2025屆新高考數(shù)學(xué)期末新題匯編：函數(shù)與導(dǎo)數(shù)解答題 學(xué)生版

2 4 在函數(shù)上某點(diǎn)的切線方程：函數(shù)y=f(x)在點(diǎn)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?yàn)榍?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(diǎn)求k的值．112.(24-25江蘇無(wú)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(shí)數(shù)c的值.223.(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(shí)，求曲線y=f(x(在點(diǎn)(1,f(1((處的切線方程；(1)若曲線y=f(x(在x=1處的切線方程為ax-by+1=0，求實(shí)數(shù)a,b的值；(2)討論函數(shù)f(x(的單調(diào)性.33(1)當(dāng)a=2時(shí)，求曲線y=f(x(在點(diǎn)(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(shí)數(shù)a的取值范圍.6.(24-25湖南長(zhǎng)沙·期末)已知函數(shù)f(x(=eaxlnx，其中a>0.(2)若x=x0是f(x(的極小值點(diǎn)，證明:f(x0(<-e.44(1)當(dāng)a=3時(shí)，求曲線y=f(x(在(0,f(0((處的切線方程；求證：當(dāng)a≤1時(shí)在(0,+∞(不存在最小值；55669.(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(diǎn)(1,-1(處的切線與曲線y=x2+2x相切．(2)若函數(shù)且曲線y=g(x(關(guān)于直線x=n對(duì)稱(chēng)，(i)求m和n的值；(ii)證明：g(x(>4．7788(1)設(shè)直線x=4與曲線y=f(x)交于點(diǎn)P，求P點(diǎn)縱坐標(biāo)的最小值；99(1)討論函數(shù)f(x(的單調(diào)性；(1)討論函數(shù)f(x(零點(diǎn)的個(gè)數(shù)；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(gè)不同的零點(diǎn)．(ii)若(xlnx-bx2((x2-cx+d(≤0恒成立，求證求參數(shù)的值或取值范圍??16.(24-25安徽阜陽(yáng)·期末)已知函數(shù)f(x(=x3+3x2，直線l:y=kx+b.(1)已知函數(shù)f(x(的圖象關(guān)于點(diǎn)P(m,n(成中心對(duì)稱(chēng)圖形的充要條件是函數(shù)y=f(x+m(-n是奇函稱(chēng)函數(shù)y=f(x(為區(qū)間I上的“M(t(函數(shù)”．+2(≥6(lnx-1(．B(x2,y2(，則A，B之間的曼哈頓距離為d(A,B)=|x1-x2|+|y1-y2|．現(xiàn)已知直線l:2x-2y+5=0，點(diǎn)P是直線l上一動(dòng)點(diǎn)．(2)若點(diǎn)A1(1,6),A2(2,5),A3(3,4)，試計(jì)算d(P,A1(+d