JEE Main 2025 Jan 28 Shift 2, Mathematics Q20: Invertible And Composite Functions

JEE Main2025Jan 28, Shift 2Mathematics

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 :

A
$1$
B
$6$
C
$7$
D
$9$

Solution

A function satisfying $f (x) f (1/ x) = f (x) + f (1/ x)$ for all relevant $x$ must have the form $f (x) = 1 \pm x^{n}$ . For a degree-2 polynomial, $n = 2$ , so $f (x) = 1 + x^{2}$ or $f (x) = 1 - x^{2}$ .

The range condition $(- \infty, 1)$ rules out $1 + x^{2}$ (whose range is $[1, \infty)$ ). So $f (x) = 1 - x^{2}$ .

Now $f (K) = - 2 K \Rightarrow 1 - K^{2} = - 2 K \Rightarrow K^{2} - 2 K - 1 = 0$ .

For the two roots $α, β$ : $α + β = 2$ , $α β = - 1$ , so $α^{2} + β^{2} = (α + β)^{2} - 2 α β = 4 + 2 = 6$ .

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JEE Main 2025 Jan 28 Shift 2, Mathematics Q20: Invertible And Composite Functions

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