《數(shù)學(xué)理高考二輪專(zhuān)題復(fù)習(xí)與測(cè)試:第二部分 專(zhuān)題六 第4講 導(dǎo)數(shù)的綜合應(yīng)用 Word版含解析》由會(huì)員分享,可在線閱讀,更多相關(guān)《數(shù)學(xué)理高考二輪專(zhuān)題復(fù)習(xí)與測(cè)試:第二部分 專(zhuān)題六 第4講 導(dǎo)數(shù)的綜合應(yīng)用 Word版含解析(10頁(yè)珍藏版)》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。
1、A級(jí)基礎(chǔ)通關(guān)一、選擇題1設(shè)f(x)是定義在R上的奇函數(shù),且f(2)0,當(dāng)x0時(shí),有0的解集是()A(2,0)(2,)B(2,0)(0,2)C(,2)(2,)D(,2)(0,2)解析:x0時(shí),0,所以(x)在(0,)為減函數(shù),又(2)0,所以當(dāng)且僅當(dāng)0x0,此時(shí)x2f(x)0.又f(x)為奇函數(shù),所以h(x)x2f(x)也為奇函數(shù)故x2f(x)0的解集為(,2)(0,2)答案:D2已知函數(shù)f(x)的定義域?yàn)?,4,部分對(duì)應(yīng)值如下表:x10234f(x)12020f(x)的導(dǎo)函數(shù)yf(x)的圖象如圖所示當(dāng)1a2時(shí),函數(shù)yf(x)a的零點(diǎn)的個(gè)數(shù)為()A1B2C3D4解析:根據(jù)導(dǎo)函數(shù)圖象,知2是函數(shù)的
2、極小值點(diǎn),函數(shù)yf(x)的大致圖象如圖所示由于f(0)f(3)2,1a2,所以yf(x)a的零點(diǎn)個(gè)數(shù)為4.答案:D3若函數(shù)f(x)在R上可導(dǎo),且滿足f(x)xf(x)0,則()A3f(1)f(3) B3f(1)f(3)C3f(1)f(3) Df(1)f(3)解析:由于f(x)xf(x),則0恒成立,因此y在R上是單調(diào)減函數(shù),所以,即3f(1)f(3)答案:B4已知函數(shù)f(x)exln x,則下面對(duì)函數(shù)f(x)的描述正確的是()Ax(0,),f(x)2Bx(0,),f(x)2Cx0(0,),f(x0)0Df(x)min(0,1)解析:因?yàn)閒(x)exln x的定義域?yàn)?0,),且f(x)ex,令
3、g(x)xex1,x0,則g(x)(x1)ex0在(0,)上恒成立,所以g(x)在(0,)上單調(diào)遞增,又g(0)g(1)(e1)0,所以x0(0,1),使g(x0)0,則f(x)在(0,x0)上單調(diào)遞減,在(x0,)上單調(diào)遞增,則f(x)minf(x0)ex0ln x0,又ex0,x0ln x0,所以f(x)minx02.答案:B5已知函數(shù)f(x),若函數(shù)g(x)f(x)a無(wú)零點(diǎn),則實(shí)數(shù)a的取值范圍為()A. B.C(2e,0 D(e,0解析:依題意,f(x),令h(x)ln x,注意到函數(shù)h(x)單調(diào)遞增,且h(e)0,故當(dāng)x(0,e)時(shí),h(x)0.故函數(shù)f(x)在(0,1)和(1,e)上
4、單調(diào)遞減,在(e,)上單調(diào)遞增,作出函數(shù)f(x)的圖象如下圖所示令f(x)a0,得f(x)a,觀察可知0ae,即ex20,f(x1)f(x2)0),因?yàn)榍€yf(x)在點(diǎn)(e,f(e)處的切線與直線x20垂直,所以f(e)0,即0,得ke,所以f(x)(x0)由f(x)0得0x0得xe.所以f(x)在(0,e)上單調(diào)遞減,在(e,)上單調(diào)遞增,當(dāng)xe時(shí),f(x)取得極小值,且f(e)ln e2.所以f(x)的極小值為2.(2)由題意知對(duì)任意的x1x20,f(x1)x10),則h(x)在(0,)上單調(diào)遞減,所以h(x)10在(0,)上恒成立,故當(dāng)x0時(shí),kx2x恒成立,又,則k,故實(shí)數(shù)k的取值范
5、圍是.9(2019天津卷節(jié)選)設(shè)函數(shù)f(x)excos x,g(x)為f(x)的導(dǎo)函數(shù)(1)求f(x)的單調(diào)區(qū)間;(2)當(dāng)x時(shí),證明:f(x)g(x)0.(1)解:由已知,有f(x)ex(cos xsin x)因此,當(dāng)x(kZ)時(shí),有sin xcos x,得f(x)0,則f(x)單調(diào)遞減;當(dāng)x(kZ)時(shí),有sin x0,則f(x)單調(diào)遞增所以f(x)的單調(diào)遞增區(qū)間為(kZ),f(x)的單調(diào)遞減區(qū)間為(kZ)(2)證明:記h(x)f(x)g(x).依題意及(1),有g(shù)(x)ex(cos xsin x),從而g(x)2exsin x.當(dāng)x時(shí),g(x)0,故h(x)f(x)g(x)g(x)(1)g(
6、x)0.因此,h(x)在區(qū)間上單調(diào)遞減,進(jìn)而h(x)hf0.所以當(dāng)x時(shí),f(x)g(x)0.B級(jí)能力提升10已知函數(shù)f(x)ln x,g(x)xm(mR)(1)若f(x)g(x)恒成立,求實(shí)數(shù)m的取值范圍;(2)已知x1,x2是函數(shù)F(x)f(x)g(x)的兩個(gè)零點(diǎn),且x1x2,求證:x1x20),則F(x)1(x0),當(dāng)x1時(shí),F(xiàn)(x)0,當(dāng)0x0,所以F(x)在(1,)上單調(diào)遞減,在(0,1)上單調(diào)遞增F(x)在x1處取得最大值1m,若f(x)g(x)恒成立,則1m0,即m1.(2)證明:由(1)可知,若函數(shù)F(x)f(x)g(x)有兩個(gè)零點(diǎn),則m1,0x11x2,要證x1x21,只需證x
7、2F,由F(x1)F(x2)0,mln x1x1,即證lnmlnx1ln x10,令h(x)x2ln x(0x0,故h(x)在(0,1)上單調(diào)遞增,h(x)h(1)0,所以x1x21.11(2018全國(guó)卷)已知函數(shù)f(x)xaln x.(1)討論f(x)的單調(diào)性;(2)若f(x)存在兩個(gè)極值點(diǎn)x1,x2,證明:2,令f(x)0,得x或x.當(dāng)x(0,)(,)時(shí),f(x)0.所以f(x)在(0,),(,)上單調(diào)遞減,在(,)上單調(diào)遞增(2)證明:由(1)知,f(x)存在兩個(gè)極值點(diǎn)當(dāng)且僅當(dāng)a2.由于f(x)的兩個(gè)極值點(diǎn)x1,x2滿足x2ax10,所以x1x21,不妨設(shè)0x11.由于1a2a2a,所以a2等價(jià)于x22ln x20.設(shè)函數(shù)g(x)x2ln x,由(1)知,g(x)在(0,)上單調(diào)遞減又g(1)0,從而當(dāng)x(1,)時(shí),g(x)0.所以x22ln x20,故a2.