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If $\tan^{-1}(x-1)+\tan^{-1}x+\tan^{-1}(x+1)=\tan^{-1}3x$ then $x$ is

$(a)\;\pm\large\frac{1}{2}$$\qquad(b)\;0,\large\frac{1}{2}\qquad$$(c)\;0,-\large\frac{1}{2}$$\qquad(d)\;0,\pm\large\frac{1}{2}$

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1 Answer

Given :
$\tan^{-1}(x-1)+\tan^{-1}(x)+\tan^{-1}(x+1)=\tan^{-1}(3x)$
$\Rightarrow \tan^{-1}(x-1)+\tan^{-1}x=\tan^{-1}(3x)-\tan^{-1}(x+1)$
$\Rightarrow \tan^{-1}\big[\large\frac{(x-1)+x}{1-(x-1)x}\big]$$=\tan^{-1}\big[\large\frac{3x-(x+1)}{1+3x(x+1)}\big]$
$\Rightarrow \large\frac{2x-1}{1-x^2+x}=\frac{2x-1}{1+3x^2+3x}$
$(2x-1)(1+3x^2+3x)=(1-x^2+x)(2x-1)$
$\Rightarrow (2x-1)(4x^2+2x)=0$
$\Rightarrow (2x-1)2x(2x+1)=0$
$\Rightarrow x=0,\pm \large\frac{1}{2}$
Hence (d) is the correct answer.
answered Nov 19, 2013 by sreemathi.v
 

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