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Question Number 148321 by mathdanisur last updated on 27/Jul/21

Answered by dumitrel last updated on 27/Jul/21

$$p\left(x\right)=x^{2021}-x-1,x>0$$$$p\left(a\right)=0;p\left(x\right)\nearrow forx\in (a,\mathrm{\infty });a>1$$$$ifb⩾a\Rightarrow f\left(b\right)⩾f\left(a\right)=0\Rightarrow b^{2021}⩾b+1\Rightarrow$$$$b^{4042}⩾b^2+2b+1\Rightarrow 3a+b⩾b^2+2b+1\Rightarrow$$$$3b⩾3a⩾b^2+b+1\Rightarrow {(b-1)}^2⩽0\Rightarrow b=1\Rightarrow 1⩾afalse$$$$\Rightarrow b<a$$

Commented by mathdanisur last updated on 27/Jul/21

$$ThankyouSer$$

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