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)=ex-ex,xR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+2cos x,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=12x2-ln x的單調(diào)遞減區(qū)間為()A.(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)=kx-ln x在區(qū)間(1,+)上單調(diào)遞增,則k的取值范圍是()A.(-,-2 B.(-,-1C.2,+) D.1,+)6.對(duì)于R上可導(dǎo)的任意函數(shù)f(x),若滿足1-xf (x)0,則必有()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)=x-2sin x,則函數(shù)f(x)在(0, f(0)處的切線方程為;在(0,)上的單調(diào)遞增區(qū)間為.8.若函數(shù)f(x)=x3-12x在區(qū)間(k-1,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)=x2-ex-ax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+12x2-2x+1.(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)當(dāng)00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(duì)任意的xR,總有f (x)3,則不等式f(x)0.討論f(x)的單調(diào)性.14.已知函數(shù)f(x)=exln x-aex(aR).(1)若f(x)的圖象在點(diǎn)(1, f(1)處的切線與直線y=1ex+1垂直,求a的值;(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)為增函數(shù).故選C.3.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.A函數(shù)y=12x2-ln x的定義域?yàn)閤|x0,y=x-1x=x2-1x,令x2-1x0,所以x2-10,解得0x1,01x1,k1,故選D.6.A當(dāng)x1時(shí), f (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;3,解析由f(x)=x-2sin x,得f (x)=1-2cos x,f (0)=1-2cos 0=-1,又f(0)=0-2sin 0=0,函數(shù)f(x)在(0, f(0)處的切線方程為y=-x.由1-2cos x0,得cos x12,又x(0,),3x0,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(k-1,k+1)上不是單調(diào)函數(shù),所以k-1-2k+1或k-12k+1,解得-3k-1或1k3.9.答案(-,2ln 2-2解析f(x)=x2-ex-ax,f (x)=2x-ex-a,函數(shù)f(x)=x2-ex-ax在R上存在單調(diào)遞增區(qū)間,f (x)=2x-ex-a0,即a2x-ex有解,令g(x)=2x-ex,則g(x)=2-ex,令g(x)=0,解得x=ln 2,則當(dāng)x0,g(x)單調(diào)遞增,當(dāng)xln 2時(shí),g(x)0,g(x)單調(diào)遞減,當(dāng)x=ln 2時(shí),g(x)取得最大值,且g(x)max=g(ln 2)=2ln 2-2,a2ln 2-2.10.解析(1)由f(x)=13x3+12x2-2x+1得f (x)=x2+x-2=(x+2)(x-1),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)+0-0+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+12x2-2x+1可得f(-2)=133.當(dāng)-a-2,即2a52時(shí),由(1)可得f(x)在-a,-2)和(1,a上單調(diào)遞增,在(-2,1)上單調(diào)遞減,所以,函數(shù)f(x)在區(qū)間-a,a上的最大值為maxf(-2), f(a),又由(1)可知f(a)f52=133,所以maxf(-2), f(a)=f(-2)=133.當(dāng)-a-2,a1,即01,即10,或由(1)可知f(-a)f(-1)=196, f(a)f(2)=53,所以f(-a)f(a)所以maxf(-a), f(a)=f(-a)=-a33+a22+2a+1.綜上,當(dāng)2a52時(shí),函數(shù)f(x)在區(qū)間-a,a上的最大值為133;當(dāng)00等價(jià)于當(dāng)x0時(shí), f (x)0,當(dāng)x0時(shí), f (x)0時(shí),函數(shù)f(x)在R上單調(diào)遞增,此時(shí)1x2,當(dāng)x0時(shí),函數(shù)f(x)在R上單調(diào)遞減,此時(shí)x0,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)-34+15=0,所以f(x)3x-15的解集為(4,+).13.解析由題意知, f(x)的定義域是(0,+),f (x)=1+2x2-ax=x2-ax+2x2.設(shè)g(x)=x2-ax+2,一元二次方程g(x)=0的判別式=a2-8.當(dāng)0,即0a0都有f (x)0.此時(shí)f(x)是(0,+)上的單調(diào)遞增函數(shù).當(dāng)=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).當(dāng)0,即a22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=a+a2-82,且0x10時(shí)恒成立.即1x-a+ln x0在x0時(shí)恒成立,即a1x+ln x(x0)恒成