C4 June 2015 Q7

Notes

M1: Forming a correct identity. For example, \(2 \equiv A(P - 2) + BP\) from \(\dfrac{2}{P(P - 2)} = \dfrac{A}{P} + \dfrac{B}{(P - 2)}\)

Note: \(A\) and \(B\) are not referred to in question.

A1: Either one of \(A = -1\) or \(B = 1\).

A1: \(\dfrac{1}{(P - 2)} - \dfrac{1}{P}\) or any equivalent form. This answer cannot be recovered from part (b).

Note: M1A1A1 can also be given for a candidate who finds both \(A = -1\) and \(B = 1\) and \(\dfrac{A}{P} + \dfrac{B}{(P - 2)}\) is seen in their working.

Note: Candidates can use ‘cover-up’ rule to write down \(\dfrac{1}{(P - 2)} - \dfrac{1}{P}\), so as to gain all three marks.

Note: Equating coefficients from \(2 \equiv A(P - 2) + BP\) gives \(A + B = 2, -2A = 2 \Rightarrow A = -1, B = 1\)

Notes

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

Note: Eg: \(\displaystyle\int \dfrac{2}{P^2 - 2P}\,\mathrm{d}P = \displaystyle\int \cos 2t\,\mathrm{d}t\) or \(\displaystyle\int \dfrac{1}{P(P - 2)}\,\mathrm{d}P = \dfrac{1}{2}\displaystyle\int \cos 2t\,\mathrm{d}t\) o.e. are also fine for B1.

1^st M1: \(\pm\lambda\ln(P - 2) \pm \mu\ln P,\ \lambda \neq 0, \mu \neq 0\). Also allow \(\pm\lambda\ln(M(P - 2)) \pm \mu\ln NP\); \(M, N\) can be 1.

Note: Condone \(2\ln(P - 2) + 2\ln P\) or \(2\ln(P(P - 2))\) or \(2\ln(P^2 - 2P)\) or \(\ln(P^2 - 2P)\)

1^st A1: Correct result of \(\ln(P - 2) - \ln P = \dfrac{1}{2}\sin 2t\) or \(2\ln(P - 2) - 2\ln P = \sin 2t\) o.e. with or without \(+c\)

2^nd M1: Some evidence of using both \(t = 0\) and \(P = 3\) in an integrated equation containing a constant of integration. Eg: \(c\) or \(A\), etc.

3^rd M1: Starting from an equation of the form \(\pm\lambda\ln(P - \beta) \pm \mu\ln P = \pm K\sin\delta t + c\), \(\lambda, \mu, \beta, K, \delta \neq 0\), applies a fully correct method to eliminate their logarithms.

4^th M1: dependent on the third method mark being awarded.
A complete method of rearranging to make \(P\) the subject. Condone sign slips or constant errors.

Note: For the 3^rd M1 and 4^th M1 marks, a candidate needs to have included a constant of integration, in their working. eg. \(c\), \(A\), \(\ln A\) or an evaluated constant of integration.

2^nd A1: Correct proof of \(P = \dfrac{6}{(3 - \mathrm{e}^{\frac{1}{2}\sin 2t})}\). Note: This answer is given in the question.

Note: \(\ln\left(\dfrac{(P - 2)}{P}\right) = \dfrac{1}{2}\sin 2t + c\) followed by \(\dfrac{(P - 2)}{P} = \mathrm{e}^{\frac{1}{2}\sin 2t} + \mathrm{e}^{c}\) is 3^rd M0, 4^th M0, 2^nd A0.

Note: \(\ln\left(\dfrac{(P - 2)}{P}\right) = \dfrac{1}{2}\sin 2t + c \to \dfrac{(P - 2)}{P} = \mathrm{e}^{\frac{1}{2}\sin 2t + c} \to \dfrac{(P - 2)}{P} = \mathrm{e}^{\frac{1}{2}\sin 2t} + \mathrm{e}^{c}\) is final M1M0A0

4^th M1 for making \(P\) the subject

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

(1) M1 for \(\dfrac{3(P - 2)}{P} = \mathrm{e}^{\frac{1}{2}\sin 2t} \Rightarrow 3(P - 2) = P\mathrm{e}^{\frac{1}{2}\sin 2t} \Rightarrow 3P - 6 = P\mathrm{e}^{\frac{1}{2}\sin 2t} \Rightarrow P(3 - \mathrm{e}^{\frac{1}{2}\sin 2t}) = 6\)
\(\Rightarrow P = \dfrac{6}{(3 - \mathrm{e}^{\frac{1}{2}\sin 2t})}\)

(2) M1 for \(\dfrac{3(P - 2)}{P} = \mathrm{e}^{\frac{1}{2}\sin 2t} \Rightarrow 3 - \dfrac{6}{P} = \mathrm{e}^{\frac{1}{2}\sin 2t} \Rightarrow 3 - \mathrm{e}^{\frac{1}{2}\sin 2t} = \dfrac{6}{P} \Rightarrow\Rightarrow P = \dfrac{6}{(3 - \mathrm{e}^{\frac{1}{2}\sin 2t})}\)

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

Method 2 for Q7(b)

Notes

M1: States \(P = 4\) or applies \(P = 4\)

M1: Obtains \(\pm\lambda\sin 2t = \ln k\) or \(\pm\lambda\sin t = \ln k\), where \(\lambda\) and \(k\) are numerical values and \(\lambda\) can be 1

A1: anything that rounds to 0.473. (Do not apply isw here)

Note: Do not apply ignore subsequent working for A1. (Eg: 0.473 followed by 473 years is A0.)

Note: Use of \(P = 4000\): Without the mention of \(P = 4\), \(\dfrac{1}{2}\sin 2t = \ln 2.9985\) or \(\sin 2t = 2\ln 2.9985\) or \(\sin 2t = 2.1912\ldots\) will usually imply M0M1A0

Note: Use of Degrees: \(t\) = awrt 27.1 will usually imply M1M1A0