2021-2022 英语周报 高二 新课标22答案

读后续写Possible version lI stared at the paper and then looked around. I was cold. Although I could not see my sistersnor the new house, which was hidden from my sight by a grey wall, I got the feeling that I was notalone. Walking along the shore was a tall familiar figure who looked lost in his thoughts. I lookedgain at the paper in my hand, closed my eyes and wishedAs /opened my eyes, I found my father was standing right beside me. Happy birthday, sonhe said, patting my shoulder gently. I smiled up at him and took his hand. Looking back, life fromthat day on marked the start of a close bond and a strong friendship with my father. What was mybirthday surprise? Wishes do come truePossible version 2I stared at the paper and then looked around. On the backside, there was a map. I noticed thatit was clearly leading me in a direction, but who could have sent this? I stood there in silence andstudied it. I decided to follow the map and it led me down the beach until I reached a staircase. Iooked back and could no longer see my siblings. I continued to follow the point on the map, untilI unexpectedly lost my footing and fell zksq down the stepsAs I opened my eyes, I found my father was standing right beside me. Happy birthday, sonhe said, patting my shoulder gently. Then he asked how I had come upon this place? I explained tohim that I had found the bottle in the ocean and when I opened it had the words on it. Shocked bythe message, my father realized what I had found I threw this bottle into the ocean many years agowhen I was alone. he said. I used to come to this exact place to get away from everyone else. NowI get to share this experience with you. What a birthday surprise

21.解:(1)题得,f(x)的定义域为(0,+∞),f(x)=(=1)(a-1)x(1分)当a≥1时,f(x)>0,函数f(x)在(0,+∞)上单调递增;(2分当a<1时,记g(x)=1+(a-1)x,则函数g(x)=1+(a-1)x在(0,+∞)上单调递减,令g(x)=0→x=1所以当x∈(0,1)时g(x)>0,即f(2>0,函数f(2)在(O,12)上单调递增当x∈(1,+∞)时g()<0,即f(x)<0,函数f(x)在(1,+∞)单调递减(5分)综上可知:当a≥1时,函数f(x)在(0,+∞)上单调递增;当a<1时,函数f(x)在(f-a)单调递增,在(12,+∞)上单调递减,(6分)2)当a=0时,(x)=lmx-x+1,/(x)=1-1一由)可知,函数∫(x)在(0,1)上单调递增,在(1,∞)上单调递减,所以x=1为函数f(x)的极大值点,所以f(x)≤f(1)=0(8分)因为g(x)=2x2-4x+m=2(x-1)2+m-2,所以g(x)≥m-2,当且仅当x=1时,g(x)m=m-2.(10分)所以m-2≥0→m≥2,所以m的最小值为2.(12分)

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