C4 June 2013 (R) Q1
1. Express in partial fractions \[\frac{5x + 3}{(2x + 1)(x + 1)^2}\] (4)
| Scheme | Marks |
|---|---|
| \(\dfrac{5x + 3}{(2x + 1)(x + 1)^2} \equiv \dfrac{A}{(2x + 1)} + \dfrac{B}{(x + 1)} + \dfrac{C}{(x + 1)^2}\) At least one of “\(A\)” or “\(C\)” are correct. | B1 |
| \(A = 2,\quad C = 2\) Breaks up their partial fraction correctly into three terms and both "\(A\)" \(= 2\) and "\(C\)" \(= 2\). | B1 cso |
| \(5x + 3 \equiv A(x + 1)^2 + B(2x + 1)(x + 1) + C(2x + 1)\) \(x = -1 \Rightarrow -2 = -C \Rightarrow C = 2\) \(x = -\dfrac{1}{2} \Rightarrow -\dfrac{5}{2} + 3 = \dfrac{1}{4}A \Rightarrow \dfrac{1}{2} = \dfrac{1}{4}A \Rightarrow A = 2\) Writes down a correct identity and attempts to find the value of either one “\(A\)” or “\(B\)” or “\(C\)”. | M1 |
| Either \(x^2: 0 = A + 2B\), constant: \(3 = A + B + C\) \(x: 5 = 2A + 3B + 2C\) leading to \(B = -1\) Correct value for “\(B\)” which is found using a correct identity and follows from their partial fraction decomposition. | A1 cso |
| So, \(\dfrac{5x + 3}{(2x + 1)(x + 1)^2} \equiv \dfrac{2}{(2x + 1)} - \dfrac{1}{(x + 1)} + \dfrac{2}{(x + 1)^2}\) | |
| (4) | |
| (4 marks) |
Notes
BE CAREFUL!: Candidates will assign their own “\(A\), \(B\) and \(C\)” for this question.
B1: At least one of “\(A\)” or “\(C\)” are correct.
B1: Breaks up their partial fraction correctly into three terms and both "\(A\)" \(= 2\) and "\(C\)" \(= 2\).
M1: Writes down a correct identity (although this can be implied) and attempts to find the value of either one of “\(A\)” or “\(B\)” or “\(C\)”.
This can be achieved by either substituting values into their identity or comparing coefficients and solving the resulting equations simultaneously.
A1: Correct value for “\(B\)” which is found using a correct identity and follows from their partial fraction decomposition.
Note: If a candidate does not give partial fraction decomposition then:
- the 2nd B1 mark can follow from a correct identity.
- the final A1 mark can be awarded for a correct “\(B\)” if a candidate goes writes out their partial fractions at the end.
Note: The correct partial fraction from no working scores B1B1M1A1.
Note: A number of candidates will start this problem by writing out the correct identity and then attempt to find “\(A\)” or “\(B\)” or “\(C\)”. Therefore the B1 marks can be awarded from this method.