If alpha in left( 0, frac{pi}{2} right),then | vedclass

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(A) Let $y = \sqrt{x^2 + x} + \frac{	an^2 \alpha}{\sqrt{x^2 + x}}$.Since $\alpha \in \left( 0, \frac{\pi}{2} ight)$,$	an^2 \alpha > 0$. Also,for $

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$2 	an \alpha$

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(A) Let $y = \sqrt{x^2 + x} + \frac{	an^2 \alpha}{\sqrt{x^2 + x}}$.Since $\alpha \in \left( 0, \frac{\pi}{2} ight)$,$	an^2 \alpha > 0$. Also,for $x^2 + x > 0$,the expression is well-defined.Using the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$:$\frac{\sqrt{x^2 + x} + \frac{	an^2 \alpha}{\sqrt{x^2 + x}}}{2} \ge \sqrt{\sqrt{x^2 + x} \cdot \frac{	an^2 \alpha}{\sqrt{x^2 + x}}}$$\frac{\sqrt{x^2 + x} + \frac{	an^2 \alpha}{\sqrt{x^2 + x}}}{2} \ge \sqrt{	an^2 \alpha}$$\sqrt{x^2 + x} + \frac{	an^2 \alpha}{\sqrt{x^2 + x}} \ge 2 	an \alpha$.

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

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