高中數(shù)學(xué)一輪復(fù)習(xí)課時作業(yè)梯級練五函數(shù)的單調(diào)性與最值課時作業(yè)理含解析新人教A版

一輪復(fù)習(xí)精品資料(高中)PAGE1-課時作業(yè)梯級練五函數(shù)的單調(diào)性與最值一、選擇題(每小題5分,共35分)1.下列四個函數(shù)中,在x∈(0,+∞)上為增函數(shù)的是 ()A.f(x)=3-x B.f(x)=x2-3xC.f(x)=-QUOTE D.f(x)=-|x|〖解析〗選C.當(dāng)x>0時,f(x)=3-x為減函數(shù);當(dāng)x∈QUOTE時,f(x)=x2-3x為減函數(shù),當(dāng)x∈QUOTE時,f(x)=x2-3x為增函數(shù);當(dāng)x∈(0,+∞)時,f(x)=-QUOTE為增函數(shù);當(dāng)x∈(0,+∞)時,f(x)=-|x|為減函數(shù).2.下列函數(shù)中,在區(qū)間(-∞,0)上是減函數(shù)的是 ()A.y=1-x2 B.y=x2+2xC.y=-QUOTE D.y=QUOTE〖解析〗選D.對于選項A,該函數(shù)是開口向下的拋物線,在區(qū)間(-∞,0)上是增函數(shù);對于選項B,該函數(shù)是開口向上的拋物線,在區(qū)間(-∞,-1〗上是減函數(shù),在區(qū)間〖-1,+∞)上是增函數(shù);對于選項C,在區(qū)間(-∞,0)上是增函數(shù);對于選項D,因為y=QUOTE=1+QUOTE.易知其在(-∞,1)上為減函數(shù).3.若函數(shù)f(x)=(m-1)x+b在R上是增函數(shù),則f(m)與f(1)的大小關(guān)系是()A.f(m)>f(1) B.f(m)<f(1)C.f(m)≥f(1) D.f(m)≤f(1)〖解析〗選A.因為f(x)=(m-1)x+b在R上是增函數(shù),則m-1>0,所以m>1,所以f(m)>f(1).4.(2021·贛州模擬)已知函數(shù)f(x)在R上單調(diào)遞減,且當(dāng)x∈〖0,2〗時,有f(x)=x2-4x,則關(guān)于x的不等式f(x)+3<0的解集為 ()A.(-∞,1) B.(1,3)C.(1,+∞) D.(3,+∞)〖解析〗選C.根據(jù)題意,當(dāng)x∈〖0,2〗時,有f(x)=x2-4x,有f(1)=1-4=-3,f(x)+3<0?f(x)<-3?f(x)<f(1),又由函數(shù)f(x)在R上單調(diào)遞減,則有x>1,即不等式的解集為(1,+∞).5.已知函數(shù)f(x)為定義在區(qū)間〖-1,1〗上的增函數(shù),則滿足f(x)<fQUOTE的實數(shù)x的取值范圍為 ()A.QUOTE B.QUOTEC.QUOTE D.QUOTE〖解析〗選C.由題設(shè)得QUOTE解得-1≤x<QUOTE.故實數(shù)x的取值范圍為QUOTE.6.(2021·臨滄模擬)若函數(shù)f(x)=2|x-a|+3在區(qū)間〖1,+∞)上不單調(diào),則a的取值范圍是 ()A.〖1,+∞) B.(1,+∞)C.(-∞,1) D.(-∞,1〗〖解析〗選B.因為函數(shù)f(x)=2|x-a|+3=QUOTE因為函數(shù)f(x)=2|x-a|+3在區(qū)間〖1,+∞)上不單調(diào),所以a>1.所以a的取值范圍是(1,+∞).7.函數(shù)f(x)=QUOTE在區(qū)間(-2,+∞)上單調(diào)遞增,則實數(shù)a的取值范圍是()A.QUOTEB.QUOTEC.(-2,+∞)D.(-∞,-1)∪(1,+∞)〖解析〗選B.因為當(dāng)a=0時,f(x)=QUOTE在區(qū)間(-2,+∞)上單調(diào)遞減,故a=0舍去,所以a≠0,此時f(x)=QUOTE=QUOTE=a+QUOTE,又因為y=QUOTE在區(qū)間(-2,+∞)上單調(diào)遞減,而函數(shù)f(x)=QUOTE在區(qū)間(-2,+∞)上單調(diào)遞增,所以1-2a<0,即a>QUOTE.〖加練備選·拔高〗若f(x)=-x2+4mx與g(x)=QUOTE在區(qū)間〖2,4〗上都是減函數(shù),則m的取值范圍是()A.(-∞,0)∪(0,1〗B.(-1,0)∪(0,1〗C.(0,+∞)D.(0,1〗〖解析〗選D.函數(shù)f(x)=-x2+4mx的圖象開口向下,且以直線x=2m為對稱軸,若在區(qū)間〖2,4〗上是減函數(shù),則2m≤2,解得m≤1;g(x)=QUOTE的圖象由y=QUOTE的圖象向左平移一個單位長度得到,若在區(qū)間〖2,4〗上是減函數(shù),則2m>0,解得m>0.綜上可得,m的取值范圍是(0,1〗.二、填空題(每小題5分,共15分)8.(2021·百色模擬)函數(shù)y=-x2+2|x|+3的單調(diào)遞減區(qū)間是________.

