1、 高考解答題專項(xiàng)一函數(shù)與導(dǎo)數(shù)中的綜合問題第2課時利用導(dǎo)數(shù)研究不等式恒(能)成立問題1.(2021湖南衡陽高三月考)已知函數(shù)f(x)=(ax+1)ex.(1)討論函數(shù)f(x)的單調(diào)性;(2)若a=2,當(dāng)x0時,f(x)kx,求實(shí)數(shù)k的取值范圍.2.(2021山東濟(jì)南高三月考)已知f(x)=ex-ax2-x-1.(1)當(dāng)a=e2時,求f(x)的極值點(diǎn)個數(shù);(2)當(dāng)x0,+)時,f(x)0,求實(shí)數(shù)a的取值范圍.3.(2021重慶八中高三月考)已知函數(shù)f(x)=ex-kx,g(x)=x2+k2-3.(1)討論函數(shù)y=f(x)的單調(diào)區(qū)間;(2)若2f(x)g(x)對任意的x0恒成立,求實(shí)數(shù)k的取值范圍.

2、4.(2021遼寧沈陽高三三模)已知函數(shù)f(x)=(x+1)e-ax,a0,若對任意的x0,不等式f(x)12x+1恒成立,求實(shí)數(shù)a的取值范圍.5.(2021浙江寧波高三期中)已知函數(shù)f(x)=-mx+ln x+1,g(x)=cos x+xsin x-1.(1)討論函數(shù)f(x)的單調(diào)區(qū)間與極值;(2)若m12,對任意的x11,2,總存在x20,使得不等式f(x1)-g(x2)1成立,試求實(shí)數(shù)m的取值范圍.第2課時利用導(dǎo)數(shù)研究不等式恒(能)成立問題1.解 (1)因?yàn)閒(x)=(ax+1)ex,所以f(x)=aex+(ax+1)ex=(ax+a+1)ex.若a=0,則f(x)0,f(x)在R上單調(diào)

3、遞增;若a0,則當(dāng)x-a-1a時,f(x)0;當(dāng)x-a-1a時,f(x)0.故f(x)的單調(diào)遞增區(qū)間為-a-1a,+,單調(diào)遞減區(qū)間為-,-a-1a;若a-a-1a時,f(x)0;當(dāng)x0,故f(x)的單調(diào)遞減區(qū)間為-a-1a,+,單調(diào)遞增區(qū)間為-,-a-1a.(2)當(dāng)a=2時,不等式f(x)kx即為(2x+1)exkx.當(dāng)x=0時,原不等式等價于1k0恒成立,此時kR.當(dāng)x0時,原不等式等價于k(2x+1)exx.令函數(shù)g(x)=(2x+1)exx(x0),則g(x)=x(2x+3)ex-(2x+1)exx2=(x+1)(2x-1)exx2.當(dāng)0 x12時,g(x)12時,g(x)0,g(x)單

4、調(diào)遞增.故g(x)min=g12=4e,所以k4e.綜上所述,實(shí)數(shù)k的取值范圍為(-,4e.2.解 (1)當(dāng)a=e2時,f(x)=ex-e2x2-x-1,所以f(x)=ex-ex-1,f(x)=ex-e,所以當(dāng)x1時,f(x)1時,f(x)0,f(x)在(1,+)上單調(diào)遞增,因?yàn)閒(0)=0,f(1)=-1,f(2)=e2-2e-10,所以存在x0(1,2),使f(x0)=0,所以,x(-,0)時,f(x)0;x(0,x0)時,f(x)0,所以0和x0是f(x)的極值點(diǎn),所以f(x)有兩個極值點(diǎn).(2)f(x)=ex-ax2-x-1,f(x)=ex-2ax-1,設(shè)h(x)=f(x)=ex-2a

