2024年新高考突破專練 考點(diǎn)突破練19　不等式恒成立或有解問題（附答案解析）

考點(diǎn)突破練19不等式恒成立或有解問題1.(2023江蘇南京模擬)已知函數(shù)f(x)=4lnx-ax+a+3x(a≥(1)當(dāng)a=12時(shí),求f(x(2)當(dāng)a≥1時(shí),設(shè)g(x)=2ex-4x+2a,若存在x1,x2∈12,2,使得f(x1)>g(x2)成立,求實(shí)數(shù)a的取值范圍.(e為自然對(duì)數(shù)的底數(shù),e=2.(2020新高考Ⅰ,21)已知函數(shù)f(x)=aex-1-lnx+lna.(1)當(dāng)a=e時(shí),求曲線y=f(x)在點(diǎn)(1,f(1))處的切線與兩坐標(biāo)軸圍成的三角形的面積;(2)若f(x)≥1恒成立,求實(shí)數(shù)a的取值范圍.3.(2023湖南郴州三模)已知函數(shù)f(x)=x2-ax+1,g(x)=lnx+a(a∈R).(1)若a=1,f(x)>g(x)在區(qū)間(0,t)上恒成立,求實(shí)數(shù)t的取值范圍;(2)若函數(shù)f(x)和g(x)有公切線,求實(shí)數(shù)a的取值范圍.4.(2023河北石家莊三模)若定義在區(qū)間I上的函數(shù)y=f(x),其圖象上存在不同兩點(diǎn)處的切線相互平行,則稱函數(shù)y=f(x)為區(qū)間I上的“曲折函數(shù)”,現(xiàn)已知函數(shù)f(x)=2a2lnx+x2(a>0).(1)證明:y=f(x)是(0,+∞)上的“曲折函數(shù)”;(2)設(shè)0<x0<a,證明:?xm∈(x0,a),使得對(duì)于?x∈(xm,a),均有(a-x0)f'(x)-f(a)+f(x0)<0.

考點(diǎn)突破練19不等式恒成立或有解問題1.解(1)f(x)的定義域?yàn)?0,+∞),當(dāng)a=12時(shí),f(x)=4lnx-x2+72x,∴f'(x令f'(x)>0,得1<x<7,令f'(x)<0,可得0<x<1或x>7,∴函數(shù)f(x)的單調(diào)遞減區(qū)間為(0,1),(7,+∞),單調(diào)遞增區(qū)間為(1,7).∴當(dāng)x=1時(shí),函數(shù)取得極小值f(1)=3;當(dāng)x=7時(shí),函數(shù)取得極大值f(7)=4ln7-3.(2)f'(x)=-ax2+4x-(a+3)x2(x>0),令h若a≥1,則Δ=16-4a2-12a=-4(a-1)(a+4)≤0,當(dāng)且僅當(dāng)a=1時(shí),等號(hào)成立,∴h(x)≤0,即f'(x)≤0且不恒為零,∴f(x)在區(qū)間(0,+∞)上單調(diào)遞減,∴當(dāng)a≥1時(shí),f(x)在12,∴f(x)在12,2上的最大值為f12=-4ln2+32a+6.g'(x)=2ex-4,令g'(x)=0,∴當(dāng)x∈12,ln2時(shí),g'(x)<0,g(x當(dāng)x∈(ln2,2)時(shí),g'(x)>0,g(x)單調(diào)遞增,∴g(x)在12,2上的最小值為g(ln2)=4-4ln2+若存在x1,x2∈12,2,使得f(x1)>g(x2)成立,則f(x)max>g(x)min成立,∴-4ln2+32a+6>4-4ln2+2a,解得a<4,又a≥1,∴實(shí)數(shù)2.解(1)當(dāng)a=e時(shí),f(x)=ex-lnx+1,∴f'(x)=ex-1x∴f'(1)=e-1.