新教材2023-2024學(xué)年高中數(shù)學(xué)第六章導(dǎo)數(shù)及其應(yīng)用綜合訓(xùn)練課件新人教B版選擇性必修第三冊(cè)

第六章綜合訓(xùn)練一、單項(xiàng)選擇題12345678910111213141516171819201.下列導(dǎo)數(shù)運(yùn)算正確的是(

)C.(sin2x)'=cos2x

D.(2x)'=x·2x-1B對(duì)于C,(sin

2x)'=2cos

2x,故C錯(cuò)誤;對(duì)于D,(2x)'=2x·ln

2,故D錯(cuò)誤.故選B.212212345678910111213141516171819202.[2023北京海淀校級(jí)期末]函數(shù)f(x)=x2在區(qū)間[0,2]上的平均變化率等于x=m時(shí)的瞬時(shí)變化率,則m=(

)B由f(x)=x2,得f'(x)=2x,所以f'(m)=2m.因?yàn)閒(x)=x2在區(qū)間[0,2]上的平均變化率等于x=m時(shí)的瞬時(shí)變化率,所以2=2m,解得m=1.故選B.212212345678910111213141516171819203.[2023湖北期中]已知直線l是曲線f(x)=ex的切線,切點(diǎn)的橫坐標(biāo)為-1,直線l與x軸和y軸分別相交于A,B兩點(diǎn),則△OAB的面積為(

)C21221234567891011121314151617181920A212212345678910111213141516171819205.[2023江蘇南京鼓樓期中]已知f(x)=x3+3ax2+bx+a2在x=-1處有極值0,則a+b=(

)A.11或4 B.-4或-11 C.11 D.4C解析

根據(jù)題意,f'(x)=3x2+6ax+b.∵函數(shù)f(x)在x=-1處有極值0,∴f'(-1)=3-6a+b=0且f(-1)=-1+3a-b+a2=0,∴a=1,b=3或a=2,b=9,當(dāng)a=1,b=3時(shí)f'(x)=3x2+6x+3≥0恒成立,函數(shù)f(x)在R上單調(diào)遞增,此時(shí)函數(shù)無極值點(diǎn),不符合題意.21221234567891011121314151617181920當(dāng)a=2,b=9時(shí),f'(x)=3x2+12x+9=3(x+1)(x+3).當(dāng)x<-3或x>-1時(shí),f'(x)>0,f(x)單調(diào)遞增,當(dāng)-3<x<-1時(shí),f'(x)<0,f(x)單調(diào)遞減,∴x=-1是極小值點(diǎn),f(-1)=-1+6-9+4=0,符合題意.∴a=2,b=9,∴a+b=11.故選C.212212345678910111213141516171819206.[2023河南洛陽月考]若函數(shù)f(x)=lnx-ax在區(qū)間(3,4)內(nèi)有極值點(diǎn),則實(shí)數(shù)a的取值范圍是(

)D21221234567891011121314151617181920D21221234567891011121314151617181920212212345678910111213141516171819208.f(x)是定義在R上的偶函數(shù),當(dāng)x<0時(shí),xf'(x)-f'(x)<0,且f(-3)=0,則不等式A.(-∞,-3)∪(3,+∞) B.(-∞,-3)∪(0,3)C.(-3,3) D.(-3,0)∪(3,+∞)B21221234567891011121314151617181920即不等式的解集為(-∞,-3)∪(0,3).21221234567891011121314151617181920二、多項(xiàng)選擇題9.如圖是函數(shù)y=f(x)的導(dǎo)函數(shù)y=f'(x)的圖象,則下面判斷正確的有(

)A.在(-2,1)內(nèi)f(x)是增函數(shù)B.在(3,4)內(nèi)f(x)是減函數(shù)C.在x=-1處取得極小值D.在x=1處取得極大值BC21221234567891011121314151617181920解析

根據(jù)導(dǎo)函數(shù)的正負(fù),得到原函數(shù)的增減性,由圖可得,當(dāng)x變化時(shí),f(x),f'(x)的變化情況如下表:x(-3,-1)-1(-1,2)2(2,4)4(4,+∞)f'(x)-0+0-0+f(x)↘極小值↗極大值↘極小值↗在(3,4)內(nèi)f(x)是減函數(shù),在x=-1處取得極小值.正確的有B,C.2122123456789101112131415161718192010.[2023安徽模擬]已知函數(shù)f(x)=x3-x(x∈R),則(

