適用于新高考新教材2024版高考數(shù)學(xué)二輪復(fù)習(xí)考點(diǎn)突破練19不等式恒成立或有解問(wèn)題課件

考點(diǎn)突破練19不等式恒成立或有解問(wèn)題12341234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)=4ln

7-3.123412342.(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.123412341234(方法二)由題意知a>0,x>0,令aex-1=t,∴l(xiāng)n

a+x-1=ln

t,∴l(xiāng)n

a=ln

t-x+1.∴f(x)=aex-1-ln

x+ln

a=t-ln

x+ln

t-x+1.∵f(x)≥1,即t-ln

x+ln

t-x+1≥1?t+ln

t≥x+ln

x,12343.(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.1234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-ln

x-1=x2-x-ln

x(x>0),∴h(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,∴h(x)min=h(1)=0.∴t的取值范圍為(0,1].1234123412344.(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)<0.12341234123412341234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)=0.12341234因?yàn)閠>x0,所以h'(t)<0,n'(t)在(x0,+∞)上單調(diào)遞減,所以n'(t)<n'(x0)=0,所以n(t)在(x0,+∞)上單調(diào)遞減,所以n(t)<n(x0)=0,所以F(a)<0.④又因?yàn)镕(x)在(x0,a)上單調(diào)遞減,結(jié)合③與④,由零點(diǎn)存在定理,