Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $f(x)=(x+1{)}^2-1,x\ge -1$ \\ \textbf{Statement-1 :} The set $\{x:f(x)=f^{-1}(x)\}=\{0,-1\}$ \\ \textbf{Statement-2 :} $f$ is a bijection.

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are exploring the elegant geometry of functions.

We are looking at the function $f(x)=(x+1{)}^2-1$, defined for $x\ge -1$. At first glance, it looks like a standard quadratic, but the moment you see that domain restriction, you should feel a spark of intuition.

This is not the full parabola; it is a carefully curated branch. Let us dive into why this matters.

Imagine you are standing at the vertex of this parabola. The vertex is at $(-1,-1)$.

Because our domain is restricted to $x\ge -1$, we are effectively ignoring the left side of the parabola. We are only looking at the right-hand branch, which climbs steadily upward.

To understand if this function is a bijection, we must ask: is it one-to-one and onto? To check if it is one-to-one, we look at the rate of change. We calculate the derivative:

Since our domain is $x\ge -1$, the term $(x+1)$ is always greater than or equal to zero. This means $f^{\prime }(x)\ge 0$.

Because the derivative is non-negative (and strictly positive for $x>-1$), the function is strictly increasing. A strictly increasing function is the gold standard for being one-to-one. It never doubles back on itself; it never takes the same $y$-value twice.

Now, what about the 'onto' part? The minimum value of our function is at the vertex, $f(-1)=-1$.

As $x$ grows toward infinity, $f(x)$ also grows toward infinity. Thus, the range is $[-1,\infty )$.

If we assume the codomain is $[-1,\infty )$, then the range equals the codomain. We have a perfect bijection. Statement-2 is undeniably true.

Now, let us tackle Statement-1. We are asked about the intersection of $f(x)$ and its inverse, $f^{-1}(x)$.

Here is where many students stumble. They try to find the inverse function algebraically. They write $y=(x+1{)}^2-1$, swap $x$ and $y$, and solve for $y$.

While that works, it is a long, winding road prone to algebraic pitfalls. Instead, let us use the geometric soul of the problem.

The graph of an inverse function is simply the reflection of the original function across the line $y=x$. If a function is strictly increasing, it can only intersect its mirror image on the line of reflection itself.

Think about it: if the function is always climbing, it can only cross the line $y=x$ to meet its inverse. Therefore, the equation $f(x)=f^{-1}(x)$ is equivalent to solving $f(x)=x$.

This is a massive simplification! We have reduced a complex functional equation to a simple quadratic equation.

Let us set up the equation:

Expanding the left side, we get:

The constants $+1$ and $-1$ cancel out, leaving us with a beautiful, clean expression:

Subtracting $x$ from both sides, we arrive at:

Factoring this is straightforward:

This gives us two roots: $x=0$ and $x=-1$. Both of these values are within our domain of $x\ge -1$.

When we plug these back into the function, we get the points $(0,0)$ and $(-1,-1)$. These are the exact points where the function, its inverse, and the line $y=x$ all meet. Statement-1 is true.

Finally, we must address the relationship between the two statements. We have proven that Statement-1 is true and Statement-2 is true.

But does the fact that $f(x)$ is a bijection explain why the roots are $0$ and $-1$? No.

Being a bijection is a property that allows the inverse to exist, but it does not dictate the specific intersection points. The roots $0$ and $-1$ are a consequence of the specific algebraic form of the function.

Therefore, Statement-2 is not the correct explanation for Statement-1.