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 arctan12−arctan13=arctana, a∈Q+, find the value of a.
[3]
a.
Hence, or otherwise, solve the equation arcsinx=arctana.
[2]
b.
Markscheme
tan(arctan12−arctan13)=tan(arctana) (M1)
a=0.14285…=17 (A1)A1
[3 marks]
a.
arctan(17)=arcsin(x)⇒x=sin(arctan17)≈0.141 (M1)A1
Note: Accept exact value of (1√50).
[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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