5、x-1(x0),則h(x)=ex-2a單調(diào)遞增,又因?yàn)閔(0)=1-2a,所以當(dāng)a12時,h(x)0,h(x)在0,+)上單調(diào)遞增,所以h(x)h(0)=0,即f(x)0,f(x)在0,+)上單調(diào)遞增,所以f(x)f(0)=0,符合題意.當(dāng)a12時,令h(x)=0,解得x=ln 2a,當(dāng)x0,ln 2a)時,h(x)0,h(x)在0,ln 2a)上單調(diào)遞減,f(x)=h(x)h(0)=0,f(x)在(0,ln 2a)上單調(diào)遞減,所以x(0,ln 2a)時,f(x)0恒成立,則y=f(x)在R上單調(diào)遞增;若k0,當(dāng)xln k時,f(x)0,f(x)的單調(diào)遞增區(qū)間為(ln k,+);當(dāng)xln k時

6、,f(x)0,f(x)的單調(diào)遞減區(qū)間為(-,ln k).(2)2ex-2kxx2+k2-3對任意的x0恒成立,即x2+2kx+k2-3ex2對任意的x0恒成立.令h(x)=x2+2kx+k2-3ex,則h(x)=-(x+k+1)(x+k-3)ex.當(dāng)k3時,h(x)0在(0,+)上恒成立,h(x)在(0,+)上單調(diào)遞減,所以只需h(0)=k2-32,即k-5,5,與k3矛盾.當(dāng)-1k3時,h(x)在(0,-k+3)上單調(diào)遞增,在(-k+3,+)上單調(diào)遞減,所以只需h(-k+3)=6e-k+32,即k3-ln 3.所以-1k3-ln 3;當(dāng)k-1時,h(x)在(0,-k-1)上單調(diào)遞減,在(-k

7、-1,-k+3)上單調(diào)遞增,在(-k+3,+)上單調(diào)遞減.因此h(0)2,h(-k+3)2-5k3-ln 3,所以-5k0,12x+10,將不等式兩邊取對數(shù)得ln(x+1)-axln12x+1,即ln(x+1)-ln12x+1-ax0.因此只需證明當(dāng)x0時,不等式ln(x+1)-ln12x+1-ax0恒成立即可.令g(x)=ln(x+1)-ln12x+1-ax,則g(x)=1x+1-1x+2-a=1(x+1)(x+2)-a.若a12,因?yàn)楫?dāng)x0時,1(x+1)(x+2)12,所以g(x)=1(x+1)(x+2)-a0,g(x)在0,+)上單調(diào)遞減,因此g(x)g(0)=0.若a0,所以g(x)

8、=1(x+1)(x+2)-a0,g(x)在0,+)上單調(diào)遞增,因此g(x)g(0)=0,不合題意.若0a0,所以g(x)在0,4+a4a-32上單調(diào)遞增,于是g(x)g(0)=0,不合題意.綜上,當(dāng)不等式f(x)12x+1恒成立時,實(shí)數(shù)a的取值范圍是12,+.5.解 (1)f(x)=-m+1x(x0).當(dāng)m0時,f(x)0,f(x)在(0,+)上單調(diào)遞增,無極值.當(dāng)m0時,令f(x)=-m+1x=0,得x=1m.令f(x)0,則0 x1m;令f(x)1m.所以f(x)在0,1m上單調(diào)遞增,在1m,+上單調(diào)遞減,此時f(x)極大值=f1m=-ln m,無極小值.綜上,當(dāng)m0時,f(x)的單調(diào)遞增

9、區(qū)間為(0,+),無單調(diào)遞減區(qū)間,無極值;當(dāng)m0時,f(x)的單調(diào)遞增區(qū)間為0,1m,單調(diào)遞減區(qū)間為1m,+,極大值為-ln m,無極小值.(2)對任意的x11,2,總存在x20,使得f(x1)-g(x2)1成立,等價于f(x)在1,2上的最小值f(x)min與g(x)在0,上的最小值g(x)min的差大于1,g(x)=xcos x,當(dāng)x0,2時,g(x)0,g(x)在0,2上單調(diào)遞增;當(dāng)x2,時,g(x)0,g(x)在2,上單調(diào)遞減.又因?yàn)間(0)=0,g()=-2,所以g(x)min=g()=-2.由(1)知,當(dāng)0-2得m1+12ln 2,所以1m1+12ln 2.當(dāng)11m212m1時,f(