第2課時(shí)導(dǎo)數(shù)的綜合應(yīng)用A組基礎(chǔ)題組時(shí)間:30分鐘分值:45分1.設(shè)函數(shù)f(x)=ex(2x1)ax+a,其中a<1,若存在唯一的整數(shù)x0使得f(x0)<0,則a的取值范圍是()A.-32e,1 B.-32e2.(2017安徽皖南八校聯(lián)考)已知x∈(0,2),若關(guān)于x的不等式xex<A.[0,e+1) B.[0,2e1) C.[0,e) D.[0,e1)3.(2017甘肅蘭州模擬)設(shè)函數(shù)f(x)在R上的導(dǎo)函數(shù)為f'(x),?x∈R,f(x)+f(x)=x2,在(0,+∞)上,f'(x)x<0,若f(4m)f(m)≥84m,則實(shí)數(shù)m的取值范圍是.

4.(2017河北唐山模擬)已知函數(shù)f(x)=1-2lnxx2,若對(duì)任意的x1,x2∈0,1e,且x1≠x5.(2017江西南昌十校第二次模擬)已知函數(shù)f(x)=(x2x5)·ex,g(x)=tx2+ex4e2(t∈R)(其中e為自然對(duì)數(shù)的底數(shù)).(1)求函數(shù)f(x)的單調(diào)區(qū)間與極值;(2)是否存在t<0,對(duì)任意的x1∈R,任意的x2∈(0,+∞),都有f(x1)>g(x2)?若存在,求出t的取值范圍;若不存在,請(qǐng)說(shuō)明理由.6.(2017課標(biāo)全國(guó)Ⅲ,21,12分)已知函數(shù)f(x)=lnx+ax2+(2a+1)x.(1)討論f(x)的單調(diào)性;(2)當(dāng)a<0時(shí),證明f(x)≤34B組提升題組時(shí)間:20分鐘分值:25分1.已知函數(shù)f(x)=1-(1)求a的取值范圍;(2)若b>0,試證明1a+b<lna2.(2017甘肅張掖模擬)設(shè)函數(shù)f(x)=x2(1)當(dāng)a=1時(shí),求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;(2)求函數(shù)f(x)的單調(diào)區(qū)間和極值;(3)若函數(shù)f(x)在區(qū)間(1,e2]內(nèi)恰有兩個(gè)零點(diǎn),試求a的取值范圍.

答案精解精析A組基礎(chǔ)題組1.D解法一:由題意可知存在唯一的整數(shù)x0,使得ex0(2x01)<ax0a,設(shè)g(x)=ex(2x1),h(x)=axa,由g'(x)=ex(2x+1),可知g(x)在-∞,-12上單調(diào)遞減,在-12,解法二:由f(x0)<0,即ex0(2x01)a(x01)<0得ex0(2x當(dāng)x0=1時(shí),得e<0,顯然不成立,所以x0≠1.若x0>1,則a>ex令g(x)=ex(2x當(dāng)x∈1,當(dāng)x∈32要滿足題意,則x0=2,此時(shí)需滿足g(2)<a≤g(3),得3e2<a≤52e3,與a<1矛盾,所以x0因?yàn)閤0<1,所以a<ex易知,當(dāng)x∈(∞,0)時(shí),g'(x)>0,g(x)為增函數(shù),當(dāng)x∈(0,1)時(shí),g'(x)<0,g(x)為減函數(shù),要滿足題意,則x0=0,此時(shí)需滿足g(1)≤a<g(0),得32e2.D依題意,知k+2xx2>0,即k>x22x對(duì)任意x∈(0,2)恒成立,從而k≥0,所以由xex<1k+2x-x2可得k<exx+x令f'(x)=0,得x=1,當(dāng)x∈(1,2)時(shí),f'(x)>0,函數(shù)f(x)在(1,2)上單調(diào)遞增,當(dāng)x∈(0,1)時(shí),f'(x)<0,函數(shù)f(x)在(0,1)上單調(diào)遞減,所以k<f(x)min=f(1)=e1,故實(shí)數(shù)k的取值范圍是[0,e1).3.答案[2,+∞)解析令g(x)=f(x)12x2,則g(x)+g(x)=f(x)12x2+f(x)12x2=0,∴函數(shù)g(x)為奇函數(shù).當(dāng)x∈(0,+∞)時(shí),g'(x)=f'(x)x<0,∴函數(shù)g(x)在(0,+∞)上是減函數(shù),故函數(shù)g(x)在(∞,0)上也是減函數(shù).由f(0)=0,可得g(x)在R上是減函數(shù),∴f(4m)f(m)=g(4m)+12(4m)2g(m)4.