A) -1
B) -2
C) 1
D) 2
Correct Answer: A
Solution :
[a] Given: \[{{\tan }^{-1}}(2x)+ta{{n}^{-1}}(3x)=\frac{\pi }{4}\] \[\Rightarrow {{\tan }^{-1}}\frac{(2x+3x)}{(1-2x.3x)}={{\tan }^{-1}}(1)\] \[\Rightarrow \frac{5x}{1-6{{x}^{2}}}=1\Rightarrow 6{{x}^{2}}+5x-1=0\] \[\Rightarrow (6x-1)(x+1)=0\Rightarrow x=\frac{1}{6}or-1.\]
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