Question
The formula $${\sin ^{ - 1}}\left\{ {2x\left( {1 - {x^2}} \right)} \right\} = 2\,{\sin ^{ - 1}}x$$ is true for all values of $$x$$ lying in the interval
A.
$$\left[ { - 1,1} \right]$$
B.
$$\left[ { 0,1} \right]$$
C.
$$\left[ { - 1,0} \right]$$
D.
$$\left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]$$
Answer :
$$\left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]$$
Solution :
$${\sin ^{ - 1}}\left\{ {2x\left( {1 - {x^2}} \right)} \right\} = 2\,{\sin ^{ - 1}}x$$ is true $$\forall x \in \left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]$$