2025版高考數(shù)學(xué)一輪總復(fù)習(xí)應(yīng)用創(chuàng)新題組3.2導(dǎo)數(shù)的應(yīng)用

.2導(dǎo)數(shù)的應(yīng)用應(yīng)用篇知行合一應(yīng)用利用導(dǎo)數(shù)探討三次函數(shù)的性質(zhì)1.(2024屆河南安陽(yáng)月考,10課程學(xué)習(xí)情境)若函數(shù)y=ax3-x在R上是減函數(shù),則()A.a≥13B.a≤0C.a=2答案B因?yàn)楹瘮?shù)y=ax3-x在R上是減函數(shù),所以y'=3ax2-1≤0在R上恒成立,當(dāng)a=0時(shí),y'=-1<0,符合,當(dāng)a≠0時(shí),由3ax2-1≤0恒成立得a<0.綜上所述,a≤0.故選B.2.(2024百師聯(lián)盟第四次測(cè)試,13)若函數(shù)f(x)=x3+2ax2+ax-1在(0,1)上存在唯一極值點(diǎn),則實(shí)數(shù)a的取值范圍是.

答案-解析由三次函數(shù)圖象特點(diǎn)知,其最多有1個(gè)極大值點(diǎn)和1個(gè)微小值點(diǎn).f'(x)=3x2+4ax+a,若f(x)在(0,1)上存在唯一極值點(diǎn),則f'(0)·f'(1)<0?a·(3+4a+a)<0?a∈-33.(2024山東淄博一模,13)已知等比數(shù)列{an}中,首項(xiàng)a1=2,公比q>1,a2,a3是函數(shù)f(x)=13x3-6x2+32x的兩個(gè)極值點(diǎn),則數(shù)列{an}的前9項(xiàng)和是答案1022解析由f(x)=13x3-6x2+32x得f'(x)=x2-12x+32,因?yàn)閍2,a3是函數(shù)f(x)=13x3-6x2+32x的兩個(gè)極值點(diǎn),所以a2,a3是函數(shù)f'(x)=x2-12x+32的兩個(gè)零點(diǎn).故a2+a3=12,a2·a3=32.因?yàn)閝>1,所以a2=4,a3=8,故q=2,則數(shù)列4.(2024屆四川綿陽(yáng)一診,20課程學(xué)習(xí)情境)已知函數(shù)f(x)=-13x3+ax2+3a2x-53(1)當(dāng)a=-1時(shí),求f(x)在區(qū)間[-4,2]上的最大值與最小值;(2)若函數(shù)f(x)僅有一個(gè)零點(diǎn),求a的取值范圍.解析(1)f'(x)=-x2+2ax+3a2=-(x-3a)(x+a).當(dāng)a=-1時(shí),f'(x)=-(x-1)(x+3),在x∈[-4,2]上,由f'(x)>0,解得-3<x<1;由f'(x)<0,解得-4≤x<-3或1<x≤2.∴函數(shù)f(x)在區(qū)間(-3,1)上單調(diào)遞增,在區(qū)間[-4,-3),(1,2]上單調(diào)遞減.又f(-4)=-253,f(-3)=-323,f(1)=0,f(2)=-73,∴函數(shù)f(x)在區(qū)間[-4,2]上的最大值為0,最小值為(2)f'(x)=-x2+2ax+3a2=-(x-3a)(x+a).i)當(dāng)a<0時(shí),由f'(x)>0,解得3a<x<-a,∴函數(shù)f(x)在區(qū)間(3a,-a)上單調(diào)遞增;由f'(x)<0,解得x<3a或x>-a,∴函數(shù)f(x)在區(qū)間(-∞,3a),(-a,+∞)上單調(diào)遞減.又f(0)=-53<0,∴只須要f(-a)<0,解得ii)當(dāng)a=0時(shí),明顯f(x)只有一個(gè)零點(diǎn).iii)當(dāng)a>0時(shí),由f'(x)>0,解得-a<x<3a,∴f(x)在區(qū)間(-a,3a)上單調(diào)遞增,由f'(x)<0,解得x<-a或x>3a,∴函數(shù)f(x)在區(qū)間(-∞,-a),(3a,+∞)上單調(diào)遞減,又f(0)=-53<0,∴只須要f(3a)<0,解得0<a<3綜上,實(shí)數(shù)a的取值范圍是-1,35.(2024屆甘肅頂級(jí)名校二模,21探究創(chuàng)新情境)已知函數(shù)f(x)=x3-x2+ax+1.(1)探討f(x)的單調(diào)性;(2)求曲線y=f(x)過(guò)坐標(biāo)原點(diǎn)的切線與曲線y=f(x)的公共點(diǎn)的坐標(biāo).解析(1)f'(x)=3x2-2x+a,Δ=4-12a,當(dāng)Δ=4-12a≤0,即a≥13時(shí),f'(x)≥0,f(x)在R上單調(diào)遞增當(dāng)Δ=4-12a>0,即a<13時(shí),令f'(x)=0,解得x1=1-1-3a3,x當(dāng)x∈-∞,1-1-3a3當(dāng)x∈1-1-3a3,當(dāng)x∈1+1-3a3,+∞綜上,當(dāng)a≥13時(shí),f(x)在R上單調(diào)遞增當(dāng)a<13時(shí),f(x)在-∞,1-1-3a3,1+1-3a3,+∞(2)設(shè)切點(diǎn)坐標(biāo)為(x0,y0),則f(x0)=x03-x02+ax0+1,f'(x0)=3x02-2x0+a,故切線方程為y-(x03-x02+ax0+1)=(3x02-2x0+a)(x-x0),又切線過(guò)坐標(biāo)原點(diǎn),則0-(整理可得(x0-1)(2x02+x0+1)=0,解得x則f(x0)=f(1)=1-1+a+1=a+1,f'(x0)=f'(1)=1+a,切線方程為y=(a+1)x,與y=x3-x2+ax+1聯(lián)立,化簡(jiǎn)得x3-x2-x+1=0,即(x-1)(x2-1)=0,解得x1=1,x2=-1,∴f(1)=a+1,f(-1)=-1-a.綜上,曲線y=f(x)過(guò)坐標(biāo)原點(diǎn)的切線與曲線y=f(x)的公共點(diǎn)的坐標(biāo)為(1,a+1)和(-1,-1-a).6.(2024屆北大附中10月月考,18)已知函數(shù)f(x)=x3-3x2+ax+4,a∈R.(1)當(dāng)a=2時(shí),求f(x)在區(qū)間12,3(2)若f(x)>0對(duì)x∈(1,2)恒成立,求a的取值范圍.解析(1)當(dāng)a=2時(shí),f(x)=x3-3x2+2x+4,f'(x)=3x2-6x+2,令f'(x)=0,得x1=1-33,x2=1+3因?yàn)?-33<12<1+所以f(x)與f'(x)在區(qū)間12,3的改變狀況x11+31+f'(x)-0+f(x)↘微小值↗又f12=358,f(3)=10,所以f所以f(x)在區(qū)間12,3(2)當(dāng)x∈(1,2)時(shí),“f(x)>0”等價(jià)于“a>-x2+3x-4x”設(shè)g(x)=-x2+3x-4x,x∈(1,2),則g'(x)=-2