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Date November 2015 Marks available 6 Reference code 15N.3ca.hl.TZ0.2
Level HL only Paper Paper 3 Calculus Time zone TZ0
Command term Find Question number 2 Adapted from N/A

Question

Let f(x)=exsinx.

Show that f.

[4]
a.

By further differentiation of the result in part (a) , find the Maclaurin expansion of f(x), as far as the term in {x^5}.

[6]
b.

Markscheme

f'(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x     M1A1

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

= 2\left( {{{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x - {{\text{e}}^x}\sin x} \right)     M1

= 2\left( {f'(x) - f(x)} \right)     AG

[4 marks]

a.

f(0) = 0,{\text{ }}f'(0) = 1,{\text{ }}f''(0) = 2(1 - 0) = 2     (M1)A1

 

Note:     Award M1 for attempt to find f(0), f'(0) and f''(0).

 

f'''(x) = 2\left( {f''(x) - f'(x)} \right)     (M1)

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

so f(x) = x + \frac{2}{{2!}}{x^2} + \frac{2}{{3!}}{x^3} - \frac{4}{{5!}} +  \ldots     (M1)A1

= x + {x^2} + \frac{1}{3}{x^3} - \frac{1}{{30}}{x^5} +  \ldots

[6 marks]

Total [10 marks]

b.

Examiners report

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a.
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b.

Syllabus sections

Topic 9 - Option: Calculus » 9.6 » Taylor polynomials; the Lagrange form of the error term.

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