June 2019 Paper 1 Q7

Notes

(a) Option 1

M1: For \(V = A\mathrm{e}^{\pm kt}\)   Do not allow if \(k\) is fixed, eg \(k = -0.5\)
Condone different variables \(V \leftrightarrow y\)   \(t \leftrightarrow x\) for this mark, but for A1 \(V\) and \(t\) must be used.

M1: Substitutes \(t = 0 \Rightarrow A = 20\,000\) into their exponential model
Candidates may start by simply writing \(V = 20\,000\mathrm{e}^{kt}\) which would be M1 M1

dM1: Substitutes \(t = 1 \Rightarrow 16\,000 = 20\,000\mathrm{e}^{-1k} \Rightarrow k = \ldots\) via the correct use of logs.
It is dependent upon both previous M’s.

A1: \(V = 20\,000\mathrm{e}^{-0.223t}\) (with accuracy to at least 3sf) or \(V = 20\,000\mathrm{e}^{t\ln 0.8}\)
A correct linking formula with correct constants must be seen somewhere in the question

(a) Option 2

M1: For \(V = Ar^t\) or equivalent such as \(V = kr^{t-1}\)
Condone different variables \(V \leftrightarrow y\)   \(t \leftrightarrow x\) for this mark, but for A1 \(V\) and \(t\) must be used.

M1: Uses \(t = 0 \Rightarrow A = 20\,000\) in their model. Alternatively uses \((0,\ 20\,000)\) and \((1,\ 16\,000)\) to give \(r = \dfrac{4}{5}\) oe
You may award if one of the number pair \((0,\ 20\,000)\) or \((1,\ 16\,000)\) works in an allowable model

dM1: \(t = 1 \Rightarrow 16\,000 = 20\,000r^1 \Rightarrow r = \ldots\)    Dependent upon both previous M’s
In the alternative it would be for using \(r = \dfrac{4}{5}\) with one of the points to find \(A = 20\,000\)
You may award if both number pairs \((0,\ 20\,000)\) or \((1,\ 16\,000)\) work in an allowable model

A1: \(V = 20\,000 \times 0.8^t\)    Note that \(V = 20\,000 \times 1.25^{-t}\)   \(V = 16\,000 \times 0.8^{t-1}\) and is also correct

(a) Option 3

M1: They may suggest an exponential model with a lower bound. For example, for \(V = A\mathrm{e}^{\pm kt} + 2000\) The bound must be stated but do not allow k to be fixed . Allow as long as the bound < 10 000

M1: \(t = 0, V = 20\,000 \Rightarrow A = 18\,000\)

dM1: \(t = 1, V = 16\,000 \Rightarrow 16\,000 = 2\,000 + 18\,000e^{k} \Rightarrow k = \ldots\)    Dependent upon both previous M’s

A1: \(V = 18\,000 \times \mathrm{e}^{-0.251t} + 2000\)

Differential equation

It is entirely possible that they start part (a) from a differential equation.

M1: \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = kV \Rightarrow \displaystyle\int \dfrac{\mathrm{d}V}{V} = \int k\,\mathrm{d}t \Rightarrow \ln V = kt + c\)    M1: \(\ln 20\,000 = c\)

dM1: Using \(t = 1, V = 16\,000 \Rightarrow k = \ldots\)    A1: \(\ln V = -\ln\left(\dfrac{5}{4}\right)t + \ln 20\,000\)

Notes

(b) Option 1

M1: Uses a model of the form \(V = A\mathrm{e}^{\pm kt}\) to find the value of \(V\) when \(t = 10\).
Alternatively substitutes \(V = 2000\) into their model and finds \(t\)

A1: This can only be scored from an acceptable model with correct constants with accuracy to at least 2sf .
Compares \(V =\) (£) 2150 with (£) 2 000 and states "reliable as \(2150 \approx 2000\) " or "reasonably good as they are close" or ""OK but a little high".
Allow a candidate to argue that it is unreliable as long as they state a suitable reason. Eg. ‘‘It is too far away from £2000’’ or ‘‘It is over £100 away, so it is not good’’
Do not allow ‘‘it is not a good model because it is not the same’’
In the alternative it is for comparing their value of \(t\) with 10 and making a suitable comment as to the reliability of their model with a reason.
\(V = 20\,000\mathrm{e}^{-0.223t} \Rightarrow 2000 = 20\,000\mathrm{e}^{-0.223t} \Rightarrow t = 10.3\) years.
Deduction Reliable model as the time is approximately the same as 10 years. A candidate can argue that the model is unreliable if they can give a suitable reason.

(b) Option 2

M1: Uses a model of the form \(V = Ar^t\) oe to find the value of \(V\) when \(t = 10\). Eg. \(20\,000 \times 0.8^{10}\)
Alternatively substitutes \(V = 2000\) into their model and finds \(t\)

A1: This can only be scored from an acceptable model with correct constants also allowing an accuracy to 2sf.
Compares (£) 2147 with (£) 2 000 and states "reliable as \(2147 \approx 2000\) " or "reasonably good as they are close" or ""OK but a little high".
Allow a candidate to argue that it is unreliable as long as they state a suitable reason. Eg. ‘‘It is too far away from £2000’’ or ‘‘It is over £100 away, so it is not good’’
Do not allow ‘‘it is not a good model because it is not the same’’

(b) Option 3

M1: Uses their model to find the value of \(V\) when \(t = 10\).
Alternatively substitutes \(V = 2000\) into their model and finds \(t\)

A1: For \(V = 18\,000 \times \mathrm{e}^{-0.251 \times 10} + 2000 = \)£3462.83 Deduction: Unreliable model as £3462.83 is not close to £2 000 This can only be scored from an acceptable model with correct constants

Notes

(c) Option 1

B1ft: For a correct statement. Eg states that the value of their ‘\(-0.223\)’ should become less negative.
Alt states that the value of their ‘0.223’ should become smaller. If they refer to \(k\) then refer to the model and apply the same principles.
Condone the fact that they don’t state their \(-0.223\) doesn’t lie in the range \((-0.223,\ 0)\)

(c) Option 2

B1ft: States a value of \(r\) in the range \((0.8,\ 1)\) or states would increase the value of "0.8"
They do not need to state that "0.8" must lie in the range \((0.8,\ 1)\)
Condone increase the 0.8. Also allow decrease the "1.25" for \(V = 20\,000 \times 1.25^{-t}\)

(c) Option 3

B1: States make the value of \(k\) or the \(-0.251\) greater (or less negative) so that it lies in the range \((-0.251,\ 0)\)
Condone ‘make the value of \(k\) or the \(-0.251\) greater (or less negative)’