3 5 22在函數(shù)上某點的切線方程：函數(shù)y=f(x)在點A(x0O，f(x0))處的切線方程為y-f(x0)=f/EQ\*jc3\*hps20\o\al(\s\up6(y),k)EQ\*jc3\*hps14\o\al(\s\up6(x0),x0)0)為：y-y0=f/(x0)(x-x0)，又因為切(x01.(24-25江蘇蘇州·期末)已知函數(shù)f(x(=ax2+(a-2(x-lnx．(2)若f(x(的圖象與直線y=kx-1切于點,Of，求k的值．(1)0(2)k=3(1)f(x(的定義域為{x|x>0{，當(dāng)a=1時，f(x(=x2-x-lnx，f/(x(==，令f/(x(=0，得x=1，11當(dāng)0<x<1時，f/(x(<0，f(x(在(0,1(上單調(diào)遞減；當(dāng)x>1時，f/(x(>0，f(x(在(1,O+∞(單調(diào)遞增，所以f(x(在x=1處取得極小值，極小值為f(1(=0；消去k得-2+ln=0，令g(x(=-2+ln，易知g(x(在(0,O+∞(上單調(diào)遞增，將a=2代入a2+(a-2(-=k，得k=3.所以k=3.2.(24-25江蘇無錫·期末)已知函數(shù)f(x(=x(x-c)2.(1)若f(x(在x=2處有極小值，求f(x(的單調(diào)遞增區(qū)間；(2)若函數(shù)y=f(x(的圖象與直線y=-x+c相切，求實數(shù)c的值.(1)(2,+∞((2)±2(1)由題意可知：f/(x(=(x-c)2+2(x-c(.x=(x-c((3x-c(，因為f(x(在x=2處有極小值，則f/(2(=(2-c((6-c(=0，解得c=2或6，/(x(=(x-2((3x-2(，令f/(x(>0，解得x<或x>2；令f/(x(<0，解得<x<2；可知f(x(在x=2處有極小值，符合題意；/(x(=(x-6((3x-6(，令f/(x(>0，解得x<2或x>6；令f/(x(<0，解得2<x<6；可知f(x(在(-∞,2(,(6,+∞(上單調(diào)遞增，在(2,6(上單調(diào)遞減，可知f(x(在x=2處有極大值，不符合題意；綜上所述：c=2，f(x(的單調(diào)增區(qū)間為(2,+∞(./(x(=(x-c((3x-c(，設(shè)f(x(與y=-x+c切于P(x0,x0(x0-c(2(，則切線斜率k=f/(x0(=(x0-c((3x0-c(，可得切線方程為y=(x0-c((3x0-c((x-x0(+x0(x0-c(2=(x0-c((3x0-c(x-2xEQ\*jc3\*hps12\o\al(\s\up3(2),0)(x0-c(，EQ\*jc3\*hps15\o\al(\s\up6(x0),2)EQ\*jc3\*hps21\o\al(\s\up6(x),c)=-1，顯然x0≠c，22整理可得(2x0-c((x0-c(=0，解得，代入(x0-c((3x0-c(=-1可得所以c=±2.3.(24-25山東濰坊·期末)已知函數(shù)f(x(=ex(x2+ax+1(.(1)當(dāng)a=0時，求曲線y=f(x(在點(1,f(1((處的切線方程；(1)4ex-y-2e=0(1)當(dāng)a=0時，函數(shù)f(x(=(x2+1(ex，得f/(x(=(x2+1(ex+2xex=ex(x+1(2，所以f/(1(=4e，f(1(=2e，所以曲線y=f(x(在點(1,f(1((處的切線方程為y-2e=4e(x-1(，即切線方程為4ex-y-2e=0；(2)當(dāng)a≠0時，f(x(=ex(x2+ax+1(，f/(x(=ex(x+1((x+a+1(，令f/(x(=0，得x=-1，x=-1-a，當(dāng)a>0時，-1>-1-a，令f/(x(>0，得x>-1或x<-1-a，令f/(x(<0，得-1-a<x<-1，33所以函數(shù)f(x(的單調(diào)增區(qū)間為(-∞,-1-a(和(-1,+∞(，單調(diào)減區(qū)間為(-1-a,-1(當(dāng)a<0時，-1<-1-a，令f/(x(>0，得x>-1-a或x<-1，令f/(x(<0，得-1<x<-1-a，所以函數(shù)f(x(的單調(diào)增區(qū)間為(-1-a,+∞(和(-∞,-1(，單調(diào)減區(qū)間為(-1,-1-a(；綜上所述，當(dāng)a>0時，f(x(的單調(diào)增區(qū)間為(-∞,-1-a(和(-1,+∞(，單調(diào)減區(qū)間為(-1-a,-1(；當(dāng)a<0時，f(x(的單調(diào)增區(qū)間為(-1-a,+∞(和(-∞,-1(，單調(diào)減區(qū)間為(-1,-1-a(．