2019-2020学年人教A版湖南省长沙市长郡中学高三第二学期(3月份)第一次段考(理科)数学试卷含解析

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湖南省长郡中学2019届高三月考试卷(一)文科数学

长郡中学2019届高三月考试卷（一）数学（文科）（考试时间：120分钟，满分150分）注意事项：1．答题前，先将自己的姓名、准考证号填写在试题卷和答题卡上，并将准考证号条形码粘贴在答题卡上的指定位置。

2．选择题的作答：每小题选出答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑，写在试题卷、草稿纸和答题卡上的非答题区域均无效。

3．非选择题的作答：用签字笔直接答在答题卡上对应的答题区域内。

写在试题卷、草稿纸和答题卡上的非答题区域均无效。

4．考试结束后，请将本试题卷和答题卡一并上交。

一、选择题：本大题共12个小题，每小题5分，共60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{|02}A x x =<<，{|1}B x x =≥，则A B =U （）A ．{|01}x x <<B ．{|1}x x ≥C ．{|12}x x ≤<D ．{|0}x x >2.复数z 满足(2)36z i i +=-（i 为虚数单位），则复数z 的虚部为（）A ．3B ．3-C ．3iD ．3i -3.已知2sin 5α=，则cos2α=（） A ．725 B ．725- C ．1725D ．1725- 4.某家具厂的原材料费支出x （单位：万元）与销售额y （单位：万元）之间有如下数据，根据表中提供的全部数据，用最小二乘法得出y 与x 的线性回归方程为$8.5y x b =+，则b 为（）A ．7.5B ．10C ．12.5D ．17.55.已知向量(2,1)a =-r ，(1,3)b =-r ，则（）A ．//a b r rB ．a b ⊥r rC ．//()a a b -r r rD ．()a a b ⊥-r r r6.执行如图所示的程序框图输出的结果是（）A ．8B ．6C ．5D ．37.已知曲线1C ：sin y x =，2C ：2sin 23y x π⎛⎫=+⎪⎝⎭，则下面结论正确的是（） A ．把1C 上各点的横坐标缩短到原来的12倍，纵坐标不变，再把得到的曲线向左平移23π个单位长度，得到曲线2CB ．把1C 上各点的横坐标缩短到原来的12倍，纵坐标不变，再把得到的曲线向左平移3π个单位长度，得到曲线2CC ．把1C 上各点的横坐标伸长到原来的2倍，纵坐标不变，再把得到的曲线向左平移23π个单位长度，得到曲线2CD ．把1C 上各点的横坐标伸长到原来的2倍，纵坐标不变，再把得到的曲线向左平移3π个单位长度，得到曲线2C8.曲线()2x f x x e =-在点(0,(0))f 处的切线方程是（）A ．210x y --=B ．10x y -+=C ．0x y -=D ．10x y --=9.平面α截球O 的球面所得圆的半径为1，球心O 到平面α2，则此球的体积为（）A ．43πB ．3πC 6πD ．46π10.已知()f x 是定义在R 上的偶函数，且在区间(,0]-∞上单调递增.若实数a 满足()(2)f a f >，则a 的取值范围是（）A ．(,2)-∞B ．2,)+∞C ．(2,2)-D ．(,2)(2,)-∞-+∞U11.已知四棱锥S ABCD -的三视图如图所示，则围成四棱锥S ABCD -的五个面中的最大面积是（）A ．3B ．6C ．8D ．1012.已知F 是抛物线C ：28y x =的焦点，M 是C 上一点，FM 的延长线交y 轴于点N .若M 为FN 的中点，则FN =（）A ．4B ．6C ．8D ．10 二、填空题：本题共4小题，每小题5分，共20分.13.已知x ，y 满足020x y x y y -≥⎧⎪+≤⎨⎪≥⎩，则2z x y =+的最大值为．14.若点(1,1)P 为圆2260x y x +-=的弦MN 的中点，则弦MN 所在直线的方程为． 15.在ABC ∆中，面积2221()4S a b c =+-，则角C 的大小为． 16.已知函数3()lg 92f x x x =+-在区间(,1)()n n n Z +∈上存在零点，则n = ．三、解答题：本大题共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，每个试题考生都必需作答.第22、23题为选考题，考生根据要求作答.（一）必考题：共60分.17.等比数列{}n a 中，已知12a =，416a =.（1）求数列{}n a 的通项公式；（2）若3a ，5a 分别为等差数列{}n b 的第3项和第5项，试求数列{}n b 的通项公式及前n 项和n S .18.已知四棱锥P ABCD -中，底面ABCD 是边长为2的正方形，2PA PD ==CD PD ⊥，E 为CD的中点.（1）求证：PD ⊥平面PAB ；（2）求三棱锥P ABE -的体积.19.某家电公司销售部门共有200名销售员，每年部门对每名销售员都有1400万元的年度销售任务.已知这200名销售员去年完成的销售额都在区间[2,22]（单位：百万元）内，现将其分成5组，第1组、第2组、第3组、第4组、第5组对应的区间分别为[2,6)，[6,10)，[10,14)，[14,18)，[18,22]，并绘制出如下的频率分布直方图.（1）求a 的值，并计算完成年度任务的人数；（2）用分层抽样的方法从这200名销售员中抽取容量为25的样本，求这5组分别应抽取的人数；（3）现从（2）中完成年度任务的销售员中随机选取2名，奖励海南三亚三日游，求获得此奖励的2名销售员在同一组的概率.20.过椭圆C ：22221(0)x y a b a b+=>>的右焦点2F 的直线交椭圆于A ，B 两点，1F 为其左焦点，已知1AF B ∆的周长为363. （1）求椭圆C 的方程；（2）设P 为椭圆C 的下顶点，椭圆C 与直线3y x m =+相交于不同的两点M 、N .当PM PN =时，求实数m 的值.21.已知函数()x xa f x e e =-. （1）当1a =时，求函数()[()'()]F x x f x f x =-的最小值；（2）若()()g x f x =在[0,1]上单调递增，求实数a 的取值范围.（二）选考题：共10分.请考生在22、23题中任选一题作答，如果多做，则按所做的第一题记分.22.选修4-4：坐标系与参数方程在平面直角坐标系xOy 中，直线l 的参数方程为12x t y a t=+⎧⎨=-⎩（其中t 为参数）.在以坐标原点为极点，x 轴的正半轴为极轴的极坐标系中，圆C 的极坐标方程为4cos ρθ=.（1）分别写出直线l 的普通方程和圆C 的直角坐标方程；（2）若直线l 与圆C 相切，求实数a 的值.23.选修4-5：不等式选讲设函数()f x x a x a =++-.（1）当1a =时，解不等式()4f x ≥；（2）若()6f x ≥在x R ∈上恒成立，求a 的取值范围.长郡中学2019届高三月考试卷（一）数学（文科）参考答案一、选择题1-5: DBCAD 6-10: ABDAC 11、12：CB二、填空题13. 4 14. 210x y --= 15. 45︒ 16. 5三、解答题17.【解析】（1）设{}n a 的公比为q 由已知得3162q =，解得2q =，所以2n n a =.（2）由（1）得38a =，532a =，则38b =，532b =，设{}n b 的公差为d ，则有1128432b d b d +=⎧⎨+=⎩，解得11612b d =-⎧⎨=⎩，从而1612(1)1228n b n n =-+-=-.所以数列{}n b 的前n 项和2(161228)6222n n n S n n -+-==-. 18.【解析】（1）∵底面ABCD 是正方形，∴//AB CD ，又CD PD ⊥，∴AB PD ⊥，∵PA PD ==2AD =，∴222PA PD AD +=，∴PD PA ⊥，又PA AB A =I ，∴PD ⊥平面PAB .（2）∵AB AD ⊥，AB PD ⊥且AD PD D =I ，∴AB ⊥平面PAD ，又AB ⊂平面ABCD ，∴平面PAD ⊥平面ABCD ，过P 作PO AD ⊥于O ，则PO ⊥平面ABCD ，∴PO 为三棱锥P ABE -的高，∴13P ABE ABE V S PO -∆=⋅112122323=⨯⨯⨯⨯=. 19.【解析】（1）∵(0.020.080.092)41a +++⨯=，∴0.03a =，完成年度任务的人数为2420048a ⨯⨯=.（2）第1组应抽取的人数为0.024252⨯⨯=，第2组应抽取的人数为0.084258⨯⨯=，第3组应抽取的人数为0.094259⨯⨯=，第4组应抽取的人数为0.034253⨯⨯=，第5组应抽取的人数为0.034253⨯⨯=.（3）在（2）中完成年度任务的销售员中，第4组有3人，记这3人分别为1A ，2A ，3A ；第5组有3人，记这3人分别为1B ，2B ，3B ；从这6人中随机选取2名，所有的基本事件为12A A ，13A A ，11A B ，12A B ，13A B ，23A A ，21A B ，22A B ，23A B ，31A B ，32A B ，33A B ，12B B ，13B B ，23B B ，共有15个基本事件.获得此奖励的2名销售员在同一组的基本事件有6个，故所求概率为62155=.20.【解析】（1）由椭圆定义知，4a =，a =3c e a ===得c =1b =，所以椭圆C 的方程为2213x y +=.（2）由方程组2213y mx y ⎧=+⎪⎪⎨⎪+=⎪⎩2223(1)0x m ⇒++-=，设11(,)M x y ，22(,)N x y ，MN 的中点为00(,)E x y，则12x x +=.∴12022x x x m +==-，02m y =，∴,2m E m ⎛⎫⎪ ⎪⎝⎭，由PM PN =得PE MN ⊥，又(0,1)P -，∴1PE k =-，∴1m =.满足221224(1)0m m ∆=-->.综上1m =.21.【解析】（1）2()x xF x e =-，2(1)'()0x x F x e -==，令'()0F x =，得1x =，所以当1x <时，'()0F x <，()F x 单调递减，当1x >时，'()0F x >，()F x 单调递增，所以当1x =时，()F x 取得最小值为2e -.（2）当0a ≤时，()0x x af x e e =->，()()g x f x =，若在[0,1]上单调递增，则'()0f x ≥恒成立，即：2max []x a e ≥-，1a ≥-，10a -≤≤；当0a >时，'()0x x af x e e =+>，()x x af x e e =-在[0,1]上是单调递增的，又()()g x f x =在[0,1]上单调递增，所以()0f x ≥在[0,1]上恒成立.2min []x a e ≤，01a <≤.综上：11a -≤≤.22.【解析】（1）直线l 的直角坐标系方程是220x y a +--=，圆C 的直角坐标方程是22(2)4x y -+=.（2）由（1）知圆心为(2,0)C ，半径2r =，设圆心到直线的距离为d ，因为直线与圆相切，所以2d ===，解得2a =±23.【解析】（1）当1a =时，不等式()4114f x x x ≥⇔++-≥，当1x >时，()24f x x =≥，解得2x ≥；当11x -≤≤时，()24f x =≥，无解；当1x <-时，()24f x x =-≥，解得2x ≤-，综上所述，不等式的解集为(,2][2,)-∞-+∞U .（2）()f x x a x a =++-()()2x a x a a ≥+--=， ∴26a ≥，解得3a ≥或3a ≤-，即a 的取值范围是(,3][3,)-∞-+∞U .21。

湖南省长郡中学2019届高三月考试卷(一)文科数学

2．选择题的作答：每小题选出答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑，写在试题卷、草稿纸和答题卡上的非答题区域均无效。

3．非选择题的作答：用签字笔直接答在答题卡上对应的答题区域内。

写在试题卷、草稿纸和答题卡上的非答题区域均无效。

4．考试结束后，请将本试题卷和答题卡一并上交。

一、选择题：本大题共12个小题，每小题5分，共60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{|02}A x x =<<，{|1}B x x =≥，则AB =（）A ．{|01}x x <<B ．{|1}x x ≥C ．{|12}x x ≤<D ．{|0}x x > 2.复数z 满足(2)36z i i +=-（i 为虚数单位），则复数z 的虚部为（） A ．3 B ．3- C ．3i D ．3i -3.已知2sin 5α=，则cos2α=（） A ．725 B ．725- C ．1725D ．1725-4.某家具厂的原材料费支出x （单位：万元）与销售额y （单位：万元）之间有如下数据，根据表中提供的全部数据，用最小二乘法得出y 与x 的线性回归方程为8.5y x b =+，则b 为（）A ．7.5B ．10C ．12.5D ．17.5 5.已知向量(2,1)a =-，(1,3)b =-，则（）A ．//a bB ．a b ⊥C ．//()a a b -D ．()a a b ⊥- 6.执行如图所示的程序框图输出的结果是（）A ．8B ．6C ．5D ．3 7.已知曲线1C ：sin y x =，2C ：2sin 23y x π⎛⎫=+ ⎪⎝⎭，则下面结论正确的是（） A ．把1C 上各点的横坐标缩短到原来的12倍，纵坐标不变，再把得到的曲线向左平移23π个单位长度，得到曲线2CB ．把1C 上各点的横坐标缩短到原来的12倍，纵坐标不变，再把得到的曲线向左平移3π个单位长度，得到曲线2CC ．把1C 上各点的横坐标伸长到原来的2倍，纵坐标不变，再把得到的曲线向左平移23π个单位长度，得到曲线2CD ．把1C 上各点的横坐标伸长到原来的2倍，纵坐标不变，再把得到的曲线向左平移3π个单位长度，得到曲线2C8.曲线()2xf x x e =-在点(0,(0))f 处的切线方程是（） A ．210x y --= B ．10x y -+= C ．0x y -= D ．10x y --=9.平面α截球O 的球面所得圆的半径为1，球心O 到平面α，则此球的体积为（）A ．B ．CD ．10.已知()f x 是定义在R 上的偶函数，且在区间(,0]-∞上单调递增.若实数a 满足()(f a f >，则a 的取值范围是（）A ．(,-∞B ．)+∞C．( D．(,(2,)-∞+∞11.已知四棱锥S ABCD -的三视图如图所示，则围成四棱锥SABCD -的五个面中的最大面积是（）A ．3B ．6C ．8D ．1012.已知F 是抛物线C ：28y x =的焦点，M 是C 上一点，FM 的延长线交y 轴于点N .若M 为FN 的中点，则FN =（）A ．4B ．6C ．8D ．10二、填空题：本题共4小题，每小题5分，共20分.13.已知x ，y 满足020x y x y y -≥⎧⎪+≤⎨⎪≥⎩，则2z x y =+的最大值为．14.若点(1,1)P 为圆2260x y x +-=的弦MN 的中点，则弦MN 所在直线的方程为．15.在ABC ∆中，面积2221()4S a b c =+-，则角C 的大小为． 16.已知函数3()lg 92f x x x =+-在区间(,1)()n n n Z +∈上存在零点，则n = ．三、解答题：本大题共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，每个试题考生都必需作答.第22、23题为选考题，考生根据要求作答. （一）必考题：共60分.17.等比数列{}n a 中，已知12a =，416a =. （1）求数列{}n a 的通项公式；（2）若3a ，5a 分别为等差数列{}n b 的第3项和第5项，试求数列{}n b 的通项公式及前n 项和n S . 18.已知四棱锥P ABCD -中，底面ABCD是边长为2的正方形，PA PD ==CD PD ⊥，E 为CD的中点.（1）求证：PD ⊥平面PAB ；（2）求三棱锥P ABE -的体积.19.某家电公司销售部门共有200名销售员，每年部门对每名销售员都有1400万元的年度销售任务.已知这200名销售员去年完成的销售额都在区间[2,22]（单位：百万元）内，现将其分成5组，第1组、第2组、第3组、第4组、第5组对应的区间分别为[2,6)，[6,10)，[10,14)，[14,18)，[18,22]，并绘制出如下的频率分布直方图.（1）求a 的值，并计算完成年度任务的人数；（2）用分层抽样的方法从这200名销售员中抽取容量为25的样本，求这5组分别应抽取的人数；（3）现从（2）中完成年度任务的销售员中随机选取2名，奖励海南三亚三日游，求获得此奖励的2名销售员在同一组的概率.20.过椭圆C ：22221(0)x y a b a b+=>>的右焦点2F 的直线交椭圆于A ，B 两点，1F 为其左焦点，已知1AF B ∆的周长为3. （1）求椭圆C 的方程；（2）设P 为椭圆C 的下顶点，椭圆C 与直线y x m =+相交于不同的两点M 、N .当PM PN =时，求实数m 的值. 21.已知函数()xxa f x e e =-. （1）当1a =时，求函数()[()'()]F x x f x f x =-的最小值；（2）若()()g x f x =在[0,1]上单调递增，求实数a 的取值范围.（二）选考题：共10分.请考生在22、23题中任选一题作答，如果多做，则按所做的第一题记分.22.选修4-4：坐标系与参数方程在平面直角坐标系xOy 中，直线l 的参数方程为12x ty a t=+⎧⎨=-⎩（其中t 为参数）.在以坐标原点为极点，x 轴的正半轴为极轴的极坐标系中，圆C 的极坐标方程为4cos ρθ=. （1）分别写出直线l 的普通方程和圆C 的直角坐标方程；（2）若直线l 与圆C 相切，求实数a 的值. 23.选修4-5：不等式选讲设函数()f x x a x a =++-. （1）当1a =时，解不等式()4f x ≥；（2）若()6f x ≥在x R ∈上恒成立，求a 的取值范围.长郡中学2019届高三月考试卷（一）数学（文科）参考答案一、选择题1-5: DBCAD 6-10: ABDAC 11、12：CB二、填空题13. 4 14. 210x y --= 15. 45︒ 16. 5三、解答题17.【解析】（1）设{}n a 的公比为q 由已知得3162q =，解得2q =，所以2n n a =.（2）由（1）得38a =，532a =，则38b =，532b =，设{}n b 的公差为d ，则有1128432b d b d +=⎧⎨+=⎩，解得11612b d =-⎧⎨=⎩，从而1612(1)1228n b n n =-+-=-. 所以数列{}n b 的前n 项和2(161228)6222n n n S n n -+-==-.18.【解析】（1）∵底面ABCD 是正方形，∴//AB CD ，又CD PD ⊥， ∴AB PD ⊥，∵PA PD ==2AD =，∴222PA PD AD +=，∴PD PA ⊥，又PA AB A =，∴PD ⊥平面PAB .（2）∵AB AD ⊥，AB PD ⊥且ADPD D =，∴AB ⊥平面PAD ，又AB ⊂平面ABCD ，∴平面PAD ⊥平面ABCD ，过P 作PO AD ⊥于O ，则PO ⊥平面ABCD ， ∴PO 为三棱锥P ABE -的高，∴13P ABE ABE V S PO -∆=⋅112122323=⨯⨯⨯⨯=. 19.【解析】（1）∵(0.020.080.092)41a +++⨯=，∴0.03a =，完成年度任务的人数为2420048a ⨯⨯=. （2）第1组应抽取的人数为0.024252⨯⨯=，第2组应抽取的人数为0.084258⨯⨯=，第3组应抽取的人数为0.094259⨯⨯=，第4组应抽取的人数为0.034253⨯⨯=，第5组应抽取的人数为0.034253⨯⨯=.（3）在（2）中完成年度任务的销售员中，第4组有3人，记这3人分别为1A ，2A ，3A ；第5组有3人，记这3人分别为1B ，2B ，3B ；从这6人中随机选取2名，所有的基本事件为12A A ，13A A ，11A B ，12A B ，13A B ，23A A ，21A B ，22A B ，23A B ，31A B ，32A B ，33A B ，12B B ，13B B ，23B B ，共有15个基本事件.获得此奖励的2名销售员在同一组的基本事件有6个，故所求概率为62155=. 20.【解析】（1）由椭圆定义知，4a =，a =3c e a ===得c =1b =，所以椭圆C 的方程为2213x y +=. （2）由方程组2213y m x y ⎧=+⎪⎪⎨⎪+=⎪⎩2223(1)0x m ⇒++-=，设11(,)M x y ，22(,)N x y ，MN 的中点为00(,)E x y，则12x x +=.∴12022x x x m +==-，02m y =，∴,22m E m ⎛⎫- ⎪ ⎪⎝⎭，由PM PN =得PE MN ⊥，又(0,1)P -，∴13PE k ⨯=-，∴1m =. 满足221224(1)0m m ∆=-->.综上1m =. 21.【解析】（1）2()x x F x e =-，2(1)'()0xx F x e -==，令'()0F x =，得1x =，所以当1x <时，'()0F x <，()F x 单调递减，当1x >时，'()0F x >，()F x 单调递增，所以当1x =时，()F x 取得最小值为2e-. （2）当0a ≤时，()0xxaf x e e =->，()()g x f x =，若在[0,1]上单调递增，则'()0f x ≥恒成立，即：2max []xa e ≥-，1a ≥-，10a -≤≤；当0a >时，'()0xx a f x e e =+>，()xx a f x e e=-在[0,1]上是单调递增的，又()()g x f x =在[0,1]上单调递增，所以()0f x ≥在[0,1]上恒成立.2min []x a e ≤，01a <≤.综上：11a -≤≤.22.【解析】（1）直线l 的直角坐标系方程是220x y a +--=，圆C 的直角坐标方程是22(2)4x y -+=. （2）由（1）知圆心为(2,0)C ，半径2r =，设圆心到直线的距离为d ，因为直线与圆相切，所以2d ===，解得2a =±23.【解析】（1）当1a =时，不等式()4114f x x x ≥⇔++-≥，当1x >时，()24f x x =≥，解得2x ≥；当11x -≤≤时，()24f x =≥，无解；当1x <-时，()24f x x =-≥，解得2x ≤-，综上所述，不等式的解集为(,2][2,)-∞-+∞. （2）()f x x a x a =++-()()2x a x a a ≥+--=， ∴26a ≥，解得3a ≥或3a ≤-，即a 的取值范围是(,3][3,)-∞-+∞.21。

