（浙江專用）2020版高考數(shù)學(xué)大一輪復(fù)習(xí) 高考解答題專項(xiàng)練1 函數(shù)與導(dǎo)數(shù).docx

高考解答題專項(xiàng)練函數(shù)與導(dǎo)數(shù)1.(2017浙江湖州改編)已知函數(shù)f(x)=ln x+1-xax,其中a為大于零的常數(shù).(1)若函數(shù)f(x)在區(qū)間1,+)內(nèi)單調(diào)遞增,求a的取值范圍;(2)求函數(shù)f(x)在區(qū)間1,2上的最小值.解(1)由題意,f(x)=1x-1ax2=ax-1ax2,a為大于零的常數(shù),若使函數(shù)f(x)在區(qū)間1,+)上單調(diào)遞增,則使ax-10在區(qū)間1,+)上恒成立,即a-10,故a1;(2)當(dāng)a1時(shí),f(x)>0在(1,2)上恒成立,這時(shí)f(x)在1,2上為增函數(shù),f(x)min=f(1)=0.當(dāng)0<a12時(shí),f(x)<0在(1,2)上恒成立,這時(shí)f(x)在1,2上為減函數(shù),f(x)min=f(2)=ln2-12a,當(dāng)12<a<1時(shí),令f(x)=0,得x=1a(1,2).又對(duì)于x1,1a時(shí),有f(x)<0,對(duì)于x1a,2時(shí),有f(x)>0,f(x)min=f1a=ln1a+1-1a,綜上,f(x)在1,2上的最小值為當(dāng)0<a12時(shí),f(x)min=ln2-12a;當(dāng)12<a<1時(shí),f(x)min=ln1a+1-1a;當(dāng)a1時(shí),f(x)min=0.2.已知函數(shù)f(x)=(t+1)ln x+tx2+3t,tR.(1)若t=0,求證:當(dāng)x0時(shí),f(x+1)x-12x2;(2)若f(x)4x對(duì)任意x1,+)恒成立,求t的取值范圍.(1)證明t=0時(shí),f(x)=lnx,f(x+1)=ln(x+1),即證ln(x+1)x-12x2;令g(x)=ln(x+1)-x+12x2(x0);則g(x)=x2x+1>0,g(x)在(0,+)遞增,g(x)g(0)=0,即ln(x+1)x-12x2;(2)解由f(x)4x(t+1)lnx+tx2+3t-4x0,令(x)=(t+1)lnx+tx2+3t-4x,首先由(1)0t1,此時(shí)(x)=2tx2-4x+t+1x,令h(x)=2tx2-4x+t+1,t1,=16-8t(t+1)<0,h(x)>0恒成立,即(x)>0,(x)在1,+)遞增,故(x)(1)=4t-40,綜上,t1.3.(2018浙江臺(tái)州一模)已知函數(shù)f(x)=2x3-3(m+1)x2+6mx,mR.(1)若m=2,寫(xiě)出函數(shù)f(x)的單調(diào)遞增區(qū)間;(2)若對(duì)于任意的x-1,1,都有f(x)<4,求m的取值范圍.解(1)若m=2,則f(x)=2x3-9x2+12x,f(x)=6x2-18x+12=6(x2-3x+2)=6(x-1)(x-2),令f(x)>0,則x<1,或x>2,函數(shù)f(x)的單調(diào)遞增區(qū)間為(-,1),(2,+).(2)f(x)=2x3-3(m+1)x2+6mx,f(x)=6x2-6(m+1)x+6m=6(x-1)(x-m),當(dāng)m1時(shí),f(x)在區(qū)間(-1,1)上遞增,f(x)max=f(1)=3m-1<4,得m<53,1m<53;當(dāng)-1<m<1時(shí),f(x)在區(qū)間(-1,m)上遞增,在區(qū)間(m,1)上遞減,f(x)max=f(m)=-m3+3m2<4,即m3-3m2+4>0,(m+1)(m-2)2>0恒成立,-1<m<1;當(dāng)m-1時(shí),f(x)在區(qū)間(-1,1)上遞減,f(x)max=f(-1)=-9m-5<4,得m>-1(舍去).綜上,m的取值范圍為-1<m<53.4.