Question

If $\alpha \in \left(0,\frac{\pi }{2}\right)$, 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. $se{c}^2\alpha$

Answer 1

The correct option is A

$2\tan \alpha$

Explanation for the correct option:

Step 1. Find the value of $\sqrt{(x^2+x)}+\frac{\tan 2\alpha }{\sqrt{(x^2+x)}}$:

Given that $\alpha \in \left(0,\frac{\pi }{2}\right)$

Thus, $\tan \alpha$ is positive.

Let $\sqrt{(x^2+x)}$ and $\frac{\tan 2\alpha }{\sqrt{(x^2+x)}}$ be the two terms.

Step 2. As we know $AM\ge GM$

$\Rightarrow$ $\frac{\sqrt{(x^2+x)}+{\displaystyle \frac{\tan 2\alpha }{\sqrt{(x^2+x)}}}}{2}\ge \sqrt{\sqrt{(x^2+x)}\left(\frac{\tan 2\alpha /}{\sqrt{(x^2+x)}}\right)}$

$\Rightarrow$ $\sqrt{(x^2+x)}+\frac{\tan 2\alpha }{\sqrt{(x^2+x)}}\ge 2\sqrt{\left({\tan }^2\alpha \right)}$

$\Rightarrow$$\sqrt{\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{x}\mathbf{)}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{2}\mathbf{\alpha }}{\sqrt{\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{x}\mathbf{)}}}\mathbf{}\mathbf{\ge }\mathbf{}\mathbf{}\mathbf{2}\mathbf{}\mathit{t}\mathit{a}\mathit{n}\mathit{\alpha }$

Hence, Option ‘A’ is Correct.