8540x2x00x10x1x2x0的必要不充分條件y2xy12x定義域相同解析式相等2-【解析】由已知fxax2bxc的圖象關(guān)于直線x3.5對稱f(1f(6fx(3.5上是減函數(shù)456f(4f(5f(6f(1x1時,lnx11xex1時,ex1x0.fx1的解集為(-∞,01,e【解析】由已知a+b+1=k+1,所以(k+1(4+9 =(a+b+1(4+9 =13 4(b+19a1312253a2(b1時等號成立49 25.251k24 3618求.60f(mf(n(mnm0n2m112n2m2n2m+n2m2n≤2m

1mn2m+n1mnmnlog27m0nn=-2,C正確3515 = -2= 【解析】由已知得：a<0,故c=6a<0,b=-5a>0,所以 +ax+c<0化簡得：5x2-x-6<0,故解集為(x-1<x< 1頁（共3頁23a=-a2,log3b=-b2a,by=-x2y3xlog3x的圖象交點橫坐標(biāo)y3xylog3xyx對稱ab【解析】(1)由x2-9x=0,解得x1=0,x2= 2由y=x2-9x的圖象,可知解集為x0<x 4 6由(1)知U= 8?UA2,4,7,8 10(?U

∩B=

13與之相鄰的鄰邊長為 2則菜園的面積為S=xy,x+2y= 5S

?x?2y≤

×x+

10當(dāng)且僅當(dāng)x=2y= 12即x=10m,y=5m時,菜園面積最大,最大面積為 15說明：用二次函數(shù)解法該處相應(yīng)的分數(shù)【解析】(1)令t=ax,因為x∈-1,1,a>1,所以t∈1 2 4yt22tt1

上為增函數(shù)故當(dāng)t=a時,y=t2+2t有最大值 6a3a=-因為a>1,故a= 8由(1y32x2令m=3x,故m2+2m-8≤ 10m22m80m1=-4,m2所以m2+2m-8≤0解為-4≤m≤ 12故-43x又3x>0,只需3x≤2,解得x≤ 14故實數(shù)x的集合為xx 15(1f

RxR,f(-

=f(x即log2(2-x

+kx=log2(2x+

42kxlog2(2x1log2(2-x1又log2(2x

-log2(2-x+

=

2x+22-x+

82頁（共3頁2kxx即(2k-1x0xR恒成立所以k= 10(2由(1fxlog2(2x11xlog2x2-x x由f(x=m得2+2-= 12t2xt0y2x2-xt yt1在

單調(diào)遞減在

上單調(diào)遞增t1時,yt

有最小值 14f

m=2m有兩個不等的實數(shù)解,故2m> 16m實數(shù)m的取值范圍為 17令x=-1,y=-1,則f(-1= 2令y=-1,得f(-x=f(x+f(-1+1=f(x 4x(-∞,0(0,+∞故y=f(x(x≠0為偶函 5x1,x2(0,+∞,x1

1f

<- 7f

=fx?

=f(

+f

+1<f(x1fx1f故y=f(x(x≠0在(0,+∞上為減函 10②由(1f(1=-1fx2x-11fx2x-1<-1fx2x-1f(1fx是偶函數(shù)且在(0,+∞故x2x-1所以x2-x-1>1或x2-x-1<- 12x2