1. Given function and its domain

We have

Simplify:

Since the domain is $[-1,\infty)$, let us check whether $f$ is one-one and onto its range.

2. Check whether $f$ is a bijection

For $x\ge -1$,

and in fact $f'(x)>0$ for $x>-1$. So $f$ is strictly increasing on $[-1,\infty)$. Hence $f$ is one-one.

Now find its range:

As $x\to\infty$, $f(x)\to\infty$. Since $f$ is continuous and increasing on $[-1,\infty)$, its range is

Thus, viewed as a function

it is onto as well. Therefore, $f$ is a bijection.

So Statement 2 is true.

3. Find the inverse function

Let

Then

Since $x\ge -1$, we have $x+1\ge 0$, so

Hence

Therefore,

4. Solve $f(x)=f^{-1}(x)$

We need

Add $1$ to both sides:

Let

Then the equation becomes

Now write $\sqrt{t}=t^{1/2}$:

For $t\ge 0$, this gives

So either

or

Thus,

or

Hence

So Statement 1 is true.

5. Check whether Statement 2 explains Statement 1

Statement 2 says only that $f$ is a bijection, so $f^{-1}$ exists. But merely being a bijection does not explain why the solution set of

is exactly $\{-1,0\}$. That required actual computation using the explicit forms of $f$ and $f^{-1}$.

So:

• Statement 1 is true
• Statement 2 is true
• Statement 2 is not the correct explanation for Statement 1

Therefore, the correct option is

6. Comparison with stored answer

Stored correct answer: $C$

But we derived $B$. The stored answer appears incorrect because Statement 2 is clearly true: on domain $[-1,\infty)$, $f(x)=(x+1)^2-1$ is strictly increasing with range $[-1,\infty)$, hence bijective.