FP2 June 2017 Q4
4. \[y = \ln\left(\frac{1}{1 - 2x}\right), \qquad |x| \lt \frac{1}{2}\]
(a) Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) and \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\) (4)
(b) Hence, or otherwise, find the series expansion of \(\ln\left(\dfrac{1}{1 - 2x}\right)\) about \(x = 0\), in ascending powers of \(x\), up to and including the term in \(x^3\). Give each coefficient in its simplest form. (3)
(c) Use your expansion to find an approximate value for \(\ln\left(\dfrac{3}{2}\right)\), giving your answer to 3 decimal places. (3)
| Scheme | Marks |
|---|---|
| \(y = \ln\left(\dfrac{1}{1 - 2x}\right)\) | |
| \(y = \ln(1 - 2x)^{-1} = (\ln 1) - \ln(1 - 2x)\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{1 - 2x}\times-2\ \left(= \dfrac{2}{1 - 2x}\right)\) M1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-1}{(1 - 2x)}\times\dfrac{\mathrm{d}(1 - 2x)}{\mathrm{d}x}\) Must use chain rule ie \(\dfrac{k}{1 - 2x}\) with \(k \neq \pm1\) needed. Minus sign may be missing. A1: Correct derivative | M1A1 |
| OR \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = (1 - 2x)\times-(1 - 2x)^{-2}\times-2\ \left(= \dfrac{2}{1 - 2x}\right)\) M1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{(1 - 2x)^{-1}}\times\dfrac{\mathrm{d}(1 - 2x)^{-1}}{\mathrm{d}x}\) Must use chain rule. Minus sign may be missing. A1: Correct derivative | M1A1 |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -2\times(1 - 2x)^{-2}\times-2\ \left(= \dfrac{4}{(1 - 2x)^2}\right)\) Correct second derivative obtained from a correct first derivative. | A1 |
| \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = -8\times(1 - 2x)^{-3}\times-2\ \left(= \dfrac{16}{(1 - 2x)^3}\right)\) Correct third derivative obtained from correct first and second derivatives | A1 |
| (4) |
Notes
Alternative by use of exponentials and implicit differentiation
| Scheme | Marks |
|---|---|
| \(y = \ln\left(\dfrac{1}{1 - 2x}\right) \Rightarrow \mathrm{e}^y = \dfrac{1}{1 - 2x} = (1 - 2x)^{-1}\) | |
| \(\mathrm{e}^y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2(1 - 2x)^{-2}\) Differentiates using implicit differentiation and chain rule. | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\mathrm{e}^{-y}(1 - 2x)^{-2}\) or \(\dfrac{2}{(1 - 2x)}\) Correct derivative in either form. Equivalents accepted. | A1 |
If \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2}{(1 - 2x)}\) has been used from here, see main scheme for second and third derivatives
| Scheme | Marks |
|---|---|
| \((y_0 = 0),\ y'_0 = 2,\ y''_0 = 4,\ y'''_0 = 16\) Attempt values at \(x = 0\) using their derivatives from (a) \(y_0 = 0\) need not be seen but other 3 values must be attempted. | M1 |
| \((y =)\,(0) + 2x + \dfrac{4x^2}{2!} + \dfrac{16x^3}{3!}\) Uses their values in the correct Maclaurin series. Must see \(x^3\) term Can be implied by a final series which is correct for their values. 2!,3! or 2 and 6 | M1 |
| \(y = 2x + 2x^2 + \dfrac{8}{3}x^3\) Correct expression. Must start \(y = \ldots\) or \(\ln\left(\dfrac{1}{1 - 2x}\right) = \ldots\) \(\mathrm{f}(x) = \ldots\) allowed only if \(\mathrm{f}(x)\) is defined to be one of these. | A1cao |
| (3) |
Notes
Alternative (b)
| Scheme | Marks |
|---|---|
| \(y = \ln\left(\dfrac{1}{1 - 2x}\right) = -\ln(1 - 2x)\) Log power law applied correctly | M1 |
| \(= -\left((-2x) - \dfrac{(-2x)^2}{2} + \dfrac{(-2x)^3}{3}\right)\) Replaces \(x\) with -2\(x\) in the expansion for \(\ln(1 + x)\) (in formula book) | M1 |
| \(y = 2x + 2x^2 + \dfrac{8}{3}x^3\) Correct expression | A1cao |
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{1 - 2x} = \dfrac{3}{2} \Rightarrow x = \dfrac{1}{6}\) Correct value for \(x\), seen explicitly or substituted in their expansion | B1 |
| \(\ln\left(\dfrac{3}{2}\right) \approx 2\left(\dfrac{1}{6}\right) + 2\left(\dfrac{1}{6}\right)^2 + \dfrac{8}{3}\left(\dfrac{1}{6}\right)^3\) Substitute their value of \(x\) into their expansion. May need to check this is correct for their expansion and their \(x\). (Calculator value for \(\ln\left(\dfrac{3}{2}\right)\) is 0.405) | M1 |
| \(= 0.401\) Must come from correct work | A1cso |
| NB: \(\ln 3 - \ln 2\) or \(\ln 3 + \ln\left(\dfrac{1}{2}\right)\) scores 0/3 as \(|x|\) must be \(\lt \dfrac{1}{2}\) | |
| Answer with no working scores 0/3 | |
| (3) | |
| (10 marks) |