人教版導(dǎo)與練總復(fù)習(xí)數(shù)學(xué)一輪教學(xué)課時(shí)作業(yè)：第二章第2節(jié)　函數(shù)的單調(diào)性與最值

1、第2節(jié)函數(shù)的單調(diào)性與最值知識(shí)點(diǎn)、方法基礎(chǔ)鞏固練綜合運(yùn)用練應(yīng)用創(chuàng)新練函數(shù)單調(diào)性的判定、求單調(diào)區(qū)間1,5,91316函數(shù)的最值2,3,7,812,1417,18函數(shù)單調(diào)性的應(yīng)用4,6,1011151.(2021江西萍鄉(xiāng)二模)下列函數(shù)中,在(0,+)上單調(diào)遞增的是(C)A.y=-x2+1 B.y=|x-1|C.y=x3 D.y=2-x解析:函數(shù)y=-x2+1在(0,+)上單調(diào)遞減,因此A不符合題意;由于函數(shù)y=|x-1|的圖象關(guān)于直線x=1對(duì)稱,在(1,+)上單調(diào)遞增,不符合題意;x(0,+)時(shí),函數(shù)y=x3的導(dǎo)數(shù)為y=3x20,因此函數(shù)在(0,+)上單調(diào)遞增,故C滿足題意;函數(shù)y=2-x=(12)

2、 x在區(qū)間(0,+)上單調(diào)遞減.故選C.2.函數(shù)y=2-x2+4x的值域是(C)A.-2,2B.1,2C.0,2D.-2,2解析:由0-x2+4x=-(x-2)2+42可知函數(shù)y=2-x2+4x的值域?yàn)?,2.故選C.3.函數(shù)y=2-xx+1,x(m,n的最小值為0,則m的取值范圍是(B)A.(1,2) B.(-1,2) C.1,2) D.-1,2)解析:函數(shù)f(x)=2-xx+1=3-(x+1)x+1=3x+1-1在區(qū)間(-1,+)上是減函數(shù),且f(2)=0,所以n=2.根據(jù)題意,x(m,n時(shí),ymin=0.所以m的取值范圍是(-1,2).故選B.4.已知函數(shù)f(x)=ex+x-1,若a(-

3、1,0),則f(a),f(2a),f2(a)的大小關(guān)系為(D)A.f(2a)f(a)f2(a)B.f(2a)f2(a)f(a)C.f2(a)f(2a)f(a)D.f2(a)f(a)f(2a)解析:顯然f(x)在R上是增函數(shù),且f(0)=0,當(dāng)a(-1,0)時(shí),2aa0,所以f(2a)f(a)0,從而f2(a)f(a)f(2a).故選D.5.(多選題)下列函數(shù)中,在(2,4)上是減函數(shù)的是(AC)A.y=(13) xB.y=log2(x2+3x)C.y=1x-2D.y=cos x解析:根據(jù)指數(shù)函數(shù)的性質(zhì)得y=(13) x在(2,4)上是減函數(shù),符合題意;根據(jù)復(fù)合函數(shù)的單調(diào)性可知y=log2(x2

4、+3x)在(2,4)上是增函數(shù),不符合題意;根據(jù)反比例函數(shù)的性質(zhì)及函數(shù)圖象的平移得y=1x-2在(2,4)上是減函數(shù),符合題意;根據(jù)余弦函數(shù)的性質(zhì)得,y=cos x在(2,4)上先減后增,不符合題意.故選AC.6.(2021陜西咸陽(yáng)高三一模)已知函數(shù)f(x)=22x+1-1,且f(4x-1)f(3),則實(shí)數(shù)x的取值范圍是(D)A.(2,+)B.(-,2)C.(1,+)D.(-,1)解析:由題意知函數(shù)f(x)=22x+1-1在R上單調(diào)遞減,由于f(4x-1)f(3),所以4x-13,解得x1)在區(qū)間(0,3)上單調(diào)遞減,則a的取值范圍為.解析:由對(duì)勾函數(shù)的性質(zhì)可知函數(shù)y=x+a-1x(a1)在(

5、0,a-1上單調(diào)遞減,在(a-1,+)上單調(diào)遞增,因?yàn)楹瘮?shù)y=x+a-1x(a1)在區(qū)間(0,3)上單調(diào)遞減,所以a-13,解得a10.答案:10,+)10.設(shè)函數(shù)f(x)=-x2+4x,x4,log2x,x4.若函數(shù)y=f(x)在區(qū)間(a,a+1)上單調(diào)遞增,則實(shí)數(shù)a的取值范圍是.解析:作出函數(shù)f(x)的圖象如圖所示,由圖象可知f(x)在(a,a+1)上單調(diào)遞增,需滿足a4或a+12,即a1或a4.答案:(-,14,+)11.已知圖象開口向上的二次函數(shù)f(x)對(duì)任意xR都滿足f(3-x)=f(x),若f(x)在區(qū)間(a,2a-1)上單調(diào)遞減,則實(shí)數(shù)a的取值范圍為(B)A.(-,54 B.(1

