A) \[\frac{\pi }{6}-\frac{1}{4}log3\]
B) \[\frac{\pi }{3}+\frac{1}{4}.log3\]
C) \[\frac{\pi }{3}+\frac{1}{4}log3\]
D) \[\frac{\pi }{3}-\frac{1}{4}log3\]
Correct Answer: D
Solution :
[d] \[\because \,f(x)=ta{{n}^{-1}}x-\frac{1}{2}logx\] \[f(x)=\frac{1}{1+{{x}^{2}}}-\frac{1}{2}.\frac{1}{x}=\frac{2x-(1+{{x}^{2}})}{2x(1+{{x}^{2}})}=\frac{-{{(1-x)}^{2}}}{2x(1+{{x}^{2}})}\] If \[{{f}^{'}}\left( x \right)=0\] \[x-1=0\Rightarrow x=1\] For maximum value of f(x) put the values of x lies in the given interval in f(x). \[\therefore f(I)={{\tan }^{-1}}1-\frac{1}{2}\log 1=\frac{\pi }{4}-0=\frac{\pi }{4}\] \[f\left( \frac{1}{\sqrt{3}} \right)={{\tan }^{-1}}\left( \frac{1}{\sqrt{3}} \right)-\frac{1}{2}\log \left( \frac{1}{\sqrt{3}} \right)\] \[=\frac{\pi }{6}+\frac{1}{4}.\log 3\] \[f(\sqrt{3})={{\tan }^{-1}}\left( \sqrt{3} \right)-\frac{1}{2}\log (\sqrt{3})=\frac{\pi }{3}-\frac{1}{4}.\log 3\]Thus, the least value of \[f\left( x \right)=\frac{\pi }{3}-\frac{1}{4}.log3\] Hence, option [d] is correct.
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