江苏省南通市海安市2027届高三上学期期初测试数学试卷（含答案）

高三年级一项是符合题目要求的.1i为虚数单位）的实部为13232．设集合A0,a，B2,3，若AB，则a的取值范围是3．已知a，b，c是实数，则“ab”是“acbc”的A．充分不必要条件B．必要不充分条件C．充分必要条件D．既不充分也不必要条件4．已知函数fx2x22x3,xa（a0，a1）在0,上单调递增，则a的取值范围为logax,xa5．已知函数fxex1，gxex2，若直线l是两个曲线的公切线，则l的方程为A．exy0B．exye0C．xy0D．xy106．关于x的方程xaxaa（a0）的实数解的个数为A．0B．1C．2D．与a的取值有关7．已知ae0.21，bln1.2，ctan0.2，则8．已知函数fxexex2sinx，不等式fax2exf2lnxx0对任意的x0,恒成立，则a的最大值为9．已知m0，n0，关于x的不等式2mtx2ntx10的解集为1,，则C．的最小值为4D．的最小值为10．设fx，gx都是定义在R上的奇函数，且fx为单调函数，f11．若对任意实数x均有fgxxa（a为常数gfx2gfx2x2，则A．g20B．f33C．fxx为周期函数D．f4k2n22nk111．若函数fxaxbx在0,上存在减区间，则a和b的取值可以为13．已知函数fx（a，bN*）的极小值点为2，则fx的极大值点为.14．已知函数fx3x23，若mn，且fmfn，则m的取值范围为，mn的取值范围为.记Sn为等差数列an的前n项和．已知a13，a3，a4成等差数列，且S10145.（1）求数列an的通项公式；（2）数列bn满足：bnbn1an，且b11a1，求数列bn的前n项和Tn.已知集合Axx2x20，集合Bx2x22k5x5k0，kR.（1）求集合A，B；（2）记MAB，且集合M中有且仅有一个整数，求实数k的取值范围.sinCb2a2c2在△ABC中，角A，BsinCb2a2c2（1）求角B的大小；（2）若c8，cosC，求边a的长；（3）求sin2Asin2C的取值范围.1817分）已知直线y2x与椭圆E：x22y21交于A，B两点（点A在第一象限点P4t,t在椭圆E的内部，射线AP，BP与椭圆E的另一交点分别为C，D.（1）求点A坐标；（2）求证：直线CD的斜率为定值；（3）求四边形ABCD面积的最大值.1917分）设fx是定义在0,的可导函数，且不恒为0，记gnx（nN*若对定义域内的每一个x，总有gnx0，则称fx为“n阶负函数”；若对定义域内的每一个x，总有gnx0，则称fx为“n阶不减函数”（gnx为函数gnx的导函数）.（1）判断函数fxx2x1（x0）是否既是“3阶负函数”又是“3阶不减函数”，并说明理由；（2）若fxx（x0）既是“1阶负函数”，又是“1阶不减函数”，求实数a的取值范围；（3）对任给的“2阶不减函数”fx，如果存在常数c，使得fxc恒成立，试判断fx是否为“2阶负函数”？并说明理由.数学答案一项是符合题目要求的.9．ACD10．BC解：设an的公差为d，因为a13，a3，a4成等差数列，所以2a3a13a4，即2a12da13a13d，所以d3.······················································································································2分.······················································································································4分（2）由（1）知bnbn13n2，且b12，24103n52n1416解1）因为x2x20，所以x1或x2，即Axx1或x2.···························2分又2x22k5x5k2x5xk0，·······································································4分当k时，Bxkx；当k时，B；当k时，Bxxk.····················································································7分当k时，Bxkx，此时MB，当k2时，Bx2xk，仅有整数2M，2故2k3，即3k2，解1）在△ABC中，由正、余弦定理得，，········································2分因为sinC0，所以sinBcosC2sinAcosBsinCcosB，所以2sinAcosBsinBcosCsinCcosBsinBCsinA，因为sinA0，所以cosB， 22（2）因为cosC，所以sinC， 22sinAsinBCsinCsinCcoscosCsin215，··························································································7分在△ABC中，ac，所以a1410，···························································9分sinAsinC277（3）Tsin2Asin2C11cos2Acos2A1cos2Asin2A1cos2A·······························································12分因为0A，所以02A，故2A，因此1cos2A，1817分）y2x1解1）由x22y,21，y2x1（2）设Cx1,y1则x14t4t，y1tt，所以4t4t22tt21，····································································5分即18t21236t218t210，故18t212210，又因为P在椭圆内部，所以10，所以18t2118t210.····························································································7分同理，设得18t2118t210，相减得18t210，因为18t210，所以，则AB//CD，所以直线CD的斜率为2.·······························································9分（3）设直线CD的方程为y2xm，根据对称性不妨设m0，联立椭圆方程得，9x28mx2m210，290m.····················································11分2由64m2362m21368m20290m.····················································11分2 368m22 368m229所以 m m m m d且,AB，点A到直线CD的距离且,535所以fmm392m2，所以所以fmm392m2，所以22m392m2922m392m294m2fm92m2令fm0，即34m29，解得m2，当0m2时，fm0，当m2时，fm0，1917分）解1）因为fxx2x1（x0所以g3xx23x1x20.且g3x0，所以既是“3阶负函数”又是“3阶不减函数”.······································································4分（2）依题意，g1x1在0,上单调递增，故g1x0恒成立，得ax2，因为x0，所以a0.而当a0时，g1x10显然在0,恒成立，（3）①先证fx0：若不存在正实数x0，使得g2x00，则g2x0恒成立.假