Consider `f(x)=1-e^((1)/(x)-1)`
Q. If `D` is the set of all real `x` such that `f(x)ge0` then `D` is equal to

Options

A. `(-infty,infty)`
B. `(2,infty)`
C. `(-infty,0)cup(1,infty)`
D. `(-1,infty)`

Explanation

To solve the problem, we need to find the set \( D \) of all real \( x \) such that \( f(x) \geq 0 \), where \( f(x) = 1 - e^{\left(\frac{1}{x} - 1\right)} \).

### Step-by-Step Solution:

1. **Set up the inequality**:
We start with the function:
\[ f(x) = 1 - e^{\left(\frac{1}{x} - 1\right)} \geq 0 \]
This means we need to solve:
\[ 1 - e^{\left(\frac{1}{x} - 1\right)} \geq 0 \]

2. **Rearranging the inequality**:
Rearranging gives us:
\[ e^{\left(\frac{1}{x} - 1\right)} \leq 1 \]

3. **Taking the natural logarithm**:
Since the exponential function is always positive, we can take the natural logarithm of both sides:
\[ \frac{1}{x} - 1 \leq 0 \]
This simplifies to:
\[ \frac{1}{x} \leq 1 \]

4. **Rearranging further**:
Rearranging gives:
\[ 1 \leq x \]
or equivalently:
\[ x \geq 1 \]

5. **Consider the case when \( x < 0 \)**:
The function \( f(x) \) is not defined at \( x = 0 \) (as \( \frac{1}{x} \) becomes undefined). Therefore, we also need to consider the behavior of \( f(x) \) for \( x < 0 \).

6. **Analyzing \( x < 0 \)**:
For \( x < 0 \):
\[ \frac{1}{x} \text{ is negative, thus } e^{\left(\frac{1}{x} - 1\right)} \text{ becomes very small.} \]
Therefore, \( f(x) \) will be positive for \( x < 0 \).

7. **Combining the intervals**:
From the analysis:
- For \( x < 0 \), \( f(x) \geq 0 \).
- For \( x \geq 1 \), \( f(x) \geq 0 \).

Thus, the set \( D \) can be expressed as:
\[ D = (-\infty, 0) \cup [1, \infty) \]

### Final Answer:
The set \( D \) is:
\[ D = (-\infty, 0) \cup [1, \infty) \]