高考數(shù)學(xué)大一輪復(fù)習(xí) 課時(shí)作業(yè)5 函數(shù)的單調(diào)性與最值 理 新人教A版-新人教A版高三數(shù)學(xué)試題

課時(shí)作業(yè)(五)第5講函數(shù)的單調(diào)性與最值時(shí)間/45分鐘分值/100分基礎(chǔ)熱身1.下列函數(shù)中,在區(qū)間(0,+∞)上為增函數(shù)的是 ()A.y=x+1B.y=sinxC.y=2-x D.y=log12(2.已知函數(shù)f(x)=ax2+2(a-3)x+3在區(qū)間(-∞,3)上是減函數(shù),則a的取值范圍是 ()A.0,B.0C.0,D.03.函數(shù)y=2xx-A.在區(qū)間(1,+∞)上單調(diào)遞增B.在區(qū)間(1,+∞)上單調(diào)遞減C.在區(qū)間(-∞,1)上單調(diào)遞增D.在定義域內(nèi)單調(diào)遞減4.[2018·貴州凱里一中月考]已知函數(shù)f(x)=2-x+1,則滿足f(log4a)>3的實(shí)數(shù)a的取值范圍是A.13,1C.14,15.若函數(shù)y=|2x+c|是區(qū)間(-∞,1)上的單調(diào)函數(shù),則實(shí)數(shù)c的取值范圍是.

能力提升6.[2018·晉城二模]若f(x)=x-2+x2-2x+4的最小值與g(x)=x+a-A.1 B.2C.2 D.227.函數(shù)f(x)滿足f(x+2)=3f(x),且x∈R,若當(dāng)x∈[0,2]時(shí),f(x)=x2-2x+2,則當(dāng)x∈[-4,-2]時(shí),f(x)的最小值為 ()A.19 B.C.-13 D.-8.能推斷出函數(shù)y=f(x)在R上為增函數(shù)的是 ()A.若m,n∈R且m<n,則f(3m)<f(3n)B.若m,n∈R且m<n,則f12mC.若m,n∈R且m<n,則f(m2)<f(n2)D.若m,n∈R且m<n,則f(m3)<f(n3)9.[2018·濰坊一中月考]已知函數(shù)f(x)=(a-3)x+5,x≤1,2ax,x>1,若對(duì)R上的任意x1,x2(x1≠x2),恒有(x1-xA.(0,3) B.(0,3]C.(0,2) D.(0,2]10.已知函數(shù)f(x)=e-|x|,設(shè)a=f(e-0.3),b=f(ln0.3),c=f(log310),則 ()A.a>b>c B.b>a>cC.c>a>b D.c>b>a11.若函數(shù)f(x)=132x2+mx-312.已知函數(shù)f(x)=(x-1)2,x≥0,2x13.函數(shù)f(x)=x2,x≥t,x,0<x14.(12分)已知函數(shù)f(x)=log4(ax2+2x+3).(1)若f(1)=1,求f(x)的單調(diào)區(qū)間.(2)是否存在實(shí)數(shù)a,使得f(x)的最小值為0?若存在,求出a的值;若不存在,請說明理由.15.(13分)已知定義域?yàn)镽的函數(shù)f(x)滿足:f-12=2,對(duì)于任意實(shí)數(shù)x,y恒有f(x+y)=f(x)f(y),且當(dāng)x>0時(shí),0<f(x)<(1)求f(0)的值,并證明:當(dāng)x<0時(shí),f(x)>1;(2)判斷函數(shù)f(x)在R上的單調(diào)性并加以證明;(3)若不等式f[(a2-a-2)x2-(2a-1)2x+2]>4對(duì)任意x∈[1,3]恒成立,求實(shí)數(shù)a的取值范圍.難點(diǎn)突破16.(5分)[2018·永州三模]已知函數(shù)f(x)=a+log2(x2+a)(a>0)的最小值為8,則 ()A.a∈(5,6) B.a∈(7,8)C.a∈(8,9) D.a∈(9,10)17.(5分)函數(shù)f(x)的定義域?yàn)镈,若滿足:①f(x)在D內(nèi)是單調(diào)函數(shù);②存在[a,b]?D,使得f(x)在[a,b]上的值域?yàn)閍2,b2.則稱函數(shù)f(x)為“成功函數(shù)”.若函數(shù)f(x)=logm(mx+2t)(其中m>0,且m≠1)是“成功函數(shù)”,則實(shí)數(shù)tA.(0,+∞) B.-C.18,1課時(shí)作業(yè)(五)1.A[解析]y=x+1在區(qū)間(0,+∞)上為增函數(shù);y=sinx在區(qū)間(0,+∞)上不單調(diào);y=2-x在區(qū)間(0,+∞)上為減函數(shù);y=log12(x+1)在區(qū)間(0,+∞)上為減函數(shù).2.D[解析]當(dāng)a=0時(shí),f(x)=-6x+3,在(-∞,3)上是減函數(shù),符合題意;若函數(shù)f(x)是二次函數(shù),由題意有a>0,對(duì)稱軸為直線x=-a-3a,則-a-3a≥3,又a>0,所以0<a≤34.所以0≤3.B[解析]y=2xx-1=2(x-1)+2x-1=4.B[解析]由題意求得函數(shù)f(x)的定義域?yàn)镽,且在R上為減函數(shù),又f(log4a)>3,f(-1)=21+1=3,則由f(log4a)>f(-1),得log4a<-1,解得0<a<14,5.c≤-2[解析]函數(shù)y=2x+c=2x+c,x≥-c2,-2x-c,x<-6.C[解析]f(x)在定義域[2,+∞)上是增函數(shù),所以f(x)的最小值為f(2)=2.