Question

If $\alpha \in (0,\frac{\pi }{2})$ 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

The correct option is A $2\tan \alpha$

We know that for positive quantities, A.M is always greater or equal to than G.M.

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

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

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