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