2019届长沙长郡中学高三月考英语试卷(一)及答案

2019届长郡中学高三月考英语试卷（一）第一部分听力（共两节，满分30分）（略）第二部分阅读理解（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A.B.C和D四个选项中，选出最佳选项。

AChina has announced it's abolishing its one-child policy.What difference has it made, statistically speaking?400million births preventedThe one-child policy,officially in place since1979,has prevented400million births.Parents have faced fines and other punishments for having more children.The majori ty of the decrease in China's fertility(生育)rate happened in the1970s.It dropped from5.8children per woman in1970t02.7in1978.Despite the one-child policy the rate had only fallen t01.7by2013.21:28baby death rateSince the one-child policy was introduced,baby girls have become more likely to die than boys.In the1970s,according to the United Nations,60males per l,000live births died under the age of one.For girls the figure was53.In the1980s,after the one-child policy became official, the rate for both was36.By the1990s,26males per l,000live births died before the age of one-and33girls.The2000s saw21boys per l,000live births dying and28girls.1.16boys born for every girlSexually selective abortions have been considered as a major cause of China's unusual sexual imbalance.Gietel-Basten,associate professor in social policy at Oxford University,says the births of many girls are not registered if parents have broken the rule by having two children,adding officials often turn a blind eye.It's estimated there are now33million more men than women in China.4：2：1familiesWith the ageing of China's population and the continuation of the one-child policy,a“4: 2:1”home is the description given to households in which there are four grandparents cared for by two working age parents,who themselves have one child.By2050,it's predicted that a quarter of China's population will be65or older.The predicted decline in the number of people of working age is thought to have persuaded the government to drop the one-child policy.21.When was the baby death rate for both boys and girls equal?A.In the1970s.B.In the1980s.C.In the1990s.D.In the2000s.22.What makes the one-child policy abolished according to the passage?A.The decline of birth rate.B.The rise of baby death rate.C.The change of family structure.D.The decline of working age people.23.The passage sums up the one-child policy by.A.numberB.exampleC.contrastD.analysisBlt's a little hard to figure out the rules for sure,but the baby African buffalo(7K午)seems to have the upper hand.The two stand a couple feet apart,staring at each other.Behind them,their mothers look on with the kind of indifference(不关心)of mothers everywhere who see their kids playing a harmless game.The alarm went off about half an hour ago,the second time tonight.Here,one bell means “elephant”,two,“rhino”(犀牛),and three and four,1really can't remember for what,because when you're awoken by bells in the middle of the night,your first thought won't be“Animal!”. But as soon as I realized it was the rhino alarm,I was running for the stairs.Yet I'm the only one watching the animals stare-down.Either everyone else died of a heart attack w hen the bells went off,or l was the only one in the entire hotel who didn't turn the alarm switch to off before going to bed.The Ark,a hotel shaped like the biblical ark(圣经的方舟),lies in the highlands of Kenya's Aberdare National Park,about100kilometers north of Nairobi.At the Ark's prow(船首)are huge windows overlooking a waterhole.Earlier tonight,I'd watched a pack of wolves,buffalos and elephants.And now,at three or so in the morning,I'm down for rhino bells.The first time,2.5 hours ago,it was a single black rhino,which came down,got a drink,and left.I'd have been sorry for the lost sleep.I've already arranged with a guide to take me out at sunrise for the so-called“should_never_miss”bird-watching,where we'll get to see30species, but really,how many times in your life will you get to wake up and say,“Wow!Rhino!”?Isn't that the definition of a pretty good night?Last week we went from Uganda into Kenya.Before I left home,1thought I'd be happy with ten elephants and five or six giraffes on the whole trip.I didn't dare to dream the rhino.24.The message that the hotel's bells sent was.A.a very wise way the author had never heard ofB.hard for the author to remember in detailC.a matter that all visitors had got used toD.like conditioned response training for animals25.What's the author's attitude toward the bell?A.It's bothering but two bells are welcome.B.It should be turned off whenever it rings.C.It lets visitors have a chance to see buffalos.D.It is a very bad way to wake the guests up.26.What can we learn from the passage?A.The author hates all the animals.B.There are lots of wild animals near the Ark.C.The author slept well during the whole trip.D.The author has no interest in thinos.27.What did the author see at about3:00a.m.?A.Two buffalos were fighting fiercely for food.B.A rhino and a buffalo were fighting for water.C.A rhino and a buffalo were staring at each other.D.A buffalo was teaching its baby to walk freely.CMany people trying to sell homes find that an increase in home prices has turned the market in their favor.But sellers can still get the short end of the deal if they aren't careful.Here are a few tips for you:Don't test your luck.Of course you think anyone who moves into your lovely home should be willing to pay top dollars,especially if you've recently invested in some improvements.But listing a home at a price that's too high above the market price could turn away some buyers.Buyers noticing that the home still hasn't sold may begin to assume there's something wrong with the house and use that as a reasonable excuse for offering a lower price.And if a home hasn't received any offers after two weeks,it might be time to reset the price.A price that's too low can bring about an undesired outcome.Listing your home at or slightly below the market price can have the effect of drawing in a large group of buyers and increase the chances that a home will receive multiple offers.But setting the price too low comes with several risks.One possibility is that buyers will get skeptical of the home that is listed for$15,000to $20,000less than similar homes in the area,especially if it's not properly marketed.Once again, people might assume there is something wrong with the home and may not bother to look at it.Spy on the competition.Going to other people's open houses can give you a better sense of how your home compares to others on the market.Check out the decoration in their kitchens,the size of their backyards and use the information to figure out where your home should fit in the range of the price.But don't set your pricing just on what you see elsewhere.28.What does the underlined part in Paragraph l probably mean?A.Be at a disadvantage.B.Get the upper hand.C.Have control over the situation.D.Be unable to fit in.29.What should home sellers do according to the passage?A.Price your house on the basis of its geographic location.B.Price your house slightly above the market price after decorating it.C.Change the price if no offer has been received within a week.D.Price your house at or slightly below the market price.30.If you set the price of your house too low,.A.buyers might think it not worthwhile to go to have a lookB.you are likely to come into conflict with the neighborhood sellersC.your house will be crowded with buyers within a couple of daysD.chances of your house being sold at a better price will be greatly increased31.What is the benefit of going to other people's open house?A.You can pick up some useful lessons on house selling.B.You can get your pricing mainly based on what you see elsewhere.C.You can work out how much money you should ask for your home.D.You can know how to make your house stand out against other houses.DWhat's your favorite type of music?Most people may prefer rock and roll,pop or jazz. These types are most publicized by television and the radio.Country music now would seem to mistakenly belong to the category of rock or pop.Publicity of this category started to drop only after the year2000;however,it did have its peak years just before falling down.What people may not know is that this type of music inspired(促成)rock and roll especially with one of the most important figures in music history-Elvis Presley-who is known as the6/King of Rock and Roll".Elvis used to be very famous at a radio program which broadcast country music back in the late1940s.Country music is influenced by jazz and blues.Rock music is also influenced by jazz and blues so that they may even seem similar.Nowadays,country music is closely related to pop music due to artists'preference and popular demand.Starting from the1980's it slowly divided itself into New and Old Country Music. Legends like Johnny Cash began to slip away because the new understanding of country music actually contains pop and his music just doesn't fit the category.Fortunately,country music is still heavily promoted through websites and reviews.Certain types of music nowadays are linked with different age groups.Classical for the people in the mid19th century,pop in the1920's,and then there's country music.All these types continue to grow and modernize while still appealing to certain age groups.More modern country music compositions tend to become different in their own ways which makes it even harder to identify them as country music while they,again,remain appealing only to certain age groups. Active promoters of country music videos,however,seem to appeal to a large number of age groups especially those active in Australia,for as long as country music isn't mixed with other types,it will eventually regain its charm as music for all ages.32.It can be learned from the first paragraph that country music____.A.is most publicized by television and the radioed to be liked by more people compared to jazzC.has become less popular since the year2000D.should be included in the category of rock or pop33.What is the relationship between rock music and country music?A.Both types influenced the formation of rock and roll.B.Presley turned pop music into country music.C.They are both influenced by jazz and blues.D.The use of the same instruments makes them sound similar.34.Some legends of country music began to slip away because____.A.their songs totally changed the style of country musicB.they began to work on pop music for more profitsC.there is no marketing space for country music anymoreD.they can't meet the popular demands on country music35.The author seems to believe that____.A.country music should keep its own style to attract more listenersB.it's necessary for country music to be mixed with other typesC.only a certain age group of people will like country music in the futuremon characteristics can be found in all modern music第二节（共5小题；每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

2020届湖南省长沙市长郡中学高三下学期3月“停课不停学”阶段性检测数学(文)试题(解析版)

