June 2025 Paper 1 Q13

Notes

M1: Uses \(h = 31 - A\mathrm{e}^{-kt}\) with \(h = 6,\ t = 10\) and \(h = 11,\ t = 20\) to create simultaneous equations and proceeds via correct work to eliminate \(A\) or \(k\).
Condone the equations appearing as e.g. \(6\text{ m} = 31 - A\mathrm{e}^{-10k},\ 11\text{ m} = 31 - A\mathrm{e}^{-20k}\)
Look for the overall method being correct but condone slips.

Examples:

Eliminates \(A\):

• \(A\mathrm{e}^{-10k} = 25,\ A\mathrm{e}^{-20k} = 20 \Rightarrow \mathrm{e}^{10k} = \dfrac{25}{20}\)
• \(A\mathrm{e}^{-10k} = 25,\ A\mathrm{e}^{-20k} = 20 \Rightarrow \ln A - 10k = \ln 25,\ \ln A - 20k = \ln 20 \Rightarrow 10k = \ln 25 - \ln 20\)

Eliminates \(k\):

• \(A\mathrm{e}^{-10k} = 25,\ A\mathrm{e}^{-20k} = 20 \Rightarrow A\left(\mathrm{e}^{-10k}\right)^2 = 20 \Rightarrow A\left(\dfrac{25}{A}\right)^2 = 20\)
• \(A\mathrm{e}^{-10k} = 25,\ A\mathrm{e}^{-20k} = 20 \Rightarrow \ln A - 10k = \ln 25,\ \ln A - 20k = \ln 20 \Rightarrow \ln A = 2\ln 25 - \ln 20\)

dM1: Proceeds to find a value for \(A\) or \(k\) using the correct order of operations but condoning slips.

A1: Achieves either \(k =\) awrt 0.0223 or \(A =\) awrt 31.3 (or \(A =\) awrt 31.2)
Note the exact values are \(k = \dfrac{1}{10}\ln\dfrac{5}{4}\) or \(A = \dfrac{125}{4}\ (31.25)\) which are acceptable.

A1: cso Proceeds correctly to \(h = 31 - 31.2\mathrm{e}^{-0.0223t}\) or \(h = 31 - 31.3\mathrm{e}^{-0.0223t}\)
Exact answer is \(h = 31 - \dfrac{125}{4}\mathrm{e}^{-\frac{1}{10}\ln\frac{5}{4}t}\) o.e., e.g., \(h = 31 - 31.25\mathrm{e}^{\frac{1}{10}\ln\frac{4}{5}t}\) which is acceptable.
Finding \(A\) and \(k\) is insufficient for this mark as the question asks for a complete equation, however, allow this mark if the complete correct equation is seen subsequently.

Notes

B1: cao 31 m and units are required. Must be seen as their answer to (b).
There must be no other values.

Notes

(c) (i)

B1ft: Accept \(31 - \text{``}A\text{''}\) with correct units. This must be evaluated correctly for their \(A\).
Only penalise the omission of units once from an otherwise acceptable answer in (b) and (c)(i) and penalise it on the first occurrence.

(c) (ii)

B1ft: Concludes that the model is unsuitable because (the initial/early) height is negative
Follow through on their (c)(i) here but it must be negative.
Alternatively, allow that the model is suitable if they conclude that the depth of the seed is \(\text{``}{-0.3}\text{''}\) m or e.g. suitable as it would be underground.

Notes

(d) (i)

M1: Achieves the form \(\left\{\dfrac{\mathrm{d}h}{\mathrm{d}t} =\right\} a\mathrm{e}^{-kt}\). It is acceptable to use \(k\) and \(A\) for this mark and allow if \(k \lt 0\) but must be consistent with their \(h\).

A1: \(\left\{\dfrac{\mathrm{d}h}{\mathrm{d}t} =\right\}\) awrt \(0.698\mathrm{e}^{-0.0223t}\) Allow awrt \(0.697\mathrm{e}^{-0.0223t}\) or awrt \(0.696\mathrm{e}^{-0.0223t}\) and allow exact e.g. \(\dfrac{25}{8}\ln\dfrac{5}{4}\mathrm{e}^{-\frac{1}{10}\ln\frac{5}{4}t}\)

(d)(ii)

B1ft: Sets their \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = 0.3\) where \(\dfrac{\mathrm{d}h}{\mathrm{d}t}\) is not their \(h\).

M1: Uses their \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = \ldots\), and proceeds from \(a\mathrm{e}^{-kt} = b\) where \(ab \gt 0\) to \(ct = \ln d\) where \(d \gt 0\) using the correct order of operations but condoning slips.
Allow equivalent work e.g.
\(0.696\mathrm{e}^{-0.0223t} = 0.3 \Rightarrow \ln 0.696 - 0.0223t = \ln 0.3 \Rightarrow -0.0223t = \ln 0.3 - \ln 0.696\)

A1: cso awrt 37.7 or 37.8 or 37.9 but note that this may be scored if \(k\) is rounded in (a).
Ignore any units if given and just look for the value.