Sets, Relations & Functions
Functions
nta_pyq_2025_jan
Grade 11
Question:
Let f : R - {0} \to (-\infty, 1) be a polynomial of degree 2, satisfying f (x)f ( 1 ) = f (x) + f ( 1 ) . If x x f (K) = -2K , then the sum of squares of all possible values of K is :
Step-by-Step Solution
Key Concept: Apply the core result for domains, ranges and functional equations and simplify using the given constraints.
as f (x) is a polynomial of degree two let it be (2) f (x) = ax 2 + bx + c (a \ne 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 2 f (x) = 1 - x now f (k) = -2k 2 2 1 - k = -2k \to k - 2k - 1 = 0 let roots of this equation be \alpha&\beta then \alpha + \beta 2 2 = (\alpha + \beta) 2 - 2\alpha\beta = 4 - 2(-1) = 6
Correct Answer: 2