.2函數(shù)的基本性質(zhì)3.2.1單調(diào)性與最大(小)值第1課時函數(shù)的單調(diào)性基礎(chǔ)過關(guān)練題組一單調(diào)性的概念1.已知函數(shù)f(x)的圖象如圖所示,則f(x)的單調(diào)遞減區(qū)間為()A.(-3,-1)∪(1,4)B.(-5,-3)∪(-1,1)C.(-3,-1),(1,4)D.(-5,-3),(-1,1)2.已知函數(shù)f(x)在R上單調(diào)遞增,則下列一定成立的是()A.y=-f(x)在R上單調(diào)遞減B.y=1f(xC.y=[f(x)]2在R上單調(diào)遞增D.y=af(x)(a為實數(shù))在R上單調(diào)遞增3.已知函數(shù)f(x)在[a,b]上單調(diào)遞增,則對于任意的x1,x2∈[a,b](x1≠x2),下列結(jié)論中正確的是()A.f(B.(x1-x2)[f(x1)-f(x2)]<0C.f(a)≤f(x1)<f(x2)≤f(b)D.f(x1)>f(x2)題組二單調(diào)性的判斷與證明4.下列函數(shù)中,在區(qū)間(0,+∞)上單調(diào)遞減的是()A.y=-1xB.y=2xC.y=x25.(教材習(xí)題改編)已知函數(shù)f(x)=x-2x-1A.f(x)在(-1,+∞)上單調(diào)遞增B.f(x)在(-∞,-1)上單調(diào)遞減C.f(x)在(1,+∞)上單調(diào)遞增D.f(x)在(-1,1)上單調(diào)遞減6.已知二次函數(shù)f(x)=ax2+bx+c(a≠0)滿足?x∈R,f(x)≤f(3)恒成立,則函數(shù)f(x)的單調(diào)遞增區(qū)間為.

7.函數(shù)f(x)=x|x-1|的單調(diào)遞減區(qū)間為.

8.函數(shù)f(x)=|x2-3x+2|的單調(diào)遞減區(qū)間是.

9.已知函數(shù)f(x)=-(1)求f(1),f(f(-6))的值;(2)畫出f(x)的圖象(無需列表);(3)根據(jù)(2)中的圖象,寫出f(x)的單調(diào)區(qū)間和值域.10.已知函數(shù)f(x)=ax2(1)求函數(shù)f(x)的解析式;(2)判斷函數(shù)f(x)在(0,2)上的單調(diào)性,并用定義證明.題組三單調(diào)性的應(yīng)用11.函數(shù)y=f(x)在R上為增函數(shù),且f(2m)>f(-m+9),則實數(shù)m的取值范圍是()A.(-∞,-3)B.(0,+∞)C.(3,+∞)D.(-∞,-3)∪(3,+∞)12.(易錯題)已知二次函數(shù)y=x2-2ax+1在區(qū)間(2,3)內(nèi)是單調(diào)函數(shù),則實數(shù)a的取值范圍是()A.(-∞,2]∪[3,+∞)B.[2,3]C.(-∞,-3]∪[-2,+∞)D.[-3,-2]13.(易錯題)若函數(shù)f(x)=ax-1,x<1,x2-2ax,A.0,23B.14.已知函數(shù)f(x)是R上的增函數(shù),A(0,-1),B(3,1)是函數(shù)圖象上的兩點,那么|f(x+1)|≥1的解集是()A.(-1,2)B.(1,4)C.(-∞,-1]∪[4,+∞)D.(-∞,-1]∪[2,+∞)15.(易錯題)若函數(shù)f(x)=x2+2(a-2)x+2的單調(diào)遞增區(qū)間為[3,+∞),則實數(shù)a的值是.

