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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We use the $AM \geq GM$ inequality for $7$ positive terms: $\frac{x^2}{2} + \frac{x^2}{2} + \frac{x^2}{2} + \frac{x^2}{2} + \frac{x^2}{2} + \frac{

Vedclass · Accepted Answer

$128$

Vedclass · Answer

(C) We use the $AM \geq GM$ inequality for $7$ positive terms: $\frac{x^2}{2} + \frac{x^2}{2} + \frac{x^2}{2} + \frac{x^2}{2} + \frac{x^2}{2} + \frac{\alpha}{2x^5} + \frac{\alpha}{2x^5} \geq 7 \sqrt[7]{\left(\frac{x^2}{2}\right)^5 \cdot \left(\frac{\alpha}{2x^5}\right)^2}$.Simplifying the expression inside the root: $7 \sqrt[7]{\frac{x^{10}}{2^5} \cdot \frac{\alpha^2}{4x^{10}}} = 7 \sqrt[7]{\frac{\alpha^2}{2^7}} = \frac{7 \cdot \alpha^{2/7}}{2}$.Given that the minimum value is $14$,we set $\frac{7 \cdot \alpha^{2/7}}{2} = 14$.$\alpha^{2/7} = \frac{14 \cdot 2}{7} = 4$.$\alpha^{2/7} = 2^2$.Raising both sides to the power of $7/2$: $\alpha = (2^2)^{7/2} = 2^7 = 128$.

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

Similar Questions

The function $f(x) = 2x^3 - 15x^2 + 36x + 4$ is maximum at $x=$ ......

The function $f(x) = x^5 - 5x^4 + 5x^3 - 10$ has a local maximum when $x$ is equal to:

The slant height of a right circular cone is $3 \text{ cm}$. The height of the cone for maximum volume is

The minimum distance of a point on the curve $y=x^2-4$ from the origin is

Let $f(x) = \int_{0}^{x} e^{x+t} dt$. Then the abscissa of the point where the tangent to $f(x)$ is parallel to the $x$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module