（新教材適用）高中數(shù)學(xué)第五章一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用習(xí)題課函數(shù)的單調(diào)性的應(yīng)用課后習(xí)題新人教A版選擇性

習(xí)題課——函數(shù)的單調(diào)性的應(yīng)用課后訓(xùn)練鞏固提升A組1.若函數(shù)y=x3+x2+mx+1是R上的單調(diào)函數(shù),則實(shí)數(shù)m的取值范圍是()A.13,+C.13,+解析:由單調(diào)性可知函數(shù)的導(dǎo)數(shù)在R上恒非負(fù)或恒非正,且不恒等于0.當(dāng)y'=3x2+2x+m≥0對所有x∈R成立時(shí),此時(shí)應(yīng)滿足Δ=44×3m≤0,解得m≥13因?yàn)?>0,所以拋物線y'=3x2+2x+m開口向上,所以y'≤0不可能恒成立.因此滿足條件的m的取值范圍是13答案:C2.已知函數(shù)f(x)=12x3+ax+4,則“a>0”是“f(x)在R上單調(diào)遞增”的(A.充分不必要條件B.必要不充分條件C.充要條件D.既不充分也不必要條件解析:f'(x)=32x2+a,當(dāng)a>0時(shí),f'(x)>0在R所以當(dāng)a>0時(shí),函數(shù)f(x)在R上單調(diào)遞增.若函數(shù)f(x)在R上單調(diào)遞增,則f'(x)=32x2+a≥0在R上恒成立,即a≥32x2恒成立,從而a故“a>0”是“函數(shù)f(x)在R上單調(diào)遞增”的充分不必要條件.答案:A3.若函數(shù)f(x)=kxlnx在區(qū)間(1,+∞)上單調(diào)遞增,則k的取值范圍是()A.(∞,2] B.(∞,1]C.[2,+∞) D.[1,+∞)解析:因?yàn)閒(x)=kxlnx,所以函數(shù)f(x)的定義域?yàn)?0,+∞),f'(x)=k1x因?yàn)楹瘮?shù)f(x)在區(qū)間(1,+∞)上單調(diào)遞增,所以當(dāng)x>1時(shí),f'(x)=k1x≥0恒成立,即k≥1x在區(qū)間(1,+∞因?yàn)閤>1,所以0<1x<1,所以k≥1.故選D答案:D4.已知函數(shù)f(x)在定義域R上可導(dǎo),若f(x)=f(2x),且當(dāng)x∈(∞,1)時(shí),(x1)f'(x)<0,設(shè)a=f(0),b=f12,c=f(3),則(A.a<b<c B.c<b<aC.c<a<b D.b<c<a解析:由題意得,當(dāng)x<1時(shí),f'(x)>0,所以f(x)在區(qū)間(∞,1)上單調(diào)遞增.由題意得f(3)=f(1),且1<0<12<1,因此f(1)<f(0)<f12,即f(3)<f(0)<f12答案:C5.若函數(shù)f(x)=x3+bx2+cx+d的單調(diào)遞增區(qū)間為(∞,1)和(2,+∞),則b=,c=.

解析:f'(x)=3x2+2bx+c,由題意知x<1或x>2是不等式3x2+2bx+c>0的解集,即1,2是方程3x2+2bx+c=0的兩個根,則1+2=2b3,1×2=c3,解得b=32答案:326.若函數(shù)f(x)=14x+3-kx+lnx在區(qū)間[1,2]上單調(diào)遞增,則實(shí)數(shù)k解析:∵函數(shù)f(x)=14x+3-kx∴f'(x)=14+k-3x2+1x≥0在區(qū)間[1,2]上恒成立,∴k設(shè)g(x)=14x2x+3,則函數(shù)g(x)圖象的對稱軸為直線x=∴g(x)=14x2x+3在區(qū)間[1,2]上單調(diào)遞減∴在區(qū)間[1,2]上,g(x)max=141+3=7∴k≥74答案:77.若函數(shù)f(x)=x3kx在區(qū)間(3,1)上不單調(diào),則實(shí)數(shù)k的取值范圍是.

解析:f'(x)=3x2k,當(dāng)k≤0時(shí),對x∈R,不等式f'(x)≥0恒成立,則f(x)在R上單調(diào)遞增,不符合題意,所以k>0.令f'(x)=0,得x=±k3因?yàn)楹瘮?shù)f(x)在區(qū)間(3,1)上不單調(diào),所以3<k3<1,即3<k<27答案:(3,27)8.已知函數(shù)f(x)=ax3+x在R上有三個單調(diào)區(qū)間,則a的取值范圍是.