答案(∞,4]解析由對(duì)任意的x1,x2∈0,1e,且x1≠x2,f(x1)-f(x2)x12-x225.解析(1)∵f(x)=(x2x5)ex,∴f'(x)=(2x1)ex+(x2x5)ex=(x2+x6)ex=(x+3)(x2)ex.當(dāng)x<3或x>2時(shí),f'(x)>0,即函數(shù)f(x)單調(diào)遞增.當(dāng)3<x<2時(shí),f'(x)<0,即函數(shù)f(x)單調(diào)遞減.∴函數(shù)f(x)的單調(diào)遞增區(qū)間為(∞,3)和(2,+∞),單調(diào)遞減區(qū)間為(3,2).故當(dāng)x=3時(shí),函數(shù)f(x)取得極大值,即f(x)極大值=f(3)=7e3.當(dāng)x=2時(shí),函數(shù)f(x)取得極小值,即f(x)極小值=f(2)=3e2.(2)存在.由題意,只需f(x)min>g(x)max.由(1)可得當(dāng)x趨近于∞時(shí),f(x)趨近于0,∴f(x)min=f(2)=3e2,∵g(x)=tx2+ex4e2=tx+e2t∴g(x)max=g-e2t=e故3e2>e24t4e2,即1>1∴存在負(fù)數(shù)t∈-∞,-16.解析(1)f(x)的定義域?yàn)?0,+∞),f'(x)=1x+2ax+2a+1=(若a≥0,則當(dāng)x∈(0,+∞)時(shí),f'(x)>0,故f(x)在(0,+∞)單調(diào)遞增.若a<0,則當(dāng)x∈0,-當(dāng)x∈-1故f(x)在0,-12(2)證明:由(1)知,當(dāng)a<0時(shí),f(x)在x=12a取得最大值,最大值為f-12a所以f(x)≤34a2等價(jià)于ln-12a114a設(shè)g(x)=lnxx+1,則g'(x)=1x當(dāng)x∈(0,1)時(shí),g'(x)>0;當(dāng)x∈(1,+∞)時(shí),g'(x)<0.所以g(x)在(0,1)單調(diào)遞增,在(1,+∞)單調(diào)遞減.故當(dāng)x=1時(shí),g(x)取得最大值,最大值為g(1)=0.所以當(dāng)x>0時(shí),g(x)≤0.從而當(dāng)a<0時(shí),ln-12a+1B組提升題組1.解析(1)f'(x)=1ax2+1因?yàn)閒'(x)≥0,且a>0,所以ax1≥0,即x≥1a,因?yàn)閤∈(1,+∞),所以1(2)證明:因?yàn)閎>0,a≥1,所以a+bb>1,又f(x)=1-xax+lnx在(1,+∞)上是增函數(shù),所以f化簡(jiǎn)得1a+blna+bb<ab等價(jià)于lna+令g(x)=ln(1+x)x(x∈(0,+∞)),則g'(x)=11+x1=所以gab=ln1+abab綜上,1a+b<lna2.解析(1)當(dāng)a=1時(shí),f(x)=x22lnx,則f'(x)=x所以f'(1)=0,又f(1)=12所以曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y12=0×(x1),即y=1(2)由f(x)=x22alnx,得f'(x)=xax①當(dāng)a≤0時(shí),f'(x)>0,函數(shù)f(x)在(0,+∞)上單調(diào)遞增,函數(shù)既無(wú)極大值,也無(wú)極小值;②當(dāng)a>0時(shí),由f'(x)=0,得x=a或x=a(舍去).當(dāng)x變化時(shí),f'(x)與f(x)的變化情況如下表:x(0,a)a(a,+∞)f'(x)0+f(x)↘a↗所以,f(x)的單調(diào)遞減區(qū)間是(0,a),單調(diào)遞增區(qū)間是(a,+∞);函數(shù)f(x)在x=a處取得極小值f(a)=a(綜上可知,當(dāng)a≤0時(shí),函數(shù)f(x)的單調(diào)遞增區(qū)間為(0,+∞),函數(shù)f(x)既無(wú)極大值也無(wú)極小值;當(dāng)a>0時(shí),函數(shù)f(x)的單調(diào)遞減區(qū)間是(0,a),單調(diào)遞增區(qū)間為(a,+∞),函數(shù)f(x)有極小值a((3)當(dāng)a≤0時(shí),由(2)知函數(shù)f(x)在區(qū)間(0,+∞)上單調(diào)遞增,故函數(shù)f