Let `f : R to R` be defined by `f(x)=(1)/(x-1)AA, x in R`. Then F is

Options

A. one-one
B. onto
C. bijective
D. f is not defined

Answer

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

### Step 1: Identify the function
The function is given as:
\[ f(x) = \frac{1}{x - 1} \]

### Step 2: Determine the domain of the function
The domain of a function consists of all the values of \( x \) for which the function is defined. In this case, the function \( f(x) \) will be undefined when the denominator is zero.

Set the denominator equal to zero:
\[ x - 1 = 0 \]
Solving for \( x \):
\[ x = 1 \]

This means that \( f(x) \) is undefined at \( x = 1 \). Therefore, the domain of \( f \) is:
\[ \text{Domain of } f = \mathbb{R} \setminus \{1\} \]

### Step 3: Check if the function is one-to-one (injective)
A function is one-to-one if different inputs produce different outputs. To check this, we can assume:
\[ f(a) = f(b) \implies \frac{1}{a - 1} = \frac{1}{b - 1} \]
Cross-multiplying gives:
\[ b - 1 = a - 1 \implies a = b \]
Since \( a = b \) holds true, the function \( f \) is one-to-one.

### Step 4: Check if the function is onto (surjective)
A function is onto if every element in the codomain (in this case \( \mathbb{R} \)) is the image of at least one element in the domain.

To check if \( f \) is onto, we need to see if for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \setminus \{1\} \) such that:
\[ f(x) = y \implies \frac{1}{x - 1} = y \]
Rearranging gives:
\[ x - 1 = \frac{1}{y} \implies x = \frac{1}{y} + 1 \]
For \( y \neq 0 \), \( x \) will be defined. However, if \( y = 0 \), \( f(x) \) can never equal zero since the function \( f(x) \) approaches zero but never reaches it. Thus, \( f \) is not onto.

### Step 5: Conclusion
Since \( f \) is one-to-one but not onto, we conclude that \( f \) is injective but not surjective.

### Final Answer
The function \( f(x) = \frac{1}{x - 1} \) is **injective** but **not surjective**.

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