| \[A=\left\{ x\ge 0:{{\tan }^{-1}}(2x)+ta{{n}^{-1}}(3x)=\frac{\pi }{4} \right\}\] |
A) contains two elements
B) contains more than two elements
C) is an empty set
D) is a singleton
Correct Answer: D
Solution :
Here, \[A=\left\{ x\ge 0:{{\tan }^{-1}}(2x)+\tan {{\,}^{-1}}(3x)=\frac{\pi }{4} \right\}\] Now,\[\tan {{\,}^{-1}}(2x)+ta{{n}^{-3}}(3x)=\frac{\pi }{4}\] \[\Rightarrow \]\[{{\tan }^{-1}}\left( \frac{2x+3x}{1-2x\times 3x} \right)=\frac{\pi }{4}\] \[\Rightarrow \]\[{{\tan }^{-1}}\left( \frac{5x}{1-6{{x}^{2}}} \right)=\frac{\pi }{4}\] \[\Rightarrow \]\[\frac{5x}{1-6{{x}^{2}}}=\tan \frac{\pi }{4}=1\]\[\Rightarrow \]\[5x=1-6{{x}^{2}}\] \[\Rightarrow \]\[6{{x}^{2}}+5x-1=0\]\[\Rightarrow \]\[(6x-1)(x+1)=0\] \[\Rightarrow \]\[x=\frac{1}{6}\] \[[\because x\ge 0]\]
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