Question 147 of 155HardMCQ
The domain of the function \[{\cos ^{ - 1}}\left( {{{2{{\sin }^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right)} \over \pi }} \right)\] is :
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Official solution
\[ - 1 \le {{2{{\sin }^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right)} \over \pi } \le 1\]
\[ \Rightarrow - {\pi \over 2} \le {\sin ^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right) \le {\pi \over 2}\]
\[ \Rightarrow - 1 \le {1 \over {4{x^2} - 1}} \le 1\]
\[\therefore\] \[{1 \over {4{x^2} - 1}} + 1 \ge 0\]
\[ \Rightarrow {{1 + 4{x^2} - 1} \over {4{x^2} - 1}} \ge 0\]
\[ \Rightarrow {{4{x^2}} \over {4{x^2} - 1}} \ge 0\]
\[ \Rightarrow \] \[{{4{x^2}} \over {\left( {2x + 1} \right)\left( {2x - 1} \right)}} \ge 0\] ...... (1)
\[\therefore\] \[x \in \left( { - \alpha , - {1 \over 2}} \right) \cup \{ 0\} \cup \left( {{1 \over 2},\alpha } \right)\] .....(2)
And \[{1 \over {4{x^2} - 1}} - 1 \le 0\]
\[ \Rightarrow {{1 - 4{x^2} + 1} \over {4{x^2} - 1}} \le 0\]
\[ \Rightarrow {{2 - 4{x^2}} \over {4{x^2} - 1}} \le 0\]
\[ \Rightarrow {{2{x^2} - 1} \over {4{x^2} - 1}} \ge 0\]
\[ \Rightarrow \] \[{{\left( {\sqrt 2 x + 1} \right)\left( {\sqrt 2 x - 1} \right)} \over {\left( {2x + 1} \right)\left( {2x - 1} \right)}} \ge 0\] ...... (3)
\[x \in \left( { - \alpha , - {1 \over {\sqrt 2 }}} \right) \cup \left( { - {1 \over 2},{1 \over 2}} \right) \cup \left( {{1 \over {\sqrt 2 }},\alpha } \right)\] .....(4)
From (3) and (4), we get
\[\therefore\] \[x \in \left[ { - \alpha , - {1 \over {\sqrt 2 }}} \right) \cup \left[ {{1 \over {\sqrt 2 }},\alpha } \right) \cup \{ 0\} \]
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