A function`f: Rvec[1,oo)` satisfies the equation `f(x y)=f(x)f(y)-f(x)-f(y)+2.` If differentiable on `R-{0}a n df(2)=5,f^(prime)(x)=(f(x)-1)/xdotlambdat h e nlambda=` `2^(prime)f(1)` b. `3f^(prime)(1)` c. `1/2f^(prime)(1)` d. `f^(prime)(1)`

Options

A. `2f'(1)`
B. `3f'(1)`
C. `(1)/(2)f'(1)`
D. `f'(1)`

Explanation

To solve the given problem step by step, we will analyze the functional equation and the conditions provided.

### Step 1: Analyze the functional equation
The functional equation given is:
\[ f(xy) = f(x)f(y) - f(x) - f(y) + 2 \]

### Step 2: Substitute specific values
To find specific values of \( f \), we can substitute \( x = 1 \) and \( y = 1 \):
\[ f(1 \cdot 1) = f(1)f(1) - f(1) - f(1) + 2 \]
This simplifies to:
\[ f(1) = f(1)^2 - 2f(1) + 2 \]
Rearranging gives:
\[ f(1)^2 - 3f(1) + 2 = 0 \]
Factoring this quadratic equation:
\[ (f(1) - 1)(f(1) - 2) = 0 \]
Thus, \( f(1) = 1 \) or \( f(1) = 2 \).

### Step 3: Use the condition \( f(2) = 5 \)
Next, we substitute \( x = 1 \) and \( y = 2 \):
\[ f(1 \cdot 2) = f(1)f(2) - f(1) - f(2) + 2 \]
This gives:
\[ f(2) = f(1) \cdot 5 - f(1) - 5 + 2 \]
Using \( f(2) = 5 \), we have:
\[ 5 = 5f(1) - f(1) - 5 + 2 \]
Simplifying:
\[ 5 = 4f(1) - 3 \]
Thus:
\[ 4f(1) = 8 \]
\[ f(1) = 2 \]

### Step 4: Determine the function \( f(x) \)
Now we know \( f(1) = 2 \). We can use this information to find \( f(x) \).

### Step 5: Differentiate \( f(x) \)
We are given that:
\[ f'(x) = \frac{f(x) - 1}{x} \]
To find \( f'(1) \):
\[ f'(1) = \frac{f(1) - 1}{1} = \frac{2 - 1}{1} = 1 \]

### Step 6: Find \( \lambda \)
We need to find \( \lambda \) where:
\[ \lambda = 2f'(1) \]
Substituting \( f'(1) = 1 \):
\[ \lambda = 2 \cdot 1 = 2 \]

### Step 7: Conclusion
The value of \( \lambda \) is \( 2 \).

### Final Answer
The answer is \( \lambda = 2 \).