第二章導(dǎo)數(shù)及其應(yīng)用一、單項(xiàng)選擇題1.已知函數(shù)f(x)=x2-2x,則f(x)在區(qū)間[1,2]上的平均變化率為()A.-1 B.3 C.2 D.42.若某種樹木的高度h(t)(單位:m)與生長(zhǎng)時(shí)間t(單位:年)之間的關(guān)系是h(t)=3t3+12,則h'(2)的實(shí)際意義為()A.t=2時(shí)樹木高度的瞬時(shí)變化率B.t=2時(shí)樹木的高度C.t=2是h(t)的極值點(diǎn)D.t=2時(shí)h(t)取最大值3.已知函數(shù)g(x)=cos6x-π3+ln(2x+5)-x,則[g'(0)]'等于().A.-12 B.0 C.35 D4.下列各式正確的是().A.(sina)'=cosa(a為常數(shù)) B.(cosx)'=sinxC.(sinx)'=cosx D.(x-5)'=-12x-5.若函數(shù)f(x)=13x3-f'(1)·x2-x,則f'(1)的值為()A.0 B.2 C.1 D.-16.函數(shù)f(x)=2x+ex-4的單調(diào)遞增區(qū)間為().A.(-∞,+∞) B.(-∞,1) C.(0,+∞) D.-13,47.函數(shù)f(x)=x3+bx2+cx+d的大致圖象如下圖所示,則x12+x2A.89 B.109 C.1698.已知函數(shù)f(x)=ax+lnx(a∈R)有兩個(gè)不相同的零點(diǎn),則a的取值范圍為()A.0,1e B.0,1e C.1e,+∞ D.(e,+∞)二、多項(xiàng)選擇題9.如圖是導(dǎo)函數(shù)y=f'(x)的圖象,則下列說法錯(cuò)誤的是().A.(-1,3)為函數(shù)y=f(x)的單調(diào)遞增區(qū)間B.(0,3)為函數(shù)y=f(x)的單調(diào)遞減區(qū)間C.函數(shù)y=f(x)在x=0處取得極大值D.函數(shù)y=f(x)在x=5處取得極小值10.設(shè)函數(shù)f(x)=13x3-x2+x的導(dǎo)函數(shù)為f'(x),則()A.f'(1)=0B.x=1是函數(shù)f(x)的極值點(diǎn)C.f(x)存在兩個(gè)零點(diǎn)D.f(x)在(1,+∞)上單調(diào)遞增11.(2023·新高考Ⅰ卷,10)已知函數(shù)f(x)=x3-x+1,則().A.f(x)有兩個(gè)極值點(diǎn)B.f(x)有三個(gè)零點(diǎn)C.點(diǎn)(0,1)是曲線y=f(x)的對(duì)稱中心D.直線y=2x是曲線y=f(x)的切線12.已知函數(shù)f(x)是定義在R上的奇函數(shù),當(dāng)x>0時(shí),f(x)=e-x(x-1).則下列結(jié)論正確的是().A.f(-1)=0B.函數(shù)f(x)有3個(gè)零點(diǎn)C.若關(guān)于x的方程f(x)=m有解,則實(shí)數(shù)m的取值范圍是{m|f(-2)≤m≤f(2)}D.?x1,x2∈R,|f(x2)-f(x1)|<2恒成立三、填空題13.曲線y=g(x)=2-12x2與y=h(x)=14x3-2在交點(diǎn)處的兩切線的斜率之和為14.法國(guó)數(shù)學(xué)家拉格朗日于1797年在其著作《解析函數(shù)論》中給出一個(gè)定理:如果函數(shù)y=f(x)滿足如下條件:(1)在閉區(qū)間[a,b]上是連續(xù)不斷的;(2)在區(qū)間(a,b)上都有導(dǎo)數(shù).那么在區(qū)間(a,b)上至少存在一個(gè)實(shí)數(shù)t,使得f(b)-f(a)=f'(t)(b-a),其中t稱為“拉格朗日中值”.函數(shù)g(x)=x2在區(qū)間[0,1]上的“拉格朗日中值”t=.

15.在直徑為d的圓木中,截取一個(gè)具有最大抗彎強(qiáng)度的長(zhǎng)方體梁,則矩形面的長(zhǎng)為.(強(qiáng)度與bh2成正比,其中h為矩形的長(zhǎng),b為矩形的寬)

16.已知函數(shù)f(x)=x-ln(x+a),若a=2,則f'(0)=;若f(x)的最小值為0,其中a>0,則a的值為.

