《(通用版)2020高考數(shù)學(xué)二輪復(fù)習(xí) 規(guī)范解答集訓(xùn)(六) 函數(shù)、導(dǎo)數(shù)、不等式 文》由會員分享,可在線閱讀,更多相關(guān)《(通用版)2020高考數(shù)學(xué)二輪復(fù)習(xí) 規(guī)范解答集訓(xùn)(六) 函數(shù)、導(dǎo)數(shù)、不等式 文(5頁珍藏版)》請在裝配圖網(wǎng)上搜索。
1、規(guī)范解答集訓(xùn)(六)函數(shù)、導(dǎo)數(shù)、不等式(建議用時:60分鐘)1(2019洛陽模擬)已知函數(shù)f(x)ex(x22xa)(其中aR,a為常數(shù),e為自然對數(shù)的底數(shù))(1)討論函數(shù)f(x)的單調(diào)性;(2)設(shè)曲線yf(x)在(a,f(a)處的切線為l,當(dāng)a1,3時,求直線l在y軸上截距的取值范圍解(1)f(x)ex(x22xa)ex(2x2)ex(x2a2),當(dāng)a2時,f(x)0恒成立,函數(shù)f(x)在區(qū)間(,)上單調(diào)遞增;當(dāng)a2時,f(x)0x22ax或x,函數(shù)f(x)在區(qū)間(,),(,)上單調(diào)遞增,在區(qū)間(,)上單調(diào)遞減(2)f(a)ea(a2a),f(a)ea(a2a2),所以直線l的方程為yea(a
2、2a)ea(a2a2)(xa)令x0,得截距bea(a3a),記g(a)ea(a3a)(1a3),則g(a)ea(a33a2a1),記h(a)a33a2a1(1a3),則h(a)3a26a10(1a3),所以h(a)在1,3上單調(diào)遞減,所以h(a)h(1)20,所以g(a)0),得4x2mx10(x0)若m0,此時4x2mx10(x0)恒成立;若m0,有m2160,即00),若m0,則x(0,m)時,F(xiàn)(x)0,故F(x),F(xiàn)(x)隨x的變化如表,x(0,m)m(m,)F(x)0F(x)極小值F(x)minF(m)m2ln m0,即m1,故0m1;若m0,則當(dāng)x時,F(xiàn)(x)取得最小值,F(xiàn)(x)
3、minFm2ln0,即ln0,2em0,根據(jù)題意,此時可以不予考慮;若m0,則F(x)x20恒成立由m0為g(x)f(x)恒成立時m取到的最大值,知m01.當(dāng)mm01時,f(x)ln x2x2x1,f(1)2,即切點(diǎn)為(1,2),又f(x)4x1,f(1)4,所以切線的斜率為4,故所求的切線方程為y4x2.3已知函數(shù)f(x)ax22(a1)x2ln x,a(0,)(1)討論函數(shù)f(x)的單調(diào)性;(2)若a4,證明:對任意的x2,都有f(x)ex(x1)axln x成立(其中e為自然對數(shù)的底數(shù),e2.718 28)解(1)f(x)的定義域?yàn)?0,),f(x)2ax2(a1),當(dāng)a(0,1)時,f
4、(x)在(0,1)和上單調(diào)遞增,在上單調(diào)遞減;當(dāng)a1時,f(x)在(0,)上單調(diào)遞增;當(dāng)a(1,)時,f(x)在和(1,)上單調(diào)遞增,在上單調(diào)遞減(2)當(dāng)a4時,即證4x26x3ln xex(x1),x2,設(shè)g(x)4x26x3ln xex(x1)(x2),則g(x)8x6xex.令h(x)8x6xex,則h(x)8(x1)ex8(x1)ex83e20,h(x)在2,)上單調(diào)遞減,h(x)8x6xexh(2)1662e22e20,即g(x)0在2,)上恒成立,g(x)在2,)上單調(diào)遞減,g(x)4x26x3ln xex(x1)g(2)16123ln 2e243e27e20,原不等式恒成立4已知
5、函數(shù)f(x)x2(a1)xaln x.(1)當(dāng)a1時,討論函數(shù)f(x)的單調(diào)性;(2)若不等式f(x)(a1)xxa1e對于任意xe1,e成立,求正實(shí)數(shù)a的取值范圍解(1)函數(shù)f(x)的定義域?yàn)?0,),f(x)x(a1).若0a1,則當(dāng)0x1時,f(x)0,f(x)單調(diào)遞增;當(dāng)ax1時,f(x)0,f(x)單調(diào)遞減若a0,則當(dāng)0x1時,f(x)1時,f(x)0,f(x)單調(diào)遞增綜上所述,當(dāng)a0時,函數(shù)f(x)在(1,)上單調(diào)遞增,在(0,1)上單調(diào)遞減;當(dāng)0a1時,函數(shù)f(x)在(a,1)上單調(diào)遞減,在(0,a)和(1,)上單調(diào)遞增(2)原題等價于對任意x,有aln xxae1成立,設(shè)g(x
6、)aln xxa,x,則g(x)maxe1.g(x)axa1,令g(x)0,得0x0,得x1.函數(shù)g(x)在上單調(diào)遞減,在(1,e上單調(diào)遞增,g(x)max為gaea與g(e)aea中的較大者設(shè)h(a)g(e)geaea2a(a0),則h(a)eaea2220,h(a)在(0,)上單調(diào)遞增,故h(a)h(0)0,g(e)g,從而g(x)maxg(e)aea.aeae1,即eaae10.設(shè)(a)eaae1(a0),則(a)ea10,(a)在(0,)上單調(diào)遞增又(1)0,滿足eaae10的a的取值范圍為(,1a0,a的取值范圍為(0,15已知函數(shù)f(x)x3x2ax2的圖象過點(diǎn)A.(1)求函數(shù)f(
7、x)的單調(diào)增區(qū)間;(2)若函數(shù)g(x)f(x)2m3有3個零點(diǎn),求m的取值范圍解(1)因?yàn)楹瘮?shù)f(x)x3x2ax2的圖象過點(diǎn)A,所以4a4a2,解得a2,即f(x)x3x22x2,所以f(x)x2x2.由f(x)0,得x2.所以函數(shù)f(x)的單調(diào)增區(qū)間是(,1),(2,)(2)由(1)知f(x)極大值f(1)22,f(x)極小值f(2)242,由數(shù)形結(jié)合(圖略),可知要使函數(shù)g(x)f(x)2m3有三個零點(diǎn),則2m3,解得m.所以m的取值范圍為.6(2019開封模擬)已知函數(shù)f(x)aln xx22x.(1)當(dāng)a4時,求函數(shù)f(x)的單調(diào)區(qū)間;(2)若函數(shù)yf(x)有兩個極值點(diǎn)x1,x2,且x1x2,求證:0,解得x2,f(x)的單調(diào)遞增區(qū)間為(2,),令f(x)0,解得0x2,f(x)的單調(diào)遞減區(qū)間為(0,2)f(x)的單調(diào)遞增區(qū)間為(2,),單調(diào)遞減區(qū)間為(0,2)(2)f(x),x(0,),設(shè)g(x)2x22xa,若f(x)有兩個極值點(diǎn)x1,x2,則g(x)0在(0,)上有兩個不同的根x1,x2,可得0a,a2x2(1x2),又0x1x2,x21,x212x2ln x2,設(shè)h(x)x12xln x,x1,h(x)2ln x0,h(x)在上單調(diào)遞減,h(x)h2ln ln 2,ln 2.- 5 -