Question

If $a\in (0,\frac{\pi }{2})$, 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 $A.M.\ge G.M.$

therefore, $\frac{a+b}{2}\ge \sqrt{ab}$

let $a=\sqrt{x^2+x},b=\frac{ta{n}^2\alpha }{\sqrt{x^2+x}}$

$⟹\frac{\sqrt{x^2+x}+\frac{ta{n}^2\alpha }{\sqrt{x^2+x}}}{2}\ge \sqrt{\sqrt{x^2+x}\ast \frac{ta{n}^2\alpha }{\sqrt{x^2+x}}}$

$⟹\frac{\sqrt{x^2+x}+\frac{ta{n}^2\alpha }{\sqrt{x^2+x}}}{2}\ge \sqrt{ta{n}^2\alpha }$

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

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