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Date May 2016 Marks available 3 Reference code 16M.1.hl.TZ2.9
Level HL only Paper 1 Time zone TZ2
Command term Verify Question number 9 Adapted from N/A

Question

Consider the equation \(\frac{{\sqrt 3  - 1}}{{\sin x}} + \frac{{\sqrt 3  + 1}}{{\cos x}} = 4\sqrt 2 ,{\text{ }}0 < x < \frac{\pi }{2}\). Given that \(\sin \left( {\frac{\pi }{{12}}} \right) = \frac{{\sqrt 6  - \sqrt 2 }}{4}\) and \(\cos \left( {\frac{\pi }{{12}}} \right) = \frac{{\sqrt 6  + \sqrt 2 }}{4}\)

verify that \(x = \frac{\pi }{{12}}\) is a solution to the equation;

[3]
a.

hence find the other solution to the equation for \(0 < x < \frac{\pi }{2}\).

[5]
b.

Markscheme

EITHER

\({\text{LHS}} = \frac{{\sqrt 3  - 1}}{{\frac{{\sqrt 6  - \sqrt 2 }}{4}}} + \frac{{\sqrt 3  + 1}}{{\frac{{\sqrt 6  + \sqrt 2 }}{4}}}\)    M1

\( = \frac{{\sqrt 3  - 1}}{{\frac{{\sqrt 3  - 1}}{{2\sqrt 2 }}}} + \frac{{\sqrt 3  + 1}}{{\frac{{\sqrt 3  + 1}}{{2\sqrt 2 }}}}\)    A1

\( = 2\sqrt 2  + 2\sqrt 2 \)    A1

\({\text{LHS}} = 4\sqrt 2  \Rightarrow x = \frac{\pi }{{12}}\) is a solution     AG

OR

\({\text{LHS}} = \frac{{\sqrt 3  - 1}}{{\frac{{\sqrt 6  - \sqrt 2 }}{4}}} + \frac{{\sqrt 3  + 1}}{{\frac{{\sqrt 6  + \sqrt 2 }}{4}}}\)    M1

\( = \frac{{\left( {\sqrt 3  - 1} \right)\left( {\frac{{\sqrt 6  + \sqrt 2 }}{4}} \right) + \left( {\sqrt 3  + 1} \right)\left( {\frac{{\sqrt 6  - \sqrt 2 }}{4}} \right)}}{{\left( {\frac{{\sqrt 6  - \sqrt 2 }}{4}} \right)\left( {\frac{{\sqrt 6  + \sqrt 2 }}{4}} \right)}}\)    A1

\( = 2\sqrt {18}  - 2\sqrt 2 \) (or equivalent)     A1

\({\text{LHS}} = 4\sqrt 2  \Rightarrow x = \frac{\pi }{{12}}\) is a solution     AG

[3 marks]

a.

\(\frac{{\sqrt 2 }}{4}\left( {\frac{{\sqrt 3  - 1}}{{\sin x}} + \frac{{\sqrt 3  + 1}}{{\cos x}}} \right) = 2 \Rightarrow \frac{{\sin \frac{\pi }{{12}}}}{{\sin x}} + \frac{{\cos \frac{\pi }{{12}}}}{{\cos x}} = 2\)    M1

\(\frac{{\sin \frac{\pi }{{12}}\cos x + \cos \frac{\pi }{{12}}\sin x}}{{\sin x\cos x}} = 2\)    M1

\(\sin \frac{\pi }{{12}}\cos x + \cos \frac{\pi }{{12}}\sin x = 2\sin x\cos x\)

\(\sin \left( {\frac{\pi }{{12}} + x} \right) = \sin 2x\)    A1

\(\frac{\pi }{{12}} + x = \pi  - 2x\) or \(\pi  - \left( {\frac{\pi }{{12}} + x} \right) = 2x\)     (M1)

\(x = \frac{{11\pi }}{{36}}\)    A1

[5 marks]

b.

Examiners report

This question proved to be the most problematic question in the paper.

Part (a) was generally well done, with competent fraction and surd manipulation seen successfully in leading to the given answer.

a.

This question proved to be the most problematic question in the paper.

The number of scripts seen where part (b) was tackled with complete success numbered in the single figures; solutions were rarely if ever seen. Some candidates scored one mark by finding, or using, the common denominator \(\sin x\cos x\).

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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