If f(x) = sqrt{x^2 + x} + frac{tan^2 alpha}{s | vedclass

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(A) We are given $f(x) = \sqrt{x^2 + x} + \frac{	an^2 \alpha}{\sqrt{x^2 + x}}$.Since $x > 0$,$\sqrt{x^2 + x} > 0$. Also,$	an^2 \alpha > 0$ for $\alp

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$2 	an \alpha$

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(A) We are given $f(x) = \sqrt{x^2 + x} + \frac{	an^2 \alpha}{\sqrt{x^2 + x}}$.Since $x > 0$,$\sqrt{x^2 + x} > 0$. Also,$	an^2 \alpha > 0$ for $\alpha \in (0, \pi/2)$.Using the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for two positive terms $a$ and $b$,we have $\frac{a+b}{2} \geq \sqrt{ab}$,or $a+b \geq 2\sqrt{ab}$.Let $a = \sqrt{x^2 + x}$ and $b = \frac{	an^2 \alpha}{\sqrt{x^2 + x}}$.Then $f(x) = a + b \geq 2\sqrt{a \cdot b}$.$f(x) \geq 2\sqrt{\sqrt{x^2 + x} \cdot \frac{	an^2 \alpha}{\sqrt{x^2 + x}}}$.$f(x) \geq 2\sqrt{	an^2 \alpha}$.Since $\alpha \in (0, \pi/2)$,$	an \alpha > 0$,so $\sqrt{	an^2 \alpha} = 	an \alpha$.Therefore,$f(x) \geq 2 	an \alpha$.

If $f(x) = \sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}$,where $\alpha \in (0, \pi/2)$ and $x > 0$,then the value of $f(x)$ is greater than or equal to:

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