2020届湖南省长沙市长郡中学高三下学期3月“停课不停学”阶段性检测数学（文）试题一、单选题1．设全集U =R ，{}(){}260,ln 1A x x x B x y x =--<==-，则()U A B =I ð（） A ．[1，3) B ．(1，3] C ．(1，3) D ．(－2，1]【答案】A【解析】首先确定集合,A B 中的元素，然后由集合运算法则计算．【详解】由题意2{|60}{|23}A x x x x x =--<=-<<，{|10}{|1}B x x x x =->=<，∴{|1}U B x x =≥ð，(){|13}[1,3)U A B x x =≤<=I ð．故选：A. 【点睛】本题考查集合的运算，考查一元二次不等式的解法，掌握集合的运算定义是解题关键．本题还考查了对数型复合函数的定义域．需要掌握对数函数的性质． 2．设复数()1331iz i i+=++-（其中i 为虚数单位），则下列说法中正确的是( ) A ．它的实部为﹣3 B ．共轭复数34z i =+C ．它的模||5z =D ．在复平面对应的点的坐标为(3,4)-【答案】C【解析】利用复数代数形式的乘除运算化简，然后逐一核对四个选项得答案．【详解】解：∵()()()21(1)333334111i i z i i i i i i ++=++=++=+--+，∴z 的实部为3，34z i =-，5z ==，在复平面对应的点的坐标为（3，4）. 故选: C. 【点睛】本题考查复数代数形式的乘除运算，以及复数的基本概念、共轭复数、复数的模和复数的几何意义，是基础题．3．已知向量a v ＝(1，0)，b v＝(-3，4)的夹角为θ，则sin2θ等于 ( )A ．725-B ．725C ．2425-D ．2425【答案】C【解析】首先根据向量夹角公式求出cos θ的值，然后求出sin θ，最后根据二倍角正弦公式即可得出结果. 【详解】33cos 155a b a bθ⋅==-=-⨯⋅r r r ，∵0θπ≤≤，∴4sin 5θ==，24sin 22sin cos 25θθθ==-，故选C. 【点睛】本题主要考查了向量夹角的计算以及二倍角正弦公式的应用，属于中档题. 4．设3log 18a =，4log 24b =，342c =，则a 、b 、c 的大小关系是( )A ．a b c <<B ．a c b <<C ．b c a <<D ．c b a <<【答案】D【解析】容易判断出2a >，2b >，2c <，并且得出1862463344log 1log ,log 1log a b ==+==+，容易判断出43log 6log 6<，从而得出a ，b ，c 的大小关系．【详解】解：由题可知，3422c =<，1892416334422a log log b log log =>==>>，，又186246334411a log log b log log ==+==+，，∵4366116643log log log log ==，且66log 4log 30>>， ∴661143log log <， 43log 24log 18∴<，即：b a <，综上得：2c b a <<<，则c b a << 故选D.【点睛】本题考查利用指数函数、对数函数的单调性来比较大小，还考查对数的运算和对数的换底公式． 5．函数()()23ln 1x f x x+=的大致图象是A ．B ．C ．D ．【答案】A【解析】利用函数的对称性及函数值的符号即可作出判断. 【详解】由题意可知函数()f x 为奇函数，可排除B 选项；当x 0<时，()0f x ＜，可排除D 选项；当x 1=时，()12f ln =，当x 3=时，ln10ln10(3),ln 22727f =>，即()()1?3f f ＞，可排除C 选项，故选：A 【点睛】本题考查了函数图象的判断，函数对称性的应用，属于中档题．6．等比数列{}n a 的前n 项和为S n ，且14a ，22a ，3a 成等差数列，若1a =1，则S 10=（） A ．512 B ．511C ．1024D ．1023【答案】D【解析】设等比数列的公比为q ，由题意，得2344a a =+，即244q q =+，解得2q =，则1010101221102312S -==-=-；故选D.7．运行如图所示的程序框图，若输出的S 的值为101，则判断框中可以填( )A ．200?i >B ．201?i ≥C ．202?i >D ．203?i >【答案】C【解析】由已知中的程序语句可知：该程序的功能是利用循环结构计算并输出变量S 的值，模拟程序的运行过程，分析循环中各变量值的变化情况，可得答案．【详解】程序的功能是计算3571sin3sin5sin 7sin 2222S ππππ=⨯+⨯+⨯+⨯+=L 1357-+-+L ，而101150213579199201=+⨯=-+-++-+L ，2012203i =+=，故条件为202?i >，故选C. 【点睛】本题考查了程序框图的应用问题，解题时应模拟程序框图的运行过程，以便得出正确的结论，是基础题．8．将函数()2sin(2)16f x x π=--的图象向左平移6π个单位长度得到函数()g x 的图象，则下列说法正确的是 ( ) A ．函数()g x 的最小正周期是2π B ．函数()g x 的图象关于直线12x π=-对称C ．函数()g x 在,62ππ⎛⎫⎪⎝⎭上单调递减 D ．函数()g x 在0,6π⎛⎫⎪⎝⎭上的最大值是1 【答案】C【解析】求出函数的周期判断A 的正误；函数的对称轴判断B 的正误；函数的单调性判断C 的正误；函数的最值判断D 的正误；【详解】由题意知：()2sin(2)16g x x π=+-，最小正周期T 22ππ==，选项A 错误; 当12x π=-时，112g π⎛⎫-=- ⎪⎝⎭，即函数()g x 的图象关于点(,1)12π--对称，选项B 错误；当(,)62x ππ∈时，72(,)626x πππ+∈， ∴函数()g x 在,62ππ⎛⎫⎪⎝⎭上单调递减，选项C 正确; ∵函数()g x 在0,6π⎛⎫⎪⎝⎭上单调递增，()()16g x g π<=，即函数()g x 在0,6π⎛⎫⎪⎝⎭上没有最大值， ∴选项D 错误，故选C. 【点睛】本题考查三角函数的简单性质，最值、单调性、周期以及单调性，考查命题的真假的判断，属于中档题．9．小王因上班繁忙，来不及做午饭，所以叫了外卖.假设小王和外卖小哥都在12：00~12：10之间随机到达小王所居住的楼下，则小王在楼下等候外卖小哥的时间不超过5分钟的概率是（） A ．12B ．45C ．38D ．34【答案】C【解析】设出两人到达小王的时间，根据题意列出不等式组，利用几何概型计算公式进行求解即可. 【详解】设小王和外卖小哥到达小王所居住的楼下的时间分别为,x y ，以12：00点为开始算起，则有5x y y x ≤⎧⎨-≤⎩，在平面直角坐标系内，如图所示：图中阴影部分表示该不等式组的所表示的平面区域，所以小王在楼下等候外卖小哥的时间不超过5分钟的概率为：11101010105532210108P ?创-创==´. 故选：C 【点睛】本题考查了几何概型中的面积型公式，考查了不等式组表示的平面区域，考查了数学运算能力.10．已知长方体1111ABCD A B C D -各个顶点都在球面上，8AB AD ==，16AA =，过棱AB 作该球的截面，则当截面面积最小时，球心到截面的距离为( ) A ．3 B ．4C ．5D ．6【答案】C【解析】过棱AB 作该球的截面，则当截面面积最小时，截面的直径为8AB =，求出球的半径，可得球心到截面的距离．【详解】解：过棱AB 作该球的截面，则当截面面积最小时，截面的直径为8AB =，Q 长方体1111ABCD A B C D -各个顶点都在球面上，8AB AD ==，16AA =，∴球的半径为2221886412++=， ∴球心到截面的距离为41165-=．故选：C. 【点睛】本题考查求球心到截面的距离，考查学生的计算能力，确定当截面面积最小时，截面的直径为8AB =是关键，是基础题．11．已知()f x 是偶函数,当0x ≥时,202()822x x f x x x ⎧≤<=⎨-≥⎩，，,若(1)(1)f a f -<-,则a 的取值范围是( )A ．()11-，B ．()()2024-，，U C ．()()()3113-∞--+∞，，，U U D ．()(2)02(4)-∞-+∞，，，U U 【答案】D【解析】因为()f x 是偶函数,当0x ≥时,202()822x x f x x x ⎧≤<=⎨-≥⎩，，,根据偶函数图像关于y ,画出其函数图像,即可求解a 的取值范围.【详解】Q 根据()f x 是偶函数(1)(1)2f f -==∴ (1)(1)f a f -<-,即(1)2f a -<因为()f x 是偶函数,当0x ≥时,202()822x x f x x x ⎧≤<=⎨-≥⎩，，,根据偶函数图像关于y 轴对称，,画出其函数图像,如图:当02x ≤<时,由()2f x =,解得11x =,根据对称可知,21x =-当2x ≥时,由()2f x =,解得33x =,根据对称可知,43x =-Q 保证(1)2f a -<,结合图像可知:∴ 41a x -<或211x a x <-<或41a x ->即: 13a -<-或111a -<-<或13a -> 故: 2a <-或02a <<或4a > 故选:D. 【点睛】本题考查了求解函数不等式,解题关键是根据函数的对称性画出其函数图像,数形结合,考查了分析能力和计算能力,属于中等题.12．已知双曲线C ：22221x y a b-=（0,0a b >>）的左、右焦点分别为12,F F ，过1F 的直线l 与双曲线C 的左支交于A 、B 两点.若22,120=∠=oAB AF BAF ，则双曲线C 的渐近线方程为（）A ．y x =±B ．y =C ．=±y xD ．)1=±y x【答案】D【解析】设2AF m =，利用余弦定理，结合双曲线的定义进行求解即可. 【详解】设22,AB AF m BF ==∴==，由双曲线的定义可知：12,AF m a =-因此12,BF a =再由双曲线的定义可知：1223BF BF a m -=⇒=，在三角形12AF F 中，由余弦定理可知： 222212222222112cos120(5(5F F AF AF AF AF c a a b a ︒=+-⋅⋅⇒=-⇒+=-2222(4(41b bb a a a⇒=-⇒=-⇒=，因此双曲线的渐近线方程为：)1=±y x .故选：D【点睛】本题考查了双曲线的定义的应用，考查了余弦定理的应用，考查了双曲线的渐近线方程，考查了数学运算能力.二、填空题13．为了解某市居民用水情况，通过抽样，获得了100位居民某年的月均用水量（单位：吨），将该数据按照[0，0.5），[0.5，1），…[4.4.5]分成9组，绘制了如图所示的频率分布直方图，政府要试行居民用水定额管理，制定了一个用水量标准a，使85%的居民用水量不超过a（假设a为整数），按平价收水费，超出a的部分按议价收费，则a的最小值为_____.【答案】3吨【解析】求出[0，0.5)的频数为4，[0.5，1)的频数为8，[1.5，2)的频数为22，[2，2.5)的频数为25，[2.5，3)的频数为4，[3，3.5)的频数为4，[4，4.5)的频数为2，+++++=，由此能求出a的最小值．481522151486【详解】解：[0，0.5)的频数为0.080.51004⨯⨯=，⨯⨯=，[0.5，1)的频数为0.160.51008⨯⨯=，[1.5，2)的频数为0.440.510022⨯⨯=，[2，2.5)的频数为0.50.510025⨯⨯=．[2.5，3)的频数为0.280.51004⨯⨯=，[3，3.5)的频数为0.080.51004⨯⨯=，[4，4.5)的频数为0.040.51002481522151486+++++=，∴前6组占86%，a 为3吨．故答案为：3吨. 【点睛】本题考查使85%的居民用水量标准的最小值的求法，以及频率分布直方图等基础知识，还考查推理能力与计算能力，属于基础题．14．已知抛物线2:2(0)C y px p =>的焦点F，点00()2pM x x >是抛物线上一点，以M 为圆心的圆与直线2px =交于A 、B 两点(A 在B 的上方），若5sin 7MFA ∠=，则抛物线C 的方程为 _____. 【答案】212y x =【解析】依题意作图，可以把5sin 7MFA ∠=放在直角三角形中，可得5sin 7MC MFA MF ∠==，由抛物线定义转化MF MD =，02p MC x =-，即可得到0x 与p 的关系，再代入方程中即可求出p ，则抛物线方程可求．【详解】解：如图所示，过M 点作CM ⊥抛物线的准线，垂足为C ，交准线于D ， 5sin 7MCMFA MF∴∠==，由抛物线定义可得：MF MD =，∴005272px MC p MF x -==+，即00575722x p x p +=+， 03x p ∴=，Q点00()2p M x x >是抛物线上一点，∴202px =，即23666p ⨯=，6p ∴=，得212y x =．故答案为：212y x =．【点睛】本题考查抛物线的标准方程，利用抛物线的定义进行线段的转化是关键，是中档题． 15．已知等差数列{}n a 的前n 项和为n S ，且43a =-，1224S =，若0(i j a a i +=，j ∈*N 且1)i j <„，则i 的取值集合是 _____. 【答案】{1，2，3，4，5，6，7，8，9，10}【解析】利用等差数列前n 项和公式和通项公式，求出19a =-，2d =，从而9(1)2211n a n n =-+-⨯=-，由此能求出i 的取值集合．【详解】解：Q 等差数列{}n a 的前n 项和为n S ，且43a =-，1224S =，∴4112133121112242a a d S a d =+=-⎧⎪⎨⨯=+=⎪⎩，解得19a =-，2d =， 9(1)2211n a n n ∴=-+-⨯=-，0(i j a a i +=Q ，j ∈*N 且1)i j <„，1i ∴=时，92110j -+-=，解得10j =；2i =时，4112110j -+-=，解得9j =； 3i =时，6112110j -+-=，解得8j =； 4i =时，8112110j -+-=，解得7j =； 5i =时，10112110j -+-=，解得6j =； 6i =时，12112110j -+-=，解得5j =；7i =时，14112110j -+-=，解得4j =； 8i =时，16112110j -+-=，解得3j =； 9i =时，18112110j -+-=，解得2j =， 10i =时，20112110j -+-=，解得1j =．综上，i 的取值集合为{1，2，3，4，5，6，7，8，9，10}．故答案为：{1，2，3，4，5，6，7，8，9，10}. 【点睛】本题考查数列的项数的取值集合的求法，运用到等差数列的通项公式和求和公式，还考查推理能力与计算能力，属于基础题．16．如图，已知在长方体1111ABCD A B C D -中，3AB =，4=AD ，15AA =，点E 为1CC 上的一个动点，平面1BED 与棱1AA 交于点F ，给出下列命题：①四棱锥11B BED F -的体积为20；②存在唯一的点E ，使截面四边形1BED F 的周长取得最小值274；③当E 点不与C ，1C 重合时，在棱AD 上均存在点G ，使得//CG 平面1BED ； ④存在唯一的点E ，使得1B D ⊥平面1BED ，且165CE =．其中正确的命题是_____（填写所有正确的序号）【答案】①②③④【解析】由题意逐个讨论所给的命题，判断它们的真假．【详解】解：①由题意可得1//D F BE ，1111111111B BED F B BED B BFD D BEB D BFB V V V V V -----=+=+ 111111111[](543543)2032232BB BC AB BB D A AB =+=⨯⨯+⨯⨯=g g g g g g g ，所以①正确； ②将长方体展开，如图所示，恰好过B 点时，截面的周长为12BD ，而在1BDD ∆中，2215(34)74BD ++，所以最小值为274由面面平行的性质，可得四边形1BED F 为平行四边形，且E 为展开图中唯一的点，所以②正确；③E 点不与C ，1C 重合，则F 不会为A ，即CG 不在面1EBD 内，可作出CG 的平面与1EBD 平行，所以在棱AD 上均有相应的G ，使得//CG 面1EBD ，故③正确；④因为1BB BD =，可得对角面11BB D D 为正方形，可得11B D BD ⊥，若1BE B C ⊥时，由三垂线定理可得1B D BE ⊥，即有1B D ⊥面1EBD ，矩形11BB C C 中，1BE B C ⊥，所以1CE BC BCCC =，所以1165BC BC CE CC ==g ，故④正确综上可得：正确为①②③④. 故答案为：①②③④. 【点睛】本题考查了正方体的性质、线面平行和垂直的判定定理、棱锥的体积、简易逻辑的判定方法，考查了推理能力与计算能力，属于中档题三、解答题17．如图，在四边形ABCD 中，45ADB o ∠=，105BAD ∠=o ，6AD =，2BC =，3AC =．()1求边AB 的长及cos ABC ∠的值；()2若记ABC ∠α=，求23sin πα⎛⎫- ⎪⎝⎭的值．【答案】（1）3AB =3 cos ABC ∠=；（25311-【解析】()1由已知可求ABD ∠，ABD V 中，由正弦定理可求AB ，ABC V 中由余弦定理，2222AB BC AC cos ABC AB BC∠+-=⋅可求．()2由()1可得cos α，进而可求sin α，进而根据二倍角公式22sin sin cos ααα=，2221cos cos αα=-可求，然后根据两角差的余弦公式即可求解.【详解】() 1由题意，因为45ADB o ∠=，105BAD ∠=o ，30ABD ∠∴=o ，AD =Q 2BC =，ABD V 中，由正弦定理可得，24530AB sin sin =o o，AB ∴=， 3AC =Q ．ABC V 中由余弦定理可得，2222AB BC AC cos ABC AB BC ∠+-===⋅()2由()1可得cos α=sin α∴=226sin sin cos ααα∴==-，252216cos cos αα=-=-122232212sin sin cos πααα⎛⎫∴-=-=⎪⎝⎭．【点睛】本题主要考查了正弦定理、余弦定理和三角恒等变换的应用，其中在解有关三角形的题目时，要有意识地考虑用哪个定理更合适，要抓住能够利用某个定理的信息．一般地，如果式子中含有角的余弦或边的二次式时，要考虑用余弦定理；如果式子中含有角的正弦或边的一次式时，则考虑用正弦定理，着重考查了运算与求解能力，属于基础题． 18．某蛋糕店每天制作生日蛋糕若干个，每个生日蛋糕成本为50元，每个蛋糕的售价为100元，如果当天卖不完，剩余的蛋糕作垃圾处理.现搜集并整理了100天生日蛋糕的日需求量（单位：个），得到如图所示的柱状图.100天记录的各需求量的频率作为每天各需求量发生的概率.（1）若该蛋糕店某一天制作生日蛋糕17个，设当天的需求量为)(N n n ∈，则当天的利润y （单位：元）是多少？（2）若蛋糕店一天制作17个生日蛋糕.①求当天的利润y （单位：元）关于当天需求量n 的函数解析式； ②求当天的利润不低于600圆的概率.（3）若蛋糕店计划一天制作16个或17个生日蛋糕，请你以蛋糕店一天利润的平均值作为决策依据，应该制作16个还是17个生日蛋糕？【答案】（1）17n ≥时，利润为850；16n ≤时，利润为100850n -；（2）①100850(16)()850(17)n n Y n N n -≤⎧=∈⎨≥⎩，②2225；（3）17．【解析】试题分析：（1）分17n ≥、16n ≤分别求得相应利润；（2）①结合（1）即可求得函数解析式；②将问题转化为“需求量不低于15个”的概率，由此利用条件概率公式求解即可；（3）分别求得一天制作16个和17个的平均利润，由此作比较可得结论．试题解析：（1）当17n ≥时，17(10050)850Y =⨯-=，当16n ≤时，1001750100850Y n n =-⨯=-．（2）①由（1）得当天的利润Y 关于当天需求量n 的函数解析式为：100850(16)()850(17)n n Y n N n -≤⎧=∈⎨≥⎩②设“当天利润不低于600”为事件A ，由①知，“当天利润不低于600”等价于“需求量不低于15个”1222()110025P A ∴=-=，所以当天的利润不低于600元的概率为：2225．（3）若一天制作16个蛋糕，则平均利润为：11(600127001880070)758100x =⨯+⨯+⨯=；若一天制作17个蛋糕，则平均利润为：21(55012650187501885052)760100x =⨯+⨯+⨯+⨯=； 12x x <Q ，∴蛋糕店一天应该制作17个生日蛋糕.【考点】1、频率分布直方图；2、分段函数；3、条件概率．19．在如图所示的五面体ABCDEF 中，四边形ABCD 为菱形，且60DAB ∠=︒，//EF 平面ABCD ，22EA ED AB EF ====，M 为BC 中点．（1）求证：//FM 平面BDE ；（2）若平面ADE ⊥平面ABCD ，求F 到平面BDE 的距离．【答案】（1）证明见解析；（215. 【解析】（1）取CD 中点N ，连接MN ，FN ，说明//MN BD ，证明//MN 平面BDE ，证明//EF AB ，//AB CD ，推出//EF CD ，//FN ED ，证明//FN 平面BDE ，转化证明//FM 平面BDE ；（2）说明F 到平面BDE 的距离等于M 到平面BDE 的距离，取AD 的中点H ，连接EH ，BH ，推出EH ⊥平面ABCD ，EH BH ⊥，设F 到平面BDE 的距离为h ，由E BDM M BDE V V --=，转化求解即可．【详解】解：（1）取CD 中点N ，连接MN ，FN ，因为N ，M 分别为CD ，BC 中点，所以//MN BD ，又BD ⊂平面BDE ，且MN ⊂平面BDE ，所以//MN 平面BDE ，因为//EF 平面ABCD ，EF ⊂平面ABEF ，平面ABCD I 平面ABEF AB =，所以//EF AB ，又222AB CD DN EF ====，//AB CD ，所以//EF CD ，EF DN =．所以四边形EFND 为平行四边形．所以//FN ED ．又ED ⊂平面BDE 且FN ⊂/平面BDE ，所以//FN 平面BDE ，又FN M N N =I ，所以平面//MFN 平面BDE ．又MF ⊂平面MFN ，所以//FM 平面BDE ．（2）由（1）得//FM 平面BDE ，所以F 到平面BDE 的距离等于M 到平面BDE 的距离，取AD 的中点H ，连接EH ，BH ，由四边形ABCD 为菱形，且60DAB ∠=︒，2EA ED AB EF ===，可得EH AD ⊥，BH AD ⊥，因为平面ADE ⊥平面ABCD ，平面ADE I 平面ABCD AD =，所以EH ⊥平面ABCD ，EH BH ⊥，因为3EH BH ==，所以6BE =，所以22161562()22BDE S ∆=⨯⨯-=，设F 到平面BDE 的距离为h ，又因为113342242BDM BCD S S ∆∆==⨯⨯=，所以由E BDM M BDE V V --=，得1311533232h ⨯⨯=⨯⨯，解得155h =．【点睛】本题考查直线与平面平行与垂直的判断与性质的应用，点、线、面距离的求法，等体积法的应用，考查空间想象能力以及计算能力． 20．已知函数21()2f x lnx ax x =--．（1）若函数()f x 在[1，)+∞上单调递增，求实数a 的取值范围；（2）若函数()f x 在1x =处的切线平行于x 轴，是否存在整数k ，使不等式[()1](2)x f x x k x +->-在1x >时恒成立？若存在，求出k 的最大值；若不存在，请说明理由．【答案】（1）a 14≤-；（2）不存在，理由见解析. 【解析】（1）对原函数求导，根据导数和函数的单调性的关系即可求出a 的取值范围；（2）问题转化为即(1)20xlnx k x k -++>在1x >时恒成立，令()(1)2g x xlnx k x k =-++，1x >求导后分0k „和0k >求函数的单调区间，进一步求得函数的最值得答案．【详解】解：（1）Q 函数()f x 在[1，)+∞上单调递增， 1()10f x ax x∴'=--… 在[1，)+∞ 上恒成立， 2211111()24a x x x ∴-=--„， ∴当2x =时，()211124x --有最小值14-，14a ∴-„；（2）1()1f x ax x'=--Q ， f ∴'（1）11a a =--=-，Q 函数()f x 在1x =处的切线平行于x 轴，0a ∴=，()f x lnx x ∴=-，Q 不等式[()1](2)x f x x k x +->-在1x >时恒成立，(2)xlnx x k x ∴->-在1x >时恒成立，即(1)20xlnx k x k -++>在1x >时恒成立，令()(1)2g x xlnx k x k =-++，1x >， ()g x lnx k ∴'=-，当0k „时，()0g x '>在(1,)+∞上恒成立，即()g x 在(1,)+∞上单调递增， ()g x g >（1）10k =->，则1k >，矛盾，当0k >时，令()0g x '=，解得k x e =，令()0g x '>，解得：k x e >，令()0g x '<，解得：1k x e <<，()g x ∴在(1,)k e 单调递减，在(k e ，)+∞单调递增，()()(1)220k k k k min g x g e ke k e k k e ∴==-++=->，令()2k h k k e =-，0k >，()2k h k e ∴'=-，当2k ln <时，()0h k '>，函数()h k 单调递增，当2k ln >时，()0h k '<，函数()h k 单调递减， ()(2)2222(21)0max h k h ln ln ln ∴==-=-<，∴不存在整数k 使得20k k e ->恒成立，综上所述不存在满足条件的整数k ．【点睛】本题考查利用导数研究函数的单调性和最值，导数的几何意义，还运用分离参数法和函数构造法解决恒成立问题，同时考查了数学转化思想方法以及推理能力和运算能力，属难题．21．已知斜率为k 的直线l 与椭圆22143x y C +=：交于A ，B 两点，线段AB 的中点为()()10M m m >，．（1）证明：12k <-；（2）设F 为C 的右焦点，P 为C 上一点,且0FP FA FB ++=u u u v u u u v u u u v．证明：FA u u u v ，FP u u u v ，FB u u u v成等差数列，并求该数列的公差．【答案】（1）12k <-（2）28或28- 【解析】分析：（1）设而不求，利用点差法进行证明．（2）解出m,进而求出点P 的坐标，得到FP u u u r ,再由两点间距离公式表示出,FA FB u u u r u u u r，得到直l 的方程，联立直线与椭圆方程由韦达定理进行求解．详解：（1）设()()1122,,,A x y B x y ，则222211221,14343x y x y +=+=.两式相减，并由1212y y k x x -=-得1212043x x y y k +++⋅=. 由题设知12121,22x x y y m ++==，于是 34k m=-.①由题设得302m <<，故12k <-.（2）由题意得()1,0F ，设()33,P x y ，则()()()()3311221,1,1,0,0x y x y x y -+-+-=.由（1）及题设得()()31231231,20x x x y y y m =-+==-+=-<. 又点P 在C 上，所以34m =，从而31,2P ⎛⎫- ⎪⎝⎭，32FP =u u u v . 于是122xFA ===-u u u v .同理222x FB =-u u u v .所以()121432FA FB x x +=-+=u u u v u u u v .故2FP FA FB u u u v u u u v u u u v =+，即,,FA FP FB u u u v u u u v u u u v成等差数列.设该数列的公差为d ，则1212||2d FB FA x x =-=-=u u u v u u u v ②将34m =代入①得1k =-. 所以l 的方程为74y x =-+，代入C 的方程，并整理得2171404x x -+=.故121212,28x x x x +==，代入②解得28d=. 所以该数列的公差为28或28-.点睛：本题主要考查直线与椭圆的位置关系，等差数列的性质，第一问利用点差法，设而不求可减小计算量，第二问由已知得到0FP FM +=u u u r u u u u r,求出m 得到直线方程很关键，考查了函数与方程的思想，考察学生的计算能力，难度较大．22．在平面直角坐标系xOy 中，曲线21240x y C x +-=：，直线l 的参数方程为cos sin x t y t αα=⎧⎨=⎩（t 为参数），其中0,6πα⎛⎫∈ ⎪⎝⎭，以坐标原点O 为极点，x 轴非负半轴为极轴，建立极坐标系.（1）求曲线1C 的极坐标方程和直线l 的普通方程；（2）设(4,0)M ，2C 的极坐标方程ρθ=，A ，B 分别为直线l 与曲线12,C C 异于原点的公共点，当30AMB ∠=︒时，求直线l 的斜率；【答案】（1）曲线的1C 极坐标方程为4cos ρθ=，直线l 的普通方程为tan ,0,6y x παα⎛⎫=⋅∈ ⎪⎝⎭（2）4【解析】(1)利用cos sin x y ρθρθ=⎧⎨=⎩将1C 的普通方程转化为极坐标方程，消去参数t 将直线l 的参数方程转化为普通方程； (2)根据题意求出||||OA OB 、及||AB ，又点M 在曲线1C上，则1||tan 4sin AM ραα==，由|||AM AB =列出方程即可得解.【详解】（1）将cos sin x y ρθρθ=⎧⎨=⎩代入曲线1C 的普通方程得极坐标方程为4cos ρθ=，直线l 的普通方程为tan ,0,6y x παα⎛⎫=⋅∈ ⎪⎝⎭；（2）由已知可得θα=，则12||4cos ,||OA OB ραρα====，||4cos ,AB αα=-因为点M 在曲线1C 上且AM AB ⊥，所以1||tan 4sin ,AM ραα==在直角三角形ABM 中30AMB ∠=︒，则|||AM AB =所以4sin )ααα=-，得直线l 的斜率tan k α==【点睛】本题考查普通方程与极坐标方程的互化，参数方程化成普通方程，直线与圆的位置关系，直径所对的圆周角是直角，属于中档题.23．函数()223f x x x =-++（1）求不等式()25f x x ≥+的解集；（2）若()f x 的最小值为k ，且实数,,a b c 满足()a b c k +=，求证：22228a b c ++≥【答案】（1）(,0][4,)-∞⋃+∞（2）证明见解析【解析】(1)分类去绝对值符号后解不等式，最后取并集；(2)求出函数的最小值k ，根据基本不等式得出结论.【详解】（1）①当3x <-时，不等式即为3125x x --≥+，解得6,35x x ≤-∴<- ②当31x -≤≤时，不等式即为525x x -≥+，030x x ≤∴-≤≤③当1x >时，不等式即为3125x x +≥+，44x x ≥∴≥综上，()25f x x ≥+的解集为(,0][4,)-∞⋃+∞ （2）由51,3()5,3131,1x x f x x x x x --<-⎧⎪=--≤≤⎨⎪+>⎩∴当1x =时，()f x 取最小值4，即4,()4k a b c =∴+=，即4ab ac += ()()22222222228a b c a b a c ab ac ∴++=+++≥+=当且仅当a b c ===时等号成立【点睛】本题考查绝对值不等式的解法，不等式的证明与基本不等式的应用，属于中档题.。

2020届长郡中学高三第3次月考试卷-英语试卷及答案

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2019-2020学年长沙市长郡中学高三英语下学期期末试题及答案解析

2019-2020学年长沙市长郡中学高三英语下学期期末试题及答案解析第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AInformation on school visits to Kew GardensEnjoy yourselves in a wonderland of science with over 50,000 living plants and a variety of educational events or amusing activities. Here is essential information about planning a school visit to Kew.Educational course pricesYou can plan a self-led visit or book one of our educational courses. Students will take part in the educational courses in groups of 15. Prices vary according to different situations.EYFS (Early Years Foundation Stage) to Key Stage 4:45-minute course: 35/group 90-minute course: 70/groupKey Stage 5:Half day (one course): 80/group Full day (two courses): 160/groupTeachers and adults:Up to required key stage proportions (比例): FreeAdults needed for 1:1 special educational needs support: FreeAdults above the required proportions: 11/personThe payment will due within 28 calendar days of making the booking.Health and safetyRequired supervising (监护) adult-student proportions:Key Stage 1: 1:5 Key Stage 2: 1:8Key stage 3: 1:10 Key Stage 4: 1:12Key Stage 5: 1:12The group sizes should be controlled if you are visiting potentially busy areas such as the glasshouse and other attractions. The maximum number of students visiting the glasshouses is 15 per group and each group to Kew shops should include no more than 10 students.If there is an emergency, please contact the nearest Kew staff member or call Constabulary on 0208 32 3333 for direct and quick support. Please do not call 999.Planning your visitYour tickets and two planning passes will be sent to you upon receipt of your payment. You can complete your risk assessment with the passes, ensure you bring your tickets and the receipt document and show them to the staff members at the gate on the day of your visit.Recommended timingsThe Kew Gardens opens at 10 am. You are recommended to spend at least three to five hours on your visit. The closing time varies throughout the year. But the earliest is 3:30 pm. We have a fixed schedule for educational courses, which is from 10:30 am to 2:20 pm.1.How much should a group of 15 Key Stage I students and 4 teachers pay for a 45-minute course?A.35B.46C.57D.812.What should one do in an emergency?A.Check the risk assessment.B.Call 999 immediately.C.Ask adults or teachers for help.D.Seek help from the staff member nearby.3.What is the purpose of the text?A.To introduce Kew Gardens.B.To give tips on visiting Kew Gardens.C.To attract potential visitors to Kew Gardens.D.To inform coming activities in Kew Gardens.BCycling has long been a sport where a competitive edge is pursued, often at any cost. Cycling shorts, however, were an area that technology had yet to enter. Little was added to them apart from, perhaps, some padding to protect a rider.Now, however, new cycling shorts are being equipped with sensors to help riders defeat opponents easily. A British start-up has created a pair of shorts that uses electric sensors and artificial intelligence to shape and strengthen cyclists' muscles while they are moving. It plans to also adapt this product next year for joggers and people at the gym.The shorts, created by Devon Lewis, a PhD student at theUniversityofSouthampton, are able to send out tiny electric currents into a user's muscles to improve their cycling technique. Mr Lewis, 27, came up with the ideaafter studying a connection between the brain and muscles that sends chemical signals from motor neurons to muscle fibres to make the muscle contract and strengthen.The shorts, created by Mr Lewis's start-up bypass this connection using electric cur rents to build muscle strength directly, without relying on the brain to send out signals. “We have a limited ability to control our muscles naturally,” Mr Lewis said. “You can control them more exactly, get more from your muscles and contract them more strongly, if you stimulate them directly with electricity.” Electric wires in the shorts send out currents that activate the muscle. “For those worried about getting a shock while cycling,” Mr Lewis said, “It's sort of the same thing as when you have spicy food and you get that little rush. It feels strange when you first start using it but then you adapt to it quite quickly.”His idea has received support from Future Worlds, a start-up accelerator atSouthamptonUniversity. The product is being shown next week at the technology conference CES, held digitally this year.4. What's the strength of the newly invented cycling shorts ?A. They help cyclists perform better.B. They last longer than normal ones.C. They can keep a balanced comfort.D. They have tools to make good figures.5. Why can the cycling shorts help the runners and joggers?A. They are made of quality cotton materials.B. They are created by world-top designers.C. They can make running or jogging less boring.D. They are made intelligently with electric sensors.6. What may be some people's worry about the shorts?A. They may spend more than they can afford.B. They get little help from the new shorts.C. They may get shocked by the electricity.D. The new shorts can't function well enough.7.What's the best title of the text?A. The Special CyclingShortsB. Waysto Be a Good RunnerC. How Shorts Benefit YouD. Some Tips about Running FastCOver the years, NASA has successfully sent several rovers (飞行器) to Mars. While the science laboratories continue to provide important information on the Red Planet, they keep the space exploration rovers staying closeto the original landing place.To get a more comprehensive (详尽的) view of Mars, the US Space Agency plans to make a small test helicopter with the Mars 2020 Rover, which will be sent off in July 2020.The football-sized “marscopter” weighs about four pounds.It can fly at about ten times the speed of helicopters on Earth. The light weight and fast speed are important for the helicopter to be able to fly in the thin Mars atmosphere, which is about just one percent of that of Earth. “To make it fly in that thin Mars atmosphere, we had to make everything ready, and make it as light as possible while being as strong and powerful as it can possibly be. “Mimi Aung, Mars helicopter project manager, said in a meeting.Upon landing on the Red Planet, the Mars 2020 Rover will find a proper location to send of the helicopter. The helicopter will start with a short climb, no higher than 10 feet, and hover (盘旋) for just 30 seconds before landing back on the Mars surface. If all goes well, five more flights of longer distances, for 90 seconds each, will be conducted over the next 30 days. The short journeys will be recorded by a small camera and sent back to the scientists on Earth.If the plan goes on well, the first helicopter to fly in another world, will open up a whole new way to explore Mars. The Mars helicopter's first flight will be a great invention. For those of us whose research is about flight, that would be a wonderful, historic moment.8. To fly in thin Mars atmosphere，we should make the helicopter_______.A. strong and heavyB. heavy and fastC. light and fastD. strong and slow9. How many times will the helicopter fly?A. Three timesB. Four timesC. Five timesD. Six times10. What can we learn from the last paragraph?A. It’s interesting to fly in another world.B. I' s important to fly in another world.C. The plan to fly the helicopter goes on well.D. The helicopters first flight is successful.11. What may be the best title for the passage?A. To send a helicopter to Mars in 2020B. To send a rover to Mars in 2020C. To live in the thin Mars atmosphereD. To send back records to scientistsDIt all happened one afternoon in Carl's backyard a few years ago. We had just finished playing stickball, and I was about to go home.“Wait a minute,” Carl yelled. He ran into his house and cameback with a book for me to take home and read.All he said was, “See if you like it.”I said ly nothing. I kept the book for a couple of weeks and then returned it unread. Carl never asked me if I liked it or not. During the following two years Carl lent me three more books. Each time I returned them unread.A few days after graduating from high school, Carl asked, “Benny, which college are you going to?”“I'm not going to college.” I said.“Why not?” he asked.“Because my father can't afford the tuition (学费).” I answered.“Is that it?” Carl asked. “Yes,” I said.I lied. I had no intention of going back to school now that I was out. The following day, Carl knocked on my door and handed me a check for seventy-five dollars from his father along with the bookMartin Eden.“I think that should do it.” he said.Once again I was in shock. I was working full-time in my brother's bakery. I attended two classes atWayneUniversitypart-time at night. Halfway through the semester, after receiving failing grades on exams and essays, I decided there was no way I would ever become a good student and get satisfactory grades. I dropped out of college.One day, curious, I picked up the book, thinking Carl was trying to tell me something. Despite difficulty, I pressed on. By the time I finished the book, I understood why: the main character, Martin Eden, had my own poor educational background, but managed to educate himself and become a published author.12. What do we know about the author?A. He often told lies.B. He quit school unwillingly.C. He had thought little of education before.D. He became a published author.13. What kind of person is Carl?A. Caring.B. Emotional.C. Stubborn.D. Cautious.14. What message does Carl want to convey?A. Reading makes a rich man.B. Reading is the journey of the soul.C. Reading makes a person better known.D.Readingopens up new opportunities.15. What is the best title for the text?A. A strong desire for collegeB. A wish for better educationC. A wise friendD. A wise book第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2020届湖南省长沙市长郡中学高三下学期三月教学质量监测理综物理试题(解析版)

