Sequences And Series Ques 5
- If $\alpha \in\left(0, \frac{\pi}{2}\right)$, then $\sqrt{x^2+x}+\frac{\tan ^2 \alpha}{\sqrt{x^2+x}}$ is always greater than or equal to
(2003, 2M)
(a) $2 \tan \alpha$
(b) $1$
(c) $2$
(d) $\sec ^2 \alpha$
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Answer:
Correct Answer: 5.(a)
Solution: (a) Here, $\alpha \in\left(0, \frac{\pi}{2}\right) \Rightarrow \tan \alpha>0$
$ \therefore \quad \frac{\sqrt{x^2+x}+\frac{\tan ^2 \alpha}{\sqrt{x^2+x}}}{2} \geq \sqrt{\sqrt{x^2+x} \cdot \frac{\tan ^2 \alpha}{\sqrt{x^2+x}}} $
[using $\mathrm{AM} \geq \mathrm{GM}$ ]
$\Rightarrow \quad \sqrt{x^2+x}+\frac{\tan ^2 \alpha}{\sqrt{x^2+x}} \geq 2 \tan \alpha$
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