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陕西省西安中学2020届高三上学期期中考试数学(理)试题

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西安中学高2020届高三期中考试 数学(理科)试题 第Ⅰ卷(选择题 共60分)‎ 一、选择题:本大题共12小题,每小题5分,在每小题给出的四个选项中,只有一项是符合题目要求的.‎ ‎1.已知集合,,则(  ) ‎ A. B. C. D.‎ ‎2.命题“对任意都有”的否定是(  ) ‎ A.对任意,都有 B.不存在,使得 C.存在,使得 D.存在,使得 ‎3.在等差数列中,a2=4,a3=6,则a10=(  )‎ A.20 B.22 C.18 D.16‎ ‎4.下列函数中,既是偶函数又有零点的是(  ) ‎ A. B. C. D. ‎ ‎5.若,则(  ) ‎ A. B. C. D.2‎ ‎6.函数f(x)=2x+3x的零点所在的一个区间是(  )‎ A. B. C. D.‎ ‎7.已知函数,则=(  ) ‎ A. B.2 C. D.4‎ ‎8.若实数满足约束条件,则的最小值是(  ) ‎ A.2 B.3 C.4 D.5‎ ‎9.已知,则实数a的取值范围是(  ) ‎ A. B. C. D.‎ ‎10.在直角中,,,,点是外接圆上任意一点,则的最大值为(  ) ‎ A.6 B.8 C.10 D. 12‎ ‎11.已知定义在上的函数在上有和两个零点,且函数与函数 都是偶函数,则在上的零点至少有(  ) 个 A.404 B.406 C.808 D.812‎ ‎12.定义在R上的函数的导函数为,若对任意实数,都有,且为奇函数,则不等式的解集为(  )‎ A. B. C. D.‎ 第Ⅱ卷(非选择题 共90分)‎ 二、填空题:本大题共4小题,每小题5分.‎ ‎13.已知,,若与平行,则 .‎ ‎14.若不等式恒成立,则实数的取值范围为 .‎ ‎15.函数的递减区间为 .‎ ‎16.已知,函数在区间上的最大值是5,则实数a的取值范围是 .‎ 三、解答题:共70分.解答应写出文字说明、证明过程或演算步骤.第17~21题为必考题,每个试题考生都必须作答.第22,23题为选考题,考生根据要求作答.‎ ‎(一)必考题:共60分.‎ ‎17.(本小题满分12分)‎ 已知向量,,函数的最大值为6.‎ ‎(Ⅰ)求A;‎ ‎(Ⅱ)将函数y=f(x)的图像向左平移 个单位,再将所得图像上各点的横坐标缩短为原来的倍,纵坐标不变,得到函数y=g(x)的图像,求g(x)在上的值域.‎ ‎18.(本小题满分12分)‎ 在角中,角A,B,C的对边分别是,若.‎ ‎(Ⅰ)求角A的大小; ‎ ‎(Ⅱ)若的面积为,,求的周长.‎ ‎19.(本小题满分12分)‎ 设函数,其中.‎ ‎(Ⅰ)当时,求的极值点;‎ ‎(Ⅱ)若在上为单调函数,求的取值范围.‎ ‎20.(本小题满分12分)‎ 以椭圆的中心为圆心,以为半径的圆称为该椭圆的“伴随”. 已知椭圆的离心率为,且过点.‎ ‎(Ⅰ)求椭圆及其“伴随”的方程; ‎ ‎(Ⅱ)过点作“伴随”的切线交椭圆于,两点,记为坐标原点)的面积为,求的最大值.‎ ‎21.(本小题满分12分)‎ 已知函数,曲线在点处的切线方程为.‎ ‎(Ⅰ)求的解析式;‎ ‎(Ⅱ)判断方程在内的解的个数,并加以证明.‎ ‎(二)选考题:共10分.请考生在第22,23题中任选一题作答.如果多做,那么按所做的第一题计分.‎ ‎22.(本小题满分10分)[选修4—4:坐标系与参数方程]‎ 在直角坐标系中,曲线,为参数,,其中 ‎,在以为极点,轴正半轴为极轴的极坐标系中,曲线:,曲线:.‎ ‎(Ⅰ)求与交点的直角坐标;‎ ‎(Ⅱ)若与相交于点,与相交于点,求的最大值.‎ ‎23.(本小题满分10分)[选修4—5:不等式选讲]‎ 已知,,.求证: ‎ ‎(Ⅰ);‎ ‎(Ⅱ). ‎ 西安中学高2020届高三期中考试 数 学(理科)参考答案 一、选择题:‎ 题号 ‎1‎ ‎2‎ ‎3‎ ‎4‎ ‎5‎ ‎6‎ ‎7‎ ‎8‎ ‎9‎ ‎10‎ ‎11‎ ‎12‎ 答案 A D A D A B C B A D C B 二、填空题:‎ ‎13. 14. 15. 16.‎ 三、解答题:‎ ‎17.解:(Ⅰ)=Asin xcos x+cos 2x ‎=A ‎=Asin.··········································(4分)‎ 因为A>0,由题意知A=6.················································(5分)‎ ‎(Ⅱ)由(Ⅰ)f(x)=6sin.‎ 将函数y=f(x)的图像向左平移个单位后得到y=6sin=6sin的图像;··········································································(7分)‎ 再将得到图像上各点横坐标缩短为原来的倍,纵坐标不变,得到y=6sin的图像.‎ 因此g(x)=6sin.······························································(9分)‎ 因为x∈,所以4x+∈,‎ 故g(x)在上的值域为[-3,6] .·········································(12分)‎ ‎18.