〖解析〗由題意知,當(dāng)x≥0時,y=-x2+2x+3=-(x-1)2+4;當(dāng)x<0時,y=-x2-2x+3=-(x+1)2+4,二次函數(shù)的圖象如圖,由圖象可知,函數(shù)y=-x2+2|x|+3的單調(diào)遞減區(qū)間為〖-1,0〗,〖1,+∞).〖答案〗:〖-1,0〗,〖1,+∞)〖加練備選·拔高〗函數(shù)f(x)=lg(x2-4)的單調(diào)遞增區(qū)間為________.

〖解析〗由復(fù)合函數(shù)的單調(diào)性,要使f(x)單調(diào)遞增,需QUOTE解得x>2.所以函數(shù)fQUOTE的單調(diào)遞增區(qū)間是(2,+∞).〖答案〗:(2,+∞)9.(2021·青島模擬)函數(shù)f(x)=QUOTE的最大值為________.

〖解析〗當(dāng)x≥1時,函數(shù)f(x)=QUOTE為減函數(shù),所以f(x)在x=1處取得最大值,為f(1)=1;當(dāng)x<1時,易知函數(shù)f(x)=-x2+2在x=0處取得最大值,為f(0)=2.故函數(shù)f(x)的最大值為2.〖答案〗:210.若函數(shù)f(x)=QUOTE的值域為R,則a的取值范圍是________.

〖解析〗當(dāng)x≥1時,f(x)=x2≥1,若a=0,x<1時,f(x)=0,f(x)的值域不是R;若a<0,x<1時,f(x)>2a,f(x)的值域不是R,若a>0,x<1時,f(x)<2a,所以當(dāng)2a≥1時,f(x)的值域為R,所以a的取值范圍是QUOTE.〖答案〗:QUOTE1.(5分)(2021·南寧模擬)已知函數(shù)f(x)=x-QUOTE+QUOTE在(1,+∞)上是增函數(shù),則實數(shù)a的取值范圍是________.

