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Date May 2017 Marks available 3 Reference code 17M.3ca.hl.TZ0.2
Level HL only Paper Paper 3 Calculus Time zone TZ0
Command term Hence and Determine Question number 2 Adapted from N/A

Question

Let the Maclaurin series for \(\tan x\) be

\[\tan x = {a_1}x + {a_3}{x^3} + {a_5}{x^5} + \ldots \]

where \({a_1}\), \({a_3}\) and \({a_5}\) are constants.

Find series for \({\sec ^2}x\), in terms of \({a_1}\), \({a_3}\) and \({a_5}\), up to and including the \({x^4}\) term

by differentiating the above series for \(\tan x\);

[1]
a.i.

Find series for \({\sec ^2}x\), in terms of \({a_1}\), \({a_3}\) and \({a_5}\), up to and including the \({x^4}\) term

by using the relationship \({\sec ^2}x = 1 + {\tan ^2}x\).

[2]
a.ii.

Hence, by comparing your two series, determine the values of \({a_1}\), \({a_3}\) and \({a_5}\).

[3]
b.

Markscheme

\(({\sec ^2}x = ){\text{ }}{a_1} + 3{a_3}{x^2} + 5{a_5}{x^4} + \ldots \)     A1

[1 mark]

a.i.

\({\sec ^2}x = 1 + {({a_1}x + {a_3}{x^3} + {a_5}{x^5} + \ldots )^2}\)

\( = 1 + a_1^2{x^2} + 2{a_1}{a_3}{x^4} + \ldots \)     M1A1

 

Note:     Condone the presence of terms with powers greater than four.

 

[2 marks]

a.ii.

equating constant terms: \({a_1} = 1\)     A1

equating \({x^2}\) terms: \(3{a_3} = a_1^2 = 1 \Rightarrow {a_3} = \frac{1}{3}\)     A1

equating \({x^4}\) terms: \(5{a_5} = 2{a_1}{a_3} = \frac{2}{3} \Rightarrow {a_5} = \frac{2}{{15}}\)     A1

[3 marks]

b.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.

Syllabus sections

Topic 9 - Option: Calculus » 9.6
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