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 functional equation.
The initial equation is:\ \[\n\frac{f(x+y)}{3} = \frac{f(x) + f(y) + f(0)}{3}\n\]\nMultiply both sides by 3:\ \[\nf(x + y) = f(x) + f(y) + f(0)\n\]\n\n
Step 2: Analyze \( f \) at \( x = 0 \).
\nSubstitute \( x = 0 \) into the equation:\ \[\nf(0 + y) = f(0) + f(y) + f(0)\n\]\nThis simplifies to:\ \[\nf(y) = f(y) + 2f(0)\n\]\nTherefore:\ \[\n2f(0) = 0 \quad \Rightarrow \quad f(0) = 0\n\]\n\n
Step 3: Determine linearity.
\nThe equation now becomes:\ \[\nf(x + y) = f(x) + f(y)\n\]\nThis is a standard functional equation with the general solution:\ \[\nf(x) = cx\n\]\nwhere \( c \) is a constant.\n\nSince \( f \) is differentiable at \( x = 0 \), it must be of the form \( f(x) = cx \), a linear function.\n\n
Final Answer: \( f \) is linear.