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Question Number 96607 by bemath last updated on 03/Jun/20
Itisgiventhatf(x)isafunctiondefinedonR,satisfyingf(1)=1andforanyx∈R,f(x+5)⩾f(x)+5andf(x+1)⩽f(x)+1.Ifg(x)=f(x)+1−x,theng(2002)=___
Commented by bobhans last updated on 03/Jun/20
wedeterminef(2002)first.fromtheconditiongiven,wehavef(x)+5⩽f(x+5)⩽f(x+4)+1⩽f(x+3)+2⩽f(x+2)+3⩽f(x+1)+4⩽f(x+5)Thustheequalityholdsforall.sowehavef(x+1)=f(x)+1.Hencefromf(1)=1wegetf(2)=2,f(3)=3,...,f(2002)=2002thereforeg(2002)=f(2002)+1−2002=1
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