湖北鄂西南三校2023-2024學(xué)年高一上學(xué)期12月聯(lián)考數(shù)學(xué)答案

0+1+?2?=1+2+2=5.............................................................................................................5分（23x=4y=36， 1=36xyxy=36xy.................................................................................................................................故AnB={2}；...................................................................................................................4分（2）由AnB=B得B?A，2?(2a+1)x+2=0，?=(2a+1)2?8a=4(a?)2≥0，a=，212故實(shí)數(shù)a的取值集合為{0,1,}....................................................所以5k?2k2>0?0<k<又k∈Z，所以k=1或k=5,22，..............................................................所以f(x)=x2........................................................................................................6分所以2x?1<2?x，..............................................................................................9分2<2?x2<1??1<x<1，201）當(dāng)x≥0時(shí)，f(x)=?x2因?yàn)閒(x)為定義在R上的奇函數(shù)，所以m=1,n=4；...........................................................................................................4分?22fx)=?x2+ax，開口向下，對稱軸為x=?a=?22所以≤0，解得a≤0，.............................故f(u?1)+f故u2+t>1?u，即t>?u2?u+1=?2+≤，故t>，實(shí)數(shù)t的取值范圍為,+∞...........................................211）當(dāng)0<x<40時(shí)，f(x)=15×100x?10x2?900x?5000=?10x2+600x?5000，xx當(dāng)0<x<40時(shí)，f(x)=?10x2+600x?5000=?10(x?30)2+4000，f(x)max=f(30)=4000，.................................8分xxx當(dāng)x≥40時(shí)，f(x)=?x?10000+4600=?(x+10000)+4600≤4600?2=4600?200=xxx當(dāng)且僅當(dāng)x=x,即x=100時(shí)取等號，此時(shí)x=100>40，又4400>4000，......11分因?yàn)間(x)=?x2+mx開口向下，對稱軸為x=，則?1?m≤，解得m≥?，不合題意；若?1<<1，即?2<m<2時(shí)，則g(x)在