2020届湖南省长沙市长郡中学高三下学期三月教学质量监测理综物理试题（解析版）第Ⅰ卷(选择题共48分)一、选择题:本大题共8小题，每小题6分。

在每小题给出的四个选项中，第1~4题只有一项符合题目要求,第5~8题有多项符合题目要求。

全部选对的得6分,选对但不全的得3分,有选错的得0分。

1.甲、乙两物体在同一直线上运动，其位移一时间图象如图所示，由图象可知A. 甲比乙运动得快B. 乙开始运动时，两物体相距20mC. 在前25s 内，两物体距离先增大后减小D. 在前25s 内，两物体位移大小相等【答案】C 【解析】【详解】A ．位移时间图象的斜率等于速度，斜率越大，速度越大，甲比乙运动得慢，故A 错误； B ．由图知：乙从原点出发，乙开始运动时，甲的位置坐标大于20m ，则两物体相距大于20m ，故B 错误； C ．在0~10s 这段时间内，乙静止在原点，甲沿正向做匀速直线运动，则两物体间的距离逐渐增大，在10~25s 这段时间内，甲的运动速率小于乙的运动速率，甲在乙的前方，则两者距离逐渐减小，故C 正确； D ．在前25s 内，甲的位移为1402020x m m =-=，乙的位移为140040x m m =-=，故D 错误；故选C ．2.图甲是上官同学站在力传感器上做下蹲、起跳动作的示意图，中间的O 表示人的重心。

图乙是根据传感器采集到的数据画出的F -t 图线，两图中a ~g 各点均对应，其中有几个点在图甲中没有画出。

取重力加速度g ＝10m/s 2，根据图象分析可知（）A. 人的重力为1500N B. c点位置人处于失重状态C. e点位置人处于超重状态D. d点的加速度小于f点的加速度【答案】C 【解析】【详解】A．开始时人处于平衡状态，人对传感器的压力是500N，根据牛顿力平衡可知，人的重力也是500N。

故A错误；B．c点时人对传感器的压力大于其重力，处于超重状态。

故B错误；C．e点时人对传感器的压力大于其重力，处于超重状态。

长郡中学2020届高三英语下学期第3次月考试卷(含答案)

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2019-2020学年长沙市长郡中学高三英语月考试卷及答案解析

2019-2020学年长沙市长郡中学高三英语月考试卷及答案解析第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AIt's time to put your two-wheeler to good use on these fun bike paths — each varying in distance and difficulty. Just choose one fit for you.• Paulinskill Valley TrailCheck out Paulinskill Valley Trail, filled with forests, wetlands, and small townsalong this 25- mile route. While the trail is mostly flat, you can do a quick ride. You're likely to catch sight of beautiful birds, considering more than 100 species find a home in the land near the path. Check out the trail in the fall - prime time for pretty sights.•AlaHele MakalaeOn theislandofKaui, you'll find a bike path with lots of beachy views that'll leave you feeling accomplished yet calm. The name translates to “The Path that Goes By Coast,” and, as you might guess, the seven-mile path hugs the shoreline. Start early enough and you'll witness an incredible sunrise to make it even more amazing,• TheCheaha RouteThis ride covers up to 126 miles, with steep climbs and extreme downhills along the way. So prepare for a thrilling ride - one that’s not necessarily for the inexperienced or those looking for an easy, casual ride. Along the route, you'll pedal through five towns. The journey is worth it, though, because you get some of the most scenic views in the state.• The Whitefish TrailFamous for its countless route options, whether you’re a new biker looking for smooth tracks or you have more experience and want to play around on rocky, more technical land, this bike path brings in lots of visitors. The 43-mile route offers beautiful green scenery. Around every comer, you’ll see a new jaw-dropping landscape, from glassy lakes to green mountains.1. When is the best time to visit Paulinskill Valley Trail?A. In Autumn.B. All year round.C. On early mornings.D. On sunny days.2. Which route is not fit for someone new to cycling?A. The Whitefish Trail.B. Ala Hele Makalae.C. Paulinskill Valley Trail.D. TheCheaha Route.3. What's the common feature of the four routes?A. They are full of challenges.B. Various route options are offered.C. They have beautiful scenery.D. Different species can be found there.BWhen you walk on a sandy beach, it takes more energy than striding down a sidewalk — because the weight of your body pushes into the sand. Turns out, the same thing is true for vehicles driving on roads. The weight of the vehicles creates a very shallow indentation (凹陷) in the pavement (路面) — and it makes it such that it’s continuously driving up a very shallow hill.Jeremy Gregory, a sustainability scientist at M.I.T. and histeam modeled how much energy could be saved — and green-house gases avoided — by simply stiffening (硬化) the nation’s roads and highways. And they found that stiffening 10 percent of the nation’s roads every year could prevent 440 megatons of carbon dioxide equivalent emissions over the next five decades — enough to offset half a percent of projected transportation sector emissions over that time period. To put those emissions savings into context — that amount is equivalent to how much CO2 you’d spare the planet by keeping a billion barrels of oil in the ground — or by growing seven billion trees — for a decade.The results are in the Transportation Research Record.As for how to stiffen roads? Gregory says you could mix small amounts of synthetic fibers orcarbon nanotubes into paving materials. Or you could pave with cement-based concrete, which is stiffer than asphalt (沥青).This system could also be a way to shave carbon emissions without some of the usual hurdles. Usually, when it comes to reducing emissions in the transportation sector, you’re talking about changing policies related to vehicles and also driver behavior, which involves millions and millions of people — as opposed to changing the way we design and maintain our pavements. That’s just on the order of thousands of people who are working in transportation agencies. And when it comes to retrofitting (翻新) our streets and highways —those agencies are where the rubber meets the road.4. Why does the author mention “walk on a sandy beach” in paragraph 1?A. To present a fact.B. To make a contrast.C. To explain a rule.D. To share an experience.5. What suggestion does the author give to reduce CO2 emissions?A. Hardening the road.B. Keeping oil in the ground.C. Growing trees for decades.D. Improving the transportation.6. What is the advantage of this suggestion?A. Gaining more support.B. Consuming less money.C. Involving more people.D. Facing fewer usual obstacles.7. What does the underlined part mean in the last paragraph?A.Those agencies are likely to make more rules.B. Those agencies will change some related policies.C. Those agenciesmight put more rubber tires on the roads.D. Those agencies will play a key role in making this happen.CScientists at the Massachusetts Institute of Technology (麻省理工学院) have turned spider webs into music——creating an strange soundtrack that could help them better understand how the spiders output their complex creations and even how they communicate.The MIT team worked with Berlin-based artist Tomas Saraceno to take 2D (two-dimensional) laser (激光) scans of a spider web, which were linked together and made into a mathematical model that could recreate the web in 3Din VR (virtual reality). They also worked with MIT’s music department to create the virtual instrument.“Even though the web looks really random (随机)，there actually are a lot of inside structures and you can visualize (可视化) them and you can look at them, but it’s really hard to grasp for the human imagination or human brain to understand all these structural details,” said MIT engineering professor Markus Buehler, who presented the work on Monday at a virtual meeting of the American Chemical Society.Listening to the music while moving through the VR spider web lets you see and hear these structural changes and gives a better idea of how spiders see the world, he told CNN. “Spiders use vibrations (振动) as a way to locate themselves, to communicate with other spiders and so the idea of thinking really like a spider would experience the world was something that was very important to us as spider material scientists,” Buehler said.Spiders are able to build their webs without shelves or supports, so having a better idea of how they work could lead to the development of advanced new 3D printing techniques. “The reason why I did that is I wanted to be able to get information really from the spider world, which is very weird and mysterious,” Buehler explained. In addition to the scientific value, Buehler said the webs are musically interesting and that you can hear the sounds the spider creates during construction. “It’s unusual and eerie and scary, but finally beautiful.” he described.8. What have MIT scientists done according to the passage?A. They have translated spider webs into sounds.B. They have made a mathematical model to produce webs.C. They have created a soundtrack to catch spiders.D. They have known how spiders communicate.9. What can we know about spider webs from paragraph 3?A. Their structures are beautiful and clear.B. Professor Markus Buehler knows them well.C. The American Chemical Society presents the result.D. They are complex for people to figure it out.10. In which field will the study be helpful?A. virtual realityB. printingC. paintingD. film-making11. What is the main idea of the passage?A. It tells us that the music created by spiders is scary.B. It shows how the researchers carry out the experiment.C. It presents a new and creative way to study spiders.D. It explains why scientists did the experiment.DJanet Fein, aged 84, received her bachelor's degree from the University last week, having waited a long time for her chance to reach that goal.Growing up in the Bronx area ofNew York City, Fein worked at a dress manufacturer after graduating early at the age of 16. After getting married, she spent 18 years staying home with her children. She held several jobs through her life, including 20 years as a secretary at a hospital until her retirement at age 77.Fein has had a full life. But even then, she was not ready to take it easy and rest during a well-earned retirement. She decided to major in sociology because she felt it was “substantial.”Fein took part in a state program that lets people who are 65 and older take free classes at public universities in Texasand kept going to class even as her health conditions worsened. During her studies, it became necessary for her to use a walker to get around and she required oxygen. She also developed knee problems. So, Fein took online classes to finish the last part of her degree requirements.Sheila Rollerson, Fein' scollege advisor, and Carol, the professor told the media that Fein never showed signs of giving up even with all of her difficulties. She would speak up a lot in class and it just made for a more interesting class.Fein has also inspired Renee Brown, one of Fein's caregivers. At 53, Renee plans to begin nursing school to further her career. “Renee, you can do it. If I can do it you can do it, and you will feel so good about it,” Brown remembers what Fein told her.12. What is the second paragraph mainly about?A. Janet Fein's growthB. Janet Fein's aimC. Janet Fein's job experiencesD. Janet Fein's regrets13. What does the underlined phrase “substantial” in paragraph 3 refer to?A. InterestingB. ValuableC. EasyD. Affordable14. What can be learned about Fein according to the passage?A. Fein lived on little money after her retirement.B. Fein had a problem with her arms while at universities.C. The Internet played a role in Fein's university education.D. The good health helped Fein get her bachelor's degree.15. What is Janet Fein like according to the passage?A. Hardworking and humorous.B. Determined and generous.C. Positive and patient.D. Inspiring and perseverant.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2019-2020学年长沙市长郡中学高三英语月考试卷及答案解析

2019-2020学年长沙市长郡中学高三英语月考试卷及答案解析第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AElectric Shocks Can Be FatalGovernment statistics recently showed that in theUK, more than 3,000 people a year experience electric shocks in their homes. A smaller number of people are killed after they touch the power lines outside their homes. Electric shocks can cause a person's heart or breath to stop and are potentially fatal. It is essential for people to learn basic techniques to deal with such emergencies.What to do?● If you are the first person to reach someone who has an electric shock, don't touch him or her!● If the victim is still holding the appliance that gives him or her the shock (e.g. a hair dryer), unplug it or turn off the power at its source.Under no circumstances can you try to move the appliance with your hands!● Ifyou can't turn off the power, use a piece of wood, like a broom handle or a chair, to separate the victim from the appliance or the power source. You may even be able to do this with a folded newspaper.● The victim must remain lying down. If he or she isunconscious, the victim should be placed on his or her side. But he or she should not be moved if there is a possibility of neck or spine injuries unless it is ly necessary.● It is essential to maintain the victim's body heat, so make sure you cover him or her with a blanket before you do anything else. If the victim is not breathing, apply mouth-to-mouth resuscitation (人工呼吸). Keep the victim's head low until professional help arrives.● If the electric shock has been caused by an external power line, the dangers to the victim and to anybody providing first aid are much greater.1. What kind of passage is it?A. An advertisement.B. A horror story.C. A news report.D. First aid emergency advice.2. The underlined sentence implies that ________.A. you should move the applianceB. you should pick up the appliance and turn off the electricityC. it is very dangerous to touch the appliance with your handsD. it is unnecessary to unplug the appliance with your hands3. When a person has got an electric shock, you should ________.A. separate the victim from the appliance and let him sit upB. keep the victim warm and help him or her breathe againC. move the victim onto his or her side if he or she has got neck injuriesD. keep the victim's head high until professional help arrivesBLight pollution is a significant but overlooked driver of the rapid decline of insect populations, according to the most comprehensive review of the scientific evidence to date.Artificial light at night can affect every aspect of insects' lives, the researchers said. "We strongly believe artificial light at night — in combination with habitat loss, chemical pollution.invasive (入侵的) species, and climate change — is driving insect declines, " the scientists concluded after assessing more than 150 studies.Insect population collapses have been reported around the world, and the first global scientific review published in February,said widespread declines threatened to cause a "catastrophic collapse of nature's ecosystems".There are thought to be millions of insect species, most still unknown to science, and about half are active at night. Those active in the day may also be disturbed by light at night when they are at rest.The most familiar impact of light pollution is moths (飞蛾) flapping around a bulb, mistaking it for the moon. Some insects use the polarisation of light to find the water they need to breed, as light waves line up after reflecting from a smooth surface. But artificial light can scupper (使泡汤) this. Insects areimportant prey (猎物) for many species, but light pollution can tip the balance in favour of the predator if it traps insects around lights. Such increases in predation risk were likely to cause the rapid extinction of affected species, the researchers said.The researchers said most human-caused threats to insects have analogues in nature, such as climate change and invasive species. But light pollution is particularly hard for insects to deal with.However, unlike other drivers of decline, light pollution is ly easy to prevent. Simply turning off lights that are not needed is the most obvious action, he said, while making lights motion-activated also cuts light pollution. Shading lights so only the area needed is lit up is important. It is the same with avoiding blue-white lights, which interfere with daily rhythms. LED lights also offer hope as they can be easily tuned to avoid harmful colours and flicker rates.4. What is discussed in the passage?A. Causes of declining insect populations.B. Consequences of insect population collapses.C. Light pollution: the key bringer of insect declines.D. Insect declines: the driver of the collapsed ecosystem.5. What is the 5th paragraph mainly about?A. How light travels in space.B. How light helps insects find food.C. How the food chain is interrelated.D. How light pollution affects insects.6. What does the underlined word"analogues"in Paragraph 6probably mean?A. Selective things.B. Similar things.C. Variations.D. Limitations.7. What is the purpose of the last paragraph?A. To offer solutions.B. To give examples.C. To make comparisons.D. To present arguments.CIt is a question people have been asking for ages. Is there a way to turn back the aging process?For centuries, people have been looking for a “fountain of youth”. The idea is that if you find a magical fountain, and drink from its waters, you will not age.Researchers in New York did not find an actual fountain of youth, but they may have found a way to turn back the aging process. It appears the answer may be hidden right between your eyes, in an area called the hypothalamus (下丘脑). The hypothalamus is part of your brain. It controls important activities within the body.Researchers at New York’s Albert Einstein College of Medicine found that hypothalamus neural (神经的) stem cells also influence how fast aging takes place in the body.What are stem cells(干细胞)? They are simple cells that can develop into specialized cells, like blood or skin cells. Stem cells can also repair damaged tissues and organs.Dongsheng Cai is a professor at the Albert Einstein College of Medicine. He was the lead researcher in a study on aging in mice. He and his team reported their findings in the journal Nature, Cai explained when hypothalamus function is in decline, particularly the loss of hypothalamus stem cells, and this protection against the agingdevelopment is lost. it eventually leads to aging.Using this information, the researchers began trying to activate, or energize, the hypothalamus laboratory mice. They did this by injecting the animals with stem cells, Later, the researchers examined tissues and tested for changes in behavior. They looked for changes in the strength and coordination (协调) of the animals muscles. They also studied the social behavior and cognitive ability of the mice. The researchers say the results show that the treatment slowed aging in the animals, Cai says injecting middle-aged mice with stem cells from younger mice helped the older animals live longer.But these results were just from studying mice in a laboratory. If the mice can live longer, does that mean people could have longer lives? The next step is to see if the anti-aging effects also work in.8. In Paragraph 2 a “fountain of youth” is mentioned to ________.A. introduce the main topicB. show a hidden secret.C. describe scientists researchD. recommend a way to stay young9. Aging takes place in the body when _______.A. stem cells develop into specialized cellsB. there are important activities within the bodyC. hypothalamus neural stem cells fail to protect against agingD. the hypothalamus fails to repair damaged tissues and organs10. What do we know about the researchers at Albert Einstein College of Medicine from the text?A. They did experiments to see how stem cells work.B. They studied mice to find their connection with humans.C. They have found a possible way to slow the aging progress.D. They have found no changes in mice s behavior during the experiment11. What will the researchers probably do next?A. They will help some animals live longer.B. They will announce the fountain of youth doesn’t existC. They will develop products to help people live a longer life immediatelyD. They will do research to see if what they have found in mice will apply to humans.DEver wondered if dogs can learn new words? Yes, say researchers as they have found that talented dogs may have the ability to grasp new words after hearing them only four times.While previous evidence seems to show that most dogs do not learn words, unless eventually very well trained, a few individuals have shown some extraordinary abilities, according to a study published in the journal Scientific Reports.“We wanted to know under which conditions the gifted dogs may learn novel words” said researcher xuekw Claudia Fugazza from theEötvösLorándUniversityinHungary. For the study, the team involved two gifted dogs, Whisky and Vicky Nina. The team exposed the dogs to the new words in two different conditions.In the exclusion-based task, presented with seven known toys and one new toy, the dogs were able to select the new toy when presented with a new name. Researchers say this proves that dogs can choose by exclusion when faced with a new word, they selected the only toy which did not have a known name.However, this was not the way they would learn the name of the toy. In fact, when they were presented with one more equally new name to test their ability to recognize the toy by its name, the dogs got totally confused and failed.The other condition, the social one, where the dogs played with their owners who pronounced the name of the toy while playing with the dog, proved to be the successful way to learn the name of the toy, even after hearing it only 4 times. “The rapid learning that we observed seems to equal children’s ability to learn many new words at a fast rate around the age of 18 months,” Fugazza says. “But we do not know whether the learning mechanisms(机制) behind this learning are the same for humans and dogs. ”To test whether most dogs would learn words this way, 20 other dogs were tested in the same condition, but none of them showed any evidence of learning the toy names, confirming that the abilityto learn words rapidly in the absence of formal training is very rare and is only present in a few gifted dogs.12. What was the purpose of the study published in Scientific Reports?A. To better train dogs’ ability to learn new words.B. To further confirm previous evidence about dogs.C. To prove extraordinary memory abilities of gifted dogs.D. To explore favorable conditions for gifted dogs’ new-word learning.13. How did the dogs react when exposed to two new names in the first condition?A. Slow to understand.B. Quick to learn.C. At a loss.D. In a panic.14. What was found about dogs’ new-word learning in the social condition?A. Learning through playing applied to most dogs.B. The social condition helped dogs learn new words.CDogs’ new-word learning turned out to be less effective.D. Dogs shared similar learning mechanisms with children.15. Which of the following is the best title for the text?A. Gifted Dogs Can Learn New Words Rapidly.B. Dogs Identify Newly-named Toys by Exclusion.C. Dogs Can Acquire Vocabulary through Tons of Training.D. Gifted Dogs Have Similar Learning Abilities to Humans.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

