JEE Main 2024 (Online) 4th April Evening ShiftQuestion 1
Consider the function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=\dfrac{2x}{\sqrt{1+9x^{2}}}$. If the composition of $f$, $\underbrace{(f\circ f\circ f\circ\cdots\circ f)}_{10\text{ times}}(x)=\dfrac{2^{10}x}{\sqrt{1+9\alpha x^{2}}}$, then the value of $\sqrt{3\alpha+1}$ is equal to $\_\_\_\_\_\_\_\_$ .
Functions
Answer$1024$
Solution
By induction $f^{(n)}(x)=\frac{2^nx}{\sqrt{1+3(4^n-1)x^2}}$, so $9\alpha=3(4^{10}-1)$, $3\alpha+1=4^{10}$ and $\sqrt{3\alpha+1}=2^{10}=1024$.