Question

Consider the function \( f : \mathbb{R} \to \mathbb{R} \) defined by \[ f(x) = \frac{2x}{\sqrt{1 + 9x^2}}. \] If the composition of \( f \), \[ (f \circ f \circ f \circ \dots \circ f)(x) \quad \text{(10 times)} = \frac{2^{10}x}{\sqrt{1 + 9\alpha x^2}}, \] then the value of \( \sqrt{3\alpha + 1} \) is equal to \( \dots \).

Answer 1

Correct Answer: 1024

To determine the value of \( \alpha \), let’s analyze the repeated composition of \( f(x) \).

1. Starting with \( f(x) = \frac{2x}{\sqrt{1 + 9x^2}} \), we compute \( f(f(x)) \): \[ f(f(x)) = \frac{2f(x)}{\sqrt{1 + 9f(x)^2}} = \frac{4x}{\sqrt{1 + 9x^2 + 9 \cdot 2^2 x^2}} = \frac{2^2 x}{\sqrt{1 + 9(1 + 2)x^2}}. \] This gives us \( \alpha_2 = 1 + 2 \) for the second composition.
2. Repeating this process, we observe a pattern: after \( n \) compositions, the denominator takes the form \[ \sqrt{1 + 9(1 + 2 + 2^2 + \cdots + 2^{n-1})x^2}. \]
3. The series \( 1 + 2 + 2^2 + \cdots + 2^{n-1} \) is a geometric series that sums to \( 2^n - 1 \). Therefore, after 10 compositions, we have: \[ \alpha = 2^{10} - 1 = 1023. \]

Now, we calculate \( \sqrt{3\alpha + 1} \):

\[ \sqrt{3\alpha + 1} = \sqrt{3 \cdot 1023 + 1} = \sqrt{3072 + 1} = \sqrt{3072} = 1024. \]

Answer: 1024

Answer 2

Given function:
\[ f(x) = \frac{2x}{\sqrt{1 + 9x^2}} \] Define: \[ f^{(n)}(x) = (f \circ f \circ \cdots \circ f)(x), n \text{ times} \] Given: \[ f^{(10)}(x) = \frac{2^{10} x}{\sqrt{1 + 9 \alpha x^2}} \] Step 1: Assume the form after \( n \) iterations
Assume for \( n \), \[ f^{(n)}(x) = \frac{2^n x}{\sqrt{1 + 9 a_n x^2}} \] where \( a_n \) depends on \( n \). Step 2: Use recursion relation
Apply \( f \): \[ f^{(n+1)}(x) = f\left( f^{(n)}(x) \right) = \frac{2 \cdot f^{(n)}(x)}{\sqrt{1 + 9 (f^{(n)}(x))^2}} = \frac{2 \cdot \frac{2^n x}{\sqrt{1 + 9 a_n x^2}}}{\sqrt{1 + 9 \left( \frac{2^n x}{\sqrt{1 + 9 a_n x^2}} \right)^2}} \] Simplify denominator inside square root: \[ 1 + 9 \left( \frac{2^{2n} x^2}{1 + 9 a_n x^2} \right) = \frac{1 + 9 a_n x^2 + 9 \cdot 2^{2n} x^2}{1 + 9 a_n x^2} \] Thus, \[ f^{(n+1)}(x) = \frac{2^{n+1} x}{\sqrt{1 + 9 a_n x^2} \cdot \sqrt{\frac{1 + 9 a_n x^2 + 9 \cdot 2^{2n} x^2}{1 + 9 a_n x^2}}} = \frac{2^{n+1} x}{\sqrt{1 + 9 a_n x^2 + 9 \cdot 2^{2n} x^2}} \] So, \[ 1 + 9 a_{n+1} x^2 = 1 + 9 a_n x^2 + 9 \cdot 2^{2n} x^2 \] Equate coefficients: \[ a_{n+1} = a_n + 2^{2n} = a_n + 4^{n} \] with initial condition \( a_1 = 1 \) because: \[ f^{(1)}(x) = f(x) = \frac{2x}{\sqrt{1 + 9 \cdot 1 \cdot x^2}} \] Step 3: Find \( a_{10} \)
\[ a_{10} = \sum_{k=0}^9 4^{k} = \frac{4^{10} - 1}{4 - 1} = \frac{4^{10} - 1}{3} \] Calculate \( 4^{10} = (2^2)^{10} = 2^{20} = 1,048,576 \), so: \[ a_{10} = \frac{1,048,576 - 1}{3} = \frac{1,048,575}{3} = 349,525 \] Step 4: Find \( \sqrt{3\alpha + 1} \)
Given \( \alpha = a_{10} = 349,525 \), so: \[ 3\alpha + 1 = 3 \times 349,525 + 1 = 1,048,575 + 1 = 1,048,576 = 2^{20} \] Hence: \[ \sqrt{3\alpha + 1} = \sqrt{2^{20}} = 2^{10} = 1024 \] Final answer: \( \sqrt{3\alpha +1} = 1024 \)