高考文數(shù)一輪夯基作業(yè)本3第三章導(dǎo)數(shù)及其應(yīng)用夯基提能作業(yè)本2

第二節(jié)導(dǎo)數(shù)與函數(shù)的單調(diào)性A組基礎(chǔ)題組1.函數(shù)f(x)=exex,x∈R的單調(diào)遞增區(qū)間是()A.(0,+∞) B.(∞,0) C.(∞,1) D.(1,+∞)2.設(shè)f'(x)是函數(shù)f(x)的導(dǎo)函數(shù),y=f'(x)的圖象如圖所示,則y=f(x)的圖象最有可能是()3.已知函數(shù)f(x)=x2+2cosx,若f'(x)是f(x)的導(dǎo)函數(shù),則函數(shù)y=f'(x)的圖象大致是()4.(2016北京臨川學(xué)校期末)函數(shù)y=12x2A.(0,1) B.(0,+∞) C.(1,+∞)D.(0,2)5.(2016北京臨川學(xué)校期末)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,+∞)6.對于R上可導(dǎo)的任意函數(shù)f(x),若滿足1-A.f(0)+f(2)>2f(1) B.f(0)+f(2)≤2f(1)C.f(0)+f(2)<2f(1) D.f(0)+f(2)≥2f(1)7.(2015北師大附屬實(shí)驗(yàn)中學(xué)期中)已知函數(shù)f(x)=x2sinx,則函數(shù)f(x)在(0,f(0))處的切線方程為;在(0,π)上的單調(diào)遞增區(qū)間為.

8.若函數(shù)f(x)=x312x在區(qū)間(k1,k+1)上不是單調(diào)函數(shù),則實(shí)數(shù)k的取值范圍是.

9.若函數(shù)f(x)=x2exax在R上存在單調(diào)遞增區(qū)間,則實(shí)數(shù)a的取值范圍是.

10.(2017北京海淀二模)已知函數(shù)f(x)=13x3+12x(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)當(dāng)0<a≤52時(shí),求函數(shù)f(x)在區(qū)間[a,aB組提升題組11.(2016北京東城期中)已知定義在R上的函數(shù)y=f(x)的圖象如圖所示,則x·f'(x)>0的解集為()A.(∞,0)∪(1,2) B.(1,2)C.(∞,1) D.(∞,1)∪(2,+∞)12.已知定義域?yàn)镽的函數(shù)f(x)滿足f(4)=3,且對任意的x∈R,總有f'(x)<3,則不等式f(x)<3x15的解集為.

