Statement -1 All points of intersection of `y=f(x)` and `y=f^(-1)(x)` lies on `y=x` only.
Statement -2 If point `P(alpha,beta)` lies on `y=f(x)` then `Q(beta,alpha)` lies on `y=f^(-1)(x)`.
Statement -3 Inverse of invertible function is unique and its range is equal to the function domain.
Which of the following option is correct for above statements in order

Options

A. a.TTF
B. b.FTT
C. c.TTT
D. d.TFT

Explanation

To analyze the statements provided in the question, we will evaluate each statement step by step.

### Step 1: Evaluate Statement 1
**Statement 1:** All points of intersection of \( y = f(x) \) and \( y = f^{-1}(x) \) lie on \( y = x \) only.

**Solution:**
To find the points of intersection between the two functions, we set them equal to each other:
\[ f(x) = f^{-1}(x) \]
If we let \( y = f(x) \), then by the definition of the inverse function, we have:
\[ f(f^{-1}(x)) = x \]
This implies that if \( f(x) = y \), then \( f^{-1}(y) = x \). Thus, if \( f(x) = f^{-1}(x) \), we can replace \( f^{-1}(x) \) with \( y \):
\[ f(x) = x \]
This means that the points of intersection must lie on the line \( y = x \). Therefore, Statement 1 is **correct**.

### Step 2: Evaluate Statement 2
**Statement 2:** If point \( P(\alpha, \beta) \) lies on \( y = f(x) \), then \( Q(\beta, \alpha) \) lies on \( y = f^{-1}(x) \).

**Solution:**
Given that point \( P(\alpha, \beta) \) lies on \( y = f(x) \), we have:
\[ f(\alpha) = \beta \]
To find if \( Q(\beta, \alpha) \) lies on \( y = f^{-1}(x) \), we need to check if:
\[ f^{-1}(\beta) = \alpha \]
By the definition of the inverse function, if \( f(\alpha) = \beta \), then:
\[ f^{-1}(\beta) = \alpha \]
This confirms that point \( Q(\beta, \alpha) \) indeed lies on \( y = f^{-1}(x) \). Therefore, Statement 2 is **correct**.

### Step 3: Evaluate Statement 3
**Statement 3:** The inverse of an invertible function is unique and its range is equal to the function's domain.

**Solution:**
This statement is a fundamental property of invertible functions. If a function \( f \) is invertible, it means that for every output \( y \) in the range of \( f \), there is a unique input \( x \) in the domain such that \( f(x) = y \). Thus, the inverse function \( f^{-1} \) is unique and its range (the set of possible outputs) is equal to the domain of the original function \( f \). Therefore, Statement 3 is **correct**.

### Conclusion
All three statements are correct. Thus, the correct option is:
- **C: All statements are correct.**