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 :
\[{{\tan }^{-1}}(x-1)+{{\tan }^{-1}}(x)+{{\tan }^{-1}}(x+1)={{\tan }^{-1}}3x\] \[\Rightarrow {{\tan }^{-1}}(x-1)+{{\tan }^{-1}}(x)={{\tan }^{-1}}3x-{{\tan }^{-1}}(x+1)\] \[\Rightarrow {{\tan }^{-1}}\left[ \frac{(x-1)+x}{1-(x-1)(x)} \right]={{\tan }^{-1}}\left[ \frac{3x-(x+1)}{1+3x(x+1)} \right]\] \[\Rightarrow \frac{2x-1}{1-{{x}^{2}}+x}=\frac{2x-1}{1+3{{x}^{2}}+3x}\] \[\Rightarrow (1-{{x}^{2}}+x)(2x-1)=(1+3{{x}^{2}}+3x)(2x-1)\] On simplification \[x=0,\,\pm \frac{1}{2}.\]
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