2025屆高考數(shù)學(xué)二輪總復(fù)習(xí)專題1函數(shù)與導(dǎo)數(shù)專題突破練3利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性極值與最值課件

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123456789101112主干知識(shí)達(dá)標(biāo)練1314151617D123456789101112131415161712345678910111213141516172.(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) A12345678910111213141516173.(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.D1234567891011121314151617C123456789101112131415161712345678910111213141516175.(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,+∞)解析

因?yàn)閒'(x)=3xln

3-2單調(diào)遞增,且f'(0)=ln

3-2<0,f'(1)=3ln

3-2>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.A12345678910111213141516176.(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.123456789101112131415161712345678910111213141516177.(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(

)BD123456789101112131415161712345678910111213141516178.(5分)(2024江西上饒一模)若函數(shù)f(x)=x3-ax2+6x在區(qū)間(1,3)內(nèi)單調(diào)遞增,則a的取值范圍為

.

12345678910111213141516179.(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=

.

123456789101112131415161710.(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,)是其一個(gè)漸緩增區(qū)間,那么實(shí)數(shù)a的取值范圍是

.

12345678910111213141516171234567891011121314151617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)BD1234567891011121314151617即f'(x)=3xln

3-2xln

2<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);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;1234567891011121314151617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;1234567891011121314151617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.123456789101112131415161712.(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(

)C1234567891011121314151617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1234567891011121314151617BCD1234567891011121314151617則g'(x)=2xcos

x-(x2+1)sin

x-2sin

x-2xcos

x=-(x2+3)sin

x,當(dāng)x∈(0,π]時(shí),可得sin

x≥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sin

x≤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),1234567891011121314151617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,時(shí),g(x)>0,即f'(x)>0,所以函數(shù)f(x)單調(diào)遞增,所以C正確;123456789101112131415161714.(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,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],f(x)[2,3],則實(shí)數(shù)a=

.

23解析

(1)由題意g(x)=x+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+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.(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.123456789101112131415161715.(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

≤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

.

12345678910111213141516171234567891011121314151617(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.1234567891011121314151617(2)由題意存在x1,x2∈(1,+∞)且x1≠x