Let `f:RtoR` be a function defined by `f(x)=(x-m)/(x-n)`, where `mnen`. Then show that f is one-one but not onto.

Answer

To show that the function \( f: \mathbb{R} \to \mathbb{R} \) defined by

\[ f(x) = \frac{x - m}{x - n} \]

is one-one but not onto, we will follow these steps:

### Step 1: Show that \( f \) is one-one

To prove that \( f \) is one-one, we need to show 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:
\[ x_1 x_2 - n x_1 - m x_2 + m n = x_1 x_2 - n x_2 - m x_1 + m n \]

4. Cancel \( x_1 x_2 \) and \( m n \) from both sides:
\[ -n x_1 - m x_2 = -n x_2 - m x_1 \]

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

6. Factor out \( x_1 \) and \( x_2 \):
\[ (m - n)x_1 = (m - n)x_2 \]

7. Since \( m \neq n \), we can divide both sides by \( m - n \):
\[ x_1 = x_2 \]

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

### Step 2: Show that \( f \) is not onto

To show that \( f \) is not onto, we need to find the range of \( f \) and compare it with its codomain.

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

2. Rearranging gives:
\[ y(x - n) = x - m \]

3. Expanding and rearranging:
\[ yx - ny = x - m \implies yx - x = ny - m \implies x(y - 1) = ny - m \]

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

5. The expression for \( x \) is valid as long as \( y - 1 \neq 0 \) (i.e., \( y \neq 1 \)). This means \( y \) can take all real values except \( 1 \).

Thus, the range of \( f \) is:
\[ \text{Range} = \mathbb{R} \setminus \{1\} \]

6. The codomain of \( f \) is \( \mathbb{R} \).

Since the range \( \mathbb{R} \setminus \{1\} \) is not equal to the codomain \( \mathbb{R} \), we conclude that \( f \) is not onto.

### Final Conclusion

Therefore, we have shown that the function \( f(x) = \frac{x - m}{x - n} \) is one-one but not onto.

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