A2 October 2021 Q6
6.
\[\left[\begin{gathered}\textit{The Taylor series expansion of}\;\; \mathrm{f}(x)\;\; \textit{about}\;\; x = a\;\; \textit{is given by}\\ \mathrm{f}(x) = \mathrm{f}(a) + (x - a)\mathrm{f}^{\prime}(a) + \frac{(x - a)^2}{2!}\mathrm{f}^{\prime\prime}(a) + \ldots + \frac{(x - a)^r}{r!}\mathrm{f}^{(r)}(a) + \ldots\end{gathered}\right]\]Given that
\[y = (1 + \ln x)^2 \qquad x \gt 0\]Give each coefficient in simplest form. (3)
| Scheme | Marks | AO |
|---|---|---|
| \(y = (1 + \ln x)^2 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = k(1 + \ln x) \times \dfrac{1}{x}\) or \(y = 1 + 2\ln x + (\ln x)^2 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{A}{x} + B\ln x \times \dfrac{1}{x}\) | M1 | 1.1b |
| \(y = (1 + \ln x)^2 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2(1 + \ln x) \times \dfrac{1}{x}\) or \(y = 1 + 2\ln x + (\ln x)^2 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2}{x} + 2\ln x \times \dfrac{1}{x}\) | A1 | 1.1b |
| \(\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \dfrac{k\left(\frac{1}{x}\right) \times x - k(1 + \ln x) \times 1}{x^2}\) or \(\dfrac{k}{x}x^{-1} + k(1 + \ln x)\left(-x^{-2}\right)\) or \(\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} = -\dfrac{A}{x^2} + \dfrac{\frac{B}{x} \times x - 2\ln x \times 1}{x^2}\) | M1 | 1.1b |
| \(\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \dfrac{\left(\frac{2}{x}\right) \times x - 2(1 + \ln x)}{x^2}\) or \(\dfrac{2}{x}x^{-1} + 2(1 + \ln x)\left(-x^{-2}\right)\) Leading to \(\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} = -\dfrac{2\ln x}{x^2}\ *\) achieved from correct work. | A1* | 2.1 |
| (4) |
Notes
M1: Attempts the first derivative, including use of the chain rule. May expand first. E.g. accept forms as shown.
A1: Correct first derivative, need not be simplified.
M1: Attempts second derivative using quotient rule or product rule – examples as shown, or equivalents accepted.
A1*: Correct result achieved from correct work.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}^3 y}{\mathrm{d}x^3} = \dfrac{\pm\frac{C}{x} \times x^2 \pm Dx\ln x}{x^4}\) or \(\dfrac{\mathrm{d}^3 y}{\mathrm{d}x^3} = (-2\ln x)\left(-Cx^{-3}\right) + \left(-\dfrac{D}{x}\right)\left(x^{-2}\right)\) | M1 | 1.1b |
| \(\dfrac{\mathrm{d}^3 y}{\mathrm{d}x^3} = -\dfrac{\frac{2}{x} \times x^2 - 4x\ln x}{x^4}\) or \(\dfrac{-\frac{2}{x} \times x^2 - (-2\ln x)(2x)}{x^4}\) or \(-\dfrac{2}{x^3} + \dfrac{4\ln x}{x^3}\) | A1 | 1.1b |
| (2) |
Notes
M1: Applies quotient rule or product rule to achieve third derivative. If formula is quoted it must be correct, if not accept derivatives of the form shown as there may be confusion with the minus sign.
A1: Correct third derivative, any form.
| Scheme | Marks | AO |
|---|---|---|
| \(y(1) = 1,\ y^{\prime}(1) = 2,\ y^{\prime\prime}(1) = 0,\ y^{\prime\prime\prime}(1) = -2\) | M1 | 1.1b |
| \([y] = 1 + 2(x - 1) + \dfrac{0}{2!}(x - 1)^2 + \dfrac{-2}{3!}(x - 1)^3 + \ldots\) | M1 | 2.5 |
| \([y] = 1 + 2(x - 1) - \dfrac{1}{3}(x - 1)^3 + \ldots\) or \([y] = -1 + 2x - \dfrac{1}{3}(x - 1)^3 + \ldots\) | A1 | 1.1b |
| (3) |
Notes
M1: Find value of derivatives at \(x = 1\).
M1: Applies Taylor series expansion
A1: Correct series, may be unsimplified, isw once correct series seen. Must be using a correct third derivative.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{2x - 1 - (1 + \ln x)^2}{(x - 1)^3} = \dfrac{2x - 1 - 1 - 2(x - 1) + \frac{1}{3}(x - 1)^3 + \ldots}{(x - 1)^3} = \dfrac{\frac{1}{3}(x - 1)^3 + \ldots}{(x - 1)^3}\) | M1 | 1.1b |
| Simplifies and realises that terms cancel to leave a constant term \(\dfrac{\frac{1}{3}\cancel{(x - 1)^3} + \ldots}{\cancel{(x - 1)^3}} = \dfrac{1}{3}\) | M1 | 3.1a |
| Hence \(\lim\limits_{x \to 1}\dfrac{2x - 1 - (1 + \ln x)^2}{(x - 1)^3} = \dfrac{1}{3}\) as all remaining terms will become zero in the limit as they are multiples of \((x - 1)^k\), which tends to 0. | A1 | 2.4 |
| (3) | ||
| (12 marks) |
Notes
M1: Applies the series to the limit and cancels terms in numerator to leave term in \((x - 1)^3\) and above only (may not see \(+\ldots\) for this mark)
M1: Simplifies and realises that the \((x - 1)^3\) cancels and achieves a constant \(A\)
A1: Correct limit deduced with reasoning given why the remaining terms disappear.