Let 'f is a function, continuous on `[0,1]` such that `f(x) leq sqrt5 AA x in [0,1/2] and f(x) leq 2/x AA x in [1/2,1]` then smallest 'a' for which `int_0^1 f(x)dx leq a` holds for all 'f' is

Options

A. `sqrt5`
B. `(sqrt5)/(2) + 2 ln 2`
C. `2+ ln ((sqrt5)/(2))`
D. `2+2ln ((sqrt5)/(2))`

Answer

To solve the problem, we need to evaluate the integral of the function \( f(x) \) over the interval \([0, 1]\) given the constraints on \( f(x) \).

### Step-by-Step Solution:

1. **Define the Function \( f(x) \)**:
Given the conditions:
- For \( x \in [0, \frac{1}{2}] \), \( f(x) \leq \sqrt{5} \)
- For \( x \in [\frac{1}{2}, 1] \), \( f(x) \leq \frac{2}{x} \)

We can define \( f(x) \) as:
\[ f(x) = \begin{cases} \sqrt{5} & \text{for } x \in [0, \frac{1}{2}] \\ \frac{2}{x} & \text{for } x \in [\frac{1}{2}, 1] \end{cases} \]

2. **Set Up the Integral**:
We need to compute the integral:
\[ \int_0^1 f(x) \, dx = \int_0^{\frac{1}{2}} f(x) \, dx + \int_{\frac{1}{2}}^1 f(x) \, dx \]
Substituting the defined function:
\[ = \int_0^{\frac{1}{2}} \sqrt{5} \, dx + \int_{\frac{1}{2}}^1 \frac{2}{x} \, dx \]

3. **Evaluate the First Integral**:
The first integral is:
\[ \int_0^{\frac{1}{2}} \sqrt{5} \, dx = \sqrt{5} \cdot \left[ x \right]_0^{\frac{1}{2}} = \sqrt{5} \cdot \left( \frac{1}{2} - 0 \right) = \frac{\sqrt{5}}{2} \]

4. **Evaluate the Second Integral**:
The second integral is:
\[ \int_{\frac{1}{2}}^1 \frac{2}{x} \, dx = 2 \cdot \left[ \ln |x| \right]_{\frac{1}{2}}^1 = 2 \cdot \left( \ln(1) - \ln\left(\frac{1}{2}\right) \right) = 2 \cdot (0 + \ln(2)) = 2 \ln(2) \]

5. **Combine the Results**:
Now, we combine both results:
\[ \int_0^1 f(x) \, dx = \frac{\sqrt{5}}{2} + 2 \ln(2) \]

6. **Determine the Smallest \( a \)**:
We need to find the smallest \( a \) such that:
\[ \int_0^1 f(x) \, dx \leq a \]
Thus, the smallest \( a \) is:
\[ a = \frac{\sqrt{5}}{2} + 2 \ln(2) \]

### Final Answer:
The smallest value of \( a \) for which \( \int_0^1 f(x) \, dx \leq a \) holds for all \( f \) is:
\[ \boxed{\frac{\sqrt{5}}{2} + 2 \ln(2)} \]