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Date May 2017 Marks available 7 Reference code 17M.1.SL.TZ2.S_7
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 2
Command term Solve Question number S_7 Adapted from N/A

Question

Solve log2(2sinx)+log2(cosx)=1, for 2π<x<5π2.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

correct application of loga+logb=logab     (A1)

eglog2(2sinxcosx), log2+log(sinx)+log(cosx)

correct equation without logs     A1

eg2sinxcosx=21, sinxcosx=14, sin2x=12

recognizing double-angle identity (seen anywhere)     A1

eglog(sin2x), 2sinxcosx=sin2x, sin2x=12

evaluating sin1(12)=π6 (30)     (A1)

correct working     A1

egx=π12+2π, 2x=25π6, 29π6, 750, 870, x=π12and x=5π12, one correct final answer

x=25π12, 29π12 (do not accept additional values)     A2     N0

[7 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.6—Pythagorean identity, double angles
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Topic 3— Geometry and trigonometry

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