Let mean value of `f(x) = 1/(x+c)` over interval `(0,2)` is `1/2 ln 3` then positive value of c is

Options

A. 1
B. `1/2`
C. `2`
D. `3/2`

Answer

To solve the problem, we need to find the positive value of \( c \) such that the mean value of the function \( f(x) = \frac{1}{x+c} \) over the interval \( (0, 2) \) is equal to \( \frac{1}{2} \ln 3 \).

### Step-by-step Solution:

1. **Understanding the Mean Value of a Function**:
The mean value of a function \( f(x) \) over the interval \( [a, b] \) is given by the formula:
\[ \text{Mean value} = \frac{1}{b-a} \int_a^b f(x) \, dx \]
In our case, \( a = 0 \) and \( b = 2 \).

2. **Setting Up the Integral**:
We need to compute:
\[ \frac{1}{2-0} \int_0^2 \frac{1}{x+c} \, dx = \frac{1}{2} \ln 3 \]
This simplifies to:
\[ \int_0^2 \frac{1}{x+c} \, dx = \ln 3 \]

3. **Calculating the Integral**:
The integral \( \int \frac{1}{x+c} \, dx \) can be calculated as:
\[ \int \frac{1}{x+c} \, dx = \ln |x+c| + C \]
Therefore, we evaluate:
\[ \int_0^2 \frac{1}{x+c} \, dx = \left[ \ln |x+c| \right]_0^2 = \ln(2+c) - \ln(c) \]

4. **Using the Properties of Logarithms**:
Using the property of logarithms \( \ln a - \ln b = \ln \frac{a}{b} \), we can rewrite the integral result:
\[ \ln(2+c) - \ln(c) = \ln\left(\frac{2+c}{c}\right) \]
Setting this equal to \( \ln 3 \):
\[ \ln\left(\frac{2+c}{c}\right) = \ln 3 \]

5. **Exponentiating Both Sides**:
By exponentiating both sides, we get:
\[ \frac{2+c}{c} = 3 \]

6. **Solving for \( c \)**:
Rearranging gives:
\[ 2+c = 3c \]
\[ 2 = 3c - c \]
\[ 2 = 2c \]
\[ c = 1 \]

### Final Answer:
The positive value of \( c \) is \( \boxed{1} \).