Sets, Relations & Functions
Polynomial Functions / Functional Equations
nta_pyq_2025_apr
Grade 11
Question:
Let $f : \mathbb{R} - \{0\} \to (-\infty, 1)$ be a polynomial of degree 2, satisfying $f(x)f\!\left(\frac{1}{x}\right) = f(x) + f\!\left(\frac{1}{x}\right)$. If $f(K) = -2K$, then the sum of squares of all possible values of $K$ is:
Step-by-Step Solution
Key Concept: From the functional equation with $f(x) = ax^2+bx+c$, derive that $c=1$ and $a=\pm1$. Since range $\subseteq(-\infty,1)$, conclude $f(x)=1-x^2$. Then solve $f(K)=-2K$.
Functional equation forces $c=1$, $a=\pm1$. Range condition: $f(x)=1-x^2$. $f(K)=-2K \Rightarrow 1-K^2=-2K \Rightarrow K^2-2K-1=0$. Sum of squares $= (K_1+K_2)^2-2K_1K_2 = 4-2(-1) = 6$.
Correct Answer: 6