C4 June 2007 Q4
4. \[\frac{2(4x^2+1)}{(2x+1)(2x-1)} \equiv A + \frac{B}{(2x+1)} + \frac{C}{(2x-1)}.\]
Way 1: A method of long division gives,
| Scheme | Marks |
|---|---|
| \(\dfrac{2(4x^2+1)}{(2x+1)(2x-1)} \equiv 2 + \dfrac{4}{(2x+1)(2x-1)}\) | B1 |
| \(\dfrac{4}{(2x+1)(2x-1)} \equiv \dfrac{B}{(2x+1)} + \dfrac{C}{(2x-1)}\) | |
| \(4 \equiv B(2x-1) + C(2x+1)\) or their remainder, \(Dx + E \equiv B(2x-1) + C(2x+1)\) | M1 |
| Let \(x = -\tfrac{1}{2}\), \(4 = -2B \Rightarrow B = -2\) | |
| Let \(x = \tfrac{1}{2}\), \(4 = 2C \Rightarrow C = 2\) | A1 A1 |
| (4) |
Notes
B1: \(A = 2\)
M1: Forming any one of these two identities. Can be implied.
A1: either one of \(B = -2\) or \(C = 2\) (see note below). A1: both \(B\) and \(C\) correct
If a candidate states one of either \(B\) or \(C\) correctly then the method mark M1 can be implied.
Aliter 4. (a) Way 2
| Scheme | Marks |
|---|---|
| \(\dfrac{2(4x^2+1)}{(2x+1)(2x-1)} \equiv A + \dfrac{B}{(2x+1)} + \dfrac{C}{(2x-1)}\) | |
| See below for the award of B1 | B1 |
| \(2(4x^2+1) \equiv A(2x+1)(2x-1) + B(2x-1) + C(2x+1)\) | M1 |
| Equate \(x^2\), \(8 = 4A \Rightarrow A = 2\) | |
| Let \(x = -\tfrac{1}{2}\), \(4 = -2B \Rightarrow B = -2\) | |
| Let \(x = \tfrac{1}{2}\), \(4 = 2C \Rightarrow C = 2\) | A1 A1 |
| (4) |
B1: decide to award B1 here!! … … for \(A = 2\)
M1: Forming this identity. Can be implied. If a candidate states one of either \(B\) or \(C\) correctly then the method mark M1 can be implied.
A1: either one of \(B = -2\) or \(C = 2\) (see note below). A1: both \(B\) and \(C\) correct
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{2(4x^2+1)}{(2x+1)(2x-1)}\,\mathrm{d}x = \int 2 - \frac{2}{(2x+1)} + \frac{2}{(2x-1)}\,\mathrm{d}x\) | |
| \(= 2x - \tfrac{2}{2}\ln(2x+1) + \tfrac{2}{2}\ln(2x-1)\ \ (+c)\) | M1* B1ft A1 cso & aef |
| \(\displaystyle\int_1^2 \frac{2(4x^2+1)}{(2x+1)(2x-1)}\,\mathrm{d}x = \Big[2x - \ln(2x+1) + \ln(2x-1)\Big]_1^2\) | |
| \(= (4 - \ln 5 + \ln 3) - (2 - \ln 3 + \ln 1)\) | depM1* |
| \(= 2 + \ln 3 + \ln 3 - \ln 5\) | |
| \(= 2 + \ln\left(\dfrac{3(3)}{5}\right)\) | M1 |
| \(= 2 + \ln\left(\dfrac{9}{5}\right)\) | A1 |
| (6) | |
| (10 marks) |
Notes
M1*: Either \(p\ln(2x+1)\) or \(q\ln(2x-1)\) or either \(p\ln 2x+1\) or \(q\ln 2x-1\)
B1ft: \(A \to Ax\)
A1 cso & aef: \(-\tfrac{2}{2}\ln(2x+1) + \tfrac{2}{2}\ln(2x-1)\) or \(-\ln(2x+1) + \ln(2x-1)\). See note below.
Some candidates may find rational values for \(B\) and \(C\). They may combine the denominator of their \(B\) or \(C\) with \((2x+1)\) or \((2x-1)\). Hence: Either \(\tfrac{a}{b(2x-1)} \to k\ln(b(2x-1))\) or \(\tfrac{a}{b(2x+1)} \to k\ln(b(2x+1))\) is okay for M1.
Candidates are not allowed to fluke \(-\ln(2x+1) + \ln(2x-1)\) for A1. Hence cso. If they do fluke this, however, they can gain the final A1 mark for this part of the question.
depM1*: Substitutes limits of 2 and 1 and subtracts the correct way round. (Invisible brackets okay.)
M1: Use of correct product (or power) and/or quotient laws for logarithms to obtain a single logarithmic term for their numerical expression. To award this M1 mark, the candidate must use the appropriate law(s) of logarithms for their ln terms to give a one single logarithmic term. Any error in applying the laws of logarithms would then earn M0. Note: This is not a dependent method mark.
A1: \(2 + \ln\left(\tfrac{9}{5}\right)\) Or \(2 - \ln\left(\tfrac{5}{9}\right)\) and k stated as \(\tfrac{9}{5}\).