八上英语周报2018-2022答案

作者:网络 来源: 2022-09-04   阅读:

第一节短文改错The moment I stepped into high school. I have madeup my mind to study hard in order to enter the gooduniversity. In the first vear. I made a lot o

八上英语周报2018-2022答案图片

第一节短文改错The moment I stepped into high school. I have madeup my mind to study hard in order to enter the gooduniversity. In the first vear. I made a lot of new friendsall of who were excellent. But my scores went up orwhomanddown, which made ne feel stressful. My teacher toldstressedmyself that I needed relaxation. She recommended meNmelisten to music and take more activity. With her helpactivitiesgradual recovered my confidence andmade mucgraaduallyprogress than before. Finally, I was admitted to my idealuniversity

八上英语周报2018-2022答案图片

八上英语周报2018-2022答案图片

八上英语周报2018-2022答案图片

18解:(1)因为f(x)=1x2-alnx,其定义域为(0,∞)所以f"(x)=x-a(x>0),当a≤0时,f(x)>0,则f(x)在(0,十∞)上单调递增;当a>0时,f(x)=x-a=x“,所以当0 √a时,f(x)>0所以f(x)在(0,√a)上单调递减在(a,十∞)上单调递增综上所述,当a≤0时,f(x)的单调递增区间为(0,∞x),无单调递减区间;当a>0时,f(x)的单调递增区间为(a, ∞),单调递减区间为(0,a)(6分)(2)当a=-1时,f(x)<2x3在(1, ∞)上恒成立证明如下:设g(x)12则(x-1)(2x2 x当x>1时,g"(x)>0,g(x)在(1, ∞)上是增函数,从而g(x)>g(1)0,即0所以bx2 nx<2x故当x>1时,1x2 1mx<2x恒成立.(12分)

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