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Question

The function $f$ is defined by $f(x) = {{\rm{e}}^x}\cos x$ .

Show that $f''(x) = - 2{{\rm{e}}^x}\sin x$ .

[2]

Determine the Maclaurin series for $f(x)$ up to and including the term in ${x^4}$ .

[5]

By differentiating your series, determine the Maclaurin series for ${{\rm{e}}^x}\sin x$ up to the term in ${x^3}$ .

[4]

$f'(x) = {{\rm{e}}^x}\cos x - {{\rm{e}}^x}\sin x$ A1

$f''(x) = {{\rm{e}}^x}\cos x - {{\rm{e}}^x}\sin x - {{\rm{e}}^x}\sin x - {{\rm{e}}^x}\cos x$ A1

$= - 2{{\rm{e}}^x}\sin x$ AG

[2 marks]

$f'''(x) = - 2{{\rm{e}}^x}\sin x - 2{{\rm{e}}^x}\cos x$ A1

${f^{IV}}(x) = - 4{{\rm{e}}^x}\cos x$ A1

$f(0) = 1$ , $f'(0) = 1$ , $f''(0) = 0$ , $f'''(0) = - 2$ , ${f^{IV}}(0) = - 4$ (A1)

the Maclaurin series is

${{\rm{e}}^x}\cos x = 1 + x - \frac{{{x^3}}}{3} - \frac{{{x^4}}}{6} + \ldots$ M1A1

Note: Accept multiplication of series method.

[5 marks]

differentiating,

${{\rm{e}}^x}\cos x - {{\rm{e}}^x}\sin x = 1 - {x^2} - \frac{{2{x^3}}}{3} + \ldots$ M1A1

${{\rm{e}}^x}\sin x = 1 + {x^{}} - \frac{{{x^3}}}{3} + \ldots - 1 + {x^2} + \frac{{2{x^3}}}{3} + \ldots$ M1

$= x + {x^2} + \frac{{{x^3}}}{3} + \ldots$ A1

[4 marks]

[N/A]

${{\text{e}}^x}$ ,

$\sin x$ ,

$\cos x$ ,

$\ln (1 + x)$ ,

${(1 + x)^p}$ ,

$P \in \mathbb{Q}$ .