Let `f(x) =int_1^x e^t/tdt,x in R^+` . Then complete set of valuesof x for which `f(x) leq In` x is

Options

A. `(0,1]`
B. `[1,oo)`
C. `(0,oo)`
D. none of these

Explanation

To solve the problem, we need to find the complete set of values of \( x \) for which \( f(x) \leq \ln x \), where \( f(x) = \int_1^x \frac{e^t}{t} dt \).

### Step-by-Step Solution:

1. **Define the Function**:
We start with the function:
\[ f(x) = \int_1^x \frac{e^t}{t} dt \]
We need to analyze the inequality \( f(x) \leq \ln x \).

2. **Define a New Function**:
Let's define a new function \( g(x) \):
\[ g(x) = f(x) - \ln x \]
We need to find the values of \( x \) such that \( g(x) \leq 0 \).

3. **Differentiate \( g(x) \)**:
To analyze \( g(x) \), we differentiate it:
\[ g'(x) = f'(x) - \frac{1}{x} \]
By the Fundamental Theorem of Calculus, we have:
\[ f'(x) = \frac{e^x}{x} \]
Therefore,
\[ g'(x) = \frac{e^x}{x} - \frac{1}{x} = \frac{e^x - 1}{x} \]

4. **Analyze \( g'(x) \)**:
The sign of \( g'(x) \) depends on \( e^x - 1 \):
- For \( x > 0 \), \( e^x > 1 \), hence \( g'(x) > 0 \).
- This implies that \( g(x) \) is an increasing function for \( x > 0 \).

5. **Evaluate \( g(1) \)**:
Now we evaluate \( g(1) \):
\[ g(1) = f(1) - \ln 1 = \int_1^1 \frac{e^t}{t} dt - 0 = 0 - 0 = 0 \]

6. **Conclusion**:
Since \( g(x) \) is increasing and \( g(1) = 0 \), it follows that:
- For \( x < 1 \), \( g(x) < 0 \) (since \( g(x) \) is increasing).
- For \( x = 1 \), \( g(x) = 0 \).
- For \( x > 1 \), \( g(x) > 0 \).

Therefore, the complete set of values of \( x \) for which \( f(x) \leq \ln x \) is:
\[ x \in (0, 1] \]

### Final Answer:
The complete set of values of \( x \) for which \( f(x) \leq \ln x \) is \( (0, 1] \).