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3.2導(dǎo)數(shù)與函數(shù)單調(diào)性A組基礎(chǔ)題組1.函數(shù)y=4x2+1x的單調(diào)遞增區(qū)間為()A.(0,+)B.12,+C.(-,-1)D.-,-12答案B由y=4x2+1x得y=8x-1x2,令y0,即8x-1x20,解得x12,函數(shù)y=4x2+1x在12,+上單調(diào)遞增.故選B.2.已知m是實(shí)數(shù),函數(shù)f(x)=x2(x-m),若f (-1)=-1,則函數(shù)f(x)的單調(diào)增區(qū)間是() A.-43,0B.0,43C.-,-43,(0,+)D.-,-43(0,+)答案C由題意得f (x)=3x2-2mx,f (-1)=3+2m=-1,解得m=-2,f (x)=3x2+4x,令f (x)0,解得x0,故f(x)的單調(diào)增區(qū)間為-,-43,(0,+).3.已知函數(shù)f(x)=x2+2cos x,若f (x)是f(x)的導(dǎo)函數(shù),則函數(shù)f (x)的圖象大致是()答案A令g(x)=f (x)=2x-2sin x,則g(x)=2-2cos x,易知g(x)0,所以函數(shù)f (x)在R上單調(diào)遞增.4.若冪函數(shù)f(x)的圖象過點(diǎn)22,12,則函數(shù)g(x)=exf(x)的單調(diào)遞減區(qū)間為()A.(-,0)B.(-,-2)C.(-2,-1)D.(-2,0)答案D設(shè)冪函數(shù)f(x)=x,因?yàn)閳D象過點(diǎn)22,12,所以12=22,=2,所以f(x)=x2,故g(x)=exx2,則g(x)=exx2+2exx=ex(x2+2x),令g(x)0,得-2x0,故函數(shù)g(x)的單調(diào)遞減區(qū)間為(-2,0).5.若函數(shù)f(x)=x+aln x不是單調(diào)函數(shù),則實(shí)數(shù)a的取值范圍是()A.0,+)B.(-,0C.(-,0)D.(0,+)答案Cf (x)=1+ax=x+ax,若f(x)=x+aln x不是單調(diào)函數(shù),則f (x)=0在(0,+)內(nèi)有解,所以a0,故選C.6.已知函數(shù)f(x)=ax+ln x,則當(dāng)a0時(shí), f(x)的單調(diào)遞增區(qū)間是,單調(diào)遞減區(qū)間是.答案0,-1a;-1a,+解析由已知得f(x)的定義域?yàn)?0,+).當(dāng)a-1a時(shí),f (x)0,當(dāng)0x0,所以f(x)的單調(diào)遞增區(qū)間為0,-1a,單調(diào)遞減區(qū)間為-1a,+.7.若f(x)=xsin x+cos x,則f(-3), f2, f(2)的大小關(guān)系是.答案f(-3)f(2)f2解析函數(shù)f(x)為偶函數(shù),因此f(-3)=f(3).又f (x)=sin x+xcos x-sin x=xcos x,當(dāng)x2,時(shí), f (x)f(2)f(3)=f(-3).8.已知函數(shù)f(x)=12x2+2ax-ln x,若f(x)在區(qū)間13,2上是增函數(shù),則實(shí)數(shù)a的取值范圍是.答案43,+解析由題意得f (x)=x+2a-1x0在13,2上恒成立,即2a-x+1x在13,2上恒成立,-x+1xmax=83,2a83,即a43.9.已知函數(shù)f(x)=ln x+a2x2-(a+1)x.若曲線y=f(x)在x=1處的切線方程為y=-2,求f(x)的單調(diào)區(qū)間.解析由已知得f (x)=1x+ax-(a+1),則f (1)=0.而f(1)=ln 1+a2-(a+1)=-a2-1,曲線y=f(x)在x=1處的切線方程為y=-a2-1.-a2-1=-2,解得a=2.f(x)=ln x+x2-3x, f (x)=1x+2x-3.由f (x)=1x+2x-3=2x2-3x+1x0,得0x1,由f (x)=1x+2x-30,得12x1,f(x)的單調(diào)遞增區(qū)間為0,12和(1,+), f(x)的單調(diào)遞減區(qū)間為12,1.10.已知函數(shù)f(x)=ax3+x2(aR)在x=-43處取得極值.