Question

Let \(f : R → R\) be a differentiable function that satisfies the relation \(f(x + y) = f(x) + f(y) – 1, ∀x, y ∈R\). If \(f'(0) = 2\), then \(|f(–2)|\) is equal to ___________.

Answer 1

Correct Answer: 3

We are given that \( f(x + y) = f(x) + f(y) - 1 \) for all \( x, y \in \mathbb{R} \), and \( f'(0) = 2 \).
Step 1: Differentiating the functional equation. To solve this, let's differentiate the given functional equation with respect to \( y \): \[ \frac{d}{dy} \left( f(x + y) \right) = \frac{d}{dy} \left( f(x) + f(y) - 1 \right). \] This simplifies to: \[ f'(x + y) = f'(y). \] Thus, we find that \( f'(x + y) = f'(y) \) for all \( x, y \in \mathbb{R} \). This implies that \( f'(x) \) is a constant function. Let \( f'(x) = c \), where \( c \) is a constant.
Step 2: Determining the constant \( c \). We are given that \( f'(0) = 2 \). Since \( f'(x) = c \), it follows that \( c = 2 \). Therefore, \( f'(x) = 2 \) for all \( x \in \mathbb{R} \).
Step 3: Finding the general form of \( f(x) \). Since \( f'(x) = 2 \), we integrate to find \( f(x) \): \[ f(x) = 2x + C, \] where \( C \) is a constant.

Step 4: Using the given functional equation to find \( C \). Substitute \( f(x) = 2x + C \) into the original functional equation \( f(x + y) = f(x) + f(y) - 1 \): \[ f(x + y) = 2(x + y) + C = 2x + 2y + C, \] \[ f(x) + f(y) - 1 = (2x + C) + (2y + C) - 1 = 2x + 2y + 2C - 1. \] Equating the two expressions: \[ 2x + 2y + C = 2x + 2y + 2C - 1. \] This simplifies to: \[ C = 2C - 1 \quad \Rightarrow \quad C = 1. \] 

Step 5: Final form of \( f(x) \). Thus, the function is: \[ f(x) = 2x + 1. \] 

Step 6: Finding \( |f(-2)| \). Substitute \( x = -2 \) into the function \( f(x) = 2x + 1 \): \[ f(-2) = 2(-2) + 1 = -4 + 1 = -3. \] Thus, \( |f(-2)| = 3 \).