∵f(1)=e+1,∴切點(diǎn)坐標(biāo)為(1,1+e),∴曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y-e-1=(e-1)(x-1),即y=(e-1)x+2,∴切線與兩坐標(biāo)軸交點(diǎn)坐標(biāo)分別為(0,2),-2∴所求三角形面積為12×2×-(2)(方法一)∵f(x)=aex-1-lnx+lna,∴f'(x)=aex-1-1x,且a>0.設(shè)g(x)=f'(x),則g'(x)=aex-1+1x∴g(x)在(0,+∞)上單調(diào)遞增,即f'(x)在(0,+∞)上單調(diào)遞增,當(dāng)a=1時(shí),f'(1)=0,∴f(x)min=f(1)=1,∴f(x)≥1成立.當(dāng)a>1時(shí),0<1a<1,∴e1∴f'1af'(1)=a(e1a-1-1)(a-1)<0,∴存在唯一x0>0,使得f'(x0)=aex0-1?1x0=0,即aex0-1=1x0,即lna+x0-1=-lnx0,且當(dāng)x∈(0,x0)時(shí),f'(x)<0,f(x)單調(diào)遞減,當(dāng)x∈(x0,+∞)時(shí),f'(x)>0,f(x)單調(diào)遞增,因此f(x)min=f(x0)=aex0-1-lnx0+lna=1x0+lna+x0-1+lna≥2lna-1+21x0·x0=2lna+∴f(1)<1,f(x)≥1不是恒成立.綜上所述,實(shí)數(shù)a的取值范圍是[1,+∞).(方法二)由題意知a>0,x>0,令aex-1=t,∴l(xiāng)na+x-1=lnt,∴l(xiāng)na=lnt-x+1.∴f(x)=aex-1-lnx+lna=t-lnx+lnt-x+1.∵f(x)≥1,即t-lnx+lnt-x+1≥1?t+lnt≥x+lnx,而y=x+lnx在(0,+∞)上單調(diào)遞增,故t≥x,即aex-1≥x,分離參數(shù)后有a≥xe令g(x)=xex-1,∴g'(x當(dāng)0<x<1時(shí),g'(x)>0,g(x)單調(diào)遞增;當(dāng)x>1時(shí),g'(x)<0,g(x)單調(diào)遞減.∴當(dāng)x=1時(shí),g(x)=xex-1取得最大值,且最大值為g(1)=1.∴a≥1.∴a的取值范圍為[1,3.解(1)由題意,當(dāng)a=1時(shí),設(shè)h(x)=f(x)-g(x),則h(x)=x2-x+1-lnx-1=x2-x-lnx(x>0),h'(x)=2x-1-1x=2x2-x-1x=(2x+1)(x-1)x,令h'(x)=0,得x=1,∴h(x)在(0,1)上單調(diào)遞減,在(1,(2)設(shè)函數(shù)f(x)在點(diǎn)(x1,f(x1))處與函數(shù)g(x)在點(diǎn)(x2,g(x2))處有相同的切線,且切線斜率存在,則f'(x1)=g'(x2)=f(x1)-g(x2∴x1=12x2+a2,代入x1-x2x2=x12-ax1+1-lnx2-a,得14x22+a2x2+lnx2+a24+a-2=0.∴問題轉(zhuǎn)化為關(guān)于x的方程1F'(x)=-12設(shè)2x02-ax0-1=0(x0>當(dāng)0<x<x0時(shí),F'(x)<0,當(dāng)x>x0時(shí),F'(x)>0.∴F(x)在(0,x0)上單調(diào)遞減,在(x0,+∞)上單調(diào)遞增,∴問題轉(zhuǎn)化為F(x)的最小值小于或等于0.∴F(x)的最小值為F(x0)=14x02+a2x0+由2x02-ax0-1=0,知a=2x0-1x0,故F(x0)=x02+2x0-1x0+lnx0-2.設(shè)φ(x)=x2+2x-1x+lnx-2(x>0),則φ'(x)=2x+2+1x2+1x>0,故φ(x)在(0,+∞)上單調(diào)遞增.