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If 1 < x < sqrt2, then number of solutio...

If `1 < x < sqrt2,` then number of solutions of the equation `tan^-1(x-1)+tan^-1 x+tan^-1(x+1)=tan^-1 3x` is

A

`+-(1)/(2)`

B

`0,(1)/(2)`

C

`0,-(1)/(2)`

D

`0+-(1)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
Given , `tan^(-1)(x-1)+tan^(-1)x=tan^(-1)3x-tan^(-1)(x+1)`
`rArrtan^(-1)[((x-1)+x)/(1-(x-1)x)]="tan"^(-1)[(3x-(x+1))/(1+3x(x+1))]`
`rArr(1+3x^(2)+3x)(2x-1)=(1-x^(2)+x)(2x-1)`
`rArr(2x-1)(4x^(2)+2x)=0`
`rArrx=0,+-(1)/(2)`
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