（浙江專用）2020版高考數(shù)學(xué)大一輪復(fù)習(xí) 第三章 導(dǎo)數(shù)及其應(yīng)用 考點(diǎn)規(guī)范練13 導(dǎo)數(shù)與函數(shù)的單調(diào)性.docx

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考點(diǎn)規(guī)范練13導(dǎo)數(shù)與函數(shù)的單調(diào)性基礎(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,+)答案D解析因?yàn)閒(x)=(x-3)ex,則f(x)=ex(x-2),令f(x)0,得x2,所以f(x)的單調(diào)遞增區(qū)間為(2,+).2.(2017浙江嘉興調(diào)研)已知函數(shù)f(x)=12x3+ax+4,則“a0”是“f(x)在R上單調(diào)遞增”的()A.充分不必要條件B.必要不充分條件C.充要條件D.既不充分也不必要條件答案A解析f(x)=32x2+a,當(dāng)a0時(shí),f(x)0恒成立,故“a0”是“f(x)在R上單調(diào)遞增”的充分不必要條件.3.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()答案C解析由y=f(x)的圖象易知當(dāng)x2時(shí),f(x)0,故函數(shù)y=f(x)在區(qū)間(-,0)和(2,+)上單調(diào)遞增;當(dāng)0x2時(shí),f(x)0,故函數(shù)y=f(x)在區(qū)間(0,2)上單調(diào)遞減.4.設(shè)函數(shù)f(x)=12x2-9ln x在區(qū)間a-1,a+1上單調(diào)遞減,則實(shí)數(shù)a的取值范圍是()A.1a2B.a4C.a2D.00),當(dāng)x-9x0,即00,且a+13,解得10時(shí),g(x)在(0,+)上為增函數(shù),則g120,解得14+a4-10,a3;當(dāng)a0(x0),可得1-lnx0,x0,解得x(0,e).7.(2017浙江麗水模擬)已知函數(shù)f(x)=ln x+2x,若f(x2+2)0,函數(shù)單調(diào)遞增,所以由f(x2+2)f(3x)得x2+23x,所以1x2.8.已知函數(shù)f(x)=x3+3x對任意的m-2,2,f(mx-2)+f(x)0,所以f(x)在R上單調(diào)遞增,所以f(mx-2)+f(x)0可化為f(mx-2)-f(x)=f(-x),由f(x)在R上單調(diào)遞增可知mx-2-x,即mx+x-20,則對任意的m-2,2,f(mx-2)+f(x)0恒成立,等價(jià)于對任意的m-2,2,mx+x-20恒成立,所以-2x+x-20,2x+x-20,解得-2x0,f(2)=92,則不等式f(lg x)0,g(x)在(0,+)遞增,而g(2)=f(2)-12=4,故由f(lgx)1lgx+4,得g(lnx)g(2),故0lnx2,解得1x100,故選D.12.已知函數(shù)f(x)(xR)滿足f(1)=1,且f(x)的導(dǎo)函數(shù)f(x)12,則f(x)x2+12的解集為()A.x|-1x1B.x|x-1C.x|x1D.x|x1答案D解析設(shè)F(x)=f(x)-x2+12,則F(1)=f(1)-12+12=1-1=0,又F(x)=f(x)-12,對任意xR,有F(x)=f(x)-120,即函數(shù)F(x)在R上單調(diào)遞減,則F(x)0的解集為(1,+),即f(x)0時(shí),f(x)=lnx+32x2,則f(x)=1xx2-lnx+322xx4=x-2xlnx-3xx4=-2x(1+lnx)x4,由f(x)0得-2x(1+lnx)0,得1+lnx0,即lnx-1,得0x1e,此時(shí)函數(shù)單調(diào)遞增,由f(x)0得-2x(1+lnx)0,即lnx-1,得x1e,此時(shí)函數(shù)單調(diào)遞減,即當(dāng)x0時(shí),x=1e時(shí),函數(shù)f(x)取得極大值f1e=ln1e+32(1e)2=-1+32e2=12e2,作出函數(shù)f(x)的圖象如圖.要使a=ln|x|+32x2有4個(gè)不同的交點(diǎn),則滿足0a12e2,故選A.14.