課時規(guī)范練15　利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性

1、 課時規(guī)范練15利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性基礎(chǔ)鞏固組1.函數(shù)f(x)=x3-ax為R上增函數(shù)的一個充分不必要條件是() A.a0B.a02.(2020山東青島二中月考)已知定義域為R的函數(shù)f(x)的導(dǎo)數(shù)為f(x),且滿足f(x)x2-1的解集是()A.(-,-1)B.(-1,+)C.(2,+)D.(-,2)3.(2020山東德州二模,8)已知函數(shù)f(x)的定義域為R,且f(x)+13ex的解集為()A.(1,+)B.(-,1)C.(0,+)D.(-,0)4.已知函數(shù)f(x)=lnxx,則()A.f(2)f(e)f(3)B.f(3)f(e)f(2)C.f(e)f(2)f(3)D.f(e)f(3)f

2、(2)5.設(shè)函數(shù)f(x)=12x2-9ln x在區(qū)間a-1,a+1上單調(diào)遞減,則實數(shù)a的取值范圍是.6.已知函數(shù)f(x)=ax2-4ax-ln x,則f(x)在(1,3)上不具有單調(diào)性的一個充分不必要條件是()A.a-,16B.a-12,+C.a-12,16D.a12,+7.已知函數(shù)f(x)=aln x-2x,若不等式f(x+1)ax-2ex在x(0,+)上恒成立,則實數(shù)a的取值范圍是()A.(-,2B.2,+)C.(-,0D.0,28.若函數(shù)f(x)=x2-4ex-ax在R上存在單調(diào)遞增區(qū)間,則實數(shù)a的取值范圍為.9.(2020河北唐山一模,文21)已知a0,函數(shù)f(x)=2ax3-3(a2

3、+1)x2+6ax-2.(1)討論f(x)的單調(diào)性;(2)若f(x)在R上僅有一個零點,求a的取值范圍.綜合提升組10.(2020湖南郴州二模,文12)已知定義在R上的函數(shù)f(x)的導(dǎo)數(shù)為f(x),滿足f(x)=f(-x).且對任意x0,2,有f(x)cos x+f(x)sin x0,若a=233f-6,b=2f-4,c=2f3,則()A.abcB.bcaC.acbD.cba11.(2020山東泰安一中期中)已知函數(shù)f(x)的定義域為-1,5,部分對應(yīng)值如下表:x-1045f(x)1221f(x)的導(dǎo)函數(shù)y=f(x)的圖象如圖所示,關(guān)于f(x)的結(jié)論正確的是()A.函數(shù)f(x)是周期函數(shù)B.函

4、數(shù)f(x)在0,2上單調(diào)遞增C.函數(shù)y=f(x)-a的零點個數(shù)可能為0,1,2,3,4D.當(dāng)1a2時,函數(shù)y=f(x)-a有4個零點12.(2020安徽皖東名校聯(lián)盟聯(lián)考)若函數(shù)f(x)=-12x+m,xe,x-lnx,xe的值域是e-1,+),其中e是自然對數(shù)的底數(shù),則實數(shù)m的最小值是.13.(2020山東濰坊臨朐模擬一,22)已知函數(shù)f(x)=mln x-x+mx(mR),討論f(x)的單調(diào)性.創(chuàng)新應(yīng)用組14.(2020山東濰坊臨朐模擬一,8)已知奇函數(shù)f(x)的定義域為-2,2,其導(dǎo)函數(shù)為f(x),當(dāng)0 x2時,有f(x)cos x+f(x)sin x0成立,則關(guān)于x的不等式f(x)2f4

5、cos x的解集為()A.4,2B.-2,-44,2C.-4,00,4D.-4,04,215.設(shè)函數(shù)f(x)=aln x+x-1x+1,其中a為常數(shù).(1)若a=0,求曲線y=f(x)在點(1,f(1)處的切線方程;(2)討論函數(shù)f(x)的單調(diào)性.參考答案課時規(guī)范練15利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性1.B函數(shù)f(x)=x3-ax為R上增函數(shù)的充要條件是f(x)=3x2-a0在R上恒成立,所以a(3x2)min.因為(3x2)min=0,所以a0.而(-,0)(-,0.故選B.2.D令g(x)=f(x)-x2,則g(x)=f(x)-2xx2-1可化為f(x)-x2-1,而g(2)=f(2)-22=3-

6、4=-1,所以不等式可化為g(x)g(2),故不等式的解集為(-,2).故選D.3.C令g(x)=f(x)+1ex,f(x)+10,故g(x)在R上單調(diào)遞增,且g(0)=3,由f(x)+13ex,可得f(x)+1ex3,即g(x)g(0),所以x0,故選C.4.Df(x)=1-lnxx2(x0),當(dāng)x(0,e)時,f(x)0;當(dāng)x(e,+)時,f(x)f(3)f(2).故選D.5.(1,2f(x)=12x2-9ln x,f(x)=x-9x(x0),當(dāng)x-9x0時,有00且a+13,解得10,g(1)g(3)0,解得a16.而12,+-,-1216,+,故選D.7.Af(ex)=ax-2ex,所

