C4 June 2012 Q1
1.\[\mathrm{f}(x) = \frac{1}{x(3x - 1)^2} = \frac{A}{x} + \frac{B}{(3x - 1)} + \frac{C}{(3x - 1)^2}\]
(a) Find the values of the constants \(A\), \(B\) and \(C\). (4)
(b)
(i) Hence find \(\displaystyle\int \mathrm{f}(x)\,\mathrm{d}x\).
(ii) Find \(\displaystyle\int_1^2 \mathrm{f}(x)\,\mathrm{d}x\), leaving your answer in the form \(a + \ln b\), where \(a\) and \(b\) are constants. (6)
| Scheme | Marks |
|---|---|
| \(1 = A(3x - 1)^2 + Bx(3x - 1) + Cx\) | B1 |
| \(x \to 0 \qquad (1 = A)\) | M1 |
| \(x \to \tfrac{1}{3} \qquad 1 = \tfrac{1}{3}C \Rightarrow C = 3\) any two constants correct | A1 |
| Coefficients of \(x^2\) \(0 = 9A + 3B \Rightarrow B = -3\) all three constants correct | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| (i) \(\displaystyle\int \left(\frac{1}{x} - \frac{3}{3x - 1} + \frac{3}{(3x - 1)^2}\right)\mathrm{d}x\) \(= \ln x - \dfrac{3}{3}\ln(3x - 1) + \dfrac{3}{(-1)3}(3x - 1)^{-1} \quad (+C)\) \(\left(= \ln x - \ln(3x - 1) - \dfrac{1}{3x - 1} \quad (+C)\right)\) | M1 A1ft A1ft |
| (ii) \(\displaystyle\int_1^2 \mathrm{f}(x)\,\mathrm{d}x = \left[\ln x - \ln(3x - 1) - \frac{1}{3x - 1}\right]_1^2\) \(= \left(\ln 2 - \ln 5 - \dfrac{1}{5}\right) - \left(\ln 1 - \ln 2 - \dfrac{1}{2}\right)\) | M1 |
| \(= \ln\dfrac{2 \times 2}{5} + \ldots\) | M1 |
| \(= \dfrac{3}{10} + \ln\left(\dfrac{4}{5}\right)\) | A1 |
| (6) | |
| (10 marks) |
Notes
In the printed scheme a bracket joins these method marks: each later M mark is dependent on the M mark before it.