(2)討論函數(shù)f(x(的單調(diào)性. 令g(x(=ax2+2(a-1(x+a,x∈(0,+∞(，f/(x(<0，所以f(x(在(0,+∞(上單調(diào)遞減；2(a-1(x+a=0的根為x1=,x2=，由于(1-a(2-(1-2a(=a2>0,1-a>0,1-2a>0,∴1-a>、1-2a，即x1>0,x2>0，f/f(x(在和上單調(diào)遞增；44綜上，當(dāng)a≤0時，f(x(在(0,+∞(上單調(diào)遞減；當(dāng)0<a<時，f(x(在(0,和,+∞(上單調(diào)遞增，當(dāng)a≥時，f(x(在(0,+∞(上單調(diào)遞增；(1)當(dāng)a=2時，求曲線y=f(x(在點(1,f(1((處的切線方程；(2)若函數(shù)f(x(有極小值，且f(x(的極小值小于1-a2，求實數(shù)a的取值范圍.(1)x+y-1=0.(2)(0,1(.所以f(1(=0，.所以在(1,f(1((切線f/(1(=-1，所以切線方程為y=-(x-1(，即x+y-1=0.則f/(x(=-=，①當(dāng)a≤0時，f/(x(>0恒成立，此時函數(shù)f(x(在(0,+∞(上單調(diào)遞增，無極小值，x(0,a(a(a,+∞(f/(x(-0+f(x(所以f(x)極小值=f(a(=lna+1-a，由題意可得lna+1-a<1-a2，即a2+lna-a<0，令g(a(=a2+lna-a(a>0(，則g(1(=0.因為g/(a(=2a+-1=>0，所以函數(shù)g(a(在(0,+∞(單調(diào)遞增，所以由g(a(<g(1(=0，得0<a<1，55(2)若x=x0是f(x(的極小值點，證明:f(x0(<-e.可；結(jié)論.所以y=f(x(在點(1,0)處的切線為y=ea(x-1)，(x)>0，0≠,故f(x0)<-e得證.66(3)(-∞,0(．則f/(0(=e0-3=-2，又f(0(=e0-0=1．故曲線y=f(x(在(0,f(0((處的切線方程為2x+y-1=0．F/(x(=(x+1(ex-ax2-ax=(x+1((ex-ax(，設(shè)m(x(=ex-ax,x∈(0,+∞(，則m/(x(=ex-a．故F(x(=xf(x(+ax2在(0,+∞(單調(diào)遞增，不存在最小值．當(dāng)x∈(-∞,0(時h/(x(=<0，所以h(x(=在(-∞,0(單調(diào)遞減77【分析】(1)根據(jù)條件求得f(0(與f/(0(，即可得到88x-2sinx，f'(x(=ex-2cosx，則f'(0(=e0-2cos0=-1，又f(0(=e0-2sin0=1，所以曲線y=f(x)在點(0,f(0((處的切線方程為x+y-1=0.(x(=，x0 π4π'(x(+0-g(x(49.(24-25山東淄博·期末)已知函數(shù)f(x(=alnx-x(a<2(，曲線y=f(x(在點(1,-1(處的切線與曲EQ\*jc3\*hps19\o\al(\s\up4(Γ),L)EQ\*jc3\*hps19\o\al(\s\up4(Γ),L)(i)求m和n的值；【詳解】(1)因為f(x(=alnx-x(a<2(，所以f'(x(=-1，f(1(=-1，所以f'(1(=a-1，所以曲線y=f(x(在點(1,-1(處的切線方程為y+1=(a-1((x-1(，設(shè)直線y+1=(a-1((x-1(與曲線y=x2+2x相切與點(x0,y0(，99EQ\*jc3\*hps15\o\al(\s\up3(2),0)EQ\*jc3\*hps18\o\al(\s\up4(Γ),L)(++1|EQ\*jc3\*hps18\o\al