16.若函數(shù)f(x)=ax-1x-1在區(qū)間(1,+∞)上單調(diào)遞減,則實數(shù)a17.已知函數(shù)f(x)=x2-(3a-2)x+b.(1)若關(guān)于x的不等式f(x)<0的解集為(-2,3),求實數(shù)a,b的值;(2)若函數(shù)f(x)在區(qū)間-103,+∞上單調(diào)遞增,求實數(shù)能力提升練題組一單調(diào)性的判斷1.若函數(shù)f(x)在定義域[-9,9]上單調(diào)遞增,則函數(shù)y=f(x2)的單調(diào)遞增區(qū)間是()A.[-9,9]B.[0,9]C.[-3,3]D.[0,3]2.(多選題)定義在(0,+∞)上的函數(shù)f(x)滿足:?x1,x2∈(0,+∞),當(dāng)x1≠x2時,恒有x2f(x1)-xA.f(x)=1B.f(x)=x2+2C.f(x)=x3-xD.f(x)=x43.函數(shù)f(x)=x2-3x4.函數(shù)f(x)的定義域為(0,+∞),?x,y∈(0,+∞),總有f(xy)=f(x)+f(y),且當(dāng)x>1時,f(x)<0.(1)證明:f(x)在(0,+∞)上單調(diào)遞減;(2)若f12=2,求不等式f(x)+f(x-1)+2>0的解集題組二單調(diào)性的應(yīng)用5.已知函數(shù)f(x)=x2-ax+5,x≤1,ax,x>1,若?x1,x2∈R,且x1A.(0,3]B.[2,+∞)C.(0,+∞)D.[2,3]6.已知函數(shù)f(x)在定義域(0,+∞)上單調(diào),若?x∈(0,+∞),都有ff(x)-1xA.12023B.2023C.2024D.27.已知函數(shù)f(x)滿足對任意x1,x2∈[0,+∞),當(dāng)x1<x2時,f(x1)-x1>f(x2)-x2恒成立,若f(4)=4,則不等式f(2x)<2xA.[0,2)B.[0,4)C.(2,+∞)D.(4,+∞)8.已知f(x)=ax2+1是定義在R上的函數(shù),若?x1,x2∈[-3,-1],當(dāng)x1<x2時,都有f(x1)-fA.{0}B.[0,+∞)C.-139.定義在(0,+∞)上的函數(shù)f(x)滿足x2f(x1)-x110.已知函數(shù)f(x)=-18x2+ax+12答案與分層梯度式解析3.2函數(shù)的基本性質(zhì)3.2.1單調(diào)性與最大(小)值第1課時函數(shù)的單調(diào)性基礎(chǔ)過關(guān)練1.C2.A3.A4.D5.C11.C12.A13.B14.D1.C由題圖知函數(shù)f(x)的圖象在區(qū)間(-3,-1)和(1,4)上是下降的,在區(qū)間(-5,-3)和(-1,1)上是上升的,因此該函數(shù)的單調(diào)遞減區(qū)間為(-3,-1),(1,4).故選C.2.A任取x1,x2∈R,且x1<x2,因為函數(shù)f(x)在R上單調(diào)遞增,所以必有f(x1)<f(x2),所以-f(x1)>-f(x2),A選項一定成立.其余三項不一定成立,如當(dāng)f(x)=x時,B,C不成立,當(dāng)a≤0時,D不成立.故選A.3.A因為f(x)在[a,b]上單調(diào)遞增,所以對于任意的x1,x2∈[a,b](x1≠x2),當(dāng)x1>x2時,f(x1)>f(x2),所以x1-x2>0,f(x1)-f(x2)>0,所以f(x1)-f(x2)當(dāng)x1<x2時,f(x1)<f(x2),所以x1-x2<0,f(x1)-f(x2)<0,所以f(x1)-f(x2)綜上,f(x1)-f(x2)x1-x2>0,(x由于x1,x2的大小關(guān)系不確定,所以f(x1)與f(x2)的大小關(guān)系不確定,故C,D不正確.故選A.4.D5.C易得f(x)=x-2x-1=x畫出函數(shù)f(x)的圖象,如圖所示.所以函數(shù)f(x)在(-∞,1)和(1,+∞)上均單調(diào)遞增,故選C.