C4 January 2006 Q5
5.
\[\mathrm{f}(x) = \frac{3x^2 + 16}{(1 - 3x)(2 + x)^2} = \frac{A}{(1 - 3x)} + \frac{B}{(2 + x)} + \frac{C}{(2 + x)^2}, \qquad |x| \lt \tfrac{1}{3}.\]
(a) Find the values of \(A\) and \(C\) and show that \(B = 0\). (4)
(b) Hence, or otherwise, find the series expansion of \(\mathrm{f}(x)\), in ascending powers of \(x\), up to and including the term in \(x^3\). Simplify each term. (7)
| Scheme | Marks |
|---|---|
| Considers \(3x^2 + 16 = A(2 + x)^2 + B(1 - 3x)(2 + x) + C(1 - 3x)\) and substitutes \(x = -2\), or \(x = 1/3\), or compares coefficients and solves simultaneous equations | M1 |
| To obtain \(A = 3\), and \(C = 4\) | A1, A1 |
| Compares coefficients or uses simultaneous equation to show \(B = 0\). | B1 |
| (4) |
| Scheme | Marks |
|---|---|
| Writes \(3(1 - 3x)^{-1} + 4(2 + x)^{-2}\) | M1 |
| \(= 3(1 + 3x, + 9x^2 + 27x^3 + \ldots\ldots) +\) | (M1, A1) |
| \(\dfrac{4}{4}\left(1 + \dfrac{(-2)}{1}\left(\dfrac{x}{2}\right) + \dfrac{(-2)(-3)}{1.2}\left(\dfrac{x}{2}\right)^2 + \dfrac{(-2)(-3)(-4)}{1.2.3}\left(\dfrac{x}{2}\right)^3 + \ldots\right)\) | (M1 A1) |
| \(= 4 + 8x, + 27\tfrac{3}{4}x^2 + 80\tfrac{1}{2}x^3 + \ldots\) | A1, A1 |
| (7) | |
| (11 marks) |
Or
| uses \((3x^2 + 16)(1 - 3x)^{-1}(2 + x)^{-2}\) | M1 |
| \((3x^2 + 16)\,(1 + 3x, + 9x^2 + 27x^3 +) \times\) | (M1A1)× |
| \(\tfrac{1}{4}\left(1 + \dfrac{(-2)}{1}\left(\dfrac{x}{2}\right) + \dfrac{(-2)(-3)}{1.2}\left(\dfrac{x}{2}\right)^2 + \dfrac{(-2)(-3)(-4)}{1.2.3}\left(\dfrac{x}{2}\right)^3\right)\) | (M1A1) |
| \(= 4 + 8x, + 27\tfrac{3}{4}x^2 + 80\tfrac{1}{2}x^3 + \ldots\) | A1, A1 |
| (7) |