高考一輪復(fù)習(xí)專項(xiàng)練習(xí)數(shù)學(xué)課時(shí)規(guī)范練6函數(shù)的單調(diào)性與最值

課時(shí)規(guī)范練6函數(shù)的單調(diào)性與最值基礎(chǔ)鞏固組1.已知函數(shù)f(x)=k(x+2),x≤0,2x+k,x>0,則A.充分不必要條件B.必要不充分條件C.充要條件D.既不充分也不必要條件2.已知函數(shù)f(x)=ax,x>1,(4-a2)A.(1,+∞) B.[4,8)C.(4,8) D.(1,8)3.已知函數(shù)f(x)=x2-2x-A.(∞,1] B.[3,+∞)C.(∞,1] D.[1,+∞)4.若2x+5y≤2y+5x,則有()A.x+y≥0 B.x+y≤0C.xy≤0 D.xy≥05.函數(shù)f(x)在(∞,+∞)上單調(diào)遞減,且為奇函數(shù),若f(1)=1,則滿足1≤f(x2)≤1的x的取值范圍是()A.[2,2] B.[1,1]C.[0,4] D.[1,3]6.(2020全國2,理11,文12)若2x2y<3x3y,則()A.ln(yx+1)>0 B.ln(yx+1)<0C.ln|xy|>0 D.ln|xy|<07.函數(shù)f(x)=12

-x2+2mx-mA.2 B.2 C.1 D.18.(多選)(2020山東滕州一中月考,6)下列四個(gè)說法,其中不正確的是()A.函數(shù)f(x)在(0,+∞)上單調(diào)遞增,在(∞,0]上單調(diào)遞增,則f(x)在R上是增函數(shù)B.若函數(shù)f(x)=ax2+bx+2與x軸沒有交點(diǎn),則b28a<0且a>0C.當(dāng)a>b>c時(shí),則有ab>ac成立D.y=|1+x|和y=(1+9.(多選)已知函數(shù)f(x)=x2x,g(x)=acosπx2+52a(a>0).給出下列四個(gè)命題,其中是真命題的為A.若?x∈[1,2],使得f(x)<a成立,則a>1B.若?x∈R,使得g(x)>0恒成立,則0<a<5C.若?x1∈[1,2],?x2∈R,使得f(x1)>g(x2)恒成立,則a>6D.若?x1∈[1,2],?x2∈[0,1],使得f(x1)=g(x2)成立,則3≤a≤410.設(shè)函數(shù)f(x)=1,x>0,0,x=0,-1,x<0,g(x11.函數(shù)f(x)=2xx+1在區(qū)間[1,2]12.已知函數(shù)f(x)=x+2x-3,x≥1,lg(x2+1),x綜合提升組13.(多選)(2020山東淄博4月模擬,12)函數(shù)f(x)在[a,b]上有定義,若對(duì)任意x1,x2∈[a,b],有fx1+x22≤12[f(x1)+f(x2)],則稱f(x)在[a,b]上具有性質(zhì)P.設(shè)f(x)在[1,3]上具有性質(zhì)P,A.f(x)在[1,3]上的圖像是連續(xù)不斷的B.f(x2)在[1,3]上具有性質(zhì)PC.若f(x)在x=2處取得最大值1,則f(x)=1,x∈[1,3]D.對(duì)任意x1,x2,x3,x4∈[1,3],有fx1+x2+x3+x44≤12[f(x1)+f(x2)14.(2020山東聊城二模,14)已知f(x)=1-lnx,0<x≤1,-1+lnx,x>1創(chuàng)新應(yīng)用組15.如果函數(shù)y=f(x)在區(qū)間I上單調(diào)遞增,且函數(shù)y=f(x)x在區(qū)間I上單調(diào)遞減,那么稱函數(shù)y=f(x)是區(qū)間I上的“緩增函數(shù)”,區(qū)間I叫做“緩增區(qū)間”.若函數(shù)f(x)=12x2x+32是區(qū)間I上的“緩增函數(shù)”,則“緩增區(qū)間A.[1,+∞) B.[0,3]C.[0,1] D.[1,3]16.(2020山東棗莊二模,8)已知P(m,n)是函數(shù)y=-x2-2x圖像上的動(dòng)點(diǎn),則|4m+3n21|A.25 B.21 C.20 D.4參考答案課時(shí)規(guī)范練6函數(shù)的單調(diào)性與最值1.D若f(x)單調(diào)遞增,則k>0且k(0+2)≤20+k,解得0<k≤1,因?yàn)椤発<1”與“0<k≤1”沒有包含的關(guān)系,所以充分性和必要性都不成立.2.B由f(x)在R上單調(diào)遞增,則有a>1,4-a3.B設(shè)t=x22x3,由t≥0,即x22x3≥0,解得x≤1或x≥3.所以函數(shù)的定義域?yàn)?∞,1]∪[3,+∞).因?yàn)楹瘮?shù)t=x22x3的圖像的對(duì)稱軸為x=1,所以函數(shù)t在(∞,1]上單調(diào)遞減,在[3,+∞)上單調(diào)遞增.所以f(x)的單調(diào)遞增區(qū)間為[3,+∞).4.B設(shè)函數(shù)f(x)=2x5x,易知f(x)為增函數(shù),又f(y)=2y5y,由已知得f(x)≤f(y),∴x≤y,∴x+y≤0.5.D由題意f(1)=f(1)=1,1≤f(x2)≤1等價(jià)于f(1)≤f(x2)≤f(1).又f(x)在(∞,+∞)上單調(diào)遞減,所以1≤x2≤1,即1≤x≤3.所以x的取值范圍是[1,3].6.A∵2x2y<3x3y,∴2x3x<2y3y.∵f(t)=2t3t在R上為增函數(shù),且f(x)<f(y),∴x<y,∴yx>0,∴yx+1>1,∴l(xiāng)n(yx+1)>ln1=0.故選A.7.B∵x2+2mxm21=(xm)21≤1,∴12

