Let f : R - {n} `rarr` R be a function defined by `f(x)=(x-m)/(x-n)`, where `m ne n`. Then,

Options

A. f is one-one onto
B. f is one-one into
C. f is many-one onto
D. is many-one into

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{x - m}{x - n} \) where \( m \neq n \). We will determine whether this function is one-to-one (injective) and onto (surjective).

### Step 1: Check if \( f \) is one-to-one (injective)

To show that \( f \) is one-to-one, we need to prove that if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \).

1. Assume \( f(x_1) = f(x_2) \):
\[ \frac{x_1 - m}{x_1 - n} = \frac{x_2 - m}{x_2 - n} \]

2. Cross-multiply:
\[ (x_1 - m)(x_2 - n) = (x_2 - m)(x_1 - n) \]

3. Expand both sides:
\[ x_1 x_2 - n x_1 - m x_2 + mn = x_2 x_1 - n x_2 - m x_1 + mn \]

4. Simplifying gives:
\[ -n x_1 - m x_2 = -n x_2 - m x_1 \]

5. Rearranging terms:
\[ m x_1 - m x_2 = n x_1 - n x_2 \]

6. Factoring out common terms:
\[ m (x_1 - x_2) = n (x_1 - x_2) \]

7. If \( x_1 \neq x_2 \), we can divide both sides by \( x_1 - x_2 \) (which is valid since \( m \neq n \)):
\[ m = n \]
This is a contradiction since \( m \neq n \). Therefore, \( x_1 = x_2 \).

Thus, \( f \) is one-to-one.

### Step 2: Check if \( f \) is onto (surjective)

To check if \( f \) is onto, we need to show that for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} - \{n\} \) such that \( f(x) = y \).

1. Set \( f(x) = y \):
\[ \frac{x - m}{x - n} = y \]

2. Cross-multiply:
\[ x - m = y(x - n) \]

3. Rearranging gives:
\[ x - m = yx - yn \]

4. Collecting terms involving \( x \):
\[ x - yx = m - yn \]

5. Factoring out \( x \):
\[ x(1 - y) = m - yn \]

6. Solving for \( x \):
\[ x = \frac{m - yn}{1 - y} \]

7. The denominator \( 1 - y \) cannot be zero, which occurs when \( y = 1 \). Therefore, \( y \) can take any real value except \( 1 \).

Thus, the range of \( f \) is \( \mathbb{R} - \{1\} \), which is not equal to the codomain \( \mathbb{R} \). Therefore, \( f \) is not onto.

### Conclusion

The function \( f(x) = \frac{x - m}{x - n} \) is one-to-one but not onto.