olymmath

OBM 2020 Problem 6

    Sejam f(x)=2x2+x−1f(x) = 2x^2 + x - 1, f0(x)=xf^0(x) = x e fn+1(x)=f(fn(x))f^{n+1}(x) = f(f^n(x)) para todo xx real e n≥0n \ge 0 inteiro (ou seja, fnf^n é a nn-ésima iterada de ff).

    (a) (2 pontos) Determine o número de soluções reais distintas da equação de f3(x)=xf^3(x) = x.

    (b) (8 pontos) Determine, para cada n≥0n \ge 0 inteiro, o número de soluções reais distintas da equação fn(x)=0f^n(x) = 0.

    Official text Brazilian Mathematical Olympiad 2020 official paper, page 2. Quoted with source.

    A model read this text from the scan of the paper by OCR, with vision/claude@2.1.318 (claude-opus-5-5). No person has checked it against the paper yet.

    Solutions

    No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

    Add a solution

    Similar problems

    These problems have the most words in common with this statement.

    Locator br-obm/2020/p6 · Tag Y43XM4MZPT · Edit this page on GitHub · Keys: j next, k previous