四、解答題17.已知函數(shù)f(x)=x3-1x+2(1)求函數(shù)f(x)在區(qū)間[1,4]上的平均變化率;(2)求曲線y=f(x)在點(diǎn)(1,2)處切線的斜率.18.求函數(shù)y=f(x)=x4-2x2+2在區(qū)間[-3,3]上的最大值和最小值.19.設(shè)函數(shù)f(x)=x3-3ax2+3bx的圖象與直線12x+y-1=0相切于點(diǎn)(1,-11).(1)求實(shí)數(shù)a,b的值;(2)討論函數(shù)f(x)的單調(diào)性.20.已知函數(shù)f(x)=1-xax+(1)若函數(shù)f(x)在區(qū)間[1,+∞)上單調(diào)遞增,求正實(shí)數(shù)a的取值范圍;(2)當(dāng)a>0時(shí),討論函數(shù)f(x)在區(qū)間12,2上的單調(diào)性.21.已知函數(shù)f(x)=lnx+ax2+(2a+1)x(a∈R).(1)討論f(x)的單調(diào)性;(2)當(dāng)a<0時(shí),求證:f(x)≤-34a-22.函數(shù)f(x)=ex-a(x2(1)當(dāng)a=e時(shí),試求f(x)的單調(diào)區(qū)間;(2)若f(x)在區(qū)間(0,1)上有極值,試求a的取值范圍.參考答案一、選擇題1.答案:D解析:平均變化率為f(2)-f(1)2-1=2.答案:A3.答案:B解析:∵g'(0)為一個(gè)常數(shù),∴[g'(0)]'=0.4.答案:C解析:由導(dǎo)數(shù)公式知選項(xiàng)A中(sina)'=0;選項(xiàng)B中(cosx)'=-sinx;選項(xiàng)D中(x-5)'=-5x-6.5.答案:A解析:f'(x)=x2-2f'(1)·x-1,則f'(1)=12-2f'(1)×1-1,解得f'(1)=0.6.答案:A解析:函數(shù)f(x)的定義域?yàn)镽.∵f'(x)=2+ex>0,∴f(x)在R上單調(diào)遞增.7.答案:C解析:由函數(shù)f(x)的圖象可知,f∴-解得b=-1,c=-2,d=0.∴f(x)=x3-x2-2x,從而f'(x)=3x2-2x-2.∵x1與x2是函數(shù)f(x)的極值點(diǎn),∴x1,x2是方程f'(x)=0的根,∴x1+x2=23,x1x2=-2∴x12+x22=(x1+x2)2-2x8.答案:A解析:函數(shù)f(x)的定義域?yàn)?0,+∞),且f'(x)=x-①當(dāng)a≤0時(shí),f'(x)>0恒成立,所以函數(shù)f(x)在區(qū)間(0,+∞)上為增函數(shù),不符合題意;②當(dāng)a>0時(shí),令f'(x)=0,得x=a;當(dāng)0<x<a時(shí),f'(x)<0,函數(shù)f(x)在區(qū)間(0,a)上單調(diào)遞減;當(dāng)x>a時(shí),f'(x)>0,函數(shù)f(x)在區(qū)間(a,+∞)上單調(diào)遞增,所以f(x)在x=a處取得極小值,也是最小值f(a),依題意知f(a)<0.由f(a)=1+lna<0,得0<a<1e由于a<1,f(1)=a>0,函數(shù)f(x)在區(qū)間(a,+∞)上單調(diào)遞增,故f(x)在區(qū)間(a,+∞)上有唯一的一個(gè)零點(diǎn).因?yàn)?<a<1e所以0<a2<a<1ef(a2)=1a+lna2=1a+2ln令g(a)=1a+2lna當(dāng)0<a<1e時(shí),g'(a)=-1a所以f(a2)=g(a)>g1e=e-2>0又f(a)<0,函數(shù)f(x)在區(qū)間(0,a)上單調(diào)遞減,所以f(x)在區(qū)間(0,a)上有唯一的一個(gè)零點(diǎn).綜上,當(dāng)a∈0,1e時(shí),函數(shù)f(x二、選擇題9.