指點(diǎn)迷津(三)　在導(dǎo)數(shù)應(yīng)用中如何構(gòu)造函數(shù)

指點(diǎn)迷津(三)在導(dǎo)數(shù)應(yīng)用中如何構(gòu)造函數(shù)第三章在導(dǎo)數(shù)應(yīng)用中如何構(gòu)造函數(shù)函數(shù)與方程思想、轉(zhuǎn)化與化歸思想是高中數(shù)學(xué)思想中比較重要的兩大思想,而構(gòu)造函數(shù)的解題思路恰好是這兩種思想的良好體現(xiàn).一、條件中f(x)與f'(x)共存問(wèn)題的函數(shù)構(gòu)造例1(1)已知f(x)是定義在R上的函數(shù),f'(x)是f(x)的導(dǎo)函數(shù),且f'(x)+f(x)>1,f(1)=2,則下列結(jié)論一定成立的是(

)答案:(1)D

(2)C

解析:(1)令φ(x)=ex[f(x)-1],則φ'(x)=ex[f'(x)+f(x)-1]>0,規(guī)律方法

解答f(x)與f'(x)共存的不等式問(wèn)題的策略是:將f(x)與f'(x)共存的不等式與導(dǎo)數(shù)運(yùn)算法則結(jié)合起來(lái),合理構(gòu)造出相關(guān)的可導(dǎo)函數(shù),然后利用函數(shù)的性質(zhì)解決問(wèn)題.幾種常見(jiàn)構(gòu)造函數(shù)的方法有:(1)對(duì)于f'(x)g(x)+f(x)g'(x)>0(或<0),構(gòu)造F(x)=f(x)·g(x);(2)對(duì)于xf'(x)+cf(x)>0(或<0)(c為常數(shù)),構(gòu)造F(x)=xcf(x);(3)對(duì)于f'(x)+cf(x)>0(或<0)(c為常數(shù)),構(gòu)造F(x)=ecxf(x),對(duì)點(diǎn)訓(xùn)練1(1)已知函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),且f'(x)<f(x)對(duì)任意的x∈R恒成立,則(

)A.f(1)<ef(0),f(2020)>e2020f(0)B.f(1)>ef(0),f(2020)>e2020f(0)C.f(1)>ef(0),f(2020)<e2020f(0)D.f(1)<ef(0),f(2020)<e2020f(0)答案:(1)D

(2)A

二、解題過(guò)程中為解決問(wèn)題而構(gòu)造函數(shù)

A.a<b<c

B.c<b<aC.c<a<b

D.a<c<b答案:(1)A

(2)C

k'(x)=(1-x2-2x)ex>0,∴k(x)在區(qū)間(0,0.1]上單調(diào)遞增.∴k(x)>k(0)=0.∴在區(qū)間(0,0.1]上,y2'

>0,∴y2=xex+ln(1-x)在區(qū)間(0,0.1]上單調(diào)遞增.∴y2>0,∴a1>c1.∴在區(qū)間(0,0.1]上,b1>a1>c1.故當(dāng)x=0.1時(shí),有b>a>c.規(guī)律方法

解題中若遇到比較大小及有關(guān)不等式、方程及最值之類的問(wèn)題,設(shè)法建立起目標(biāo)函數(shù),并確定變量的限制條件,通過(guò)研究函數(shù)的單調(diào)性、最值等問(wèn)題,常可使問(wèn)題變得明了.準(zhǔn)確構(gòu)造出符合題意的函數(shù)是解題的關(guān)鍵,構(gòu)造函數(shù)時(shí)往往從觀察所求對(duì)象的特征入手,常把某常數(shù)看成變量抽象出函數(shù)來(lái).答案:(1)D

(2)D

解析:(1)設(shè)f(x)=xex,則f'(x)=ex+xex=(x+1)ex.當(dāng)x>0時(shí),f'(x)>0,所以f(x)=xex在(0,+∞)上單調(diào)遞增.因?yàn)?2x+y)ey=3e3-2x,所以(2x+y)ey+2x=3e3,即f(2x+y)=f(3),三、對(duì)題設(shè)條件同構(gòu)后構(gòu)造出函數(shù)例3(2022江蘇南京、鹽城二模)已知實(shí)數(shù)a,b∈(1,+∞),且2(a+b)=e2a+2lnb+1,e為自然對(duì)數(shù)的底數(shù),則(

)A.1<b<a

B.a<b<2aC.2a<b<ea

D.ea<b<e2a答案:D

解析:由2(a+b)=e2a+2ln

b+1,得e2a-2a-1=2(b-ln

b-1)=2(eln

b-ln

b-1),構(gòu)造函數(shù)f(x)=ex-x-1(x>0),則f'(x)=ex-1>0,函數(shù)f(x)在(0,+∞)上單調(diào)遞增,且f(0)=0,所以f(2a)=2f(ln

b)>f(ln

b),所以2a>ln

b,即b<e2a,又e2a-2a-1>2(ea-a-1),所以f(2a)=2f(ln

b)>2f(a),所以a<ln

b,即ea<b,則ea<b<e2a,故選D.規(guī)律方法

對(duì)于含有ex,ln

x的超越不等式,是不容易求出其解集的,經(jīng)常采用對(duì)不等式的結(jié)構(gòu)變形,在結(jié)構(gòu)上符合某一函數(shù)的兩個(gè)函數(shù)值,抽象出這個(gè)函數(shù)即構(gòu)造出函數(shù),然后利用函數(shù)的性質(zhì)求解.對(duì)點(diǎn)訓(xùn)練3(2023江蘇南通調(diào)研)已知α,β均為銳角,且