(1)確定a的值;(2)若g(x)=f(x)ex,討論g(x)的單調(diào)性.解析(1)對f(x)求導(dǎo)得f (x)=3ax2+2x.因?yàn)閒(x)在x=-43處取得極值,所以f -43=3a169+2-43=16a3-83=0,解得a=12,經(jīng)檢驗(yàn),滿足題意.(2)由(1)知,g(x)=12x3+x2ex,所以g(x)=32x2+2xex+12x3+x2ex=12x3+52x2+2xex=12x(x+1)(x+4)ex.令g(x)=0,解得x=0或x=-1或x=-4.當(dāng)x-4時(shí),g(x)0,故g(x)在(-,-4)上為減函數(shù);當(dāng)-4x0,故g(x)在(-4,-1)上為增函數(shù);當(dāng)-1x0時(shí),g(x)0時(shí),g(x)0,故g(x)在(0,+)上為增函數(shù).綜上,g(x)在(-,-4)和(-1,0)上為減函數(shù),在(-4,-1)和(0,+)上為增函數(shù).11.已知函數(shù)f(x)=x2+aln x.(1)當(dāng)a=-2時(shí),求函數(shù)f(x)的單調(diào)遞減區(qū)間;(2)若函數(shù)g(x)=f(x)+2x在1,+)上單調(diào),求實(shí)數(shù)a的取值范圍.解析(1)由題意知,函數(shù)的定義域?yàn)?0,+),當(dāng)a=-2時(shí), f (x)=2x-2x=2(x+1)(x-1)x,由f (x)0得0x1,故f(x)的單調(diào)遞減區(qū)間是(0,1).(2)由題意得g(x)=2x+ax-2x2,因函數(shù)g(x)在1,+)上單調(diào),故:若g(x)為1,+)上的單調(diào)增函數(shù),則g(x)0在1,+)上恒成立,即a2x-2x2在1,+)上恒成立,設(shè)(x)=2x-2x2.(x)在1,+)上單調(diào)遞減,在1,+)上,(x)max=(1)=0,a0.若g(x)為1,+)上的單調(diào)減函數(shù),則g(x)0在1,+)上恒成立,易知其不可能成立.實(shí)數(shù)a的取值范圍是0,+).B組提升題組1.已知f(x)的定義域?yàn)?0,+), f (x)為f(x)的導(dǎo)函數(shù),且滿足f(x)(x-1)f(x2-1)的解集是() A.(0,1)B.(1,+)C.(1,2)D.(2,+)答案D因?yàn)閒(x)+xf (x)0,所以(xf(x)(x-1)f(x2-1),所以(x+1)f(x+1)(x2-1)f(x2-1),所以0x+12.2.若定義在R上的函數(shù)f(x)滿足f(x)+f (x)1, f(0)=4,則不等式f(x)3ex+1(e為自然對數(shù)的底數(shù))的解集為()A.(0,+)B.(-,0)(3,+)C.(-,0)(0,+)D.(3,+)答案A由f(x)3ex+1,得exf(x)3+ex,構(gòu)造函數(shù)F(x)=exf(x)-ex-3,得F (x)=exf(x)+exf (x)-ex=exf(x)+f (x)-1,由f(x)+f (x)1,ex0,可知F (x)0,所以F(x)在R上單調(diào)遞增,又因?yàn)镕(0)=e0f(0)-e0-3=f(0)-4=0,所以F(x)0的解集為(0,+),即f(x)3ex+1的解集為(0,+).3.已知函數(shù)f(x)=a(x-ln x)+2x-1x2(aR),討論f(x)的單調(diào)性.解析易知f(x)的定義域?yàn)?0,+),f (x)=a-ax-2x2+2x3=(ax2-2)(x-1)x3.當(dāng)a0時(shí),若x(0,1),則f (x)0, f(x)單調(diào)遞增,若x(1,+),則f (x)0時(shí), f (x)=a(x-1)x3x-2ax+2a,當(dāng)0a1,當(dāng)x(0,1)或x2a,+時(shí), f (x)0, f(x)單調(diào)遞增,當(dāng)x1,2a時(shí), f (x)2時(shí),02a0, f(x)單調(diào)遞增,當(dāng)x2a,1時(shí), f (x)0,f(x)單調(diào)遞減.綜上所述,當(dāng)a0時(shí), f(x)在(0,1)上單調(diào)遞增,在(1,+)上單調(diào)遞減;當(dāng)0a2時(shí), f(x)在0,2a上單調(diào)遞增,在2a,1上單調(diào)遞減,在(1,+)上單調(diào)遞增.