〖解析〗方法一:設(shè)1<x1<x2,所以x1x2>1.因為函數(shù)f(x)在(1,+∞)上是增函數(shù),所以f(x1)-f(x2)=x1-QUOTE+QUOTE-QUOTE=(x1-x2)QUOTE<0.因為x1-x2<0,所以1+QUOTE>0,即a>-x1x2.因為1<x1<x2,x1x2>1,所以-x1x2<-1,所以a≥-1.所以a的取值范圍是〖-1,+∞).方法二:由f(x)=x-QUOTE+QUOTE得f′(x)=1+QUOTE,由題意得1+QUOTE≥0(x>1),可得a≥-x2,當(dāng)x∈(1,+∞)時,-x2<-1.所以a的取值范圍是〖-1,+∞).〖答案〗:〖-1,+∞)2.(5分)(2020·瀘州模擬)已知f(x)=x|x|,則滿足f(2x-1)+f(x)≥0的x的取值范圍為________.

〖解析〗根據(jù)題意得,f(x)=x|x|=QUOTE則f(x)為奇函數(shù)且在R上為增函數(shù),則f(2x-1)+f(x)≥0?f(2x-1)≥-f(x)?f(2x-1)≥f(-x)?2x-1≥-x,解得x≥QUOTE,即x的取值范圍為QUOTE.〖答案〗:QUOTE3.(5分)若函數(shù)f(x)=x2+a|x-1|在〖0,+∞)上單調(diào)遞增,則實數(shù)a的取值范圍是________.

