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

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(A) We are given the expression $E = \sqrt{x^2 + x} + \frac{	an^2 \alpha}{\sqrt{x^2 + x}}.$ Since $\alpha \in \left( 0, \frac{\pi}{2} ight),$ $	an

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$2 	an \alpha$

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(A) We are given the expression $E = \sqrt{x^2 + x} + \frac{	an^2 \alpha}{\sqrt{x^2 + x}}.$ Since $\alpha \in \left( 0, \frac{\pi}{2} ight),$ $	an \alpha > 0,$ and for $x^2 + x > 0,$ both terms are positive. By the Arithmetic Mean-Geometric Mean Inequality $(AM \ge GM)$,for any two positive real numbers $a$ and $b,$ $a + b \ge 2\sqrt{ab}.$ Let $a = \sqrt{x^2 + x}$ and $b = \frac{	an^2 \alpha}{\sqrt{x^2 + x}}.$ Then $a + b \ge 2 \sqrt{\sqrt{x^2 + x} \cdot \frac{	an^2 \alpha}{\sqrt{x^2 + x}}}.$ $a + b \ge 2 \sqrt{	an^2 \alpha}.$ Since $	an \alpha > 0$ for $\alpha \in \left( 0, \frac{\pi}{2} ight),$ we have $a + b \ge 2 	an \alpha.$ Thus,the expression is always greater than or equal to $2 	an \alpha.$

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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