C4 January 2007 Q4
4.
\[(2x - 3)(x - 1)\frac{\mathrm{d}y}{\mathrm{d}x} = (2x - 1)\,y.\]
(5)| Scheme | Marks |
|---|---|
| \(\dfrac{2x - 1}{(x - 1)(2x - 3)} \equiv \dfrac{A}{(x - 1)} + \dfrac{B}{(2x - 3)}\) | |
| \(2x - 1 \equiv A(2x - 3) + B(x - 1)\) | M1 |
| Let \(x = \frac{3}{2}\), \(2 = B\left(\frac{1}{2}\right) \Rightarrow B = 4\) Let \(x = 1\), \(1 = A(-1) \Rightarrow A = -1\) | A1 A1 |
| giving \(\dfrac{-1}{(x - 1)} + \dfrac{4}{(2x - 3)}\) | |
| (3) |
Notes
M1 Forming this identity. NB: \(A\) & \(B\) are not assigned in this question
A1 either one of \(A = -1\) or \(B = 4\).
A1 both correct for their \(A\), \(B\).
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{\mathrm{d}y}{y} = \int \frac{(2x - 1)}{(2x - 3)(x - 1)}\,\mathrm{d}x\) | B1 |
| \(\displaystyle= \int \frac{-1}{(x - 1)} + \frac{4}{(2x - 3)}\,\mathrm{d}x\) | M1ft |
| \(\therefore \ln y = -\ln(x - 1) + 2\ln(2x - 3) + c\) | M1 A1ft A1 |
| (5) |
Notes
B1 Separates variables as shown. Can be implied
M1ft Replaces RHS with their partial fraction to be integrated.
M1 At least two terms in ln’s
A1ft At least two ln terms correct
A1 All three terms correct and ‘+ c’
Parts (b) and (c) are marked together: see (c) for the alternative ways.
| Scheme | Marks |
|---|---|
| \(y = 10,\ x = 2\) gives \(c = \ln 10\) | B1 |
| \(\therefore \ln y = -\ln(x - 1) + 2\ln(2x - 3) + \ln 10\) | |
| \(\ln y = -\ln(x - 1) + \ln(2x - 3)^2 + \ln 10\) | M1 |
| \(\ln y = \ln\left(\dfrac{(2x - 3)^2}{(x - 1)}\right) + \ln 10\) or \(\ln y = \ln\left(\dfrac{10(2x - 3)^2}{(x - 1)}\right)\) | M1 |
| \(\underline{y = \dfrac{10(2x - 3)^2}{(x - 1)}}\) | A1 aef |
| (4) | |
| (12 marks) |
Notes
B1 \(c = \ln 10\)
M1 Using the power law for logarithms
M1 Using the product and/or quotient laws for logarithms to obtain a single RHS logarithmic term with/without constant c.
A1 aef \(\underline{y = \frac{10(2x - 3)^2}{(x - 1)}}\) or aef. isw
Aliter (b) & (c) Way 2
| \(\displaystyle\int \frac{\mathrm{d}y}{y} = \int \frac{(2x - 1)}{(2x - 3)(x - 1)}\,\mathrm{d}x\) | B1 |
| \(\displaystyle= \int \frac{-1}{(x - 1)} + \frac{4}{(2x - 3)}\,\mathrm{d}x\) | M1ft |
| \(\therefore \ln y = -\ln(x - 1) + 2\ln(2x - 3) + c\) | M1 A1ft A1 |
| See below for the award of B1 decide to award B1 here!! | B1 |
| \(\ln y = -\ln(x - 1) + \ln(2x - 3)^2 + c\) | M1 |
| \(\ln y = \ln\left(\dfrac{(2x - 3)^2}{x - 1}\right) + c\) | M1 |
| \(\ln y = \ln\left(\dfrac{A(2x - 3)^2}{x - 1}\right)\) where \(c = \ln A\) or \(\mathrm{e}^{\ln y} = \mathrm{e}^{\ln\left(\frac{(2x - 3)^2}{x - 1}\right) + c} = \mathrm{e}^{\ln\left(\frac{(2x - 3)^2}{x - 1}\right)}\mathrm{e}^c\) \(y = \dfrac{A(2x - 3)^2}{(x - 1)}\) | |
| \(y = 10,\ x = 2\) gives \(A = 10\) (\(A = 10\) for B1) | award above |
| \(\underline{y = \dfrac{10(2x - 3)^2}{(x - 1)}}\) or aef & isw | A1 aef |
| [5] & [4] |
Note: The B1 mark (part (c)) should be awarded in the same place on ePEN as in the Way 1 approach.
Aliter (b) & (c) Way 3
| \(\displaystyle\int \frac{\mathrm{d}y}{y} = \int \frac{(2x - 1)}{(2x - 3)(x - 1)}\,\mathrm{d}x\) | B1 |
| \(\displaystyle= \int \frac{-1}{(x - 1)} + \frac{2}{\left(x - \frac{3}{2}\right)}\,\mathrm{d}x\) | M1ft |
| \(\therefore \ln y = -\ln(x - 1) + 2\ln\left(x - \tfrac{3}{2}\right) + c\) | M1 A1ft A1 |
| [5] | |
| \(y = 10,\ x = 2\) gives \(c = \underline{\ln 10 - 2\ln\left(\tfrac{1}{2}\right)} = \underline{\ln 40}\) | B1 oe |
| \(\therefore \ln y = -\ln(x - 1) + 2\ln\left(x - \tfrac{3}{2}\right) + \ln 40\) | |
| \(\ln y = -\ln(x - 1) + \ln\left(x - \tfrac{3}{2}\right)^2 + \ln 40\) | M1 |
| \(\ln y = \ln\left(\dfrac{\left(x - \frac{3}{2}\right)^2}{(x - 1)}\right) + \ln 40\) or \(\ln y = \ln\left(\dfrac{40\left(x - \frac{3}{2}\right)^2}{(x - 1)}\right)\) | M1 |
| \(\underline{y = \dfrac{40\left(x - \frac{3}{2}\right)^2}{(x - 1)}}\) or aef. isw | A1 aef |
| [4] |
In Way 3: B1 oe for \(c = \ln 10 - 2\ln\left(\frac{1}{2}\right)\) or \(c = \ln 40\).
CHECK (corrected from the printed mark scheme: in Way 3 the power-law line is printed ending \(+ \ln 10\); with \(c = \ln 40\) it ends \(+ \ln 40\).)
Note: Please mark parts (b) and (c) together for any of the three ways.