)A.f(x)是奇函數(shù)AB21221234567891011121314151617181920解析

對(duì)于A,因?yàn)閷?duì)?x∈R,f(-x)=-x3+x=-f(x),所以f(x)是奇函數(shù),故A正確;是函數(shù)f(x)的極小值點(diǎn),極值點(diǎn)是實(shí)數(shù),故D錯(cuò)誤.故選AB.2122123456789101112131415161718192011.[2023湖北武漢青山校級(jí)月考]已知函數(shù)f(x)=-1+lnx,則(

)A.f(x)在點(diǎn)(1,0)處的切線為x軸B.f(x)在(0,+∞)內(nèi)單調(diào)遞減C.x=1為f(x)的極值點(diǎn)D.f(x)的最小值為0ACD21221234567891011121314151617181920故f(x)在點(diǎn)(1,0)處的切線的斜率為f'(1)=0,而f(1)=1-1+ln

1=0,故f(x)在點(diǎn)(1,0)處的切線方程為y-0=0(x-1),即y=0,所以f(x)在點(diǎn)(1,0)處的切線為x軸,A正確;當(dāng)0<x<1時(shí),f'(x)<0,當(dāng)x>1時(shí),f'(x)>0,故f(x)在(0,1)內(nèi)單調(diào)遞減,在(1,+∞)內(nèi)單調(diào)遞增,B錯(cuò)誤;由此可得x=1為f(x)的極小值點(diǎn),C正確;由于在(0,+∞)上f(x)只有一個(gè)極小值點(diǎn),故函數(shù)的極小值也為函數(shù)的最小值,最小值為f(1)=0,D正確,故選ACD.21221234567891011121314151617181920ACD21221234567891011121314151617181920令f'(x)=0得x=e,所以在(0,e)內(nèi),f'(x)>0,f(x)單調(diào)遞增,在(e,+∞)內(nèi),f'(x)<0,f(x)單調(diào)遞減.對(duì)于A,由上可知f(x)在x=e處取得極大值,f(e)=,故A正確;對(duì)于B,因?yàn)楹瘮?shù)f(x)在(0,e)內(nèi)單調(diào)遞增,且f(1)=0,所以函數(shù)f(x)在(0,e)內(nèi)只有一個(gè)零點(diǎn).21221234567891011121314151617181920當(dāng)x∈[e,+∞)時(shí),f(x)>0恒成立,所以函數(shù)f(x)在[e,+∞)內(nèi)沒有零點(diǎn).所以函數(shù)f(x)只有一個(gè)零點(diǎn).因?yàn)閒(x)在(e,+∞)內(nèi)單調(diào)遞減,所以f(3)>f(π)>f(4),2122123456789101112131415161718192021221234567891011121314151617181920所以在(0,1)內(nèi),g'(x)>0,g(x)單調(diào)遞增,在(1,+∞)內(nèi),g'(x)<0,g(x)單調(diào)遞減,所以g(x)max=g(1)=1,所以k>1,故D正確,故選ACD.2122123456789101112131415161718192013.[2023江西豐城校級(jí)月考]函數(shù)f(x)=lnx+的極值點(diǎn)為

.

三、填空題1當(dāng)x∈(0,1)時(shí),f'(x)<0,此時(shí)函數(shù)f(x)單調(diào)遞減;當(dāng)x∈(1,+∞)時(shí),f'(x)>0,此時(shí)函數(shù)f(x)單調(diào)遞增.∴1是函數(shù)f(x)的極小值點(diǎn).2122123456789101112131415161718192014.[2023河南周口項(xiàng)城月考]已知函數(shù)f(x)=asinx+cosx在區(qū)間

上單調(diào)遞減,則實(shí)數(shù)a的取值范圍是

.

212212345678910111213141516171819202122123456789101112131415161718192015.用長為18cm的鋼條圍成一個(gè)長方體形狀的框架,要求長方體的長與寬之比為2∶1,則該長方體的長、寬、高分別為

時(shí),其體積最大.

2122123456789101112131415161718192016.若函數(shù)f(x)=mlnx-x3+x2-4x+4在(2,+∞)內(nèi)單調(diào)遞減,則實(shí)數(shù)m的取值范圍為

.

