f: (0, 2 ) A_ 1 f(x)= _ 2 ( x^ x +1 ) , then the minimum value of f(x) is
Mathematics · JEE Main · NTA Exams — Sets, Relations and Functions
\(f:\left(0, \frac{\pi}{2}\right) \rightarrow A_{1} f(x)=\log _{2}\left(\sin x^{\sin x}+1\right)\), then the minimum value of \(f(x)\) is
- loge2
- \(\log _{e}\left(\left(\frac{1}{e}\right)^{1 / e}+1\right)\)
- \(\log _{2}\left[(e)^{2}+1\right]\)
- 2
Answer
(B) _ e ( ( 1 e )^ 1 / e +1 )
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