Let f(x) = (x + 1)² - 1, x ≥ -1. Then the set {x : f(x) = f⁻¹(x)} is (a) {0, -1, (-3+i√3)/2, (-3-i√3

Q: Wait, why can't I just find the inverse $f^{-1}(x)$ and solve $f(x) = f^{-1}(x)$ directly?

You absolutely can, but in the heat of a JEE exam, efficiency is king! Calculating the inverse involves square roots and potential sign errors. Using the property $f(f(x)) = x$ keeps everything in polynomial form, which is much safer and faster.

Q: Why do we reject the complex roots from the equation $(x + 1)^3 = 1$?

Great catch! The problem explicitly states the domain is $x \geq -1$. This implies we are working within the real number system. Complex numbers do not exist on the standard Cartesian plane where we plot $f(x)$ and $f^{-1}(x)$, so they are extraneous to our geometric intersection.

Q: Is it always true that $f(x) = f^{-1}(x)$ implies $f(f(x)) = x$?

Yes, for any invertible function, if $f(x) = y$, then $f^{-1}(y) = x$. If $f(x) = f^{-1}(x)$, then $f(x) = y$ implies $f(y) = x$. Substituting $y = f(x)$ gives $f(f(x)) = x$. However, always check if your solutions satisfy the original domain constraints!

Question

Let $f(x) = (x + 1)^2 - 1, x \geq -1$. Then the set $\{x : f(x) = f^{-1}(x)\}$ is

Options:

(A) $\{0, -1, \frac{-3+i\sqrt{3}}{2}, \frac{-3-i\sqrt{3}}{2}\}$

(B) $\{0, 1, -1\}$

(C) $\{0, -1\}$

(D) empty

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given function is $f(x) = (x + 1)^2 - 1$ with the domain constraint $x \geq -1$. This restriction ensures that the function is strictly increasing, which is a necessary condition for the function to be invertible.

The vertex of this parabola is located at $(-1, -1)$. By restricting the domain to $x \geq -1$, we focus exclusively on the right-hand branch of the parabola, ensuring it passes the horizontal line test.

### The Mirror Symmetry

To find the intersection points of $f(x)$ and its inverse $f^{-1}(x)$, we rely on the property that these functions are symmetric about the line $y = x$. While one could solve for the inverse explicitly, a more elegant approach exists.

If $f(x) = f^{-1}(x)$, then applying $f$ to both sides yields $f(f(x)) = f(f^{-1}(x))$. Since $f(f^{-1}(x)) = x$, we arrive at the fundamental equation:

$$f(f(x)) = x$$

### The Algebraic Symphony

We substitute $f(x) = (x + 1)^2 - 1$ into the equation $f(f(x)) = x$. This results in the following nested expression:

$$[((x + 1)^2 - 1) + 1]^2 - 1 = x$$

Simplifying the inner terms, the $-1$ and $+1$ cancel out, leaving us with:

$$[(x + 1)^2]^2 - 1 = x$$

Applying the power of a power rule, $(a^m)^n = a^{mn}$, the expression simplifies to a fourth-degree polynomial:

$$(x + 1)^4 - 1 = x$$

### Factoring for Victory

To solve for $x$, we rearrange the equation into a standard polynomial form:

$$(x + 1)^4 - (x + 1) = 0$$

By factoring out the common term $(x + 1)$, we obtain:

$$(x + 1) [ (x + 1)^3 - 1 ] = 0$$

This gives us two primary cases to solve:
1. $x + 1 = 0 \Rightarrow x = -1$
2. $(x + 1)^3 = 1 \Rightarrow x + 1 = 1 \Rightarrow x = 0$

### The Final Resolution

We have identified two real candidates for the intersection points: $x = -1$ and $x = 0$. Both values satisfy the domain constraint $x \geq -1$.

While the cubic equation $(x + 1)^3 = 1$ possesses complex roots, they are excluded as we are restricted to the real domain. Thus, the intersection points of the function and its inverse are found at:

**$x \in \{0, -1\}$**

Animated Solution for Mathematics - Functions: Let $f (x) = (x + 1)^{2} - 1, x \geq - 1$ . Then the set ${x : f (x) = f^{- 1} (x)}$ is

Select Answer:

Visualized Solution

Visualizing the Function $f (x)$

The Inverse Property

Setting up $f (f (x)) = x$

Simplifying the Inner Terms

Applying Power Rules

Transposing for Comparison

Factoring the Equation

Solving for $x$

Checking Domain and Complex Roots

The Sigma Insight: Inverse of a Function

Analyzing the Setup

The Mirror Symmetry

The Algebraic Symphony

Factoring for Victory

The Final Resolution

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