北师大版选修45 综合法放缩法 课时作业.docx

2018-2019学年北师大版选修4-5 综合法放缩法 课时作业1.已知a,br+,a=a+b,b=a+b,则a,b的大小关系是()a.a=bb.abc.abd.a0.又a0,b0,ab.答案:c2.已知0a10b.logab+logba+20c.logab+logba+20d.logab+logba+20解析:0a1b,logab0.(-logab)+1-logab2当且仅当b=1a时取“=”号,-logab+1-logab-2,即logab+1logab-2,logab+logba-2.logab+logba+20.答案:d3.已知a,br+,则下列各式成立的是()a.cos2lg a+sin2lg blg(a+b)c.acos2bsin2=a+bd.acos2bsin2a+b解析:cos2lg a+sin2lg bcos2lg(a+b)+sin2lg(a+b)=lg(a+b).答案:a4.已知s=1+112+1123+1123n(n是大于2的自然数),则有()a.s1b.2s3c.1s2d.3s4解析:由1123k2),得s=11+112+1123+1123n1+12+122+123+12n-1=1-12n1-12=2-12n-11.故选c.答案:c5.下列四个命题中,不正确的是()a.若012,则cos(1+)cos(1-)b.若0a1+a2ac.若实数x,y满足y=x2,则log2(2x+2y)的最小值是78d.若a,br,则a2+b2+ab+1a+b解析:若012,则0121-1+322,又函数y=cos x在0,2内是减少的,故选项a正确.当0a0,11-a1+a.1+a2a当且仅当a=1时取“=”号,但0a0,b0,c0,且a,b,c不全相等,求证:bca+acb+abca+b+c.分析利用综合法证明,注意条件a,b,c不全相等的使用.证明a0,b0,c0,bca+acb2bcaacb=2c,bca+abc2bcaabc=2b,acb+abc2acbabc=2a.又a,b,c不全相等,上面三式“=”号不能同时成立,三式相加后两边除以2,得bca+acb+abca+b+c.8.已知an=4n-2n,tn=2na1+a2+an,求证:t1+t2+t3+tn32.证明a1+a2+an=41+42+43+4n-(21+22+2n)=4(1-4n)1-4-2(1-2n)1-2=43(4n-1)+2(1-2n),tn=2n43(4n-1)+2(1-2n)=2n4n+13-43+2-2n+1=2n4n+13+23-2n+1=32n4n+1-32n+1+2=322n2(2n)2-32n+1=322n(22n-1)(2n-1)=3212n-1-12n+1-1.t1+t2+t3+tn=321-13+13-17+12n-1-12n+1-132(x+y+z).证明x2+xy+y2=x+y22+34y2x+y22=x+y2x+y2,同理可得y2+yz+z2y+z2,z2+zx+x2z+x2.又x,y,z不全为零,上述三个不等式不能同时取“=”号,x2+xy+y2+y2+yz+z2+z2+zx+x2x+y+z+x2+y2+z2=32(x+y+z).10.求证:32-1n+11+122+132+1n2k2k(k-1)(kn+,且k2),1k(k+1)1k21k(k-1),即1k-1k+11k21k-1-1k(kn+,且k2).分别令k=2,3,n,得12-131221-12,13-1413212-13,1n-1n+11n21n-1-1n.将这些不等式相加,得12-13+13-14+1n-1n+1122+132+1n21-12+12-13+1n-1-1n,即12-1n+11