If f(x) is defined on domain [0, 1], then f(2 sin x) is defined on
If f(x) is defined on domain [0, 1], then f(2 sin x) is defined on
\(\bigcup\limits_{n∈I}\){[2nπ,2nπ+π/6]\(\bigcup\)[2nπ+\(\frac{5π}{6}\),(2n+1)π]}
\(\bigcup\limits_{n∈I}\)[2nπ,2nπ+\(\fracπ6\)]
\(\bigcup\limits_{n∈I}\)[2nπ+\(\frac{5π}{6}\),(2n+1)π]
None of these
The Correct Option is A
Solution and Explanation
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