Edexcel FP2 2004 June — Question 11 11 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2004
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTaylor series
TypeDirect multiplication of series
DifficultyStandard +0.3 This is a straightforward application of multiplying two known Maclaurin series (e^x and cos x) and collecting terms up to x^4. While it requires careful algebraic manipulation and organization of terms, it's a standard textbook exercise in FP2 with no conceptual difficulty or novel insight required—students are explicitly told to use multiplication ('Hence') from part (a).
Spec4.08a Maclaurin series: find series for function4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n
11. (b) Hence find the Maclaurin series expansion of \(\mathrm { e } ^ { x } \cos x\), in ascending powers of \(x\), up to and including the term in \(x ^ { 4 }\).
(Total 11 marks)
(a) \(n=1\): \(\frac{d}{dx}(e^x\cos x) = e^x\cos x - e^x\sin x\) M1
(Use of product rule)
\(\cos\left(x+\frac{\pi}{4}\right) = \cos x\cos\frac{\pi}{4} - \sin x\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}(\cos x - \sin x)\) M1
\(\frac{d}{dx}(e^x\cos x) = 2^{1/2} \cdot 2^{-1/2} e^x\cos\left(x+\frac{\pi}{4}\right)\) True for \(n=1\) (c.s.o. + comment) A1
Suppose true for \(n=k\).
\(\frac{d^{k+1}}{dx^{k+1}}(e^x\cos x) = \frac{d}{dx}\left[2^{k/2} e^x \cos\left(x+\frac{k\pi}{4}\right)\right]\) M1
\(= 2^{k/2} e^x \cos\left(x+\frac{k\pi}{4}\right) - e^x\sin\left(x+\frac{k\pi}{4}\right)\) A1
\(= 2^{(k+1)/2} e^x \cos\left(x+\frac{k\pi}{4}+\frac{\pi}{4}\right) = 2^{(k+1)/2} e^x \cos\left(x+\frac{(k+1)\pi}{4}\right)\) M1 A1
\(\therefore\) True for \(n=k+1\), so true (by induction) for all \(n \geq 1\) A1 (cso)
(b) \(1 + 2\cos\frac{\pi}{4} \cdot x + \frac{1}{2} \cdot 2\cos\frac{\pi}{2} \cdot x^2 + \frac{1}{6} \cdot 2^{3/2}\cos\frac{3\pi}{4} \cdot x^3 + \frac{1}{24}\cos\pi \cdot x^4\) M1
\((1) \quad (0) \quad (-\sqrt{2}) \quad (-4)\)
\(e^x\cos x = 1 + x - \frac{1}{3}x^3 - \frac{1}{6}x^4\) (or equiv. fractions) A2 (1,0)
(a) $n=1$: $\frac{d}{dx}(e^x\cos x) = e^x\cos x - e^x\sin x$ M1

(Use of product rule)

$\cos\left(x+\frac{\pi}{4}\right) = \cos x\cos\frac{\pi}{4} - \sin x\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}(\cos x - \sin x)$ M1

$\frac{d}{dx}(e^x\cos x) = 2^{1/2} \cdot 2^{-1/2} e^x\cos\left(x+\frac{\pi}{4}\right)$ True for $n=1$ (c.s.o. + comment) A1

Suppose true for $n=k$.

$\frac{d^{k+1}}{dx^{k+1}}(e^x\cos x) = \frac{d}{dx}\left[2^{k/2} e^x \cos\left(x+\frac{k\pi}{4}\right)\right]$ M1

$= 2^{k/2} e^x \cos\left(x+\frac{k\pi}{4}\right) - e^x\sin\left(x+\frac{k\pi}{4}\right)$ A1

$= 2^{(k+1)/2} e^x \cos\left(x+\frac{k\pi}{4}+\frac{\pi}{4}\right) = 2^{(k+1)/2} e^x \cos\left(x+\frac{(k+1)\pi}{4}\right)$ M1 A1

$\therefore$ True for $n=k+1$, so true (by induction) for all $n \geq 1$ A1 (cso)

(b) $1 + 2\cos\frac{\pi}{4} \cdot x + \frac{1}{2} \cdot 2\cos\frac{\pi}{2} \cdot x^2 + \frac{1}{6} \cdot 2^{3/2}\cos\frac{3\pi}{4} \cdot x^3 + \frac{1}{24}\cos\pi \cdot x^4$ M1

$(1) \quad (0) \quad (-\sqrt{2}) \quad (-4)$

$e^x\cos x = 1 + x - \frac{1}{3}x^3 - \frac{1}{6}x^4$ (or equiv. fractions) A2 (1,0)

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11. (b) Hence find the Maclaurin series expansion of $\mathrm { e } ^ { x } \cos x$, in ascending powers of $x$, up to and including the term in $x ^ { 4 }$.\\
(Total 11 marks)\\

\hfill \mbox{\textit{Edexcel FP2 2004 Q11 [11]}}
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