新教材蘇教版選擇性必修第一冊(cè)第5章導(dǎo)數(shù)及其應(yīng)用專題強(qiáng)化練12導(dǎo)數(shù)與函數(shù)的單調(diào)性及其應(yīng)用作業(yè)

專題強(qiáng)化練12導(dǎo)數(shù)與函數(shù)的單調(diào)性及其應(yīng)用一、選擇題1.(2020江蘇徐州三校高二下聯(lián)考,)已知f(x)=12x2-cosx,f'(x)為f(x)的導(dǎo)函數(shù),則f'(x)的大致圖象是()2.(2020廣西百色高二上期末,)定義域?yàn)镽的函數(shù)f(x)滿足f(-3)=6,且f'(x)>x2+1對(duì)x∈R恒成立,則f(x)>13x3+15的解集為 ()A.(-3,+∞)B.(-∞,-3)C.(-∞,3)D.(3,+∞)3.(2020江蘇南京秦淮中學(xué)高二下階段測(cè)試,)已知函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),在(0,+∞)上滿足xf'(x)>f(x),則下列結(jié)論一定成立的是 ()A.2019f(2020)>2020f(2019)B.f(2019)>f(2020)C.2019f(2020)<2020f(2019)D.f(2019)<f(2020)4.(2020江蘇揚(yáng)州大學(xué)附屬中學(xué)高二下階段測(cè)試,)若函數(shù)f(x)=x+asin2x在0,π4上單調(diào)遞增,則a的取值范圍是 (A.-12C.-12,5.(多選)(2020江蘇鎮(zhèn)江呂叔湘中學(xué)高二下月考,)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)=ex(x-1),則下列判斷正確的是 ()x<0時(shí),f(x)=-e-x(x+1)B.f(x)<0的解集為(-∞,-1)∪(0,1)f(x)有3個(gè)零點(diǎn)6.(多選)()已知函數(shù)f(x)=xlnx,若0<x1<x2,則下列結(jié)論正確的是 ()A.x2f(x1)<x1f(x2)B.x1+f(x1)<x2+f(x2)C.f(xD.當(dāng)lnx>-1時(shí),x1f(x1)+x2f(x2)>2x2f(x1)二、填空題7.(2020江蘇江陰高級(jí)中學(xué)高二下月考,)已知函數(shù)f(x)=3x-2sinx,若f(a2-3a)+f(3-a)<0,則實(shí)數(shù)a的取值范圍是.?8.(2021江蘇南通如皋中學(xué)高二上階段測(cè)試,)已知函數(shù)f(x)=(x2-a)e-x的圖象過(guò)點(diǎn)(3,0),若函數(shù)f(x)在(m,m+1)上是增函數(shù),則實(shí)數(shù)m的取值范圍為.?三、解答題9.(2020江蘇連云港高二下期末,)已知函數(shù)f(x)=xln(x+1)-ax2+1(a∈R).(1)若h(x)=f'(x)在(-1,+∞)上是單調(diào)遞增函數(shù),求實(shí)數(shù)a的取值范圍;(2)若f(x)在(-1,+∞)上是單調(diào)遞減函數(shù),求實(shí)數(shù)a的取值集合.10.(2020山東德州高三上期末,)已知函數(shù)f(x)=lnx+ax2-(a+2)x+2(a為常數(shù)).(1)若曲線y=f(x)在(1,f(1))處的切線與直線x+3y=0垂直,求a的值;(2)若a>0,討論函數(shù)f(x)的單調(diào)性;(3)若a為正整數(shù),且函數(shù)f(x)恰有兩個(gè)零點(diǎn),求a的值.答案全解全析專題強(qiáng)化練12導(dǎo)數(shù)與函數(shù)的單調(diào)性及其應(yīng)用一、選擇題1.A依題意得f'(x)=x+sinx,令h(x)=x+sinx,則h'(x)=1+cosx.