Question

If $\alpha ϵ[0,\frac{\pi }{2})$, 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, $\alpha ϵ[0,\frac{\pi }{2})\Rightarrow \tan \alpha$ is $(+ve)$

As, we know, if $a,b>0\Rightarrow \frac{a+b}{2}\ge \sqrt{ab}$.

i.e., $AM\ge GM$

$\frac{\sqrt{x^2+x}+\frac{{\tan }^2\alpha }{\sqrt{x^2+x}}}{2}\ge \sqrt{\sqrt{x^2+x}\cdot \frac{{\tan }^2\alpha }{\sqrt{x^2+x}}}$ (using $AM\ge GM)$

$\Rightarrow \sqrt{x^2+x}+\frac{{\tan }^2\alpha }{\sqrt{x^2+x}}\ge 2\tan \alpha$