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Find the RANGE of the function attached in image..................…...

ABC , 7 Years ago
Grade 11
anser 1 Answers
Aditya Gupta
note that 16sin^2x+1 can lie in [1, 17]
also, maximum value of 2 – log2(16sin^2x+1) is 2 – log21=2 – 0=2
also, 16sin^2x+1 can attain value 4 (which lies between 1 and 17), which means that 2 – log2(16sin^2x+1) can attain 0 value.
when it approaches zero, the value of log2^1/2[2 – log2(16sin^2x+1)]= 2log2[2 – log2(16sin^2x+1)] approaches negative infinity
when it attains 2, log2^1/2[2 – log2(16sin^2x+1)]= 2log2[2 – log2(16sin^2x+1)] attains 2
so range is (–infinity, 2].
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