答案:BC解析:由題圖可知,當(dāng)x<-1時(shí),f'(x)<0,故f(x)單調(diào)遞減;當(dāng)x∈(-1,3)時(shí),f'(x)>0,故f(x)單調(diào)遞增;當(dāng)x∈(3,5)時(shí),f'(x)<0,故f(x)單調(diào)遞減;當(dāng)x>5時(shí),f'(x)>0,故f(x)單調(diào)遞增,且f'(-1)=0,f'(3)=0,f'(5)=0,則該函數(shù)在x=-1和x=5處取得極小值;在x=3處取得極大值.10.答案:AD解析:f'(x)=x2-2x+1=(x-1)2≥0,所以函數(shù)f(x)在R上單調(diào)遞增,所以函數(shù)不存在極值點(diǎn),故B錯(cuò)誤,D正確;f'(1)=0,故A正確;f(x)=13x3-x2+x=0,得x(x2-3x+3)=0,方程x2-3x+3=0中,Δ=9-12<0,所以x2-3x+3>0恒成立,即方程只有一個(gè)實(shí)數(shù)根,即x=0,故C錯(cuò)誤11.答案AC解析∵f(x)=x3-x+1,∴f'(x)=3x2-1.由3x2-1=0,得x=33或x=-33.∴f(x)有2個(gè)極值點(diǎn)-33與33,且在區(qū)間-∞,-33內(nèi)單調(diào)遞增,在區(qū)間-33,33內(nèi)單調(diào)遞減,在區(qū)間33,+∞內(nèi)單調(diào)遞增.又f33=1-239>0,當(dāng)x=-2時(shí),f(-2)=-5<0,而f-33>f33>0,∴f(x)只在區(qū)間-2,-∵f(x)+f(-x)=2,∴點(diǎn)(0,1)是曲線y=f(x)的對(duì)稱中心.由f'(x)=3x2-1=2,解得x=±1,∴曲線y=f(x)在點(diǎn)(1,1)處的切線方程為y-1=2(x-1),即y=2x-1;曲線y=f(x)在點(diǎn)(-1,1)處的切線方程為y-1=2(x+1),即y=2x+3.∴直線y=2x與曲線y=f(x)不相切.故選AC.12.答案:ABD解析:f(-1)=-f(1)=0,故A正確.當(dāng)x>0時(shí),f(x)=x-1ex,所以f'(x令f'(x)=0,解得x=2,當(dāng)0<x<2時(shí),f'(x)>0;當(dāng)x>2時(shí),f'(x)<0,所以函數(shù)f(x)在(0,2)上單調(diào)遞增,在(2,+∞)上單調(diào)遞減,故當(dāng)x=2時(shí),函數(shù)f(x)取得極大值e-2,且e-2>0.當(dāng)0<x<2時(shí),f(0)·f(2)<0,又f(1)=0,故函數(shù)f(x)在(0,2)上僅有一個(gè)零點(diǎn)1.當(dāng)x>2時(shí),f(x)=x-1ex>0,所以函數(shù)f(x)在(2,+∞)上沒有零點(diǎn),所以函數(shù)f(x)在(0,+∞)上僅有一個(gè)零點(diǎn),函數(shù)f(x)是定義在R上的奇函數(shù),故函數(shù)f(x)在(-∞,0)上僅有一個(gè)零點(diǎn)-1.又f(0)=0,故函數(shù)f(x)在R上有3個(gè)零點(diǎn).故B正確.作出函數(shù)f((第12題)由圖可知,若關(guān)于x的方程f(x)=m有解,則實(shí)數(shù)m的取值范圍是{m|-1<m<1}.故C錯(cuò)誤.由圖可知,對(duì)任意x1,x2∈R,|f(x2)-f(x1)|<|1-(-1)|=2.故D正確.三、填空題13.答案1解析由y=2-12x2,y=14x3-2,得x3+2x2解得x=2.所以兩曲線只有一個(gè)交點(diǎn),交點(diǎn)坐標(biāo)為(2,0).設(shè)兩條切線的斜率分別為k1,k2,則k1=g'(2)=-2,k2=h'(2)=3.所以k1+k2=1.14.答案1解析因?yàn)間(x)=x2,所以g'(x)=2x,結(jié)合“拉格朗日中值”的定義可得g'(t)=g(1)-g(0)1-0=1,所以g'(15.答案63解析截面如圖所示,設(shè)抗彎強(qiáng)度系數(shù)為k,抗彎強(qiáng)度為ω,(第15題)則ω=kbh2(k>0).∵h(yuǎn)2=d2-b2,∴ω=kb(d2-b2)=-kb3+kd2b,ω'=-3kb2+kd2.