Let `f :R to R` is not identically zero, differentiable function and satisfy the equals `f(xy)= f(x) f(y) and f (x+z) = f(x) + f (z),` then `f (5)=`

Options

A. 3
B. 5
C. 10
D. 15

Answer

To solve the problem, we need to analyze the given functional equations step by step.

### Step 1: Analyze the equations
We have two functional equations:
1. \( f(xy) = f(x)f(y) \)
2. \( f(x+z) = f(x) + f(z) \)

### Step 2: Substitute specific values
Let's start by substituting \( x = 1 \) in the first equation:
\[ f(1 \cdot y) = f(1)f(y) \implies f(y) = f(1)f(y) \]
Since \( f \) is not identically zero, \( f(y) \neq 0 \) for some \( y \). Thus, we can divide both sides by \( f(y) \) (which is valid since \( f(y) \neq 0 \)):
\[ 1 = f(1) \]
This means \( f(1) = 1 \).

### Step 3: Use the second equation
Now, substituting \( z = 0 \) in the second equation:
\[ f(x + 0) = f(x) + f(0) \implies f(x) = f(x) + f(0) \]
This implies:
\[ f(0) = 0 \]

### Step 4: Differentiate the second equation
Now, we differentiate the second equation \( f(x + z) = f(x) + f(z) \) with respect to \( x \):
\[ \frac{d}{dx}[f(x + z)] = \frac{d}{dx}[f(x) + f(z)] \]
Using the chain rule on the left side:
\[ f'(x + z) = f'(x) \]
This indicates that \( f' \) is constant, as it does not depend on \( z \).

### Step 5: Conclude that \( f' \) is constant
Let \( f'(x) = c \) for some constant \( c \). Integrating this gives:
\[ f(x) = cx + r \]
where \( r \) is a constant of integration.

### Step 6: Substitute back into the equations
Now, we substitute \( f(x) = cx + r \) into the first equation \( f(xy) = f(x)f(y) \):
\[ f(xy) = c(xy) + r \]
And:
\[ f(x)f(y) = (cx + r)(cy + r) = c^2xy + crx + cry + r^2 \]
Setting these equal gives:
\[ c(xy) + r = c^2xy + crx + cry + r^2 \]
Comparing coefficients, we find:
1. Coefficient of \( xy \): \( c = c^2 \) implies \( c(c - 1) = 0 \) which gives \( c = 0 \) or \( c = 1 \).
2. Since \( f \) is not identically zero, \( c \neq 0 \), hence \( c = 1 \).

### Step 7: Determine the value of \( r \)
Substituting \( c = 1 \) into the equation:
\[ f(x) = x + r \]
Substituting into the second equation \( f(x + z) = f(x) + f(z) \):
\[ (x + z + r) = (x + r) + (z + r) \implies x + z + r = x + z + 2r \]
This implies \( r = 0 \).

### Final Function
Thus, we have:
\[ f(x) = x \]

### Step 8: Calculate \( f(5) \)
Finally, we can find \( f(5) \):
\[ f(5) = 5 \]

### Conclusion
The answer is:
\[ \boxed{5} \]