備戰(zhàn)2025年高考數(shù)學(xué)一輪復(fù)習(xí)(世紀(jì)金榜高中全程復(fù)習(xí)方略數(shù)學(xué)人教A版基礎(chǔ)版)課時(shí)作業(yè)拓展拔高練四

PAGE溫馨提示：此套題為Word版，請(qǐng)按住Ctrl,滑動(dòng)鼠標(biāo)滾軸，調(diào)節(jié)合適的觀看比例，答案解析附后。板塊。拓展拔高練四(時(shí)間:45分鐘分值:40分)1.(10分)已知函數(shù)f(x)=aex-x,a∈R.若f(x)有兩個(gè)不同的零點(diǎn)x1,x2.證明:x1+x2>2.【證明】由f(x)=aex-x=0,得xex-令g(x)=xex-a,則g'(x)=由g'(x)=1-xe由g'(x)=1-xex所以g(x)在(-∞,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,由于x1,x2是方程g(x)=0的兩個(gè)不相等的實(shí)根,不妨設(shè)x1<1<x2,方法一(對(duì)稱構(gòu)造函數(shù)法):要證x1+x2>2,只要證x2>2-x1>1.由于g(x)在(1,+∞)上單調(diào)遞減,因此只要證g(x2)<g(2-x1).由于g(x1)=g(x2)=0,因此只要證g(x1)<g(2-x1),令H(x)=g(x)-g(2-x)=xex-2-則H'(x)=1-xex-因?yàn)閤<1,所以1-x>0,2-x>x,所以e2-x>ex,即e2-x所以H'(x)>0,所以H(x)在(-∞,1)上單調(diào)遞增.所以H(x1)<H(1)=0,即有g(shù)(x1)<g(2-x1)成立,所以x1+x2>2.方法二(比值代換法):設(shè)0<x1<1<x2,由g(x1)=g(x2),得x1e-x1=x等式兩邊取對(duì)數(shù)得lnx1-x1=lnx2-x2.令t=x2x1>1,則x2=tx1,代入上式得lnx1-x1=lnt+lnx1-tx1,得x1=lntt-所以x1+x2=(t+1)lntt-設(shè)k(t)=lnt-2(t-所以k'(t)=1t-2(t所以當(dāng)t>1時(shí),k(t)單調(diào)遞增,所以k(t)>k(1)=0,所以lnt-2(t-1)t+1>0,故2.(10分)(2023·六安模擬)已知函數(shù)f(x)=xlnx-ax2+x(a∈R).(1)證明:曲線y=f(x)在點(diǎn)(1,f(1))處的切線l恒過定點(diǎn);【證明】(1)函數(shù)f(x)的定義域?yàn)?0,+∞),由f(x)=xlnx-ax2+x,得f'(x)=lnx-2ax+2,則f'(1)=2(1-a),又f(1)=1-a,則曲線y=f(x)在點(diǎn)(1,f(1))處的切線l的方程為y-(1-a)=2(1-a)(x-1),即y=2(1-a)(x-12),顯然恒過定點(diǎn)(12,0(2)若f(x)有兩個(gè)零點(diǎn)x1,x2,且x2>2x1,證明:x1x2>8e【證明】(2)若f(x)有兩個(gè)零點(diǎn)x1,x2,則x1lnx1-ax12+x1=0,x2lnx2-ax22+x2=0,得a=lnx1因?yàn)閤2>2x1>0,令x2=tx1(t>2),則lnx1x1+1x1=ln(則lnx2=ln(tx1)=lnt+lnx1=tln所以ln(x1x2)=lnx1+lnx2=lntt-1-1+t令h(t)=(t+1)lntt-1-2(令φ(t)=-2lnt+t-1t(t>2),則φ'(t)=-2t+1+1t2=(t-1)2t2>0,則φ(所以h'(t)=φ(t)(t-1)2>0,則h(t即ln(x1x2)>ln8e2,故x1x2>【加練備選】(2022·全國(guó)甲卷)已知函數(shù)f(x)=exx-lnx+x(1)若f(x)≥0,求a的取值范圍;【解析】(1)f(x)的定義域?yàn)?0,+∞),f'(x)=ex(x=(e令f'(x)>0,解得x>1,故函數(shù)f(x)在(0,1)上單調(diào)遞減,(1,+∞)上單調(diào)遞增,故f(x)min=f(1)=e+1-a,要使得f(x)≥0恒成立,僅需e+1-a≥0,故a≤e+1,故a的取值范圍是(-∞,e+1];(2)證明:若f(x)有兩個(gè)零點(diǎn)x1,x2,則x1x2<1.