1、課時(shí)規(guī)范練15導(dǎo)數(shù)與函數(shù)的小綜合基礎(chǔ)鞏固組1.函數(shù)f(x)=(x-3)ex的單調(diào)遞增區(qū)間是()A.(-,2)B.(0,3)C.(1,4)D.(2,+)2.(2017山東煙臺(tái)一模,文9)已知函數(shù)f(x)=ax3+bx2+cx+d的圖象如圖所示,則下列結(jié)論成立的是()A.a0,b0,c0,d0,b0,c0,d0C.a0,b0,d0D.a0,b0,c0,d03.已知函數(shù)f(x)=x3-3x2+x的極大值點(diǎn)為m,極小值點(diǎn)為n,則m+n=()A.0B.2C.-4D.-24.定義域?yàn)镽的可導(dǎo)函數(shù)y=f(x)的導(dǎo)函數(shù)f(x),滿足f(x)2ex的解集為()A.(-,0)B.(-,2)C.(0,+)D.(2,

2、+)5.(2017遼寧大連一模,文8)函數(shù)f(x)=的圖象大致為()6.(2017河南濮陽一模,文12)設(shè)f(x)是函數(shù)f(x)定義在(0,+)上的導(dǎo)函數(shù),滿足xf(x)+2f(x)=,則下列不等式一定成立的是()A.B.C.D.導(dǎo)學(xué)號(hào)7.已知函數(shù)f(x)=x(ln x-ax)有兩個(gè)極值點(diǎn),則實(shí)數(shù)a的取值范圍是()A.(-,0)B.C.(0,1)D.(0,+)8.已知函數(shù)f(x)=-x2+4x-3ln x在t,t+1上不單調(diào),則t的取值范圍是.9.(2017河北保定二模)已知定義在(0,+)上的函數(shù)f(x)的導(dǎo)函數(shù)f(x)的圖象是連續(xù)不斷的,若方程f(x)=0無解,且x(0,+),f(f(x)

3、-log2 015x)=2 017,設(shè)a=f(20.5),b=f(log43),c=f(log3),則a,b,c的大小關(guān)系是.10.設(shè)函數(shù)f(x)是奇函數(shù)f(x)(xR)的導(dǎo)函數(shù),f(-1)=0,當(dāng)x0時(shí),xf(x)-f(x)0成立的x的取值范圍是.11.(2017山東泰安一模,文14)已知函數(shù)f(x)是定義在R上的奇函數(shù),若g(x)=f(x+1)+5,g(x)為g(x)的導(dǎo)函數(shù),對(duì)xR,總有g(shù)(x)2x,則g(x)x2+4的解集為.綜合提升組12.(2017廣西南寧一模)已知函數(shù)f(x)=-x2-6x-3,g(x)=2x3+3x2-12x+9,m0,且對(duì)x(0,+),2f(x)xf(x)0,

4、則a的取值范圍是.課時(shí)規(guī)范練15導(dǎo)數(shù)與函數(shù)的小綜合1.D函數(shù)f(x)=(x-3)ex的導(dǎo)數(shù)為f(x)=(x-3)ex=ex+(x-3)ex=(x-2)ex.由導(dǎo)數(shù)與函數(shù)單調(diào)性的關(guān)系,得當(dāng)f(x)0時(shí),函數(shù)f(x)單調(diào)遞增,此時(shí)由不等式f(x)=(x-2)ex0,解得x2.2.C由題圖可知f(0)=d0,排除選項(xiàng)A,B;f(x)=3ax2+2bx+c,且由題圖知(-,x1),(x2,+)是函數(shù)的遞減區(qū)間,可知a0,排除D.故選C.3.B因?yàn)楹瘮?shù)f(x)=x3-3x2+x的極大值點(diǎn)為m,極小值點(diǎn)為n,所以m,n為f(x)=3x2-6x+1=0的兩根.由根與系數(shù)的關(guān)系可知m+n=-=2.4.C設(shè)g(

