Q57 / 90 · JEE Main 2024 (04 Apr Shift 2) · Mathematics
Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x)=\frac{2 x}{\sqrt{1+9 x^2}}$. If the composition of $f, \underbrace{(f \circ f \circ f \circ \cdots \circ f)}_{10 \text { times }}(x)=\frac{2^{10} x}{\sqrt{1+9 \alpha x^2}}$, then the value of $\sqrt{3 \alpha+1}$ is equal to ______
Correct answer: 1024
Solution
$\begin{aligned}
& f(f(x))=\frac{2 f(x)}{\sqrt{1+9 f^2(x)}}=\frac{4 x}{\sqrt{1+9 x^2+9.2^2 x^2}} \\
& f(f(f(x)))=\frac{2^3 x / \sqrt{1+9 x^2}}{\sqrt{1+9\left(1+2^2\right) \frac{2^2 x^2}{1+9 x^2}}}=\frac{2^3 x}{\sqrt{1+9 x^2\left(1+2^2+2^4\right)}}
\end{aligned}$
$\therefore$ By observation
$\begin{aligned}
& \alpha=1+2^2+2^4+\ldots .+2^{18}=1\left(\frac{\left(2^2\right)^{10}-1}{2^2-1}\right)=\frac{2^{20}-1}{3} \\
& 3 \alpha+1=2^{20} \rightarrow \sqrt{3 \alpha+1}=2^{10}=1024
\end{aligned}$
Hence, the correct answer is 1024.