C4 January 2011 Q5
5.
(a) Use the binomial theorem to expand \[(2 - 3x)^{-2}, \qquad |x| < \frac{2}{3},\] in ascending powers of \(x\), up to and including the term in \(x^3\). Give each coefficient as a simplified fraction. (5)
\[\mathrm{f}(x) = \frac{a + bx}{(2 - 3x)^2}, \qquad |x| < \frac{2}{3}, \quad \text{where } a \text{ and } b \text{ are constants.}\]
In the binomial expansion of \(\mathrm{f}(x)\), in ascending powers of \(x\), the coefficient of \(x\) is 0 and the coefficient of \(x^2\) is \(\dfrac{9}{16}\). Find
(b) the value of \(a\) and the value of \(b\), (5)
(c) the coefficient of \(x^3\), giving your answer as a simplified fraction. (3)
| Scheme | Marks |
|---|---|
| \((2 - 3x)^{-2} = 2^{-2}\left(1 - \dfrac{3}{2}x\right)^{-2}\) | B1 |
| \(\left(1 - \dfrac{3}{2}x\right)^{-2} = 1 + (-2)\left(-\dfrac{3}{2}x\right) + \dfrac{-2.-3}{1.2}\left(-\dfrac{3}{2}x\right)^2 + \dfrac{-2.-3.-4}{1.2.3}\left(-\dfrac{3}{2}x\right)^3 + \ldots\) | M1 A1 |
| \(= 1 + 3x + \dfrac{27}{4}x^2 + \dfrac{27}{2}x^3 + \ldots\) | |
| \((2 - 3x)^{-2} = \dfrac{1}{4} + \dfrac{3}{4}x + \dfrac{27}{16}x^2 + \dfrac{27}{8}x^3 + \ldots\) | M1 A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = (a + bx)\left(\dfrac{1}{4} + \dfrac{3}{4}x + \dfrac{27}{16}x^2 + \dfrac{27}{8}x^3 + \ldots\right)\) | |
| Coefficient of \(x\); \(\dfrac{3a}{4} + \dfrac{b}{4} = 0\quad (3a + b = 0)\) | M1 |
| Coefficient of \(x^2\); \(\dfrac{27a}{16} + \dfrac{3b}{4} = \dfrac{9}{16}\quad (9a + 4b = 3)\) A1 either correct | M1 A1 |
| Leading to \(a = -1,\ b = 3\) | M1 A1 |
| (5) |
| Scheme | Marks |
|---|---|
| Coefficient of \(x^3\) is \(\dfrac{27a}{8} + \dfrac{27b}{16} = \dfrac{27}{8} \times (-1) + \dfrac{27}{16} \times 3\) | M1 A1ft |
| \(= \dfrac{27}{16}\) cao | A1 |
| (3) | |
| (13 marks) |