備戰(zhàn)2025年高考二輪復(fù)習(xí)數(shù)學(xué)專題突破練3　利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性、極值與最值(提升篇)

專題突破練(分值:113分)學(xué)生用書P139主干知識(shí)達(dá)標(biāo)練1.(2024陜西西安二模)函數(shù)f(x)=xx2+1在[-3,3]上的最大值和最小值分別是A.613,-613 B.25C.310,-310 D.12答案D解析f'(x)=1-x2(x2+1)2,x∈[-3,3],令f'(x)>0,解得-1<x<1,即f(x)在(-1,1)內(nèi)單調(diào)遞增,令f'(x)<0,解得x∈[-3,-1)∪(1,3],所以f(x)在[-3,-1)和(1,3]上單調(diào)遞減,又f(-3)=-310,f(-1)=-12,f(1)=12,f(3)=310,所以函數(shù)f(x)2.(2024山西晉城模擬)若f(x)=alnx+x2在x=1處有極值,則函數(shù)f(x)的單調(diào)遞增區(qū)間是()A.(1,+∞) B.(0,1) C.(1,3) D.12,1答案A解析由f(x)=alnx+x2,得f'(x)=ax+2x,f'(1)=a+2=0,解得a=-故f'(x)=-2x+2x=2x2-2x=2(x+1)(x-1)x(x>0),當(dāng)x∈(0,1)時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(1,+∞)時(shí),f'(x)>0,f3.(2024陜西渭南模擬)已知函數(shù)f(x)=xex+a在區(qū)間[0,1]上的最小值為1,則實(shí)數(shù)a的值為()A.-2 B.2 C.-1 D.1答案D解析由題意可知f'(x)=(x+1)ex,所以當(dāng)x∈[0,1]時(shí),f'(x)>0,則f(x)在[0,1]上單調(diào)遞增,所以f(x)min=f(0)=a=1.故選D.4.(2024江蘇一模)用min{x,y}表示x,y中的最小數(shù).已知函數(shù)f(x)=xex,則min{f(x),f(x+ln2)}的最大值為(A.2e2 B.1e答案C解析∵f(x)=xex,∴f'(x)=1-xex,易知f(x)在(-∞,1)內(nèi)單調(diào)遞增,在(1,+∞)內(nèi)單調(diào)遞減,由題意令f(x)=f(x+ln2),即xex=x+ln2則min{f(x),f(x+ln2)}的最大值為兩函數(shù)圖象交點(diǎn)處函數(shù)值,為ln22.故選5.(2024北京延慶一模)已知函數(shù)f(x)=3x-2x-1,則不等式f(x)<0的解集是()A.(0,1) B.(0,+∞) C.(-∞,0) D.(-∞,0)∪(1,+∞)答案A解析因?yàn)閒'(x)=3xln3-2單調(diào)遞增,且f'(0)=ln3-2<0,f'(1)=3ln3-2>0,所以存在唯一x0∈(0,1),使得f'(x0)=0,所以當(dāng)x<x0時(shí),f'(x)<0,當(dāng)x>x0時(shí),f'(x)>0,所以函數(shù)f(x)在(-∞,x0)內(nèi)單調(diào)遞減,在(x0,+∞)內(nèi)單調(diào)遞增,又f(0)=f(1)=0,且0<x0<1,所以由f(x)<0可得0<x<1,故選A.6.(多選題)(2024江蘇蘇州模擬)函數(shù)f(x)=ax3-bx2+cx的圖象如圖,且f(x)在x=x0與x=1處取得極值,給出下列判斷,其中正確的是()A.c<0B.a<0C.f(1)+f(-1)>0D.函數(shù)f'(x)在(0,+∞)內(nèi)單調(diào)遞減答案AC解析f'(x)=3ax2-2bx+c=3a(x-x0)(x-1),由圖知x>1時(shí),f(x)單調(diào)遞增,可知f'(x)>0,所以a>0,故B錯(cuò)誤;又f'(x)=3ax2-2bx+c=3a(x-x0)(x-1)=3ax2-3a(1+x0)x+3ax0,∴2b=3a(1+x0),c=3ax0.∵x0<-1<0,∴c=3ax0<0,故A正確;∵x0<-1<0,∴1+x0<0,∴f(1)+f(-1)=-2b=-3a(1+x0)>0,故C正確;f'(x)=3ax2-2bx+c,其圖象開口向上,對稱軸小于0,函數(shù)f'(x)在(0,+∞)上單調(diào)遞增,故D錯(cuò)誤.