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

Solution

## The correct option is C none of theseFor 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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