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=2−1, sinxcosx=14, sin2x=12
recognizing double-angle identity (seen anywhere) A1
eglog(sin2x), 2sinxcosx=sin2x, sin2x=12
evaluating sin−1(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]