Animated Solution for Mathematics - Binomial Theorem: If the term independent of $x$ in the expansion of $(\sqrt{a{x}^2}+\frac{1}{2x^3}{)}^{10}$ is 105 , then $a^2$ is equal to :

A Journey into the Independent Term

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of a binomial expression to find a hidden treasure—the term independent of $x$.

This is a classic JEE Advanced challenge, one that tests your precision, your algebraic stamina, and your ability to see the structure within the chaos.

The DNA of the Expansion

We are presented with the expression $(\sqrt{a}{x}^2+\frac{1}{2x^3}{)}^{10}$. At first glance, it looks like a daunting mountain of variables and powers.

But remember, the Binomial Theorem is our map. The general term formula, $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)(A{)}^{n-r}(B{)}^r$, is the DNA of this expansion. It allows us to look at any single term without having to expand the entire expression.

Here, our $n=10$, $A=\sqrt{a}{x}^2$, and $B=\frac{1}{2x^3}$. By substituting these into our formula, we get:

The Hunt for $x^0$

Now, we must be surgical. We need to isolate the variable $x$. Let us separate the constants from the variables.

We pull out the binomial coefficient $\left(\genfrac{}{}{0ex}{}{10}{r}\right)$, the constant $\sqrt{a}$, and the fraction $\frac{1}{2}$. What remains are the powers of $x$:

Focus your attention on the $x$ terms. We have $x^{2(10-r)}$ in the numerator and $x^{3r}$ in the denominator. Using the laws of exponents, we combine them: $x^{20-2r}\cdot x^{-3r}=x^{20-5r}$.

This is the moment of truth. The problem demands a term 'independent of $x$'. As we discussed, this means the exponent must be zero.

So, we set $20-5r=0$, which elegantly gives us $r=4$. We have found our target! The fifth term ($T_5$) is the one we are looking for.

The Final Calculation

With $r=4$ in our pocket, the rest is a beautiful dance of arithmetic. We substitute $r=4$ back into our coefficient expression:

Calculating $\left(\genfrac{}{}{0ex}{}{10}{4}\right)$ is straightforward:

And $(\sqrt{a}{)}^6$ simplifies beautifully to $a^3$. So, our coefficient is $\frac{210\cdot a^3}{16}$. The problem states this value is $105$. Thus:

Dividing both sides by $105$, we get $\frac{2}{16}{a}^3=1$, which simplifies to $\frac{a^3}{8}=1$, or $a^3=8$. This gives us $a=2$.

Finally, the question asks for $a^2$. Since $a=2$, $a^2=4$.

Look at what we have achieved. We took a complex binomial expression and, through the systematic application of the general term formula, reduced it to a simple algebraic equation.

This is the essence of JEE mathematics—breaking down the intimidating into the manageable. Keep this clarity of thought, and no problem will ever be too large for you to solve.