浙江地區(qū)2020版高考數(shù)學(xué)第二章函數(shù)考點(diǎn)規(guī)范練4函數(shù)的單調(diào)性與最值.docx

考點(diǎn)規(guī)范練4函數(shù)的單調(diào)性與最值基礎(chǔ)鞏固組1.下列函數(shù)f(x)滿足“對于任意x1,x2(0,+),當(dāng)x1x2,都有f(x1)f(x2)”的是()A.f(x)=1xB.f(x)=(x-1)2C.f(x)=1exD.f(x)=ln(x+1)答案D解析f(x)滿足“對于任意x1,x2(0,+),當(dāng)x1x2,都有f(x1)f(x2)”,所以f(x)在(0,+)上單調(diào)遞增;f(x)=1x在(0,+)上單調(diào)遞減,f(x)=(x-1)2在(0,1)上單調(diào)遞減,(1,+)上單調(diào)遞增,f(x)=1ex在(0,+)上單調(diào)遞減,f(x)=ln(x+1)在(0,+)上單調(diào)遞增;故選D.2.若函數(shù)y=ax與y=-bx在(0,+)上都是減函數(shù),則y=ax2+bx在(0,+)上()A.單調(diào)遞增B.單調(diào)遞減C.先增后減D.先減后增答案B解析因為函數(shù)y=ax與y=-bx在(0,+)上都是減函數(shù),所以a0,b0,則y=ax2+bx圖象的對稱軸方程x=-b2af3f(-)B.f3f(-1)f(-)C.f(-)f(-1)f3D.f(-1)f(-)f3答案A解析由題意得,013f3f()=f(-),故選A.4.設(shè)偶函數(shù)f(x)在0,+)上單調(diào)遞增,則使得f(x)f(2x-1)成立的x的取值范圍是()A.13,1B.-,13(1,+)C.-13,13D.-,-1313,+答案A解析由f(x)為偶函數(shù),f(x)f(2x-1)可化為f(|x|)f(|2x-1|),又f(x)在0,+)上單調(diào)遞增,所以|x|2x-1|,解得13x-14B.a-14C.-14a0D.-14a0答案D解析當(dāng)a=0時,f(x)=2x-3,在定義域R上是單調(diào)遞增的,故在(-,4)上單調(diào)遞增;當(dāng)a0時,二次函數(shù)f(x)的對稱軸為x=-1a,因為f(x)在(-,4)上單調(diào)遞增,所以a0,且-1a4,解得-14a0.綜合上述得-14a0.故選D.6.已知f(x)=(a2-2a-2)x是增函數(shù),則實(shí)數(shù)a的取值范圍是.答案a3解析由f(x)=(a2-2a-2)x是增函數(shù)可得a2-2a-21,解得a3.7.設(shè)f(x)=-x2+2x,x2,log2x-1,x2,則f(f(4)=,函數(shù)f(x)的單調(diào)遞減區(qū)間是.答案11,2解析f(4)=log24-1=1,f(f(4)=f(1)=-12+21=1;當(dāng)x2時,f(x)=-x2+2x,對稱軸為x=1;所以f(x)在1,2單調(diào)遞減,所以f(x)的單調(diào)遞減區(qū)間是1,2.8.已知函數(shù)f(x)=x2-2,xa恒成立,則實(shí)數(shù)a的取值范圍是.答案(-,-1解析因當(dāng)x-1;當(dāng)x-1時,f(x)-12.即所以實(shí)數(shù)a的取值范圍是(-,-1,故應(yīng)填(-,-1.能力提升組9.已知f(x)是(0,+)的增函數(shù),若ff(x)-ln x=1,則f(e)=()A.2B.1C.0D.e答案A解析由題意得f(x)-lnx為常數(shù),設(shè)為a,則f(a)-lna=a,又因為f(a)=1,所以1-lna=a,可解得a=1,因此f(e)=lne+1=2,故選A.10.已知f(x)是定義在R上的偶函數(shù),且在區(qū)間(-,0)上單調(diào)遞減,若實(shí)數(shù)a滿足f(3|2a+1|)f(-3),則a的取值范圍是()A.-,-34-14,+B.-,-34C.-14,+D.-34,-14答案A解析函數(shù)f(x)是偶函數(shù),f(3|2a+1|)f(-3),等價為f(3|2a+1|)f(3),偶函數(shù)f(x)在區(qū)間(-,0)上單調(diào)遞減,f(x)在區(qū)間0,+)上單調(diào)遞增,3|2a+1|3,即2a+112,解得a-14,故選A.11.若函數(shù)f(x)=x2+a|x|+2,xR在區(qū)間3,+)和-2,-1上均為增函數(shù),則實(shí)數(shù)a的取值范圍是()A.-113,-3B.-6,-4C.-3,-22D.