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)一、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)2.(2024·云南昆明高三期中)設(shè)a=1e,b=ln33,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>a3.(2024·福建寧德模擬)已知a=ln22,b=1e(e=2.718…為自然對(duì)數(shù)的底數(shù)),c=2ln39,則a,b,cA.a>b>c B.a>c>bC.b>a>c D.b>c>a4.(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)5.(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)6.(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.37.(多選題)(2023·河北滄州高二校考階段練習(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<1.18.(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.

9.(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).

二、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<b211.(2024·陜西咸陽高三聯(lián)考)已知函數(shù)f(x)=exx-ax,當(dāng)x2>x1>0時(shí),不等式f(x1)A.(-∞,e) B.(-∞,e]C.(-∞,e2) D.(-∞,12.(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<c13.(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)=-14,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)14.(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()A.13ln3 B.3-C.-13ln3 D.3+15.(2024·山東濰坊高三期末)定義在(0,+∞)內(nèi)的函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),當(dāng)x>0時(shí),xf'(x)<1,且f(e)=3,則不等式f(x2)-2lnx<2的解集為.

16.(2024·浙江寧波高三模擬)若對(duì)任意的x1,x2∈[1,π2],x1<x2,x2sinx1-

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)1.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.2.A解析構(gòu)造函數(shù)f(x)=xex,則f'(x)=1-xex,當(dāng)x>1時(shí),f'(x)<0,函數(shù)f(x)=xex在[1,+∞)內(nèi)為減函數(shù),而a=1e=f(1),b=ln33=ln3eln3=f(ln3),c=e-2+ln2=2e2=f(2),又因?yàn)?<ln3<3.C解析設(shè)f(x)=lnxx,所以f'(x)=1-lnxx2(x>0),當(dāng)x∈(0,e)時(shí),f'(x)>0,f(x)=lnxx單調(diào)遞增;當(dāng)x∈(e,+∞)時(shí),f'(x)<0,f(x)=lnxx單調(diào)遞減.因?yàn)閍=ln22=2ln24=ln44=f(4),b=1e4.C解析由已知得f'(x)-f(x)<0,設(shè)g(x)=f(x)ex,則g'(x)=f'(x)-f(x)ex<0,所以g(x)在R上單調(diào)遞減,所以5.A解析設(shè)g(x)=f(x)x,x>0,因?yàn)閤f'(x)-f(x)>0,所以g'(x)=xf'(x)-f(x)x2>0,所以g(x)在(0,+∞)內(nèi)單調(diào)遞增,因?yàn)閒(1)=2,所以g(1)=f(1)1=2,由f(ex)>2ex,且ex>0得f(ex)ex>2,則g(ex)=f(ex)ex>2=g(1),所以e6.BCD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(lna)=b+eb-(lna+elna)=a+lna-(lna+a)=0,所以b=lna,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.7.ABC解析對(duì)于A,∵15625=56>65=7776,∴l(xiāng)n56>ln65,因此6ln5>5ln6,故A正確;對(duì)于B,∵5<4e,∴5e5<4e6,故B正確;對(duì)于C,令函數(shù)f(x)=lnx-x+1(x>1),f'(x)=1x-1=1-xx<0,故f(x)在(1,+∞)內(nèi)單調(diào)遞減,∴f(1.1)<f(1)=0,即ln1.1<0.1,故C正確;對(duì)于D,令函數(shù)g(x)=ex-x-1(1>x>0),g'(x)=ex-1>0,故g(x)在(0,1)內(nèi)單調(diào)遞增,∴g(0.1)>g(0)=0,即e0.1>1.1,故D錯(cuò)誤8.(ln23,+∞)解析令g(x)=f(x)+1e3x,因?yàn)閒(0)=1,故g(0)=f(0)+1e0=2,所以g'(x)=f'(x)e3x-3(f(x)+1)e3x(e3x)2=f'(x)-3f(x)-3e3x=0,所以g(x)為常數(shù)函數(shù),則9.(23,+∞)解析令g(x)=f(x)ex,則g'(x)=f'(x)-f(x)ex>0,所以g(x)在R上單調(diào)遞增,由f(2x-1)-e3x-2·f(1-x)>0,兩端同除以e2x-1,并移項(xiàng)得f(2x-1)e2x-1>f(1-x)e1-x,即g(2x-1)>g(1-x),又因?yàn)間(x)在R上單調(diào)遞增10.B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;又f(a)-f(b2)=3a+log3a-(3b2+log3b2)=32b+log3b-(3b2+log3b2)=32b-3b2-log3b,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11.D解析因?yàn)楫?dāng)x2>x1>0時(shí),不等式f(x1)x2?f(x2)x1<0恒成立,所以f(x1)x2<f(x2)x1,即x1f(x1)<x2f(x2).令g(x)=xf(x)=ex-ax2,則g(x1)<g(x2),又因?yàn)閤1<x2,所以g(x)在(0,+∞)內(nèi)單調(diào)遞增,所以g'(x)=ex-2ax≥0在(0,+∞)內(nèi)恒成立,分離參數(shù)得2a≤exx恒成立,令h(x)=exx(x>0),則只需2a≤h(x)min,而h'(x)=ex·x-1x2,令h'(x)>0,得x>1,令h'(x)<0,得0<x<1,12.C解析因?yàn)閥=tanx在(0,π2)內(nèi)單調(diào)遞增,所以b=tan1.1>tanπ3=3>1.7=a,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>ln2.12=2ln2.1,即a>c,故c<a<b13.ABD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.又因?yàn)閒(0)=-14,可得g(0)=e0f(0)=-14,由g(-1)>g(0),即e-2f(-1)>-14,可得f(-1)>-14e2>-2,又由g(1)>g(0),即e2f(1)>-14,可得f(1)>-14e2>-1,因?yàn)間(x)=e2xf(x),可得f(x)=g(x)e2x,可得f'(x)=g'(x)-2g(x)e2x,設(shè)h(x)=g'(x)-2g(x),可得h'(x)=(所以函數(shù)h(x)為增函數(shù),又因?yàn)閔(0)=g'(0)-2g(0)=-e0f(0)=14>0,所以f'(x)>所以f(x)在(0,+∞)內(nèi)是增函數(shù),所以D正確.故選ABD.14.B解析因?yàn)閤lnx+xlny=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>ln3;令f'(x)=ex-3<0,解得0<x<ln3.所以f(x)=ex-3x在(0,ln3)內(nèi)單調(diào)遞減,在(ln3,+∞)內(nèi)單調(diào)遞增,所以f(x)min=eln3-3×ln3=3-3ln3,即xy-3y的最小值為3-3ln3,故選B.15.(e,+∞)解析構(gòu)造函數(shù)g(x)=f(x)-lnx(x>0),則g'(x)=f'(x)-1x=xf'(x)-1x<0,所以g(x)在(0,+∞)內(nèi)單調(diào)遞減,由f(x2)-2lnx<2,得f(x2)-lnx2<f(e)-lne,即g(x2)<g(e),所以x2>e,x>0,解得x>e,所以不等式f(x216.cos1-sin