令ω'=0,則b2=d2解得b=33d或b=-33d(舍去由函數(shù)ω的單調(diào)性,得當(dāng)b=33d時(shí),函數(shù)ω取得極大值也是最大值,此時(shí),h=d216.答案12解析f(x)的定義域?yàn)?-a,+∞),f'(x)=1-1x當(dāng)a=2時(shí),f'(x)=1-1x故f'(0)=1-12由f'(x)=0,解得x=1-a>-a.當(dāng)-a<x<1-a時(shí),f'(x)<0,f(x)在區(qū)間(-a,1-a)上單調(diào)遞減;當(dāng)x>1-a時(shí),f'(x)>0,f(x)在區(qū)間(1-a,+∞)上單調(diào)遞增.因此,f(x)在x=1-a處取得最小值,由題意知f(1-a)=1-a=0,故a=1.四、解答題17.解(1)平均變化率為f((2)由f(x)=x3-1x+2,得f'(x)=3x2+12x-32,所以切線的斜率k=f'(1)=318.解由y=x4-2x2+2,得y'=4x3-4x=4x(x+1)(x-1),令y'=0,得x=-1或x=0或x=1.∵f(-1)=1,f(0)=2,f(1)=1,f(-3)=65,f(3)=65,∴函數(shù)y=f(x)=x4-2x2+2在區(qū)間[-3,3]上的最大值為65,最小值為1.19.解:(1)對(duì)函數(shù)f(x)求導(dǎo),得f'(x)=3x2-6ax+3b.因?yàn)楹瘮?shù)f(x)的圖象與直線12x+y-1=0相切于點(diǎn)(1,-11),所以f(1)=-11,f'(1)=-12,即1解得a(2)由a=1,b=-3,得f'(x)=3x2-6ax+3b=3(x2-2x-3)=3(x+1)(x-3).令f'(x)>0,解得x<-1或x>3;令f'(x)<0,解得-1<x<3.所以函數(shù)f(x)在區(qū)間(-∞,-1),(3,+∞)上單調(diào)遞增;在區(qū)間(-1,3)上單調(diào)遞減.20.解:(1)由已知得函數(shù)f(x)的定義域?yàn)?0,+∞),f'(x)=ax-1ax依題意ax-1ax2≥0對(duì)x∈[1,+∞)恒成立,即ax-1≥0對(duì)x∈[1,+∞)恒成立,也即a≥1x對(duì)x∈[1,+∞)恒成立,故a≥1x(2)f'(x)=ax-1a令f'(x)<0,得x<1a令f'(x)>0,得x>1a故函數(shù)f(x)在區(qū)間0,1a上單調(diào)遞減,在區(qū)間①當(dāng)1a≤12,即a≥2時(shí),函數(shù)f(x)在區(qū)間12②當(dāng)1a≥2,即0<a≤12時(shí),函數(shù)f(x)在區(qū)間12,2上單調(diào)遞減③當(dāng)12<1a<2,即12<a<2時(shí),函數(shù)f(x)在區(qū)間12,1a上單調(diào)遞減,21.(1)解由題意得函數(shù)f(x)的定義域?yàn)?0,+∞),f'(x)=1x+2ax+2a+1=(若a≥0,則f'(x)>0,故f(x)在(0,+∞)上單調(diào)遞增;若a<0,則當(dāng)x∈0,-12a時(shí),f'(x)>0;當(dāng)x∈-12a,+∞時(shí),f'(x)<0.故函數(shù)f(x)在區(qū)間0,-12a上單調(diào)遞增,在區(qū)間-12a,+∞上單調(diào)遞減.(2)證明:由(1)知,當(dāng)a<0時(shí),函數(shù)f(x)在x=-12a處取得極大值,也是最大值,所以f(x)≤f-12a,最大值為f-12a設(shè)g(a)=f-12a--34a-2=ln-12a+12則g'(a)=-1令g'(a)=0,得a=-12令g'(a)<0,得-12<a<0;令g'(a)>0,得a<-12,所以函數(shù)g(a)在區(qū)間-∞,-12上單調(diào)遞增在區(qū)間-12,0上單調(diào)遞減.所以函數(shù)g(a)在a=-12處取得極大值,也是最大值,最大值為g-12所以g(a)≤0在a<0時(shí)恒成立,即f-12a≤-34a-2所以f(x)≤f-12a≤-34a-22.解:(1)由題意知