2022英语周报第三十七答案

22.解:(1)∵f(x)=e2-ax,定义域为R且f(x)=①当a≤0时,则f(x)>0,函数y=f(x)在R上单调递增(1分)②当a>0时,由f(x)=0,得2e2=当x ln时,f(x)>0,函数y=f(x)单调递增.此时,函数y=f(x)的单调减区间为n2),单调增区间为(ln2,+∞).(3分)综上所述,当a≤0时,函数y=f(x)的单调递增区间为(-∞当a>0时,函数y=f(x)的单调减区间为∞}n号)单调增区间为(2m,+∞)(4分)(2)f(x)>ax2+1变形为e2-ax2-ax-1>0,令g(x)=er-ax2-ax-1,定义域为(0,+∞),且g(0)=0,g'(x)=2e2-2ax-a=2f(x)-a①当a≤0时,对任意的x>0,g(x)>0,函数y=g(x)在区间(0,+∞)上为增函数,此时,g(x)>g(0)=0,合乎题意;(7分)②当a>0时,则函数y=g(x)在R上的单调减区间为(-∞h号)单调增区间为(2l号②当a>0时,则函数y=g(x)在R上的单调减区间为(-∞h),单调增区间为((1)当11na≤0时,即当0 g(0)=2-a≥0,则函数y=g(x)在区间(0,+∞)上为增函数此时,g(x)>g(0)=0,合乎题意;(9分)()当lna>0时,即当a>2时,则函数y=8(x)在区间(o2m=)上单调递减,在区间(ln5,+∞)上单调递增,所以,g(x)m=8(ll2)=-a<0(10分)又g(0)=2-a<0,所以函数y=g(x)在区间(o)上单调递减,当x∈(0,hn2)时,g(x)

书面表达One possible versionDear PeterI'm writing to invite you to see the Chinese Painting Exhibition to be held in our city.The exhibition will start at 8: 30 am on Saturday and last till 5: 00 pm on Sunday in the city museum It'ssaid that a large number of Chinese paintings, some of which are original works by famous painters like ZhangDaqian and Qi Baishi, will be on display. Besides, there will be various souvenirs to be sold. I know you re veryinterested in traditional Chinese culture, so I can t wait to tell you the news and hope we can go together.Looking forward to your reply.You

以上就是2022英语周报第三十七答案，更多英语周报答案请关注本网站。