9、x(0,)恒成立,則實數(shù)a可取的值組成的集合是 A.a|4a0 B.a|a4 C.a|0a4 D.a|a4,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,,12,解析由題意得ax2xln xx23,,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,當(dāng)x(0,1)時,g(x)0,g(x)單調(diào)遞增, 當(dāng)x(1,)時,g(x)<0,g(x)單調(diào)遞減, 函數(shù)g(x)maxg(1)4, 所以ag(x)max4,即a|a4.,12,4.若函數(shù)f(x)2x39x212xa恰好有兩個不同的零點,則a可能的值為 A.4 B.6 C.7 D.8,,1,2,3,
10、4,5,6,7,8,9,10,11,13,14,15,16,,解析由題意得f(x)6x218x126(x1)(x2), 由f(x)0,得x2,由f(x)<0,得1
11、6,7,8,9,10,11,13,14,15,16,解析由題意得|AB||et1(2t1)| |et2t2|,令h(t)et2t2, 則h(t)et2,所以h(t)在(,ln 2)上單調(diào)遞減, 在(ln 2,)上單調(diào)遞增, 所以h(t)minh(ln 2)42ln 20, 即|AB|的最小值是42ln 2,故選C.,12,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,,12,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,因為f(x)在x2處有最小值,且x1,4, 所以f(2)0,即b8, 所以c5,經(jīng)檢驗,b8,c5符合題意.,12,,1
12、,2,3,4,5,6,7,8,9,10,11,13,14,15,16,所以f(x)在1,2)上單調(diào)遞減,在(2,4上單調(diào)遞增,,所以函數(shù)f(x)在M上的最大值為5,故選B.,12,7.已知函數(shù)f(x)x1(e1)ln x,其中e為自然對數(shù)的底數(shù),則滿足f(ex)<0的x的取值范圍為________.,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,(0,1),解析令g(x)f(ex)ex1(e1)x, 則g(x)ex(e1), 當(dāng)xln(e1)時,g(x)0. 當(dāng)x(,ln(e1))時,g(x)0,g(x)單調(diào)遞增. 又g(x)有0和1兩個零點,所以f(ex)<0的x
13、的取值范圍為(0,1).,12,8.已知x(0,2),若關(guān)于x的不等式 恒成立,則實數(shù)k的取值范圍為____________.,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,(0,e1),12,解析由題意,知k2xx20. 即kx22x對任意x(0,2)恒成立,從而k0,,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,令f(x)0,得x1, 當(dāng)x(1,2)時,f(x)0,函數(shù)f(x)在(1,2)上單調(diào)遞增, 當(dāng)x(0,1)時,f(x)<0,函數(shù)f(x)在(0,1)上單調(diào)遞減, 所以k
14、0,e1).,12,9.已知函數(shù)f(x)ax33x21,若f(x)存在唯一的零點x0,且x00,則實數(shù)a的取值范圍是____________.,解析當(dāng)a0時,f(x)3x21有兩個零點,不合題意, 故a0,f(x)3ax26x3x(ax2),,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,(,2),若a0,由三次函數(shù)圖象知f(x)有負(fù)數(shù)零點,不合題意,故a<0.,又a<0,所以a<2.,12,10.定義在R上的奇函數(shù)yf(x)滿足f(3)0,且不等式f(x)xf(x)在(0,)上恒成立,則函數(shù)g(x)xf(x)lg|x1|的零點個數(shù)為________.,3,,1,2
15、,3,4,5,6,7,8,9,10,11,13,14,15,16,12,解析定義在R上的奇函數(shù)f(x)滿足: f(0)0f(3)f(3),f(x)f(x), 當(dāng)x0時,f(x)xf(x),即f(x)xf(x)0, xf(x)0,即h(x)xf(x)在x0時是增函數(shù), 又h(x)xf(x)xf(x), h(x)xf(x)是偶函數(shù), 當(dāng)x<0時,h(x)是減函數(shù),結(jié)合函數(shù)的定義域為R, 且f(0)f(3)f(3)0, 可得函數(shù)yxf(x)與ylg|x1|的大致圖象如圖, 由圖象可知,函數(shù)g(x)xf(x)lg|x1|的零點的個數(shù)為3.,,1,2,3,4,5,6,7,8,9,10,11,13,14,
