2019屆高考數(shù)學(xué)一輪復(fù)習(xí) 第二章 函數(shù)、導(dǎo)數(shù)及其應(yīng)用 第13節(jié) 導(dǎo)數(shù)的綜合應(yīng)用 第一課時(shí)練習(xí) 新人教A版.doc
第二章 第13節(jié) 導(dǎo)數(shù)的綜合應(yīng)用 第一課時(shí)1(導(dǎo)學(xué)號(hào)14577225)(2018銀川市模擬)設(shè)f(x)xln xax2,a為常數(shù)(1)若曲線yf(x)在x1處的切線過(guò)點(diǎn)A(0,2),求實(shí)數(shù)a的值;(2)若f(x)有兩個(gè)極值點(diǎn)x1,x2且x1x2求證:a0求證:f (x2)f (x1).解:(1)f(x)xln xax2的導(dǎo)數(shù)為f(x)ln x12ax,在x1處的切線斜率為k12a,切點(diǎn)為(1,a),在x1處的切線過(guò)點(diǎn)A(0,2),則k12aa2,解得a1;(2)證明:由題意可得f(x)0有兩個(gè)不等的實(shí)根x1,x2,且0x1x2,設(shè)g(x)ln x12ax,g(x)2a,x0.當(dāng)a0,則g(x)0,g(x)在(0,)遞增,不合題意;當(dāng)a0時(shí),g(x)0解得x,g(x)0解得x,即有g(shù)(x)在遞增,在遞減即有g(shù)ln0,解得a0;由上可知,f(x)在(x1,x2)遞增,即有f(x2)f(x1),f(1)g(1)12a0,則x1(0,1),由可得ax1,即有f(x1)x1ln x1ax(x1ln x1x1),設(shè)h(x)(xln xx),0x1,h(x) ln x0在(0,1)恒成立,故h(x)在(0,1)遞減,故h(x)h(1),由此可得f(x1),綜上可得f (x2)f (x1).1(導(dǎo)學(xué)號(hào)14577225)(2018銀川市模擬)設(shè)f(x)xln xax2,a為常數(shù)(1)若曲線yf(x)在x1處的切線過(guò)點(diǎn)A(0,2),求實(shí)數(shù)a的值;(2)若f(x)有兩個(gè)極值點(diǎn)x1,x2且x1x2求證:a0求證:f (x2)f (x1).解:(1)f(x)xln xax2的導(dǎo)數(shù)為f(x)ln x12ax,在x1處的切線斜率為k12a,切點(diǎn)為(1,a),在x1處的切線過(guò)點(diǎn)A(0,2),則k12aa2,解得a1;(2)證明:由題意可得f(x)0有兩個(gè)不等的實(shí)根x1,x2,且0x1x2,設(shè)g(x)ln x12ax,g(x)2a,x0.當(dāng)a0,則g(x)0,g(x)在(0,)遞增,不合題意;當(dāng)a0時(shí),g(x)0解得x,g(x)0解得x,即有g(shù)(x)在遞增,在遞減即有g(shù)ln0,解得a0;由上可知,f(x)在(x1,x2)遞增,即有f(x2)f(x1),f(1)g(1)12a0,則x1(0,1),由可得ax1,即有f(x1)x1ln x1ax(x1ln x1x1),設(shè)h(x)(xln xx),0x1,h(x) ln x0在(0,1)恒成立,故h(x)在(0,1)遞減,故h(x)h(1),由此可得f(x1),綜上可得f (x2)f (x1).2(導(dǎo)學(xué)號(hào)14577226)已知函數(shù)f(x)xln xmx(mR)的圖象在點(diǎn)(1,f(1)處的切線的斜率為2.(1)求實(shí)數(shù)m的值;(2)設(shè)g(x),討論g(x)的單調(diào)性;(3)已知m,nN*且m>n>1,證明 >.解:(1)因?yàn)閒(x)xln xmx,所以f(x)1ln xm.由題意f(1)1ln 1m2,得m1.(2)g(x)(x>0,x1),所以g(x).設(shè)h(x)x1ln x,h(x)1.當(dāng)x>1時(shí),h(x)1>0,h(x)是增函數(shù),h(x)>h(1)0,所以g(x)>0,故g(x)在(1,)上為增函數(shù);當(dāng)0<x<1時(shí),h(x)1<0,h(x)是減函數(shù),h(x)>h(1)0,所以g(x)>0,故g(x)在(0,1)上為增函數(shù);所以g(x)在區(qū)間(0,1)和(1,)上都是單調(diào)遞增的(3)證明:由已知可知要證>,即證>ln nln m,即證ln m>ln n,即證>,即證g(m)>g(n),又m>n>1(m,nN*),由(2)知g(m)>g(n)成立,所以>.3(導(dǎo)學(xué)號(hào)14577227)(理科)函數(shù)f(x)ln(xm)nln x.