For any non-zero real number x, let$ f(x) + 2f\left(\frac{1}{x}\right) = 3x.$Then, the sum of all possible values of x for which f(x) = 3, is

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:

Answer 1

We are given the functional equation:
\[ f(x) + 2f\left(\frac{1}{x}\right) = 3x \]
We need to find the sum of all $x$ values where $f(x) = 3$.
If $f(x) = 3$, the equation becomes:
\[ 3 + 2f\left(\frac{1}{x}\right) = 3x \]
Solving for $f\left(\frac{1}{x}\right)$:
\[ 2f\left(\frac{1}{x}\right) = 3x - 3 \]
\[ f\left(\frac{1}{x}\right) = \frac{3x - 3}{2} \]
Now, let's replace $x$ with $\frac{1}{x}$ in the original equation:
\[ f\left(\frac{1}{x}\right) + 2f(x) = \frac{3}{x} \]
We now have a system of equations. Solving this system reveals that the sum of all possible $x$ values for which $f(x) = 3$ is -3.

For any non-zero real number x, let\( f(x) + 2f\left(\frac{1}{x}\right) = 3x.\)Then, the sum of all possible values of x for which f(x) = 3, is

The Correct Option is B

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