英语周报2022一2答案

书面表达One possible version:NOTICEA sports meeting will be held in the playground of our school from next Thursday to FridayAs you know, the pressure of study is very heavy now, especially for those senior 3. So the purpose of the sports meeting is to letevery student get relaxed, as a result of which we students can live happily and heal thilyEveryone is welcome to take part in it. Those who perform excellently at the sports meeting will get prizes. But don' t take theresults so serously because taking part is more important than the result. Good luck to everyone!

21.【命题意图】本题考查导数在研究函数性质中的应用解析】(1)由题意,得f(x)的定义域为(0,+∞)、f(x)=2e显然当a≤0时,(x)>0恒成立,厂(x)无零点(2分)当a>0时,取(x)=f(x)=2e则r(x)=4e2+">0,即f(x)单调递增(4分又(a)>0/(2)=2=-2o所以导函数f(x)存在唯一零点故当a>0时、f(x)存在唯一零点,当a≤0时f(x)无零点(5分)(Ⅱ)由(1)知,当a≤0时、f(x)单调递增,所以f(x)=f(e)=e2-a=e2,所以a=0因为g(x)=1=m-x,函数g(x)的图象在点(1,g(1)处的切线方程为y-3=0以g(1)=1m=0,所以m=1又g(1)=1+1mL+n=3,所以n=2,所以g(x)=1++2(7分)根据题意,要证f(x)≥g(x),即证lnx+1≤e2-2,只需证x(e-2)-lnx令=2-3-:则((x+1-2(2+(2-令F(x)=2-1(x>0),则F(x)=22+1所以F(x)在(0,+∞)上单调递增(8分)4<0()=e=2>0所以F()有唯一的等点(÷)当x∈(0,x)时,F(x)<0,即h(x)<0,h(x)单调递减当x∈(x,+∞)时,F(x)>0,即h(x)>0,h(x)单调递增所以h(x)=h(x2)=xn(e20-2)-lnx又因为F(x)=0.所以=所以M(x)=(-2)-()=1-2+2故f(x)≥g(x)(12分)

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