(1)當(dāng)m1,n0時(shí),求f(x)的單調(diào)減區(qū)間;(2)n1時(shí),函數(shù)g(x)(m2x)f(x)am,若存在m0,使得g(x)0恒成立,求實(shí)數(shù)a的取值范圍解:(1)f(x)ln(x1)nln x,定義域?yàn)?0,),f(x),當(dāng)n1時(shí),f(x)0,此時(shí)f(x)的單調(diào)減區(qū)間為(0,);當(dāng)0n1時(shí),0x時(shí),f(x)0,此時(shí)f(x)的單調(diào)減區(qū)間為;當(dāng)n1時(shí),x時(shí),f(x)0,此時(shí)減區(qū)間為.(2)n1時(shí),g(x)(m2x)ln(xm)lnxam,g(x)0,0,即lna0,設(shè)t1,(t1)ln ta(t1)0,ln t0.設(shè)h(t)ln t,h(t),h(1)0,當(dāng)a2時(shí),t22(1a)t1t22t10,故h(t)0,h(t)在(1,)上單調(diào)遞增,因此h(t)0;當(dāng)a2時(shí),令h(t)0,得:t1a1,t2a1,由t21和t1t21,得:t11,故h(t)在(1,t2)上單調(diào)遞減,此時(shí)h(t)h(1)0.綜上所述,a2.3(文科)(2018西安市三模)已知函數(shù)f(x)x26ax1,g(x)8a2ln x2b1,其中a0.(1)設(shè)兩曲線yf(x),yg(x)有公共點(diǎn),且在該點(diǎn)處的切線相同,用a表示b,并求b的最大值;(2)設(shè)h(x)f(x)g(x),證明:若a1,則對(duì)任意x1,x2(0,),x1x2,有>14.解:(1):設(shè)f(x)與g(x)的圖象交于點(diǎn)P(x0,y0)(x00),則有f(x0)g(x0),即x6ax018a2ln x02b1又由題意知f(x0)g(x0),即2x06a,由解得x0a或x04a(舍去),將x0a代入整理得ba24a2ln a,令K(a)a24a2ln a,則K(a)a(38ln a),當(dāng)a時(shí),K(a)單調(diào)遞增,當(dāng)a時(shí)K(a)單調(diào)遞減,所以K(a)K()2e,即b2e,b的最大值為2e;(2)證明:不妨設(shè)x1,x2(0,),x1x2,>14,變形得h(x2)14x2h(x1)14x1,令T(x)h(x)14x,T(x)2x6a14,a1,T(x)2x6a148a6a140,則T(x)在(0,)上單調(diào)遞增,T(x2)T(x1),即>14成立,同理可證,當(dāng)x1x2時(shí),命題也成立綜上,對(duì)任意x1,x2(0,),x1x2,不等式>14成立4(導(dǎo)學(xué)號(hào)14577229)(理科)(2018大慶市一模)已知函數(shù)f(x)ln (xa)x2x在x0處取得極值(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)若關(guān)于x的方程f(x)xb在區(qū)間(0,2)有兩個(gè)不等實(shí)根,求實(shí)數(shù)b的取值范圍;(3)對(duì)于nN*,證明:>ln(n1)解:(1)由已知得f(x)2x1,f(0)0,0,a1.f(x)ln (x1)x2x(x1),于是f(x)(x1),由f(x)0得1x0;由f(x)0,得x0,f(x)的單調(diào)遞增區(qū)間是(1,0),單調(diào)遞減區(qū)間是(0,)(2)令g(x)f(x)ln (x1)x2xb,x(0,2),則g(x)2x,令g(x)0,得x1或x(舍去)當(dāng)0x1時(shí),g(x)0;當(dāng)1x2時(shí)g(x)0,即g(x)在(0,1)上單調(diào)遞增,在(1,2)上單調(diào)遞減方程f(x)xb在區(qū)間(0,2)有兩個(gè)不等實(shí)根等價(jià)于函數(shù)g(x)在(0,2)上有兩個(gè)不同的零點(diǎn),即;亦即,ln 31bln 2,故所求實(shí)數(shù)b的取值范圍為.