Consider the function f: mathbb{R} rightarrow | vedclass

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Question

$ \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

Vedclass · Accepted Answer

$1024$

Vedclass · Answer

$ \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ight) \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ight)}} $

$ 	herefore 	ext { By observation } $

$ \alpha=1+2^2+2^4+\ldots+2^{18}=1\left(\frac{\left(2^2ight)^{10}-1}{2^2-1}ight)=\frac{2^{20}-1}{3} $

$ 3 \alpha+1=2^{20} ightarrow \sqrt{3 \alpha+1}=2^{10}=1024$

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