A2 June 2022 Q8
8.
\[\left[\begin{gathered}\textit{The Taylor series expansion of}\;\; \mathrm{f}(x)\;\; \textit{about}\;\; x = a\;\; \textit{is given by}\\ \mathrm{f}(x) = \mathrm{f}(a) + (x - a)\mathrm{f}^{\prime}(a) + \frac{(x - a)^2}{2!}\mathrm{f}^{\prime\prime}(a) + \ldots + \frac{(x - a)^r}{r!}\mathrm{f}^{(r)}(a) + \ldots\end{gathered}\right]\](Solutions relying entirely on calculator technology are not acceptable.)
(4)| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{f}(x) = \ln x \Rightarrow \mathrm{f}(1) = 0\) \(\mathrm{f}^{\prime}(x) = \dfrac{1}{x} \Rightarrow \mathrm{f}^{\prime}(1) = 1\) \(\mathrm{f}^{\prime\prime}(x) = -\dfrac{1}{x^2} \Rightarrow \mathrm{f}^{\prime\prime}(1) = -1\) | M1 A1 | 1.1b 1.1b |
| \((\ln x) = (0 +)(x - 1) - \dfrac{1}{2}(x - 1)^2 + \ldots\) | M1 A1 | 2.5 1.1b |
| (4) |
Notes
Notes: ignore extra terms throughout.
M1: Differentiates \(\mathrm{f}(x) = \ln x\) twice and finds \(\mathrm{f}(1)\), \(\mathrm{f}^{\prime}(1)\) and \(\mathrm{f}^{\prime\prime}(1)\)
A1: Correct differentiation and values for \(\mathrm{f}(1)\), \(\mathrm{f}^{\prime}(1)\) and \(\mathrm{f}^{\prime\prime}(1)\).
M1: Uses correct mathematical notation to find the Taylor series for \(\ln x\) in powers of \((x - 1)\) up to \((x - 1)^2\)
A1: Correct expansion with simplified coefficients. Do not be concerned with the left hand side.
(Corrected from the printed mark scheme: the third line of working is garbled in the printed scheme; it is shown here as \(\mathrm{f}^{\prime\prime}(x) = -\dfrac{1}{x^2} \Rightarrow \mathrm{f}^{\prime\prime}(1) = -1\), which the expansion uses.)
| Scheme | Marks | AO |
|---|---|---|
| \(\lim\limits_{x \to 1}\left(\dfrac{\ln x}{x - 1}\right) = \lim\limits_{x \to 1}\left(\dfrac{(x - 1) - \frac{1}{2}(x - 1)^2 + \ldots}{x - 1}\right)\) \(= \lim\limits_{x \to 1}\left(1 - \dfrac{1}{2}(x - 1) + \ldots\right)\) | M1 | 2.1 |
| \(= \lim\limits_{x \to 1}\left(1 - \dfrac{1}{2}(x - 1) + \ldots\right) = 1\ *\) cso | A1* | 2.2a |
| (2) |
Notes
Question says “hence” so the result of (a) must be used. No marks for l’Hosptial’s rule on the original functions (send to review if attempted with their part (a)).
M1: Substitutes their Taylor series for \(\ln x\) in powers of \((x - 1)\) up to \((x - 1)^2\) into the limit and cancels a factor \((x - 1)\) from each term. Allow for the cancelling seeing a relevant strikethrough in all \(x - 1\) terms.
A1*: \(\lim\limits_{x \to 1}\left(\dfrac{\ln x}{x - 1}\right) = \lim\limits_{x \to 1}\left(1 - \dfrac{1}{2}(x - 1) + \ldots\right) = 1\) cso Must have come from a correct expansion. Must see the \(1 - \dfrac{1}{2}(x - 1)\)
| Scheme | Marks | AO |
|---|---|---|
| Writes as an indeterminate form For example \(\dfrac{\sin(2x)}{(x + 3)\tan(6x)}\) or \(\dfrac{\sin(2x)\cos(6x)}{(x + 3)\sin(6x)}\) | M1 | 3.1a |
| Differentiates numerator and denominator using appropriate rules \(\dfrac{2\cos(2x)}{\tan(6x) + 6(x + 3)\sec^2(6x)}\) or \(\dfrac{2\cos(2x)\cos(6x) - 6\sin(2x)\sin(6x)}{\sin(6x) + 6(x + 3)\cos(6x)}\) | M1 A1 | 1.1b 1.1b |
| \(\lim\limits_{x \to 0}\left(\dfrac{1}{(x + 3)\tan(6x)\operatorname{cosec}(2x)}\right) = \dfrac{2}{18} = \dfrac{1}{9}\) o.e | A1cso | 2.2a |
| (4) | ||
| (10 marks) |
Notes
M1: Writes the fraction in an indeterminate form \(\dfrac{f(x)}{g(x)}\) where \(\dfrac{\boldsymbol{f}(0)}{\boldsymbol{g}(0)} = \dfrac{0}{0}\) or \(\dfrac{\boldsymbol{f}(0)}{\boldsymbol{g}(0)} = \dfrac{\text{“}\infty\text{”}}{\text{“}\infty\text{”}}\)
M1: Differentiates numerator and denominator using appropriate rules, ie product rule for a product etc. Allow slips in coefficients but the form should be correct. This mark is available as long as written as \(\dfrac{f(x)}{g(x)}\) even if not an indeterminate form. May be seen written as separate from the fraction, ie \(\mathrm{f}^{\prime}(x) = \ldots\) and \(\mathrm{g}^{\prime}(x) = \ldots\)
A1: Depend on both Ms. Correct differentiation for derivatives that lead to a limit. Must have a derivative for which \(\mathrm{g}^{\prime}(0) \neq 0\) and is finite.
A1cso: Deduces the correct limit from fully correct work.