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It-is-given-that-f-x-is-a-function-defined-on-R-satisfying-f-1-1-and-for-any-x-R-f-x-5-f-x-5-and-f-x-1-f-x-1-If-g-x-f-x-1-x-then-g-2002-

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Question Number 96607 by bemath last updated on 03/Jun/20
It is given that f(x) is a function defined on R, satisfying f(1)=1 and for any x∈R, f(x+5) ≥f(x)+5 and f(x+1) ≤f(x)+1. If g(x)= f(x)+1−x, then g(2002) = ___
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
we determine f(2002) first. from the condition given , we have f(x)+5 ≤ f(x+5)≤f(x+4)+1 ≤ f(x+3)+2≤ f(x+2)+3 ≤ f(x+1)+4≤f(x+5) Thus the equality holds for all . so we have f(x+1)=f(x)+1. Hence from f(1)=1 we get f(2)=2, f(3)=3 , ..., f(2002)=2002 therefore g(2002)=f(2002)+1−2002=1
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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