Fie \( a,b \in (0, \infty) \) cu \( a+b<1 \) si \( f: [0,\infty) \rightarrow [0,\infty) \) o functie crescatoare astfel incat \( \int_{0}^{x}f(t)dt=\int_{0}^{ax}f(t)dt + \int_{0}^{bx}f(t)dt,\forall x \in [0,\infty) \). Sa se arate ca \( f(x)=0, \forall x>0 \).
Mihai Piticari, ONM 1998
Functie crescatoare
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bogdanl_yex
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Functie crescatoare
"Don't worry about your difficulties in mathematics; I can assure you that mine are still greater"(Albert Einstein)
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Edgar Dobriban
- Euclid
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Concluzia ramane adevarata pt orice functie continua, nu neaparat crescatoare. Aceasta problema mai generala a fost data la OJM 2004.
Aceasta e solutia din barem:
Derivand relatia rezulta \( f(x)=af(ax)+bf(bx),\ \forall x \in[0,\infty) \).
Fie \( n>0 \) si \( M=\sup{|f(x)|,\ x\in[0,n]} \).
Avem \( |f(x)|\leq(a+b)M,\ \forall x\in[0,n] \), deci \( M\leq(a+b)M \).
Rezulta \( M=0 \).
Aceasta e solutia din barem:
Derivand relatia rezulta \( f(x)=af(ax)+bf(bx),\ \forall x \in[0,\infty) \).
Fie \( n>0 \) si \( M=\sup{|f(x)|,\ x\in[0,n]} \).
Avem \( |f(x)|\leq(a+b)M,\ \forall x\in[0,n] \), deci \( M\leq(a+b)M \).
Rezulta \( M=0 \).