注意事項(xiàng):44 235264522高一年級(jí)學(xué)生得分A.fx與gx的圖象關(guān)于直線對(duì)稱fx>gx)11.已知函數(shù)fx的定義域?yàn)镽,且fxf1-y+fyf1-x)=2fx+y,f0)=0fx0<0____.棋子,則恰好每列都有兩枚棋子的概率為_(kāi)___.1243.11A1111A1刻畫(huà)隨機(jī)變量X蘊(yùn)含的信息量的大小.2)題.充分條件.EQ\*jc3\*hps23\o\al(\s\up0(-),e)EQ\*jc3\*hps23\o\al(\s\up0(→),1)EQ\*jc3\*hps23\o\al(\s\up0(-),e)EQ\*jc3\*hps23\o\al(\s\up0(→),2)EQ\*jc3\*hps23\o\al(\s\up0(-→),a)EQ\*jc3\*hps23\o\al(\s\up0(-),e)EQ\*jc3\*hps23\o\al(\s\up0(→),1)EQ\*jc3\*hps23\o\al(\s\up0(-),e)EQ\*jc3\*hps23\o\al(\s\up0(→),2)EQ\*jc3\*hps23\o\al(\s\up0(-),e)EQ\*jc3\*hps23\o\al(\s\up0(→),1)EQ\*jc3\*hps23\o\al(\s\up0(-),e)EQ\*jc3\*hps23\o\al(\s\up0(→),2)EQ\*jc3\*hps23\o\al(\s\up0(-→),a)【分析】本題考查向量的坐標(biāo)運(yùn)算及向量夾角的計(jì)EQ\*jc3\*hps23\o\al(\s\up0(-→),a)EQ\*jc3\*hps23\o\al(\s\up0(-),e)EQ\*jc3\*hps23\o\al(\s\up0(→),1)EQ\*jc3\*hps23\o\al(\s\up0(-),e)EQ\*jc3\*hps23\o\al(\s\up0(→),2)EQ\*jc3\*hps23\o\al(\s\up0(-),e)EQ\*jc3\*hps23\o\al(\s\up0(→),1)EQ\*jc3\*hps23\o\al(\s\up0(-),e)EQ\*jc3\*hps23\o\al(\s\up0(→),2)EQ\*jc3\*hps23\o\al(\s\up0(-→),a)EQ\*jc3\*hps23\o\al(\s\up0(-→),a)EQ\*jc3\*hps23\o\al(\s\up2(-→),b)24422A.fx與gx的圖象關(guān)于直線對(duì)稱A.設(shè)fx上一點(diǎn)為x1,fx1)),B.設(shè)fx上一點(diǎn)為x2,fx2)),:fx>gx,C正確;:fx與gx的圖象恰有四個(gè)交點(diǎn),D正確.11.已知函數(shù)fx的定義域R,fxf1-y+fyf1-x)=2fx+y,f0)=1,則0fx0<0令x=y=0,f0f1+f0f1=2f1=2f0,則f1=f0=1令x=y=1,f1f0+f1f0=2f2,則f2=f1=1,A正確;令y=0,則fxf1+f0f1—x)=2fx,fx+f(1—x)=2fx,所以f(1—x)=fx令y=1,則fxf0+f1f1—x)=2fx+1),所以f(1所以f2x=f2x,fx=f2()≥0,B錯(cuò)誤;【分析】本題考查分段函數(shù)、函數(shù)的單調(diào)性2.2；EQ\*jc3\*hps23\o\al(\s\up7(≤),2)EQ\*jc3\*hps23\o\al(\s\up7(≤),a)2f(1)=42a;行都放置兩枚棋子,則恰好每列都有兩枚棋子的概率為_(kāi)___.EQ\*jc3\*hps16\o\al(\s\up4(2),4)EQ\*jc3\*hps16\o\al(\s\up4(2),4)EQ\*jc3\*hps16\o\al(\s\up4(2),4)EQ\*jc3\*hps16\o\al(\s\up4(1),2)EQ\*jc3\*hps16\o\al(\s\up4(1),2)EQ\*jc3\*hps16\o\al(\s\up4(1),2)EQ\*jc3\*hps16\o\al(\s\up4(2),4)EQ\*jc3\*hps16\o\al(\s\up4(2),4)EQ\*jc3\*hps16\o\al(\s\up4(1),2)EQ\*jc3\*hps16\o\al(\s\up4(1),2)EQ\*jc3\*hps16\o\al(\s\up4(1),2)111111C平面11A111EQ\*jc3\*hps23\o\al(\s\up0(-→),n)1AEQ\*jc3\*hps23\o\al(\s\up2(-),C)EQ\*jc3\*hps23\o\al(\s\up2(-),B)EQ\*jc3\*hps23\o\al(\s\up2(-→),1)EQ\*jc3\*hps23\o\al(\s\up2(-),C)EQ\*jc3\*hps23\o\al(\s\up2(-→),B)EQ\*jc3\*hps23\o\al(\s\up2(-),C)EQ\*jc3\*hps23\o\al(\s\up2(-),C)EQ\*jc3\*hps23\o\al(\s\up2(→),1)1A(2)由(1)知,fx=(x-1)ex-x-1,所以f’x=xex-1,-1,+∞)上單調(diào)遞增,0,所以fx≥fx0,0x00所以fx僅有兩個(gè)零點(diǎn).·11分設(shè)x1是函數(shù)fx的零點(diǎn),則fx1=(x1-1)ex1-又f(-x1=-x1-1)e-x1+x1-1=e-x1I(x1-1)ex1-x1-1=0,EQ\*jc3\*hps16\o\al(\s\up4(2),0)1,2,3.1,3,所以2即(3)由題意知fp=HX0