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 $(x\ne 0,-1)$

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

Answer 1

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

Clearly $\sqrt{x^2+x}$ and $\frac{{\tan }^2\alpha }{\sqrt{x^2+x}}$ are positive non zero numbers
Thus using $A.M\ge G.M$
$\sqrt{x^2+x}+\frac{{\tan }^2\alpha }{\sqrt{x^2+x}}\ge 2{(\sqrt{x^2+x}.\frac{{\tan }^2\alpha }{\sqrt{x^2+x}})}^{1/2}=2\tan \alpha$