高考專題復(fù)習(xí)1-第1練　函數(shù)的性質(zhì)新課標(biāo)試卷

1、熱點(diǎn)題型強(qiáng)化速練第1練函數(shù)的性質(zhì)一、選擇題 1.下列函數(shù)中,是偶函數(shù),且在(-,0)上單調(diào)遞增的是()A.f(x)=x2B.f(x)=2|x|C.f(x)=log21|x+1|D.f(x)=-2|x|答案Df(x)=x2是偶函數(shù),在(-,0)上單調(diào)遞減; f(x)=2|x|的定義域?yàn)镽, f(-x)=2|-x|=2|x|=f(x),所以f(x)是偶函數(shù),在(-,0)上單調(diào)遞減; f(x)=log21|x+1|的定義域?yàn)閤R|x-1,不關(guān)于原點(diǎn)對(duì)稱,所以f(x)不是偶函數(shù); f(x)=-2|x|的定義域?yàn)镽, f(-x)=-2|-x|=-2|x|=f(x),所以f(x)是偶函數(shù),在(-,0)上單

2、調(diào)遞增.故選D.2.已知函數(shù)f(x)=ax-1-b,x1,-log2(x+1),x>1(a>0,a1)在定義域上單調(diào),則ab的值不可能是()A.-1B.1C.-2D.2答案D函數(shù)f(x)在定義域上單調(diào),且f(x)在(1,+)上單調(diào)遞減,0<a<1且-log2(1+1)a1-1-b,即0<a<1且b2,ab<2.3.已知f(x)為奇函數(shù)且滿足f(x+1)=-f(x),當(dāng)0x1時(shí), f(x)=2x(1-x),則f-52=()A.-12B.14C.14D.12答案A根據(jù)題意可得f-52=f52=-f32=f12,當(dāng)0x1時(shí), f(x)=2x(1-x),所以f

3、-52=f12=2×12×12=12.4.設(shè)定義在R上的奇函數(shù)f(x)滿足對(duì)任意x1,x2(0,+),x1x2,都有f(x2)-f(x1)x2-x1<0,且f(2)=0,則不等式3f(-x)-2f(x)x0的解集為()A.(-,-2(0,2 B.-2,02,+)C.(-,-22,+)D.-2,0)(0,2答案C因?yàn)閷?duì)任意x1,x2(0,+),且x1x2都有f(x2)-f(x1)x2-x1<0,所以函數(shù)f(x)在(0,+)上單調(diào)遞減,又因?yàn)閒(x)為R上的奇函數(shù),所以f(x)在(-,0上單調(diào)遞減.由f(2)=0,得f(-2)=0,因?yàn)?f(-x)-2f(x)x=-

4、3f(x)-2f(x)x=-5f(x)x0,所以當(dāng)x>0時(shí), f(x)0,即x2,當(dāng)x<0時(shí), f(x)0,即x-2,綜上,不等式的解集為(-,-22,+),故選C.5.(多選題)若函數(shù)f(x)=ex-e2-x,則下列敘述正確的是()A.f(x)在(-,+)上單調(diào)遞增B.f(x)的值域?yàn)?0,+)C.y=f(x)的圖象關(guān)于直線x=1對(duì)稱D.y=f(x)的圖象關(guān)于點(diǎn)(1,0)對(duì)稱答案ADy=ex在(-,+)上單調(diào)遞增,y=-e2-x在(-,+)上單調(diào)遞增,f(x)在(-,+)上單調(diào)遞增,故A中敘述正確; f(0)=1-e2<0,故B中敘述錯(cuò)誤; f(0)=1-e2, f(2)=

5、e2-1, f(0)f(2),所以y=f(x)的圖象不關(guān)于直線x=1對(duì)稱,故C中敘述錯(cuò)誤;f(1+x)=e1+x-e1-x, f(1-x)=e1-x-e1+x,f(1+x)+f(1-x)=0,則y=f(x)的圖象關(guān)于點(diǎn)(1,0)對(duì)稱,故D中敘述正確.6.設(shè)函數(shù)f(x)=cos2-x-(x+2)2x2+4的最大值為M,最小值為m,則(M+m+1)2 020的值是()A.0B.1C.22 019D.22 020答案Bf(x)=cos2-x-(x+2)2x2+4=sinx-4xx2+4-1.設(shè)g(x)=sinx-4xx2+4,因?yàn)間(-x)=sin(-x)-4(-x)(-x)2+4=4x-sinxx

6、2+4=-g(x),所以g(x)為奇函數(shù),所以g(x)max+g(x)min=0,則M+m=-2,所以(M+m+1)2 020=1.7.已知函數(shù)f(x)=x,0x1,ln2x,1<x2,若存在實(shí)數(shù)x1,x2,滿足0x1<x22,且f(x1)=f(x2),則x2-x1的最大值為()A.e2B.e2-1C.1-ln 2D.2-ln 4答案B函數(shù)f(x)=x,0x1,ln2x,1<x2的圖象如下:存在實(shí)數(shù)x1,x2滿足0x1<x22,且f(x1)=f(x2),即x1=ln 2x2,由圖知,x21,e2,則x2-x1=x2-ln 2x2.令g(x)=x-ln 2x,x1,e2,

