全品高考备战2027年数学一轮学生用书12第21讲双变量不等式的证明【答案】作业手册

第21讲双变量不等式的证明1.解:(1)当a=0时,f(x)=lnx+x,则f(1)=1,又因为f'(x)=1x+1,所以切线的斜率k=f'故所求切线方程为y-1=2(x-1),即2x-y-1=0.(2)证明:当a=-2时,f(x)=lnx+x2+x(x>0).由f(x1)+f(x2)+x1x2=0,得lnx1+x12+x1+lnx2+x22+x2+x即(x1+x2)2+(x1+x2)=x1x2-ln(x1x2),令t=x1x2(t>0),φ(t)=t-lnt,则φ'(t)=1-1t=t-1t,易知φ(t)在(0,1)上单调递减,在(1,+∞)上单调递增,所以φ(t)≥φ(1)=1,所以(x1+x2)2+(x1+x2)≥1,又x1>0,x2>0,所以x1+x2.解:(1)函数f(x)的定义域为(0,+∞),f'(x)=1-①当m≤0时,f'(x)>0在(0,+∞)上恒成立,所以f(x)在(0,+∞)上单调递增.②当m>0时,令f'(x)=1-mxx>0,得0<x<1m,所以f(令f'(x)=1-mxx<0,得x>1m,所以f(x综上,当m≤0时,f(x)在(0,+∞)上单调递增;当m>0时,f(x)在0,1m上单调递增,在(2)易得f(1)=0,由(1)知,当m≤0时,不满足题意,故m>0,则f(x)在0,1m所以f(x)max=f1m=-lnm-1+m,故只需-lnm-1+m≤0即可令g(m)=-lnm-1+m,则g'(m)=1-1m=m-1m,所以当m∈(0,1)时,g'(m)<0,当m∈(1,+∞)时,g'(m)>0,所以g(m)在(0,1)上单调递减,在(1,+∞)上单调递增,所以g(m)≥g(1)=0,即-lnm-1+m≥0.又因为-ln所以-lnm-1+m=0,解得m=1.(3)证明:f(b)-f(a)b-a=lnb-lnab-a-1=1a·lnbaba-1-1,因为b因为t>1,所以lntt-1<1,即lnbaba-1<1.因为1a>0,所以3.解:(1)由f(x)=x-aex+b(a>0,b∈R)得f'(x)=1-aex,由f'(x)>0得x<ln1a,由f'(x)<0得x>ln1a,所以f(x)在-∞,ln1a上单调递增,在ln1a,+∞上单调递减,所以f(x)的最大值为fln(2)证明:由题知x两式相减得x1-x2=a(ex1-ex2),即故要证x1+x2<-2lna,只需证x1+x2<-2lnx1即证ex1+即证(x1-x2不妨设x1<x2,令x2-x1=t>0,则只需证t2<e-t-2+et.设g(t)=t2-e-t+2-et,t>0,则g'(t)=2t+e-t-et,t>0,设h(t)=2t+e-t-et,t>0,则h'(t)=2-e-t-et=2-et所以h(t)在(0,+∞)上单调递减,所以h(t)<h(0)=0,所以g(t)在(0,+∞)上单调递减,所以g(t)<g(0)=0,故原不等式得证.4.解:(1)令f(x)=0,若lnx=0,则x=1是一个零点,但不是题设区间内的零点,所以x2-kx+1=0的两个根分别在区间(0,1)和(1,+∞)内,故只需12-k+1=2-k<0,所以k>2.(2)由题可得f'(x)=(2x-k)lnx+x2-kx+1x,结合(1)知f'(x1)=(2x1-k)lnx1,所以曲线y=f(x)在点(x1,0)处的切线方程为y=(2x1-k)lnx1·(x-x1令x=0,则y=(2x1-k)lnx1·(-x1)=(kx1-2x12)lnx1,又kx1=x12+1,所以y=(1-x12)lnx1,所以S=12x1|(1-x12)lnx1|,易知x1x2=1,则x2所以Sx2-x1=12x1(x令t=x12∈(0,1),则Sx2-x1=h(t)=-14tlnt,则当0<t<1e时,h'(t)>0,当1e<t<1时,h'(t)<0,所以h(t)在0,1e上单调递增,在1e,1上单调递减,所以h所以Sx2-(3)证明:方法一:记f'(x)的导函数为f″(x),则f″(x)=3+2lnx-kx-1x2,显然f″(x)在(0,+∞)上单调递增,又f″(1)=2-k<0,当x→+∞时f″(x)→+∞,所以存在x0∈(1,+∞),使得f″(当0<x<x0时,f″(x)<0,则f'(x)在(0,x0)上单调递减,当x>x0时,f″(x)>0,则f'(x)在(x0,+∞)上单调递增.