由于f'(0)=0,因此排除C選項(xiàng).由于h'(0)=1+1=2>0,因此f'(x)在x=0處的導(dǎo)數(shù)大于零,故排除B,D選項(xiàng).故選A.2.A構(gòu)造函數(shù)F(x)=f(x)-13x3則F'(x)=f'(x)-x2,由f'(x)>x2+1對(duì)x∈R恒成立得,F'(x)>1>0,∴F(x)是R上的單調(diào)遞增函數(shù).又f(-3)=6,∴F(-3)=f(-3)-13×(-3)3=15∴f(x)>13x3+15?f(x)-13x3>15?F(x)>由F(x)的單調(diào)性知此不等式的解集為(-3,+∞),故選A.3.A令g(x)=f(x)x(x>0),則g'(由已知得,當(dāng)x>0時(shí),g'(x)>0,故函數(shù)g(x)在(0,+∞)上是增函數(shù),所以g(2020)>g(2019),即f(2020)2020>f(2019)2019,所以20194.C若函數(shù)f(x)=x+asin2x在0,π4上單調(diào)遞增,則f'(x)=1+2acos2x≥0在所以2a≥-1cos2x在又當(dāng)x∈0,π4時(shí),0<cos2x≤1,所以-1cos2x≤-1,即-1cos2x有最大值-1,則a≥-5.BD當(dāng)x<0時(shí),-x>0,所以f(-x)=e-x(-x-1),又f(x)是定義在R上的奇函數(shù),所以f(-x)=-f(x),所以f(x)=e-x(x+1),故A錯(cuò)誤;易得f(x)=e令f(x)<0,可得x<0,e-x(x+1)<0或x>0,ex(x-1)<0,解得x<-1或0<x<1,所以f(x)<0的解集為(-∞,-1)∪(0,1),故B正確;當(dāng)x>0時(shí),f(x)=ex(x-1),f'(x)=ex(x-1)+ex=xex>0,所以f(x)在(0,+∞)上為增函數(shù),又因?yàn)閒(x)是定義在R上的奇函數(shù),所以函數(shù)f(x)在(-∞,0),(0,+∞)上單調(diào)遞增,但當(dāng)x>0且x→0時(shí),f(x)→-1;當(dāng)x<0且x→0時(shí),f(x)→1;當(dāng)x=0時(shí),f(x)=0,所以f(x)在R上不單調(diào)遞增,故C錯(cuò)誤;令f(x)=0,6.AD對(duì)于A,令g(x)=f(x)x=lnx,易知g(x)在(0,+∞)上單調(diào)遞增,所以g(x1)<g(x2),即f(x1)x1<f(x2)x2,所以x2對(duì)于B,令h(x)=x+f(x)=x+xlnx,則h'(x)=1+lnx+1=2+lnx,當(dāng)x∈(0,e-2)時(shí),h'(x)<0,h(x)單調(diào)遞減,當(dāng)x∈(e-2,+∞)時(shí),h'(x)>0,h(x)單調(diào)遞增,所以x1+f(x1)<x2+f(x2)不恒成立,故B錯(cuò)誤;對(duì)于C,由f(x)=xlnx可得f'(x)=lnx+1,當(dāng)x∈(0,e-1)時(shí),f'(x)<0,f(x)單調(diào)遞減,當(dāng)x∈(e-1,+∞)時(shí),f'(x)>0,f(x)單調(diào)遞增,所以f(x1)-f(x對(duì)于D,當(dāng)x∈(e-1,+∞),即lnx>-1時(shí),f(x)單調(diào)遞增,由e-1<x1<x2得f(x2)>f(x1),所以x1[f(x1)-f(x2)]-x2[f(x1)-f(x2)]=(x1-x2)[f(x1)-f(x2)]>0,則x1f(x1)+x2f(x2)>x1f(x2)+x2f(x1),由A中分析可知x2f(x1)<x1f(x2),利用不等式的傳遞性可得x1f(x1)+x2f(x2)>2x2·f(x1),故D正確.故選AD.二、填空題7.