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`        Find the RANGE of the function attached in image..................…...`
6 months ago

1071 Points
```							note that 16sin^2x+1 can lie in [1, 17]also, maximum value of 2 – log2(16sin^2x+1) is 2 – log21=2 – 0=2also, 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 infinitywhen it attains 2, log2^1/2[2 – log2(16sin^2x+1)]= 2log2[2 – log2(16sin^2x+1)] attains 2so range is (–infinity, 2].
```
6 months ago
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### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions