A) \[\frac{1}{\sqrt{3}}\]
B) \[\frac{1}{3}\]
C) \[3\]
D) \[\sqrt{3}\]
Correct Answer: B
Solution :
We have, \[A+B=\frac{\pi }{3}\] \[\therefore \,B=\frac{\pi }{3}-A\Rightarrow \tan B=\frac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A}\] Let \[Z=tan\text{ }A.\text{ }tan\text{ }B.\]Then, \[Z=\tan A.\frac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A}=\frac{\sqrt{3}\tan A-{{\tan }^{2}}A}{1+\sqrt{3}\tan A}\] \[\Rightarrow \,\,Z=\frac{\sqrt{3}x-{{x}^{2}}}{1+\sqrt{3}x},\] where \[x=\tan A\] \[\Rightarrow \,\,\frac{dZ}{dx}=-\frac{(x+\sqrt{3})(\sqrt{3}x-1)}{{{(1+\sqrt{3}x)}^{2}}}\] For max Z, \[\frac{dZ}{dx}=0\Rightarrow x=\frac{1}{\sqrt{3},}-\sqrt{3}\] \[x\ne -\sqrt{3}\] because \[A+B=\pi /3\] which implies that \[x=\tan A>0.\] It can be easily checked that \[\frac{{{d}^{2}}Z}{d{{x}^{2}}}<0\] for \[x=\frac{1}{\sqrt{3}}.\] Hence, Z is maximum for \[x=\frac{1}{\sqrt{3}}\] i.e. \[\tan A=\frac{1}{\sqrt{3}}\] or \[A=\pi /6\] For this value of x, \[Z=\frac{1}{3}\]
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