Let f : R - {0} → (-∞,1) be a polynomial of degree 2, satisfying f(x)f(1/x) = f(x) + f(1/x). If f(K)

Q: Wait, why do we add 1 to both sides? That seems like a random guess.

That is a fantastic question! It is not a random guess; it is a classic technique known as 'Simon's Favorite Factoring Trick' (SFFT). When you see an expression like $xy - x - y = 0$, it is difficult to factor. But if you add $1$ to both sides, it becomes $xy - x - y + 1 = 1$, which factors beautifully into $(x-1)(y-1) = 1$. In functional equations, we often look for ways to transform sums into products to simplify the structure. It is a powerful tool in your JEE arsenal!

Q: Why must $g(x) = \pm x^n$? Can't it be something else?

For a polynomial $g(x)$ satisfying $g(x)g(\frac{1}{x}) = 1$, let $g(x) = a_n x^n + ... + a_0$. If you multiply $g(x)$ by $g(1/x)$, the highest power of $x$ will be $x^n$ and the lowest will be $x^{-n}$. For the product to be a constant ($1$), the coefficients must align such that all intermediate terms cancel out. This forces the polynomial to be a simple monomial of the form $c x^n$. Since $g(x)g(1/x) = 1$, the constant $c$ must be $\pm 1$.

Q: Why do we reject $f(x) = 1 + x^2$?

We reject it because of the range constraint. The problem states the range of $f(x)$ is $(-\infty, 1)$. If $f(x) = 1 + x^2$, the minimum value is $1$ (at $x=0$), so the range would be $[1, \infty)$. This contradicts the given range. However, $f(x) = 1 - x^2$ has a maximum value of $1$ and extends to $-\infty$, which perfectly matches the given range. Always check your constraints!

Question

Let $f:\mathbb{R}-\{0\}\rightarrow(-\infty,1)$ be a polynomial of degree 2, satisfying $f(x)f(\frac{1}{x})=f(x)+f(\frac{1}{x})$. If $f(K)=-2K$, then the sum of squares of all possible values of K is:

Options:

(A) 7

(B) 6

(C) 1

(D) 9

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the edge of a vast mathematical landscape. You are presented with a riddle: a polynomial $f(x)$ of degree $2$ that obeys the strange rule $f(x)f(\frac{1}{x}) = f(x) + f(\frac{1}{x})$.

This is not just an equation; it is a structural constraint that defines the very soul of the function. Let us break it down with the precision of a master architect.

### The Algebraic Transformation

Our first step is to tame the expression. We see $f(x)f(\frac{1}{x}) = f(x) + f(\frac{1}{x})$.

Let us bring everything to one side:
$$f(x)f\left(\frac{1}{x}ight) - f(x) - f\left(\frac{1}{x}ight) = 0$$

This looks like a mess, but here is where the magic happens. We apply the 'Simon's Favorite Factoring Trick'. By adding $1$ to both sides, we transform the equation into:
$$\left(f(x) - 1ight)\left(f\left(\frac{1}{x}ight) - 1ight) = 1$$

Suddenly, the complexity vanishes. We have turned a sum into a product, finding the hidden symmetry in the chaos.

### The Polynomial Constraint

Now, let us define a new function $g(x) = f(x) - 1$. The equation becomes:
$$g(x)g\left(\frac{1}{x}ight) = 1$$

This is a profound property. If a polynomial $g(x)$ satisfies this, it must be a simple power function, $g(x) = \pm x^n$.

Because the degree of $f(x)$ is $2$, the degree of $g(x)$ must also be $2$. Thus, $n=2$. We are left with two possibilities:
$$f(x) - 1 = x^2 \quad 	ext{or} \quad f(x) - 1 = -x^2$$

This gives us $f(x) = 1 + x^2$ or $f(x) = 1 - x^2$.

### The Range Filter

Now, we must be detectives. The problem gives us a crucial clue: the range of $f(x)$ is $(-\infty, 1]$.

If $f(x) = 1 + x^2$, the graph is an upward-opening parabola with a minimum value of $1$. Its range is $[1, \infty)$, which does not fit.

If $f(x) = 1 - x^2$, the graph is a downward-opening parabola with a maximum value of $1$. It stretches down to $-\infty$, which is the perfect match. Our function is **$f(x) = 1 - x^2$**.

### The Intersection and the Final Calculation

We are told that $f(K) = -2K$. This means the parabola $y = 1 - x^2$ intersects the line $y = -2x$ at $x = K$.

Substituting $K$ into our function, we get $1 - K^2 = -2K$. Rearranging this, we arrive at the quadratic equation:
$$K^2 - 2K - 1 = 0$$

We need the sum of the squares of the roots, $K_1^2 + K_2^2$. Instead of solving for $K$ directly, we use Vieta's formulas.

For the equation $K^2 - 2K - 1 = 0$, the sum of the roots is $K_1 + K_2 = 2$ and the product is $K_1K_2 = -1$. Using the identity $K_1^2 + K_2^2 = (K_1 + K_2)^2 - 2K_1K_2$, we calculate:
$$(2)^2 - 2(-1) = 4 + 2 = 6$$

The final answer is **$6$**.

Animated Solution for Mathematics - Quadratic Equations: Let $f : R - {0} \to (- \infty, 1)$ be a polynomial of degree 2, satisfying $f (x) f (\frac{1}{x}) = f (x) + f (\frac{1}{x})$ . If $f (K) = - 2 K$ , then the sum of squares of all possible values of K is:

Select Answer:

Visualized Solution

Introduction to the Functional Equation

Rearranging the Equation

The Factorization Trick

Defining a New Function $g (x)$

Finding the General Form of $f (x)$

Using the Range to Finalize $f (x)$

Setting up the Condition $f (K) = - 2 K$

Forming the Quadratic Equation

Sum and Product of Roots

Calculating the Sum of Squares

Final Computation

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Algebraic Transformation

The Polynomial Constraint

The Range Filter

The Intersection and the Final Calculation

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