新教材老高考適用2024高考數(shù)學(xué)一輪總復(fù)習(xí)高考解答題專項(xiàng)一第1課時(shí)利用導(dǎo)數(shù)證明不等式北師大版

PAGEPAGE8高考解答題專項(xiàng)一函數(shù)與導(dǎo)數(shù)中的綜合問題第1課時(shí)利用導(dǎo)數(shù)證明不等式1.(2024吉林長春診斷測試)已知函數(shù)f(x)=aex-ex.(1)若對隨意的實(shí)數(shù)x都有f(x)≥0成立,求實(shí)數(shù)a的取值范圍;(2)當(dāng)a≥1且x≥0時(shí),證明:f(x)≥(x-1)2.2.(2024浙江寧波高三期末)已知函數(shù)f(x)=aex-4x,a∈R.(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)當(dāng)a=1時(shí),證明:f(x)+x2+1>0.3.(2024遼寧朝陽高三一模)已知函數(shù)f(x)=ex-asinx-x,曲線f(x)在點(diǎn)(0,f(0))處的切線方程為x+y-1=0.(1)求實(shí)數(shù)a的值;(2)證明:?x∈R,f(x)>0.4.(2024河北石家莊高三三模)已知函數(shù)f(x)=alnx-x2+x+3a.若0<a<14,證明:f(x)<exx-x5.(2024福建泉州高三二模)已知函數(shù)f(x)=a-lnxx(1)求實(shí)數(shù)a的值,并求函數(shù)f(x)的單調(diào)區(qū)間;(2)證明:f(x)+x+23>06.(2024湖南郴州高三三模)已知函數(shù)f(x)=(x+1)lnx.(1)求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;(2)證明:ln21+ln76+…+ln(n2-2)

高考解答題專項(xiàng)一函數(shù)與導(dǎo)數(shù)中的綜合問題第1課時(shí)利用導(dǎo)數(shù)證明不等式1.(1)解若對隨意的實(shí)數(shù)x都有f(x)≥0,即aex-ex≥0,所以a≥ex令g(x)=exex,則g'(x)=1-xex-1.令g'當(dāng)x<1時(shí)g'(x)>0;當(dāng)x>1時(shí)g'(x)<0,所以g(x)在x=1處取得極大值亦即最大值g(1)=1,即a≥1.故實(shí)數(shù)a的取值范圍是[1,+∞).(2)證明由于當(dāng)a≥1且x≥0時(shí),f(x)=aex-ex≥ex-ex,因此只需證明ex-ex≥(x-1)2.只需證明(x-1設(shè)h(x)=(x-1)2則h'(x)=(x所以當(dāng)0≤x<3-e時(shí),h'(x)<0,h(x)單調(diào)遞減;當(dāng)3-e<x<1時(shí),h'(x)>0,h(x)單調(diào)遞增;當(dāng)x>1時(shí),h'(x)<0,h(x)單調(diào)遞減.又因?yàn)閔(0)=0,h(1)=0,且x=1是h(x)的極大值,因此當(dāng)x≥0時(shí),必有h(x)≤0,故原不等式成立.2.(1)解f'(x)=aex-4.當(dāng)a≤0時(shí),f'(x)<0,f(x)在R上單調(diào)遞減;當(dāng)a>0時(shí),令f'(x)<0,可得x<ln4a令f'(x)>0,可得x>ln4a,所以f(x)在-∞,ln4a上單調(diào)遞減,綜上所述,當(dāng)a≤0時(shí),f(x)的單調(diào)遞減區(qū)間為(-∞,+∞);當(dāng)a>0時(shí),f(x)的單調(diào)遞增區(qū)間為ln4a,+(2)證明當(dāng)a=1時(shí),f(x)=ex-4x,令g(x)=f(x)+x2+1=ex-4x+x2+1.g'(x)=ex-4+2x,令h(x)=ex-4+2x,則h'(x)=ex+2>0恒成立,所以g'(x)在R上單調(diào)遞增,又因?yàn)間'(0)=-3<0,g'(1)=e-2>0,由函數(shù)零點(diǎn)存在定理可得存在x0∈(0,1),使得g'(x0)=0,即ex0-4+2x0=當(dāng)x∈(-∞,x0)時(shí),g'(x)<0,g(x)單調(diào)遞減;當(dāng)x∈(x0,+∞)時(shí),g'(x)>0,g(x)單調(diào)遞增.