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Question: 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

A

⋃n∈I{[2nπ,2nπ+π6]⋃[2nπ+5π6,(2n+1)π]}\bigcup_{n \in I}^{}\left\{ \left\lbrack 2n\pi,2n\pi + \frac{\pi}{6} \right\rbrack\bigcup_{}^{}\left\lbrack 2n\pi + \frac{5\pi}{6},(2n + 1)\pi \right\rbrack \right\}

B

⋃n∈I[2nπ,2nπ+π6]\bigcup_{n \in I}^{}\left\lbrack 2n\pi,2n\pi + \frac{\pi}{6} \right\rbrack

C

⋃n∈I[2nπ+5π6,(2n+1)π]\bigcup_{n \in I}^{}\left\lbrack 2n\pi + \frac{5\pi}{6},(2n + 1)\pi \right\rbrack

D

None of these

Answer

⋃n∈I{[2nπ,2nπ+π6]⋃[2nπ+5π6,(2n+1)π]}\bigcup_{n \in I}^{}\left\{ \left\lbrack 2n\pi,2n\pi + \frac{\pi}{6} \right\rbrack\bigcup_{}^{}\left\lbrack 2n\pi + \frac{5\pi}{6},(2n + 1)\pi \right\rbrack \right\}

Explanation

Solution

f(x) is defined on [0,1] ⇒ 0 ≤ x ≤ 1

Now f(2sin x) shall be defined , if 0 ≤ 2 sin x ≤ 1 ⇒ 0 ≤ sin x ≤ 12\frac{1}{2} ⇒ x ∈ ⋃n∈I{[2nπ,2nπ+π6] 6mu⋃[2nπ+5π6,(2n+1)π]}\bigcup_{n \in I}^{}\left\{ \left\lbrack 2n\pi,2n\pi + \frac{\pi}{6} \right\rbrack\mspace{6mu}\bigcup_{}^{}\left\lbrack 2n\pi + \frac{5\pi}{6},(2n + 1)\pi \right\rbrack \right\}