適用于新高考新教材備戰(zhàn)2025屆高考數(shù)學(xué)一輪總復(fù)習(xí)課時(shí)規(guī)范練21在導(dǎo)數(shù)應(yīng)用中如何構(gòu)造函數(shù)新人教A版

課時(shí)規(guī)范練21在導(dǎo)數(shù)應(yīng)用中如何構(gòu)造函數(shù)基礎(chǔ)鞏固練1.(2024·重慶江北高三模擬)若函數(shù)y=f(x)滿意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)函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),對(duì)隨意的x∈R,都有f'(x)>f(x)ln2成立,則()A.4f(3)>f(5)B.4f(3)<f(5)C.4f(3)=f(5)D.4f(3)與f(5)大小關(guān)系不確定4.(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)隨意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)滿意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,若b+eb=a+lna,則a-b的值可以是()A.-1 B.1 C.2 D.37.(多選題)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),且滿意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.

綜合提升練10.設(shè)a,b為正數(shù),且lnab=2b-a,則(A.12<ab<1 B.1C.12<ab<1 D.1<ab<11.(2024·陜西咸陽(yáng)高三聯(lián)考)已知函數(shù)f(x)的定義域是(-5,5),其導(dǎo)函數(shù)為f'(x),且f(x)+xf'(x)>2,則不等式(2x-3)f(2x-3)-(x-1)f(x-1)>2x-4的解集是()A.(2,+∞) B.(2,6)C.(-4,6) D.(2,4)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·海南海口模擬)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)上單調(diào)遞減D.f(x)在(0,+∞)上單調(diào)遞增14.(2024·山東濰坊高三期末)定義在(0,+∞)上的函數(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.

創(chuàng)新應(yīng)用練15.(2024·四川瀘州期末)已知正數(shù)x,y滿意xlnx+xlny=ey,則xy-3y的最小值為()A.13ln3 B.3-3ln3 C.-13ln3 D.316.(2024·浙江寧波高三模擬)若對(duì)隨意的x1,x2∈[1,π2],x1<x2,x2sinx1-x1sin

課時(shí)規(guī)范練21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a>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,+∞)上單調(diào)遞減,而a=1e=f(1),b=ln33=ln3eln3=f(ln3),c=e-2+ln2=2e3.B解析構(gòu)造函數(shù)h(x)=f(x)2x,則h'(x)=2xf'(x)-2xln2·f(x)24.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(x5.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,+∞)上單調(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)=f6.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,故選BCD.7.ABC解析對(duì)于A,由于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),則f'(x)=1x-1=1-xx<0,故f(x)在(1,+∞)上單調(diào)遞減,所以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,所以g(0.1)>g(0)=0,即e0.1>8.(ln23,+∞)解析令g(x)=f(x)+1e3x,因?yàn)閒(0)=1,故g(0)=f'(x)e所以g(x)為常函數(shù),則g(x)=f(x)+1e3x=2,所以f(x)=2e3x-1,f'(x)=6e3x,又4f(x)>f'(x),所以8e3x-49.(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-2f(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)10.D解析由a,b為正數(shù),且lnab=2b-a可得a+lna=2b+ln1b>1b+ln1b,因?yàn)楹瘮?shù)f(x)=x+lnx在(0,+∞)上單調(diào)遞增,且f(a)>f)1b),所以a>1b,所以ab>1,所以2b-a=lnab>0,ab<2,故1<ab<2,故C錯(cuò)誤,D正確;設(shè)a=b,則2lna=2a-a,設(shè)g(x)=2lnx+x-2x,x>0,則g'(x)=2x+1+2x2>0,g(x)在(0,+∞)上單調(diào)遞增,且g(1)=-1<0,g(2)=2ln2+1>0,所以存在a∈(1,2)使得g11.D解析設(shè)g(x)=xf(x)-2x,則g'(x)=f(x)+xf'(x)-2.∵f(x)+xf'(x)>2,∴g'(x)>0,則g(x)在(-5,5)內(nèi)單調(diào)遞增.不等式(2x-3)f(2x-3)-(x-1)f(x-1)>2x-4等價(jià)于(2x-3)f(2x-3)-2(2x-3)>(x-1)f(x-1)-2(x-1),即g(2x-3)>g(x-1),則-解得2<x<4,故選D.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,+∞)上單調(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>c13.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,所以g(0)=e0f(0)=-14,由g(-1)>g(0),即e-2f(-1)>-14,可得f(-1)>-14e2>-2,所以A正確;又由g(1)>g(0),即e2f(1)>-14,可得f(1)>-14e2>-1,所以B正確;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)=(xe2x)'-2xe2x=e2x>0,所以函數(shù)h(x)為單調(diào)遞增函數(shù),又因?yàn)閔(0)=g'(0)-14.(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)在區(qū)間(0,+∞)上單調(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>15.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,+∞)上單調(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,+∞)上單調(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.16.co