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If  $$\displaystyle \:f\left ( x \right )= \sin ^{-1}\left \{ \frac{\sqrt{3}}{2}x-\frac{1}{2}\sqrt{1-x^{2}} \right \},-\frac{1}{2}\leq x\leq 1,$$ then $$\displaystyle \:f\left ( x \right )$$ is equal to


A
sin112sin1x
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B
sin1xπ6
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C
sin1x+π6
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D
none of these
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Solution

The correct option is A $$\displaystyle \:\sin ^{-1}x-\frac{\pi}{6}$$
$$\displaystyle \:f\left ( x \right )= \sin ^{-1}\left \{ \frac{\sqrt{3}}{2}x-\frac{1}{2}\sqrt{1-x^{2}} \right \}$$
Substitute $$x=\sin \theta$$
$${ \: f\left( x \right) =\sin ^{ -1 } \left\{ \dfrac { \sqrt { 3 }  }{ 2 } \sin { \theta  } -\dfrac { 1 }{ 2 } \cos { \theta  }  \right\}  }$$
$${ \: f\left( x \right) =\sin ^{ -1 } \left\{ \sin { (\theta -\dfrac { \pi  }{ 6 } ) }  \right\}  }$$
$$=\theta -\dfrac { \pi  }{ 6 } $$

$$=\sin ^{ -1 } x-\dfrac { \pi  }{ 6 } $$

Mathematics

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