C4 January 2012 Q8

Notes

M1: Forming a correct identity. For example, \(1 = A(5 - P) + BP\). Note \(A\) and \(B\) not referred to in question.

A1: Either one of \(A = \dfrac{1}{5}\) or \(B = \dfrac{1}{5}\).

A1: \(\dfrac{\frac{1}{5}}{P} + \dfrac{\frac{1}{5}}{(5 - P)}\) or any equivalent form, eg: \(\dfrac{1}{5P} + \dfrac{1}{25 - 5P}\), etc. Ignore subsequent working.
This answer must be stated in part (a) only.
A1 can also be given for a candidate who finds both \(A = \dfrac{1}{5}\) and \(B = \dfrac{1}{5}\) and \(\dfrac{A}{P} + \dfrac{B}{5 - P}\) is seen in their working.

Candidate can use ‘cover-up’ rule to write down \(\dfrac{\frac{1}{5}}{P} + \dfrac{\frac{1}{5}}{(5 - P)}\), as so gain all three marks.
Candidate cannot gain the marks for part (a) in part (b).

Notes

B1: Separates variables as shown. \(\mathrm{d}P\) and \(\mathrm{d}t\) should be in the correct positions, though this mark can be implied by later working. Ignore the integral signs.

M1*: Both \(\pm\lambda\ln P\) and \(\pm\mu\ln(\pm 5 \pm P)\), where \(\lambda\) and \(\mu\) are constants.
Or \(\pm\lambda\ln mP\) and \(\pm\mu\ln(n(\pm 5 \pm P))\), where \(\lambda\), \(\mu\), \(m\) and \(n\) are constants.

A1ft: Correct follow through integration of both sides from their \(\displaystyle\int \frac{\lambda}{P} + \frac{\mu}{(5 - P)}\,\mathrm{d}P = \int K\,\mathrm{d}t\) with or without \(+\,c\)

dM1*: Use of \(t = 0\) and \(P = 1\) in an integrated equation containing \(c\)
dM1*: Using ANY of the subtraction (or addition) laws for logarithms CORRECTLY.
dM1*: Apply logarithms (or take exponentials) to eliminate ln’s CORRECTLY from their equation.
dM1*: A full ACCEPTABLE method of rearranging to make \(P\) the subject. (See below for examples!)

A1: \(P = \dfrac{5}{(1 + 4\mathrm{e}^{-\frac{1}{3}t})}\ \{\text{where } a = 5, b = 1, c = 4\}\).
Also allow any “integer” multiples of this expression. For example: \(P = \dfrac{25}{(5 + 20\mathrm{e}^{-\frac{1}{3}t})}\)

Note: If the first method mark (M1*) is not awarded then the candidate cannot gain any of the six remaining marks for this part of the question.

Note: \(\displaystyle\int \frac{1}{P(5 - P)}\,\mathrm{d}P = \int 15\,\mathrm{d}t \Rightarrow \int \frac{\frac{1}{5}}{P} + \frac{\frac{1}{5}}{(5 - P)}\,\mathrm{d}P = \int 15\,\mathrm{d}t \Rightarrow \ln P - \ln(5 - P) = 15t\) is B0M1A1ft.

dM1* for making P the subject
Note there are three type of manipulations here which are considered acceptable to make \(P\) the subject.

(1) M1 for \(\dfrac{P}{5 - P} = \mathrm{e}^{\frac{1}{3}t} \Rightarrow P = 5\mathrm{e}^{\frac{1}{3}t} - P\mathrm{e}^{\frac{1}{3}t} \Rightarrow P(1 + \mathrm{e}^{\frac{1}{3}t}) = 5\mathrm{e}^{\frac{1}{3}t} \Rightarrow P = \dfrac{5}{(1 + \mathrm{e}^{-\frac{1}{3}t})}\)

(2) M1 for \(\dfrac{P}{5 - P} = \mathrm{e}^{\frac{1}{3}t} \Rightarrow \dfrac{5 - P}{P} = \mathrm{e}^{\frac{1}{3}t} \Rightarrow \dfrac{5}{P} - 1 = \mathrm{e}^{\frac{1}{3}t} \Rightarrow \dfrac{5}{P} = \mathrm{e}^{\frac{1}{3}t} + 1 \Rightarrow P = \dfrac{5}{(1 + \mathrm{e}^{\frac{1}{3}t})}\)

(3) M1 for \(P(5 - P) = 4\mathrm{e}^{\frac{1}{3}t} \Rightarrow P^2 - 5P = -4\mathrm{e}^{\frac{1}{3}t} \Rightarrow \left(P - \dfrac{5}{2}\right)^2 - \dfrac{25}{4} = -4\mathrm{e}^{\frac{1}{3}t}\) leading to \(P = \ldots\)

Note: The incorrect manipulation of \(\dfrac{P}{5 - P} = \dfrac{P}{5} - 1\) or equivalent is awarded this dM0*.
Note: \((P) - (5 - P) = \mathrm{e}^{\frac{1}{3}t} \Rightarrow 2P - 5 = \dfrac{1}{3}t\) leading to \(P = \ldots\) or equivalent is awarded this dM0*

Alternative method for part (b)

then award the final M1A1 in the same way.

Notes

B1: \(1 + 4\mathrm{e}^{-\frac{1}{3}t} \gt 1\) and \(P \lt 5\) and a conclusion relating population (or even \(P\)) or meerkats to 5000.
For \(P = \dfrac{25}{(5 + 20\mathrm{e}^{-\frac{1}{3}t})}\), B1 can be awarded for \(5 + 20\mathrm{e}^{-\frac{1}{3}t} \gt 5\) and \(P \lt 5\) and a conclusion relating population (or even \(P\)) or meerkats to 5000.
B1 can only be obtained if candidates have correct values of \(a\) and \(b\) in their \(P = \dfrac{a}{(b + c\mathrm{e}^{-\frac{1}{3}t})}\).

Award B0 for: As \(t \to \infty\), \(\mathrm{e}^{-\frac{1}{3}t} \to 0\). So \(P \to \dfrac{5}{(1 + 0)} = 5\), so population cannot exceed 5000,
unless the candidate also proves that \(P = \dfrac{5}{(1 + 4\mathrm{e}^{-\frac{1}{3}t})}\) oe. is an increasing function.
If unsure here, then send to review!