JEE MATHEMATICS MCQ: MAXIMA & MINIMA

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Application +4 | -1

\alpha \in\left(0, \frac{\pi}{2}\right)

; then minimum value of

\sqrt{x^2+x}+\frac{\tan ^2 \alpha}{\sqrt{x^2+x}}

View answer and explanation

Correct option: C

2 \tan \alpha

Option C (

2 \tan \alpha

) is correct because it represents the minimum value obtained by applying the AM-GM inequality to the given expression.

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