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Date November 2011 Marks available 2 Reference code 11N.2.hl.TZ0.4
Level HL only Paper 2 Time zone TZ0
Command term Hence or otherwise and Solve Question number 4 Adapted from N/A

Question

Given that \(\arctan \frac{1}{2} - \arctan \frac{1}{3} = \arctan a,{\text{ }}a \in {\mathbb{Q}^ + }\), find the value of a.

[3]
a.

Hence, or otherwise, solve the equation \(\arcsin x = \arctan a\).

[2]
b.

Markscheme

\(\tan \left( {\arctan \frac{1}{2} - \arctan \frac{1}{3}} \right) = \tan (\arctan a)\)     (M1)

\(a = 0.14285 \ldots  = \frac{1}{7}\)     (A1)A1

[3 marks]

a.

\(\arctan \left( {\frac{1}{7}} \right) = \arcsin (x) \Rightarrow x = \sin \left( {\arctan \frac{1}{7}} \right) \approx 0.141\)     (M1)A1

Note: Accept exact value of \(\left( {\frac{1}{{\sqrt {50} }}} \right)\).

 

[2 marks]

b.

Examiners report

Many candidates failed to give the answer for (a) in rational form. The GDC can render the answer in this form as well as the decimal approximation, but this was obviously missed by many candidates.

a.

(b) was generally answered successfully.

b.

Syllabus sections

Topic 3 - Core: Circular functions and trigonometry » 3.6 » Algebraic and graphical methods of solving trigonometric equations in a finite interval, including the use of trigonometric identities and factorization.
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