所以g(x)min=g(x0)=ex0-4x0+x02+1=4-2x0-4x0+x02+1=x由于x0∈(0,1),所以由二次函數(shù)性質(zhì)可得g(x)min>g(1)=0,所以g(x)>0,故f(x)+x2+1>0.3.(1)解依據(jù)題意,f(x)=ex-asinx-x?f'(x)=ex-acosx-1,因?yàn)榍€f(x)在點(diǎn)(0,f(0))處的切線方程為x+y-1=0,所以f'(0)=-1?1-a-1=-1?a=1.故實(shí)數(shù)a的值為1.(2)證明由于f(x)=ex-sinx-x,要證明?x∈R,f(x)>0,需證明ex-x>sinx.因?yàn)閟inx∈[-1,1],故需證明ex-x>1.令g(x)=ex-x,g'(x)=ex-1,令g'(x)=0?x=0.g'(x)>0?x>0,g'(x)<0?x<0,所以函數(shù)g(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,故g(x)min=g(0)=1,即?x∈R,ex-x≥1,所以ex-x-sinx≥1-sinx≥0,所以?x∈R,f(x)>0.4.證明由已知得需證a(lnx+3)<exx.因?yàn)閍>0,x>0,所以exx>0,當(dāng)lnx+3<0時(shí),當(dāng)lnx+3>0時(shí),由于0<a<14所以a(lnx+3)<14(lnx+因此只需證14(lnx+3)<exx,即證lnx+34x<exx2.令g(x)=lnx+34x,所以g'(x)當(dāng)x∈(0,e-2)時(shí),g'(x)>0,當(dāng)x∈(e-2,+∞)時(shí),g'(x)<0,即g(x)在(0,e-2)上單調(diào)遞增,在(e-2,+∞)上單調(diào)遞減.所以g(x)max=g(e-2)=e2令h(x)=exx2,則h'(x)=ex(x-2)x3,當(dāng)x∈(0,2)時(shí),h'(x)<0,當(dāng)x∈(2,所以h(x)在(0,2)上單調(diào)遞減,在(2,+∞)上單調(diào)遞增,所以h(x)min=h(2)=e24.所以g(x)≤h(x),但兩邊取得最值的條件不相等,即證得a(lnx+3)<exx,故f(x)<e5.(1)解f'(x)=-1-a+lnxx2,由題意得f'(1)=-于是f'(x)=lnxx2當(dāng)x∈(0,1)時(shí),f'(x)<0;當(dāng)x∈(1,+∞)時(shí),f'(x)>0,所以實(shí)數(shù)a的值為-1,f(x)的單調(diào)遞減區(qū)間為(0,1),單調(diào)遞增區(qū)間為(1,+∞).(2)證明要證f(x)+x+23>0,即證-1-lnxx+x+23>0,因?yàn)閤>0,即證x2+23x-lnx-1>0.令g(x)=x-1-lnx,則g'所以當(dāng)x∈(0,1)時(shí),g(x)單調(diào)遞減,當(dāng)x∈(1,+∞)時(shí),g(x)單調(diào)遞增,所以g(x)≥g(1)=0,即lnx≤x-1,則ln2x≤2x-1,即ln2+lnx≤2x-1,所以lnx≤2x-1-ln2,則x2+23x-lnx-1≥x2+23x-2x+1+ln2-1=x2-43令h(x)=x2-43x+ln2=x-232+ln2-49,又因?yàn)閘n2>lne=12,所以ln2-49>0,則h(x)>0,故x2+23x-lnx-1>0成立6.(1)解函數(shù)f(x)的定義域?yàn)?0,+∞),f'(x)=lnx+x+1x,所以曲線y=f(x)在點(diǎn)(1,f(1))處的切線斜率為k=f'(1)=2,又因?yàn)閒(1)=0,所以該切線方程為y=2(x-(2)證明設(shè)F(x)=(x+1)lnx-2x+2(x>1),則F'(x)=lnx+1x-1,令g(x)=F'(x),則g'(x)=1x?1x2=x-1x2,當(dāng)x>1時(shí),g'(x)>0,所以g(x)=F'(x)在(1,+∞)上單調(diào)遞增,又因?yàn)間(1)=即F(x)在(1,+∞)上單調(diào)遞增,