解題模板對“一次分式”函數(shù),常利用分離常數(shù)的方法轉(zhuǎn)化解析式,然后通過反比例函數(shù)的圖象結(jié)合平移變換解決問題.6.答案(-∞,3]解析因為?x∈R,都有f(x)≤f(3),所以f(x)的圖象開口向下,對稱軸是直線x=3,因此函數(shù)f(x)的單調(diào)遞增區(qū)間為(-∞,3].7.答案1解析由已知得f(x)=x畫出函數(shù)f(x)的圖象,如圖.由圖可得f(x)的單調(diào)遞減區(qū)間是128.答案(-∞,1)和3解析令x2-3x+2=0,得x=1或x=2,則f(x)=|x2-3x+2|=x2-3x+2,x≥2或由圖可知f(x)的單調(diào)遞減區(qū)間為(-∞,1)和32易錯警示求函數(shù)的單調(diào)區(qū)間,若單調(diào)遞減(增)區(qū)間有多個,則寫出單調(diào)區(qū)間時不能用“∪”連接.9.解析(1)f(1)=12-2×1=-1;f(-6)=-(-6)-3=3,f(f(-6))=f(3)=32-2×3=3.(2)畫出f(x)的圖象,如圖:(3)由(2)中圖象可知函數(shù)f(x)的單調(diào)遞減區(qū)間為(-∞,0),(0,1);單調(diào)遞增區(qū)間為(1,+∞).函數(shù)f(x)的值域為(-3,+∞).10.解析(1)∵f(x)的圖象過點A(1,5),B(2,4),∴f(1)=a+b+4=5,(2)函數(shù)f(x)=x+4x在(0,2)上單調(diào)遞減證明:任取x1,x2∈(0,2),且x1<x2,則f(x1)-f(x2)=x1+=(x∵0<x1<x2<2,∴x1-x2<0,x1x2>0,x1x2-4<0,∴f(x1)-f(x2)>0,即f(x1)>f(x2),∴函數(shù)f(x)在(0,2)上單調(diào)遞減.11.C∵函數(shù)y=f(x)在R上為增函數(shù),且f(2m)>f(-m+9),∴2m>-m+9,解得m>3,故選C.12.A易知二次函數(shù)y=x2-2ax+1的圖象的對稱軸為直線x=a,若y=x2-2ax+1在區(qū)間(2,3)內(nèi)是單調(diào)遞增函數(shù),則有a≤2;若y=x2-2ax+1在區(qū)間(2,3)內(nèi)是單調(diào)遞減函數(shù),則有a≥3.故選A.易錯警示解決二次函數(shù)的單調(diào)性問題,其關(guān)鍵是確定二次函數(shù)圖象的對稱軸與單調(diào)區(qū)間之間的位置關(guān)系,由此對參數(shù)進行分類討論.13.B根據(jù)題意,結(jié)合y=x2-2ax(x≥1)的圖象,知f(x)必定為增函數(shù),則有a>0,a≤1,a-1≤1-2a,解得0<a≤23,易錯警示由分段函數(shù)的單調(diào)性確定參數(shù)的值時,不僅要分別利用每段函數(shù)的單調(diào)性列出不等式,還要根據(jù)在分界點處的函數(shù)值的大小關(guān)系列出不等式.14.D|f(x+1)|≥1可化為f(x+1)≤-1或f(x+1)≥1,因為f(0)=-1,f(3)=1,所以f(x+1)≤f(0)或f(x+1)≥f(3),又f(x)為R上的增函數(shù),所以x+1≤0或x+1≥3,解得x≤-1或x≥2,即不等式的解集為(-∞,-1]∪[2,+∞).故選D.15.答案-1解析易得f(x)=x2+2(a-2)x+2的單調(diào)遞增區(qū)間為[2-a,+∞),∴2-a=3,∴a=-1.易錯警示注意函數(shù)在某區(qū)間上單調(diào)遞增(減)與函數(shù)的單調(diào)遞增(減)區(qū)間為某區(qū)間的區(qū)別,對“某區(qū)間”:前者為單調(diào)遞增(減)區(qū)間的子集,后者即為單調(diào)遞增(減)區(qū)間.16.答案(1,+∞)解析函數(shù)f(x)=ax-1x-1=a+a-1x-1,由當(dāng)x∈(1,+∞)時,f(x)單調(diào)遞減,得a-1>0(17.