28867: Difference between revisions

Revision as of 12:32, 5 August 2025

28867 (Natalia Fărcaș)

Fie funcția injectivă Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f:\mathbb{R} \to \mathbb{R}} , cu proprietatea că există numerele reale Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} și Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} astfel încât Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f\left(x\right) \cdot f\left(1-x\right) = f\left(ax+b\right)} oricare ar fi Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\in \mathbb{R}} .

Demonstrați că Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f\left(1-b\right)=1} .
Dați un exemplu de șir Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(f_n\right)_{n\ge 1}} de funcții injective Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_n:\mathbb{R} \to \mathbb{R}} , cu proprietatea că există Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a,b \in \mathbb{R}} , astfel încât pentru orice Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\in \mathbb{R}} , avem Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_n\left(x\right) \cdot f_n\left(1-x\right) = f_n\left(ax+b\right)} și Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \log_{n+1} f_n\left(x\right) = a - \log_{n+1} f_n\left(-x\right).}

Soluție Pentru Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0} se obține egalitatea Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f\left(0\right) \cdot f\left(1\right) = f\left(b\right)} , iar pentru Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=1} se obține Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f\left(1\right)\cdot f\left(0\right) = f\left(a+b\right)} . Cum Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} este injectivă, rezultă Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a+b=b} , deci Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a=0} . Egalitatea din ipoteza problemei devine Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f\left(x\right) \cdot f\left(1-x\right) = f\left(b\right)\ne 0, \forall x\in \mathbb{R}.}

Dacă presupunem că Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f\left(b\right) = 0} , atunci din Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f\left(x\right) \cdot f\left(1-x\right) = 0} rezultă că există Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0\in \mathbb{R}\setminus\left\{b\right\}} cu Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f\left(x_0\right) = 0 = f\left(b\right)} , contradicție cu proprietatea de injectivitate a func\c tiei Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} . Așadar Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f\left(b\right) \ne 0} .

a) Pentru Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=b} , din Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f\left(x\right) \cdot f\left(1-x\right) = f\left(b\right)\ne 0} rezultă Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f\left(b\right)\cdot f\left(1-b\right) = f\left(b\right)} . Deoarece Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f\left(b\right) \ne 0} , se obține Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f\left(1-b\right) = 1} .

b) Pentru orice Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \in \mathbb{N}^\ast} , fie Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_n : \mathbb{R} \to \mathbb{R}} , cu Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_n\left(x\right) = \left( n+1 \right)^x} . Evident, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_n} este injectivă și dacă Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a=0} și Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b=1} , funcția Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_n} verifică egalitățile din enunț.

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 '''Soluție'''
+Pentru <math>x=0</math> se obține egalitatea <math>f\left(0\right) \cdot f\left(1\right) = f\left(b\right)</math>, iar pentru <math> x=1</math> se obține <math>f\left(1\right)\cdot f\left(0\right) = f\left(a+b\right)</math>. Cum <math>f</math> este injectivă, rezultă <math>a+b=b</math>, deci <math>a=0</math>. Egalitatea din ipoteza problemei devine <math> f\left(x\right) \cdot f\left(1-x\right) = f\left(b\right)\ne 0, \forall x\in \mathbb{R}.</math>
+Dacă presupunem că  <math>f\left(b\right) = 0</math>, atunci din <math>f\left(x\right) \cdot f\left(1-x\right) = 0</math> rezultă că există <math>x_0\in \mathbb{R}\setminus\left\{b\right\}</math> cu <math>f\left(x_0\right) = 0 = f\left(b\right)</math>, contradicție cu proprietatea de injectivitate a func\c tiei <math>f</math>. Așadar <math>f\left(b\right) \ne 0</math>.
+a) Pentru <math>x=b</math>, din <math>f\left(x\right) \cdot f\left(1-x\right) = f\left(b\right)\ne 0</math> rezultă <math>f\left(b\right)\cdot f\left(1-b\right) = f\left(b\right)</math>. Deoarece <math>f\left(b\right) \ne 0</math>, se obține <math>f\left(1-b\right) = 1</math>.
+b) Pentru orice <math> n \in \mathbb{N}^\ast</math>, fie <math> f_n : \mathbb{R} \to \mathbb{R}</math>, cu <math>f_n\left(x\right) = \left( n+1 \right)^x</math>. Evident, <math>f_n</math> este injectivă și dacă <math>a=0</math> și <math>b=1</math>, funcția <math>f_n</math> verifică egalitățile din enunț.