The domain of the function {\cos ^{ - 1}}\left( {{{2{{\sin }^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right)} \over \pi }} \right) is :

Question

Accepted Answer

Answer: D. \left( { - \infty ,{{ - 1} \over {\sqrt 2 }}} ight] \cup \left[ {{1 \over {\sqrt 2 }},\infty } ight) \cup \{ 0\} - 1 \le {{2{{\sin }^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} ight)} \over \pi } \le 1 \Rightarrow - {\pi \over 2} \le {\sin ^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} ight) \le {\pi \over 2} \Rightarrow - 1 \le {1 \over {4{x^2} - 1}} \le 1 	herefore {1 \over {4{x^2} - 1}} + 1 \ge 0 \Rightarrow {{1 + 4{x^2} - 1} \over {4{x^2} - 1}} \ge 0 \Rightarrow {{4{x^2}} \over {4{x^2} - 1}} \ge 0 \Rightarrow {{4{x^2}} \over {\left( {2x + 1} ight)\left( {2x - 1} ight)}} \ge 0 ...... (1) 	herefore x \in \left( { - \alpha , - {1 \over 2}} ight) \cup \{ 0\} \cup \left( {{1 \over 2},\alpha } ight) .....(2) And {1 \over {4{x^2} - 1}} - 1 \le 0 \Rightarrow {{1 - 4{x^2} + 1} \over {4{x^2} - 1}} \le 0 \Rightarrow {{2 - 4{x^2}} \over {4{x^2} - 1}} \le 0 \Rightarrow {{2{x^2} - 1} \over {4{x^2} - 1}} \ge 0 \Rightarrow {{\left( {\sqrt 2 x + 1} ight)\left( {\sqrt 2 x - 1} ight)} \over {\left( {2x + 1} ight)\left( {2x - 1} ight)}} \ge 0 ...... (3) x \in \left( { - \alpha , - {1 \over {\sqrt 2 }}} ight) \cup \left( { - {1 \over 2},{1 \over 2}} ight) \cup \left( {{1 \over {\sqrt 2 }},\alpha } ight) .....(4) From (3) and (4), we get 	herefore x \in \left[ { - \alpha , - {1 \over {\sqrt 2 }}} ight) \cup \left[ {{1 \over {\sqrt 2 }},\alpha } ight) \cup \{ 0\}

The domain of the function $${\cos ^{ - 1}}\left( {{{2{{\sin }^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right)} \over \pi }} \right)$$ is :

Answer

Explanation

More JEE Math questions