Animated Solution for Mathematics - Sequence and Series: 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 :

Beyond Calculus

Imagine you are standing on a graph, looking at the function $f(x)=\frac{5x^2}{2}+\frac{\alpha }{x^5}$. It is a beautiful, U-shaped curve in the first quadrant, dipping down to a precise, absolute minimum value of $14$.

Your mission is to uncover the hidden constant $\alpha$. While your first instinct might be to reach for the derivative, let us pause. There is a more powerful, more elegant tool in our mathematical arsenal: the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

The AM-GM inequality is a fundamental principle that states for any set of positive real numbers $a_1,a_2,\dots ,a_n$, the arithmetic mean is always greater than or equal to the geometric mean:

This is not just a formula; it is a bridge between addition and multiplication. Our goal is to transform our function $f(x)$ into a sum of terms such that when we multiply them, the $x$ variables vanish, leaving us with a constant.

This is the secret to unlocking the minimum value without ever touching a derivative.

We have two terms: $\frac{5x^2}{2}$ and $\frac{\alpha }{x^5}$. If we multiply them directly, we get $\frac{5\alpha }{2x^3}$, which still depends on $x$. We need to eliminate $x$.

The powers are $2$ and $-5$. To make them cancel, we need to balance them. The least common multiple of $2$ and $5$ is $10$.

This tells us exactly what to do: we need $5$ copies of the $x^2$ term and $2$ copies of the $x^{-5}$ term. We split $\frac{5x^2}{2}$ into $5$ equal parts:

And we split $\frac{\alpha }{x^5}$ into $2$ equal parts:

Now, we have $7$ terms in total. The sum of these $7$ terms is exactly our function $f(x)$.

Applying the AM-GM inequality to these $7$ terms, we get:

Watch the magic happen as we simplify the product inside the root:

The $x^{10}$ terms cancel out perfectly! We are left with a constant. Now, the inequality becomes:

Multiplying by $7$, we find the minimum value of $f(x)$ is $7\cdot {\left(\frac{{\alpha }^2}{2^7}\right)}^{\frac{1}{7}}$.

We are given that the minimum value is $14$. So, we set our expression equal to $14$:

Dividing by $7$, we get $2={\left(\frac{{\alpha }^2}{2^7}\right)}^{\frac{1}{7}}$. Raising both sides to the power of $7$ yields $2^7=\frac{{\alpha }^2}{2^7}$.

Finally, ${\alpha }^2=2^7\cdot 2^7=2^{14}$. Taking the square root, we find $\alpha =2^7=128$.

And there it is! Through the power of AM-GM, we have navigated the problem with precision and grace. Keep practicing this technique—it is a powerful tool that will serve you well in your journey toward mastering JEE mathematics.