If `f:R rarr R` is a function such that `f(5x)+f(5x+1)+f(5x+2)=0, AA x in R`, then the period of f(x) is

Answer

To find the period of the function \( f: \mathbb{R} \to \mathbb{R} \) given the equation:

\[ f(5x) + f(5x + 1) + f(5x + 2) = 0 \quad \forall x \in \mathbb{R} \]

we will follow these steps:

### Step 1: Substitute \( x \) with \( \frac{x + 1}{5} \)

We start by substituting \( x \) in the original equation to find another relation for \( f \):

\[ f\left(5\left(\frac{x + 1}{5}\right)\right) + f\left(5\left(\frac{x + 1}{5}\right) + 1\right) + f\left(5\left(\frac{x + 1}{5}\right) + 2\right) = 0 \]

This simplifies to:

\[ f(x + 1) + f(x + 2) + f(x + 3) = 0 \]

### Step 2: Set up the equations

Now we have two equations:

1. \( f(5x) + f(5x + 1) + f(5x + 2) = 0 \) (Equation 1)
2. \( f(x + 1) + f(x + 2) + f(x + 3) = 0 \) (Equation 2)

### Step 3: Subtract Equation 1 from Equation 2

Next, we can manipulate these equations. We will subtract Equation 1 from Equation 2.

From Equation 2, we can express \( f(x + 1) + f(x + 2) + f(x + 3) \) and from Equation 1, we can express \( f(5x) + f(5x + 1) + f(5x + 2) \).

Subtracting these gives:

\[ (f(x + 1) - f(5x)) + (f(x + 2) - f(5x + 1)) + (f(x + 3) - f(5x + 2)) = 0 \]

### Step 4: Rearranging the terms

This implies:

\[ f(x + 3) - f(5x + 2) + f(x + 2) - f(5x + 1) + f(x + 1) - f(5x) = 0 \]

### Step 5: Establish periodicity

From the above, we can deduce that:

\[ f(5x + 3) = f(5x) \]

This means that \( f \) is periodic with period \( 3 \) when evaluated at \( 5x \).

### Step 6: Generalize the periodicity

Now, if we replace \( x \) with \( \frac{x}{5} \):

\[ f(x + 3) = f(x) \]

This indicates that \( f \) is periodic with period \( 3 \).

### Conclusion

Thus, the period of the function \( f(x) \) is:

\[ \text{Period} = 3 \]