Let `f:[-1,oo] in [-1,oo]` be a function given `f(x)=(x+1)^(2)-1, x ge -1`
Statement-1: The set `[x:f(x)=f^(-1)(x)]={0,1}`
Statement-2: f is a bijection.

Options

A. Statement 1 is ture, statement 2 is true, statement 2 is a correct explanation for statement 1.
B. Statement 1 is ture, statement 2 is true, statement 2 is not a correct explanation for statement 1.
C. Statement 1 is ture, statement 2 is false.
D. Statement 1 is false, statement 2 is true.

Explanation

To solve the problem, we need to analyze the function \( f(x) = (x + 1)^2 - 1 \) for \( x \geq -1 \) and check the validity of both statements provided.

### Step 1: Determine the function \( f(x) \)

Given:
\[ f(x) = (x + 1)^2 - 1 \]

### Step 2: Simplify the function

Let's simplify \( f(x) \):
\[ f(x) = (x + 1)^2 - 1 = x^2 + 2x + 1 - 1 = x^2 + 2x \]

### Step 3: Check if \( f \) is a bijection

A function is a bijection if it is both one-to-one (injective) and onto (surjective).

#### Check for injectivity (one-to-one):

To check if \( f \) is one-to-one, we can find the derivative \( f'(x) \):
\[ f'(x) = 2x + 2 \]

Since \( f'(x) \) is positive for \( x \geq -1 \) (as \( 2x + 2 \geq 0 \)), the function is increasing in this interval. Therefore, \( f \) is injective.

#### Check for surjectivity (onto):

To check if \( f \) is onto, we need to see if every \( y \) in the codomain has a corresponding \( x \) in the domain such that \( f(x) = y \). The range of \( f(x) \) can be found by evaluating the function at the endpoints of the domain.

- At \( x = -1 \):
\[ f(-1) = (-1 + 1)^2 - 1 = 0 - 1 = -1 \]

- As \( x \to \infty \):
\[ f(x) \to \infty \]

Thus, the range of \( f \) is \( [-1, \infty) \). However, since we are not given the codomain explicitly, we cannot conclude that \( f \) is onto.

### Conclusion for Statement 2:

Since we cannot confirm that \( f \) is onto, Statement 2 is **false**.

### Step 4: Solve \( f(x) = f^{-1}(x) \)

To find the set \( \{ x : f(x) = f^{-1}(x) \} \), we first need to find \( f^{-1}(x) \).

1. Set \( y = f(x) \):
\[ y = x^2 + 2x \]

2. Rearranging gives:
\[ x^2 + 2x - y = 0 \]

3. Using the quadratic formula:
\[ x = \frac{-2 \pm \sqrt{(2)^2 + 4y}}{2} = -1 \pm \sqrt{1 + y} \]

Thus,
\[ f^{-1}(y) = -1 + \sqrt{1 + y} \quad \text{(since we take the positive root for } x \geq -1\text{)} \]

### Step 5: Set up the equation \( f(x) = f^{-1}(x) \)

Now we set:
\[ f(x) = f^{-1}(x) \]
This gives:
\[ x^2 + 2x = -1 + \sqrt{1 + x} \]

### Step 6: Solve the equation

Squaring both sides:
\[ (x^2 + 2x + 1)^2 = 1 + x \]
This leads to:
\[ x^2 + 2x + 1 = \sqrt{1 + x} \]

Now let's find the roots:
1. Rearranging gives:
\[ x^2 + 2x + 1 - x = 0 \implies x^2 + x + 1 = 0 \]

This quadratic has roots:
\[ x = 0, -1 \]

### Conclusion for Statement 1:

The valid solutions are \( x = 0 \) and \( x = 1 \). Therefore, Statement 1 is **true**.

### Final Answer:

- Statement 1 is true: \( \{ x : f(x) = f^{-1}(x) \} = \{0, 1\} \)
- Statement 2 is false: \( f \) is not necessarily a bijection.