数学+答案【江苏名校】江苏南通市海安高级中学2026-2027学年2027届高三年级上学期开学考试(8月底)

高三年级21i为虚数单位）的实部为(2,3)，若AB=g，则a的取值范围是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f(xa（a>0，a≠1）在(0,+∞)上单调递增，则a的取值范围为5．已知函数f(x)=ex_1，gex2，若直线l是两个曲线的公切线，则l的方程为A．ex_y=0B．ex_y_e=0C．x_y=06．关于x的方程(x_a)x_a=a（a≠0）的实数解的个数为A．0B．1C．28．已知函数f(x)=ex_e__x_2sinx，不等式f(a_x2ex)+f(2lnx+x)≤0对任意的x∈(0,+∞)恒成立，则a的最大值为18C的最小值为4D．的最小值为10．设f(x)，g(x)都是定义在R上的奇函数，且f(x)为单调函数，f(1)>1．若对任意实数x均有C．f(x)-x为周期函数D．>2n2+2n11．若函数f(x)=ax+bx在(0,+∞)上存在减区间，则a和b的取值可以为1.813．已知函数fa，b∈N*）的极小值点为2，则f(x)的极大值点为．14．已知函数f(x)=3-x2-3，若m<n，且f(m)=f(n)，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a1+3，a3，a4成等差数列，且S10=145．（1）求数列{an}的通项公式；（2）数列{bn}满足：bn+bn+1=an，且b1-1=a1，求数列{bn}的前n项和Tn．（1）求集合A，B；（2）记M=AB，且集合M中有且仅有一个整数，求实数k的取值范围．在△ABC中，角A，B，C所对的边分别为a，b，c．已知（1）求角B的大小；（3）求sin2A+sin2C的取值范围．1817分）已知直线y=2x与椭圆E：x2+2y2=1交于A，B两点（点A在第一象限点P(-4t,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f(x)是定义在(0,+∞)的可导函数，且不恒为0，记gn．若对定义域内的每一个x，总有gn(x)<0，则称f(x)为“n阶负函数”；若对定义域内的每一个x，总有gn(x),≥0，则称f(x)（1）判断函数f(x)=-x2+x-1（x>0）是否既是“3阶负函数”又是“3阶不减函数”，并说明理由；（2）若f既是“1阶负函数”，又是“1阶不减函（3）对任给的“2阶不减函数”f(x)，如果存在常数c，使得f(x)<c恒成立，试判断f(x)是否为“2阶负函数”？并说明理由．数学答案9．ACD10．BC(-3,3)解：设{an}的公差为d，因为a1+3所以d=3．······················································································································2分又S10=1．······················································································································4分2)+xx<1或x>2}．···························2分当k<时，B····················································································7分当k时，B此时M=B，解1）在△ABC中，由正、余弦定理得，·······································2分因为sinC≠0，所以sinBcosC=2sinAcosB一sinCcosB，所以2sinAcosB=sinBcosC+sinCcosB因为sinA≠0，所以cosBsinA=sin=sinsinCcoscosCsin··························································································7分 cos···························································12分2π4π故因此-1≤cos1817分）解1）由又因为点A在第一象限，所以····································································5分18t2+1+36t2λ+18t2-1=0，故18t2(λ+1)2+λ2-1=0，又因为P在椭圆内部，所以λ+1>0，所以18t2+1λ+18t2-1=0．····························································································7分18t2+1所以λ=μ,则AB//CD，所以直线CD的斜率为2．·······························································9分（3）设直线CD的方程为y=2x+m，根据对称性不妨设m>0，联立椭圆方程得，9x2+8mx+2m2-1=0，22-36(2m2-1)=36-8m2>0，得0<m2<9．····················································11分2且AB点A到直线CD的距离d令f=m所以f,+m=4m2-9，解得m21917分）解1）因为f(x)=-x2+x-1（x>0所以g3且所以既是“3阶负函数”又是“3阶不减函数”．······································································4分故g1≥0恒成立，得a≤x2，而当a≤0时，g显然在(0,+∞)恒成立，（3）①先证f(x)≤0：若不存在正实数x0，使得g2(x0)>0，则g2(x)≤0恒成立．假设