2022英语周报外研版高一第30期答案

第四部分写作第一节短文改错事Offering help is always a good tradition in ourcountries. It is believed that by heelp anothercountnhelpingperson, you are helping yourself. Helping others dodohave a significant effect on a person's happiness. Eventhough we usually feel the urge A compete with othertotaktime to assist otherseneficial to us. Besides, helping people is a goodbenefitway to build trust, particular if you never expect orparticularldemand something to be given to you in retum.anythingNowadays, a rising number of individaIs areready to assist tht others. This has resulted from awonderful community, which people live in harmonynIwherewith each other

21.(1)①因为f(x)=e-ax2,所以f(x)=e-2ax因为曲线f(x)与直线l4y=(e-2)x+b在x=1处相切所以f(1)=e-2a=e-2,所以a=1(2分)所以f(x)=e2-x2,所以f(1)=e-1,又切点(1,e-1)在直线l上,所以e-1=e-2+b,(3分)所以b=1,所以a+b=2②由①知a=1,b=1,可设h(x)=e4-x2-(e-2)x-1(x≥0),(4分)则g(x)=h'(x)=e-2x-(e-2),g'(x)=e-2,当x ln2时,g'(x)>0所以h(x)在(0,n2)上单调递减,在(ln2,+∞)上单调递增,由h'(0)=3-e>0,h(1)=0,0<1n2<1,所以h'(ln2)<0,所以存在x0∈(0,ln2),使得h'(x。)=0(5分)所以当x∈(0,x0)U(1,+∞)时,h(x)>0,当x∈(x0,1)时,h'(x)<0所以h(x)在(0,x)上单调递增,在(x0,1)上单调递减,在(1,+∞)上单调递增.(6分)因为h(0)=h(1)=0,所以h(x)≥0,即f(x)≥(e-2)x+1,当且仅当x=1时取等号,所以当x≥0时,e2-x2≥(e-2)x+1,故当x≥0时,f(x)≥(e-2)x+b(7分)(2)先证e2≥x+1构造函数p(x)=e-x-1,则p'(x)=e-1,故当x∈(0,+∞)时,p(x)>0,p(x)在(0,+∞)上递增,当x∈(-∞,0)时,p(x)<0,p(x)在(-∞,0)上递减,所以p(x)≥p(0)=0,即e≥x+1又当a=0,且x∈(0,+∞)时,xf(x)≤mx+2nx+1等价于m>2-2(9分)故原题等价于x∈(0,+∞)时,m≥x'e-2Int-1有解10分因为e-2nx-1e+m-2nx-1x+lnx2+1-2lmx-(当x+lnx2=0时取等号),所以m≥1

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