◆答案:C
★解析:asinx?bcosx?c?a2?b2sin(x??)?c?a2?b2,c?a2?b2. 只需c?a2?b2?0即可,得a2?b2?c,故选C.
x0?x1?x2?x3?01993*11、设任意实数
x1x2x3x3??,要使
logx01993?logx11993?logx21993?k?logx01993恒成立,则k的最大值是_____ __.◆答案:9 ★解析:显然即
x0?1,从而logx01993?0. x3x3111k???,即
lgx0?lgx1lgx1?lgx2lgx2?lgx3lgx0?lgx3??111????lgx0?lgx1???lgx1?lgx2???lgx2?lgx3???????klgx?lgxlgx?lgxlgx?lgx011223??.
又lgx0?lgx1?0,lgx1?lgx2?0,lgx2?lgx3?0,由柯西不等式,知k?9.即k的最大值为9.
1992*13、 (本题满分20 分)求证:16?★证明:因为
?k?1801?17. k122???2k?k?1, kk?kk?1?k122???2k?1?k. 同时kk?kk?1?k8080801?1??2k?k?1 于是得?2k?1?k??kk?1k?1k?1801?1?280?1?1?2??9?1??17. 即16??kk?1??????????
xy1991*15.已知0?a?1,x?y?0,求证:loga(a?a)?loga2?xy18xy21. 8x?y2★证明:由于0?a?1,不等式即证a?a?2a.由于a?a?2a.
1而x?y?x?x?x?1?x??.于是a42x?y2?a.
18∴a?a?2a
xyx?y2?2a.故证.
181990*7.设n为自然数,a,b为正实数,且满足a?b?2,则是 . ◆答案:1
11?的最小值nn1?a1?b - 16 -
★解析:由题意得
?a?b?ab????1?2?2,从而
anbn?1,故
111?an?1?bn???1.注意以上式子的等号当且仅当a?b?1时成立.即nnnnnn1?a1?b1?a?b?ab所求最小值为1.
1990*9.设n为自然数,对于任意实数x,y,z,恒有x2?y2?z2则n的最小值是 . ◆答案:3
★解析:由于x2?y2?z2??2?n(x4?y4?z4)成立,
?x4?y4?z4?2x2y2?2y2z2?2x2z2
?x4?y4?z4??x4?y4???y4?z4???x4?z4??3?x4?y4?z4?.等号当且仅当x?y?z时成立.故n?3.
1989*7.若loga◆答案:?0,1????22,??
?0?a?1?a?1★解析:不等式等价于?或?.解得a??0,1??a?2a?2??
1989*13. (本题满分20分)
已知a1,a2,?,an是n个正数,满足a1a2?an?1. 求证:?2?a1??2?a2???2?an??3.
n?2?1,则a的取值范围是 .
??2,??
?★证明:∵2?ai?1?1?ai?33ai,(i?1,2,?,n)
∴?2?a1??2?a2???2?an???1?1?a1??1?1?a2???1?1?an??3
1989*二、(本题满分35分)已知xi?R,(i?1,2,?,n,n?2)满足求证:
n3a1a2?an?3n.
?xi?1ni?1,?xi?0,
i?1nxi11??。 ?22ni?1in1,2,?,n?,且i?j). ★证明:由已知可知,必有xi?0,也必有xj?0 (i,j??设xi1,xi2,?,xil为诸xi中所有?0的数,xj1,xj2,?,xjm为诸xi中所有?0的数.由已知得
11A?xi1?xi2???xil?,B?xj1?xj2???xjm??.
22kxmxilj???h于是当时,?l?1lh?1hnkxkxkxi1mA?B11iljh???x?x????; ?????iljhilhn2222ni?1l?1h?1l?1h?1kxmxnkxkxkxi1m?BA11iljhiljh???当?时,? ????????xil??xjh????.lhnh?12222nl?1h?1i?1il?1lh?1hl?1 - 17 -
总之,
xi11??成立. ?22ni?1i11??1,试证:对每一个n?N?, abkkn1988*12、(本题满分15分)已知a、b为正实数,且
?a?b?n?an?bn?22n?2n?1
2k★证明:由已知得a?b?ab?2ab,故a?b?ab?4.于是?a?b???ab??2.
?2?a?b??2k?1.下面用数学归纳法证明:
1° 当n?1时,左?右?0.左?右成立.
kkk2kk?1 2° 设当n?k(k?1,k?N)时结论成立,即?a?b??a?b?2?2成立.
又 a?b?2kkk?ab?k则?a?b?k?1k??a?b??a?b??ak?bk?abak?1?bk?1?422k?2k?1?4?2k?1?4?2k?22?k?1??2?k?1??1.即命题对于n?k?1也成立.
?故对于一切n?N,命题成立.
??ak?1?bk?1??a?b??a?b???a?b?ak?bk?abak?1?bk?1
k??????????
?1984*五、(本题满分15分) 设x1,x2,?,xn都是正数,
222xnxnx12x2?1求证:??????x1?x2???xn.
x2x3xnx1222xnxnx12x2?1★证明:由于?x2?2x1,?x3?2x2,???xn?2xn?1,?x1?2xn.
x2x3xnx1上述各式相加即得.
1983*5、已知函数数f(x)?ax?c,满足?4?f(1)??1,?1?f(2)?5,那么f(3)应满
足( )
A.7?f(3)?26 B. ?4?f(3)?15 C. ?1?f(3)?20 D.
2?2835?f(3)? 33◆答案:
★解析:由于f(1)?a?c,f(2)?4a?c,f(3)?9a?c.令9a-c=λ(a-c)+μ(4a-c),
5855408840,??f(2)?, f(1)?f(2).但??f(1)?33333333∴?1?f(3)?20.选C.
∴f(3)??
bd?,那么( ) mnA.P?Q B.P?Q C.P?Q D.P、Q的大小关系不确定,而与m,n的大小
1983*6、设a,b,c,d,m,n都是正实数,P?ab?cd,Q?ma?nc?有关. ◆答案:B
★解析:由柯西不等式,P?Q.选B.
1983*二、(本题满分16分)函数f(x)在?0,1?上有定义,f(0)?f(1).如果对于任意不同的
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x1,x2??0,1?,都有f(x1)?f(x2)?x1?x2.求证:f(x1)?f(x2)?★证明:不妨取0?x1?x2?1,若x1?x2?1. 211,则必有f(x1)?f(x2)?x1?x2?. 221111若x1?x2?,则x2?x1?,于是1??x2?x1??,即1?x2?x1?0?.
2222而f(x1)?f(x2)?f(x1)?f(0)??f(x2)?f(1)??f(x1)?f(0)?f(x2)?f(1)
?x1?0?1?x2?1?x2?x1?0?
1.故证. 21982*4、由方程x?1?y?1?1确定的曲线所围成的图形的面积是( )
A.1 B.2 C.? D.4 ◆答案:B
★解析:此曲线的图形是一个正方形,顶点为?0,1?,?1,0?,?2,1?,?1,2?;其面积为2.选B.
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