考點(diǎn)11 導(dǎo)數(shù)在研究函數(shù)中的應(yīng)用與生活中的優(yōu)化問(wèn)題舉例

考點(diǎn)11導(dǎo)數(shù)在研究函數(shù)中的應(yīng)用與生活中的優(yōu)化問(wèn)題舉例1.(2022·全國(guó)乙卷理科·T16)已知x=x1和x=x2分別是函數(shù)f(x)=2ax-ex2(a>0且a≠1)的極小值點(diǎn)和極大值點(diǎn).若x1<x2,則a的取值范圍是.

【命題意圖】考查導(dǎo)數(shù)的幾何意義,函數(shù)的單調(diào)性、極值,考查轉(zhuǎn)化思想、分類討論思想以及分析問(wèn)題解決問(wèn)題的能力.【解析】因?yàn)閤1,x2分別是函數(shù)f(x)=2ax-ex2的極小值點(diǎn)和極大值點(diǎn),所以函數(shù)f(x)在(-∞,x1)和(x2,+∞)上遞減,在(x1,x2)上遞增,所以當(dāng)x∈(-∞,x1)∪(x2,+∞)時(shí),f'(x)<0,當(dāng)x∈(x1,x2)時(shí),f'(x)>0,若a>1,當(dāng)x<0時(shí),2lna·ax>0,2ex<0,則此時(shí)f'(x)>0,與前面矛盾,故a>1不符合題意,若0<a<1,則方程2lna·ax-2ex=0的兩個(gè)根為x1,x2,即方程lna·ax=ex的兩個(gè)根為x1,x2,即函數(shù)y=lna·ax與函數(shù)y=ex的圖象有兩個(gè)不同的交點(diǎn),令g(x)=lna·ax,則g'(x)=ln2a·ax,0<a<1,設(shè)過(guò)原點(diǎn)且與函數(shù)y=g(x)的圖象相切的直線的切點(diǎn)為(x0,lna·ax則切線的斜率為g'(x0)=ln2a·ax0,故切線方程為y-lna·ax0=ln2a·a則有-lna·ax0=-x0ln2a·解得x0=1lna,則切線的斜率為ln2a·a1lna因?yàn)楹瘮?shù)y=lna·ax與函數(shù)y=ex的圖象有兩個(gè)不同的交點(diǎn),所以eln2a<e,解得1e<a<e,又因?yàn)?<a<1,所以1e<a<1,綜上所述,a的取值范圍為1e,1答案:1e,12.(2022·全國(guó)乙卷理科·T21)(12分)已知函數(shù)f(x)=ln(1+x)+axe-x.(1)當(dāng)a=1時(shí),求曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程;(2)若f(x)在區(qū)間(-1,0),(0,+∞)各恰有一個(gè)零點(diǎn),求a的取值范圍.【命題意圖】考查導(dǎo)數(shù)的幾何意義、切線的求法、運(yùn)用導(dǎo)數(shù)研究函數(shù)的性質(zhì)及零點(diǎn)個(gè)數(shù),考查分類討論思想,數(shù)學(xué)運(yùn)算能力、邏輯推理能力等.【解析】(1)f(x)的定義域?yàn)?-1,+∞),當(dāng)a=1時(shí),f(x)=ln(1+x)+xex,f(0)所以切點(diǎn)為(0,0),f'(x)=11+x+1?xex所以曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為y=2x;(2)f(x)=ln(1+x)+axex,f'(x)=11+x+設(shè)g(x)=ex+a(1-x2),①若a≥0,當(dāng)x∈(-1,0),g(x)=ex+a(1-x2)>0,即f'(x)>0,所以f(x)在(-1,0)上單調(diào)遞增,f(x)<f(0)=0,故f(x)在(-1,0)上沒(méi)有零點(diǎn),不合題意.②若-1≤a<0,當(dāng)x∈(0,+∞),則g'(x)=ex-2ax>0,所以g(x)在(0,+∞)上單調(diào)遞增,所以g(x)>g(0)=1+a>0,即f'(x)>0,所以f(x)在(0,+∞)上單調(diào)遞增,f(x)>f(0)=0,故f(x)在(0,+∞)上沒(méi)有零點(diǎn),不合題意;③若a<-1.a.當(dāng)x∈(0,+∞),則g'(x)=ex-2ax>0,所以g(x)在(0,+∞)上單調(diào)遞增,g(0)=1+a<0,g(1)=e>0,所以存在m∈(0,1),使得g(m)=0,即f'(m)=0,當(dāng)x∈(0,m),f'(x)<0,f(x)單調(diào)遞減,當(dāng)x∈(m,+∞),f'(x)>0,f(x)單調(diào)遞增,所以,當(dāng)x∈(0,m),f(x)<f(0)=0,當(dāng)x→+∞,f(x)→+∞,所以f(x)在(m,+∞)上有唯一零點(diǎn).又(0,m)上沒(méi)有零點(diǎn),即f(x)在(0,+∞)上有唯一零點(diǎn).b.當(dāng)x∈(-1,0),g(x)=ex+a(1-x2),設(shè)h(x)=g'(x)=ex-2ax,h'(x)=ex-2a>0,所以g'(x)在(-1,0)上單調(diào)遞增,g'(-1)=1e+2a<0,g'(0)=1>所以存在n∈(-1,0),使得g'(n)=0.當(dāng)x∈(-1,n),g'(x)<0,g(x)單調(diào)遞減,當(dāng)x∈(n,0),g'(x)>0,g(x)單調(diào)遞增,g(x)<g(0)=1+a<0,又g(-1)=1e>所以存在t∈(-1,n),使得g(t)=0,即f'(t)=0,當(dāng)x∈(-1,t),f(x)單調(diào)遞增,當(dāng)x∈(t,0),f(x)單調(diào)遞減,有x→-1,f(x)→-∞,而f(0)=0,所以當(dāng)x∈(t,0),f(x)>0,所以f(x)在(-1,t)上有唯一零點(diǎn),(t,0)上無(wú)零點(diǎn),即f(x)在(-1,0)上有唯一零點(diǎn),所以a<-1,符合題意.所以若f