Question

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

• A. 2 $\tan \alpha$
• B. 1
• C. 2
• D. ${\sec }^2\alpha$

Answer 1

Answer: (A)

2 $\tan \alpha$

Here $\alpha \in (0,\frac{\pi }{2})\Rightarrow \tan \alpha >0$

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