Consider the function f: R R defined by f(x)=2 x 1+9 x^2 . If the… - Vidyadip

Question

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 $f, \underbrace{(f \circ f \circ f \circ \ldots \circ f)}_{10 \text { times }}(x)=\frac{2^{10} x}{\sqrt{1+9 \alpha x^2}}$, then the value of $\sqrt{3 \alpha+1}$ is equal to....................

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Answer

d
$ \mathrm{f}(\mathrm{f}(\mathrm{x}))=\frac{2 \mathrm{f}(\mathrm{x})}{\sqrt{1+9 \mathrm{f}^2(\mathrm{x})}}=\frac{4 \mathrm{x}}{\sqrt{1+9 \mathrm{x}^2+9.2^2 \mathrm{x}^2}} $

$ \mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{x})))=\frac{2^3 \mathrm{x} / \sqrt{1+9 \mathrm{x}^2}}{\sqrt{1+9\left(1+2^2\right) \frac{2^2 \mathrm{x}^2}{1+9 \mathrm{x}^2}}}=\frac{2^3 \mathrm{x}}{\sqrt{1+9 \mathrm{x}^2\left(1+2^2+2^4\right)}} $

$ \therefore \text { By observation } $

$ \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$

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