June 2018 Paper 1 Q7
7. Given that \(k \in \mathbb{Z}^{+}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \frac{2}{(3x - k)}\,\mathrm{d}x = \frac{2}{3}\ln(3x - k)\) | M1 A1 | 1.1a 1.1b |
| \(\displaystyle\int_k^{3k} \frac{2}{(3x - k)}\,\mathrm{d}x = \frac{2}{3}\ln(9k - k) - \frac{2}{3}\ln(3k - k)\) | dM1 | 1.1b |
| \(= \dfrac{2}{3}\ln\left(\dfrac{8\cancel{k}}{2\cancel{k}}\right) = \dfrac{2}{3}\ln 4\) oe | A1 | 2.1 |
| (4) |
Notes
M1: \(\displaystyle\int \frac{2}{(3x - k)}\,\mathrm{d}x = A\ln(3x - k)\) Condone a missing bracket
A1: \(\displaystyle\int \frac{2}{(3x - k)}\,\mathrm{d}x = \dfrac{2}{3}\ln(3x - k)\)
Allow recovery from a missing bracket if in subsequent work \(A\ln 9k - k \rightarrow A\ln 8k\)
dM1: For substituting \(k\) and \(3k\) into their \(A\ln(3x - k)\) and subtracting either way around
A1: Uses correct ln work and notation to show that I \(= \dfrac{2}{3}\ln\left(\dfrac{8}{2}\right)\) or \(\dfrac{2}{3}\ln 4\) oe (ie independent of \(k\))
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \frac{2}{(2x - k)^2}\,\mathrm{d}x = -\frac{1}{(2x - k)}\) | M1 | 1.1b |
| \(\displaystyle\int_k^{2k} \frac{2}{(2x - k)^2}\,\mathrm{d}x = -\frac{1}{(4k - k)} + \frac{1}{(2k - k)}\) | dM1 | 1.1b |
| \(= \dfrac{2}{3k} \quad \left(\propto \dfrac{1}{k}\right)\) | A1 | 2.1 |
| (3) | ||
| (7 marks) |
Notes
M1: \(\displaystyle\int \frac{2}{(2x - k)^2}\,\mathrm{d}x = \dfrac{C}{(2x - k)}\)
dM1: For substituting \(k\) and \(2k\) into their \(\dfrac{C}{(2x - k)}\) and subtracting
A1: Shows that it is inversely proportional to \(k\) Eg proceeds to the answer is of the form \(A/k\) with \(A = 2/3\)
There is no need to perform the whole calculation. Accept from \(-\dfrac{1}{(3k)} + \dfrac{1}{(k)} = \left(-\dfrac{1}{3} + 1\right) \times \dfrac{1}{k} \propto \dfrac{1}{k}\)
If the calculation is performed it must be correct.
Do not isw here. They should know when they have an expression that is inversely proportional to \(k\).
You may see substitution used but the mark is scored for the same result. See below
\(u = 2x - k \rightarrow \left[\dfrac{C}{u}\right]\) for M1 with limits \(3k\) and \(k\) used for dM1