(3)證明:由(1)可得,當(dāng)x0時(shí)ln (x1)x2x(當(dāng)且僅當(dāng)x0時(shí)等號(hào)成立)設(shè)x,則ln ,即ln ln,ln,ln,ln ,將上面n個(gè)式子相加得:ln ln ln ln ln (n1),故>ln(n1)4(導(dǎo)學(xué)號(hào)14577230)(文科)(2018天津河北區(qū)三模)已知函數(shù)f(x)axbln x表示的曲線在點(diǎn)(2,f(2)處的切線方程x2y2ln 20(1)求a,b的值;(2)若f(x)kx2對(duì)于x(0,)恒成立,求實(shí)數(shù)k的取值范圍;(3)求證:nN*時(shí),n(n1)2.解:(1)函數(shù)f(x)axbln x的導(dǎo)數(shù)為f(x)a,在點(diǎn)(2,f(2)處的切線方程x2y2ln 20,即有a,解得a1,f(2)2abln 21ln 2,解得b1,則有a1,b1;(2)f(x)kx2對(duì)于x(0,)恒成立,即有x1ln xkx2對(duì)于x(0,)恒成立,即有k1對(duì)于x(0,)恒成立令g(x),g(x),當(dāng)xe2時(shí),g(x)0,g(x)遞增;當(dāng)0xe2時(shí),g(x)0,g(x)遞減則xe2處g(x)取得極小值,也為最小值,且為,即有k1,解得k1;(3)證明:f(x)x1ln x(x0),f(x)1,當(dāng)x1時(shí),f(x)0,f(x)遞增,當(dāng)0x1時(shí),f(x)0,f(x)遞減則x1處f(x)取得極小值,也為最小值,且為0,則有f(x)0,即為x1ln x,取xn,則n1ln n,即有nen1.即有12n1ee2en1.則有n(n1),即有nN*時(shí),n(n1)2.2(導(dǎo)學(xué)號(hào)14577226)已知函數(shù)f(x)xln xmx(mR)的圖象在點(diǎn)(1,f(1)處的切線的斜率為2.(1)求實(shí)數(shù)m的值;(2)設(shè)g(x),討論g(x)的單調(diào)性;(3)已知m,nN*且m>n>1,證明 >.解:(1)因?yàn)閒(x)xln xmx,所以f(x)1ln xm.由題意f(1)1ln 1m2,得m1.(2)g(x)(x>0,x1),所以g(x).設(shè)h(x)x1ln x,h(x)1.當(dāng)x>1時(shí),h(x)1>0,h(x)是增函數(shù),h(x)>h(1)0,所以g(x)>0,故g(x)在(1,)上為增函數(shù);當(dāng)0<x<1時(shí),h(x)1<0,h(x)是減函數(shù),h(x)>h(1)0,所以g(x)>0,故g(x)在(0,1)上為增函數(shù);所以g(x)在區(qū)間(0,1)和(1,)上都是單調(diào)遞增的(3)證明:由已知可知要證>,即證>ln nln m,即證ln m>ln n,即證>,即證g(m)>g(n),又m>n>1(m,nN*),由(2)知g(m)>g(n)成立,所以>.3(導(dǎo)學(xué)號(hào)14577227)(理科)函數(shù)f(x)ln(xm)nln x.(1)當(dāng)m1,n0時(shí),求f(x)的單調(diào)減區(qū)間;(2)n1時(shí),函數(shù)g(x)(m2x)f(x)am,若存在m0,使得g(x)0恒成立,求實(shí)數(shù)a的取值范圍解:(1)f(x)ln(x1)nln x,定義域?yàn)?0,),f(x),當(dāng)n1時(shí),f(x)0,此時(shí)f(x)的單調(diào)減區(qū)間為(0,);當(dāng)0n1時(shí),0x時(shí),f(x)0,此時(shí)f(x)的單調(diào)減區(qū)間為;當(dāng)n1時(shí),x時(shí),f(x)0,此時(shí)減區(qū)間為.(2)n1時(shí),g(x)(m2x)ln(xm)lnxam,g(x)0,0,即lna0,設(shè)t1,(t1)ln ta(t1)0,ln t0.設(shè)h(t)ln t,h(t),h(1)0,當(dāng)a2時(shí),t22(1a)t1t22t10,故h(t)0,h(t)在(1,)上單調(diào)遞增,因此h(t)0;當(dāng)a2時(shí),令h(t)0,得:t1a1,t2a1,由t21和t1t21,得:t11,故h(t)在(1,t2)上單調(diào)遞減,此時(shí)h(t)h(1)0.綜上所述,a2.3(文科)(2018西安市三模)已知函數(shù)f(x)x26ax1,g(x)8a2ln x2b1,其中a0.