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Question

Which of the following is the solution set of equation $$\sin ^{ -1 }{ x } =\cos ^{ -1 }{ x } +\sin ^{ -1 }{ \left( 3x-1 \right)  } $$?


A
[0,13]
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B
[13,23]
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C
[0,23]
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D
none of these
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Solution

The correct option is C none of these
For the existence of the given, we must have 
$$-1\le x\le 1\quad and\quad -1\le 3x-1\le 1\Rightarrow 0\le x\le \cfrac { 2 }{ 3 } $$
Now,
$$\sin ^{ -1 }{ x } =\cos ^{ -1 }{ x } +\sin ^{ -1 }{ \left( 3x-1 \right)  } $$
$$\Rightarrow \sin ^{ -1 }{ x } -\cos ^{ -1 }{ x } =\sin ^{ -1 }{ \left( 3x-1 \right)  } $$
$$\Rightarrow 2\sin ^{ -1 }{ x } -\cfrac { \pi  }{ 2 } =\sin ^{ -1 }{ \left( 3x-1 \right)  } $$
$$\Rightarrow \sin { \left( 2\sin ^{ -1 }{ x } -\cfrac { \pi  }{ 2 }  \right)  } x=\sin { \left( \sin ^{ -1 }{ \left( 3x-1 \right)  }  \right)  } $$
$$\Rightarrow -\cos { \left( 2\sin ^{ -1 }{ x }  \right)  } =3x-1$$
$$\Rightarrow -\left\{ 1-2\sin ^{ 2 }{ \sin ^{ -1 }{ x }  }  \right\} =3x-1$$
$$\Rightarrow -\left( 1-{ 2x }^{ 2 } \right) =3x-1$$
$$\Rightarrow 2{ x }^{ 2 }-3x=0\Rightarrow x=0,\cfrac { 3 }{ 2 }$$
$$ \Rightarrow x=0\quad \left[ \because \quad 0\le x\le { 2 }/{ 3 } \right] $$

Mathematics

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