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Question: If \(\tan ^ { - 1 } ( x - 1 ) + \tan ^ { - 1 } x + \tan ^ { - 1 } ( x + 1 ) = \tan ^ { - 1 } 3 x\),t...

If tan⁡−1(x−1)+tan⁡−1x+tan⁡−1(x+1)=tan⁡−13x\tan ^ { - 1 } ( x - 1 ) + \tan ^ { - 1 } x + \tan ^ { - 1 } ( x + 1 ) = \tan ^ { - 1 } 3 x,then x =

A

±12\pm \frac { 1 } { 2 }

B

0,120 , \frac { 1 } { 2 }

C

0,−120 , - \frac { 1 } { 2 }

D

0,±120 , \pm \frac { 1 } { 2 }

Answer

0,±120 , \pm \frac { 1 } { 2 }

Explanation

Solution

tan⁡−1(x−1)+tan⁡−1(x)+tan⁡−1(x+1)=tan⁡−13x\tan ^ { - 1 } ( x - 1 ) + \tan ^ { - 1 } ( x ) + \tan ^ { - 1 } ( x + 1 ) = \tan ^ { - 1 } 3 x

⇒tan⁡−1(x−1)+tan⁡−1(x)=tan⁡−13x−tan⁡−1(x+1)\Rightarrow \tan ^ { - 1 } ( x - 1 ) + \tan ^ { - 1 } ( x ) = \tan ^ { - 1 } 3 x - \tan ^ { - 1 } ( x + 1 )

⇒tan⁡−1[(x−1)+x1−(x−1)(x)]=tan⁡−1[3x−(x+1)1+3x(x+1)]\Rightarrow \tan ^ { - 1 } \left[ \frac { ( x - 1 ) + x } { 1 - ( x - 1 ) ( x ) } \right] = \tan ^ { - 1 } \left[ \frac { 3 x - ( x + 1 ) } { 1 + 3 x ( x + 1 ) } \right]

⇒2x−11−x2+x=2x−11+3x2+3x\Rightarrow \frac { 2 x - 1 } { 1 - x ^ { 2 } + x } = \frac { 2 x - 1 } { 1 + 3 x ^ { 2 } + 3 x }

⇒(1−x2+x)(2x−1)=(1+3x2+3x)(2x−1)\Rightarrow \left( 1 - x ^ { 2 } + x \right) ( 2 x - 1 ) = \left( 1 + 3 x ^ { 2 } + 3 x \right) ( 2 x - 1 )

On simplification x=0,±12x = 0 , \pm \frac { 1 } { 2 }