7、則g(x)=x-1x,在1,e2上g(x)>0,g(x)在1,e2上單調(diào)遞增,故g(x)max=ge2=e2-1.8.(多選題)已知函數(shù)f(x)的定義域?yàn)镽,且對(duì)任意xR,都有f(x)=f(-x)及f(x+4)=f(x)+f(2)成立,當(dāng)x1,x20,2,且x1x2時(shí),有f(x1)-f(x2)(x1-x2)>0成立,下列四個(gè)結(jié)論中正確的是()A.f(2)=0B.函數(shù)f(x)在區(qū)間-6,-4上為增函數(shù)C.直線x=-1是函數(shù)f(x)圖象的一條對(duì)稱軸D.方程f(x)=0在區(qū)間-6,6上有4個(gè)不同的實(shí)數(shù)根答案AD由題意知函數(shù)f(x)為偶函數(shù),且在區(qū)間0,2上單調(diào)遞增,由f(x+4)=f(x

8、)+f(2),令x=-2,得f(2)=f(-2)+f(2),解得f(-2)=f(2)=0,所以A中結(jié)論正確,且f(x+4)=f(x),即函數(shù)f(x)是周期為4的周期函數(shù),則函數(shù)f(x)的圖象如圖所示,由圖象可知函數(shù)f(x)在區(qū)間-6,-4上為減函數(shù),所以B中結(jié)論不正確;直線x=-1不是函數(shù)f(x)圖象的一條對(duì)稱軸,所以C中結(jié)論不正確;方程f(x)=0在區(qū)間-6,6上有-6,-2,2,6,共4個(gè)不同的實(shí)數(shù)根,所以D中結(jié)論正確.9.已知函數(shù)f(x)是R上的偶函數(shù),且f(x)的圖象關(guān)于點(diǎn)(1,0)對(duì)稱,當(dāng)x0,1時(shí), f(x)=2-2x,則f(0)+f(1)+f(2)+f(2 020)的值為()A.

9、-2B.-1C.0D.1答案D因?yàn)閒(x)是R上的偶函數(shù),所以f(-x)=f(x),又f(x)的圖象關(guān)于點(diǎn)(1,0)對(duì)稱,則f(x)=-f(2-x),所以f(-x)=-f(2-x),則f(x)=-f(2+x),得f(4+x)=-f(2+x)=f(x),所以f(x)是周期函數(shù),且周期T=4,由x0,1時(shí), f(x)=2-2x,得f(0)=1, f(1)=0, f(2)=-f(0)=-1, f(3)=f(-3)=f(1)=0,則f(0)+f(1)+f(2)+f(3)=0,則f(0)+f(1)+f(2)+f(2 020)=f(0)+505×0=f(0)=1.二、填空題10. 若函數(shù)f(x)

10、=x+ax2+bx+1在-1,1上是奇函數(shù),則f(x)的解析式為. 答案f(x)=xx2+1解析f(x)在-1,1上是奇函數(shù),f(0)=0,a=0,f(x)=xx2+bx+1.又f(-1)=-f(1),-12-b=12+b,得b=0,f(x)=xx2+1.11.已知函數(shù)f(x)=(2-a)x+1(x<1),ax(x1)滿足對(duì)任意x1,x2R,且x1x2,都有f(x1)-f(x2)x1-x2>0成立,那么a的取值范圍是. 答案32,2解析由題意知,函數(shù)y=f(x)在定義域上為增函數(shù),所以2-a>0,a>1,a2-a+1,解得32a<2.12.有下

11、列命題:函數(shù)y=f(-x+2)與y=f(x-2)的圖象關(guān)于y軸對(duì)稱;若函數(shù)f(x)=ex,則x1,x2R,都有f x1+x22f(x1)+f(x2)2;若函數(shù)f(x)=loga|x|(a>0,a1)在(0,+)上單調(diào)遞增,則f(-2)>f(a+1);若函數(shù)f(x+2 010)=x2-2x-1(xR),則函數(shù)f(x)的最小值為-2.其中真命題的序號(hào)是. 答案解析函數(shù)y=f(-x)與y=f(x)的圖象關(guān)于y軸對(duì)稱,將函數(shù)y=f(-x)與y=f(x)的圖象都向右平移2個(gè)單位長(zhǎng)度,便得函數(shù)y=f(-x+2)與y=f(x-2)的圖象,所以函數(shù)y=f(-x+2)與y=f(x-2)的圖象關(guān)于直線x=2對(duì)稱,故錯(cuò)誤;若函數(shù)f(x)=ex,則x1,x2R,都有fx1+x22=ex1+x22=ex12·ex22ex1+ex22=f(x1