数学-2025届安徽省合肥二模试题+答案

注意事项:44 235264522高一年级学生得分A.fx与gx的图象关于直线对称fx>gx)11.已知函数fx的定义域为R,且fxf1-y+fyf1-x)=2fx+y,f0)=0fx0<0____.棋子,则恰好每列都有两枚棋子的概率为____.1243.11A1111A1刻画随机变量X蕴含的信息量的大小.2)题.充分条件.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】本题考查向量的坐标运算及向量夹角的计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与gx的图象关于直线对称A.设fx上一点为x1,fx1)),B.设fx上一点为x2,fx2)),:fx>gx,C正确;:fx与gx的图象恰有四个交点,D正确.11.已知函数fx的定义域R,fxf1-y+fyf1-x)=2fx+y,f0)=1,则0fx0<0令x=y=0,f0f1+f0f1=2f1=2f0,则f1=f0=1令x=y=1,f1f0+f1f0=2f2,则f2=f1=1,A正确;令y=0,则fxf1+f0f1—x)=2fx,fx+f(1—x)=2fx,所以f(1—x)=fx令y=1,则fxf0+f1f1—x)=2fx+1),所以f(1所以f2x=f2x,fx=f2()≥0,B错误;【分析】本题考查分段函数、函数的单调性2.2；EQ\*jc3\*hps23\o\al(\s\up7(≤),2)EQ\*jc3\*hps23\o\al(\s\up7(≤),a)2f(1)=42a;行都放置两枚棋子,则恰好每列都有两枚棋子的概率为____.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,+∞)上单调递增,0,所以fx≥fx0,0x00所以fx仅有两个零点.·11分设x1是函数fx的零点,则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)由题意知fp=HX0