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lf \(\alpha\in (0,\ \displaystyle \frac{\pi}{2})\) then \(\displaystyle \sqrt{\mathrm{x}^{2}+\mathrm{x}}+\frac{\tan^{2}\alpha}{\sqrt{\mathrm{x}^{2}+\mathrm{x}}}\) is always greater than or equal to



(A) 2 \(\tan\alpha\)
(B) 1
(C) 2
(D) \(\sec^{2}\alpha\)

Answer

Answer: A
Here \(\displaystyle\alpha \in \left( 0,\frac { \pi }{ 2 } \right) \Rightarrow \tan { \alpha } >0\)

\(\displaystyle\therefore \frac { \sqrt { { x }^{ 2 }+x+ } \frac { \tan ^{ 2 }{ \alpha } }{ \sqrt { { x }^{ 2 }+x } } }{ 2 } \ge \sqrt { \sqrt { { x }^{ 2 }+x}.\frac { \tan ^{ 2 }{ \alpha } }{ \sqrt { { x }^{ 2 }+x } } } \)
(Using A.M>G.M)
\(\displaystyle\Rightarrow \sqrt { { x }^{ 2 }+x } +\frac { \tan ^{ 2 }{ \alpha } }{ \sqrt { { x }^{ 2 }+x } } \ge 2\tan { \alpha } \)
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