解:(Ⅰ)由正弦定理得:,························(2分)‎ ‎,‎ ‎,··········································································(4分)‎ 是的内角,‎ ‎ .··············································································(6分)(Ⅱ)的面积为,‎ ‎,‎ 由(Ⅰ)知,‎ ‎,··············································································(8分)由余弦定理得:,·····(10分)‎ ‎,‎ 得:,‎ 的周长为.·······························································(12分)‎ ‎19.解:对求导得 ·····························(1分)‎ ‎ (Ⅰ)若,由 ‎ 令,因为,则, ·······(2‎ 分)‎ 所以随x变化而变化的情况为:‎ ‎+‎ ‎0‎ ‎-‎ ‎0‎ ‎+‎ ‎↗‎ 极大值 ‎↘‎ 极小值 ‎↗‎ 所以,是极大值点,是极小值点.····························(5分)‎ ‎(注:未注明极大、极小值扣1分)‎ ‎ (Ⅱ)若为上的单调函数,又,所以当时,即在上恒成立. ···················(6分)‎ ‎ (1)当时,,符合题意;···················(8分)‎ ‎ (2)当时,抛物线开口向上,‎ ‎ 则的充要条件是,‎ ‎ 即,所以. ‎ ‎ 综合(1)(2)知的取值范围是.···································(12分)‎ ‎20.解:(Ⅰ)椭圆的离心率为, 则, ‎ ‎ 设椭圆的方程为 ‎ ‎ ∵椭圆过点,∴,‎ ‎ ∴, ‎ ‎ ∴椭圆的标准方程为,················································( 3‎ 分)‎ ‎ 椭圆的“伴随”方程为.···············································(4分)‎ ‎(Ⅱ)由题意知,. ······················································(5分)‎ 易知切线的斜率存在,设切线的方程为 由得 ‎ 设, 两点的坐标分别为, , 则 ‎, .··················································(7分)‎ ‎ 又由与圆相切, 所以,.·············(8分)‎ ‎ 所以 ‎ ‎ ‎······························································(10分)‎ ‎ .‎ ‎ 令,则,代入上式得:‎ ‎ (当且仅当,即时等号成立)‎ 所以的最大值为1.···············································(12分)‎ ‎21.解:(Ⅰ)直线的斜率为,过点,‎ ‎ ,则,即,‎ ‎ ,所以.··············································(4分)‎ ‎ (Ⅱ)方程在上有3个解.‎ ‎ 证明:令,则,‎ ‎ 又,,‎ ‎ 所以在上至少有一个零点,‎ ‎ 又在上单调递减,故在上只有一个零点.····················(6分)‎ ‎ 当时,,故,‎ ‎ 所以函数在上无零点.················································(8分)‎ ‎ 当时,令,,‎ ‎ 所以在上单调递增,,,‎ ‎ 所以,使得在上单调递增,在上单调递减.‎ ‎ 又,,所以函数在上有2个零点.‎ ‎ 综上,方程在上有3个解.··································(12分)‎ ‎22.解:(Ⅰ)曲线的直角坐标方程为,························(1分)‎ 曲线的直角坐标方程为.·································(2分)‎ 联立,解得,或.‎ 所以与的交点的直角坐标为和.···························(4分)‎ ‎(Ⅱ)曲线的极坐标方程为,其中.····(5分)‎ ‎ 因此的极坐标为,的极坐标为,·············(7分)‎ 所以==,·································(9分)‎ 当时,取得最大值,最大值为4. ·······························(10分)‎ ‎23.证明:(Ⅰ)‎ ‎(当且仅当时等号成立)····(5分)‎ 也可以用柯西不等式直接证明. ‎ ‎(Ⅱ)‎ ‎······················································(7分)‎ ‎···················································································(9分)‎ ‎(当且仅当时等号成立)‎ 从而.···························································(10分)‎ 也可以用分析法证明.‎

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