〖解析〗f(x)=x2+a|x-1|=QUOTE要使f(x)在〖0,+∞)上單調(diào)遞增,則QUOTE得-2≤a≤0,所以實數(shù)a的取值范圍是〖-2,0〗.〖答案〗:〖-2,0〗4.(10分)已知函數(shù)f(x)=2a+2x-QUOTE(a>0,x>0).(1)求證:f(x)在(0,+∞)上是增函數(shù);(2)若f(x)在〖1,3〗上的最大值是最小值的2倍,求a的值.〖解析〗(1)設(shè)0<x1<x2,則f(x1)-f(x2)=QUOTE-QUOTE=2(x1-x2)+QUOTE,由于0<x1<x2,故x1-x2<0,QUOTE-QUOTE<0,據(jù)此可得:f(x1)-f(x2)<0,f(x1)<f(x2),即函數(shù)f(x)在區(qū)間(0,+∞)上是增函數(shù).(2)由函數(shù)的單調(diào)性結(jié)合題意可得:f(3)=2f(1),即2a+6-QUOTE=2(2a+2-1),解得:a=QUOTE.5.(10分)已知函數(shù)g(x)=QUOTE+1,h(x)=QUOTE(x∈(-3,a〗),其中a為常數(shù)且a>0,令函數(shù)f(x)=g(x)·h(x).(1)求函數(shù)f(x)的表達(dá)式,并求其定義域;(2)當(dāng)a=QUOTE時,求函數(shù)f(x)的值域.〖解析〗(1)f(x)=QUOTE,x∈〖0,a〗(a>0).(2)函數(shù)f(x)的定義域為QUOTE,令QUOTE+1=t,則x=(t-1)2,t∈QUOTE,f(x)=F(t)=QUOTE=QUOTE.又t∈QUOTE時,t+QUOTE單調(diào)遞減,F(t)單調(diào)遞增,F(t)∈QUOTE,即函數(shù)f(x)的值域為QUOTE.課時作業(yè)梯級練五函數(shù)的單調(diào)性與最值一、選擇題(每小題5分,共35分)1.下列四個函數(shù)中,在x∈(0,+∞)上為增函數(shù)的是 ()A.f(x)=3-x B.f(x)=x2-3xC.f(x)=-QUOTE D.f(x)=-|x|〖解析〗選C.當(dāng)x>0時,f(x)=3-x為減函數(shù);當(dāng)x∈QUOTE時,f(x)=x2-3x為減函數(shù),當(dāng)x∈QUOTE時,f(x)=x2-3x為增函數(shù);當(dāng)x∈(0,+∞)時,f(x)=-QUOTE為增函數(shù);當(dāng)x∈(0,+∞)時,f(x)=-|x|為減函數(shù).2.下列函數(shù)中,在區(qū)間(-∞,0)上是減函數(shù)的是 ()A.y=1-x2 B.y=x2+2xC.y=-QUOTE D.y=QUOTE〖解析〗選D.對于選項A,該函數(shù)是開口向下的拋物線,在區(qū)間(-∞,0)上是增函數(shù);對于選項B,該函數(shù)是開口向上的拋物線,在區(qū)間(-∞,-1〗上是減函數(shù),在區(qū)間〖-1,+∞)上是增函數(shù);對于選項C,在區(qū)間(-∞,0)上是增函數(shù);對于選項D,因為y=QUOTE=1+QUOTE.易知其在(-∞,1)上為減函數(shù).3.若函數(shù)f(x)=(m-1)x+b在R上是增函數(shù),則f(m)與f(1)的大小關(guān)系是()A.f(m)>f(1) B.f(m)<f(1)C.f(m)≥f(1) D.f(m)≤f(1)〖解析〗選A.因為f(x)=(m-1)x+b在R上是增函數(shù),則m-1>0,所以m>1,所以f(m)>f(1).4.(2021·贛州模擬)已知函數(shù)f(x)在R上單調(diào)遞減,且當(dāng)x∈〖0,2〗時,有f(x)=x2-4x,則關(guān)于x的不等式f(x)+3<0的解集為 ()A.(-∞,1) B.(1,3)C.(1,+∞) D.(3,+∞)〖解析〗選C.根據(jù)題意,當(dāng)x∈〖0,2〗時,有f(x)=x2-4x,有f(1)=1-4=-3,f(x)+3<0?f(x)<-3?f(x)<f(1),又由函數(shù)f(x)在R上單調(diào)遞減,則有x>1,即不等式的解集為(1,+∞).5.已知函數(shù)f(x)為定義在區(qū)間〖-1,1〗上的增函數(shù),則滿足f(x)<fQUOTE的實數(shù)x的取值范圍為 ()A.QUOTE B.QUOTEC.QUOTE D.QUOTE〖解析〗選C.由題設(shè)得QUOTE解得-1≤x<QUOTE.故實數(shù)x的取值范圍為QUOTE.6.(2021·臨滄模擬)若函數(shù)f(x)=2|x-a|+3在區(qū)間〖1,+∞)上不單調(diào),則a的取值范圍是 ()A.〖1,+∞) B.(1,+∞)C.(-∞,1) D.(-∞,1〗〖解析〗選B.因為函數(shù)f(x)=2|x-a|+3=QUOTE因為函數(shù)f(x)=2|x-a|+3在區(qū)間〖1,+∞)上不單調(diào),所以a>1.所以a的取值范圍是(1,+∞).7.函數(shù)f(x)=QUOTE在區(qū)間(-2,+∞)上單調(diào)遞增,則實數(shù)a的取值范圍是()A.QUOTEB.QUOTEC.(-2,+∞)D.(-∞,-1)∪(1,+∞)〖解析〗選B.因為當(dāng)a=0時,f(x)=QUOTE在區(qū)間(-2,+∞)上單調(diào)遞減,故a=0舍去,所以a≠0,此時f(x)=QUOTE=QUOTE=a+QUOTE,又因為y=QUOTE在區(qū)間(-2,+∞)上單調(diào)遞減,而函數(shù)f(x)=QUOTE在區(qū)間(-2,+∞)上單調(diào)遞增,所以1-2a<0,即a>QUOTE.〖加練備選·拔高〗若f(x)=-x2+4mx與g(x)=QUOTE在區(qū)間〖2,4〗上都是減函數(shù),則m的取值范圍是()A.(-∞,0)∪(0,1〗B.(-1,0)∪(0,1〗C.(0,+∞)D.(0,1〗〖解析〗選D.函數(shù)f(x)=-x2+4mx的圖象開口向下,且以直線x=2m為對稱軸,若在區(qū)間〖2,4〗上是減函數(shù),則2m≤2,解得m≤1;g(x)=QUOTE的圖象由y=QUOTE的圖象向左平移一個單位長度得到,若在區(qū)間〖2,4〗上是減函數(shù),則2m>0,解得m>0.綜上可得,m的取值范圍是(0,1〗.二、填空題(每小題5分,共15分)8.(2021·百色模擬)函數(shù)y=-x2+2|x|+3的單調(diào)遞減區(qū)間是________.

