Consider that `f : A rarr B`
(i) If f(x) is one-one `f(x_(1)) = f(x_(2)) hArr x_(1) = x_(2)` or `f'(x) ge 0` or `f'(x) le 0`.
(ii) If f(x) is onto the range of f(x) = B.
(iii) If f(x) and g(x) are inverse of each other then f(g(x)) = g(f(x)) = x.
Now consider the answer of the following questions.
Let f be one-oe function with domain {x, y, z} and range {1, 2, 3}. It is given that exactly one of the following statement is true and the remaining two are false. f(x) = 1, `f(y) != 1, f(z) != 2`, then the value of `f^(-1)(1)` is

Options

A. x
B. z
C. y
D. Does not exist

Answer

To solve the problem step by step, we need to analyze the given statements about the function \( f \) and determine which one is true, given that \( f \) is a one-to-one function with a domain of \( \{x, y, z\} \) and a range of \( \{1, 2, 3\} \).

### Step 1: Analyze the statements
We have three statements:
1. \( f(x) = 1 \)
2. \( f(y) \neq 1 \)
3. \( f(z) \neq 2 \)

We know that exactly one of these statements is true, and the other two are false.

### Step 2: Assume each statement is true and check for contradictions

#### Case 1: Assume \( f(x) = 1 \) is true
- If \( f(x) = 1 \), then \( f(y) \) and \( f(z) \) must be different from 1 (since \( f \) is one-to-one).
- Thus, \( f(y) \) and \( f(z) \) must take values from \( \{2, 3\} \).
- This means both \( f(y) \) and \( f(z) \) can only be 2 or 3, which leads to a contradiction because \( f \) would not be one-to-one (two different inputs would map to the same output).

**Conclusion**: \( f(x) = 1 \) cannot be true.

#### Case 2: Assume \( f(y) \neq 1 \) is true
- If \( f(y) \neq 1 \), then \( f(x) \) must be 1 (since it is the only remaining option).
- However, this means \( f(z) \) must also take a value from \( \{2, 3\} \).
- If \( f(y) \) is not 1, it must be either 2 or 3, which again leads to a contradiction because \( f(x) = 1 \) and \( f(y) \) would have to map to the same output.

**Conclusion**: \( f(y) \neq 1 \) cannot be true.

#### Case 3: Assume \( f(z) \neq 2 \) is true
- If \( f(z) \neq 2 \), then \( f(x) \) and \( f(y) \) must take the remaining values in the range.
- Let’s assign \( f(y) = 1 \) (since it can be true) and \( f(x) = 3 \).
- This assignment satisfies the condition of being one-to-one, as each input maps to a unique output.

### Step 3: Determine \( f^{-1}(1) \)
Since we have established that \( f(y) = 1 \), we can conclude that:
\[ f^{-1}(1) = y \]

### Final Answer
Thus, the value of \( f^{-1}(1) \) is \( y \).