Animated Solution for Mathematics - Functions: Consider the function $f:\mathbb{R}\to \mathbb{R}$ defined by $f(x)=\frac{2x}{\sqrt{1+9x^2}}$. If the composition of $f,\underset{10\text{ times}}{\underbrace{(f\circ f\circ f\circ \cdots \circ f)}}(x)=\frac{2^{10}x}{\sqrt{1+9\alpha x^2}}$, then the value of $\sqrt{3\alpha +1}$ is equal to ______

Imagine you are standing before a mountain. If you try to climb it in one giant leap, you will fail. But if you take it step by step, the summit becomes inevitable.

This problem is exactly like that mountain. We are asked to find the $10^{\text{th}}$ composition of the function:

At first glance, the idea of composing this function ten times feels like a nightmare of algebra. But in the world of JEE Advanced, whenever you see a repeated operation, there is a hidden pattern waiting to be discovered. We do not calculate $f_{10}(x)$ directly; we observe the evolution of the function.

Unveiling the Pattern

Let us calculate $f_2(x)=f(f(x))$. When we substitute $f(x)$ into itself, we get:

Substituting the expression for $f(x)$, the numerator becomes $\frac{4x}{\sqrt{1+9x^2}}$. The denominator is where the magic happens:

When we combine the numerator and denominator, the $\sqrt{1+9x^2}$ terms cancel out beautifully, leaving us with:

Notice that $45=9(1+4)$. This is the spark! We can write this as $f_2(x)=\frac{2^{2}x}{\sqrt{1+9x^2(1+2^2)}}$.

Seeing the Infinite in the Finite

If we repeat this for $f_3(x)$, the numerator will become $2^{3}x$, and the denominator will accumulate another term, $2^4$. The pattern is now clear.

For any $n$, the $n^{\text{th}}$ composition is:

This is not just algebra; it is a symphony of numbers. The term inside the bracket is a geometric progression with first term $a=1$, common ratio $r=2^2=4$, and $n$ terms.

The sum of this GP is:

Solving for Alpha

We are given that $f_{10}(x)=\frac{2^{10}x}{\sqrt{1+9\alpha x^2}}$. By comparing our derived formula with the given form, we identify that $\alpha$ is the sum of our GP for $n=10$.

Thus, $\alpha =\frac{2^{20}-1}{3}$. The question asks for $\sqrt{3\alpha +1}$.

Substituting our value of $\alpha$, we get:

The $3$ cancels out, leaving $\sqrt{2^{20}-1+1}$, which simplifies to $\sqrt{2^{20}}$. The square root of $2^{20}$ is simply $2^{10}$, which is 1024.

We have reached the summit. The complexity vanished, replaced by the elegance of a geometric series. Remember, in mathematics, patience is your greatest tool.