Question

Let \( f(x) = 3 + 2x \) and \( g_n(x) = (f \circ f \circ \dots \text{n times}) (x) \). If for all \( n \in \mathbb{N} \), the lines \( y = g_n(x) \) pass through a fixed point \( (a, \beta) \), then \( \alpha + \beta = ? \)

• A. \( -5 \)
• B. \( -4 \)
• C. \( -3 \)
• D. \( -6 \)

Hint:

To solve problems involving repeated application of functions, look for patterns in the outputs after applying the function multiple times. For linear functions, you can derive a general expression for \( g_n(x) \) by recognizing the structure of the function.

Answer 1

The Correct Option is D

We are given that \( f(x) = 3 + 2x \) and \( g_n(x) = (f \circ f \circ \dots \text{n times}) (x) \), and the lines \( y = g_n(x) \) pass through a fixed point \( (a, \beta) \). We need to determine the value of \( \alpha + \beta \). 
Step 1: 
First, let us calculate \( g_n(x) \). Since \( f(x) = 3 + 2x \), we apply this function \( n \) times. For \( g_1(x) = f(x) \), we have: \[ g_1(x) = 3 + 2x. \] Now, applying \( f(x) \) again, we get \( g_2(x) = f(f(x)) \): \[ g_2(x) = f(3 + 2x) = 3 + 2(3 + 2x) = 3 + 6 + 4x = 9 + 4x. \] Next, applying \( f(x) \) once more for \( g_3(x) \), we get: \[ g_3(x) = f(9 + 4x) = 3 + 2(9 + 4x) = 3 + 18 + 8x = 21 + 8x. \] Observing the pattern, we can generalize the expression for \( g_n(x) \) as: \[ g_n(x) = 3 \cdot (2^n - 1) + 2^n x. \] 
Step 2: 
We are told that the lines \( y = g_n(x) \) pass through a fixed point \( (a, \beta) \), which implies that for all \( n \), the equation \( g_n(a) = \beta \) must hold. Substituting \( x = a \) into the general expression for \( g_n(x) \), we get: \[ g_n(a) = 3 \cdot (2^n - 1) + 2^n a. \] Since this equation holds for all \( n \), we can find the value of \( a \) and \( \beta \) by setting the coefficients of \( 2^n \) equal and constant for all \( n \). From the structure of \( g_n(x) \), we observe that the only possibility for the values of \( a \) and \( \beta \) to satisfy the condition for all \( n \) is \( a = -3 \) and \( \beta = -3 \). Thus, \( \alpha + \beta = -6 \).

Answer 2

Given \( f(x) = 3 + 2x \) and \( g_n(x) = (f \circ f \circ \dots \text{n times})(x) \), i.e., \( g_n \) is the \( n \)-th iterate of \( f \).

Step 1: Express \( g_n(x) \) in the form:
\[ g_n(x) = A_n x + B_n \] where \( A_n \) and \( B_n \) are constants depending on \( n \).

Step 2: For \( n = 1 \):
\[ g_1(x) = f(x) = 2x + 3 \] So, \[ A_1 = 2, \quad B_1 = 3 \]

Step 3: For \( n = 2 \):
\[ g_2(x) = f(g_1(x)) = f(2x + 3) = 2(2x + 3) + 3 = 4x + 6 + 3 = 4x + 9 \] So, \[ A_2 = 4, \quad B_2 = 9 \]

Step 4: Notice the pattern for \( A_n \):
\[ A_n = 2^n \]

Step 5: Find the recurrence relation for \( B_n \):
\[ g_n(x) = f(g_{n-1}(x)) = 2 g_{n-1}(x) + 3 = 2 (A_{n-1} x + B_{n-1}) + 3 = 2 A_{n-1} x + 2 B_{n-1} + 3 \] So, \[ A_n = 2 A_{n-1} \Rightarrow A_n = 2^n \] \[ B_n = 2 B_{n-1} + 3 \]

Step 6: Solve for \( B_n \):
This is a non-homogeneous linear recurrence:
Homogeneous part:
\[ B_n^h = 2 B_{n-1}^h \] General solution to homogeneous:
\[ B_n^h = C 2^n \]
Particular solution \( B_n^p = K \) (constant):
\[ K = 2K + 3 \Rightarrow K = -3 \]

Step 7: General solution:
\[ B_n = C 2^n - 3 \] Use initial condition \( B_1 = 3 \):
\[ 3 = C 2^1 - 3 \Rightarrow C = 3 \] So, \[ B_n = 3 \cdot 2^n - 3 = 3 (2^n - 1) \]

Step 8: Hence,
\[ g_n(x) = 2^n x + 3(2^n - 1) \]

Step 9: The line \( y = g_n(x) \) passes through fixed point \( (\alpha, \beta) \), so:
\[ \beta = g_n(\alpha) = 2^n \alpha + 3 (2^n - 1) \] \[ \Rightarrow \beta = 2^n \alpha + 3 \cdot 2^n - 3 \] \[ \beta + 3 = 2^n (\alpha + 3) \]

Step 10: For all \( n \), this must hold, so the right side grows exponentially unless:
\[ \alpha + 3 = 0 \quad \Rightarrow \quad \alpha = -3 \] Then, \[ \beta + 3 = 0 \quad \Rightarrow \quad \beta = -3 \]

Step 11: Therefore,
\[ \alpha + \beta = -3 + (-3) = -6 \]

Hence,
\[ \boxed{-6} \]