(1)設(shè)兩曲線yf(x),yg(x)有公共點(diǎn),且在該點(diǎn)處的切線相同,用a表示b,并求b的最大值;(2)設(shè)h(x)f(x)g(x),證明:若a1,則對(duì)任意x1,x2(0,),x1x2,有>14.解:(1):設(shè)f(x)與g(x)的圖象交于點(diǎn)P(x0,y0)(x00),則有f(x0)g(x0),即x6ax018a2ln x02b1又由題意知f(x0)g(x0),即2x06a,由解得x0a或x04a(舍去),將x0a代入整理得ba24a2ln a,令K(a)a24a2ln a,則K(a)a(38ln a),當(dāng)a時(shí),K(a)單調(diào)遞增,當(dāng)a時(shí)K(a)單調(diào)遞減,所以K(a)K()2e,即b2e,b的最大值為2e;(2)證明:不妨設(shè)x1,x2(0,),x1x2,>14,變形得h(x2)14x2h(x1)14x1,令T(x)h(x)14x,T(x)2x6a14,a1,T(x)2x6a148a6a140,則T(x)在(0,)上單調(diào)遞增,T(x2)T(x1),即>14成立,同理可證,當(dāng)x1x2時(shí),命題也成立綜上,對(duì)任意x1,x2(0,),x1x2,不等式>14成立4(導(dǎo)學(xué)號(hào)14577229)(理科)(2018大慶市一模)已知函數(shù)f(x)ln (xa)x2x在x0處取得極值(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)若關(guān)于x的方程f(x)xb在區(qū)間(0,2)有兩個(gè)不等實(shí)根,求實(shí)數(shù)b的取值范圍;(3)對(duì)于nN*,證明:>ln(n1)解:(1)由已知得f(x)2x1,f(0)0,0,a1.f(x)ln (x1)x2x(x1),于是f(x)(x1),由f(x)0得1x0;由f(x)0,得x0,f(x)的單調(diào)遞增區(qū)間是(1,0),單調(diào)遞減區(qū)間是(0,)(2)令g(x)f(x)ln (x1)x2xb,x(0,2),則g(x)2x,令g(x)0,得x1或x(舍去)當(dāng)0x1時(shí),g(x)0;當(dāng)1x2時(shí)g(x)0,即g(x)在(0,1)上單調(diào)遞增,在(1,2)上單調(diào)遞減方程f(x)xb在區(qū)間(0,2)有兩個(gè)不等實(shí)根等價(jià)于函數(shù)g(x)在(0,2)上有兩個(gè)不同的零點(diǎn),即;亦即,ln 31bln 2,故所求實(shí)數(shù)b的取值范圍為.(3)證明:由(1)可得,當(dāng)x0時(shí)ln (x1)x2x(當(dāng)且僅當(dāng)x0時(shí)等號(hào)成立)設(shè)x,則ln ,即ln ln,ln,ln,ln ,將上面n個(gè)式子相加得:ln ln ln ln ln (n1),故>ln(n1)4(導(dǎo)學(xué)號(hào)14577230)(文科)(2018天津河北區(qū)三模)已知函數(shù)f(x)axbln x表示的曲線在點(diǎn)(2,f(2)處的切線方程x2y2ln 20(1)求a,b的值;(2)若f(x)kx2對(duì)于x(0,)恒成立,求實(shí)數(shù)k的取值范圍;(3)求證:nN*時(shí),n(n1)2.解:(1)函數(shù)f(x)axbln x的導(dǎo)數(shù)為f(x)a,在點(diǎn)(2,f(2)處的切線方程x2y2ln 20,即有a,解得a1,f(2)2abln 21ln 2,解得b1,則有a1,b1;(2)f(x)kx2對(duì)于x(0,)恒成立,即有x1ln xkx2對(duì)于x(0,)恒成立,即有k1對(duì)于x(0,)恒成立令g(x),g(x),當(dāng)xe2時(shí),g(x)0,g(x)遞增;當(dāng)0xe2時(shí),g(x)0,g(x)遞減則xe2處g(x)取得極小值,也為最小值,且為,即有k1,解得k1;(3)證明:f(x)x1ln x(x0),f(x)1,當(dāng)x1時(shí),f(x)0,f(x)遞增,當(dāng)0x1時(shí),f(x)0,f(x)遞減則x1處f(x)取得極小值,也為最小值,且為0,則有f(x)0,即為x1ln x,取xn,則n1ln n,即有nen1.即有12n1ee2en1.則有n(n1),即有nN*時(shí),n(n1)2.