consider `f:R-{0}toR` defined by `f(x)=1-e^((1)/(x)-1)`
Q. `f(x)` is a/an

Options

A. a. one-one function
B. b. many-one function
C. c. onto function
D. d. into function

Explanation

To determine the nature of the function \( f(x) = 1 - e^{\left(\frac{1}{x} - 1\right)} \) defined for \( x \in \mathbb{R} - \{0\} \), we will analyze its properties step by step.

### Step 1: Find the derivative of \( f(x) \)

To check if \( f(x) \) is one-one, we can find its derivative \( f'(x) \):

\[ f'(x) = \frac{d}{dx} \left( 1 - e^{\left(\frac{1}{x} - 1\right)} \right) \]

Using the chain rule, we differentiate:

\[ f'(x) = -e^{\left(\frac{1}{x} - 1\right)} \cdot \frac{d}{dx}\left(\frac{1}{x} - 1\right) \]

The derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \), so:

\[ f'(x) = -e^{\left(\frac{1}{x} - 1\right)} \cdot \left(-\frac{1}{x^2}\right) = \frac{e^{\left(\frac{1}{x} - 1\right)}}{x^2} \]

### Step 2: Analyze the sign of \( f'(x) \)

Since \( e^{\left(\frac{1}{x} - 1\right)} > 0 \) for all \( x \neq 0 \) and \( x^2 > 0 \) for all \( x \neq 0 \), we conclude that:

\[ f'(x) > 0 \quad \text{for all } x \in \mathbb{R} - \{0\} \]

### Step 3: Conclusion about the function's monotonicity

Since the derivative \( f'(x) \) is positive for all \( x \neq 0 \), the function \( f(x) \) is strictly increasing on its domain.

### Step 4: Determine if \( f(x) \) is one-one

A function that is strictly increasing is one-one. Therefore, we can conclude that:

\[ f(x) \text{ is a one-one function.} \]

### Step 5: Check for onto (surjective) and into (not surjective)

Next, we need to determine the range of \( f(x) \):

1. As \( x \to 0^+ \), \( \frac{1}{x} \to +\infty \) and \( f(x) \to 1 - e^{+\infty} = 1 - \infty = -\infty \).
2. As \( x \to +\infty \), \( \frac{1}{x} \to 0 \) and \( f(x) \to 1 - e^{(0 - 1)} = 1 - \frac{1}{e} \).

Thus, the range of \( f(x) \) is \( (-\infty, 1 - \frac{1}{e}) \).

Since the codomain is \( \mathbb{R} \) and the range does not cover all of \( \mathbb{R} \), we conclude that:

\[ f(x) \text{ is not onto (surjective).} \]

### Final Conclusion

Based on the analysis, we can summarize:

- \( f(x) \) is a one-one function.
- \( f(x) \) is not onto (it is into).

Thus, the answer is that \( f(x) \) is a **one-one function**.

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