6、,54C.-32,+)D.(-,2)解析:由題意知函數(shù)圖象的對(duì)稱軸是直線x=32,且開口向上,若f(x)在區(qū)間(a,2a-1)上單調(diào)遞減,則只需322a-1,解得a54,而a1.所以實(shí)數(shù)a的取值范圍為(1,54.故選B.12.(多選題)若函數(shù)f(x)=1ax2-4ax+3的值域?yàn)?0,+),則實(shí)數(shù)a的取值可能是(CD)A.0 B.12 C.34 D.1解析:當(dāng)a=0時(shí),f(x)=33,不符合題意;當(dāng)a0時(shí),因?yàn)楹瘮?shù)f(x)=1ax2-4ax+3的值域?yàn)?0,+),所以a0,(-4a)2-4a30,解得a34.故選CD.13.(多選題)(2021山東威海高三期中)函數(shù)f(x)對(duì)任意x,yR總有f

7、(x+y)=f(x)+f(y),當(dāng)x0時(shí),f(x)0,f(1)=13,則下列命題中正確的是(BCD)A.f(x)是R上的減函數(shù)B.f(x)在-6,6上的最小值為-2C.f(-x)=-f(x)D.若f(x)+f(x-3)-1,則實(shí)數(shù)x的取值范圍為0,+)解析:取x=0,y=0,則f(0)=f(0)+f(0),解得f(0)=0.令y=-x,則f(0)=f(x)+f(-x),即-f(x)=f(-x),C正確;令x1,x2R,且x1x2,則x1-x20,因?yàn)楫?dāng)x0時(shí),f(x)0,所以f(x1-x2)0,則f(x1)-f(x2)=f(x1)+f(-x2)=f(x1-x2)0,即f(x1)f(x2),所以

8、函數(shù)f(x)是R上的增函數(shù),A錯(cuò)誤;因?yàn)楹瘮?shù)f(x)是R上的增函數(shù),所以函數(shù)f(x)在-6,6上的最小值為f(-6),f(-6)=f(-3)+f(-3)=2f(-3),f(-3)=-f(3),f(3)=f(2)+f(1)=f(1)+f(1)+f(1)=313=1,故f(-6)=-2,所以f(x)在-6,6上的最小值為-2,B正確;f(x)+f(x-3)-1,即f(2x-3)f(-3),因?yàn)楹瘮?shù)f(x)是R上的增函數(shù),所以2x-3-3,解得x0,所以實(shí)數(shù)x的取值范圍為0,+),D正確.故選BCD.14.已知函數(shù)f(x)=|x2-4x|,x2,5,則f(x)的最小值是,最大值是.解析:因?yàn)楹瘮?shù)f(

9、x)=|x2-4x|=-(x2-4x),2x4,x2-4x,4x5,對(duì)應(yīng)圖象如圖所示,故f(x)的最小值為f(4)=0,最大值為f(5)=5.答案:0515.已知函數(shù)f(x)=1+lnxx,則(C)A.f(12)f(1)f(32)B.f(32)f(1)f(12)C.f(12)f(32)f(1)D.f(32)f(12)0-ln x00 x1,f(x)0-ln x1,即函數(shù)f(x)在(0,1)上單調(diào)遞增,在(1,+)上單調(diào)遞減,所以f(32)f(1).又因?yàn)閒(12)=2(1+ln 12)=2(1-ln 2),f(32)=23(1+ln 32)=23(1+ln 3-ln 2),因?yàn)閒(12)-f(

10、32)=2-2ln 2-23-23ln 3+23ln 2=23(2-2ln 2-ln 3)=23(2-ln 22-ln 3)=23(2-ln 12)0,所以f(12)f(32),即得f(12)f(32)0,所以f(x)=1-(12)x為R上的增函數(shù),且f(x)=1-(12)x0,y0,若(x+1x)(y+1y)(x+y2+2x+y)2,則(x+y)2的最大值是.解析:令xy=t,則0t(x+y)24,令f(t)=t+1+(x+y)2t,因?yàn)?x+1x)(y+1y)(x+y2+2x+y)2xy+1+(x+y)2xy-2(x+y2+2x+y)2,等價(jià)于f(t)f(x+y)24),所以題意可轉(zhuǎn)化為函數(shù)f(t)=t+1+(x+y)2t在(0