解析(1)由關(guān)于x的不等式f(x)<0的解集為(-2,3),可得關(guān)于x的一元二次方程f(x)=0的兩根為-2和3,所以3a-2=-2+3,當(dāng)a=1,b=-6時,f(x)=x2-x-6=(x-3)(x+2),符合題意,故實數(shù)a,b的值分別為1,-6.(2)由已知得二次函數(shù)y=f(x)的圖象開口向上,且對稱軸方程為x=3a若函數(shù)f(x)在-103則3a-22≤-103,解得a故實數(shù)a的取值范圍為-∞,-14能力提升練1.D2.CD5.D6.D7.C8.C1.D由題意知0≤x2≤9,解得-3≤x≤3,∴函數(shù)y=f(x2)的定義域為[-3,3],設(shè)g(x)=x2,則函數(shù)g(x)的單調(diào)遞增區(qū)間為[0,3],單調(diào)遞減區(qū)間為[-3,0],由復(fù)合函數(shù)的單調(diào)性知,函數(shù)y=f(x2)在區(qū)間[0,3]上單調(diào)遞增,故選D.2.CD不妨設(shè)x1>x2>0,由x2f(x1)-x1f(x2)x1-x2>0,可得x2f(x對于A,y=f(x)x=1x,易知函數(shù)y=1x在(0,+∞)對于B,y=f(x)x=x+2x,易知函數(shù)y=x+2x在(0,2)上單調(diào)遞減,在(2,+∞)對于C,y=f(x)x=x2-1,易知函數(shù)y=x2-1在(0,+∞)上單調(diào)遞增,對于D,y=f(x)x=x3,易知函數(shù)y=x3在(0,+∞)上單調(diào)遞增,所以D符合題意3.答案[2,+∞)解析由x2-3x+2≥0得x≤1或x≥2,因此函數(shù)f(x)的定義域為(-∞,1]∪[2,+∞)易錯點.又y=u在[0,+∞)上單調(diào)遞增,u=x2-3x+2在-∞,32上單調(diào)遞減,在3∴f(x)的單調(diào)遞增區(qū)間為[2,+∞).4.解析(1)證明:任取x1,x2∈(0,+∞),且x1<x2,則x2x1>1,故因為f(x2)-f(x1)=fx2x1·x1-f(x1)=fx2所以f(x2)<f(x1),即f(x)在(0,+∞)上單調(diào)遞減.(2)因為f12所以f(x)+f(x-1)+2=f(x)+f(x-1)+f12=f1在f(xy)=f(x)+f(y)中,令x=y=1,得f(1)=f(1)+f(1),即f(1)=0,所以f12因為f(x)在(0,+∞)上單調(diào)遞減,所以x>0,x-1>0,0<名師點評抽象函數(shù)問題的解決常用賦值法,賦值應(yīng)以結(jié)論為依據(jù).5.D由已知得函數(shù)f(x)在R上單調(diào)遞減,所以a2≥1,a>0,-a+6≥a,6.D由題意設(shè)f(x)-1x=c(c>0),故f(x)=c+1x,且f(c)=c+1c=2,所以c=1,所以f(x)=1x+1,則f12024=27.C由f(x1)-x1>f(x2)-x2,得f(x1)-x1-2>f(x2)-x2-2,可知?x1,x2∈[0,+∞),當(dāng)x1<x2時,F(x1)>F(x2),故F(x)在[0,+∞)上單調(diào)遞減,∵f(2x)<2x+2,f(4)=4,∴F(2x)=f(2x)-2x-2<0,且F(4)=f(4)-∴2x>4,解得x>2,因此不等式f(2x)<2x+2的解集為(2,+∞).故選C8.C由已知得f(x1)-f(x2)>2(x1-x2),即f(x1)-2x1>f(x2)-2x2對任意x1,x2∈[-3,-1],x1<x2恒成立,設(shè)g(x)=f(x)-2x=ax2-2x+1,則g(x)在[-3,-1]上單調(diào)遞減,當(dāng)a=0時,g(x)=-2x+1,符合題意;當(dāng)a>0時,易得1a≥-1,解得a≤-1或a>0,所以當(dāng)a<0時,易得1a≤-3,解得-13綜上,a≥-13.故選C9