C4 June 2011 Q1
1.\[\frac{9x^2}{(x-1)^2(2x+1)} = \frac{A}{(x-1)} + \frac{B}{(x-1)^2} + \frac{C}{(2x+1)}\]
Find the values of the constants \(A\), \(B\) and \(C\). (4)
| Scheme | Marks |
|---|---|
| \(9x^2 = A(x-1)(2x+1) + B(2x+1) + C(x-1)^2\) | B1 |
| \(x \to 1 \qquad 9 = 3B \Rightarrow B = 3\) | M1 |
| \(x \to -\dfrac{1}{2} \qquad \dfrac{9}{4} = \left(-\dfrac{3}{2}\right)^2 C \Rightarrow C = 1\) Any two of \(A\), \(B\), \(C\) | A1 |
| \(x^2\) terms \(\qquad 9 = 2A + C \Rightarrow A = 4\) All three correct | A1 |
| (4) | |
| (4 marks) |
Notes
Alternatives for finding \(A\).
\(x\) terms \(\qquad 0 = -A + 2B - 2C \Rightarrow A = 4\)
Constant terms \(\qquad 0 = -A + B + C \Rightarrow A = 4\)