∵φ(1)=0,∴當(dāng)x∈(0,1]時(shí),φ(x)≤0,∴F(x)的最小值F(x0)≤0等價(jià)于0<x0≤1.又函數(shù)y=2x-1x在(0,1]上單調(diào)遞增,∴a=2x0-4.證明(1)(方法一)要證y=f(x)是(0,+∞)上的曲折函數(shù),即證存在兩個(gè)不同的x1,x2∈(0,+∞),使得f'(x1)=f'(x2),令g(x)=f'(x)=2a2x+即證?x1,x2∈(0,+∞),x1≠x2,使得g(x1)=g(x2).任取m>4a>0,考慮方程g(x)=m的正數(shù)解的情況.2a2x+2x=m?2x2-mx+2a判別式Δ=m2-16a2>0,故方程有兩個(gè)不相等的實(shí)根x1,x2,則x1+x2=m2>0,x1x2=a2>0,從而x1>0,x2>0即g(x)=m有兩個(gè)不相等的正實(shí)數(shù)解x1,x2,所以g(x1)=g(x2),即y=f(x)是(0,+∞)上的曲折函數(shù).(方法二)設(shè)P(x1,y1),Q(x2,y2)是函數(shù)f(x)圖象上兩點(diǎn),x1≠x2,f'(x)=2a2x+2x(x>0),f'(x1)=f'(x2)等價(jià)于2a2x1+2x1=2a2x2+2x2,即2(x1-x2)a2x1x2-1=0(x1≠x2),即x1x2=a2>0,即存在兩個(gè)不相等的正實(shí)數(shù)x1,x2,使f'(x1)=f'((2)設(shè)函數(shù)F(x)=(a-x0)f'(x)-f(a)+f(x0),則F(x)=(a-x0)2a2x+2x-2a2lna則F'(x)=(a-x0)x2(2x2-2a所以當(dāng)x∈(0,a)時(shí),F'(x)<0,當(dāng)x∈(a,+∞)時(shí),F'(x)>0,故F(x)在(0,a)上單調(diào)遞減,在(a,+∞)上單調(diào)遞增.(方法一)取x=x0代入函數(shù)y=F(x),可得F(x0)=(a-x0)2a2x0+2x0-2a2lnax0-a2+x令ax0=t,其中t>1,故F(x0)=a2(2t-3-1t2+構(gòu)造函數(shù)H(t)=2t-3-1t2+2t-2lnt(t>1),則H'(t)=2+2t3?2t2?2t=2(t-1)2(t+1)t3>0,從而H(t)在再取x=a代入函數(shù)y=F(x),可得F(a)=4a(a-x0)-2a2lnax0-a2+x02=a2(3-4x令x0a=n,其中0<n<1,則F(a)=a2(3-4n+n2+2ln構(gòu)造函數(shù)S(n)=3-4n+n2+2lnn,n∈(0,1),則S'(n)=2(n-1)2n>0,故S(n)在(0,1)上單調(diào)遞增,即S(n)<S(1)=0,所以F(又因?yàn)镕(x)在(x0,a)上單調(diào)遞減,結(jié)合①與②,由零點(diǎn)存在定理,必存在唯一的xm∈(x0,a),使得F(xm)=0,且對(duì)任意的x∈(xm,a),均有F(x)<F(xm)=0.(方法二)取x=x0代入函數(shù)y=F(x),可得F(x0)=(a-x0)2a2x0+2x0-2a2設(shè)m(x)=(a-x)2a2x+2x-2a2lnax-a2+x2則m'(x)=-2(a-x)2(x+a)x2<0,所以m(x)在(0,a)上單調(diào)遞減,m(x)>m(a)再取x=a代入函數(shù)y=F(x),可得F(a)=4a(a-x0)-2a2lnax0-a2+x02=3a2-4ax0+x02-設(shè)n(t)=3t2-4tx0+x02-2t2lntx0,t∈(x0,+∞),n'(t)=4t-4x0-4tlnt+4tlnx0