For a suitably chosen real constant a, let a function, f : R - {-a} → R be defined by f(x) = (a-x)/(

Q: Wait, why do we set the coefficients of the quadratic to zero? Can't we just solve for $x$?

Great question! The problem states that $(f \circ f)(x) = x$ for *any* real number $x$ in the domain. If it were just an equation to solve for $x$, we would find specific roots. But because it is an identity (true for all $x$), the equation must hold regardless of the value of $x$. For a quadratic $Ax^2 + Bx + C = 0$ to be true for all $x$, the coefficients $A$, $B$, and $C$ must all be zero.

Q: Is there a faster way to find $a$ without doing all that algebra?

You could try testing a specific value of $x$ to find $a$, but be careful! If you pick a value like $x=0$, you might find a candidate for $a$, but you must verify that this $a$ works for *all* $x$. The algebraic method of equating coefficients is the most rigorous way to ensure the condition holds universally.

Q: Why is $a=1$ the only solution? What happened to $a=-1$ or $a=0$?

When we solved the system of equations derived from the coefficients, we found that $a=1$ was the only value that satisfied *all* the conditions simultaneously. For example, while $a=-1$ might have satisfied one part of the equation, it failed to satisfy the others. In mathematics, we must satisfy the entire system of constraints, not just one.

Question

For a suitably chosen real constant $a$, let a function, $f : \mathbb{R} - \{-a\} 	o \mathbb{R}$ be defined by $f(x) = \frac{a-x}{a+x}$. Further suppose that for any real number $x 
e -a$ and $f(x) 
e -a$, $(fof)(x) = x$. Then $f\left(-\frac{1}{2}ight)$ is equal to

Options:

(A) $-3$

(B) $\frac{1}{3}$

(C) $-\frac{1}{3}$

(D) $3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of the mathematical universe! Today, we are diving into a problem that might look like a simple algebraic manipulation at first glance, but it hides a beautiful, deep concept within its structure.

We are dealing with a function $f(x) = \frac{a-x}{a+x}$ and a mysterious condition: $(f \circ f)(x) = x$. This is the definition of a **self-inverse function**.

Imagine a mirror. If you look into a mirror, you see yourself; if you look into that reflection, you see yourself again. That is exactly what $(f \circ f)(x) = x$ represents—it is a mathematical mirror.

### The Algebraic Odyssey

Let us begin by confronting the beast. We need to compute $f(f(x))$.

$$f(f(x)) = \frac{a - f(x)}{a + f(x)}$$

Now, we substitute the definition of $f(x)$ into this expression:

$$f(f(x)) = \frac{a - \left( \frac{a-x}{a+x} \right)}{a + \left( \frac{a-x}{a+x} \right)}$$

To simplify this, we multiply the numerator and the denominator by the common term $(a+x)$. This is a classic JEE trick to clear out complex fractions.

The numerator becomes $a(a+x) - (a-x)$, which simplifies to $a^2 + ax - a + x$, or $(a^2 - a) + (a + 1)x$. Similarly, the denominator becomes $a(a+x) + (a-x)$, which simplifies to $a^2 + ax + a - x$, or $(a^2 + a) + (a - 1)x$.

Now we have a much cleaner expression:

$$f(f(x)) = \frac{(a^2 - a) + (a + 1)x}{(a^2 + a) + (a - 1)x}$$

### The Identity Trap

Here is where many students stumble. We are given that $(f \circ f)(x) = x$. So, we set our simplified expression equal to $x$:

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

Expanding the right side, we get:

$$(a^2 - a) + (a + 1)x = (a^2 + a)x + (a - 1)x^2$$

Now, we bring everything to one side to form a quadratic equation:

$$(a - 1)x^2 + (a^2 + a - a - 1)x - (a^2 - a) = 0$$

This simplifies to:

$$(a - 1)x^2 + (a^2 - 1)x - a(a - 1) = 0$$

Because this must hold for **all** $x$, this is an identity. For a quadratic to be zero for all $x$, the coefficient of $x^2$, the coefficient of $x$, and the constant term must all be zero independently.

This gives us three conditions: $a-1=0$, $a^2-1=0$, and $-a(a-1)=0$. Solving these, we find that $a=1$ is the only value that satisfies all three conditions simultaneously.

### The Final Victory

With $a=1$, our function becomes $f(x) = \frac{1-x}{1+x}$. This is a beautiful rectangular hyperbola, symmetric about the line $y=x$.

The final step is simple arithmetic. We need to find $f\left(-\frac{1}{2}\right)$. Substituting $x = -\frac{1}{2}$ into our function:

$$f\left(-\frac{1}{2}\right) = \frac{1 - \left(-\frac{1}{2}\right)}{1 + \left(-\frac{1}{2}\right)} = \frac{1 + \frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{3}{2}}{\frac{1}{2}} = 3$$

And there we have it! The final answer is **3**. This problem taught us that behind every complex-looking algebraic expression lies a simple, elegant truth.

Select Answer:

Visualized Solution

Defining the Function $f (x)$

The Concept of Self-Inverse

Setting up $f (f (x))$

Simplifying the Numerator

Simplifying the Denominator

Equating to $x$

Cross-Multiplication

Rearranging into Quadratic Form

The Identity Property

Solving for the Constant $a$

The Final Function

Calculating $f (- \frac{1}{2})$

Final Answer

The Sigma Insight: Composite Functions

Analyzing the Setup

The Algebraic Odyssey

The Identity Trap

The Final Victory

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Select Answer:

Visualized Solution

Defining the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The Concept of Self-Inverse

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(f(x))

Simplifying the Numerator

Simplifying the Denominator

Equating to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Cross-Multiplication

Rearranging into Quadratic Form

The Identity Property

Solving for the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

The Final Function

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(−21​)