5、x)=,則g(x)=.f(x)0,即函數(shù)g(x)在定義域內(nèi)單調(diào)遞增.f(0)=2,g(0)=f(0)=2,不等式f(x)2ex等價(jià)于g(x)g(0).函數(shù)g(x)在定義域內(nèi)單調(diào)遞增,x0,不等式的解集為(0,+),故選C.5.B函數(shù)f(x)=的定義域?yàn)閤0,xR,當(dāng)x0時(shí),函數(shù)f(x)=,可得函數(shù)的極值點(diǎn)為x=1,當(dāng)x(0,1)時(shí),函數(shù)是減函數(shù),當(dāng)x1時(shí),函數(shù)是增函數(shù),并且f(x)0,選項(xiàng)B,D滿足題意.當(dāng)x0時(shí),函數(shù)f(x)=0,函數(shù)g(x)在(0,+)內(nèi)單調(diào)遞增.g(2)=4f(2)g(e)=e2f(e)g(3)=9f(3),.故選B.7.Bf(x)=x(ln x-ax),f(x)=ln

6、x-2ax+1,由題意可知f(x)在(0,+)內(nèi)有兩個(gè)不同的零點(diǎn),令f(x)=0,得2a=,設(shè)g(x)=,則g(x)=,g(x)在(0,1)內(nèi)單調(diào)遞增,在(1,+)內(nèi)單調(diào)遞減.當(dāng)x0時(shí),g(x)-,當(dāng)x+時(shí),g(x)0,而g(x)max=g(1)=1,只需02a1,即0a.8.(0,1)(2,3)由題意知f(x)=-x+4-=-.由f(x)=0得x1=1,x2=3,可知1,3是函數(shù)f(x)的兩個(gè)極值點(diǎn).則只要這兩個(gè)極值點(diǎn)有一個(gè)在區(qū)間(t,t+1)內(nèi),函數(shù)f(x)在區(qū)間t,t+1上就不單調(diào),由t1t+1或t3t+1,得0t1或2tcb方程f(x)=0無解,f(x)0或f(x)0恒成立,f(x)是

7、單調(diào)函數(shù);由題意得x(0,+),f(f(x)-log2 015x)=2 017,且f(x)是定義在(0,+)的單調(diào)函數(shù),則f(x)-log2 015x是定值.設(shè)t=f(x)-log2 015x,則f(x)=t+log2 015x,f(x)是增函數(shù).又0log43log31cb.故答案為acb.10.(-,-1)(0,1)當(dāng)x0時(shí),令F(x)=,則F(x)=0時(shí),F(x)=為減函數(shù).f(x)為奇函數(shù),且由f(-1)=0,得f(1)=0,故F(1)=0.在區(qū)間(0,1)內(nèi),F(x)0;在(1,+)內(nèi),F(x)0,即當(dāng)0x0;當(dāng)x1時(shí),f(x)0;當(dāng)x(-1,0)時(shí),f(x)0的解集為(-,-1)(

8、0,1).11.(-,-1)f(x)是定義在R上的奇函數(shù),f(x)的圖象過原點(diǎn),g(x)=f(x+1)+5,g(x)的圖象過點(diǎn)(-1,5).令h(x)=g(x)-x2-4,h(x)=g(x)-2x.對(duì)xR,總有g(shù)(x)2x,h(x)在R上是增函數(shù),又h(-1)=g(-1)-1-4=0,g(x)x2+4的解集為(-,-1).12.Ag(x)=2x3+3x2-12x+9,g(x)=6x2+6x-12=6(x+2)(x-1),則當(dāng)0x1時(shí),g(x)1時(shí),g(x)0,函數(shù)g(x)遞增,當(dāng)x0時(shí),g(x)min=g(1)=2.f(x)=-x2-6x-3=-(x+3)2+66,作函數(shù)y=(x)的圖象,如圖所示,當(dāng)f(x)=2時(shí),方程兩根分別為-5和-1,則m的最小值為-5,故選A.13.B令g(x)=,x(0,+),則g(x)=.x(0,+),2f (x)xf(x)3f(x)恒成立,00,函數(shù)g(x)在(0,+)內(nèi)單調(diào)遞增,又f(x)0,.令h(x)=,x(0,+),則h(x)=.x(0,+),2f(x)xf(x)3f(x)恒成立,h(x)=0,.綜上可得,故選B.14.x-y+6=0f(x)=exx2+(2-a)x+1,若f(x)在(1,3)內(nèi)只有1個(gè)極值點(diǎn),f(1)f(3)0,即(