Question

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

• A. 32
• B. 64
• C. 128
• D. 256

Answer 1

Answer: (C)

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