2025優(yōu)化設(shè)計(jì)一輪課時(shí)規(guī)范練25　在導(dǎo)數(shù)應(yīng)用中如何構(gòu)造函數(shù)

課時(shí)規(guī)范練25在導(dǎo)數(shù)應(yīng)用中如何構(gòu)造函數(shù)高考總復(fù)習(xí)優(yōu)化設(shè)計(jì)GAOKAOZONGFUXIYOUHUASHEJI202512345678910111213141516基礎(chǔ)鞏固練1.(2024·重慶江北高三模擬)若函數(shù)y=f(x)滿足xf'(x)>-f(x)在R上恒成立,且a>b,則(

)A.af(b)>bf(a) B.af(a)>bf(b)C.af(a)<bf(b) D.af(b)<bf(a)B解析

由xf'(x)>-f(x),設(shè)g(x)=xf(x),則g'(x)=xf'(x)+f(x)>0,所以g(x)在R上是增函數(shù),又因?yàn)閍>b,所以g(a)>g(b),即af(a)>bf(b),故選B.123456789101112131415162.(2024·云南昆明高三期中)設(shè)

,c=e-2+ln2,設(shè)a,b,c的大小關(guān)系為(

)A.a>b>c B.a>c>bC.b>c>a D.c>b>aA12345678910111213141516A.a>b>c B.a>c>bC.b>a>c D.b>c>aC123456789101112131415164.(2024·甘肅蘭州期末)已知函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),對(duì)任意x∈R,都有f'(x)<f(x)成立,則(

)A.ef(2)<f(3) B.ef(2)≤f(3)C.ef(2)>f(3) D.f(2)<e2f(3)C123456789101112131415165.(2024·山東威海期中)已知定義在R上的函數(shù)f(x)滿足xf'(x)-f(x)>0,且f(1)=2,則f(ex)>2ex的解集為(

)A.(0,+∞) B.(ln2,+∞)C.(1,+∞) D.(0,1)A123456789101112131415166.(多選題)(2024·河北邯鄲聯(lián)考)已知a>0,b∈R,e是自然對(duì)數(shù)的底數(shù),若b+eb=a+lna,則a-b的值可以是(

)A.-1 B.1

C.2

D.3BCD解析

設(shè)函數(shù)f(x)=x+ex,則f(x)在R上單調(diào)遞增,所以f(b)-f(ln

a)=b+eb-(ln

a+eln

a)=a+ln

a-(ln

a+a)=0,所以b=ln

a,即a=eb,所以a-b=eb-b,令g(x)=ex-x,則g'(x)=ex-1,當(dāng)x<0時(shí),g'(x)<0,g(x)單調(diào)遞減;當(dāng)x>0時(shí),g'(x)>0,g(x)單調(diào)遞增,所以g(x)≥g(0)=1,從而a-b≥1,結(jié)合選項(xiàng),選項(xiàng)BCD符合題意,故選BCD.123456789101112131415167.(多選題)(2023·河北滄州高二?？茧A段練習(xí))下列不等關(guān)系成立的有(

)A.6ln5>5ln6 B.5e5<4e6C.ln1.1<0.1 D.e0.1<1.1ABC

123456789101112131415168.(2024·四川眉山高三期中)設(shè)函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),f(0)=1且3f(x)=f'(x)-3,則4f(x)>f'(x)的解集是

.

123456789101112131415169.(2024·云南曲靖高三模擬)已知f'(x)是函數(shù)f(x)的導(dǎo)函數(shù),且滿足f'(x)>f(x)在R上恒成立,則不等式f(2x-1)-e3x-2f(1-x)>0的解集是

(用區(qū)間表示).

12345678910111213141516綜合提升練10.(2023·河南鄭州模擬)若3a+log3a=9b+3log27b,則下列結(jié)論正確的是(

)A.a>2b

B.a<2bC.a>b2

D.a<b2B解析

設(shè)f(x)=3x+log3x,易知f(x)在(0,+∞)內(nèi)單調(diào)遞增,∵3a+log3a=9b+3log27b=32b+log3b,∴f(2b)=32b+log3(2b)>32b+log3b=3a+log3a=f(a),∴2b>a,故A錯(cuò)誤,B正確;當(dāng)b=1時(shí),f(a)-f(b2)=6>0,此時(shí)f(a)>f(b2),有a>b2,當(dāng)b=2時(shí),f(a)-f(b2)=-log32<0,此時(shí)f(a)<f(b2),有a<b2,所以C,D錯(cuò)誤.故選B.12345678910111213141516D1234567891011121314151612.(2024·黑龍江大慶模擬)設(shè)a=1.7,b=tan1.1,c=2ln2.1,則(

)A.a<b<c B.a<c<bC.c<a<b D.b<a<cC所以b>a.令f(x)=ex-ex,則f'(x)=ex-e,當(dāng)x>1時(shí),f'(x)>0,所以f(x)在(1,+∞)內(nèi)單調(diào)遞增,則f(1.7)=e1.7-1.7e>f(1)=0,則e1.7>1.7e>1.7×2.7=4.59>4.41=2.12,因此1.7>ln

2.12=2ln

2.1,即a>c,故c<a<b,故選C.1234567891011121314151613.(多選題)(2024·海南?？谀M)已知函數(shù)f(x)的定義域?yàn)镽,其導(dǎo)函數(shù)為f'(x),且2f(x)+f'(x)=x,f(0)=-,則(

)A.f(-1)>-2B.f(1)>-1C.f(x)在(-∞,0)內(nèi)是減函數(shù)D.f(x)在(0,+∞)內(nèi)是增函數(shù)ABD12345678910111213141516解析

令g(x)=e2xf(x),可得g'(x)=e2x[2f(x)+f'(x)],因?yàn)?f(x)+f'(x)=x,所以g'(x)=e2xx,當(dāng)x>0時(shí),g'(x)>0,g(x)單調(diào)遞增;當(dāng)x<0時(shí),g'(x)<0,g(x)單調(diào)遞減.1234567891011121314151614.(2024·四川瀘州期末)已知正數(shù)x,y滿足xlnx+xlny=ey,則xy-3y的最小值為(

)B解析

因?yàn)閤ln

x+xln

y=ey,即xln(xy)=ey,所以(xy)ln(xy)=yey,所以ln(xy)eln(xy)=yey.令g(x)=xex(x>0),則g'(x)=(x+1)ex>0,所以g(x)=xex在(0,+∞)內(nèi)單調(diào)遞增,所以ln(xy)=y,即xy=ey,所以xy-3y=ey-3y,令f(x)=ex-3x(x>0),則f'(x)=ex-3,令f'(x)=ex-3>0,解得x>ln

3;令f'(x)=ex-3<0,解得0<x<ln

3.所以f(x)=ex-3x在(0,ln

3)內(nèi)單調(diào)遞減,在(ln

3,+∞)內(nèi)單調(diào)遞增,所以f(x)min=eln

3-3×ln

3=3-3ln

3,即xy-3y的最小值為3-3ln

3,故選B.12345678910