Question

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

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

Answer 1

Answer: (A)

$2\tan \alpha$

Applying $A.M\ge G.M$ we get
$\frac{1}{2}[\sqrt{x^2+x}+\frac{{\tan }^2\alpha }{\sqrt{x^2+x}}]\ge \sqrt{\sqrt{x^2+x}.\frac{{\tan }^2\alpha }{\sqrt{x^2+x}}}$
Or
$\frac{1}{2}[\sqrt{x^2+x}+\frac{{\tan }^2\alpha }{\sqrt{x^2+x}}]\ge |\tan \alpha |$.
Or
$[\sqrt{x^2+x}+\frac{{\tan }^2\alpha }{\sqrt{x^2+x}}]\ge 2|\tan \alpha |$.