(\s\up4(Γ),L)(x+m(，由(1)f(x(=lnx-x，EQ\*jc3\*hps18\o\al(\s\up4(Γ),L)(++1|EQ\*jc3\*hps18\o\al(\s\up4(Γ),L)(x+m(=(x+m(ln，所以函數(shù)y=g(x(的定義域為(-∞,-4(∪(0,+∞(，所以n=-2，所以函數(shù)y=g(x-2(關(guān)于y軸對稱，故函數(shù)y=g(x-2(為偶函數(shù)，所以g(-2-x(=g(-2+x(，故g(t(=g(-4-t(所以(t+m(ln=(-4-t+m(ln-，所以(t+m(ln=(4+t-m(ln，所以(4-2m(ln=0，函數(shù)y=g(x(的定義域為(-∞,-4(∪(0曲線y=g(x(關(guān)于直線x=-2對稱，要證明g(x(>4，所以函數(shù)h(t(=ln(1+t(+-2在(0,+∞(單調(diào)遞增，所以g(x(>4.(3)由題意轉(zhuǎn)化為xlna+ln(1-2x(≤0恒成立，令φ(x(=xlna+ln(1-2x(，按照0<a<1和a>1分類a(2)設(shè)點P(m,n(為函數(shù)f(x(的對稱中心，則f(x(+f(2m-x(=2n，所以ax+ka-x+a2m-x+ka-2m+x=2n，即ax(1+ka-2m(+a-x(k+a2m(=2n，所以a2x(1+ka-2m(-2nax+(k+a2m(=0，2m=-k，n=0，當(dāng)k<0時，m=loga(-k(，此時函數(shù)f(x(圖象的對稱中心為Ploga(-k(,0(.x≤在上恒成立，即xlna+ln(1-2x(≤0．令φ(x(=xlna+ln(1-2x(，則φ(0(=0，所以令則所以(x(在上單調(diào)遞減，2x(<0，則φ(x(單調(diào)遞減；時，->0，≥g(x(恒成立，將問題轉(zhuǎn)化成證明g(x(≥2lnx+2，再利用導(dǎo)數(shù)分段求證g(x(-2lnx-2≥0成立即可得證.【詳解】(1)x=4時，f(x(=2ea+4e-a+ea-e-a=ea+3e-a.令h(a(=0<x≤3.x≥3(3)由第(2)問可知f(x(≥g(x(恒成立，所以只需證明g(x(≥2lnx+2即可.①若x∈[1,3[，構(gòu)造u(x(=3x--2lnx-2u/(x(=+-=(3x-4x+1(=(3x-1((x-1(x(≥0在[1,3[上恒成立，u(x(在[1,3[上單調(diào)遞增，所以u(x(≥u(1(=0，x即3x-1≥2lnx+2在[1,3[上恒成立；xEQ\*jc3\*hps21\o\al(\s\up5(1),x)構(gòu)造t(x(=22x--2lnx-2，則t/(x(= 2x=- 2x=-x- 2x-2x-xx-.令則所以φ(x(在[3,+∞(單調(diào)遞增，t(x(在[3,+∞(單調(diào)遞增，t(x(≥t(3(=3-2ln3-2.因為3<e>t(3(=3-2ln3-2>3--2>0，g(x(=22x-≥22x->2lnx+2，所以g(x(≥2lnx+2，而f(x(≥g(x(，即證f(x(≥2lnx+2在x∈[1,+∞(上恒成立.12.(24-25深圳市龍崗區(qū)·期末)已知函數(shù)f(x(=e(1)討論函數(shù)f(x(的單調(diào)性；EQ\*jc3\*hps13\o\al(\s\up8(1),2)EQ\*jc3\*hps13\o\al(\s\up8(1),3)【詳解】(1)f/(x(=ex-m，當(dāng)m≤0時，f/(x(>0,f(x(在(-∞,+∞(上單調(diào)遞增，當(dāng)x∈(-∞,lnm(時，f/(x(<0，f(x(在(-∞,lnm(上單調(diào)遞減，，f(x(在(lnm,+∞(上單調(diào)遞增．綜上，當(dāng)m≤0時，f(x(在(-∞,+∞(上單調(diào)遞增；當(dāng)m>0時，f(x(在(-∞,lnm(上單調(diào)遞減，在(lnm,+∞(上單調(diào)遞增．