解析:f(x)的導(dǎo)數(shù)f'(x)=3ax2+1.若a>0,則f'(x)>0對x∈R恒成立,此時(shí),f(x)只有一個單調(diào)區(qū)間,與已知矛盾;若a=0,則f(x)=x,此時(shí),f(x)也只有一個單調(diào)區(qū)間,亦與已知矛盾;若a<0,則f'(x)=3a·x+1-3a·x-1-3a,f(x答案:(∞,0)9.已知函數(shù)f(x)=x2+ax(x≠0,常數(shù)a∈R),若函數(shù)f(x)在區(qū)間[2,+∞)上單調(diào)遞增,求a的取值范圍解:f'(x)=2xax若函數(shù)f(x)在區(qū)間[2,+∞)上單調(diào)遞增,則f'(x)≥0當(dāng)x∈[2,+∞)時(shí)恒成立,即2x3a≥0當(dāng)x∈[2,+∞)時(shí)恒成立,∴a≤2x3當(dāng)x∈[2,+∞)時(shí)恒成立.∵y=2x3在區(qū)間[2,+∞)上單調(diào)遞增,∴(2x3)min=16.∴a≤16.當(dāng)a=16時(shí),只有f'(2)=0.∴a的取值范圍是(∞,16].10.已知函數(shù)f(x)=x3ax1.(1)若f(x)在R上單調(diào)遞增,求實(shí)數(shù)a的取值范圍.(2)是否存在實(shí)數(shù)a,使f(x)在區(qū)間(1,1)內(nèi)單調(diào)遞減?若存在,請求出a的取值范圍;若不存在,請說明理由.解:(1)由已知得f'(x)=3x2a.∵函數(shù)f(x)在R上單調(diào)遞增,∴f'(x)=3x2a≥0在R上恒成立,即a≤3x2對x∈R恒成立.∵3x2≥0,∴a≤0.∴實(shí)數(shù)a的取值范圍是(∞,0].(2)存在.證明如下:若函數(shù)f(x)在區(qū)間(1,1)內(nèi)單調(diào)遞減,則對x∈(1,1),不等式f'(x)=3x2a≤0恒成立,即a≥3x2對x∈(1,1)恒成立.當(dāng)x∈(1,1)時(shí),3x2<3,∴a≥3.∴存在實(shí)數(shù)a,使函數(shù)f(x)在區(qū)間(1,1)內(nèi)單調(diào)遞減,實(shí)數(shù)a的取值范圍是[3,+∞).B組1.若函數(shù)f(x)=x3ax2x+6在區(qū)間(0,1)內(nèi)單調(diào)遞減,則實(shí)數(shù)a的取值范圍是()A.a≥1 B.a=1 C.a≤1 D.0<a<1解析:f'(x)=3x22ax1,∵函數(shù)f(x)在區(qū)間(0,1)內(nèi)單調(diào)遞減,∴不等式f'(x)=3x22ax1≤0對x∈(0,1)恒成立.∴f'(0)≤0,且f'(1)≤0,解得a≥1.故選A.答案:A2.設(shè)函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),且當(dāng)x∈0,π2時(shí),f'(x)cosx+f(x)sinx<0,f(0)=A.fπ6>2fπ3 B.C.f(ln2)>0 D.fπ解析:設(shè)g(x)=f(則g'(x)=f'(x∴g(x)在區(qū)間0,π∴f(ln2)cos(ln2)=g(ln2)∴f(ln2)<0.∵0<π6∴0>fπ即0>23fπ6>∴fπ4>2fπ3,fπ3∴fπ6>3fπ3>2fπ3,f答案:A3.已知函數(shù)f(x)(x∈R)滿足f(1)=1,且f(x)的導(dǎo)函數(shù)f'(x)<12,則不等式f(x)<x2+A.{x|1<x<1} B.{x|x<1}C.{x|x<1或x>1} D.{x|x>1}解析:設(shè)g(x)=f(x)x2-12,則g'(x)=f'(x)12<0,故g(∵f(1)=1,∴g(1)=f(1)12-∴g(x)=f(x)x2-12<0=g(1)的解集為{x|x>答案:D4.(多選題)若函數(shù)exf(x)(e為自然對數(shù)的底數(shù))在f(x)的定義域上單調(diào)遞增,則稱函數(shù)f(x)具有M性質(zhì).給出下列函數(shù),不具有M性質(zhì)的為()A.f(x)=lnx B.f(x)=x2+1C.f(x)=sinx D.f(x)=x3解析:對于A,f(x)=lnx,令g(x)=exlnx,則g'(x)=exlnx+1x,令h(x)=lnx+1x,則h'(x)=1x-1x2=x-1x2,則h(x)在區(qū)間(0,1)上單調(diào)遞減,在區(qū)間(1,+∞)上單調(diào)遞增,故h(x)≥h(1)=1,所以g'(x)>0,從而g(x對于B,f(x)=x2+1,令g(x)=exf(x)=ex(x2+1),則g'(x)=ex(x2+1)+2xex=ex(x+1)2≥0在R上恒成立,因此g(x)=exf(x)在f(x)的定義域R上單調(diào)遞增,則f(x)=x2+1具有M性質(zhì);對于C,f(x)=sinx,令g(x)=exsinx,則g'(x)=ex(sinx+cosx)=2exsinx+π4,顯然g(x)在f(x)的定義域R上不單調(diào),故f(x)=對于D,f(x)=x3,令g(x)=exf(x)=exx3,則g'(x)=exx3+3exx2=exx2(x+3),當(dāng)x<3時(shí),g'(x)<0,當(dāng)x>3時(shí),g'(x)≥0,因此g(x)=exf(x)在f(x)的定義域R上先單調(diào)遞減后單調(diào)遞增,故f(x)=x3不具有M性質(zhì).故選CD.答案:CD5.已知函數(shù)f(x)=x3+ax2+(2a3)x1.(1)若f(x)的單調(diào)遞減區(qū)間為(1,1),則a的取值集合為;

