If the minimum value of f(x)=frac{5 x^{2}}{2} | vedclass

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Question

$\frac{x^{2}}{2}+\frac{x^{2}}{2}+\frac{x^{2}}{2}+\frac{x^{2}}{2}+\frac{x^{2}}{2}+\frac{\alpha}{2 x^{5}}+\frac{\alpha}{2 x^{5}}$

$\geq 7\left(\frac{

Vedclass · Accepted Answer

$128$

Vedclass · Answer

$\frac{x^{2}}{2}+\frac{x^{2}}{2}+\frac{x^{2}}{2}+\frac{x^{2}}{2}+\frac{x^{2}}{2}+\frac{\alpha}{2 x^{5}}+\frac{\alpha}{2 x^{5}}$

$\geq 7\left(\frac{\alpha^{2}}{2^{7}}\right)^{\frac{1}{7}}$

$\frac{7 \cdot(\alpha)^{2 / 7}}{2}=14$

$\left(\alpha^{2}\right)^{1 / 7}=2^{2}$

$\alpha=\left(2^{2}\right)^{7 / 2}=2^{7}$

$\alpha=128$

If the minimum value of $f(x)=\frac{5 x^{2}}{2}+\frac{\alpha}{x^{5}}, x>0$, is 14 , then the value of $\alpha$ is equal to.

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