Question

$Fie~expresia~E(x)=ax+b \\ x=numar~real \\ a,b=numere~intregi \\ Concluzia~:Sa~se~determine~numerele~a~si~b~stiind~ca~E(1)~si~E(2) \\ apartin~intervalului~[1,2014]$

Answer 1

$Avem~1 \leq a+b \leq 2014~si~1 \leq 2014a+b \leq 2014~(*). \\ \\ Din ~prima~relatie~rezulta~-1 \geq-a-b \geq-2014,~adica \\ \\ -2014 \leq-a-b \leq -1~(**). \\ \\ Adunand~relatiile~marcate~cu~stelute,~obtinem: \\ \\ -2013 \leq 2013a \leq 2013 \Leftrightarrow -1 \leq a \leq 1 \Rightarrow a \in \{-1;0;1 \}. \\ \\ Daca~a=-1,~obtinem~(inlocuind~in~primele~doua~relatii): \\ \\ \left \{ {{2 \leq b \leq 2015} \atop {2015 \leq b \leq 4028}} \right. ,~de~unde~b=2015.$

$Daca~a=0,~obtinem: \\ \\ \left \{ {{1 \leq b \leq 2014} \atop {1 \leq b \leq 2014}},~deci~b \in \{1;2;3;...;2014 \}. \right. \\ \\ Daca~a=1,~obtinem: \\ \\ \left \{ {{0 \leq b \leq 2013} \atop {-2013 \leq b \leq0}} \right. \Rightarrow b=0.$