(2)若f(x)在區(qū)間(1,1)內(nèi)單調(diào)遞減,則a的取值范圍為.

解析:函數(shù)f(x)的導(dǎo)數(shù)f'(x)=3x2+2ax+2a3=(x+1)(3x+2a3).(1)∵函數(shù)f(x)的單調(diào)遞減區(qū)間為(1,1),∴1和1是方程f'(x)=0的兩根,將x=1代入3x+2a3=0,解得a=0,∴a的取值集合為{0}.(2)∵f(x)在區(qū)間(1,1)內(nèi)單調(diào)遞減,∴f'(x)≤0對x∈(1,1)恒成立.又二次函數(shù)y=f'(x)的圖象開口向上,一個零點(diǎn)為1,∴3-2a3∴a的取值范圍為{a|a≤0}.答案:(1){0}(2){a|a≤0}6.設(shè)函數(shù)f'(x)是奇函數(shù)f(x)(x∈R)的導(dǎo)函數(shù),f(1)=0,當(dāng)x>0時(shí),xf'(x)f(x)<0,則使得f(x)>0成立的x的取值范圍是.

解析:∵f(x)(x∈R)為奇函數(shù),且f(1)=0,∴f(0)=0,f(1)=f(1)=0.當(dāng)x≠0時(shí),令g(x)=f(x)x,則g(x)為偶函數(shù),g(1)=g∵當(dāng)x>0時(shí),xf'(x)f(x)<0,∴g'(x)=f(x)x∴g(x)在區(qū)間(0,+∞)上單調(diào)遞減,在區(qū)間(∞,0)上單調(diào)遞增.∴當(dāng)x>0時(shí),要使f(x)>0,則g(x)>0,故0<x<1;當(dāng)x<0時(shí),要使f(x)>0,則g(x)<0,故x<1.綜上,使得f(x)>0成立的x的取值范圍是(∞,1)∪(0,1).答案:(∞,1)∪(0,1)7.若函數(shù)f(x)=12x2alnx在其定義域內(nèi)的一個子區(qū)間(a2,a+2)上不單調(diào),則實(shí)數(shù)a的取值范圍是.解析:函數(shù)f(x)的定義域是(0,+∞),故a2≥0,解得a≥2,而f'(x)=xax,令xax=0,解得x=由題意得a2<a<a+2,解得0≤a<4.因此,a∈[2,4).答案:[2,4)8.已知函數(shù)f(x)=2axx3,x∈(0,1],a>0,若f(x)在區(qū)間(0,1]上是增函數(shù),則a的取值范圍是.

解析:由題意知f'(x)=2a3x2≥0對x∈(0,1]恒成立,所以a≥32x2對x∈(0,1]恒成立因?yàn)閤∈(0,1],所以32x2∈0,32.所以故a的取值范圍是32答案:39.已知函數(shù)f(x)=3xa2x2+lnx(a≠0)在區(qū)間[1,2]上為單調(diào)函數(shù),求a解:函數(shù)f(x)的導(dǎo)數(shù)f'(x)=3a4x+1x(a≠若函數(shù)f(x)在區(qū)間[1,2]上為單調(diào)函數(shù),則當(dāng)x∈[1,2]時(shí),f'(x)=3a4x+1x≥0或f'(x)=3a4x+1x≤0,即3a≥4x1x或3a≤4x1x對x∈[1,2]恒成立.設(shè)h(x)=4x1因?yàn)閔'(x)=4+1x2>0對x∈[1,2]恒成立,所以函數(shù)h(x所以3a≥h(2)或3a≤h(1),即解得a<0或0<a≤25或a≥1故a的取值范圍是(∞,0)∪0,25∪[1,+10.已知函數(shù)f(x)=lnx,g(x)=12ax2+2x(a≠0)(1)若函數(shù)h(x)=f(x)g(x)存在單調(diào)遞減區(qū)間,求a的取值范圍;(2)若函數(shù)h(x)=f(x)g(x)在區(qū)間[1,4]上單調(diào)遞減,求a的取值范圍.解:(1)h(x)=lnx12ax22x(a≠0),則定義域?yàn)?0,+∞),h'(x)=1xax因?yàn)閔(x)在定義域(0,+∞)上存在單調(diào)遞減區(qū)間,所以h'(x