故選AC.7.(多選題)(2024云南昆明模擬預(yù)測)已知函數(shù)f(x)=sinx·cosx-2sinx+x,則下列說法正確的是()A.f(x)在0,π2上單調(diào)遞增B.f(x)在π2,π上單調(diào)遞增C.f(x)在[0,π]上有唯一零點(diǎn)D.f(x)在[0,π]上有最小值π2-答案BD解析f'(x)=2cos2x-2cosx=2(cosx當(dāng)x∈0,π2時(shí),0≤cosx≤1,易知f'(x)≤0,僅當(dāng)x=0或x=π2時(shí),f'(x)=0,所以f(x)在0,π2上單調(diào)遞減.當(dāng)x∈π2,π時(shí),-1≤cosx≤0,易知f'(x)≥0,僅當(dāng)x=π2時(shí),f'(x)=0,所以f(x)在π2,π上單調(diào)遞增.f(x)在x=π2上取極小值為fπ2=π2-2,又f(0)=0,f(π)=π,所以f(x)在[0,π]上有兩個(gè)零點(diǎn)x1=0,x2∈π2,π,所以A,C錯(cuò)誤;B,D正確.故選BD.8.(5分)(2024江西上饒一模)若函數(shù)f(x)=x3-12ax2+6x在區(qū)間(1,3)內(nèi)單調(diào)遞增,則a的取值范圍為答案(-∞,62]解析因?yàn)閒(x)=x3-12ax2+6x,所以f'(x)=3x2-ax+6,因?yàn)楹瘮?shù)f(x)=x3-12ax2+6x在區(qū)間(1,3)內(nèi)單調(diào)遞增,所以f'(x)=3x2-ax+6≥0在(1,3)內(nèi)恒成立,即x∈(1,3)時(shí),a≤3x+6x恒成立.因?yàn)?x+6x≥23x×6x=62,當(dāng)且僅當(dāng)x=2時(shí)等號(hào)成立,即(3x+6x)min=62,9.(5分)(2024河北石家莊模擬)已知函數(shù)f(x)=ax-lnx的最小值為0,則a=.

答案1解析因?yàn)閒(x)=ax-lnx,所以f'(x)=a-1x若a≤0,則f(x)在(0,+∞)內(nèi)單調(diào)遞減,無最小值.若a>0,則f(x)在0,1a內(nèi)單調(diào)遞減,在1a,+∞內(nèi)單調(diào)遞增,所以f(x)min=f1a=1+lna=0,解得a=1e.10.(5分)(2024湖北襄陽模擬)函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),若在f(x)的定義域內(nèi)存在一個(gè)區(qū)間D,f(x)在區(qū)間D上單調(diào)遞增,f'(x)在區(qū)間D上單調(diào)遞減,則稱區(qū)間D為函數(shù)f(x)的一個(gè)“漸緩增區(qū)間”.若對于函數(shù)f(x)=aex-x2,區(qū)間0,12是其一個(gè)漸緩增區(qū)間,那么實(shí)數(shù)a的取值范圍是.

答案ee,解析對于函數(shù)f(x)=aex-x2,x∈0,12,有f'(x)=aex-2x,令g(x)=aex-2x,則g'(x)=aex-2,因?yàn)閒'(x)在區(qū)間0,12上單調(diào)遞減,所以aex-2≤0恒成立,即a≤2ex恒成立,又2ex>2又f(x)在區(qū)間0,12上單調(diào)遞增,所以f'(x)=aex-2x≥0恒成立,即a≥2xex恒成立.設(shè)h(x)=2xex,則h'(x)=2(1-x)ex,在0,12內(nèi)h'(x)>0,則h(x)單調(diào)遞增,則h(x)<ee所以a的取值范圍是ee,2關(guān)鍵能力提升練11.(多選題)(2024山東泰安模擬預(yù)測)已知函數(shù)f(x)=3x-2x,則()A.f(x)是R上的增函數(shù)B.函數(shù)h(x)=f(x)+x有且僅有一個(gè)零點(diǎn)C.