(-∞,20]由于f(x)在(2,+∞)內(nèi)單調(diào)遞減,即f'(x)≤0在(2,+∞)內(nèi)恒成立,設(shè)g(x)=3x3-3x2+4x(x>2),則m≤3x3-3x2+4x在(2,+∞)內(nèi)恒成立,21221234567891011121314151617181920即m≤g(x)min在(2,+∞)內(nèi)恒成立,g'(x)=9x2-6x+4,知Δ=36-4×9×4<0,∴x∈(2,+∞)時(shí),g'(x)>0,g(x)單調(diào)遞增,∴m≤g(x)min=g(2)=3×23-3×22+4×2=20,∴m≤20,即實(shí)數(shù)m的取值范圍為(-∞,20].21221234567891011121314151617181920四、解答題17.設(shè)函數(shù)f(x)=2x3-3(a+1)x2+6ax+8,其中a∈R.已知f(x)在x=3處取得極值.(1)求f(x)的解析式;(2)求曲線y=f(x)在點(diǎn)A(1,16)處的切線方程.解(1)f'(x)=6x2-6(a+1)x+6a.∵f(x)在x=3處取得極值,∴f'(3)=6×9-6(a+1)×3+6a=0,解得a=3.∴f(x)=2x3-12x2+18x+8.(2)A點(diǎn)在f(x)上,由(1),可知f'(x)=6x2-24x+18,f'(1)=6-24+18=0,∴切線方程為y=16.2122123456789101112131415161718192018.[2023山西太原期末]已知函數(shù)f(x)=(x-2)ex.(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)求f(x)在[-1,2]上的值域.解

(1)函數(shù)f(x)=(x-2)ex,則f'(x)=(x-1)ex.當(dāng)x>1時(shí),f'(x)>0,當(dāng)x<1時(shí),f'(x)<0,故函數(shù)f(x)在(1,+∞)內(nèi)單調(diào)遞增,在(-∞,1)內(nèi)單調(diào)遞減.(2)由(1)可得函數(shù)f(x)在(1,2]上單調(diào)遞增,在[-1,1)內(nèi)單調(diào)遞減,則f(x)在[-1,2]上的最大值f(x)max=f(2)=0,最小值f(x)min=f(1)=-e,故f(x)在[-1,2]上的值域?yàn)閇-e,0].2122123456789101112131415161718192019.已知函數(shù)f(x)=x2-2lnx.(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)求證:當(dāng)x>2時(shí),f(x)>3x-4.解

(1)依題意,知函數(shù)的定義域?yàn)閧x|x>0},由f'(x)>0,得x>1;由f'(x)<0,得0<x<1.∴f(x)的單調(diào)遞增區(qū)間為[1,+∞),單調(diào)遞減區(qū)間為[0,1].21221234567891011121314151617181920(2)設(shè)g(x)=f(x)-3x+4=x2-2ln

x-3x+4,∵當(dāng)x>2時(shí),g'(x)>0,∴g(x)在(2,+∞)內(nèi)為增函數(shù),∴g(x)>g(2)=4-2ln

2-6+4=2-2ln

2>0,∴當(dāng)x>2時(shí),x2-2ln

x>3x-4,即當(dāng)x>2時(shí),f(x)>3x-4.得證.2122123456789101112131415161718192020.甲、乙兩地相距400千米,汽車從甲地勻速行駛到乙地,速度不得超過100千米/時(shí),已知該汽車每小時(shí)的運(yùn)輸成本P(單位:元)關(guān)于速度v(單位:千(1)求全程運(yùn)輸成本Q(單位:元)關(guān)于速度v的函數(shù)關(guān)系式.(2)為使全程運(yùn)輸成本最少,汽車應(yīng)以多大速度行駛?并求此時(shí)全程運(yùn)輸成本的最小值.21221234567891011121314151617181920令Q'=0,則v=0(舍去)或v=80,當(dāng)0<v<80時(shí),Q'<0,當(dāng)80<v≤100時(shí),Q'>0,2122123456789101112131415161718192021.已知函數(shù)f(x)=aex-1-lnx+lna.(1)當(dāng)a=e時(shí),求曲線y=f(x)在點(diǎn)(1,f(1))處的切線與兩坐標(biāo)軸圍成的三角形的面積;(2)若f(x)≥1,求a的取值范圍.2122(1)當(dāng)a=e時(shí),f(x)=ex-ln

x+1,f'(1)=e-1,曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y-(e+1)=(e-1)(x-1),即y=(e-1)x+2.12345678910111213141516171819202122(2)由題意a>0,當(dāng)0<a<1時(shí),f(1)=a+ln

a<1.當(dāng)x∈(0,1)時(shí),f'(x)<0;當(dāng)x∈(1,+∞)時(shí),f'(x)>0.所以當(dāng)x=1時(shí),f(x)取得最小值,最小值為f(1)=1,從而f(x)≥1.當(dāng)a>1時(shí),f(x)=aex-1-ln

x+ln

a≥ex-1-ln

x≥1.綜上,a的取值范圍是[1,+∞).1234567891011121314151617181922.已知函數(shù)f(x)=ex-a(x+