Question

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:

• 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}{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$