函數(shù)f(x)的最小值為-1D.f(x)存在唯一極值點(diǎn)答案BD解析對于選項(xiàng)A,因?yàn)閒(x)=3x-2x,則f'(x)=3xln3-2xln2=2x32xln3-ln2,當(dāng)x=log3212時(shí),32xln3=12ln3=ln3,可得32xln3-ln2=即f'(x)=3xln3-2xln2<0,所以f(x)=3x-2x不是R上的增函數(shù),故A錯(cuò)誤;對于選項(xiàng)B,因?yàn)閔(x)=f(x)+x,當(dāng)x=0時(shí),h(0)=f(0)+0=0,可知x=0是h(x)的零點(diǎn);當(dāng)x>0時(shí),h(x)=f(x)+x=3x-2x+x>0,可知h(x)在(0,+∞)內(nèi)無零點(diǎn);當(dāng)x<0時(shí),0<32x<1,則f(x)=2x32x-1可得h(x)=f(x)+x<0,可知h(x)在(-∞,0)內(nèi)無零點(diǎn).綜上所述,函數(shù)h(x)=f(x)+x有且僅有一個(gè)零點(diǎn),故B正確;對于選項(xiàng)C,當(dāng)x>0時(shí),f(x)=3x-2x>0;當(dāng)x=0時(shí),f(0)=30-20=0;當(dāng)x<0時(shí),0<3x<1,0<2x<1,可得f(x)=3x-2x>-2x>-1.綜上所述,f(x)>-1,所以-1不是函數(shù)f(x)的最小值,故C錯(cuò)誤;對于選項(xiàng)D,因?yàn)閒'(x)=3xln3-2xln2=2x32xln3-ln2,2x>0,所以f'(x)的符號(hào)決定于32xln3-ln顯然y=32xln3-ln2是R又因?yàn)楫?dāng)x=0時(shí),32xln3-ln2=ln3-ln2當(dāng)x=log3212時(shí),32xln3-ln2=ln3所以?x0∈R,使f'(x0)=0,在(-∞,x0)內(nèi),f'(x)<0,在(x0,+∞)內(nèi),f'(x)>0,所以f(x)在(-∞,x0)內(nèi)為減函數(shù),在(x0,+∞)內(nèi)為增函數(shù).所以f(x)有唯一極值點(diǎn),故D正確.故選BD.12.(2024江蘇南通二模)若函數(shù)f(x)=eax+2x有大于零的極值點(diǎn),則實(shí)數(shù)a的取值范圍為()A.(-2,+∞) B.-12,+∞C.(-∞,-2) D.-∞,-12答案C解析函數(shù)f(x)=eax+2x,可得f'(x)=aeax+2,若a≥0,則f'(x)>0,此時(shí)f(x)單調(diào)遞增,無極值點(diǎn).故a<0,令f'(x)=aeax+2=0,解得x=1aln-2a,當(dāng)x>1aln-2a時(shí),f'(x)>0,當(dāng)x<1aln-2a時(shí),f'(x)<0,故x=1aln-2a是f(x)=eax+2x的極值點(diǎn).由于函數(shù)f(x)=eax+2x有大于零的極值點(diǎn),∴1aln-2a>0,又a<0,∴l(xiāng)n-2a<0,∴0<-2a<1,解得a<-2.故選C.13.(多選題)(2024江蘇常州模擬)已知x∈[-π,π],函數(shù)f(x)=sinxx2+1,A.f(x)的圖象關(guān)于y軸對稱B.f(x)恰有2個(gè)極值點(diǎn)C.f(x)在-π4,πD.f(x)的最小值小于-2答案BCD解析對于A,由f(x)=sinxx2+1,可得其定義域?yàn)镽,且f(-x)=sin(-x)(-x)2+1=-sinxx2對于B,由f'(x)=(x2+1)cosx-2xsinx(x2+1)2,令g則g'(x)=2xcosx-(x2+1)sinx-2sinx-2xcosx=-(x2+3)sinx,當(dāng)x∈(0,π]時(shí),可得sinx≥0,所以g'(x)≤0,且僅當(dāng)x=π時(shí),g'(x)=0,g(x)單調(diào)遞減;當(dāng)x∈[-π,0)時(shí),可得sinx≤0,所以g'(x)≥0,且僅當(dāng)x=-π時(shí),g'(x)=0,g(x)單調(diào)遞增,由g(π)=-(π2+1)<0,g(-π)=-(π2+1)<0且g(0)=1>0,可得g(0)g(π)<0,g(-π)g(0)<0,所以g(x)在(-π,0)和(0,π)內(nèi)各有一個(gè)零點(diǎn),設(shè)兩個(gè)零點(diǎn)分別為x1,x2,不妨設(shè)x1<x2.