A2 June 2024 Q4
4.
\[\left[\begin{gathered}\textit{The Taylor series expansion of}\;\; \mathrm{f}(x)\;\; \textit{about}\;\; x = a\;\; \textit{is given by}\\ \mathrm{f}(x) = \mathrm{f}(a) + (x - a)\mathrm{f}^{\prime}(a) + \frac{(x - a)^2}{2!}\mathrm{f}^{\prime\prime}(a) + \ldots + \frac{(x - a)^r}{r!}\mathrm{f}^{(r)}(a) + \ldots\end{gathered}\right]\]The curve with equation \(y = \mathrm{f}(x)\) satisfies the differential equation
\[\cos x\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + y^2\frac{\mathrm{d}y}{\mathrm{d}x} + \sin x = 0\]Given that \(\left(\dfrac{\pi}{4}, 1\right)\) is a stationary point of the curve,
| Scheme | Marks | AO |
|---|---|---|
| \(\cos\left(\dfrac{\pi}{4}\right)\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + (1)^2(0) + \sin\left(\dfrac{\pi}{4}\right) = 0 \Rightarrow \dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \ldots\) | M1 | 3.1a |
| \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = -1 \lt 0\) therefore a (local) maximum | A1 | 2.4 |
| (2) |
Notes
M1: A complete method to find the value for \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\). Substitutes into the differential equation \(x = \dfrac{\pi}{4},\ y = 1,\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) and rearranges to find a value for \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\).
A1: \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = -1 \lt 0\) therefore a maximum. Must achieve the correct value, or at least a correct expression and give reason “\(\lt 0\)” (oe) and conclusion and no contradictory statements.
| Scheme | Marks | AO |
|---|---|---|
| Achieves \(\pm\sin x\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + \cos x\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} + \alpha y\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^{1\text{ or }2} + y^2\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} \pm \cos x = 0\) or \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = -\tan x - y^2\sec x\dfrac{\mathrm{d}y}{\mathrm{d}x} \to\) \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = \pm\sec^2 x \pm \alpha y\sec x\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^{1\text{ or }2} \pm y^2\sec x\tan x\dfrac{\mathrm{d}y}{\mathrm{d}x} \pm y^2\sec x\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\) | M1 | 1.1b |
| \(-\sin x\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + \cos x\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} + 2y\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + y^2\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + \cos x = 0\) or \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = -\sec^2 x - 2y\sec x\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 - y^2\sec x\tan x\dfrac{\mathrm{d}y}{\mathrm{d}x} - y^2\sec x\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\) | A1 | 1.1b |
| \(-\sin\left(\dfrac{\pi}{4}\right)(-1) + \cos\left(\dfrac{\pi}{4}\right)\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} + 2(1)(0)^2 + (1)^2(-1) + \cos\left(\dfrac{\pi}{4}\right) = 0 \to \dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = \ldots\) Alt: \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = -2 - 2(1)\sqrt{2}(0) - 1^2\sqrt{2}(1)(0) - 1^2\sqrt{2}(-1)\) | M1 | 1.1b |
| \(\left(\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} - 1 + \dfrac{\sqrt{2}}{2} = 0 \Rightarrow \dfrac{\sqrt{2}}{2}\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = 1 - \sqrt{2}\right)\) \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\left(= \dfrac{2 - 2\sqrt{2}}{\sqrt{2}} = \dfrac{2\sqrt{2} - 4}{2}\right) = \sqrt{2} - 2\ *\) | A1* | 2.1 |
| (4) |
Notes
M1: Differentiates to the form \(\pm\sin x\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + \cos x\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} + \alpha y\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^{1\text{ or }2} + y^2\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} \pm \cos x = 0\).
Alternatively, divides through by \(\cos x\) first and differentiates to achieve the form shown in scheme.
A1: Correct differentiation, any alternative form is acceptable.
M1: Substitutes in \(x = \dfrac{\pi}{4},\ y = 1,\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\), their value of \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \text{‘}-1\text{’}\) as appropriate for their expression, and rearranges to find a value for \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\) (if not already done so).
A1*: Shows that \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = \sqrt{2} - 2\) from correct working with at least one suitable intermediate line – being either an unsimplified \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = \ldots\) or \(\alpha\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = \ldots\), before the final answer. Must have come from a correct derivative expression. Note that in the Alt the correct form comes out much easier, so look for at least one unsimplified line before the final answer.
| Scheme | Marks | AO |
|---|---|---|
| \((y =)\,1 + \left(x - \dfrac{\pi}{4}\right)(0) + \dfrac{\left(x - \dfrac{\pi}{4}\right)^2}{2!}(\text{their ‘}-1\text{’}) + \dfrac{\left(x - \dfrac{\pi}{4}\right)^3}{3!}\left(\sqrt{2} - 2\right) + \ldots\) | M1 | 2.5 |
| \((y =)\,1 - \dfrac{\left(x - \dfrac{\pi}{4}\right)^2}{2} + \dfrac{\left(x - \dfrac{\pi}{4}\right)^3\left(\sqrt{2} - 2\right)}{6} + \ldots\) | A1 | 1.1b |
| (2) | ||
| (8 marks) |
Notes
M1: Uses the Taylor’s series expansion with the correct values for \(y_{\frac{\pi}{4}} = 1,\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = 0,\ \dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = \sqrt{2} - 2\) (or their \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\) if they had a different answer to (b)) and their value of \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\). Must include the 1 at the start.
A1: Correct simplified expansion. The “\(y =\)” may be missing. Accept with \(0\left(x - \dfrac{\pi}{4}\right)\) as a second term. ISW after a correct simplified expression if further attempts to simplify are made.