已知f(x)=2xln x,g(x)=-x2+ax-3.(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)若存在x(0,+),使f(x)g(x)成立,求實(shí)數(shù)a的取值范圍.解(1)f(x)的定義域?yàn)?0,+),f(x)=2(lnx+1),令f(x)=0,得x=1e,當(dāng)x0,1e時(shí),f(x)<0,當(dāng)x1e,+時(shí),f(x)>0,所以f(x)在0,1e上單調(diào)遞減;在1e,+上單調(diào)遞增.(2)存在x(0,+),使f(x)g(x)成立,即2xlnx-x2+ax-3在x(0,+)成立,等價(jià)于a2lnx+x+3x在x(0,+)成立,等價(jià)于a2lnx+x+3xmin.記h(x)=2lnx+x+3x,x(0,+),則h(x)=2x+1-3x2=x2+2x-3x2=(x+3)(x-1)x2.當(dāng)x(0,1)時(shí),h(x)<0,當(dāng)x(1,+)時(shí),h(x)>0,所以當(dāng)x=1時(shí),h(x)取最小值為4,故a4.5.已知函數(shù)f(x)=ln x+ax.(1)若函數(shù)f(x)在x=1處的切線方程為y=2x+m,求實(shí)數(shù)a和m的值;(2)若函數(shù)f(x)在定義域內(nèi)有兩個(gè)不同的零點(diǎn)x1,x2,求實(shí)數(shù)a的取值范圍.解函數(shù)定義域?yàn)?0,+)(1)f(x)=lnx+ax,f(x)=1x+a.函數(shù)f(x)在x=1處的切線方程為y=2x+m,f(1)=1+a=2,得a=1.又f(1)=ln1+a=1,函數(shù)f(x)在x=1處的切線方程為y-1=2(x-1),即y=2x-1,m=-1.(2)由(1)知f(x)=1x+a=1+axx(x>0).當(dāng)a0時(shí),f(x)=1+axx>0,函數(shù)f(x)=lnx+ax在(0,+)上單調(diào)遞增,從而函數(shù)f(x)至多有一個(gè)零點(diǎn),不符合題意;當(dāng)a<0時(shí),f(x)=ax+1ax(x>0),函數(shù)f(x)在0,-1a上單調(diào)遞增,在-1a,+上單調(diào)遞減,函數(shù)f(x)max=f-1a=ln-1a+a-1a=ln-1a-1.要滿足函數(shù)f(x)在定義域內(nèi)有兩個(gè)不同的零點(diǎn)x1,x2,必有f(x)max=ln-1a-1>0,得a>-1e.實(shí)數(shù)a的取值范圍是-1e,0.6.(2018浙江湖州2模)已知函數(shù)f(x)=1-e-xx(x>0).(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)求證:f(x)>e-x2(x>0).(1)解已知函數(shù)f(x)=1-e-xx(x>0),其導(dǎo)函數(shù)為f(x)=1+x-exx2ex.令h(x)=ex-x-1,則h(x)=ex-1,當(dāng)x<0時(shí),h(x)=ex-1<0;當(dāng)x>0時(shí),h(x)=ex-1>0,所以h(x)min=h(0)=0,即exx+1,當(dāng)且僅當(dāng)x=0時(shí)等號(hào)成立.由已知x>0,得ex>x+1,1+x-exx2ex<0,所以f(x)<0.所以函數(shù)f(x)的單調(diào)遞減區(qū)間為(0,+).(2)證明f(x)>e-x2(x>0)等價(jià)于e-x+xe-x2-1<0(x>0).令g(x)=e-x+xe-x2-1,x>0,g(x)=-e-x+e-x2+x-12e-x2=-e-x2e-x2-x2+1,由(1)易得e-x2>-x2+1,所以g(x)<0.所以當(dāng)x>0時(shí),有g(shù)(x)<g(0)=0,即e-x+xe-x2-1<0(x>0).故f(x)>e-x2(x>0).