JEE MAIN - Mathematics (2024 - 4th April Evening Shift - No. 26)
Explanation
$$f(x)=\frac{2 x}{\sqrt{1+9 x^2}}$$
$$(f \circ f)(x)=\frac{2 f(x)}{\sqrt{1+9(f(x))^2}}=\frac{\frac{4 x}{\sqrt{1+9 x^2}}}{\sqrt{1+9 \times \frac{4 x^2}{1+9 x^2}}}=\frac{4 x}{\sqrt{1+45 x^2}}$$
$$(f \circ f \circ f)(x)=\frac{4 \times \frac{2 x}{\sqrt{1+9 x^2}}}{\sqrt{1+45 \times \frac{4 x^2}{1+9 x^2}}}=\frac{8 x}{\sqrt{1+21 \times 9 x^2}}$$
$$(f \circ f \circ f \circ f)(x)=\frac{16 x}{\sqrt{1+85 \times 9 x^2}}$$
$$\Rightarrow \alpha$$ is $$10^{\text {th }}$$ term of $$1,5,21,85, \ldots \alpha$$ is $$10^{\text {th }}$$ term of
$$\begin{aligned} & \frac{\left(2^1\right)^2-1}{3}, \frac{\left(2^2\right)^2-1}{3}, \frac{\left(2^3\right)^2-1}{3}, \frac{\left(2^4\right)^2-1}{3}, \ldots \\ \Rightarrow \quad & \alpha=\frac{\left(2^{10}\right)^2-1}{3} \\ \Rightarrow \quad & \sqrt{3 \alpha+1}=2^{10}=1024 \end{aligned}$$
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