又g(x)=2ax+a+x-a所以g(x)的最大值為g(a)=2a,所以2a=2,即a=2.故選7.A[解析]因?yàn)閒(x+2)=3f(x),所以f(x)=13f(x+2)=19f(x+因?yàn)楫?dāng)x∈[0,2]時(shí),f(x)=x2-2x+2,所以當(dāng)x∈[-4,-2],即x+4∈[0,2]時(shí),f(x)=19f(x+4)=19(x+3)2+19,故當(dāng)x=-3時(shí),f(x)取得最小值198.D[解析]若m,n∈R且m<n,則0<3m<3n,不能得到函數(shù)y=f(x)在R上為增函數(shù),故A錯(cuò)誤;若m,n∈R且m<n,則12m>12n>0,不能得到函數(shù)y=f(x)在R上為增函數(shù),若m,n∈R且m<n,則0<m<n時(shí),0<m2<n2,m<n<0時(shí),m2>n2>0,m<0<n時(shí),m2與n2的大小關(guān)系不確定,所以不能得到函數(shù)y=f(x)在R上為增函數(shù),故C錯(cuò)誤;若m,n∈R且m<n,則m3∈R,n3∈R,且m3<n3,又f(m3)<f(n3),所以函數(shù)y=f(x)在R上為增函數(shù),故D正確.9.D[解析]由題意可知函數(shù)f(x)是R上的減函數(shù),∴當(dāng)x≤1時(shí),f(x)單調(diào)遞減,即a-3<0,①當(dāng)x>1時(shí),f(x)單調(diào)遞減,即a>0,②且(a-3)×1+5≥2a聯(lián)立①②③,解得0<a≤2,故選D.10.A[解析]∵0<|e-0.3|=e-0.3<1,1<|ln0.3|=ln103<2,log310>∴0<|e-0.3|<|ln0.3|<|log310|.當(dāng)x>0時(shí),f(x)=e-|x|=1ex∴f(e-0.3)>f(ln0.3)>f(log310).故a>b>c.11.[4,+∞)[解析]由復(fù)合函數(shù)的單調(diào)性知,本題等價(jià)于y=2x2+mx-3在(-1,1)上單調(diào)遞增,所以-m4≤-1,得m≥4,即實(shí)數(shù)m的取值范圍是[4,+∞)12.-12,0[解析]f(∵f(x)在a,a∴a<0,a+32>1,13.t≥1[解析]若函數(shù)f(x)=x2,x≥t,x,0<x<t(t>0)是區(qū)間(0,+∞)上的增函數(shù)14.解:(1)∵f(x)=log4(ax2+2x+3)且f(1)=1,∴l(xiāng)og4(a·12+2×1+3)=1?a+5=4?a=-1,可得函數(shù)f(x)=log4(-x2+2x+3).由-x2+2x+3>0?-1<x<3,∴函數(shù)f(x)的定義域?yàn)?-1,3).令t=-x2+2x+3=-(x-1)2+4,可得當(dāng)x∈(-1,1)時(shí),該函數(shù)為增函數(shù);當(dāng)x∈(1,3)時(shí),該函數(shù)為減函數(shù).∴函數(shù)f(x)=log4(-x2+2x+3)的單調(diào)遞增區(qū)間為(-1,1),單調(diào)遞減區(qū)間為(1,3).(2)假設(shè)存在實(shí)數(shù)a,使得f(x)的最小值為0.由底數(shù)4>1,可得真數(shù)t=ax2+2x+3≥1恒成立,且真數(shù)t的最小值恰好是1,則a為正數(shù),且當(dāng)x=-22a=-1a時(shí),∴a>0,a·-因此存在實(shí)數(shù)a=12,使得f(x)的最小值為015.解:(1)令x=1,y=0,可得f(1)=f(1)f(0),因?yàn)楫?dāng)x>0時(shí),0<f(x)<1,所以f(1)≠0,故f(0)=1.證明:令y=-x,x<0,則f(0)=f(x)f(-x),即f(x)=1f因?yàn)?x>0,所以0<f(-x)<1,所以f(x)>1.(2)函數(shù)f(x)在R上為減函數(shù).證明如下:設(shè)x1<x2,則x1-x2<0,又f(x1)-f(x2)=f[(x1-x2)+x2]-f(x2)=f(x1-x2)f(x2)-f(x2)=f(x2)[f(x1-x2)-1],由(1)知f(x2)>0,f(x1-x2)>1,所以f(x1)-f(x2)>0,即f(x1)>f(x2),所以函數(shù)f(x)在R上為減函數(shù).(3)由f-12=2得f(-1)所以f[(a2-a-2)x2-(2a-1)2x+2]>4=f(-1),即(a2-a-2)x2-(2a-1)2x+2<-1,即(a2-a)(x2-4x)<2x2+x-3對(duì)任意x∈[1,3]恒成立.因?yàn)閤∈[1,3],所以x2-4x<0,所以a2-a>2x2+x-3x2-4設(shè)3x-1=t∈[2,8],則2+3(3x-1)x2-4x=2+27所以a2-a>0,解得a<0或a>1.16.A[解析]因?yàn)閒(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,所以f(x)min=f(0)=a+log2a=8.令g(a)=a+log2a-8,則g(a)在(0,+∞)上單調(diào)遞增,