Let f: R-\ 0\ arrow(-∞, 1) be a polynomial of degree 2, satisfying f(x) f((1)/(x))=f(x)+f((1)/(x)) . If f(K)=-2 K , then the sum of squares of all possible values of K is :

Correct answer: B. 6 as f(x) is a polynomial of degree two let it be f(x)=a x^2+b x+c (a ≠ 0) on satisfying given conditions we get C=1 \& a= ± 1 hence f(x)=1 ± x^2 also range ∈(-∞, 1] hence f(x)=1-x^2 now f(k)=-2 k 1-k^2=-2 k arrow k^2-2 k-1=0 let roots of this equation be α \& β then α^2+β^2=(α+β)^2-2 α β =4-2(-1)=6

JEE Main 202528 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Functions

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Functions

Let f: R-\ 0\ arrow(-∞, 1) be a polynomial of degree 2, satisfying...

Let

f: \mathbf{R}-\{0\} \rightarrow(-\infty, 1)

be a polynomial of degree 2, satisfying

f(x) f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)

. If

f(K)=-2 K

, then the sum of squares of all possible values of

K

is :

Options

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

Correct answer

B. $6$

Step-by-step solution

f(x)

is a polynomial of degree two let it be

f(x)=a x^2+b x+c \quad(a \neq 0)

on satisfying given conditions we get

C=1 \& a= \pm 1

hence

f(x)=1 \pm x^2

also range

\in(-\infty, 1]

hence

f(x)=1-x^2

now

f(k)=-2 k

1-\mathrm{k}^2=-2 \mathrm{k} \rightarrow \mathrm{k}^2-2 \mathrm{k}-1=0

let roots of this equation be

\alpha \& \beta

then

\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta

=4-2(-1)=6

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