2021-2022英语周报HBJ答案

21什么命题人考查函数与导数的综合问题么(1)当m=1时,f(x)=lnx,则f(x)=所以厂(1)=1,又因为f(1)=0曲线y=f(x)在x=1处的切线方程为y-0=1×(x-1),即x2分(2)F(x)=f(x)-g(x)= mIn x则F'(x)m-当m≤0时,F'(x)<0,所以函数F(x)在(0,+∞)上单调递减当m>0时,由F'(x)<0得0 0得x>所以画数F()在(0.n)上单调递减,在(,+四)上单调递增5分(3)函数fx)=mlnx的图象在点(a,mlna)处的切线方程为ymma=m(x-),即y=mx+mna一m;画戴g(x)=2的图象在点(6-1)处的切线方程为y-(1-由y=f(x)与y=g(x)的图象有唯条公切线m=①mIn a-m=②由①得,m=1,代入②消去m,整理得b2-2b-alna+a=0③3则关于b(b>0)的方程③有唯一解Ag(b)=6-2b-aln a+a=(b-1)2-aln a+a-Ia h(a)=-aln a +a-I, h(a)=-In a,由h'(a)>0得0 1,所以函数h(a)在(0,1)上单调递增,在(]、+∞)上单调递减,则h(a)≤h(1)=0.(i)当h(a)=0时,方程③有唯一解b=1,由h(a)=-ana+a1=0得a=1,此时m=(i)当h(a)<0时,二次函数g(b)=(b-1)2-alna+a-1在b∈(1,+∞)上显然有一个零点,当b∈(0,1)时,由方程②可得m(lna-1)=<0,而m>0,所以hna-1<0A g(0)=-aIn u +a=-a(In 4-1)>0所以二次函数g(b)=(b-1)2-alna+4-1在b∈(0,1)上也有一个零点,不合题意综上,存在正实数m=1使y=f(x)与y=g(x)的图象有唯一一条公切线……………12分

第三节书面表达(满分25分One possible versionDear Lucy,I am Li Hua, chairman of the Students'Union. I am writing here to extend our sinceregratitude to youDuring the COVID-19 pandemic last year, you were trapped in your country and couldntreturn to our school. However, it was lucky that you used the Internet to give lessons. All thestudents said that you put your heart into making the lessons interesting. As a consequence.though you were still in England, we had no trouble in continuing new lessons this term, whichmade us so proud of being taught by youLooking forward to your early returnLi Hua

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