June 2018 Paper 2 Q14

Notes

(b) Way 2

(b) Way 3

(b) Way 4

M1: Attempts to differentiate using

• the chain rule to give \(\dfrac{\mathrm{d}N}{\mathrm{d}t} = \pm A\mathrm{e}^{-0.25t}\left(3 + 7\mathrm{e}^{-0.25t}\right)^{-2}\) or \(\dfrac{\pm A\mathrm{e}^{-0.25t}}{\left(3 + 7\mathrm{e}^{-0.25t}\right)^2}\) o.e.
• the quotient rule to give \(\dfrac{\mathrm{d}N}{\mathrm{d}t} = \dfrac{\left(3 + 7\mathrm{e}^{-0.25t}\right)(0) \pm A\mathrm{e}^{-0.25t}}{\left(3 + 7\mathrm{e}^{-0.25t}\right)^2}\)
• implicit differentiation to give \(N\left(3 + 7\mathrm{e}^{-0.25t}\right) = 900 \Rightarrow \left(3 + 7\mathrm{e}^{-0.25t}\right)\dfrac{\mathrm{d}N}{\mathrm{d}t} \pm AN\mathrm{e}^{-0.25t} = 0\), o.e.

where \(A \neq 0\)

Note: Condone a slip in copying \(\left(3 + 7\mathrm{e}^{-0.25t}\right)\) for the M mark

A1: A correct differentiation statement

Note: Implicit differentiation gives \(\left(3 + 7\mathrm{e}^{-0.25t}\right)\dfrac{\mathrm{d}N}{\mathrm{d}t} - 1.75N\mathrm{e}^{-0.25t} = 0\)

dM1: Way 1: Complete attempt, by eliminating \(t\), to form an equation linking \(\dfrac{\mathrm{d}N}{\mathrm{d}t}\) and \(N\) only
Way 2: Complete substitution of \(N = \dfrac{900}{3 + 7\mathrm{e}^{-0.25t}}\) into \(\dfrac{\mathrm{d}N}{\mathrm{d}t} = \dfrac{N(300 - N)}{1200}\)

Note: Way 1: e.g. substitutes \(3 + 7\mathrm{e}^{-0.25t} = \dfrac{900}{N}\) and \(7\mathrm{e}^{-0.25t} = \dfrac{900}{N} - 3\) or substitutes \(\mathrm{e}^{-0.25t} = \dfrac{\frac{900}{N} - 3}{7}\) into their \(\dfrac{\mathrm{d}N}{\mathrm{d}t} = \ldots\) to form an equation linking \(\dfrac{\mathrm{d}N}{\mathrm{d}t}\) and \(N\)
(corrected from the printed mark scheme: it prints \(\mathrm{e}^{-0.25t} = \dfrac{900}{N}\) as the second substitution)

A1*: Way 1: Correct algebra leading to \(\dfrac{\mathrm{d}N}{\mathrm{d}t} = \dfrac{N(300 - N)}{1200}\) *
Way 2: See scheme

(b) Way 3 M1: Separates the variables, an attempt to form and apply partial fractions and integrates to give \(\ln\) terms \(= kt\ \{+c\},\ k \neq 0\), with or without a constant of integration \(c\)

A1: \(\dfrac{1}{300}\ln N - \dfrac{1}{300}\ln(300 - N) = \dfrac{1}{1200}t\ \{+c\}\) or equivalent with or without a constant of integration \(c\)

dM1: Uses \(t = 0, N = 90\) to find their constant of integration and obtains an expression of the form \(\lambda\mathrm{e}^{\frac{1}{4}t} = \mathrm{f}(N);\ \lambda \neq 0\) or \(\lambda\mathrm{e}^{-\frac{1}{4}t} = \mathrm{f}(N);\ \lambda \neq 0\)

A1*: Correct manipulation leading to \(N = \dfrac{900}{3 + 7\mathrm{e}^{-0.25t}}\) *

(b) Way 4 M1: Valid attempt to make \(t\) the subject, followed by an attempt to find two \(\ln\) derivatives, condoning sign errors and constant errors.

A1: \(\dfrac{\mathrm{d}t}{\mathrm{d}N} = -4\left(\dfrac{-3}{900 - 3N} - \dfrac{7}{7N}\right)\) or equivalent

dM1: Forms a common denominator to combine their fractions

A1*: Correct algebra leading to \(\dfrac{\mathrm{d}N}{\mathrm{d}t} = \dfrac{N(300 - N)}{1200}\) *

Notes

B1: Deduces or shows that \(\dfrac{\mathrm{d}N}{\mathrm{d}t}\) is maximised when \(N = 150\)

M1: Uses the model \(N = \dfrac{900}{3 + 7\mathrm{e}^{-0.25t}}\) with their \(N = 150\) and proceeds as far as \(\mathrm{e}^{-0.25T} = k,\ k \gt 0\) or \(\mathrm{e}^{0.25T} = k,\ k \gt 0\). Condone \(t \equiv T\)

dM1: Correct method of using logarithms to find a value for \(T\). Condone \(t \equiv T\)

A1: see scheme

Note: \(\dfrac{\mathrm{d}^2N}{\mathrm{d}t^2} = \dfrac{\mathrm{d}N}{\mathrm{d}t}\left(\dfrac{300}{1200} - \dfrac{2N}{1200}\right) = 0 \Rightarrow N = 150\) is acceptable for B1

Note: Ignore units for \(T\)

Note: Applying \(300 = \dfrac{900}{3 + 7\mathrm{e}^{-0.25t}} \Rightarrow t = \ldots\) or \(0 = \dfrac{900}{3 + 7\mathrm{e}^{-0.25t}} \Rightarrow t = \ldots\) is M0 dM0 A0

Note: M1 dM1 can only be gained in (c) by using an \(N\) value in the range \(90 \lt N \lt 300\)

Notes

B1: \(300\) (or accept \(299\))