16、15,16,12,11.已知函數(shù)f(x)exmx,其中m為常數(shù). (1)若對任意xR有f(x)0恒成立,求m的取值范圍;,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,解由題意,可知f(x)exm1, 令f(x)0,得xm. 故當(dāng)x(,m)時,exm1,f(x)0,f(x)單調(diào)遞增. 故當(dāng)xm時,f(m)為極小值也為最小值. 令f(m)1m0,得m1, 即對任意xR,f(x)0恒成立時, m的取值范圍是(,1.,12,(2)當(dāng)m1時,判斷f(x)在0,2m上零點的個數(shù),并說明理由.,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,12,,解
17、f(x)在0,2m上有兩個零點,理由如下: 當(dāng)m1時,f(m)1m0,f(0)f(m)1, 則g(m)em2, 當(dāng)m1時,g(m)em20, g(m)在(1,)上單調(diào)遞增. g(m)e20, 即f(2m)0.,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,12,,f(m)f(2m)<0, 又f(x)在(m,2m)上單調(diào)遞增, f(x)在(m,2m)上有一個零點. 故f(x)在0,2m上有兩個零點.,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,12,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,,令g(x)0
18、,得x0, 則g(x)在區(qū)間0,1上單調(diào)遞增.,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,12,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,,知存在x0(0,1),使得h(x0)0. 易知h(x)在0,1上是增函數(shù), 所以f(x)在區(qū)間(0,x0)上單調(diào)遞減,在區(qū)間(x0,1)上單調(diào)遞增,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,12,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,12,13.已知a,bR,直線yaxb 與函數(shù)f(x)tan x的圖象在x 處相切,設(shè)g(
19、x)exbx2a,若在區(qū)間1,2上,不等式mg(x)m22恒成立,則實數(shù)m有 A.最大值e B.最大值e1 C.最小值e D.最小值e,,技能提升練,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,,12,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,因此a2,b1,g(x)exx22, 所以當(dāng)x1,2時,g(x)ex2x0, g(x)exx22單調(diào)遞增, 所以g(x)mine1,g(x)maxe22. 所以me1且m22e22,,12,14.已知函數(shù)f(x)3ln x x22x3ln 3 ,則方程f(x)0的解的個數(shù)是____.,1,,1,
20、2,3,4,5,6,7,8,9,10,11,13,14,15,16,12,當(dāng)x(0,3)時,f(x)0,f(x)單調(diào)遞增, 當(dāng)x(3,)時,f(x)<0,f(x)單調(diào)遞減, 當(dāng)x0時,f(x),當(dāng)x時,f(x),,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,12,所以方程f(x)0只有一個解.,拓展沖刺練,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,12,(,2)(2,),,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,12,解得m2或m<2.,,1,2,3,4,5,6,7,8,9,10,11,13,14,1
21、5,16,解得m2或m2.,12,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,16.(2018浙江第二次聯(lián)盟校聯(lián)考)已知a為實數(shù),函數(shù)f(x)ex2ax. (1)討論函數(shù)f(x)的單調(diào)性;,解f(x)ex2a. 當(dāng)a0時,f(x)0,函數(shù)f(x)在R上單調(diào)遞增. 當(dāng)a0時,由f(x)ex2a0,得x2ln a. 若x2ln a,則f(x)0,函數(shù)f(x)在(2ln a,)上單調(diào)遞增; 若x<2ln a,則f(x)<0,函數(shù)f(x)在(,2ln a)上單調(diào)遞減.,12,,1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,(2)若函數(shù)f(x)有兩
22、個不同的零點x1,x2(x1