2023届高考二轮总复习试题（适用于老高考新教材） 数学 考点突破练18　利用导数证明不等式 WORD版含解析.docx

1、考点突破练18利用导数证明不等式1.(2022福建福州质检)已知函数f(x)=eln x-ax(aR).(1)讨论f(x)的单调性;(2)当a=e时,证明:xf(x)-ex+2ex0.2.(2022江西鹰潭二模(文)已知函数f(x)=ax2-(a+2)x+ln x.(1)当a0时,讨论函数f(x)的单调区间;(2)当a=0时,证明:f(x)ex-2x-2(其中e为自然对数的底数).3.(2022陕西咸阳二模)已知函数f(x)=ln x-kx+1.(1)若f(x)0恒成立,求实数k的取值范围;(2)证明:1+1221+1321+1n21).4.(2021新高考22)已知函数f(x)=x(1-ln

2、 x).(1)讨论f(x)的单调性;(2)设a,b为两个不相等的正数,且bln a-aln b=a-b,证明:21a+1be.5.(2022江苏泰州一模)已知函数f(x)=ax+ln x.(1)讨论f(x)的单调性;(2)若f(x1)=f(x2)=2(x1x2),证明:a2x1x20),若a0,则f(x)0,f(x)在(0,+)上为增函数;若a0,则当x0;当xea时,f(x)0,所以只需证f(x)exx-2e,由(1)知,当a=e时,f(x)在(0,1)上单调递增,在(1,+)上单调递减,所以f(x)max=f(1)=-e.记g(x)=exx-2e(x0),则g(x)=(x-1)exx2,所

3、以,当0x1时,g(x)1时,g(x)0,g(x)单调递增,所以g(x)min=g(1)=-e.所以当x0时,f(x)g(x),即f(x)exx-2e,即xf(x)-ex+2ex0.2.(1)解 f(x)的定义域为(0,+),f(x)=2ax-(a+2)+1x=(ax-1)(2x-1)x(a0),当01a2时,在0,1a,12,+上,f(x)0,f(x)单调递增;在1a,12上,f(x)12,即0a0,f(x)单调递增;在12,1a上,f(x)2时,f(x)的单调递增区间为0,1a,12,+,单调递减区间为1a,12.当a=2时,f(x)的单调递增区间为(0,+).当0a2时,f(x)的单调递

4、增区间为0,12,1a,+,单调递减区间为12,1a.(2)证明 当a=0时,由f(x)0,构造函数h(x)=ex-ln x-2(x0),h(x)=ex-1x,h(x)=ex+1x20,h(x)在(0,+)上单调递增,h12=e-20,故存在x012,1,使得h(x0)=0,即ex0=1x0.当x(0,x0)时,h(x)0,h(x)单调递增.所以x=x0时,h(x)取得极小值,也即是最小值.h(x0)=ex0-ln x0-2=1x0-ln1ex0-2=1x0+x0-221x0x0-2=0,所以h(x)=ex-ln x-20,故f(x)0,则g(x)=1-lnx-1x2=-lnxx2,当0x0,

5、g(x)单调递增,当x1时,g(x)0,g(x)单调递减,g(x)max=g(1)=1,k1.(2)证明 由(1)知,k=1时,有不等式ln xx-1对任意x(0,+)恒成立,当且仅当x=1时,取“=”号,当x(1,+),ln x1,且nN*),则ln1+1n21n21n2-1=121n-1-1n+1,ln1+122+ln1+132+ln1+1n2122+1212-14+1n-2-1n+1n-1-1n+1=14+1212+13-1n-1n+114+1212+13=23,即ln1+1221+1321+1n21),1+1221+1321+1n21).4.(1)解 由条件知,函数f(x)的定义域为(

