安徽省皖江名校联盟2025-2026学年高一上学期11月期中考试数学试卷

数学参考答案8540 ,P∩Q={0}.故选 x0时x0f(-xx(-x1x(x1),即-f(xx(-x1x(x1因f(xx(x1)B.的充要条件C.a0,b0时a+

2a0,b0a0,b0时a+

0a0,b0时a+

=-2,C对.(x,y)y

是点集,yy

是数集,D错.axb时yxbax)2知,y0A,D.xa时yxba知,y0

解得(a=-

f(x2x1.定义域是(-

∪(0,+∞因为f(-x)=2x+1=-(-2x+ =-f(x),所以f(x)是奇函数.①错.因为-2x,1均在(-- 和(0内单减f(x在(-∞,0内单减,f(x没有最小值.②对③错.f(x2x1得f-1

12=-1.④对f(x(ax-1)xax2xa0时f(x-xx在(0内单调递增满足题意当a>0时,f(x)在(0,1,(1+∞内单调递增,在(1, 内单调递减,不合题意.当a<0,x 2a(0时f(xax2+xf(x在(0内单调递增,满足题意.综上知a(-∞,03618求.60CRBxx5且x2,A错t4时,A4,B2,5,AB2,4,5,B对t时,t25ABØ,C错D对-30f(xx-3在(0内单减,A对f(xx

在(06,B错.f(xf(2xx32x)322x)3x324f(x的图象关于点2对称,C对.对aR,f(-xx)4x)-4ax4x-4a=f(x),f(x均为偶函数,D对故ACD.xy0f(0)f(01=f2(0f2(0),f2(01,f(01.A对.xyf(2x)f12f2(x所以-f(2x12f2(x).B错.xy,yxf(yx)f(yx1=f2(yff(yx)f(yx1=f(xy)f(xy1,f(yx)f(yxf(xy)f(xy).f(y0f(x0f(01不符f(yx0.f(yxf(xyf(-xf(x),f为偶函数35151x⋅ .1x⋅ 300

=1x+1+

≥

31x300x300时取等号.300台机器人 最低

【解析】因为fx-1)R上的奇函数,所以fx)的图象关于(-1,0)对称.f-3

=-f-

=-1【解析】(1)m2-m-1=1得,m2-m-2=0,解得m=2或m=- 3m2时,f(xx2在区间(0内是增函数舍去当m=-1时,f(x)=x-1,在区间(0,+∞)内是减函数,符合.故m=- 7(2)f(2x-

=1就是 =1,2x-等价于2x-1=1,2x-1 10解得x=1或x=0,故方程f(2x-1=1的解集是0,1 13(11-

0

x1A=1

2因为g(x)= ,1<x<m,所以值域B= ,1 5(x-1)2+

(m2-2m+ 7于是 ≤1,即m2-2m≥0,解得m≤0或m≥2 14m2-2m+ 又m>1,所以m≥2.故实数m的最小值是 15【解析】(1)设x2=t,则f(x)=g(t)=t2-2t+3=(t-1)2+ 4因为x∈-1,1,所以t∈0,1.当t=1时,f(x)min=2；当t=0时,f(x)max= 8由(1知,f(x0t22tb0x2方程t2-2⋅t+b=0有两个不相等的正实数根即可,则b>0,且(-2)2-4b> 13解得0<b<1.故b的取值范围是(0， 15(1f(xx

的图象如图所示 .................................................................................5x1,x21x1f(x1f(x2x11x21x1x211 1x1x2x1x20,x1x2111则f(x1)-f(x2)<0,即f(x1)<f(x2),所以f(x)在(1,+∞)上单调递增 10f(x在(1上单调递增f(x在[2,5上单调递增所以f(x)max=f(5)=5+1=26,f(x)min=f(2)=2+1= 17 x<- -1≤x<3 2 xf(x)min4.m4f(xm的解集是空集故实数m的取值范围是(-∞,4 4b2⋅(ⅰb2⋅1

=1+

(a2+b2)=

+

+

≥

=9

(

129a3,b6时取等号 因此1+2的最小值是9 8 于是x1x-39所以x<-1时得-

≤x<-当-1x3时49x3时3x13

12 (ⅱ)(1+1(a5+b