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

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Relations And Functions XII

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 :

Answer: (B)

Let

f (x) = a x^{2} + b x + c

a \neq = 0

, so

f (1/ x) = \frac{a}{x ^{2}} + \frac{b}{x} + c

.

The condition

f (x) f (1/ x) = f (x) + f (1/ x)

can be written as

(f (x) - 1) (f (1/ x) - 1) = 1

Comparing coefficients on both sides forces

b = 0

c = 1

a = \pm 1

. Since the range must be

(- \infty, 1)

, the parabola opens downward, so

a = - 1

, giving

f (x) = 1 - x^{2}

.

From

f (K) = - 2 K

1 - K^{2} = - 2 K ⟹ K^{2} - 2 K - 1 = 0

α, β

are the roots,

α + β = 2

α β = - 1

, so

α^{2} + β^{2} = (α + β)^{2} - 2 α β = 4 + 2 = 6

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option