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QJEEM25Mac2c3: JEE Main 2025 Mathematics - Class 11 JEEM25Mac2c3 Solution: Let be a polynomial of degree 2 , satisfying . If , then the…

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

Multiple Choice Options

  • A 9
  • B 1
  • C 6
  • D 7

Step-by-Step Verified Solution

as f(x)f(x) is a polynomial of degree two let it be

f(x)=ax2+bx+c(a≠0)f(x)=a x^2+b x+c \quad(a \neq 0)

on satisfying given conditions we get

C=1&a=±1C=1 \& a= \pm 1

hence f(x)=1±x2f(x)=1 \pm x^2

also range ∈(−∞,1]\in(-\infty, 1] hence

f(x)=1−x2 now f(k)=−2k\begin{gathered}f(x)=1-x^2 \\\text { now } f(k)=-2 k\end{gathered}

1−k2=−2k→k2−2k−1=01-\mathrm{k}^2=-2 \mathrm{k} \rightarrow \mathrm{k}^2-2 \mathrm{k}-1=0

let roots of this equation be α&β\alpha \& \beta then α2+β2=(α+β)2−2αβ\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta

=4−2(−1)=6=4-2(-1)=6

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