A) - 1
B) \[\frac{1}{6}\]
C) \[-1,\,\frac{1}{6}\]
D) None of these
Correct Answer: B
Solution :
\[{{\tan }^{-1}}2x+{{\tan }^{-1}}3x=\frac{\pi }{4}\]\[\Rightarrow \,\,{{\tan }^{-1}}\,\left( \frac{2x+3x}{1-(2x)\,(3x)} \right)=\frac{\pi }{4}\,\] \[\Rightarrow \,\,{{\tan }^{-1}}\,\left( \frac{5x}{1-6{{x}^{2}}} \right)\,={{\tan }^{-1}}(1)\]\[\Rightarrow \,\,\frac{5x}{1-6{{x}^{2}}}=1\,\] \[\Rightarrow \,\,1-6{{x}^{2}}=5x\]\[\,\Rightarrow \,\,6{{x}^{2}}+5x-1=0\] \[\Rightarrow \,\,(x+1)\,\left( x-\frac{1}{6} \right)=0\,\]\[\Rightarrow \,\,x=-1,\,\,\frac{1}{6}\] But - 1 does not hold. Trick: Check with the options. Obviously the equation holds for\[x=\frac{1}{6}\], but not for - 1.
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