又f'(x0)<f'(1)=2-k<0,当x→0或x→+∞时,f'(x)→+∞,所以存在m∈(0,x0),n∈(x0,+∞),使得f'(m)=f'(n)=0,当0<x<m或x>n时,f'(x)>0,则f(x)在(0,m),(n,+∞)上单调递增,当m<x<n时,f'(x)<0,则f(x)在(m,n)上单调递减,结合题设,可知t1,t0,t2依次在区间(0,m),(m,n),(n,+∞)上,如图.设曲线y=f(x)在点(x1,0)处的切线l1的方程为y=l1(x),则l1(x)=f'(x1)·(x-x1),设g1(x)=f(x)-l1(x)=f(x)-f'(x1)·(x-x1),0<x<m,则g'1(x)=f'(x)-f'(x1),记g'1(x)的导函数为g″1(x),则g″1(x)=f″(x)<0,故g'1(x)在(0,m)上单调递减,又g'1(x1)=0,所以当0<x<x1时,g'1(x)>0,当x1<x<m时,g'1(x)<0,所以g1(x)在(0,x1)上单调递增,在(x1,m)上单调递减,则g1(x)≤g1(x1)=0,所以g1(t1)=f(t1)-l1(t1)≤0.若直线y=a与直线l1的交点横坐标为x'1,则l1(t1)≥f(t1)=l1(x'1)=a,由k>2,0<x1<1<x2,x1+x2=k,得l1的斜率f'(x1)=(2x1-k)lnx1=(x1-x2)lnx1>0,故t1≥x'1.设曲线y=f(x)在点(x2,0)处的切线l2的方程为y=l2(x),则l2(x)=f'(x2)·(x-x2),设g2(x)=f(x)-l2(x)=f(x)-f'(x2)·(x-x2),且x>n,则g'2(x)=f'(x)-f'(x2),记g'2(x)的导函数为g″2(x),则g″2(x)=f″(x)>0,故g'2(x)在(n,+∞)上单调递增,又g'2(x2)=0,所以当n<x<x2时,g'2(x)<0,当x>x2时g'2(x)>0,所以g2(x)在(n,x2)上单调递减,在(x2,+∞)上单调递增,则g2(x)≥g2(x2)=0,所以g2(t2)=f(t2)-l2(t2)≥0.若直线y=a与直线l2的交点横坐标为x'2,则l2(t2)≤f(t2)=l2(x'2)=a,由k>2,x1,x2是关于x的方程x2-kx+1=0的根,满足x1+x2=k和x1x2=1,则l2的斜率f'(x2)=(2x2-k)lnx2=(x2-x1)lnx2>0,故t2≤x'2,综上,t2-t1≤x'2-x'1.由上知f'(x1)=(x1-x2)lnx1,f'(x2)=(x2-x1)lnx2,又x1x2=1,所以x2=1x1,所以f'(x1)=f'(x所以l1∥l2,则x'2-x'1=x2-x1,故t2-t1≤x2-x1,得证.方法二:f'(x)=(2x-k)lnx+x2-kx+1x,设m(x)=f'(x),则m'(x)=2lnx+3-kx-1x2,因为k>2,所以m'(x)在(0,+∞)上单调递增,注意到m'(1)=2-k<0,m'(k)>0,所以存在唯一的x0∈(1,k),使m'(x)=0,所以当0<x<x0时,m'(x)<0,当x>x0时,m'(x)>0,所以f'(所以f'(x)至多有两个零点,且当x>0,x→0时,f'(x)→+∞,f'(1)=2-k<0,当x→+∞时,f'(x)→+∞,∴f'(x)在(0,1)和(1,+∞)内各有一个零点x'1,x'2,且f(x)在(0,x'1)上单调递增,在(x'1,x'2)上单调递减,在(x'2,+∞)上单调递增,f(1)=0,作出f(x)的大致图象如图,当x>0且x→0时,f(x)→-∞,由(2)知曲线y=f(x)在(x