Question:
Considering only the principal values of inverse functions, the set
$A=\left\{x \geq 0: \tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}\right\}$
Correct Option: , 4
Solution:
$\tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\pi / 4$
$\Rightarrow \frac{5 x}{1-6 x^{2}}=1$
$\Rightarrow 6 x^{2}+5 x-1=0$
$x=-1$ or $x=\frac{1}{6}$
$x=\frac{1}{6} \quad \because x>0$
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