Question

Let $f\left(x\right)=\sqrt{x^2+x}+\frac{{\tan }^2\alpha }{\sqrt{x^2+x}},\alpha ϵ(0,\frac{\pi }{2}).$ The value of $f\left(x\right)$ is always greater than or equal to

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

Answer 1

Answer: (A)

$2\tan \alpha$

Now using A.M $\ge$ G.M
$\frac{\sqrt{x^2+x}+\frac{{\tan }^2\alpha }{\sqrt{x^2+x}}}{2}\ge {(\sqrt{x^2+x}.\frac{{\tan }^2\alpha }{\sqrt{x^2+x}})}^{1/2}=\tan \alpha$
$\therefore \sqrt{x^2+x}+\frac{{\tan }^2\alpha }{\sqrt{x^2+x}}\ge 2\tan \alpha$