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

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(B) Given the function $f(x) = \sqrt{x^2 + x} + \frac{	an^2 \alpha}{\sqrt{x^2 + x}}$.Since $x > 0$,the term $\sqrt{x^2 + x}$ is always positive.We us

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$2 	an \alpha$

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(B) Given the function $f(x) = \sqrt{x^2 + x} + \frac{	an^2 \alpha}{\sqrt{x^2 + x}}$.Since $x > 0$,the term $\sqrt{x^2 + x}$ is always positive.We use the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality,which states that for positive real numbers $a$ and $b$,$\frac{a + b}{2} \ge \sqrt{ab}$,or $a + b \ge 2\sqrt{ab}$.Let $a = \sqrt{x^2 + x}$ and $b = \frac{	an^2 \alpha}{\sqrt{x^2 + x}}$.Then $f(x) = a + b \ge 2\sqrt{a \cdot b}$.$f(x) \ge 2\sqrt{\sqrt{x^2 + x} \cdot \frac{	an^2 \alpha}{\sqrt{x^2 + x}}}$.$f(x) \ge 2\sqrt{	an^2 \alpha}$.Since $\alpha \in (0, \pi/2)$,$	an \alpha > 0$,so $f(x) \ge 2 	an \alpha$.The minimum value of $f(x)$ is $2 	an \alpha$.

If $f(x) = \sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}$,where $\alpha \in (0, \pi/2)$ and $x > 0$,find the minimum value of $f(x)$.

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