Consider the function f: R rightarrow R defin | vedclass

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(D) Let $f_n(x)$ denote the $n$-th composition of $f(x)$.$f_1(x) = \frac{2x}{\sqrt{1+9x^2}}$$f_2(x) = f(f(x)) = \frac{2 \cdot \frac{2x}{\sqrt{1+9x^2}}

Vedclass · Accepted Answer

$1024$

Vedclass · Answer

(D) Let $f_n(x)$ denote the $n$-th composition of $f(x)$.$f_1(x) = \frac{2x}{\sqrt{1+9x^2}}$$f_2(x) = f(f(x)) = \frac{2 \cdot \frac{2x}{\sqrt{1+9x^2}}}{\sqrt{1+9 \cdot \frac{4x^2}{1+9x^2}}} = \frac{4x}{\sqrt{1+9x^2+36x^2}} = \frac{2^2x}{\sqrt{1+9x^2(1+2^2)}}$$f_3(x) = f(f_2(x)) = \frac{2 \cdot \frac{2^2x}{\sqrt{1+9x^2(1+2^2)}}}{\sqrt{1+9 \cdot \frac{2^4x^2}{1+9x^2(1+2^2)}}} = \frac{2^3x}{\sqrt{1+9x^2(1+2^2+2^4)}}$By induction,$f_n(x) = \frac{2^nx}{\sqrt{1+9x^2(1+2^2+2^4+\ldots+2^{2n-2})}}$.For $n=10$,the denominator contains $9\alpha x^2$,where $\alpha = 1+2^2+2^4+\ldots+2^{18}$.This is a geometric progression with $a=1$,$r=4$,and $n=10$ terms.$\alpha = \frac{1(4^{10}-1)}{4-1} = \frac{2^{20}-1}{3}$.Thus,$3\alpha + 1 = 3(\frac{2^{20}-1}{3}) + 1 = 2^{20} - 1 + 1 = 2^{20}$.Therefore,$\sqrt{3\alpha+1} = \sqrt{2^{20}} = 2^{10} = 1024$.

Consider the function $f: R \rightarrow R$ defined by $f(x)=\frac{2x}{\sqrt{1+9x^2}}$. If the composition of $f$,$\underbrace{(f \circ f \circ \ldots \circ f)}_{10 \text{ times }}(x) = \frac{2^{10}x}{\sqrt{1+9\alpha x^2}}$,then the value of $\sqrt{3\alpha+1}$ is equal to:

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