2022英语周报二单元答案

As a matter of fact, every day in every way that little elf was a nag(喋喋不休的人! Patrickwas working harder than ever and it was a drag! He was staying up nights, had never felt so weary,and was going to school with his eyes puffed and bleary. Finally, the last day of school arrived andthe elf was free to go. As for homework, there was no more, so he quietly and slyly slipped out theback doorPatrick got his As. His classmates were amazed His teachers smiled and were full of praiseAnd his parents? They wondered what had happened to Patrick. He was now the model kid. Hecleaned his room, did his chores, cheerful and never rude, like he had developed a whole newattitude. You see, in the end, Patrick still thought he'd made that tiny man do all his homeworkBut I'll share a secret just between you and me. It wasnt the elf. Patrick had done it himself

21.【解析】(1)f(x)=ae+2.……①当a≥0时,(x)>0,函数f(x)在R上单调递增;…2分②当a<0时,由f0得x(-2),()=0解得h(-2)故(2在(一(一2)上单两增在((一2)+可上单综上所述,当a≥0时,f(x)在R上单调递增4分当a<0时,f(x)在n2上单调增在(n(-2),+叫)上单调速5分(2)证法一:原不等式等价于-ax+a06分令g(x)-1+2e,则g'((x-1)(a72当a≥1时,ae-x-1≥e-x-1,……令h(x)=c-x-1,则当x>0时,h(x)=g-1>0∴8分∴当x>0时,h(x)单调递增,即h(x)>h(0)=0,当0 1时,(x)>09分即1+2e0,故f(x)≥(x+ae)x11分证法二:原不等式等价于a(e-ex)≥(x-1)212分令g(x)=e-ex,则g(x)=e2-e当x<1时,g(x)<0;当x>1时,g(x)>0g(x)≥g(1)=0,即e-ex≥0,当且仅当x-1时等号成立当x=1时,a(e-ex)≥(x-1)2显然成立;6分当x>0且x≠1时,e-ec≥0欲证对任意的c≥1a(-e)≥(x-1)2成立,只需证-x≥(x-1)2思路1:x>0不等式-(x-1)可化为一x-1-e+2=0,令h(x)=三x-c+2,则h(x)=x=1)(c-x-1)易证当x>0时,e-x-1>0,10分∴当0 1时,h(x)>0函数h(x)在(0,1)上单词递减,在(1,十)上单调递增,∴h(x)m=h(1)=0,h(x)≥0,即x-x2-e+2=0从而,对任意的≥1,当x>0时,f(x)≥(+ac)x思路2:令g(x)=(x-1)2+则φg…12分从而,对任意的≥1,当x>0时,f(x)≥(+ae)2思路2:令(x)=5x-1)2+gx,则g'(x)=二(x-1)(x+e-312分((x)>03-c<1甲()<0÷x>1或0 0时/(x)(x+ae)x证法三:原不等等价于c+2x-1--ar≥0……12分令g(x)x2-(2)x-1,则g(x)今(x)=a-2x-(ae-2),则h(x)=c-2,其中>0①当a≥2时,(x)>0,(x)在(0,+∞上单调递增主意到h(1)=0,故当x∈(0,1)时,g(x)=h(x)<0;当x∈(1,+∞)时,g(x)=h(x)>0g(x)在(0,1)上单调递减,在(1,+∞)上单调递增g(x)m=g(1)=0,即f(x)≥(x+ae)x8分②当1≤a<2时,0<1n(2) <当0 h(2)时4(2)=0A(2)单调(1)若;≤a<2,则h(0)=a(1-e)+2≤0∵h(1n2) 0与①同,不等式成立10分()若1≤a<-;,则h(0)=a(1-e)+2>0h(lna) 01当x∈(x01)时,g(x)=h(x)<0;当x∈(1,+∞)时,g(x)=h(x)>0g(x)在(0,x0)上单调递增,在(x0,1)上单调递减,在(1,+∞)上单调递增g(0)此时,g(x)≥0,即f(x)≥(x+ae)x综上所述,结论得证12分

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