湖南省长沙市长郡中学2019-2020学年高三上学期第二次月考英语试题(解析版)

长郡中学2020届高三月考试卷（二）英语试题注意事项：1.答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。

2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。

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第一节（共5小题;每小题1.5分，满分7. 5分）听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

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例：How much is the shirt?A.£ 19.15.B. £9.18.C. £9.15.答案是C.1.What will the man do?A.Tie his shoes.B.Rise from a fall.C.Leave his shoes untied.2.How will the woman probably feel after the conversation?A.Glad.B.Worried.C.Disappointed.3.Who pushed the woman into the pool?A.Jane's brother.B.The man's brother.C.Amy's brother.4.Why will the man need to find another present?A.The store is closed.B.The store is too far away.C.The jacket is too expensive.5.Where are the speakers?A.At a concert.B.At Kate's house.C.At a restaurant.第二节（共15小题;每小题L 5分，满分22.5分）听下面5段对话或独白。

2020届湖南省长沙市长郡中学高三下学期3月检测数学（文）试题

2020届湖南省长沙市长郡中学高三下学期3月“停课不停学”阶段性检测数学（文）试题一、单选题1．设全集U =R ，{}(){}260,ln 1A x x x B x y x =--<==-，则()UAB =（）A ．[1，3)B ．(1，3]C ．(1，3)D ．(－2，1]【答案】A【解析】首先确定集合,A B 中的元素，然后由集合运算法则计算．【详解】由题意2{|60}{|23}A x x x x x =--<=-<<，{|10}{|1}B x x x x =->=<， ∴{|1}UB x x =≥，(){|13}[1,3)U A B x x =≤<=．故选：A. 【点睛】本题考查集合的运算，考查一元二次不等式的解法，掌握集合的运算定义是解题关键．本题还考查了对数型复合函数的定义域．需要掌握对数函数的性质． 2．设复数()1331iz i i+=++-（其中i 为虚数单位），则下列说法中正确的是( ) A ．它的实部为﹣3 B ．共轭复数34z i =+C ．它的模||5z =D ．在复平面对应的点的坐标为(3,4)-【答案】C【解析】利用复数代数形式的乘除运算化简，然后逐一核对四个选项得答案．【详解】解：∵()()()21(1)333334111i i z i i i i i i ++=++=++=+--+，∴z 的实部为3，34z i =-，5z ==，在复平面对应的点的坐标为（3，4）. 故选: C. 【点睛】本题考查复数代数形式的乘除运算，以及复数的基本概念、共轭复数、复数的模和复数的几何意义，是基础题．3．已知向量a ＝(1，0)，b ＝(-3，4)的夹角为θ，则sin2θ等于 ( ) A ．725-B ．725C ．2425-D ．2425【答案】C【解析】首先根据向量夹角公式求出cos θ的值，然后求出sin θ，最后根据二倍角正弦公式即可得出结果. 【详解】33cos 155a b a bθ⋅==-=-⨯⋅， ∵0θπ≤≤， ∴4sin 5θ==，24sin 22sin cos 25θθθ==-，故选C. 【点睛】本题主要考查了向量夹角的计算以及二倍角正弦公式的应用，属于中档题. 4．设3log 18a =，4log 24b =，342c =，则a 、b 、c 的大小关系是( ) A ．a b c << B ．a c b << C ．b c a << D ．c b a <<【答案】D【解析】容易判断出2a >，2b >，2c <，并且得出1862463344log 1log ,log 1log a b ==+==+，容易判断出43log 6log 6<，从而得出a ，b ，c 的大小关系．【详解】解：由题可知，3422c =<，1892416334422a log log b log log =>==>>，，又186246334411a log log b log log ==+==+，，∵4366116643log log log log ==，且66log 4log 30>>， ∴661143log log <， 43log 24log 18∴<，即：b a <，综上得：2c b a <<<，则c b a << 故选D. 【点睛】本题考查利用指数函数、对数函数的单调性来比较大小，还考查对数的运算和对数的换底公式． 5．函数()()23ln 1x f x x+=的大致图象是A ．B ．C ．D ．【答案】A【解析】利用函数的对称性及函数值的符号即可作出判断. 【详解】由题意可知函数()f x 为奇函数，可排除B 选项；当x 0<时，()0f x ＜，可排除D 选项；当x 1=时，()12f ln =，当x 3=时，ln10ln10(3),ln 22727f =>，即()()1?3f f ＞，可排除C 选项，故选：A 【点睛】本题考查了函数图象的判断，函数对称性的应用，属于中档题．6．等比数列{}n a 的前n 项和为S n ，且14a ，22a ，3a 成等差数列，若1a =1，则S 10=（） A ．512 B ．511 C ．1024 D ．1023【答案】D【解析】设等比数列的公比为q ，由题意，得2344a a =+，即244q q =+，解得2q，则1010101221102312S -==-=-；故选D.7．运行如图所示的程序框图，若输出的S 的值为101，则判断框中可以填( )A ．200?i >B ．201?i ≥C ．202?i >D ．203?i >【答案】C【解析】由已知中的程序语句可知：该程序的功能是利用循环结构计算并输出变量S 的值，模拟程序的运行过程，分析循环中各变量值的变化情况，可得答案．【详解】程序的功能是计算3571sin3sin5sin 7sin 2222S ππππ=⨯+⨯+⨯+⨯+=1357-+-+，而101150213579199201=+⨯=-+-++-+，2012203i =+=，故条件为202?i >，故选C. 【点睛】本题考查了程序框图的应用问题，解题时应模拟程序框图的运行过程，以便得出正确的结论，是基础题．8．将函数()2sin(2)16f x x π=--的图象向左平移6π个单位长度得到函数()g x 的图象，则下列说法正确的是 ( ) A ．函数()g x 的最小正周期是2π B ．函数()g x 的图象关于直线12x π=-对称C ．函数()g x 在,62ππ⎛⎫⎪⎝⎭上单调递减 D ．函数()g x 在0,6π⎛⎫⎪⎝⎭上的最大值是1 【答案】C【解析】求出函数的周期判断A 的正误；函数的对称轴判断B 的正误；函数的单调性判断C 的正误；函数的最值判断D 的正误；【详解】由题意知：()2sin(2)16g x x π=+-，最小正周期T 22ππ==，选项A 错误; 当12x π=-时，112g π⎛⎫-=- ⎪⎝⎭，即函数()g x 的图象关于点(,1)12π--对称，选项B 错误；当(,)62x ππ∈时，72(,)626x πππ+∈， ∴函数()g x 在,62ππ⎛⎫⎪⎝⎭上单调递减，选项C 正确; ∵函数()g x 在0,6π⎛⎫⎪⎝⎭上单调递增，()()16g x g π<=，即函数()g x 在0,6π⎛⎫⎪⎝⎭上没有最大值， ∴选项D 错误，故选C. 【点睛】本题考查三角函数的简单性质，最值、单调性、周期以及单调性，考查命题的真假的判断，属于中档题．9．小王因上班繁忙，来不及做午饭，所以叫了外卖.假设小王和外卖小哥都在12：00~12：10之间随机到达小王所居住的楼下，则小王在楼下等候外卖小哥的时间不超过5分钟的概率是（） A ．12B ．45C ．38D ．34【答案】C【解析】设出两人到达小王的时间，根据题意列出不等式组，利用几何概型计算公式进行求解即可. 【详解】设小王和外卖小哥到达小王所居住的楼下的时间分别为,x y ，以12：00点为开始算起，则有5x y y x ≤⎧⎨-≤⎩，在平面直角坐标系内，如图所示：图中阴影部分表示该不等式组的所表示的平面区域，所以小王在楼下等候外卖小哥的时间不超过5分钟的概率为：11101010105532210108P. 故选：C 【点睛】本题考查了几何概型中的面积型公式，考查了不等式组表示的平面区域，考查了数学运算能力.10．已知长方体1111ABCD A B C D -各个顶点都在球面上，8AB AD ==，16AA =，过棱AB 作该球的截面，则当截面面积最小时，球心到截面的距离为( ) A ．3 B ．4C ．5D ．6【答案】C【解析】过棱AB 作该球的截面，则当截面面积最小时，截面的直径为8AB =，求出球的半径，可得球心到截面的距离．【详解】解：过棱AB 作该球的截面，则当截面面积最小时，截面的直径为8AB =，长方体1111ABCD A B C D -各个顶点都在球面上，8AB AD ==，16AA =，∴球的半径为2221886412++=， ∴球心到截面的距离为41165-=．故选：C. 【点睛】本题考查求球心到截面的距离，考查学生的计算能力，确定当截面面积最小时，截面的直径为8AB =是关键，是基础题．11．已知()f x 是偶函数,当0x ≥时,202()822x x f x x x ⎧≤<=⎨-≥⎩，，,若(1)(1)f a f -<-,则a的取值范围是( )A ．()11-，B ．()()2024-，，C ．()()()3113-∞--+∞，，， D ．()(2)02(4)-∞-+∞，，，【答案】D【解析】因为()f x 是偶函数,当0x ≥时,202()822x x f x x x ⎧≤<=⎨-≥⎩，，,根据偶函数图像关于y ,画出其函数图像,即可求解a 的取值范围.【详解】根据()f x 是偶函数(1)(1)2f f -==∴ (1)(1)f a f -<-,即(1)2f a -<因为()f x 是偶函数,当0x ≥时,202()822x x f x x x ⎧≤<=⎨-≥⎩，，,根据偶函数图像关于y 轴对称，,画出其函数图像,如图:当02x ≤<时,由()2f x =,解得11x =,根据对称可知,21x =-当2x ≥时,由()2f x =,解得33x =,根据对称可知,43x =- 保证(1)2f a -<,结合图像可知:∴ 41a x -<或211x a x <-<或41a x ->即: 13a -<-或111a -<-<或13a -> 故: 2a <-或02a <<或4a > 故选:D. 【点睛】本题考查了求解函数不等式,解题关键是根据函数的对称性画出其函数图像,数形结合,考查了分析能力和计算能力,属于中等题.12．已知双曲线C ：22221x y a b-=（0,0a b >>）的左、右焦点分别为12,F F ，过1F 的直线l 与双曲线C 的左支交于A 、B 两点.若22,120=∠=AB AF BAF ，则双曲线C 的渐近线方程为（）A ．y x =B ．y =C ．=±y xD ．)1=±y x【答案】D【解析】设2AF m =，利用余弦定理，结合双曲线的定义进行求解即可. 【详解】设22,3AB AF m BF m ==∴==，由双曲线的定义可知：12,AF m a =-因此12,BF a =再由双曲线的定义可知：1223BF BF a m -=⇒=，在三角形12AF F 中，由余弦定理可知： 222212222222112cos120(5(5F F AF AF AF AF c a a b a ︒=+-⋅⋅⇒=-⇒+=-2222(4(41b bb a a a⇒=-⇒=-⇒=，因此双曲线的渐近线方程为：)1=±y x .故选：D【点睛】本题考查了双曲线的定义的应用，考查了余弦定理的应用，考查了双曲线的渐近线方程，考查了数学运算能力.二、填空题13．为了解某市居民用水情况，通过抽样，获得了100位居民某年的月均用水量（单位：吨），将该数据按照[0，0.5），[0.5，1），…[4.4.5]分成9组，绘制了如图所示的频率分布直方图，政府要试行居民用水定额管理，制定了一个用水量标准a，使85%的居民用水量不超过a（假设a为整数），按平价收水费，超出a的部分按议价收费，则a的最小值为_____.【答案】3吨【解析】求出[0，0.5)的频数为4，[0.5，1)的频数为8，[1.5，2)的频数为22，[2，2.5)的频数为25，[2.5，3)的频数为4，[3，3.5)的频数为4，[4，4.5)的频数为2，+++++=，由此能求出a的最小值．481522151486【详解】⨯⨯=，解：[0，0.5)的频数为0.080.51004[0.5，1)的频数为0.160.51008⨯⨯=，[1.5，2)的频数为0.440.510022⨯⨯=，[2，2.5)的频数为0.50.510025⨯⨯=，[2.5，3)的频数为0.280.51004⨯⨯=．[3，3.5)的频数为0.080.51004⨯⨯=，[4，4.5)的频数为0.040.51002⨯⨯=，481522151486+++++=，∴前6组占86%，a 为3吨．故答案为：3吨. 【点睛】本题考查使85%的居民用水量标准的最小值的求法，以及频率分布直方图等基础知识，还考查推理能力与计算能力，属于基础题．14．已知抛物线2:2(0)C y px p =>的焦点F ，点00()2p M x x >是抛物线上一点，以M 为圆心的圆与直线2p x =交于A 、B 两点(A 在B 的上方），若5sin 7MFA ∠=，则抛物线C 的方程为 _____. 【答案】212y x =【解析】依题意作图，可以把5sin 7MFA ∠=放在直角三角形中，可得5sin 7MC MFA MF ∠==，由抛物线定义转化MF MD =，02p MC x =-，即可得到0x 与p 的关系，再代入方程中即可求出p ，则抛物线方程可求．【详解】解：如图所示，过M 点作CM ⊥抛物线的准线，垂足为C ，交准线于D ， 5sin 7MCMFA MF∴∠==，由抛物线定义可得：MF MD =，∴005272px MC p MF x -==+，即00575722x p x p +=+， 03x p ∴=，点00()2pM x x >是抛物线上一点，∴202px =，即23666p ⨯=，6p ∴=，得212y x =．故答案为：212y x =．【点睛】本题考查抛物线的标准方程，利用抛物线的定义进行线段的转化是关键，是中档题． 15．已知等差数列{}n a 的前n 项和为n S ，且43a =-，1224S =，若0(i j a a i +=，j ∈*N 且1)i j <，则i 的取值集合是 _____. 【答案】{1，2，3，4，5，6，7，8，9，10}【解析】利用等差数列前n 项和公式和通项公式，求出19a =-，2d =，从而9(1)2211n a n n =-+-⨯=-，由此能求出i 的取值集合．【详解】解：等差数列{}n a 的前n 项和为n S ，且43a =-，1224S =，∴4112133121112242a a d S a d =+=-⎧⎪⎨⨯=+=⎪⎩，解得19a =-，2d =， 9(1)2211n a n n ∴=-+-⨯=-，0(i j a a i +=，j ∈*N 且1)i j <，1i ∴=时，92110j -+-=，解得10j =；2i =时，4112110j -+-=，解得9j =； 3i =时，6112110j -+-=，解得8j =； 4i =时，8112110j -+-=，解得7j =； 5i =时，10112110j -+-=，解得6j =； 6i =时，12112110j -+-=，解得5j =；7i =时，14112110j -+-=，解得4j =； 8i =时，16112110j -+-=，解得3j =； 9i =时，18112110j -+-=，解得2j =， 10i =时，20112110j -+-=，解得1j =．综上，i 的取值集合为{1，2，3，4，5，6，7，8，9，10}．故答案为：{1，2，3，4，5，6，7，8，9，10}. 【点睛】本题考查数列的项数的取值集合的求法，运用到等差数列的通项公式和求和公式，还考查推理能力与计算能力，属于基础题．16．如图，已知在长方体1111ABCD A B C D -中，3AB =，4=AD ，15AA =，点E 为1CC 上的一个动点，平面1BED 与棱1AA 交于点F ，给出下列命题：①四棱锥11B BED F -的体积为20；②存在唯一的点E ，使截面四边形1BED F 的周长取得最小值274；③当E 点不与C ，1C 重合时，在棱AD 上均存在点G ，使得//CG 平面1BED ； ④存在唯一的点E ，使得1B D ⊥平面1BED ，且165CE =．其中正确的命题是_____（填写所有正确的序号）【答案】①②③④【解析】由题意逐个讨论所给的命题，判断它们的真假．【详解】解：①由题意可得1//D F BE ，1111111111B BED F B BED B BFD D BEB D BFB V V V V V -----=+=+ 111111111[](543543)2032232BB BC AB BB D A AB =+=⨯⨯+⨯⨯=，所以①正确； ②将长方体展开，如图所示，恰好过B 点时，截面的周长为12BD ，而在1BDD ∆中，2215(34)74BD ++，所以最小值为274由面面平行的性质，可得四边形1BED F 为平行四边形，且E 为展开图中唯一的点，所以②正确；③E 点不与C ，1C 重合，则F 不会为A ，即CG 不在面1EBD 内，可作出CG 的平面与1EBD 平行，所以在棱AD 上均有相应的G ，使得//CG 面1EBD ，故③正确；④因为1BB BD =，可得对角面11BB D D 为正方形，可得11B D BD ⊥，若1BE B C ⊥时，由三垂线定理可得1B D BE ⊥，即有1B D ⊥面1EBD ，矩形11BB C C 中，1BE B C ⊥，所以1CE BC BCCC =，所以1165BC BC CE CC ==，故④正确综上可得：正确为①②③④. 故答案为：①②③④. 【点睛】本题考查了正方体的性质、线面平行和垂直的判定定理、棱锥的体积、简易逻辑的判定方法，考查了推理能力与计算能力，属于中档题三、解答题17．如图，在四边形ABCD 中，45ADB ∠=，105BAD ∠=，6AD =，2BC =，3AC =．()1求边AB 的长及cos ABC ∠的值；()2若记ABC ∠α=，求23sin πα⎛⎫- ⎪⎝⎭的值．【答案】（1）3AB =3 6cos ABC ∠=-；（2）531112．【解析】()1由已知可求ABD ∠，ABD 中，由正弦定理可求AB ，ABC 中由余弦定理，2222AB BC AC cos ABC AB BC∠+-=⋅可求．()2由()1可得cos α，进而可求sin α，进而根据二倍角公式22sin sin cos ααα=，2221cos cos αα=-可求，然后根据两角差的余弦公式即可求解.【详解】() 1由题意，因为45ADB ∠=，105BAD ∠=，30ABD ∠∴=，6AD =2BC =，ABD 中，由正弦定理可得，24530ABsin sin =，AB ∴=， 3AC =．ABC 中由余弦定理可得，2222AB BC AC cos ABC AB BC ∠+-===⋅()2由()1可得cos α=sin α∴=226sin sin cos ααα∴==-，252216cos cos αα=-=-122232212sin sin cos πααα⎛⎫∴-=-=⎪⎝⎭．【点睛】本题主要考查了正弦定理、余弦定理和三角恒等变换的应用，其中在解有关三角形的题目时，要有意识地考虑用哪个定理更合适，要抓住能够利用某个定理的信息．一般地，如果式子中含有角的余弦或边的二次式时，要考虑用余弦定理；如果式子中含有角的正弦或边的一次式时，则考虑用正弦定理，着重考查了运算与求解能力，属于基础题． 18．某蛋糕店每天制作生日蛋糕若干个，每个生日蛋糕成本为50元，每个蛋糕的售价为100元，如果当天卖不完，剩余的蛋糕作垃圾处理.现搜集并整理了100天生日蛋糕的日需求量（单位：个），得到如图所示的柱状图.100天记录的各需求量的频率作为每天各需求量发生的概率.（1）若该蛋糕店某一天制作生日蛋糕17个，设当天的需求量为)(N n n ∈，则当天的利润y （单位：元）是多少？（2）若蛋糕店一天制作17个生日蛋糕.①求当天的利润y （单位：元）关于当天需求量n 的函数解析式； ②求当天的利润不低于600圆的概率.（3）若蛋糕店计划一天制作16个或17个生日蛋糕，请你以蛋糕店一天利润的平均值作为决策依据，应该制作16个还是17个生日蛋糕？【答案】（1）17n ≥时，利润为850；16n ≤时，利润为100850n -；（2）①100850(16)()850(17)n n Y n N n -≤⎧=∈⎨≥⎩，②2225；（3）17．【解析】试题分析：（1）分17n ≥、16n ≤分别求得相应利润；（2）①结合（1）即可求得函数解析式；②将问题转化为“需求量不低于15个”的概率，由此利用条件概率公式求解即可；（3）分别求得一天制作16个和17个的平均利润，由此作比较可得结论．试题解析：（1）当17n ≥时，17(10050)850Y =⨯-=，当16n ≤时，1001750100850Y n n =-⨯=-．（2）①由（1）得当天的利润Y 关于当天需求量n 的函数解析式为：100850(16)()850(17)n n Y n N n -≤⎧=∈⎨≥⎩②设“当天利润不低于600”为事件A ，由①知，“当天利润不低于600”等价于“需求量不低于15个”1222()110025P A ∴=-=，所以当天的利润不低于600元的概率为：2225．（3）若一天制作16个蛋糕，则平均利润为：11(600127001880070)758100x =⨯+⨯+⨯=；若一天制作17个蛋糕，则平均利润为：21(55012650187501885052)760100x =⨯+⨯+⨯+⨯=； 12x x <，∴蛋糕店一天应该制作17个生日蛋糕.【考点】1、频率分布直方图；2、分段函数；3、条件概率．19．在如图所示的五面体ABCDEF 中，四边形ABCD 为菱形，且60DAB ∠=︒，//EF 平面ABCD ，22EA ED AB EF ====，M 为BC 中点．（1）求证：//FM 平面BDE ；（2）若平面ADE ⊥平面ABCD ，求F 到平面BDE 的距离．【答案】（1）证明见解析；（215【解析】（1）取CD 中点N ，连接MN ，FN ，说明//MN BD ，证明//MN 平面BDE ，证明//EF AB ，//AB CD ，推出//EF CD ，//FN ED ，证明//FN 平面BDE ，转化证明//FM 平面BDE ；（2）说明F 到平面BDE 的距离等于M 到平面BDE 的距离，取AD 的中点H ，连接EH ，BH ，推出EH ⊥平面ABCD ，EH BH ⊥，设F 到平面BDE 的距离为h ，由E BDM M BDE V V --=，转化求解即可．【详解】解：（1）取CD 中点N ，连接MN ，FN ，因为N ，M 分别为CD ，BC 中点，所以//MN BD ，又BD ⊂平面BDE ，且MN ⊂平面BDE ，所以//MN 平面BDE ，因为//EF 平面ABCD ，EF ⊂平面ABEF ，平面ABCD平面ABEF AB =，所以//EF AB ，又222AB CD DN EF ====，//AB CD ，所以//EF CD ，EF DN =．所以四边形EFND 为平行四边形．所以//FN ED ．又ED ⊂平面BDE 且FN ⊂/平面BDE ，所以//FN 平面BDE ，又FNM N N =，所以平面//MFN 平面BDE ．又MF ⊂平面MFN ，所以//FM 平面BDE ．（2）由（1）得//FM 平面BDE ，所以F 到平面BDE 的距离等于M 到平面BDE 的距离，取AD 的中点H ，连接EH ，BH ，由四边形ABCD 为菱形，且60DAB ∠=︒，2EA ED AB EF ===，可得EH AD ⊥，BH AD ⊥，因为平面ADE ⊥平面ABCD ，平面ADE 平面ABCD AD =，所以EH ⊥平面ABCD ，EH BH ⊥，因为3EH BH ==，所以6BE =，所以22161562()22BDE S ∆=⨯⨯-=，设F 到平面BDE 的距离为h ，又因为1133422BDM BCD S S ∆∆==⨯⨯=，所以由E BDM M BDE V V --=，得13115333h ⨯⨯=⨯⨯，解得155h =．【点睛】本题考查直线与平面平行与垂直的判断与性质的应用，点、线、面距离的求法，等体积法的应用，考查空间想象能力以及计算能力． 20．已知函数21()2f x lnx ax x =--．（1）若函数()f x 在[1，)+∞上单调递增，求实数a 的取值范围；（2）若函数()f x 在1x =处的切线平行于x 轴，是否存在整数k ，使不等式[()1](2)x f x x k x +->-在1x >时恒成立？若存在，求出k 的最大值；若不存在，请说明理由．【答案】（1）a 14≤-；（2）不存在，理由见解析. 【解析】（1）对原函数求导，根据导数和函数的单调性的关系即可求出a 的取值范围；（2）问题转化为即(1)20xlnx k x k -++>在1x >时恒成立，令()(1)2g x xlnx k x k =-++，1x >求导后分0k 和0k >求函数的单调区间，进一步求得函数的最值得答案．【详解】解：（1）函数()f x 在[1，)+∞上单调递增， 1()10f x ax x∴'=-- 在[1，)+∞ 上恒成立， 2211111()24ax x x ∴-=--， ∴当2x =时，()211124x --有最小值14-，14a ∴-；（2）1()1f x ax x'=--， f ∴'（1）11a a =--=-，函数()f x 在1x =处的切线平行于x 轴，0a ∴=，()f x lnx x ∴=-，不等式[()1](2)x f x x k x +->-在1x >时恒成立， (2)xlnx x k x ∴->-在1x >时恒成立，即(1)20xlnx k x k -++>在1x >时恒成立，令()(1)2g x xlnx k x k =-++，1x >， ()g x lnx k ∴'=-，当0k 时，()0g x '>在(1,)+∞上恒成立，即()g x 在(1,)+∞上单调递增， ()g x g >（1）10k =->，则1k >，矛盾，当0k >时，令()0g x '=，解得k x e =，令()0g x '>，解得：k x e >，令()0g x '<，解得：1k x e <<，()g x ∴在(1,)k e 单调递减，在(k e ，)+∞单调递增，()()(1)220k k k k min g x g e ke k e k k e ∴==-++=->，令()2k h k k e =-，0k >，()2k h k e ∴'=-，当2k ln <时，()0h k '>，函数()h k 单调递增，当2k ln >时，()0h k '<，函数()h k 单调递减， ()(2)2222(21)0max h k h ln ln ln ∴==-=-<，∴不存在整数k 使得20k k e ->恒成立，综上所述不存在满足条件的整数k ．【点睛】本题考查利用导数研究函数的单调性和最值，导数的几何意义，还运用分离参数法和函数构造法解决恒成立问题，同时考查了数学转化思想方法以及推理能力和运算能力，属难题．21．已知斜率为k 的直线l 与椭圆22143x y C +=：交于A ，B 两点，线段AB 的中点为()()10M m m >，．（1）证明：12k <-；（2）设F 为C 的右焦点，P 为C 上一点,且0FP FA FB ++=．证明：FA ，FP ，FB 成等差数列，并求该数列的公差．【答案】（1）12k <-（2321或【解析】分析：（1）设而不求，利用点差法进行证明．（2）解出m,进而求出点P 的坐标，得到FP ,再由两点间距离公式表示出,FA FB ，得到直l 的方程，联立直线与椭圆方程由韦达定理进行求解．详解：（1）设()()1122,,,A x y B x y ，则222211221,14343x y x y +=+=.两式相减，并由1212y y k x x -=-得 1212043x x y y k +++⋅=. 由题设知12121,22x x y y m ++==，于是 34k m=-.① 由题设得302m <<，故12k <-.（2）由题意得()1,0F ，设()33,P x y ，则()()()()3311221,1,1,0,0x y x y x y -+-+-=.由（1）及题设得()()31231231,20x x x y y y m =-+==-+=-<. 又点P 在C 上，所以34m =，从而31,2P ⎛⎫- ⎪⎝⎭，32FP =.于是(122x FA x ===-. 同理222x FB =-. 所以()121432FA FB x x +=-+=. 故2FP FA FB =+，即,,FA FP FB 成等差数列. 设该数列的公差为d ，则()212112||2d FB FA x x x x =-=-=+②将34m =代入①得1k =-. 所以l 的方程为74y x =-+，代入C 的方程，并整理得2171404x x -+=.故121212,28x x x x +==，代入②解得d =. 所以该数列的公差为28或28-. 点睛：本题主要考查直线与椭圆的位置关系，等差数列的性质，第一问利用点差法，设而不求可减小计算量，第二问由已知得到0FP FM +=,求出m 得到直线方程很关键，考查了函数与方程的思想，考察学生的计算能力，难度较大．22．在平面直角坐标系xOy 中，曲线21240x y C x +-=：，直线l 的参数方程为cos sin x t y t αα=⎧⎨=⎩（t 为参数），其中0,6πα⎛⎫∈ ⎪⎝⎭，以坐标原点O 为极点，x 轴非负半轴为极轴，建立极坐标系.（1）求曲线1C 的极坐标方程和直线l 的普通方程；（2）设(4,0)M ，2C 的极坐标方程ρθ=，A ，B 分别为直线l 与曲线12,C C 异于原点的公共点，当30AMB ∠=︒时，求直线l 的斜率；【答案】（1）曲线的1C 极坐标方程为4cos ρθ=，直线l 的普通方程为tan ,0,6y x παα⎛⎫=⋅∈ ⎪⎝⎭（2）4【解析】(1)利用cos sin x y ρθρθ=⎧⎨=⎩将1C 的普通方程转化为极坐标方程，消去参数t 将直线l 的参数方程转化为普通方程； (2)根据题意求出||||OA OB 、及||AB ，又点M 在曲线1C上，则1||tan 4sin AM ραα==，由|||AM AB =列出方程即可得解. 【详解】（1）将cos sin x y ρθρθ=⎧⎨=⎩代入曲线1C 的普通方程得极坐标方程为4cos ρθ=，直线l 的普通方程为tan ,0,6y x παα⎛⎫=⋅∈ ⎪⎝⎭；（2）由已知可得θα=，则12||4cos ,||OA OB ραρα====，||4cos ,AB αα=-因为点M 在曲线1C 上且AM AB ⊥，所以1||tan 4sin ,AM ραα==在直角三角形ABM 中30AMB ∠=︒，则|||AM AB =所以4sin )ααα=-，得直线l 的斜率tan k α==【点睛】本题考查普通方程与极坐标方程的互化，参数方程化成普通方程，直线与圆的位置关系，直径所对的圆周角是直角，属于中档题.23．函数()223f x x x =-++（1）求不等式()25f x x ≥+的解集；（2）若()f x 的最小值为k ，且实数,,a b c 满足()a b c k +=，求证：22228a b c ++≥【答案】（1）(,0][4,)-∞⋃+∞（2）证明见解析【解析】(1)分类去绝对值符号后解不等式，最后取并集；(2)求出函数的最小值k ，根据基本不等式得出结论.【详解】（1）①当3x <-时，不等式即为3125x x --≥+，解得6,35x x ≤-∴<- ②当31x -≤≤时，不等式即为525x x -≥+，030x x ≤∴-≤≤③当1x >时，不等式即为3125x x +≥+，44x x ≥∴≥综上，()25f x x ≥+的解集为(,0][4,)-∞⋃+∞ （2）由51,3()5,3131,1x x f x x x x x --<-⎧⎪=--≤≤⎨⎪+>⎩∴当1x =时，()f x 取最小值4，即4,()4k a b c =∴+=，即4ab ac += ()()22222222228a b c a b a c ab ac ∴++=+++≥+=当且仅当a b c ===时等号成立【点睛】本题考查绝对值不等式的解法，不等式的证明与基本不等式的应用，属于中档题.。

