江苏省泰兴中学高三期末模拟试卷.doc

江苏省泰兴中学高三期末模拟试卷（1）一、填空题（本大题共14小题,每小题5分,计70分）1已知集合，若，则实数的值为_2 _.2已知为等差数列，则_ 15 _.3过点P（1，2）且与曲线y=3x24x+2在点M（1，1）处的切线平行的直线方程是 2xy+4=0 4 设为不重合的两条直线，为不重合的两个平面，给出下列命题：若，则； 若，则；若，则； 若，则；上面命题中，所有真命题的序号为_ _.5已知点和在直线的两侧，则的取值范围是 6已知集合，设函数的值域为，若，则实数的取值范围是_.7. 一个正三棱锥的四个顶点都在半径为1的球面上，其中底面的三个顶点在该球的一个大圆上，则该正三棱锥的体积是 . 8已知正项等比数列满足，若存在两项使得，则的最小值为_.9已知，sin()= sin则cos= .10在等式的括号中，填写一个锐角，使得等式成立，这个锐角是_10 _.11已知二次函数的图像为开口向下的抛物线，且对任意都有若向量，则满足不等式的的取值范围为 12已知椭圆的短轴长为，那么直线截圆所得的弦长等于8/5。ks*5*u13已知抛物线上的两点A、B的横坐标恰是方程（是实常数）的两个实根，则直线的方程是 .14我们知道，如果定义在某区间上的函数满足对该区间上的任意两个数、，总有不等式成立，则称函数为该区间上的向上凸函数（简称上凸）. 类比上述定义，对于数列，如果对任意正整数，总有不等式：成立，则称数列为向上凸数列（简称上凸数列）. 现有数列满足如下两个条件：（1）数列为上凸数列，且；（2）对正整数（），都有，其中. 则数列中的第五项的取值范围为 .解1:由am+an=am+1+an-1 m+2=n，可得：a1+a3a2+a2 a1+a4a2+a3 . . . . . a1+a10a2+a9 把以上8个不等式相加：8a1+a104 a2+a4a3+a3 a2+a5a3+a4 . a2+a10a3+a9 把以上7个不等式相加：7a2+a107 a3+a5a4+a4 a3+a6a4+a5 . a3+a10a4+a9 把以上6个不等式相加：6a1+a1010 a4+a6a5+a5 a4+a7a5+a6 . a4+a10a5+a9 把以上5个不等式相加：5a1+a1013 再用条件2求上限 b5=25-30+10=5 a5-5=20 a5=a513 解2:假设an=An2+Bn+C(A0)由a1=1,a10=28得到B=11-3A,C=10A-2.所以an=An2+(11-3A)n+10A-2 a5=13-20A,|a5-b5|20得到-3/5A0,所以1313-20A25. a5的取值范围为13,25解3：在线段（1，a1）（10，a10）上取横坐标为5的点有a5=13，又|a5-b5|20得到a5的取值范围为13,2515（本题满分14分）在中， 所对边分别为.已知,且.（）求大小；（）若求的面积S的大小.15. 解：（I），0. 2分4分 6分 8分（II） 中，. 10分 12分的面积 14分16（本题满分14分）如图，在正方体ABCDA1B1C1D1中，M、N、G分别是A1A，D1C，AD的中点求证：（）MN/平面ABCD；（）MN平面B1BG16.证明：（1）取CD的中点记为E，连NE，AE由N，E分别为CD1与CD的中点可得NED1D且NE=D1D， 2分又AMD1D且AM=D1D4分所以AMEN且AM=EN，即四边形AMNE为平行四边形所以MNAE， 6分又AE面ABCD,所以MN面ABCD8分（）由AGDE，DAAB可得与全等10分所以， 11分又，所以所以， 12分又,所以， 13分又MNAE，所以MN平面B1BG 14分7、某房地产开发公司计划在一楼区内建造一个长方形公园ABCD，公园由长方形的休闲区A1B1C1D1和环公园人行道（阴影部分）组成已知休闲区A1B1C1D1的面积为4000平方米，人行道的宽分别为4米和10米（如图）.（1）若设休闲区的长和宽的比，求公园ABCD所占面积关于的函数 的解析式；（2）要使公园所占面积最小，休闲区A1B1C1D1的长和宽该如何设计？17、解：（1）设休闲区的宽为米，则其长为米， ， 8分 （2），当且仅当时，公园所占面积最小， 14分 此时，即休闲区的长为米，宽为米。16分18. 已知半椭圆和半圆组成曲线，其中；如图，半椭圆内切于矩形，且交轴于点，点是半圆上异于A、B的任意一点，当点位于点时，的面积最大。 （1）求曲线的方程； （2）连、交分别于点，求证：为定值.18 解：（1）已知点在半圆上，所以，又，所以， （2分）当半圆在点处的切线与直线平行时，点到直线的距离最大，此时的面积取得最大值，故半圆在点处的切线与直线平行，所以，又，所以，又，所以，（4分）所以曲线的方程为或。 （6分）（2）点，点，设，则有直线的方程为，令，得， 所以； （9分）直线的方程为，令，得， 所以； （11分）则，又由，得，代入上式得，所以为定值。 （15分）19、设函数，其中（）当时，讨论函数的单调性；（）若函数仅在处有极值，求的取值范围；（）若对于任意的，不等式在上恒成立，求的取值范围19、解：（）当时，令，解得，当变化时，的变化情况如下表：极小值极大值极小值所以在，内是增函数，在，内是减函数（）解：，显然不是方程的根为使仅在处有极值，必须恒成立，即有解此不等式，得这时，是唯一极值因此满足条件的的取值范围是（）解：由条件可知，从而恒成立当时，；当时，因此函数在上的最大值是与两者中的较大者为使对任意的，不等式在上恒成立，当且仅当 即在上恒成立所以，因此满足条件的的取值范围是20(本小题满分16分)已知数列、中,对任何正整数都有：(1)若数列是首项和公差都是1的等差数列,求证：数列是等比数列；(2)若数列是等比数列,数列是否是等差数列,若是请求出通项公式,若不是请说明理由；(3)若数列是等差数列,数列是等比数列,求证：20解：（1）依题意数列的通项公式是，故等式即为，两式相减可得 -3分得，数列是首项为1，公比为2的等比数列 -4分（2）设等比数列的首项为，公比为，则，从而有：，又，故 -6分，要使是与无关的常数，必需， -8分即当等比数列的公比时，数列是等差数列，其通项公式是；当等比数列的公比不是2时，数列不是等差数列 -9分（3）由（2）知， - -10分显然时,当时 -14分 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