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Quadratic polynomial divides [0,1] in two parts

Source: Belarus TST 2024

July 17, 2024
quadraticsalgebrapolynomial

Problem Statement

Let f(x)=x2+bx+cf(x)=x^2+bx+c, where b,c∈Rb,c \in \mathbb{R} and b>0b>0 Do there exist disjoint sets AA and BB, whose union is [0,1][0,1] and f(A)=Bf(A)=B, where f(X)={f(x),x∈X}f(X)=\{f(x), x \in X\} D. Zmiaikou
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