their expression

E1: AG must argue below the axis from correct \(y\) value. Must be clear that \(-0.81\) is a \(y\)-coordinate

Notes

(ii) B1: \(P = 0.3\mathrm{e}^{0.5 \times 9} = 27.0051\ldots\)

Notes

B1: Very large prediction in 2025 (80 501MWh) in unrealistic
“Extrapolation” alone does not score, it would need explaining/clarifying

Notes

E1: Correct reasoning (answer given)

Notes

(i) B1: Gradient increasing from near zero to maximum for value of \(t\) somewhere between 10 and 20

(i) B1: Gradient decreasing to near zero from max value

(ii) B1: Answer in range 13 to 15

Notes

B1: Accept answer in range 300 to 305

Notes

M1: Forming an equation for \(k\) and attempt to solve

A1: Allow for exact value or evaluated to at least 2 s.f.

Notes

E1: Must imply to the model tends to zero and this does not match the real situation.

Notes

M1: Taking logs and attempting to use laws of logs
Do not award for values of \(a\) and \(b\) obtained directly from the data and the natural log form of the model.

A1: FT their values for \(\theta_0\) and \(k\)
Accept as part of equation or \(a\) and \(b\) clearly stated

Notes

B1: Accept \(30^\circ\) www Must be evaluated

M1: Attempt to differentiate their exponential expression for \(\theta\)

A1: Any form eg \(\mathrm{e}^{3.4} \times -0.08\mathrm{e}^{-0.08t}\) or \(-0.08\mathrm{e}^{3.4-0.08t}\)

A1: Allow for correct negative value for \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\) or a clear statement that the rate of cooling is \(2.4^\circ\) per minute. Accept \(= -0.08\mathrm{e}^{3.4}\)

Alternative method

B1: Accept \(30^\circ\) www

M1: Uses implicit differentiation w.r.t \(t\)

A1: Correct derivative

A1: Allow for correct negative value for \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\) or a clear statement that the rate of cooling is \(2.4^\circ\) per minute

Notes

M1: Attempting to find the intersection of their (c) and the given line
Could be BC

A1: Accept awrt 7.0, 7.1 or 7.2

A1: Must be the value for \(\theta\)

Alternative method

M1: Equate their expressions for temperature and attempts to solve for \(t\)

A1: Accept awrt 7.0, 7.1 or 7.2

A1: Cao

Notes

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

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

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

Notes

M1: taking logarithms in any base

M1: 3^rd law of logs used correctly

A1: if M0M0 allow SC1 for 1.44 unsupported

Alternatively

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

Notes

(ii) M1: Straight line (+ve gradient) through origin (may stop short of origin)

A1: First quadrant only

Notes

(i) E1: Allow e.g. wolves give birth to more wolves than they started with
Do not allow e.g. population is proportional to time

(ii) A1: Allow 0.89

(iii) E1: Allow e.g. lack or resources or deforestation