Key Concepts and Formulas

• Bijective Function: A function is bijective if it is both injective (one-to-one) and surjective (onto).
• Inverse Function: For a bijective function $f: A \to B$, the inverse function $f^{-1}: B \to A$ exists such that $f(f^{-1}(x)) = x$ for all $x$ in the domain of $f^{-1}$ (range of $f$), and $f^{-1}(f(x)) = x$ for all $x$ in the domain of $f$.
• Solving $f(x) = f^{-1}(x)$: If $f$ is strictly monotonic (increasing or decreasing), the solutions to $f(x) = f^{-1}(x)$ are the same as the solutions to $f(x) = x$.

Step-by-Step Solution

Step 1: Analyze Statement - 2: "$f$ is a bijection."

To determine if $f(x) = (x+1)^2 - 1$ with $x \ge -1$ is a bijection, we need to check if it's injective and surjective.

Step 1a: Check for Injectivity We take $x_1, x_2 \in [-1, \infty)$ such that $f(x_1) = f(x_2)$. $(x_1+1)^2 - 1 = (x_2+1)^2 - 1$ $(x_1+1)^2 = (x_2+1)^2$ Taking the square root of both sides, we get $|x_1+1| = |x_2+1|$. Since $x_1 \ge -1$ and $x_2 \ge -1$, we have $x_1+1 \ge 0$ and $x_2+1 \ge 0$. Therefore, $|x_1+1| = x_1+1$ and $|x_2+1| = x_2+1$. So, $x_1+1 = x_2+1$, which implies $x_1 = x_2$. Thus, $f(x)$ is injective on its domain $[-1, \infty)$.

Step 1b: Determine the Range of $f(x)$ The domain of $f$ is $x \in [-1, \infty)$. For $x \ge -1$, we have $x+1 \ge 0$. Squaring this, we get $(x+1)^2 \ge 0$. Subtracting 1, we get $(x+1)^2 - 1 \ge -1$. So, the range of $f(x)$ is $[-1, \infty)$.

Step 1c: Check for Surjectivity For a function to be surjective onto a codomain, its range must be equal to the codomain. If the codomain is not explicitly stated, it is usually assumed to be the set of all real numbers, $\mathbb{R}$. The range of $f(x)$ is $[-1, \infty)$, which is not equal to $\mathbb{R}$. Therefore, $f(x)$ is not surjective onto $\mathbb{R}$. However, if we consider the codomain to be the range of the function, i.e., $[-1, \infty)$, then $f: [-1, \infty) \to [-1, \infty)$ is surjective. For the purpose of finding the inverse function and solving $f(x) = f^{-1}(x)$, it is standard to consider the codomain to be the range. Thus, $f$ is considered a bijection from $[-1, \infty)$ to $[-1, \infty)$. Statement - 2: "$f$ is a bijection." is true (when considering the codomain as its range).

Step 2: Find the Inverse Function $f^{-1}(x)$

To find the inverse function, we set $y = f(x)$ and solve for $x$ in terms of $y$. $y = (x+1)^2 - 1$ $y+1 = (x+1)^2$ Taking the square root, we get $\sqrt{y+1} = |x+1|$. Since $x \ge -1$, $x+1 \ge 0$, so $|x+1| = x+1$. $\sqrt{y+1} = x+1$ $x = \sqrt{y+1} - 1$ Now, we swap $x$ and $y$ to get the inverse function: $f^{-1}(x) = \sqrt{x+1} - 1$ The domain of $f^{-1}(x)$ is $x+1 \ge 0$, which means $x \ge -1$. This is the range of $f(x)$, as expected.

Step 3: Analyze Statement - 1: "The set $\{x:f(x) = f^{-1}(x)\} = \{0, -1\}$."

Since $f(x) = (x+1)^2 - 1$ is strictly increasing on its domain $x \ge -1$ (its derivative $f'(x) = 2(x+1) \ge 0$ for $x \ge -1$), we can solve $f(x) = f^{-1}(x)$ by solving $f(x) = x$.

Step 3a: Solve $f(x) = x$ $(x+1)^2 - 1 = x$ $x^2 + 2x + 1 - 1 = x$ $x^2 + 2x = x$ $x^2 + x = 0$ $x(x+1) = 0$ This gives two possible solutions: $x=0$ or $x=-1$. Both of these values are in the domain $x \ge -1$. Let's verify these solutions in $f(x) = f^{-1}(x)$: For $x=0$: $f(0) = (0+1)^2 - 1 = 1 - 1 = 0$. $f^{-1}(0) = \sqrt{0+1} - 1 = 1 - 1 = 0$. So, $f(0) = f^{-1}(0) = 0$.

For $x=-1$: $f(-1) = (-1+1)^2 - 1 = 0 - 1 = -1$. $f^{-1}(-1) = \sqrt{-1+1} - 1 = 0 - 1 = -1$. So, $f(-1) = f^{-1}(-1) = -1$.

The set of solutions is indeed $\{0, -1\}$. Statement - 1: "The set $\{x:f(x) = f^{-1}(x)\} = \{0, -1\}$." is true.

Step 4: Evaluate the Relationship Between the Statements

Statement - 2 asserts that $f$ is a bijection. Statement - 1 provides the solution set for $f(x) = f^{-1}(x)$. For a strictly monotonic function, the fact that it is a bijection (onto its range) allows for the existence of its inverse. Furthermore, the property that solutions to $f(x) = f^{-1}(x)$ lie on $y=x$ is applicable because $f$ is strictly monotonic. The bijection property is therefore a prerequisite and a foundational reason why the inverse exists and the method of solving $f(x)=x$ is valid. If $f$ were not a bijection, its inverse might not exist or might not be unique, and the equation $f(x)=f^{-1}(x)$ might not be solvable in this manner or might have different properties. The fact that $f$ is a bijection ensures that $f^{-1}(x)$ is well-defined, and its monotonicity allows us to equate $f(x)$ with $x$ to find the intersection points.

Common Mistakes & Tips

• Domain and Range: Always pay close attention to the domain and range of the function, especially when dealing with inverse functions and square roots.
• Monotonicity: Recognize when a function is strictly monotonic. This is a key property that simplifies solving $f(x) = f^{-1}(x)$ to $f(x) = x$. If the function is not monotonic, solutions can also exist where $f(x) \ne x$.
• Bijection Requirement: Remember that a function must be bijective for its inverse to exist. If the codomain is not specified, assume it to be the range for the function to be considered a bijection for inverse calculation.

Summary

We first analyzed Statement - 2 and confirmed that the given function $f(x) = (x+1)^2 - 1$ with domain $x \ge -1$ is indeed a bijection from $[-1, \infty)$ to $[-1, \infty)$. We then proceeded to find the inverse function $f^{-1}(x) = \sqrt{x+1} - 1$. Finally, we analyzed Statement - 1 by solving the equation $f(x) = f^{-1}(x)$. Since $f(x)$ is strictly increasing on its domain, we solved the equivalent equation $f(x) = x$, which yielded the solutions $x=0$ and $x=-1$. Both statements were found to be true. The fact that $f$ is a bijection is a necessary condition for $f^{-1}(x)$ to exist, and its monotonicity allows us to solve $f(x)=f^{-1}(x)$ by solving $f(x)=x$, thus Statement - 2 is a correct explanation for Statement - 1.

The final answer is $\boxed{A}$.