EQ\*jc3\*hps13\o\al(\s\up8(1),2)EQ\*jc3\*hps13\o\al(\s\up8(1),3)x-ex≥ax3-x+2a恒成立，令h(x(=ex+(1-e(x-ax3-2a，則h(x(≥0在[0,+∞)恒成立，x+(1-e(x-ax3-2a≥ex+(1-e(x-x3-，+(1-e(-x2，令u(x(=ex+(1-e(-x2，則u/(x(=ex-2x，x-ex≥0，故u/(x(=ex-2x>ex-ex≥0，所以g/(x(在(0,+∞(上為增函數(shù)，又g/(1(=0所以g(x(min=g(1(=0，x+(1-e(x-x3-≥0恒成立．x-ex≥ax3-x+2a恒成立，令h(x(=ex+(1-e(x-ax3-2a，則h(x(≥0在[0,+∞)恒成立，x+(1-e(x-ax3-2a≥ex+(1-e(x-x3-，只要證明ex+(1-e(x-x3-≥0即可．則g(x(min=g(1(=0，h/(x(=[ex-(e-1([(x3+2(-3x2[ex-(e=ex(x3-3x2+2(+(e-1((2x3-2(= (x-1([ex(x2-2x-2(+(e-1((2x2+2x+2([=ex(x2-4((x2+x+1(-ex(x2-2x故g(x(min=g(1(=e-2>0，令h/(x(=0明；(3)小問，由f(x(≥ax3-x+2a恒成立構(gòu)造函數(shù)h(x(=ex+(1-e(x-ax3-2a，則h(x(≥0在[0,(1)討論函數(shù)f(x(零點的個數(shù)；xx(<0，g(x(單調(diào)遞減；x(>0，g(x(單調(diào)遞增；當(dāng)a=-時，函數(shù)f(x(有2個零點；當(dāng)a<-時，函數(shù)f(x(有3個零點.因為f/(x(-3ax≥b，則ex-ax-b≥0，令h(x(=ex-ax-b，則h/(x(=ex-a，x與y=-ax-b的性質(zhì)可知，x→0，y=-ax-b→-∞,所以h(x(=ex-ax-b≥0不恒成立，不符合題意；，此時h(x(單調(diào)遞減，所以h(x(≥h(lna(=a-alna-b≥0，即b≤a-alna，所以ab≤a2-a2lna，令φ(x(=x2-x2lnx，x>0，則φ/(x(=x-2xlnx=x(1-2lnx(， 數(shù)形結(jié)合的方法求解.14.(24-25江蘇常州·期末)已知函數(shù)f(x)=xlnx．(2)若函數(shù)g(x)=f(x)-bx2有兩個不同的零點．(ii)若(xlnx-bx2((x2-cx+d(≤0恒成立，求證：數(shù)證明不等式.而f(x)在區(qū)間(a,+∞)上單調(diào)，所以a≥;(2)(i)g(x(=xlnx-bx2=x(lnx-bx(，且x∈(0,+∞(，令h(x(=lnx-bx且x∈(0,+∞(，則h/(x(=-b， 又h(1(=-b<0，h=-2lnb-，(ⅱ)記兩個零點為x1,x2(0<x1<x2(，結(jié)合g(x((x2-cx+d(≤0恒成立，令m=<1，則n(m(=2lnm-m+，則nI(m(=-1-1-2<0，證不等式為關(guān)鍵.2、通過等價變形，可將“函數(shù)F(x)=f(x)-g(x)的零點”與“方程f(x)=g(x)的解”問題相互轉(zhuǎn)化。(1)若f(x(≤g(x(恒成立，求a的取值范圍；【分析】(1)不等式等價于a≥lnx-x，利用導(dǎo)數(shù)求得h(x(=lnx-x的最大值即可得到a的取值范圍.(2)令F(x(=af(x+1(-g(x(=a(x+1(ln(x+1(-x2-ax，則FI(x(=aln(x+1(-2x，令m(x(=x(=0得x=-1，結(jié)合m(x(的單調(diào)性，可得F(x(max=F-1(=aln-a+2，令G(a(解的個數(shù).【詳解】(1)由題f(x(的定義域為(0,+∞(，f(x(≤g(x(即xlnx≤x2+ax，即a≥lnx-x恒成立，令h(x(=lnx-x，則h/(x(=-1，故h(x(在(0,+∞(上有最大值h(1(=-1，所以a≥-1，即a的取值范圍是[-1,+∞(.