∴f(x)的值域?yàn)閇2,+∞),∵y=12x是減函數(shù),y=(xm)21的單調(diào)遞減區(qū)間為[m,+∞),∴f(x)的單調(diào)遞增區(qū)間為[m,+∞).由條件知m=2.8.ABCf(x)=x,x≤0,lnx,x>0,滿足在(0,+∞)上單調(diào)遞增,在(∞,0]上單調(diào)遞增,但f(x)在R上不是增函數(shù),故A錯(cuò)誤;當(dāng)a=b=0時(shí),f(x)=2,它的圖像與x軸無交點(diǎn),不滿足b28a<0且a>0,故B錯(cuò)誤;當(dāng)a>b>c,但a=0時(shí),ab=ac,不等式ab>ac不成立,故C錯(cuò)誤;y=(1+x)2=|x+1|與y=|x+19.ACD對(duì)于A,由f(x)在[1,2]上單調(diào)遞增,則f(x)min=f(1)=1,所以a>1,故A正確;對(duì)于B,只需g(x)min>0,由g(x)min=a+52a=53a>0,得0<a<53,故B錯(cuò)誤;對(duì)于C,只需在給定的范圍內(nèi)f(x)min>g(x)max,即1>5a,解得a>6,故C正確;對(duì)于D,只需g(x)min≤f(x)min,g(x)max≥f(x)max,f(x)max=f(2)=222=1,所以x1∈[1,2],f(x1)∈[1,1],當(dāng)x∈[0,1]時(shí),πx2∈0,π2,所以g(x)在[0,1]上單調(diào)遞減,g(x)min=g(1)=52a,g(x)max=g(0)=5a,所以g(x)∈[52a,5a],由題意,可得5-2a≤-1,10.[0,1)∵g(x)=x2,x>1,0,x=1,-11.1,43∵f(x)=2xx+1=2(x+1)-2x+1=22x+1,∴f(x)在區(qū)間[1,2]上單調(diào)遞增,即f(x)max=f(2)=412.0223因?yàn)閒(3)=lg[(3)2+1]=lg10=1,所以f[f(3)]=f(1)=1+23=0.當(dāng)x≥1時(shí),x+2x3≥2x·2x3=223,當(dāng)且僅當(dāng)x=2x,即x=2時(shí),等號(hào)成立,此時(shí)f(x)min=當(dāng)x<1時(shí),lg(x2+1)≥lg(02+1)=0,此時(shí)f(x)min=0.所以f(x)的最小值為223.13.ABD對(duì)于A,函數(shù)f(x)=x2,1≤x<3,11,x=3在[1,3]上具有性質(zhì)P,但f(x)在[1,3]上的圖像不連續(xù),故A錯(cuò)誤;對(duì)于B,f(x)=x在[1,3]上具有性質(zhì)P,但f(x2)=x2在[1,3]上不滿足性質(zhì)P,故B錯(cuò)誤;對(duì)于C,因?yàn)閒(x)在x=2處取得最大值1,所以f(x)≤1,由性質(zhì)P可得1=f(2)≤12[f(x)+f(4x)],即f(x)+f(4x)≥2,因?yàn)閒(x)≤1,f(4x)≤1,所以f(x)=1,x∈[1,3],故C正確;對(duì)于D,fx1+x2+x3+x44=fx1+x22+x3+x422≤12fx1+x14.2e因?yàn)閒(x)=1-lnx,0<x≤1,由f(a)=f(b),得1lna=1+lnb,0<a≤1,b>1,所以lnab=2,即ab=e2.設(shè)y=1a+1b=be2+1b,令y'=1e2-1b2=b2-e2(eb)15.D因?yàn)楹瘮?shù)f(x)=12x2x+32的對(duì)稱軸為x=1,所以函數(shù)y=f(x)在區(qū)間[1,+∞)上單調(diào)遞增.又因?yàn)楫?dāng)x≥1時(shí),f(x)x=12x1+32x,令g(x)=1