7、以f(x+1)ax-2ex在(0,+)上恒成立,等價于f(x+1)f(ex)在(0,+)上恒成立.因為當(dāng)x(0,+)時,1x+11時,f(x)0恒成立,即當(dāng)x1時,ax2恒成立,所以a2.故選A.8.(-,-2-2ln 2)因為f(x)=x2-4ex-ax,所以f(x)=2x-4ex-a.由題意,f(x)=2x-4ex-a0有解,即a2x-4ex有解.令g(x)=2x-4ex,則g(x)=2-4ex.令g(x)=0,解得x=-ln 2.函數(shù)g(x)=2x-4ex在(-,-ln 2)上單調(diào)遞增;在(-ln 2,+)上單調(diào)遞減.所以當(dāng)x=-ln 2時,g(x)=2x-4ex取得最大值-2-2ln

8、2,所以a-2-2ln 2.9.解 (1)f(x)=6ax2-6(a2+1)x+6a=6(x-a)(ax-1),由f(x)=0,得x=a或x=1a.當(dāng)0aa.所以當(dāng)x1a時,f(x)0,從而f(x)在(-,a),1a,+上單調(diào)遞增;當(dāng)ax1a時,f(x)1時,a1a.所以當(dāng)xa時,f(x)0,從而f(x)在-,1a,(a,+)上單調(diào)遞增;當(dāng)1axa時,f(x)0,從而f(x)在1a,a上單調(diào)遞減.綜上,當(dāng)0a1時,f(x)在-,1a,(a,+)上單調(diào)遞增,在1a,a上單調(diào)遞減.(2)f(a)=-a4+3a2-2=(a2-1)(2-a2),f1a=1-1a2.當(dāng)0a1時,f(a)0,f1a0,所

9、以f(x)僅在1a,+上有一個零點,因此0a1時,f1a0,所以要滿足題設(shè)須有f(a)0,從而2-a20,解得1a2,因此1a0,g(x)在0,2上單調(diào)遞增.由f(x)=f(-x),得f(x)為偶函數(shù),a=233f-6=233f6=f(6)32=f(6)cos6=g6,b=2f-4=2f4=f(4)22=f(4)cos 4=g4,c=2f3=f(3)12=f(3)cos3=g3.06432,且g(x)在0,2上單調(diào)遞增,g6g4g3,即ab0,所以f(x)=x-ln x在e,+)上單調(diào)遞增,則f(x)f(e)=e-ln e=e-1,值域是e-1,+).又當(dāng)x-e2+m,值域是-e2+m,+.由

10、題設(shè)f(x)的值域為e-1,+),所以-e2+m,+e-1,+).于是-e2+me-1,解得m3e2-1.故實數(shù)m的最小值為3e2-1.13.解 由題意得x(0,+),f(x)=mx-1-mx2=-x2-mx+mx2.令g(x)=x2-mx+m,=m2-4m=m(m-4).當(dāng)0m4時,0,g(x)0恒成立,則f(x)0,f(x)在(0,+)上單調(diào)遞減.當(dāng)m0,函數(shù)g(x)與x軸有兩個不同的交點x1,x2(x1x2),x1+x2=m0,x1x2=m0,則x10.所以當(dāng)x0,m+m2-4m2時,g(x)0,則f(x)在0,m+m2-4m2上單調(diào)遞增;當(dāng)xm+m2-4m2,+時,g(x)0,f(x)

11、4時,0,函數(shù)g(x)與x軸有兩個不同的交點x1,x2(x10,x1x2=m0,則x10,x20.所以f(x)在0,m-m2-4m2,m+m2-4m2,+上單調(diào)遞減;在m-m2-4m2,m+m2-4m2上單調(diào)遞增.綜上所述,當(dāng)0m4時,f(x)在(0,+)上單調(diào)遞減;當(dāng)m4時,f(x)在0,m-m2-4m2上單調(diào)遞減,在m-m2-4m2,m+m2-4m2,m+m2-4m2,+上單調(diào)遞減.14.A根據(jù)題意,設(shè)g(x)=f(x)cosx,其導(dǎo)數(shù)為g(x)=f(x)cosx+f(x)sinxcos2x.因為當(dāng)0 x2時,f(x)cos x+f(x)sin x0,所以當(dāng)0 x2時,g(x)0,則函數(shù)g

12、(x)在0,2上單調(diào)遞減.又因為f(x)為定義域為-2,2的奇函數(shù),則g(-x)=f(-x)cos(-x)=-f(x)cosx=-g(x),則函數(shù)g(x)為奇函數(shù),所以函數(shù)g(x)在-2,2上為減函數(shù).f(x)2f4cos x,即f(x)cosx2f4,即f(x)cosxf(4)cos4,即g(x)g4.所以4x0,所以函數(shù)f(x)在(0,+)上單調(diào)遞增.當(dāng)a0時,令g(x)=ax2+2(a+1)x+a,則=4(a+1)2-4a2=4(2a+1).()當(dāng)a-12時,0,所以g(x)0,于是f(x)0,所以函數(shù)f(x)在(0,+)上單調(diào)遞減.()當(dāng)-12a0,此時g(x)=0有兩個不相等的實數(shù)根,分別是x1=-(a+1)+2a+1a,x2=-(a+1)-2a+1a,x10,x1x2=10,可得0 x1x2.當(dāng)0 xx2時,有g(shù)(x)0,f(x)0,所以函數(shù)