13.已知函數(shù)f(x)=x2x+1alnx,a14.已知函數(shù)f(x)=exlnxaex(a∈R).(1)若f(x)的圖象在點(diǎn)(1,f(1))處的切線與直線y=1e(2)若f(x)在(0,+∞)上是單調(diào)函數(shù),求實(shí)數(shù)a的取值范圍.答案精解精析A組基礎(chǔ)題組1.D2.C由f'(x)的圖象知,當(dāng)x∈(∞,0)時(shí),f'(x)>0,f(x)為增函數(shù),當(dāng)x∈(0,2)時(shí),f'(x)<0,f(x)為減函數(shù),當(dāng)x∈(2,+∞)時(shí),f'(x)>0,f(x)為增函數(shù).故選C.3.A令g(x)=f'(x)=2x2sinx,則g'(x)=22cosx,易知g'(x)≥0,所以函數(shù)f'(x)在R上單調(diào)遞增.4.A函數(shù)y=12x2lnx的定義域?yàn)閧x|x>0},y'=x1x=x2-1x,令x25.D依題意得f'(x)=k1x≥0在(1,+∞)上恒成立,即k≥1x在(1,+∞)∵x>1,∴0<1x6.A當(dāng)x<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,∴當(dāng)x=1時(shí),函數(shù)f(x)取得極小值同時(shí)是最小值,所以f(0)>f(1),f(2)>f(1),則f(0)+f(2)>2f(1).7.答案y=x;π3解析由f(x)=x2sinx,得f'(x)=12cosx,∴f'(0)=12cos0=1,又f(0)=02sin0=0,∴函數(shù)f(x)在(0,f(0))處的切線方程為y=x.由12cosx>0,得cosx<12又∵x∈(0,π),∴π3∴f(x)在(0,π)上的單調(diào)遞增區(qū)間為π38.答案(3,1)∪(1,3)解析由題意知f'(x)=3x212,由f'(x)>0,得函數(shù)的增區(qū)間是(∞,2),(2,+∞),由f'(x)<0,得函數(shù)的減區(qū)間是(2,2),由于函數(shù)在(k1,k+1)上不是單調(diào)函數(shù),所以k1<2<k+1或k1<2<k+1,解得3<k<1或1<k<3.9.答案(∞,2ln22]解析∵f(x)=x2exax,∴f'(x)=2xexa,∵函數(shù)f(x)=x2exax在R上存在單調(diào)遞增區(qū)間,∴f'(x)=2xexa≥0,即a≤2xex有解,令g(x)=2xex,則g'(x)=2ex,令g'(x)=0,解得x=ln2,則當(dāng)x<ln2時(shí),g'(x)>0,g(x)單調(diào)遞增,當(dāng)x>ln2時(shí),g'(x)<0,g(x)單調(diào)遞減,∴當(dāng)x=ln2時(shí),g(x)取得最大值,且g(x)max=g(ln2)=2ln22,∴a≤2ln22.10.解析(1)由f(x)=13x3+12x22x+1得f'(x)=x2+x2=(x+2)(x令f'(x)=0,得x1=2,x2=1,所以x,f'(x),f(x)的變化情況如下表:x(∞,2)2(2,1)1(1,+∞)f'(x)+00+f(x)↗極大值↘極小值↗所以函數(shù)f(x)的單調(diào)遞增區(qū)間為(∞,2),(1,+∞),單調(diào)遞減區(qū)間為(2,1).(2)由f(x)=13x3+12x22x+1可得f(2)=當(dāng)a<2,即2<a≤52時(shí),由(1)可得f(x)在[a,2)和(1,a]上單調(diào)遞增,在(所以,函數(shù)f(x)在區(qū)間[a,a]上的最大值為max{f(2),f(a)},又由(1)可知f(a)≤f52=13所以max{f(2),f(a)}=f(2)=133當(dāng)-a≥-2,a≤1,即0<a≤1時(shí),由(1)可得f(x)在[a,a當(dāng)-a≥所以,函數(shù)f(x)在區(qū)間[a,a]上的最大值為max{f(a),f(a)},因?yàn)閒(a)f(a)=23a(a2或由(1)可知f(a)>f(1)=196,f(a)≤f(2)=53,所以f(a)>f(a)所以max{f(a),f(a)}=f(a)=a33+綜上,當(dāng)2<a≤52時(shí),函數(shù)f(x)在區(qū)間[a,a]上的最大值為13當(dāng)0<a≤2時(shí),函數(shù)f(x)在區(qū)間[a,a]上的最大值為f(a)=a33+B組提升題組11.A不等式x·f'(x)>0等價(jià)于當(dāng)x>0時(shí),f'(x)>0,當(dāng)x<0時(shí),f'(x)<0,所以當(dāng)x>0時(shí),函數(shù)f(x)在R上單調(diào)遞增,此時(shí)1<x<2,當(dāng)x<0時(shí),函數(shù)f(x)在R上單調(diào)遞減,此時(shí)x<0,故選A.12.答案(4,+∞)解析令g(x)=f(x)3x+15,則g'(x)=f'(x)3,由題意知g'(x)<0,所以g(x)在R上是減函數(shù).又g(4)=f(4)3×4+15=0,所以f(x)<3x15的解集為(4,+∞).13.解析由題意知,f(x)的定義域是(0,+∞),f'(x)=1+2x2ax設(shè)g(x)=x2ax+2,一元二次方程g(x)=0的判別式Δ=a28.①當(dāng)Δ<0,即0<a<22時(shí),對一切x>0都有f'(x)>0.此時(shí)f(x)是(0,+∞)上的單調(diào)遞增函數(shù).②當(dāng)Δ=0,即a=22時(shí),f'(x)≥0.此時(shí)f(x)是(0,+∞)上的單調(diào)遞增函數(shù).③當(dāng)Δ>0,即a>22時(shí),方程g(x)=0有兩個(gè)不同的實(shí)根,分別為x1=a-a2-82,x2=f(x),f'(x)隨x的變化情況如下表:x(0,x1)x1(x1,x2)x2(x2,+∞)f'(x)+00+f(x)↗極大值↘極小值↗此時(shí)f(x)在0,a-a214.解析(1)f'(x)=exlnx+ex·1xaex=1x-a+ln所以f'(1)=(1a)e,由題意知(1a)e·1e=(2)由(1)知f'(x)=1x-a+ln若f(x)為單調(diào)遞減函數(shù),則f'(x)≤0在x>0時(shí)恒成立.即1xa+lnx≤0在x>0時(shí)恒即a≥1x+lnx(x>0)恒成立令g(x)=1x則g'(x)=1x2+1x由g'(x)>0,得x>1;由g'(x)<0,得0<x<1.故g