6、0,+),f(x)=-ln x.当x(0,1)时,f(x)0,f(x)单调递增;当x(1,+)时,f(x)0,从而f1b0,得1b(1,e),令g(x)=f(2-x)-f(x),则g(x)=-f(2-x)-f(x)=ln1-(x-1)2,当x(0,1)时,g(x)g(1)=0,从而f(2-x)f(x),所以f2-1af1a=f1b,由(1)得2-1a1b即20,h(x)在区间(1,e)内单调递增,h(x)h(e)=e,从而x+f(x)e,所以1b+f1be.又由1a(0,1),可得1a1a1-ln1a=f1a=f1b,所以1a+1bf1b+1b=e.由得21a+1be.(方法二)bln a-a

7、ln b=a-b变形为lnaa-lnbb=1b-1a,所以lna+1a=lnb+1b.令1a=m,1b=n.则上式变为m(1-ln m)=n(1-ln n),于是命题转换为证明2m+ne.令f(x)=x(1-ln x),则有f(m)=f(n),不妨设mn.令f(x)=0,得x=1,且f(e)=0.结合(1)知0m1,1n2.要证m+n2n2-mf(n)f(2-m)f(m)f(2-m)f(m)-f(2-m)0,g(x)在区间(0,1)内单调递增,所以g(x)2.再证m+nm,所以n(1-ln n)+nem+n0,故h(x)在区间(1,e)内单调递增.所以h(x)h(e)=e.故h(n)e,即m+

8、ne.综合可知21a+1b2同方法二.以下证明x1+x21,由x1(1-ln x1)=x2(1-ln x2)得x1(1-ln x1)=tx11-ln(tx1),ln x1=1-tlntt-1,要证x1+x2e,只需证(1+t)x1e,两边取对数得ln(1+t)+ln x11,即ln(1+t)+1-tlntt-11,即证ln(1+t)tlntt-1.记g(s)=ln(1+s)s,s(0,+),则g(s)=s1+s-ln(1+s)s2.记h(s)=s1+s-ln(1+s),则h(s)=1(1+s)2-11+s0,所以h(s)在区间(0,+)内单调递减.h(s)h(0)=0,则g(s)0,所以g(s

9、)在区间(0,+)内单调递减.由t(1,+)得t-1(0,+),所以g(t)g(t-1),即ln(1+t)tlntt-1.(方法四)由已知得lnaa-lnbb=1b-1a,令1a=x1,1b=x2,不妨设x1x2,所以f(x1)=f(x2).令f(x)=0,得x=1,且f(e)=0.由(1)及前面分析知,0x11x2e,只需证2x1+x22同方法二.再证明x1+x2e.令h(x)=1-lnxx-e(0xe),h(x)=-2+ex+lnx(x-e)2.令(x)=ln x+ex-2(0xe),则(x)=1x-ex2=x-ex2(e)=0,h(x)0,h(x)在区间(0,e)内单调递增.因为0x1x

10、2e,所以1-ln x1x1-ex1-ex2-e.又因为f(x1)=f(x2),所以1-ln x11-ln x2=x2x1,x2x1x1-ex2-e,即x22-ex20.因为x1x2,所以x1+x2e,即1a+1be.综上,有21a+1b0恒成立,则f(x)在(0,+)上单调递增,当a0时,f(x)0的解集为(a,+),即f(x)的单调递增区间为(a,+),单调递减区间为(0,a),所以当a0时,f(x)在(0,+)上单调递增,当a0时,f(x)在(a,+)上单调递增,在(0,a)上单调递减.(2)证明 因为f(x1)=f(x2)=2(x1x2),由(1)知,a0,且f(x)min=f(a)=

11、ln a+12,解得a(0,e),设x1x2,则0x1aa2,即证x2a2x1a,即证f(x2)fa2x1,即证f(x1)fa2x1,设g(x)=f(x)-fa2x=2ln x+ax-xa-2ln a,x(0,a),则g(x)=2x-ax2-1a=-(x-a)2ax20,即f(x)fa2x(x(0,a),则f(x1)fa2x1成立,因此x1x2a2成立,要证x1x2ae,即证ax2aex1,即证f(x2)faex1,即证f(x1)faex1,即证2x1e-ln x1+ln a+1,x1(0,a),而ax1+ln x1=2a=x1(2-ln x1),即证10,即(x)在(0,e)上单调递增,当xe时,(x)e,