Consider the function f:R R defined by f(x)=2x√1+9x^2.

Question

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

Answer: 1024

Accepted Answer

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

$(f\circ f)(x)=\frac{2f(x)}{\sqrt{1+9(f(x){)}^{2}}}=\frac{\frac{4x}{\sqrt{1+9{x}^{2}}}}{\sqrt{1+9	imes \frac{4{x}^{2}}{1+9{x}^{2}}}}=\frac{4x}{\sqrt{1+45{x}^{2}}}$

$(f\circ f\circ f)(x)=\frac{4	imes \frac{2x}{\sqrt{1+9{x}^{2}}}}{\sqrt{1+45	imes \frac{4{x}^{2}}{1+9{x}^{2}}}}=\frac{8x}{\sqrt{1+21	imes 9{x}^{2}}}$

$(f\circ f\circ f\circ f)(x)=\frac{16x}{\sqrt{1+85	imes 9{x}^{2}}}$

$\Rightarrow \alpha $ is ${10}^{}$ term of $1,5,21,85,\ldots \alpha $ is ${10}^{}$ term of

$\begin{gathered} \frac{{({2}^{1})}^{2}-1}{3},\frac{{({2}^{2})}^{2}-1}{3},\frac{{({2}^{3})}^{2}-1}{3},\frac{{({2}^{4})}^{2}-1}{3},\ldots \ \Rightarrow \, \alpha =\frac{{({2}^{10})}^{2}-1}{3} \ \Rightarrow \, \sqrt{3\alpha +1}={2}^{10}=1024 \end{gathered}$

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