〖解析〗由題意知,當(dāng)x≥0時,y=-x2+2x+3=-(x-1)2+4;當(dāng)x<0時,y=-x2-2x+3=-(x+1)2+4,二次函數(shù)的圖象如圖,由圖象可知,函數(shù)y=-x2+2|x|+3的單調(diào)遞減區(qū)間為〖-1,0〗,〖1,+∞).〖答案〗:〖-1,0〗,〖1,+∞)〖加練備選·拔高〗函數(shù)f(x)=lg(x2-4)的單調(diào)遞增區(qū)間為________.

〖解析〗由復(fù)合函數(shù)的單調(diào)性,要使f(x)單調(diào)遞增,需QUOTE解得x>2.所以函數(shù)fQUOTE的單調(diào)遞增區(qū)間是(2,+∞).〖答案〗:(2,+∞)9.(2021·青島模擬)函數(shù)f(x)=QUOTE的最大值為________.

〖解析〗當(dāng)x≥1時,函數(shù)f(x)=QUOTE為減函數(shù),所以f(x)在x=1處取得最大值,為f(1)=1;當(dāng)x<1時,易知函數(shù)f(x)=-x2+2在x=0處取得最大值,為f(0)=2.故函數(shù)f(x)的最大值為2.〖答案〗:210.若函數(shù)f(x)=QUOTE的值域為R,則a的取值范圍是________.

〖解析〗當(dāng)x≥1時,f(x)=x2≥1,若a=0,x<1時,f(x)=0,f(x)的值域不是R;若a<0,x<1時,f(x)>2a,f(x)的值域不是R,若a>0,x<1時,f(x)<2a,所以當(dāng)2a≥1時,f(x)的值域為R,所以a的取值范圍是QUOTE.〖答案〗:QUOTE1.(5分)(2021·南寧模擬)已知函數(shù)f(x)=x-QUOTE+QUOTE在(1,+∞)上是增函數(shù),則實數(shù)a的取值范圍是________.

〖解析〗方法一:設(shè)1<x1<x2,所以x1x2>1.因為函數(shù)f(x)在(1,+∞)上是增函數(shù),所以f(x1)-f(x2)=x1-QUOTE+QUOTE-QUOTE=(x1-x2)QUOTE<0.因為x1-x2<0,所以1+QUOTE>0,即a>-x1x2.因為1<x1<x2,x1x2>1,所以-x1x2<-1,所以a≥-1.所以a的取值范圍是〖-1,+∞).方法二:由f(x)=x-QUOTE+QUOTE得f′(x)=1+QUOTE,由題意得1+QUOTE≥0(x>1),可得a≥-x2,當(dāng)x∈(1,+∞)時,-x2<-1.所以a的取值范圍是〖-1,+∞).〖答案〗:〖-1,+∞)2.(5分)(2020·瀘州模擬)已知f(x)=x|x|,則滿足f(2x-1)+f(x)≥0的x的取值范圍為________.

〖解析〗根據(jù)題意得,f(x)=x|x|=QUOTE則f(x)為奇函數(shù)且在R上為增函數(shù),則f(2x-1)+f(x)≥0?f(2x-1)≥-f(x)?f(2x-1)≥f(-x)?2x-1≥-x,解得x≥QUOTE,即x的取值范圍為QUOTE.〖答案〗:QUOTE3.(5分)若函數(shù)f(x)=x2+a|x-1|