當(dāng)x∈[-π,x1)時(shí),g(x)<0,可得f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(x1,x2)時(shí),g(x)>0,可得f'(x)>0,f(x)單調(diào)遞增;當(dāng)x∈(x2,π]時(shí),g(x)<0,可得f'(x)<0,f(x)單調(diào)遞減,所以x1,x2為函數(shù)f(x)的2個(gè)極值點(diǎn),且只有2個(gè)極值點(diǎn),所以B正確;對于C,由B知,g(x)在[-π,0)內(nèi)單調(diào)遞增,在(0,π]上單調(diào)遞減,又由g-π4=22π216+1-π2=232(π-4)2>0,且gπ4=232(π-4)2>0,則當(dāng)x∈-π4,π4時(shí),g(x)>0,即f'(x)>0,所以函數(shù)f(x)對于D,由f-π4=sin(-π4)π42+1=-14.(5分)(2024江蘇鎮(zhèn)江模擬)如果函數(shù)f(x)在區(qū)間[a,b]上為增函數(shù),則記為f(x)[a,b],函數(shù)f(x)在區(qū)間[a,b]上為減函數(shù),則記為f(x)[a,b].已知(x+4x)[m,3],則實(shí)數(shù)m的最小值為;函數(shù)f(x)=2x3-3ax2+12x+1,且f(x)[1,2],答案23解析(1)由題意g(x)=x+4x在[m,3]上單調(diào)遞增,首先有0<m<3(若m≤0,則當(dāng)x=0時(shí),g(x)無意義),由對勾函數(shù)性質(zhì)得當(dāng)x>0時(shí),g(x)=x+4x的單調(diào)遞增區(qū)間為(2,+∞),所以2≤m<3,即實(shí)數(shù)m的最小值為(2)f(x)顯然可導(dǎo),f'(x)=6x2-6ax+12,由題意f(x)在[1,2]上單調(diào)遞減,在[2,3]上單調(diào)遞增,即x=2是函數(shù)f(x)的極值點(diǎn),所以f'(2)=6(4-2a+2)=0,解得a=3,經(jīng)檢驗(yàn)a=3滿足題意.15.(5分)(2024上海靜安模擬)記f(x)=lnx+x2-2kx+k2,若存在實(shí)數(shù)a,b,滿足12≤a<b≤2,使得函數(shù)y=f(x)在區(qū)間[a,b]上單調(diào)遞增,則實(shí)數(shù)k的取值范圍是答案-∞,94解析由題意知f'(x)=1x+2x-2k=2x2-2kx+1x>即2k<2x+1x在區(qū)間12,2內(nèi)有解.設(shè)g(x)=2x+1x,則該函數(shù)在12,22內(nèi)單調(diào)遞減,在22,2上單調(diào)遞增,且g12=3,g(2)=92,故g(x)=2x+1x在12,2上的最大值為故2k<92,即實(shí)數(shù)k的取值范圍是-∞,94.16.(17分)(2024浙江金華模擬)已知函數(shù)f(x)=1+2lnx(1)求f(x)的單調(diào)區(qū)間;(2)若存在x1,x2∈(1,+∞)且x1≠x2,使|f(x1)-f(x2)|≥k|lnx1-lnx2|成立,求k的取值范圍.解(1)由題意得f'(x)=-4lnxx3,令f'(x)=0,得x=1,當(dāng)x∈(0,1)時(shí),f'(x)>0,f(x)當(dāng)x∈(1,+∞)時(shí),f'(x)<0,f(x)在(1,+∞)上單調(diào)遞減.綜上可知,f(x)的單調(diào)遞增區(qū)間為(0,1),單調(diào)遞減區(qū)間為(1,+∞).(2)由題意存在x1,x2∈(1,+∞)且x1≠x2,不妨設(shè)x1>x2>1,由(1)知當(dāng)x∈(1,+∞)時(shí),f(x)單調(diào)遞減.|f(x1)-f(x2)|≥k|lnx1-lnx2|等價(jià)于f(x2)-f(x1)≥k(lnx1-lnx2),即f(x2)+klnx2≥f(x1)+klnx1,即存在x1,x2∈(1,+∞)且x1>x2,使f(x2)+klnx2≥f(x1)+klnx1成立.令h(x)=f(x)+klnx,則h(x)在(1,+∞)上存在單調(diào)遞減區(qū)間.即h'(x)=kx2-4lnxx3<0在(1,+∞)上有解,即k<4lnxx2在(1,+∞)上有解,即k<4ln令t(x)=4lnxx2,x∈(1,+∞),t'(x)=4(1-2lnx)x3,當(dāng)x∈