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24利用導(dǎo)數(shù)證明不等式新人教A版

課時(shí)規(guī)范練24利用導(dǎo)數(shù)證明不等式1.已知函數(shù)f(x)=(1x+12)ln((1)求曲線y=f(x)在(2,f(2))處的切線斜率;(2)當(dāng)x>0時(shí),求證:f(x)>1.2.(2024·山東濰坊模擬)已知函數(shù)f(x)=lnx+a2x2(a∈(1)討論f(x)的單調(diào)性;(2)當(dāng)a>1時(shí),證明:f(x)>3a3.(2024·河北石家莊模擬)已知函數(shù)f(x)=aex-12x2-x(1)若f(x)在R上單調(diào)遞增,求實(shí)數(shù)a的取值范圍;(2)當(dāng)a=1時(shí),證明:?x∈(-2,+∞),f(x)>sinx.4.(2024·湖南益陽模擬)已知函數(shù)f(x)=12ax2-xlnx(1)若函數(shù)f(x)在(0,+∞)上單調(diào)遞增,求實(shí)數(shù)a的取值范圍;(2)若a=2e,證明:f(x)<xex+1.5.(2024·安徽合肥模擬)已知函數(shù)f(x)=alnx+x2,其中a∈R.(1)討論f(x)的單調(diào)性;(2)當(dāng)a=1時(shí),證明:f(x)≤x2+x-1;(3)求證:對任意的n∈N*且n≥2,都有(1+122)(1+132)(1+142)…(1+1n2)

課時(shí)規(guī)范練24利用導(dǎo)數(shù)證明不等式1.(1)解f'(x)=-1x2ln(x+1)+(1x+12)1(2)證明當(dāng)x>0時(shí),欲證明f(x)>1,只需證明ln(x+1)-2xx+2設(shè)u(x)=ln(x+1)-2xx+2,則u'(x)=x2(x+1)(x+2)2>0在(0,+∞)上恒成立,所以u(x)在(0,+∞)上單調(diào)遞增,即?x>0,u(x)>u(0)=0,2.(1)解f(x)的定義域?yàn)?0,+∞),f'(x)=1當(dāng)a≤0時(shí),f'(x)>0對任意的x∈(0,+∞)恒成立,所以f(x)在(0,+∞)上單調(diào)遞增.當(dāng)a>0時(shí),令f'(x)<0,解得0<x<a;令f'(x)>0,解得x>a,所以f(x)在(0,a)內(nèi)單調(diào)遞減,在(a,+∞)上單調(diào)遞增.(2)證明由(1)可知,當(dāng)a>1時(shí),f(x)min=f(a)=lna+a2要證f(x)>3a-12a+2,只需證12lna+12>3令g(x)=lnx+4x+1-2(x>1),所以g'(x)=1因此g(x)在(1,+∞)上單調(diào)遞增,所以g(x)>g(1)=0,即lna+4a+1-2>0,故f(x)>33.(1)解因?yàn)閒(x)=aex-12x2-x,所以f'(x)=aex-x-1,由f(x)在R上單調(diào)遞增,得f'(x)≥0在R上恒成立,即aex-x-1≥0在R上恒成立,所以a≥x+1ex在R上恒成立,令h(x)=x+1ex,可得h'(x)=-xex,當(dāng)x>0時(shí),h'(x)<0,h(x)單調(diào)遞減;當(dāng)x<0時(shí),h'(x)>0,h(x)單調(diào)遞增,所以當(dāng)x=0時(shí),函數(shù)h(x)取得極大值且為最大值,最大值h(0)=(2)證明當(dāng)a=1時(shí),f(x)=ex-12x2-x,所以f'(x)=ex-x-1,且f(0)=1要證f(x)>sinx,只需證f(x)>1.①當(dāng)x∈(0,+∞)時(shí),令g(x)=f'(x)=ex-x-1,可得g'(x)=ex-1>0,所以g(x)單調(diào)遞增,所以g(x)>g(0)=0,因此f(x)單調(diào)遞增,所以f(x)>f(0)=1;②當(dāng)x=0時(shí),可得f(0)=1且sin0=0,所以f(0)>sin0,滿足f(x)>sinx;③當(dāng)-2<x<0時(shí),可得sinx<0,因?yàn)閑x>0,且-12x2-x=-12(x+1)2+12>0,所以f(x)>0,因此f(x)綜上可得,?x∈(-2,+∞),都有f(x)>sinx.4.(1)解由已知得f'(x)=ax-lnx-1.因?yàn)閒(x)在(0,+∞)上單調(diào)遞增,所以當(dāng)x>0時(shí),f'(x)≥0,即a≥lnx令h(x)=lnx+1x(x>0),則h'(x)=-lnxx2,所以當(dāng)0<x<1時(shí),h'(x)>0,當(dāng)x>1時(shí),h'(x)<0,即h(x)在(0,1)內(nèi)單調(diào)遞增,在(1,+∞)上單調(diào)遞減,因此h(x)max=h(1)=1,所以a≥1,故實(shí)數(shù)(2)證明若a=2e,要證f(x)<xex+1,只需證ex-lnx<ex+1x,即ex-ex<lnx+令t(x)=lnx+1x(x>0),則t'(x)=x-1x2,所以當(dāng)0<x<1時(shí),t'(x)<0,當(dāng)x>1時(shí),t'(x)>0,所以t(x)在(0,1)內(nèi)單調(diào)遞減,在(1,+∞)上單調(diào)遞增,則t(x)min=t(1)=1,所以令φ(x)=ex-ex(x>0),則φ'(x)=e-ex,所以當(dāng)0<x<1時(shí),φ'(x)>0,當(dāng)x>1時(shí),φ'(x)<0,所以φ(x)在(0,1)內(nèi)單調(diào)遞增,在(1,+∞)上單調(diào)遞減,則φ(x)max=φ(1)=0,所以ex-ex≤0.所以ex-ex<lnx+1x,f(x)<xex+1得證5.(1)解函數(shù)f(x)的定義域?yàn)?0,+∞),f'(x)=ax+2x=a①當(dāng)a≥0時(shí),f'(x)>0,所以f(x)在(0,+∞)上單調(diào)遞增.②當(dāng)a<0時(shí),令f'(x)=0,解得x=-當(dāng)0<x<-a2時(shí),f'(x)<0,f(x)在(0,-a2)內(nèi)單調(diào)遞減;當(dāng)x>-a2時(shí),f'(x)>0,f(x)在(綜上,當(dāng)a≥0時(shí),函數(shù)f(x)在(0,+∞)上單調(diào)遞增;當(dāng)a<0時(shí),函數(shù)f(x)在(0,-a2)內(nèi)單調(diào)遞減,在(-a2,(2)證明當(dāng)a=1時(shí),f(x)=lnx+x2,要證明f(x)≤x2+x-1,即證lnx≤x-1,即證lnx-x+1≤0.設(shè)g(x)=lnx-x+1,則g'(x)=1-xx,令g'(x)=0,得當(dāng)x∈(0,1)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.所以g(x)在x=1處取得極大值,且極大值為最大值,所以g(x)≤g(1)=0,即lnx-x+1≤0.f(x)≤x2+x-1得證.(3)證明由(2)l