Let's analyze the given functional equation: $f(xy) = f(x)f(y) + f(x) + f(y)$.
This equation can be rewritten as:
$f(xy+1) = (f(x)+1)(f(y)+1)$
Now, we are given a value: $160000$. Let's factorize it into prime factors:
$160000 = 2^6 \times 5^5$
We need to find $f(160000)$. Using the functional equation:
$f(xy) = f(x)f(y) + f(x) + f(y)$
So,
$f(160000) = f(2^6 \cdot 5^5) = f(2^6)f(5^5) + f(2^6) + f(5^5)$
Now, compute $f(2^6)$ and $f(5^5)$ recursively using the same rule.
Start with $f(2^6)$:
$f(2^6) = f(2)f(2^5) + f(2) + f(2^5)$
Similarly, compute $f(2^5), f(2^4)$, and so on down to $f(2)$. Likewise for powers of 5.
Assuming $f(2) = f(5) = 1$ (as given), we find the recursive structure forms a pattern. Each time we apply the equation, we multiply and add previous results. Using this process repeatedly:
We get:
$f(2^6) = 63$
$f(5^5) = 65$
Now substitute back:
$f(160000) = 63 \cdot 65 + 63 + 65 = 4095$
Therefore, the value of $f(160000)$ is $\boxed{4095}$.
Let \( f \) be an injective map with domain x, y, z and range 1, 2, 3 such that exactly one of the following statements is correct and the remaining are false.}
[I.] \( f(x) = 1 \Rightarrow f(y) = 1, f(z) = 2 \)
[II.] \( f(x) = 2 \Rightarrow f(y) = 1, f(z) = 1 \)
[III.] \( f(x) = 1, f(y) = 1, f(z) = 2 \) Then the value of \( f(1) \) is:
The function \( f(x) \) is defined as follows:
\[ f(x) = \begin{cases} x & \text{for } 0 \le x \le 1 \\ 1 & \text{for } x \ge 1 \\ 0 & \text{otherwise} \end{cases} \]
The properties of the function are as follows:
\[ f_1(x) = f(-x) \quad \text{for all } x \] \[ f_2(x) = -f(x) \quad \text{for all } x \] \[ f_3(x) = f(f(x)) \quad \text{for all } x \]