A) \[\pm \frac{1}{2}\]
B) \[0,\frac{1}{2}\]
C) \[0,-\frac{1}{2}\]
D) \[0,\pm \frac{1}{2}\]
Correct Answer: D
Solution :
\[{{\tan }^{-1}}(x-1)+{{\tan }^{-1}}(x)+{{\tan }^{-1}}(x+1)={{\tan }^{-1}}3x\] \[\Rightarrow \] \[{{\tan }^{-1}}(x-1)+{{\tan }^{-1}}x={{\tan }^{-1}}3x-{{\tan }^{-1}}(x+1)\] \[\Rightarrow \] \[{{\tan }^{-1}}\left[ \frac{x-1+x}{1-(x-1)x} \right]\] \[={{\tan }^{-1}}\left[ \frac{3x-(x+1)}{1+3x(x+1)} \right]\] \[\Rightarrow \] \[\frac{2x-1}{1-{{x}^{2}}+x}=\frac{2x-1}{1+3{{x}^{2}}+3x}\] \[\Rightarrow \] \[(2x-1)(1+3{{x}^{2}}+3x)=(1-{{x}^{2}}-x)(2x-1)\] \[\Rightarrow \] \[(2x-1)(4{{x}^{2}}+2x)=0\] \[\Rightarrow \] \[(2x-1)2x(2x+1)=0\] \[\Rightarrow \] \[x=0,\pm \frac{1}{2}\]
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