2019-2020学年长沙市长郡中学高三英语第一次联考试卷及参考答案

2019-2020学年长沙市长郡中学高三英语第一次联考试卷及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AIt looks like 2017 is shaping up to be a record-breaking year in movie History. Here is a list of some of the year’s biggest blockbusters so far.Kong: Skull IslandA reboot (重启) of King Kong would normally get laughed at in this day and age, but it looks like this modem version of the story will be worth watching. With US actress Brie Larson and UK actor Tom Hiddleston in the mix, this film is set to be this year’s biggest monster tale.Release Date:3/10/17Beauty and the BeastDirector Bill Condon is bringing back a part of our childhood in live action, complete with the songs we all remember and love. With British stars Emma Watson and Dan Stevens leading thecast, the classic Walt Disney story already sounds like it’ll be a delight for both children and adults.Release Date:3/17/17The Fate of the FuriousThere was doubt that we’d even get a Fast 8, especially after the fitting ending US actor Paul Walker’s character was given at the end of Furious 7. Fans thought they’d never see Dom and the rest of the crew back in action, but thankfully, US star Vin Diesel himself confirmed that The Fate of the Furious is fueling up for another go.Release Date:4/14/17Spider-Man: HomecomingSpider Man is heading home to Marvel Studios for the first time. The movie will show us Peter Parker’s high school days, and will continue the threads we saw formed during his initial appearance in Captain America: Civil War. We know that Michael Keaton is playing The Vulture in this story, and that both Happy Hogan and Tony Stark, played by US actors Jon Favreau and Robert Downey Jr., are along for the ride.Release Date:7/7/171.Which of the following movies can’t you see on May Day?A.Kong: Skull Island.B.Beauty and the Beast.C.The Fate of the Furious.D.Spider-Man: Homecoming.2.What can we know about Beauty and the Beast?A.It is fueling up for another go.B.It is produced by Marvel StudiosC.It’ll show us Peter Parker’s school days.D.It’ll be enjoyable for both children and adults.3.What does the underlined word “cast” probably mean?A.Characters.B.Actors.C.Directors.D.Teachers.BHaley Curfman, 25, of Blackwell, Oklahoma, is a teacher at Blackwell Public School and last yearshe bought a plain(朴素的) white dress, which she set up a station in her classroom for her students to decorate, encouraging them to go and draw on the dress whenever they had free time. Haley set up a station at a table with the dress and some color1 ed markers so that her students could draw on it, having first seen the idea on Pinterest.After the kids had finished the design, she then surprised them all by wearing the dress to class, sharing pictures of herself in it on Facebook in a post that has since been shared over 200, 000 times.For teachers looking to do something similar for their own students, Haley said that she bought the dress off Amazon for less than $ 20 about six months ago and pre-washed the dress before she started the project. She used markers from Walmart. But she said these things can be bought in other places. She said, “To do the project, I set up a station at a table with the dress and markers. It takes anywhere from two weeks to a month to complete as we just work on it here and there when time allows. You'd better give the students enough time so they don't have to hurry.”“Teachers have been sharing their dresses, T-shirts, etc. with me that they've been creating since the 1950s with the same idea. It is amazing, and I love the fact that you are sharing them with me! Thank you all so much for your kindness and support,” she wrote on Facebook.Asked why she came up with the idea, Haley told Scary Mommy. “We don't have art in our school, so, I always try to do little creative projects when possible.”4. What did Haley use the plain white dress to do?A. To teach her students painting.B. To prepare for her presentation.C. To ask her students to draw on it.D. To help her students with their homework.5. What do we know about the finished dress?A. It took Haley by surprise.B. It is kept by Haley at home.C. It is the product of a new idea.D. It has enjoyed popularity online.6. What should a teacher do if he or she wants to follow Haley's example?A. Buy the same dress online.B. Give the kids enough time.C. Buy the same markers in Walmart.D. Leave the finished work untouched.7. Why did Haley carry out the project?A. To teach art to her students.B. To show her idea on Pinterest.C. To share it with other teachers.D. To exciteher students' creativity.CConcerns about health, animals and the environment are leading more people to remove meat from their diet. Plant-based meat alternatives (替代品) increasingly appear in supermarkets and restaurants. But what some people call clean meat-meat grown from cells in a laboratory—is still an idea that is just beginning.More than 24 companies are testing lab-grown fish, beef and chicken. These businesses hope to enter the alternative meat market, which could be worth $140 billion by 2029. One of the companies, Shiok Meats, grows its product by taking shrimp cells and keeping them at a fixed temperature. They are then given nutrients in a solution (溶液). The cells become meat in four to six weeks.This lab-grown meat’s price is high. One kilogram of it now costs $5,000, said Shiok Meats’ chief executive Sandhya Sriram. At that cost, a single pork and shrimp dumpling could be as much as $300. Sriram, avegetarian, hopes to cut the cost to $50 for one kilogram by the end of this year. “We are looking at next year, so we might be the first ever company to launch a cell-based meat product in the world,” Sriram said. “Shiok Meats still needs approval from the city’s food regulator, and that matters the most at present.”Although people increasingly demand meat alternatives, cell-based meat companies still faceresistance(抵制) to their products. In Singapore, some people said they would give lab grown meat a second thought. “I may not exactly dare to eat it, but I do find the idea appealing because the animals in the oceans are declining,” said 60-year-old Pet Loh, while sheshopped for shrimps in a Singapore market.Any alternative way of making animal protein without harming the environment is positive, said Paul Teng, aspecialist in agriculture technology at Nanyang Technological University. But, he added, more studies are needed to understand any negative result of making cellular protein.8. Why are more people eating less meat?A. Because lab-grown meat has more nutrition.B. Because plant-based food is getting popular.C. Because meat in the market is increasingly expensive.D. Because health and the environment are their concerns.9. What is the most important for Shiok Meats at present according to Sriram?A. Bringing down the price.B. Gaining consumers’ acceptance.C. Obtaining official permission.D. Getting ahead of other companies.10. Which word best describes Pet Loh’s attitude towards lab-grown meat?A. Doubtful.B. Uncaring.C. Negative.D. Positive.11. What is the text mainly about?A. A new way to make a fortune.B. Negative results of lab-grown meat.C. New research findings on healthy diet.D. A meat alternative grown in labs.DIt was the first day of spring here but it didn't feel like it. The COVID-19 epidemic (流行病) was sweeping the globe and everywhere there were feelings of fear and loneliness. “Social Distancing” had become the new norm. Here the schools had been closed, the restaurant dining rooms had been shut, and people had been told to work from home whenever they could. Even the sheltered workshop where my oldest son worked had been closed until further notice. People had made a run on the stores and large areas of the shelves were bare. On the news the numbers of the sick and dead continued to rise. It felt like there was a weight on the souls of everyone in the world.My family were staying at home as much as possible and as I looked out of my window I wondered how long this crisis would last. It was then, however, that I saw something that lifted that weight off of my soul, made me smile, and made my heart feel happy again. On the street below my house there was an old friend of mine from high school who was a teacher there. With him was my younger son's former aide from the high school as well. They were delivering the school lunches door to door to the hungry children who were stuck at home. Watching them made me think of something everyone's childhood television neighbor, Mr. Roger's once said: "In the badtimes, always look for the helpers."I have no doubt that this crisis will pass as all the crisis before. But it is our choice on whether it brings out the best in us or the worst in us. Let it bring out the best in you. Use it to strengthen your faith. Use it to free yourself from fear. Use it to grow kinder, more giving, and more loving. Become a helper to all those in need and you will be a happy person today and all the days to come.12. Why didn't the author like the first day of spring?A. He had to work at home instead of at office.B. There wasn't social distance between each other.C. There was an epidemic everywhere in the world.D. His own company had been closed for a long time.13. How did people feel in the face of the epidemic?A. Disappointed.B. TenseC. Confident.D. Cheerful.14. What made the author happy?A. Seeing someone delivering lunches from door to door.B. Talking with his old friend in the street.C. Watching a childhood television at home.D. Staying together with his family.15. What do Mr. Roger's words actually mean?A. It's not necessary for us to find helpers in bad times.B. In good times we needn't do much to help others.C. We should ask more people to help us in bad times.D. We are supposed to help each other in bad times.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2019-2020学年长沙市长郡中学高三英语下学期期末考试试卷及答案