答案(1,3)解析易知f(x)=3x-2sinx的定義域?yàn)镽,∴f(-x)=-3x+2sinx=-f(x),∴函數(shù)f(x)為奇函數(shù),易得f'(x)=3-2cosx>0,∴函數(shù)f(x)在R上單調(diào)遞增,∵f(a2-3a)+f(3-a)<0,即f(a2-3a)<-f(3-a)=f(a-3),∴a2-3a<a-3,解得1<a<3.8.答案[-1,2]解析因?yàn)楹瘮?shù)f(x)=(x2-a)e-x的圖象過(guò)點(diǎn)(3,0),所以3-a=0,解得a=3,所以函數(shù)f(x)=(x2-3)e-x,f'(x)=-(x-3)·(x+1)e-x,令f'(x)≥0,可得-1≤x≤3,所以函數(shù)f(x)的單調(diào)遞增區(qū)間為[-1,3],因?yàn)楹瘮?shù)f(x)在(m,m+1)上是增函數(shù),所以m≥-1,m+1≤3,解得-1≤m三、解答題9.解析(1)因?yàn)閒(x)=xln(x+1)-ax2+1,所以f'(x)=ln(x+1)+xx+1-2又h(x)=f'(x)在(-1,+∞)上是單調(diào)遞增函數(shù),所以h'(x)≥0在(-1,+∞)上恒成立,且無(wú)連續(xù)區(qū)間使h'(x)恒為0,即h'(x)=1x+1+1(x+1)2-2所以1x+1+1(x+1)2≥2a在(-1,+∞)上恒成立,因?yàn)閠+t2>0,所以2a≤0,即a≤0.故實(shí)數(shù)a的取值范圍為(-∞,0].(2)因?yàn)閒(x)=xln(x+1)-ax2+1,所以f'(x)=ln(x+1)+xx+1-2又f(x)在(-1,+∞)上是單調(diào)遞減函數(shù),所以f'(x)≤0在(-1,+∞)上恒成立,且無(wú)連續(xù)區(qū)間使f'(x)恒為0.因?yàn)閒'(0)=0,所以f'(0)是f'(x)的一個(gè)極大值,令g(x)=f'(x),則g'(x)=1x+1+1(x=-2則g'(0)=0,即2-2a=0,解得a=1.當(dāng)a=1時(shí),有g(shù)'(x)=1x+1+1(當(dāng)x∈(-1,0)時(shí),g'(x)>0,則f'(x)在(-1,0)上單調(diào)遞增;當(dāng)x∈(0,+∞)時(shí),g'(x)<0,則f'(x)在(0,+∞)上單調(diào)遞減.所以f'(x)≤f'(0)=0,所以當(dāng)a=1時(shí),f(x)在(-1,+∞)上單調(diào)遞減.故實(shí)數(shù)a的取值集合為{1}.10.解析(1)由題意知x>0,f'(x)=1x+2ax-(a+2)=(ax-1)(2由于曲線y=f(x)在(1,f(1))處的切線與直線x+3y=0垂直,所以f'(1)·-1所以f'(1)=a-1=3,解得a=4.(2)因?yàn)閍>0,所以1a>0①若0<a<2,則1a>1當(dāng)0<x<12或x>1a時(shí),f'(當(dāng)12<x<1a時(shí),f'(所以函數(shù)f(x)在0,12和1a,+②若a=2,則1a=12,此時(shí)f'(x)≥0在(0,+∞)上恒成立,則函數(shù)f(x)在(0,+∞)③若a>2,則1a<1當(dāng)0<x<1a或x>12時(shí),f'(當(dāng)1a<x<12時(shí),f'(所以函數(shù)f(x)在0,1a和12,+綜上,當(dāng)0<a<2時(shí),f(x)在0,12和1a,+∞上單調(diào)遞增,在12,1a上單調(diào)遞減;當(dāng)a=2時(shí),f(x)在(0,+∞)上單調(diào)遞增;當(dāng)a>2時(shí),f(x(3)因?yàn)閍為正整數(shù),所以若0<a<2,則a=1,f(x)=lnx+x2-3x+2,由(2)知此時(shí)函數(shù)f(x)在0,12和(1,+∞)上單調(diào)遞增,在又f(1)=0,所以函數(shù)f(x)在區(qū)間12,+∞內(nèi)有且僅有一個(gè)零點(diǎn),f又f(e-2)=e-4-3e-2=e-2(e