Find the domain of the inverse trigonometric function \({\sin ^{ - 1}

1.	Find the domain of the inverse trigonometric function \({\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)\) is,1. \(\left[ { - 1,1} \right]\)2. \(\left[ {0,\frac{1}{2}} \right]\)3. \(\left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]\)4. \(\left[ { - \frac{1}{2},\frac{1}{2}} \right]\)
Answer» Correct Answer - Option 3 : \(\left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]\) Concept: The domain of inverse sine function, sin x is \(x \in \left[ { - 1,1} \right]\) Calculation: Domain of the function is calculated as follows: \({\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)\) \(- 1 \le 2x\sqrt {1 - {x^2}} \le 1\) \( - \frac{1}{2} \le x\sqrt {1 - {x^2}} \le \frac{1}{2}\) \({x^2}\left( {1 - {x^2}} \right) \le \frac{1}{4}\) \(t - {t^2} - \frac{1}{4} \le 0\) \({\left( {t - \frac{1}{2}} \right)^2} \le 0\) \(t \le \frac{1}{2}\) \({x^2} \le \frac{1}{{\sqrt 2 }}\) \(x \in \left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]\)

Find the domain of the inverse trigonometric function \({\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)\) is,1. \(\left[ { - 1,1} \right]\)2. \(\left[ {0,\frac{1}{2}} \right]\)3. \(\left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]\)4. \(\left[ { - \frac{1}{2},\frac{1}{2}} \right]\)

Answer» Correct Answer - Option 3 : \(\left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]\)

Concept:

The domain of inverse sine function, sin x is \(x \in \left[ { - 1,1} \right]\)

Calculation:

Domain of the function is calculated as follows:

\({\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)\)

\(- 1 \le 2x\sqrt {1 - {x^2}} \le 1\)

\( - \frac{1}{2} \le x\sqrt {1 - {x^2}} \le \frac{1}{2}\)

\({x^2}\left( {1 - {x^2}} \right) \le \frac{1}{4}\)

\(t - {t^2} - \frac{1}{4} \le 0\)

\({\left( {t - \frac{1}{2}} \right)^2} \le 0\)

\(t \le \frac{1}{2}\)

\({x^2} \le \frac{1}{{\sqrt 2 }}\)

\(x \in \left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]\)