【解析】(2)由已知若函數(shù)f(x)有兩個(gè)零點(diǎn),故f(1)=e+1-a<0,即a>e+1,不妨設(shè)0<x1<1<x2,要證明x1x2<1,即證明x2<1x1,因?yàn)?<x1<1,所以即證明1<x2<1x又因?yàn)閒(x)在(1,+∞)上單調(diào)遞增,即證明:f(x2)<f(1x1)?f(x1)<f(1構(gòu)造函數(shù)h(x)=f(x)-f(1x),0<xh'(x)=f'(x)+1x2f'(1(x令k(x)=ex+x-x·e1x-1,0<k'(x)=(ex+1)+(1x-1)e1x>0,所以k(x)在(0,1)上單調(diào)遞增,k(x又因?yàn)閤-1<0,x2>0,故h'(x)>0在(0,1)上恒成立,故h(x)在(0,1)上單調(diào)遞增,又因?yàn)閔(1)=0,故h(x)<h(1)=0,故f(x1)<f(1x1),即x1x23.(10分)已知函數(shù)f(x)=lnx-ax2+(2-a)x.(1)討論f(x)的單調(diào)性;【解析】(1)函數(shù)f(x)=lnx-ax2+(2-a)x的定義域?yàn)?0,+∞),f'(x)=1x-2ax+(2-a)=-(①當(dāng)a≤0時(shí),f'(x)>0,則f(x)在(0,+∞)上單調(diào)遞增;②當(dāng)a>0時(shí),若x∈(0,1a),則f'(x)>0,若x∈(1a,+∞),則f'(則f(x)在(0,1a)上單調(diào)遞增,在(1a,+∞)(2)設(shè)f(x)的兩個(gè)零點(diǎn)是x1,x2,求證:f'(x1+x2【解析】(2)方法一:構(gòu)造差函數(shù)法由(1)易知a>0,且f(x)在(0,1a)上單調(diào)遞增,在(1a,+∞)上單調(diào)遞減,不妨設(shè)0<x1<1a<f'(x1+x22)<0?x1+x22>1a?x1+x2>2a,故要證f'(構(gòu)造函數(shù)F(x)=f(x)-f(2a-x),x∈(0,1aF'(x)=f'(x)-[f(2a-x)]'=f'(x)+f'(2a-x)=2ax因?yàn)閤∈(0,1a),所以F'(x)=2(ax-1)2x所以F(x)<F(1a)=f(1a)-f(2a-1a)=0,即f(x)<f(2a-x),x∈又x1,x2是函數(shù)f(x)的兩個(gè)零點(diǎn)且0<x1<1a<x2,所以f(x1)=f(x2)<f(2a-x1而x2,2a-x1均大于1a,所以x2>2a-x1,所以x1+x2>方法二:比值代換法因?yàn)閒(x)的兩個(gè)零點(diǎn)是x1,x2,不妨設(shè)0<x1<x2,所以lnx1-ax12+(2-a)x1=lnx2-ax22+(2-a)x2,所以a(x22-x12)+(a-2)(x2-x所以lnx2-lnx1x2-f'(x)=1x-2ax+2-af'(x1+x22)=2x1+x2-a(x1+x2)-(a-2)=2令t=x2x1(t>1),g(t)=2(t-1)1+t-ln所以g(t)在(1,+∞)上單調(diào)遞減,所以當(dāng)t>1時(shí),g(t)<g(1)=0,所以f'(x1+x24.(10分)已知函數(shù)f(x)=sinxex,x(1)求函數(shù)f(x)的單調(diào)區(qū)間;【解析】(1)f'(x)=cosx-sin由f'(x)=0得x=π4當(dāng)0<x<π4時(shí),f'(x)>0;當(dāng)π4<x<π時(shí),f'(x所以f(x)的單調(diào)遞增區(qū)間是(0,π4),單調(diào)遞減區(qū)間是(π4,π(2)若x1≠x2,且f(x1)=f(x2),證明:x1+x2>π2【解析】(2)因?yàn)閤1≠x2,且f(x1)=f(x2),所以由(1)知,不妨設(shè)0<x1<π4<x2<π要證x1+x2>π2,只需證x2>π2-x而π4<π2-x1<π2,f(x)在(π故只需證f(x2)<f(π2-x1)又f(x1)=f(x2),所以只需證f(x1)<f(π2-x1)令函數(shù)g(x)=f(x)-f(π2-x)=sinxex-s