-4,-3答案B解析由函數(shù)f(x)為R上的偶函數(shù)知,只需考察f(x)在(0,+)上的單調(diào)性,因為函數(shù)f(x)=x2+a|x|+2,xR在區(qū)間3,+)和-2,-1上均為增函數(shù),所以f(x)在3,+)上為增函數(shù),在1,2上為減函數(shù),則只需函數(shù)y=x2+ax+2的對稱軸x=-a22,3,故a-6,-4,故選B.12.已知f(x)是定義在(0,+)的函數(shù).對任意兩個不相等的正數(shù)x1,x2,都有x2f(x1)-x1f(x2)x1-x20,記a=f(30.2)30.2,b=f(0.32)0.32,c=f(log25)log25,則()A.abcB.bacC.cabD.cb0,函數(shù)f(x)x是(0,+)上的增函數(shù),130.23,00.322,0.3230.2log25,ba0,a0)的定義域和值域都是0,1,則loga56+loga485=()A.1B.2C.3D.4答案C解析當(dāng)a1時,函數(shù)在0,1上單調(diào)遞減,所以a-1=1,且a-a=0,解得a=2.當(dāng)0a0,則滿足f(x)+fx-121的x的取值范圍是.答案-14,+解析f(x)=x+1,x0,2x,x0,f(x)+fx-121,即fx-121-f(x),利用圖象變換,在同一平面直角坐標(biāo)系中畫出y=fx-12與y=1-f(x)的圖象,如圖所示.由數(shù)形結(jié)合可知,滿足fx-121-f(x)的解為-14,+.15.已知函數(shù)f(x)=ex-x-1(x0),g(x)=-x2+4x-3,若f(a)=g(b),則b的最大值是.答案3解析由f(x)=ex-x-1(x0),知f(x)=ex-10,即函數(shù)f(x)=ex-x-1在(0,+)上單調(diào)遞增,即函數(shù)f(x)=ex-x-1(x0)的最小值為f(0)=0,值域為(0,+),二次函數(shù)g(x)=-x2+4x-3開口朝下,對稱軸為x=2,與x軸的交點(diǎn)為(1,0),(3,0),要使f(a)=g(b),則b1,3有解.故答案為3.16.(2017浙江杭州高級中學(xué)模擬)設(shè)a0,此時當(dāng)x=0時,3x2+a=a0不成立,若2x+b0在(a,b)上恒成立,則2b+b0,即b0,若3x2+a0在(a,b)上恒成立,則3a2+a0,即-13a0,故b-a的最大值為13,故答案為:13.17.已知函數(shù)f(x)=2xk+12x-1,k0,kR.(1)討論f(x)的奇偶性,并說明理由;(2)已知f(x)在(-,0上單調(diào)遞減,求實(shí)數(shù)k的取值范圍.解(1)函數(shù)f(x)=2xk+12x-1,k0,kR的定義域為R,f(-x)=2-xk+12-x-1=1k2x+2x-1.則當(dāng)k=1時,有f(-x)=f(x),此時f(x)為偶函數(shù),當(dāng)k1時,f(-x)f(x)且f(-x)-f(x),此時f(x)為非奇非偶函數(shù).(2)設(shè)t=2x,x(-,0,則有0t1,y=tk+1t-1,當(dāng)k0,y=tk+1t-1在(0,1單調(diào)遞減,t=2x在(-,0單調(diào)遞增,所以復(fù)合函數(shù)f(x)=2xk+12x-1單調(diào)遞減,所以k0,y=tk+1t-1在(0,k單調(diào)遞減,在(k,+)上單調(diào)遞增,已知f(x)在(-,0上單調(diào)遞減,必有k1,即k1.綜上所述,實(shí)數(shù)k的取值范圍為(-,0)1,+).18.已知函數(shù)f(x)=lg1-mx1-x為奇函數(shù).(1)求m的值,并求f(x)的定義域;(2)判斷函數(shù)f(x)的單調(diào)性,并證明;(3)若對于任意0,2,是否存在實(shí)數(shù),使得不等式fcos2+sin -13-lg 30.若存在,求出實(shí)數(shù)的取值范圍;若不存在,請說明理由.解(1)函數(shù)f(x)=lg1-mx1-x為奇函數(shù),f(-x)=-f(x)在定義域內(nèi)恒成立,即lg1+mx1+x=-lg1-mx1-x,即lg1+mx1+x+lg1-mx1-x=0,則1+mx1+x1-mx1-x=1,即1-m2x2=1-x2,在定義域內(nèi)恒成立,m=-1或m=1,當(dāng)m=1時,f(x)=lg1-mx1-x=lg1=0,定義域為x|x1,故不為奇函數(shù),故舍去.當(dāng)m=-1時,此時f(x)=lg1+x1-x,由1+x1-x0,解得-1x1,故函數(shù)的定義域是(-1,1).(2)f(x)=lg1+x1-x,-1x1,任取-1x1x21,設(shè)u(x)=1+x1-x,-1x1,則u(x1)-u(x2)=1+x11-x1-1+x21-x2=2(x1-x2)(1-x1)(1-