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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) = ___

$$\mathrm{It}\:\mathrm{is}\:\mathrm{given}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{function} \\ $$$$\mathrm{defined}\:\mathrm{on}\:\mathbb{R},\:\mathrm{satisfying}\:\mathrm{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{for}\:\mathrm{any}\:\mathrm{x}\in\mathbb{R},\:\mathrm{f}\left(\mathrm{x}+\mathrm{5}\right)\:\geqslant\mathrm{f}\left(\mathrm{x}\right)+\mathrm{5} \\ $$$$\mathrm{and}\:\mathrm{f}\left(\mathrm{x}+\mathrm{1}\right)\:\leqslant\mathrm{f}\left(\mathrm{x}\right)+\mathrm{1}.\:\mathrm{If}\:\mathrm{g}\left(\mathrm{x}\right)= \\ $$$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{1}−\mathrm{x},\:\mathrm{then}\:\mathrm{g}\left(\mathrm{2002}\right)\:=\:\_\_\_ \\ $$

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

$$\mathrm{we}\:\mathrm{determine}\:\mathrm{f}\left(\mathrm{2002}\right)\:\mathrm{first}.\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{condition}\:\mathrm{given}\:,\:\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{x}\right)+\mathrm{5}\:\leqslant\:\mathrm{f}\left(\mathrm{x}+\mathrm{5}\right)\leqslant\mathrm{f}\left(\mathrm{x}+\mathrm{4}\right)+\mathrm{1} \\ $$$$\leqslant\:\mathrm{f}\left(\mathrm{x}+\mathrm{3}\right)+\mathrm{2}\leqslant\:\mathrm{f}\left(\mathrm{x}+\mathrm{2}\right)+\mathrm{3} \\ $$$$\leqslant\:\mathrm{f}\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{4}\leqslant\mathrm{f}\left(\mathrm{x}+\mathrm{5}\right) \\ $$$$\mathrm{Thus}\:\mathrm{the}\:\mathrm{equality}\:\mathrm{holds}\:\mathrm{for}\:\mathrm{all}\:.\:\mathrm{so}\:\mathrm{we}\: \\ $$$$\mathrm{have}\:\mathrm{f}\left(\mathrm{x}+\mathrm{1}\right)=\mathrm{f}\left(\mathrm{x}\right)+\mathrm{1}.\:\mathrm{Hence}\:\mathrm{from}\:\mathrm{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{f}\left(\mathrm{2}\right)=\mathrm{2},\:\mathrm{f}\left(\mathrm{3}\right)=\mathrm{3}\:,\:...,\:\mathrm{f}\left(\mathrm{2002}\right)=\mathrm{2002} \\ $$$$\mathrm{therefore}\:\mathrm{g}\left(\mathrm{2002}\right)=\mathrm{f}\left(\mathrm{2002}\right)+\mathrm{1}−\mathrm{2002}=\mathrm{1} \\ $$

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