Question:

To find the value of $\sqrt{3\alpha+1}$, we start by examining the given function $f(x)=\dfrac{2x}{\sqrt{1+9x^2}}$ and its composition $\underbrace{(f\circ f\circ\cdots\circ f)}_{10}(x)=\dfrac{2^{10}x}{\sqrt{1+9\alpha x^2}}$. Step 1: Let's begin by computing the composition of $f$ with itself, denoted as $f\circ f(x)$, to look for a pattern that can be generalized for $10$ compositions. $$ \begin{aligned} (f\circ f)(x) &= f(f(x)) \\ &= f\left(\dfrac{2x}{\sqrt{1+9x^2}}\right) \\ &= \dfrac{2\cdot\dfrac{2x}{\sqrt{1+9x^2}}}{\sqrt{1+9\left(\dfrac{2x}{\sqrt{1+9x^2}}\right)^2}} \\ &= \dfrac{\dfrac{4x}{\sqrt{1+9x^2}}}{\sqrt{1+\dfrac{36x^2}{1+9x^2}}} \\ &= \dfrac{\dfrac{4x}{\sqrt{1+9x^2}}}{\sqrt{\dfrac{1+9x^2+36x^2}{1+9x^2}}} \\ &= \dfrac{\dfrac{4x}{\sqrt{1+9x^2}}}{\sqrt{\dfrac{1+45x^2}{1+9x^2}}} \\ &= \dfrac{4x}{\sqrt{1+45x^2}} \end{aligned} $$ Step 2: From Step 1, we observe that after two compositions, the denominator under the square root in the function becomes $1+45x^2$. Let's hypothesize that after $n$ compositions, the function takes the form $\dfrac{2^n x}{\sqrt{1+9\cdot(2^{2n-2}+2^{2n-4}+\cdots+2^2+1)x^2}}$. We need to verify this hypothesis by checking if it holds for $n=3$ and then generalize. Step 3: For $n=3$, composing $f$ with itself three times: $$ \begin{aligned} (f\circ f\circ f)(x) &= f((f\circ f)(x)) \\ &= f\left(\dfrac{4x}{\sqrt{1+45x^2}}\right) \\ &= \dfrac{2\cdot\dfrac{4x}{\sqrt{1+45x^2}}}{\sqrt{1+9\left(\dfrac{4x}{\sqrt{1+45x^2}}\right)^2}} \\ &= \dfrac{\dfrac{8x}{\sqrt{1+45x^2}}}{\sqrt{1+\dfrac{144x^2}{1+45x^2}}} \\ &= \dfrac{\dfrac{8x}{\sqrt{1+45x^2}}}{\sqrt{\dfrac{1+45x^2+144x^2}{1+45x^2}}} \\ &= \dfrac{\dfrac{8x}{\sqrt{1+45x^2}}}{\sqrt{\dfrac{1+189x^2}{1+45x^2}}} \\ &= \dfrac{8x}{\sqrt{1+189x^2}} \end{aligned} $$ This result supports our hypothesis, as $1+189x^2 = 1+9\cdot(2^4+2^2+1)x^2$. Step 4: Generalizing for $10$ compositions, we have: $$ \begin{aligned} \underbrace{(f\circ f\circ\cdots\circ f)}_{10}(x) &= \dfrac{2^{10}x}{\sqrt{1+9\alpha x^2}} \end{aligned} $$ Given that after $n$ compositions, the coefficient of $x^2$ in the denominator is $9\cdot(2^{2n-2}+2^{2n-4}+\cdots+2^2+1)$, for $n=10$, this becomes: $$ \begin{aligned} 9\cdot(2^{18}+2^{16}+\cdots+2^2+1) &= 9\cdot\frac{2^{20}-1}{2^2-1} \\ &= 9\cdot\frac{2^{20}-1}{3} \\ &= 3\cdot(2^{20}-1) \end{aligned} $$ Therefore, $3\alpha = 3\cdot(2^{20}-1)$, which implies $\alpha = 2^{20}-1$. Step 5: Finally, to find $\sqrt{3\alpha+1}$: $$ \begin{aligned} \sqrt{3\alpha+1} &= \sqrt{3\cdot(2^{20}-1)+1} \\ &= \sqrt{3\cdot2^{20}-3+1} \\ &= \sqrt{3\cdot2^{20}-2} \\ &= \sqrt{2\cdot(3\cdot2^{19}-1)} \\ &= \sqrt{2}\cdot\sqrt{3\cdot2^{19}-1} \\ &= \sqrt{2}\cdot\sqrt{(2^{20}-1)-1} \\ &= \sqrt{2}\cdot\sqrt{2^{20}-2} \\ &= \sqrt{2}\cdot\sqrt{2\cdot(2^{19}-1)} \\ &= \sqrt{2}\cdot\sqrt{2}\cdot\sqrt{2^{19}-1} \\ &= 2\cdot\sqrt{2^{19}-1} \\ &= 2\cdot\sqrt{(2^{19}-1)} \\ &= 2\cdot\sqrt{2^{19}-1} \\ \text{However, simplification in step 4 leads to:} \\ 3\alpha + 1 &= 3\cdot2^{20} \\ \sqrt{3\alpha+1} &= \sqrt{3\cdot2^{20}} \\ &= \sqrt{3}\cdot\sqrt{2^{20}} \\ &= \sqrt{3}\cdot2^{10} \\ &= \sqrt{3\cdot2^{20}} \\ &= 2^{10}\cdot\sqrt{3} \text{ is incorrect based on the initial simplification mistake.} \\ \text{Correcting the calculation from step 4:} \\ 3\alpha + 1 &= 2^{20} \\ \sqrt{3\alpha+1} &= \sqrt{2^{20}} \\ &= 2^{10} \\ &= 1024 \end{aligned} $$ Therefore: $\boxed{1024}$ <div class="key-concept"><strong>Key Concept:</strong> By induction, after $n$ compositions: $f^n(x)=\frac{2^nx}{\sqrt{1+9(1+2^2+2^4+\cdots+2^{2(n-1)})x^2}}$.</div> <div class="trap-box"><strong>Trap:</strong> $\alpha=1+2^2+2^4+\cdots+2^{18}=\frac{4^{10}-1}{3}=\frac{2^{20}-1}{3}$. $3\alpha+1=2^{20}$. $\sqrt{3\alpha+1}=2^{10}=1024$.</div>