令F(x(=a(x+1(ln(x+1(-x2-ax，(x>-1(，則F/(x(=aln(x+1(-2x，令m(x(=aln(x+1(-2x，則①當(dāng)a≤0時m/(x(<0，m(x(在(-1，+∞(上單調(diào)遞減，即F/(x(在(-1,+∞(上單調(diào)遞減，-1,0(時F/(x(>0，F(xiàn)(x(單調(diào)遞增；當(dāng)x∈(0,+∞(時F/(x(<0，F(xiàn)(x(單調(diào)遞故F(x(在x=0處取得最大值，即F(x(max=F(0(=0，所以F(x(只有一個零點，即原方程只有一個解；0解得x=-1，當(dāng)-1<x<-1時m/(x(>0，m(x(在(-1,-1(上單調(diào)遞增，即F/(x(在(-1,-1(上單調(diào)遞增；當(dāng)x>-1時m/(x(<0，m(x(在-1,+∞(上單調(diào)遞減，即F/(x(在-1,+∞(上單調(diào)遞減，所以F(x(在x=-1處取得最大值，即F(x(max=F-1(=aln-2-1(=aln-a+2，所以F(x(在(-1,+∞(上單調(diào)遞增，又F(0(=0，所以F(x(只有一個零點，即原方程只有一個解.(1)已知函數(shù)f(x(的圖象關(guān)于點P(m,n(成中心對稱圖形的充要條件是函數(shù)y=f(x+m(-n是奇函【分析】(1)由y=f(x+m(-n是奇函數(shù)，則f(-x+m(-n+[f(x+m(-n[=0，然后求解即可；g(x(=x3+3x2-kx-b有3個不相等零點x1,x2,x3，從而得到g(x(=x3+3x2+(2-b(x-b，然后利用導(dǎo)數(shù)【詳解】(1)f(x(關(guān)于(m,n(對稱，令g(x(=f(x+m(-n，則g(-x(=-g(x(，即f(-x+m(-n+[f(x+m(-n[=0，3(m+x)2+(m-x)3+3(m-x)2=2n，化簡得(3m+3(x2+(m3+3m2-n(=0，得m=-1,n=2，得對稱中心坐標(biāo)為(-1,2(；若l:y=kx+b與f(x(圖像交于A(x1,y1(,B(x2,y2(,C(x3,y3(，即g(x(=x3+3x2-kx-b有3個不相等零點x1,x2,x3，由于x3+ax2-kx-b=0=(x-x1((x-x2((x-x3(得x3-(x1+x2+x3(x2+(x1x2+x2x3+x1x3(x-x1x2x3=x3+ax2-kx-b，所以x1+x2+x3=-3，即g(x(=x3+3x2+(2-b(x-b，(i)當(dāng)Δ=12(b+1(≤0，即b≤-1時函數(shù)g(x(在R上單調(diào)遞增，令g/(x(>0(-1-3+3b,-1+3+3b(，得函數(shù)g(x(在(-∞,-1-3+3b(遞增，(-綜上可得，b>-1符合題意，即b>-1為g(x(在R上存在三個零點的必要條件得必要性成立；17.(24-25山東濱州·期末)設(shè)函數(shù)y=f(x(的定義域為D，其導(dǎo)函數(shù)為f/(x(，區(qū)間I是D的一個非空子稱函數(shù)y=f(x(為區(qū)間I上的“M(t(函數(shù)”．(2)若函數(shù)g(x(=x2-ax是[0,2[上的“M(2(函數(shù)”．(ⅱ)證明：?x∈[1,2[，g(x+2( 3(2x-4(≥6(lnx-1(，即證x-1-lnx≥0，構(gòu)造函數(shù)G(x(=x-1-lnx，利用導(dǎo)數(shù)分析函數(shù)G(x(在區(qū)【詳解】(1)因為f(x(=cosx，則f/(x(=-sinx，因為f(x+=cos(x+=-sinx+1(.f/(x(+1(.sinx．所以f(x+≥+1(f/(x(對于任意x∈[0,π[恒成立．故f(x(=cosx是[0,π[上的“M函數(shù)”．(2)(i)g/(x(=2x-a，由條件得(x+2(2-a(x+2(≥3(2x-a(對任意的x∈[0,2[恒成立，即a(x-1(≤(x-1(2+3任意的x∈[0,2[恒成立．所以F(x(在[0,1(上單調(diào)遞減，可得a≥F(x(max=F(0(=-4．設(shè)F(x(=x-1+-，則F/(x(=1--<0，所以F(x(在(1,2[上單調(diào)遞減，可得a≤F(x(min=F(2(=4．對?x∈[1,2[，g(x+2(≥3g/(x(=3(2x-a(≥3(2x-4(．-1