A) \[\frac{1}{\sqrt{3}}\]
B) \[\frac{1}{3}\]
C) \[3\]
D) \[\sqrt{3}\]
Correct Answer: B
Solution :
| [b] 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 \[Z=\tan A.\tan B\]. 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 \[x=\frac{1}{\sqrt{3}}i.e.,\tan A=\frac{1}{\sqrt{3}}orA=\pi /6\]. |
| For this value of \[x,Z=\frac{1}{3}.\] |
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