2021-2022英语周报高三课标第二期答案

书面表达One possible versionI求Dear PeterI heve good news for you Our school Food Club will hold a dumpling- making activity. As you're interestedin Chinese food, I'm writing to invite you to join in the activity with me.As a kind of traditional Chinese food, dumplings are very popular during festivals, especially in the SpringFestival, when people will get together to make dumplings. The activity will begin at g am next Saturday in theschool canteen. First, (chefs) will teach us how to make dumplings step by step. Then, there will be a dumplingmaking competition in groups. Finally comes the most exciting part--enyoying our dumplings.Looking forward to your coming.Your

21.解:(1)当a=-1时,f(x)=hx-立x?所以f(x)=1-x=1-x2=1+x)(1-x)…1分令f(x)=0,则1+x)(1-x=0,所以x=-1(舍)或x=1.…2分分析知,当x∈(0,1)时,f(x)>0;当x∈(1,+∞)时,f(x)<03分所以函数f(x)在区间(0,1)上单调递增,在区间(1,+∞)上单调递减,………4分(2)因为f(x)=hnx+ax2+(a+1)x+3,所以f(x)=1(a+1)x+1)(ax+1)x讨论:若a=0,则f(x)>0对任意的x∈(0,+∞)成立,所以函数f(x)在区间(0,+∞)上单调递增又因为f(1)<0,(1>0,函数f(的图象在区间0,+=)上为连续不间断曲线,所以函数f(x)在区间(,1)上有零点,即关于x的方程f(x)=0有实数根;…若a>0,则当x∈(0,+∞0)时,f(x)>0,所以f(x)在区间(0,+∞)上单调递增又f(1)(2+2>0)取x0=e(x0<1),则f(x0)=-2(a+1)+ar+(a+1)xo+2+D+1+a+1+2(at<0函数f(x)的图象在区间(0,+∞)上为连续不间断曲线,所以函数f(x)在区间(x0,1)上有零点,即关于x的方程f(x)=0有实数根;………8分若<0则当x∈(0,-)时,()>当x∈(-2,+)时,(2)<0所以函数f(x)在区间(,上单两递增,在区(_1,+∞)上单调递减,此时f(x)=lnx+x++ax2+ar 0,引人函数k()=hx+号一2.则(x)=1+1(1+>0),所以函数g(x)在区间(0,+∞)上单调递增,且g(1)=0,所以要保证g(x)≥0,只需满足x≥1,所以只需满足-1≥1,即-1≤a<0综上,所求实数a的取值范围是[-1,+∞0).…

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