Question

A function \( f : \mathbb{R} \to \mathbb{R} \), satisfies \[ \frac{f(x+y)}{3} = \frac{f(x) + f(y) + f(0)}{3} \quad \text{for all} \, x, y \in \mathbb{R}. \] If the function \( f \) is differentiable at \( x = 0 \), then \( f \) is:

• A. linear
• B. quadratic
• C. cubic
• D. biquadratic

Hint:

For functional equations of the form \( f(x+y) = f(x) + f(y) \), the solution is always a linear function \( f(x) = cx \) where \( c \) is constant, provided the function is differentiable.

Answer 1

The Correct Option is A

Step 1: Simplify the given functional equation.
The given equation is: \[ \frac{f(x+y)}{3} = \frac{f(x) + f(y) + f(0)}{3} \] Multiplying both sides by 3: \[ f(x + y) = f(x) + f(y) + f(0) \]
Step 2: Check the behavior of \( f \) at \( x = 0 \).
Substitute \( x = 0 \) into the equation: \[ f(0 + y) = f(0) + f(y) + f(0) \] Thus: \[ f(y) = f(y) + 2f(0) \] This implies: \[ 2f(0) = 0 \quad \Rightarrow \quad f(0) = 0 \]
Step 3: Check for linearity.
Now, the equation simplifies to: \[ f(x + y) = f(x) + f(y) \] This is a well-known functional equation, whose general solution is: \[ f(x) = cx \] where \( c \) is a constant. Since \( f \) is differentiable at \( x = 0 \), it must be of the form \( f(x) = cx \), which is a linear function.
Final Answer: \( f \) is linear.