2019-2020学年长沙市长郡中学高三英语下学期期末考试试卷及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项ASongkran, ThailandThis festival marks the Thai New Year, which is celebrated in the second week of April. This is the hottest time of the year in Thailand, so it makes sense that the biggest Songkran tradition is to throw water on people. Thais of all ages join in water fights, using buckets, hoses (软管)and water guns to celebrate the event.Boryeong Mud Festival, South KoreaDon't expect to come home clean after experiencing the Boryeong Mud Festival in South Korea. Famous for its mineral-rich mudflats, visitors rush to Boryeong in July to make themselves in mud, swim in grey pools and enjoy the party. There are even mudslides and a mud skiing competition for those who are seeking the extreme mud adventure!La Tomatina, SpainThe festival dates back to a parade in which some naughty teenagers knocked one performer off his stilts (高跷)and caused a fight of throwing vegetables accidentally. It was once banned until 1957 when the locals held a protest with a funeral. They carried a coffin containing a huge tomato as bands played a funeral march. In the following decades, La Tomatina has become a popular event. If you join the event, be aware that you squash (压扁) the tomato before throwing it. Have a great time but avoid causing any injury.Dia de los Muertos, MexicoBeginning at the midnight of October 31 and lasting through November 2,it is a festival when families gather together to remember those loved ones who have died, aiming to help them on their spiritual journey. On these days, Mexican families prepare special tables in their homes. On top of them they'll put photos of the dead and their favorite food. They also visit the graves of their beloved ones to show their respect to the dead.1.If you want to experience a special new year in a country, you can choose to visit ________.A.ThailandB.South KoreaC.SpainD.Mexico2.What may happen to you as a visitor in La Tomatina?A.You can perform stilts.B.Your can play funeral music in bands.C.You may be covered with mud.D.You may be attacked with tomatoes.3.Which festival is similar to the Tomb Sweeping Festival in China?A.Songkran.B.Boryeong Mud Festival. Tomatina.D.Dia de los Muertos.BAt first glance, there is nothing unusual about BingoBox’s convenience store–shelves packed with snacks line the walls, attracting passers-by through the glass windows. But upon closer look, BingoBox is no ordinary store. The door unlocks only after customers scan (扫描) aQR code to enter, and there is no cashier — just a lone checkout counter (柜台) in a corner. The Shanghai-based company is one of many unmanned store operators (运营者) opening outlets all over China, hoping to improve slim profit by reducing staff costs.“Ifstaff costs rise quickly, that puts greater pressure on low-profit businesses like convenience stores and supermarkets,” said Andrew Song, an analyst at Guotai Junan Securities. “InChina, manpower costs have been rising ly quickly.”However, the future vision of shopping without a check-out person is still a work in progress. A Post reporter who visited a BingoBox store inShanghaiwas briefly locked in when trying to exit without buying anything. Although a sign near the exit stated that empty-handed customers can leave by scanning a QR code, no QR code was to be found. Repeated calls to the customer service hotline went unanswered.The idea of unmanned stores first caught the world’s attention in December last year. Equipped with technology such as RFID tags, mobile payment systems and facial and movement recognition, such stores collect large amounts of data that give operators a better idea of consumer preferences and buying habits, which can then be used to optimize (使最优化) operations and make more efficient inventory decisions. For companies like BingoBox, lower operating costs also mean it can afford to expand its reach to areas with less foot traffic or fewer people, according to its founder and chief executive ChenZilin.4. What makes BingoBox store look like an ordinary convenience store?A. No cashier to check out.B. A lone checkout counter.C. Shelves packed with goods.D. Entering by scanning a QR code.5. Why are unmanned stores popular with operators?A. The customers prefer mobile payment systems.B. The unmanned stores help improve profit with lower labor costs.C. The employees focus on consumer preferences and buying habits.D. The operators care more about operations and inventory decisions.6. Why is the reporter’s case mentioned in the passage?A. To show his anger and dissatisfaction.B. To warn people not to go to a BingoBox store.C. To explain unmanned stores still have a long way to go.D. To complain that QR code service is not convenient at all..7. What can we infer from the chief executive Chen Zilin?A. Nowadays all stores should be equipped with advanced technology.B. The operators collect data about consumer preferences and buying habits.C. BingoBox made wiser decisions based on the data collected in those unmanned stores.D. The operators can open unmanned supermarkets in more distant places with low cost.CThis year researchers expect the world to snap 1.35 trillion photographs, or about 3.7 billion per day. All those pixels (像素) take up a lot of room if they are stored on personal computers or s phones, which is one reason why many people store their images in the cloud. But unlike a hard on drive which can be encrypted to protect its data, cloud storage users have to trust that a tech platform will keep their private pictures safe. Now a team of Columbia University computer scientists has developed a tool to encrypt (加密) images stored on many popular cloud services while allowing authorized users to browse and display their photographs as usual.Malicious (恶意的) attempts to access or leak cloud-based photographs can expose personal information. In November 2019, for example, a bug in the popular photograph storage app Google Photos mistakenly shared some users' private videos with strangers. Security experts also worry about employees at cloud storage companies on purpose accessing users' images.So the Columbia researchers came up with a system called Easy Secure Photos (ESP), which they presented at a recent conference. “We wanted to see if we could make it possible to encrypt data while using existing services,” says computer scientist Jason Nieh, one of the developers of ESP. “Everyone wants to stay with Google Photos and not have to register on a new encrypted-image cloud storage service.”To overcome this problem, they created a tool that preserves blocks of pixels but moves them around to effectively hide the photograph. First, ESP's algorithm (算法) divides a photograph into three separate files, each one containing the image's red, green or blue color1 data. Then the system hides the pixel blocks around amongthese three files (allowing a block from the red file, for instance, to hide out in the green or blue ones). But the program does nothing within the pixel blocks, where all the image processing happens. As a result, the files remain unchanged images but end up looking like grainy black-and-white ones to anyone who accesses them without the decryption (解密) key.8. What's probably the main purpose for people to store images in the cloud?A. To save storage room.B. To make photos beautiful.C. To try a new storage way.D. To keep their privacy safe.9. Why might employees in cloud storage companies be distrusted by experts?A. They sell users' passwords.B. They have invented new tools.C. They often let out personal information.D. They may steal a glance at users' images.10. What's the advantage of ESP?A. It can provide clear images.B. It can decrease the upload time.C. It can classify images automatically.D. It can encrypt data on the original platform.11. What does paragraph 4 mainly talk about?A. Method of decryption.B. Image-processing technique.C. Separate files of images.D. Data analysisof color1 s.DByteDance（字节跳动）Group’s TikTok, an overseas version of Chinese short video sharing app Douyin, faces an existential crisis in the United States, as murmurs of a “crackdown”（强制取缔）from the White House forced the Chinese company toengage in talks on selling its US business to Microsoft.TikTok is the fastest-growing registered global mobile internet app, with more than 100 million users, and its rapid growth, especially in the US, is seen as a threat to Facebook. The US government has long viewed globally competitive Chinese high-tech companies including Huawei as a threat and done whatever it could to crack down on（打击）them in the name of “national security”.The US government has not introduced any specific policy against TikTok only threatened it through a number of unclear statements.According to the latest media reports, Microsoft is prepared to press ahead withthe negotiations to take over TikTok’s US operations and complete the negotiations by Sept 15, following talks between Microsoft CEO Satya Nadella and US President Donald Trump.This means that TikTok will have to hand its fate over to some unpredictable power, and even have to sell its assets without the option of setting a price.The US government has failed to find a reasonable legal excuse to deal with TikTok. All private data of TikTok’s US users are stored in the US and are unlikely to be transferred. Considering the US government is trying to deal with TikTok in a political way, TikTok should consider incorporating（合并）the dispute into the US legal process to assert（坚持）its legal rights and interests.TiKTok’s core value lies in its unique algorithms,a product of artificial intelligence that represents the expertise of Chinese engineers and programmers with high-value intellectual property.The US government’s move, which has forced ByteDance to sell TikTok to a US company, is similar to a forced technology transfer and an example of the US’ openseizureof Chinese intellectual property.If ByteDance sells TikTok to a US company for “security reasons", that would set a dangerous precedent, motivating other countries where TikTok operates to follow the US administration's example and cause a chain reaction.ByteDance is a young Chinese private company that cannot deal with a political game played by the US. But as a Chinese company that has gone global, ByteDance has reasons to take up legal means to defend its legal rights. The Chinese government can also consider examining whether the technology transfer in the deal violates China’s law and harms the country’s national interests.12. What is the most valuable as for TikTok?A. Its global popularity.B. Its artificial intelligence.C. Its registered global mobile internet app.D. Its private data of TikTok’s worldwide users.13. Why does the American government force the Chinese company to hand over TikTok’s US operations to a US company?A. Out of so-called political reasons.B. Out of so-called debt reasons.C. Out of so-called technical reasons.D. Out of so-called security reasons.14. What does the underlined word “seizure”probably mean in Para. 8?A. An untrue spoken statement about someone.B. The use of legal authority to take sth from sb.C. The crime of stealing sth from a person or place.D. The act of trying to hurt somebody using physical violence.15. What can be the best title for the news report?A. TikTok must defend its rights legallyB. TikTok is seen as a threat to FacebookC. ByteDance has to sell TikTok to a US companyD. ByteDance agrees to transfer technology第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2019-2020学年长沙市长郡中学高三英语下学期期末考试试卷及答案

2019-2020学年长沙市长郡中学高三英语下学期期末考试试卷及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项A阅读下列短文，从每题所给的A、B、C和D四个选项中，选出最佳选项，并在答题卡上将该项涂黑。

The opening ceremony for the 2020 Tokyo Olympic Games was held at the Olympic Stadium in Tokyo on July 23, 2021 after a year of waiting due to the pandemic. Let's take a look at the highlights which make this opening ceremony unique.No spectatorsThere are no spectators during the opening ceremony in an effort to prevent the spread of the coronavirus infections. About 950 people, including officials and journalists, will watch in the main stadium, organizers said. Fewer athletes are expected to attend the ceremony because many are not allowed to stay for the entire Games due to COVID-19 restrictions.Male and female flag-bearersThe International Olympic Committee has actively encouraged all delegations toselect two flag-bearers at the opening ceremony, including one woman, as women's participation in the Tokyo Games reaches 48.8 percent, making it the most gender-balanced Olympics in the history of the event.Oath (宣誓) session updatedOne of the most symbolic parts of the opening ceremony—the Olympic oath—has been significantly adapted in order to “highlight the importance of solidarity, inclusion, non-discrimination and equality”. The number of oath-takers is also extended from three to six—two athletes, two coaches and two judges.Simplified ceremonyIn the 30-second video showing the rehearsal of the opening ceremony released to the public, the performances are integrated with dances, scenery and other display elements used in Japanese traditional festival activities.In order to prevent and control the COVID-19 pandemic and manage the cost, the Tokyo 2020 Organizing Committee said that the performances of the ceremony will be greatly simplified, with more light and shadow technology used instead of real decorations at the scene.1.Why isn't there any spectator during the opening ceremony?A.To limit the number of people present.B.To highlight the spirit of the Olympic Games.C.To control the spread of an infectious disease.D.To make room for the athletes all over the world.2.Why is each delegation encouraged to select one woman as a flag-bearer in Tokyo Olympics?A.Women will be in a leading position in sports.B.Women athletes make up a larger percentage than ever.C.Women will play a more important part in society.D.More and more women will take part in the Olympic Games.3.Which of the following statements is True according to the passage?A.Oath-takers are made up of six athletes.B.Tokyo uses more real decorations to cut the cost.C.About 950 people watch the entire games in the Tokyo Games.D.The opening ceremony presents the traditional culture of Japan.BUnderstanding the link between a clean environment and human life is not a new concept. In fact, it was noticed as early as ancient Rome. Today we see how green living has infiluenced our everyday lives. There is a growing community of people who embrace a zero waste lifestyle and make changes to the way they live to reduce their carbon footprint.Living a zero waste lifestyle means doing one’s best to achieve the aim of not sending anything to a landfill. People who adopt this lifestyle ultimately cut down on their waste by reducing what they need and want. They reuse what they own, sending few things to be recycled.Many people who adopt the zero waste lifestyle claim to be frustrated by the many harmful chemical substances found in beauty and cleaning products. They also find the uses of disposable items and excessive packaging. For example, how many times have we had to peel away layers of plastic wrap and cardboard before finally taking out the item which we had bought? Instead of buying pre-packed food and goods, those who identify with the zero waste philosophy tend to shop in stores that allow them to make purchases and bring their own cloth bags and glass jars to store their purchases.Many people may have the misconception that it is easier to live a zero waste lifestyle in the West. Nevertheless, Malaysian environmental journalist, Ms. Aurora Tin, has proven that a zero waste lifestyle is possibleeven in the Asian context. Instead of going to the supermarket to buy pre-packaged foods, Ms. Tin now visits the wet market and brings her own bags for vegetables. She has even stopped using store-bought toothpaste and makes her own toothpaste from coconut oil and baking soda. This lifestyle may be too big a change for the average person, but we could follow her suit to make gradual changes to our own lives.4. Which of the following is a zero waste lifestyle?A. Bringing a resuable container to take away food.B. Choosing appliances that cost less money.C. Turning off a device to stop using power.D. Classifying the garbage before throwing it away.5. What may disappoint a person who adopts a zero waste lifestyle?A. Recycable carboard.B. Excessive packaging.C. Glass jars to store purchases.D. Natural substances in cleaning products.6. What is the main idea of the last paragraph?A. How do people live a zero waste lifestyle.B. Why Ms. Tin chooses to live a zero waste lifestyle..C. We can also practice a zero waste lifestyle in Asia.D. It is easy to live a zero waste lifestyle in the West.7. What is the best title of the passage?A. Living a zero waste lifestyle.B. Going green ismore than a fashion.C. A zero waste lifetyle is easy to achieve.D. Making environmentally-conscious decisions.CIs homeschooling good or bad? That is a question that many parents ask themselves. In my opinion, homeschooled kids can give reliable answers.Kayla Murray, who lives inGreensboroNorth Carolina, started homeschooling almost eight years ago, after attending two local schools. “Homeschooling is school at home,” like the name says, but it’s not just that/ she says. “You're able to learn more and do more.”Having experienced learning at both school and home, Kayla says that she prefers the latter. “With homeschooling, I can do schoolwork in my pajamas, and no one cares,” she says, “you also have more chances. Earlier this year, my family and I were able to take a short vacation and go to an air show. My siblings and I get to take the day off for our birthdays. This year, I spent that time reading and playing the piano and guitar.”There are, however,drawbacksof being homeschooled, Kayla says, especially “the lack of people”. While she doesn't miss homework, al school, she did enjoy getting to see my friends every day.”Matthew Dimmette, also fromGreensboro, used to be a homeschooler. He now attends public high school. He says being homeschooled is quite different from going to public school. The differences depend on how you arc homeschooled and the things you do. “For me, it wasn't really different in terms of the classes but the school environment was a big change atter hearting at home. says Matthew.Talking about his homeschooling experience, Matthew says the best part of it was being able to learn what he wanted to. Yet the main disadvantage was that he couldn't do many experiments in homeschooling unless taking classes at different places. “My interest in science has been growing quickly in recent years, so I chose to go to public school,” he says.8. What does Kayla like about learning at home?A. She can spend her time more freely.B. She can always stay with her siblings.C. She can play the instruments she likes.D. She can celebrate her birthday at home.9. What does the underlined word “drawbacks” in Paragraph 4 probably mean?A. Concerns.B. Reasons.C. Changes.D. Problem.10. Why did Matthew choose to go to public school?A. He wanted to make more friends.B. He wanted to do experiments in labs.C. He wanted to take all kinds of classesD.He wanted to experience the school environment11. What in the authors attitude cowards homeschooling?A. Supportive.B. Doubtful.C. Uninterested.D. Unclear.DLike all cultural institutions, galleries and art fairs are adapting to a new reality.Art Basel Hong Kong, Asia's biggest contemporary-art fair, was cancelled because of covid-19, but anyone he who had planned to visit last week could enjoy an experimental alternative：the viewing room. At the click of a keyboard) you could enter an overall view but private visual salon, without having to brave the airless Hong Kong Convention and Exhibition Centre.There, on one webpage, was Jeff Koons riffing(翻唱)on Botticelli's "Primavera" in a tribute to the history of painting at David Zwirner Gallery. Ota Fine Arts offered one collector the chance to acquire an "infinity(无限)room“, one of the most Instagrammed artworks of recent years- the creation of the fantasies,nonagenarian (九十多岁) Japanese artist, Yayoi Kusama. White Cube presented a large collection of international works by Andreas Gursky (German), Theaster Gates (American) and Beatriz Milhazes (Brazilian). But not every artist, gallery and form showed to equal advantage in this alternative fair. Not surprisingly, simple two-dimensional works in brightcolours came across best No sculpture or conceptual art was includedBesides depth and texture, there are aspects of gallery hopping that a website is unlikely to copy. One isserendipity—the sense of wandering between artworks and encountering the unexpected. Another is sociability. Art is a communion between artist and viewer, but galleries and fairs are also places to swap opinions and share enthusiasms.Thereare ways to compensate (or these inevitable(不可避免的)disadvantages. As they shut their physical doors, some of the world's nest galleries and museums are offering advanced interactive visits, 360 degree videos and walk around tours of their collections, allwithout queues and high ticket prices. One of the best is laid on by the Rijksmuseum in Amsterdam；its tour allows visitors to view its Vermeers and Rembrandts, including the magnicent "Night Watch”, far more closely than would normally he possible. Another standout offering is from the Museu de Arte de SPaulo, which has an even broader collection. On its virtual platform, its pain tings, spanning 700 years, appear to be hanging in an open-plan space, seemingly hung on glass panels, or "crystal easels” as the. museum calls them, ideal for close-up inspection,12. From the paragraph 3, we may knowA. "Primavera" was originally painted by JeKoonsB.“innity room” was the most popular artworks of recent years on social mediaC. White Cube displayed a great many works by artists from different nationsD. Various paintings and statues showed on the webpage are well-received13. What does the underlined word "serendipity" in Paragraph 4 probably mean?A. surpriseB. treasureC. expectationD. compromise14. What is the purpose of writing the last paragraph?A.To introduce some popular art show in the world.B. To show that there are some alternative ways to promote the display during covid-19 epidemic.C. To prove that paintings hung on glass panels can be idealfor close-up inspection,D. To release some information about some extraordinary art shows.15. What is the best title?A. Art beyond the internet.B. Fascinating art.C. Art under covid-19.D. gallery hopping第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2019-2020学年湖南省长沙市长郡中学高三第二学期月考(文科)数学试卷解析版