已知函數(shù)f(x)是定義在R上的奇函數(shù),當(dāng)x0時(shí),f(x)=-xe-x;函數(shù)f(x)的單調(diào)遞減區(qū)間是(-,-1)和(1,+);對x1,x2R,都有|f(x1)-f(x2)|2e.其中正確的序號是.答案解析當(dāng)x0時(shí),f(x)=-f(-x)=-(-x)e-x=xe-x,x0,故錯(cuò)誤;當(dāng)x0時(shí),f(x)=ex(x+1),則在(-,-1)單調(diào)遞減,由奇函數(shù)對稱性可知,在(1,+)也單調(diào)遞減,故正確;由導(dǎo)函數(shù)分析可知,f(x)min=f(-1)=-1ef(x)max=f(1)=1e,所以|f(x1)-f(x2)|f(x)max-f(x)min|=2e,故正確.所以正確的命題是.15.已知函數(shù)f(x)=x2ex,若f(x)在t,t+1上不單調(diào),則實(shí)數(shù)t的取值范圍是.答案(-3,-2)(-1,0)解析由題意,得f(x)=ex(x2+2x),f(x)在(-,-2),(0,+)上單調(diào)遞增,(-2,0)上單調(diào)遞減,又f(x)在t,t+1上不單調(diào),t-2或t0,即實(shí)數(shù)t的取值范圍是(-3,-2)(-1,0),故填:(-3,-2)(-1,0).16.(2017江蘇高考)已知函數(shù)f(x)=x3-2x+ex-1ex,其中e是自然對數(shù)的底數(shù).若f(a-1)+f(2a2)0,則實(shí)數(shù)a的取值范圍是.答案-1,12解析因?yàn)閒(-x)=-x3+2x+1ex-ex=-f(x),所以函數(shù)f(x)是奇函數(shù),因?yàn)閒(x)=3x2-2+ex+e-x3x2-2+2exe-x0,所以f(x)在R上單調(diào)遞增,又f(a-1)+f(2a2)0,即f(2a2)f(1-a),所以2a21-a,即2a2+a-10,解得-1a12,故實(shí)數(shù)a的取值范圍為-1,12.17.設(shè)f(x)=ex1+ax2,其中a為正實(shí)數(shù).(1)當(dāng)a=43時(shí),求f(x)單調(diào)區(qū)間;(2)若f(x)為R上的單調(diào)函數(shù),求實(shí)數(shù)a的取值范圍.解對f(x)求導(dǎo)得f(x)=ex1+ax2-2ax(1+ax2)2.(1)當(dāng)a=43時(shí),若f(x)=0,則4x2-8x+3=0,解得x1=32,x2=12.結(jié)合,可知x-,121212,323232,+f(x)+0-0+f(x)極大值極小值所以增區(qū)間為-,12,32,+;減區(qū)間為12,32.(2)若f(x)為R上的單調(diào)函數(shù),則f(x)在R上不變號,結(jié)合與條件a0,知ax2-2ax+10在R上恒成立,即=4a2-4a=4a(a-1)0,又a0,得0a1.所以實(shí)數(shù)a的取值范圍為a|00,函數(shù)f(x)在(0,+)上單調(diào)遞增.當(dāng)a0時(shí),令g(x)=ax2+(2a+2)x+a,由于=(2a+2)2-4a2=4(2a+1).當(dāng)a=-12時(shí),=0,f(x)=-12(x-1)2x(x+1)20,函數(shù)f(x)在(0,+)上單調(diào)遞減.當(dāng)a-12時(shí),0,g(x)0,f(x)0,函數(shù)f(x)在(0,+)上單調(diào)遞減.當(dāng)-12a0.設(shè)x1,x2(x10,所以x(0,x1)時(shí),g(x)0,f(x)0,f(x)0,函數(shù)f(x)單調(diào)遞增;x(x2,+)時(shí),g(x)0,f(x)0,函數(shù)f(x)單調(diào)遞減.綜上可得:當(dāng)a0時(shí),函數(shù)f(x)在(0,+)上單調(diào)遞增;當(dāng)a-12時(shí),函數(shù)f(x)在(0,+)上單調(diào)遞減;當(dāng)-12a0時(shí),f(x)在0,-(a+1)+2a+1a,-(a+1)-2a+1a,+上單調(diào)遞減,在-(a+1)+2a+1a,-(a+1)-2a+1a上單調(diào)遞增.