2019-2020学年高三第二学期月考（文科）数学试卷一、选择题（共12小题）.1．若集合A＝{x||x+2|＝x+2}，B＝{x|x2＜9}，则A∩B＝（）A．（﹣3，3）B．（﹣2，3）C．（﹣3，2]D．[﹣2，3）2．已知实数a，b满足（i为虚数单位）则复数z＝a+bi的共轭复数为（）A．1﹣2i B．2﹣i C．2+i D．1+2i3．“双曲线的方程为x2﹣y2＝1”是“双曲线的渐近线方程为y＝±x”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．如果函数f（x）的图象与函数g（x）＝e x的图象关于直线y＝x对称，则f（4x﹣x2）的单调递增区间为（）A．（0，+∞）B．（2，+∞）C．（0，2）D．（2，4）5．如图茎叶图记录的是甲、乙两个班级各5名同学在一次数学小题训练测试中的成绩（单位：分，每题5分，共16题）．已知两组数据的平均数相等，则x、y的值分别为（）A．0，0B．0，5C．5，0D．5，56．《九章算术》是中国古代张苍，耿寿昌所撰写的一部数学专著，成书于公元一世纪左右，内容十分丰富．书中有如下问题：“今有圆堢瑽，周四丈八尺，高一丈一尺，问积几何？答曰：二千一百一十二尺．术曰：周自相乘，以高乘之，十二而一．”这里所说的圆堢瑽就是圆柱体，它的体积（底面的圆周长的平方×高），则该问题中的体积为估算值，其实际体积（单位：立方尺）应为（）A．528πB．C．704πD．7．已知向量，，，若△OAB是以O为直角顶点的等腰直角三角形，则△OAB的面积为（）A．1B．2C．D．8．如图，在平面直角坐标系xOy中，角α与角β均以Ox为始边，终边分别是射线OA和射线OB．射线OA，OC与单位圆的交点分别为，C（﹣1，0）．若∠BOC ＝，则cos（β﹣α）的值是（）A．B．C．D．9．在棱长为1的正方体ABCD﹣A1B1C1D1中，E，F分别为线段CD和A1B1上的动点，且满足CE＝A1F，则四边形D1FBE所围成的图形（如图所示阴影部分）分别在该正方体有公共顶点的三个面上的正投影的面积之和（）A．有最小值B．有最大值C．为定值3D．为定值2 10．为了解学生课外使用手机的情况，某研究学习小组为研究学校学生一个月使用手机的总时间，收集了500名学生2019年12月课余使用手机的总时间（单位：小时）的数据．从中随机抽取了50名学生，将数据进行整理，得到如图所示的频率分布直方图．已知这50人中，恰有2名女生的课余使用手机总时间在[18，20]区间，现在从课余使用手总时间在[18，20]样本对应的学生中随机抽取2人，则至少抽到1名女生的概率为（）A．B．C．D．11．已知F为抛物线C：y2＝4x的焦点，过F作两条互相垂直的直线l1，l2，直线l1与C交于A、B两点，直线l2与C交于D、E两点，则|AB|+|DE|的最小值为（）A．10B．12C．14D．1612．如图，函数f（x）＝A sin（ωx+φ）（其中A＞0，ω＞0，|φ|≤）与坐标轴的三个交点P、Q、R满足P（2，0），∠PQR＝，M为QR的中点，PM＝2，则A的值为（）A．B．C．8D．16二、填空题13．已知数列{a n}满足na n+1＝（n+1）a n，且a6＝12，则a12＝．14．已知直线3x+4y+5＝0与圆O：x2+y2＝r2（r＞0）相交于A，B两点，且∠AOB＝120°，则r＝．15．在平行四边形ABCD中，BD⊥CD，AB⊥BD，AB＝CD＝2，．沿BD把△ABD 翻折起来，形成三棱锥A﹣BCD，且平面ABD⊥平面BCD，则该三棱锥外接球的体积为．16．设函数f（x）＝，函数g（x）＝[f（x）]2﹣mf（x）+2，若函数g（x）恰有4个零点，则整数m的最小取值为．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．已知公差不为零的等差数列{a n}，满足a3＝7，且a1﹣1，a2﹣1，a4﹣1成等比数列．（1）求{a n}的通项公式；（2）在平面直角坐标系中，设A k（k，a k），B k（k，0），k∈N*，记以A k，A k+1，B k，B k+1四点为顶点的四边形面积为S k，求S1+S3+…+S2n﹣1．18．如图所示，四棱柱ABCD﹣A1B1C1D1中，底面ABCD为直角梯形，AB∥CD，AB⊥BC，平面ABCD⊥平面ABB1A1，∠BAA1＝60°，AB＝AA1＝2BC＝3CD＝6．（1）求该四棱柱的体积；（2）在线段DB1上是否存在点M，使得CM∥平面DAA1D1？若存在，求的值；若不存在，说明理由．19．已知△ABC的内角A，B，C的对边分别为a，b，c，且满足．（1）证明：3a2＝c2﹣b2；（2）若，且△ABC的面积为，求c．20．已知函数．（1）若该函数在（1，f（1））处的切线为y＝ex，求a，b的值；（2）若该函数在x1，x2处取得极值（0＜x1＜x2），且，求实数a的取值范围．21．已知椭圆的离心率为，与x轴交于点A1，A2，过x轴上一点Q引x轴的垂线，交椭圆C于点P1，P2，当Q与椭圆右焦点重合时，|P1P2|＝1．（1）求椭圆C的方程；（2）设直线A1P1与直线A2P2交于点P，是否存在定点M和N，使||PM|﹣|PN||为定值．若存在，求M、N点的坐标；若不存在，说明理由．（二）选考题：共10分.请考生在22、23两题中任选一题作答，如果多做，则按所做的第一题计分.[选修4-4：坐标系与参数方程]22．已知过点P（x0，0）的直线l的倾斜角为，以平面直角坐标系的原点为极点，x轴的非负半轴为极轴建立极坐标系，曲线C的极坐标方程为ρ＝2cosθ．（1）求曲线C的直角坐标方程并写出直线l的一个参数方程；（2）若直线l和曲线C交于A、B两点，且|PA|•|PB|＝2，求实数x0的值．[选修4-5：不等式选讲]23．设函数f（x）＝|2x﹣1|．（1）若函数F（x）＝f（x）+ax有最小值，求a的取值范围；（2）若关于x的不等式f（x）≤|2x+1|﹣|x+m|的解集为A，且，求实数m 的取值范围．参考答案一、选择题：共12小题，每小题5分，共60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．若集合A＝{x||x+2|＝x+2}，B＝{x|x2＜9}，则A∩B＝（）A．（﹣3，3）B．（﹣2，3）C．（﹣3，2]D．[﹣2，3）【分析】可以求出集合A，B，然后进行交集的运算即可．解：A＝{x|x≥﹣2}，B＝{x|﹣3＜x＜3}，∴A∩B＝[﹣2，3）．故选：D．2．已知实数a，b满足（i为虚数单位）则复数z＝a+bi的共轭复数为（）A．1﹣2i B．2﹣i C．2+i D．1+2i【分析】把已知等式变形，再由复数代数形式的乘除运算化简，然后利用复数相等的条件求得a，b的值，则答案可求．解：由，的a＝（1+bi）（1﹣i）＝1+b+（b﹣1）i，∴，即a＝2，b＝1．∴z＝2+i，则．故选：B．3．“双曲线的方程为x2﹣y2＝1”是“双曲线的渐近线方程为y＝±x”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【分析】等轴双曲线x2﹣y2＝1的渐近线为y＝±x，反之渐近线为y＝±x的双曲线为x2﹣y2＝±1，然后结合充分必要条件的判定得答案．解：双曲线的方程为x2﹣y2＝1，则a＝b＝1，其渐近线方程为y＝±x；由双曲线的渐近线方程为y＝±x，可得双曲线方程为x2﹣y2＝±1．则“双曲线的方程为x2﹣y2＝1”是“双曲线的渐近线方程为y＝±x”的充分不必要条件．故选：A．4．如果函数f（x）的图象与函数g（x）＝e x的图象关于直线y＝x对称，则f（4x﹣x2）的单调递增区间为（）A．（0，+∞）B．（2，+∞）C．（0，2）D．（2，4）【分析】由条件求得f（4x﹣x2）＝ln（4x﹣x2），令t＝4x﹣x2＞0，解得0＜x＜4．故f （4x﹣x2）的定义域为（0，4），本题即求函数f（4x﹣x2）在（0，4）上的增区间．再利用二次函数的性质可得结论．解：由题意可得函数f（x）与g（x）＝e x的互为反函数，故f（x）＝lnx，f（4x﹣x2）＝ln（4x﹣x2），令t＝4x﹣x2＞0，解得0＜x＜4．故f（4x﹣x2）的定义域为（0，4），本题即求函数f（4x﹣x2）在（0，4）上的增区间．再利用二次函数的性质可得函数f（4x﹣x2）在（0，4）上的增区间为（0，2），故选：C．5．如图茎叶图记录的是甲、乙两个班级各5名同学在一次数学小题训练测试中的成绩（单位：分，每题5分，共16题）．已知两组数据的平均数相等，则x、y的值分别为（）A．0，0B．0，5C．5，0D．5，5【分析】根据茎叶图中数据，利用两组数据的平均数相等列方程得出x与y的关系，结合题意求出x与y的值．解：根据茎叶图中数据，利用两组数据的平均数相等，得×（65+75+70+x+80+80）＝×（70+70+y+70+75+80），即5+x＝y；所以x＝0，y＝5．故选：B．6．《九章算术》是中国古代张苍，耿寿昌所撰写的一部数学专著，成书于公元一世纪左右，内容十分丰富．书中有如下问题：“今有圆堢瑽，周四丈八尺，高一丈一尺，问积几何？答曰：二千一百一十二尺．术曰：周自相乘，以高乘之，十二而一．”这里所说的圆堢瑽就是圆柱体，它的体积（底面的圆周长的平方×高），则该问题中的体积为估算值，其实际体积（单位：立方尺）应为（）A．528πB．C．704πD．【分析】利用圆柱体积计算公式即可得出．解：由题意可得：2πr＝48，解得r＝，∴这个圆柱的体积＝πr2×11＝．故选：B．7．已知向量，，，若△OAB是以O为直角顶点的等腰直角三角形，则△OAB的面积为（）A．1B．2C．D．【分析】由等腰直角三角形的性质，可得||＝||，且•＝0，应用向量的平方即为模的平方，以及向量模的公式，可得|OA|，再由等腰直角三角形的面积公式，计算可得所求值．解：由△OAB是以O为直角顶点的等腰直角三角形，可得||＝||，且•＝0，由已知条件可得|﹣|＝|+|，（﹣）•（+）＝0，化为2﹣2•+2＝2+2•+2，2＝2，即•＝0，且2＝2，即||＝||＝＝，可得|OA|＝＝＝＝2，则△OAB的面积为||•||＝×2×2＝2．故选：B．8．如图，在平面直角坐标系xOy中，角α与角β均以Ox为始边，终边分别是射线OA和射线OB．射线OA，OC与单位圆的交点分别为，C（﹣1，0）．若∠BOC ＝，则cos（β﹣α）的值是（）A．B．C．D．【分析】由三角函数的定义可知，cos，sinα＝，β＝，然后结合两角差的余弦公式即可求解解：由三角函数的定义可知，cos，sinα＝，β＝，∴cos（β﹣α）＝cos cosα+sin sinα＝＝故选：C．9．在棱长为1的正方体ABCD﹣A1B1C1D1中，E，F分别为线段CD和A1B1上的动点，且满足CE＝A1F，则四边形D1FBE所围成的图形（如图所示阴影部分）分别在该正方体有公共顶点的三个面上的正投影的面积之和（）A．有最小值B．有最大值C．为定值3D．为定值2【分析】分别在后，上，左三个平面得到该四边形的投影，求其面积和即可．【解答】解：依题意，设四边形D1FBE的四个顶点在后面，上面，左面的投影点分别为D'，F'，B'，E'，则四边形D1FBE在上面，后面，左面的投影分别如上图．所以在后面的投影的面积为S后＝1×1＝1，在上面的投影面积S上＝D'E'×1＝DE×1＝DE，在左面的投影面积S左＝B'E'×1＝CE×1＝CE，所以四边形D1FBE所围成的图形（如图所示阴影部分）分别在该正方体有公共顶点的三个面上的正投影的面积之和S＝S后+S上+S左＝1+DE+CE＝1+CD＝2．故选：D．10．为了解学生课外使用手机的情况，某研究学习小组为研究学校学生一个月使用手机的总时间，收集了500名学生2019年12月课余使用手机的总时间（单位：小时）的数据．从中随机抽取了50名学生，将数据进行整理，得到如图所示的频率分布直方图．已知这50人中，恰有2名女生的课余使用手机总时间在[18，20]区间，现在从课余使用手总时间在[18，20]样本对应的学生中随机抽取2人，则至少抽到1名女生的概率为（）A．B．C．D．【分析】课余使用手总时间在[18，20]样本对应的学生共有5人，其中2名女生，3名男生，从课余使用手总时间在[18，20]样本对应的学生中随机抽取2人，基本事件总数n ＝＝10，至少抽到1名女生包含的基本事件个数m＝＝7，由此能求出至少抽到1名女生的概率．解：这50人中，恰有2名女生的课余使用手机总时间在[18，20]区间，课余使用手总时间在[18，20]样本对应的学生共有：50×0.05×2＝5，∴课余使用手总时间在[18，20]样本对应的学生有2名女生，3名男生，现在从课余使用手总时间在[18，20]样本对应的学生中随机抽取2人，基本事件总数n＝＝10，至少抽到1名女生包含的基本事件个数m＝＝7，则至少抽到1名女生的概率为p＝＝．故选：B．11．已知F为抛物线C：y2＝4x的焦点，过F作两条互相垂直的直线l1，l2，直线l1与C 交于A、B两点，直线l2与C交于D、E两点，则|AB|+|DE|的最小值为（）A．10B．12C．14D．16【分析】根据题意可判断当A与D，B，E关于x轴对称，即直线DE的斜率为1，|AB|+|DE|最小，根据弦长公式计算即可．解：如图，l1⊥l2，直线l1与C交于A、B两点，直线l2与C交于D、E两点，要使|AB|+|DE|最小，则A与D，B，E关于x轴对称，即直线DE的斜率为1，又直线l2过点（1，0），则直线l2的方程为y＝x﹣1，联立方程组，则y2﹣4y﹣4＝0，∴y1+y2＝4，y1y2＝﹣4，∴|DE|＝•|y1﹣y2|＝×＝8，∴|AB|+|DE|的最小值为2|DE|＝16，故选：D．12．如图，函数f（x）＝A sin（ωx+φ）（其中A＞0，ω＞0，|φ|≤）与坐标轴的三个交点P、Q、R满足P（2，0），∠PQR＝，M为QR的中点，PM＝2，则A的值为（）A．B．C．8D．16【分析】由题意设出Q（2a，0）a＞0，求出R坐标以及M坐标，利用距离公式求出Q 坐标，通过五点法求出函数的解析式，即可求出A．解：∵函数f（x）＝A sin（ωx+φ）（其中A＞0，ω＞0，|φ|≤）与坐标轴的三个交点P、Q、R满足P（2，0），∠PQR＝，M为QR的中点，∴设Q（2a，0），a＞0，则R（0，﹣2a），∴M（a，﹣a），∵PM＝2，∴＝2，解得a＝4，∴Q（8，0），又P（2，0），∴T＝8﹣2＝6，∴T＝＝12，解得ω＝．∵函数经过P（2，0），R（0，﹣8），∴，∵|φ|≤，∴φ＝﹣，解得A＝，故选：A．二、填空题：共4小题，每小题5分，共20分.13．已知数列{a n}满足na n+1＝（n+1）a n，且a6＝12，则a12＝24．【分析】根据数列的递推关系，利用累乘法求出结论解：因为数列{a n}满足na n+1＝（n+1）a n，且a6＝12，∴数列{a n}各项均不为0；∴＝；∴a12＝××…××a6＝×…××12＝24；故答案为：2414．已知直线3x+4y+5＝0与圆O：x2+y2＝r2（r＞0）相交于A，B两点，且∠AOB＝120°，则r＝2．【分析】求出弦的中点C与圆心O连接，可得OC垂直于弦所在的直线，进而求出圆心到直线的距离OC，再由圆心角可得OC与半径的关系，进而求出半径．解：取AB的中点C，连接OC可得OC⊥AB，因为∠AOB＝120°，所以可得∠AOC＝60°，所以＝cos60°＝，而O到直线3x+4y+5＝0距离OC＝＝1，所以OA＝2，即半径r＝2，故答案为：2．15．在平行四边形ABCD中，BD⊥CD，AB⊥BD，AB＝CD＝2，．沿BD把△ABD 翻折起来，形成三棱锥A﹣BCD，且平面ABD⊥平面BCD，则该三棱锥外接球的体积为．【分析】将折起的三棱锥放在长方体中，由长方体的对角线等于外接球的直径，由题知可求出长方体的对角线，进而求出直径再求出球的体积．解：由题意将折起放在图3的长方体中，长宽高分别为：2，2，2，可得长方体的对角线为：＝4，再由长方体的对角线等于外接球的直径2R，所以2R＝4，R＝2，所以外接球的体积为V＝＝，故答案为：．16．设函数f（x）＝，函数g（x）＝[f（x）]2﹣mf（x）+2，若函数g（x）恰有4个零点，则整数m的最小取值为4．【分析】求函数f′（x），研究函数的单调性和极值，作出函数f（x）的图象，设t＝f （x），若函数g（x）恰有4个零点，则等价为函数h（t）＝t2﹣mt+2有两个零点，满足t＞1或0＜t＜1，利用一元二次函数根的分布进行求解即可．解：当x＞0时，f′（x）＝，由f′（x）＞0得1﹣lnx＞0得lnx＜1，得0＜x＜e时，f（x）单调递增；由f′（x）＜0得1﹣lnx＜0得lnx＞1，得x＞e时，f（x）单调递减；即当x＝e时，函数f（x）取得极大值，同时也是最大值，f（e）＝1，当x→+∞，f（x）→0，当x→0，f（x）→﹣∞，作出函数f（x）的图象如图，设t＝f（x），由图象知当t＞1或t＜0，方程t＝f（x）有一个根，当t＝0或t＝1时，方程t＝f（x）有2个根，当0＜t＜1时，方程t＝f（x）有3个根，则g（x）＝f2（x）﹣mf（x）+2，等价为h（t）＝t2﹣mt+2，当t＝0时，h（0）＝2≠0，∴若函数g（x）恰有4个零点，则等价为函数h（t）＝t2﹣mt+2有两个零点，满足t＞1或0＜t＜1，则，即h（1）＝1﹣m+2＝3﹣m＜0得m＞3，即实数m的取值范围是m＞3，则整数m的最小取值为4．故答案为：4．三、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．已知公差不为零的等差数列{a n}，满足a3＝7，且a1﹣1，a2﹣1，a4﹣1成等比数列．（1）求{a n}的通项公式；（2）在平面直角坐标系中，设A k（k，a k），B k（k，0），k∈N*，记以A k，A k+1，B k，B k+1四点为顶点的四边形面积为S k，求S1+S3+…+S2n﹣1．【分析】（1）设公差为d，且d不为零的等差数列{a n}，运用等差数列的通项公式和等比数列的中项性质，解方程可得首项和公差，即可得到所求通项公式；（2）分别求得A k（k，2k+1），A k+1（k+1，2k+3），B k（k，0），B k+1（k+1，0），运用梯形的面积公式可得S k，再由等差数列的求和公式，计算可得所求和．解：（1）设公差为d，且d不为零的等差数列{a n}，满足a3＝7，且a1﹣1，a2﹣1，a4﹣1成等比数列，可得a1+2d＝7，（a2﹣1）2＝（a1﹣1）（a4﹣1），即（a1+d﹣1）2＝（a1﹣1）（a1+3d ﹣1），解得a1＝3，d＝2，则a n＝a1+（n﹣1）d＝3+2（n﹣1）＝2n+1；（2）由A k（k，2k+1），A k+1（k+1，2k+3），B k（k，0），B k+1（k+1，0），可得S k＝（k+1﹣k）（2k+1+2k+3）＝2k+2，则S1+S3+…+S2n﹣1＝4+8+…+4n＝n（4+4n）＝2n2+2n．18．如图所示，四棱柱ABCD﹣A1B1C1D1中，底面ABCD为直角梯形，AB∥CD，AB⊥BC，平面ABCD⊥平面ABB1A1，∠BAA1＝60°，AB＝AA1＝2BC＝3CD＝6．（1）求该四棱柱的体积；（2）在线段DB1上是否存在点M，使得CM∥平面DAA1D1？若存在，求的值；若不存在，说明理由．【分析】（1）由题意可知：四边形AA1B1B是菱形，△AA1B是正三角形．取线段AB 的中点E，连接A1E．AB边上的高A1E＝3，根据面面垂直的性质定理可得：A1E⊥底面ABCD．进而得出该四棱柱的体积V．（2）假设在线段DB1上存在点M，使得CM∥平面DAA1D1．设＝k．k∈[0，1]．设平面DAA1D1的法向量为＝（x，y，z），则•＝•＝0，可得．利用•＝0即可得出k．解：（1）由题意可知：四边形AA1B1B是菱形，△AA1B是正三角形．取线段AB的中点E，连接A1E．AB边上的高A1E＝3，∵平面ABCD⊥平面ABB1A1，则A1E⊥底面ABCD．∴该四棱柱的体积V＝×3＝36．（2）假设在线段DB1上存在点M，使得CM∥平面DAA1D1．设＝k．k∈[0，1]．B（0，0，0），A（6，0，0），D（2，0，4），A1（3，3，0），C（0，0，4），B1（﹣3，3，0），＝（﹣4，0，4），＝A1（﹣3，3，0），设平面DAA1D1的法向量为＝（x，y，z），则•＝•＝0，可得：﹣4x+4z ＝﹣3x+3y＝0，取＝（，1，），＝+k＝（2﹣5k，3k，4﹣4k），则＝（2﹣5k，3k，﹣4k），∴•＝（2﹣5k）+3k﹣4k＝0，解得k＝．∴在线段DB1上存在点M，使得CM∥平面DAA1D1，＝．19．已知△ABC的内角A，B，C的对边分别为a，b，c，且满足．（1）证明：3a2＝c2﹣b2；（2）若，且△ABC的面积为，求c．【分析】（1）由已知结合同角基本关系及正弦与余弦定理进行化简即可证明；（2）由已知结合余弦定理及三角形的面积公式可得bc的关系，联立方程即可求解．【解答】（1）证明：由可得sin C cos B＝﹣2sin B cos C，所以c＝﹣2b，整理可得，3a2＝c2﹣b2，（2）解：由可得sin A＝，∵S△ABC＝＝＝，所以bc＝6①，由余弦定理可得，cos A＝＝＝×，整理可得，2b2+c2＝30②，因为3a2＝c2﹣b2＞0，则c＞b，①②联立可得，c4﹣30c2+216＝0，即（c2﹣12）（c2﹣18）＝0，解可得或，所以c＝2或c＝3．20．已知函数．（1）若该函数在（1，f（1））处的切线为y＝ex，求a，b的值；（2）若该函数在x1，x2处取得极值（0＜x1＜x2），且，求实数a的取值范围．【分析】（1）先求出导函数f'（x），由题意可知，即可求出a，b的值；（2）令f'（x）＝0得，a＝，所以x1，x2是方程a＝的两个根，令g（x）＝，利用导数画出函数g（x）的大致图象，由图可知，a＞e，0＜x1＜x2，所以x2≥3x1，令x2＝3x1＝3m，则x1＝m，x2＝3m，所以，解得m＝ln3，此时a＝，要使x2≥3x1，则a．解：（1）∵函数，∴f'（x）＝ax﹣e x，∵函数在（1，f（1））处的切线为y＝ex，∴，即，解得：a＝2e，b＝e；（2）∵f'（x）＝ax﹣e x，令f'（x）＝0得，a＝，∴x1，x2是方程a＝的两个根，令g（x）＝，则g'（x）＝，∴函数g（x）在（﹣∞，0）和（0，1）上单调递减，在（1，+∞）上单调递增，且g （1）＝e，∴函数g（x）的图象如图所示：，由图可知，a＞e，0＜x1＜x2，∵，∴x2≥3x1，令x2＝3x1＝3m，则x1＝m，x2＝3m，∴，∴，∴3e m＝e3m，∴e m[（e m）2﹣3]＝0，∴，∴m＝ln＝ln3，此时a＝＝，要使x2≥3x1，则a，∴实数a的取值范围为：[，+∞）．21．已知椭圆的离心率为，与x轴交于点A1，A2，过x轴上一点Q引x轴的垂线，交椭圆C于点P1，P2，当Q与椭圆右焦点重合时，|P1P2|＝1．（1）求椭圆C的方程；（2）设直线A1P1与直线A2P2交于点P，是否存在定点M和N，使||PM|﹣|PN||为定值．若存在，求M、N点的坐标；若不存在，说明理由．【分析】（1）由离心率和过焦点的直线与x轴垂直于椭圆的交点弦长及a，b，c之间的关系求出椭圆的方程；（2）设直线x＝m代入椭圆可得P1，P2的坐标进而求出直线A1P1和A2P2的方程，两个方程联立求出P的坐标，消参数可得P的轨迹方程，为双曲线，要使||PM|﹣|PN||为定值，则需M，N分别为双曲线的焦点即可，即M，N为定点且是双曲线的焦点．解：（1）由题意可得离心率e＝＝，x＝c代入椭圆方程可得|y|＝，所以＝1，c2＝a2﹣b2可得a2＝2，b2＝1，所以椭圆的方程为：+y2＝1；（2）假设存在定点M和N满足条件，由（1）可得A1（﹣2，0），A2（2，0），设Q（m，0）且﹣2＜m＜2，则x＝m代入椭圆中可得y2＝1﹣，所以P1（m，），P2（m，﹣），所以直线A1P1的方程为：y＝（x+2），直线A2P2的方程而：y＝﹣（x﹣2），两个方程联立解得：x＝，y＝，即P（，）由消参数m可得：（）2﹣y2＝﹣＝1，即P的轨迹方程为：﹣y2＝1，所以P的轨迹方程为中心在原点，焦点在x轴上，实轴长为4，虚轴长为2的双曲线，所以要使|PM|﹣|PN|为定值，只需要M，N为双曲线的焦点坐标即可，即M，N分别为（﹣，0），（，0）．（二）选考题：共10分.请考生在22、23两题中任选一题作答，如果多做，则按所做的第一题计分.[选修4-4：坐标系与参数方程]22．已知过点P（x0，0）的直线l的倾斜角为，以平面直角坐标系的原点为极点，x轴的非负半轴为极轴建立极坐标系，曲线C的极坐标方程为ρ＝2cosθ．（1）求曲线C的直角坐标方程并写出直线l的一个参数方程；（2）若直线l和曲线C交于A、B两点，且|PA|•|PB|＝2，求实数x0的值．【分析】（1）直接利用转换关系，把参数方程极坐标方程和直角坐标方程之间进行转换．（2）利用一元二次方程根和系数关系式的应用求出结果．解：（1）曲线C的极坐标方程为ρ＝2cosθ．转换为直角坐标方程为x2+y2＝2x，整理得（x﹣1）2+y2＝1．点P（x0，0）的直线l的倾斜角为，转换为参数方程为（t为参数）．（2）把直线的参数方程代入x2+y2＝2x，得到：，整理得：．所以|PA|•|PB|＝|t1t2|＝2，整理得，解得：．[选修4-5：不等式选讲]23．设函数f（x）＝|2x﹣1|．（1）若函数F（x）＝f（x）+ax有最小值，求a的取值范围；（2）若关于x的不等式f（x）≤|2x+1|﹣|x+m|的解集为A，且，求实数m 的取值范围．【分析】（1）由绝对值的意义，去绝对值，化简F（x），再由一次函数的单调性，结合条件，解不等式可得所求范围；（2）由题意可得f（x）≤|2x+1|﹣|x+m|在[，2]恒成立，转化为|x+m|≤2在[，2]恒成立，再由参数分离和一次函数的单调性，可得所求范围．解：（1）函数F（x）＝f（x）+ax＝|2x﹣1|+ax，当x≥时，F（x）＝（a+2）x﹣1，当x＜时，F（x）＝（a﹣2）x+1，由F（x）＝f（x）+ax有最小值，结合一次函数的单调性可得a+2≥0且a﹣2≤0，解得﹣2≤a≤2；（2）由，可得关于x的不等式f（x）≤|2x+1|﹣|x+m|在[，2]恒成立，即|2x﹣1|≤|2x+1|﹣|x+m|，即有|x+m|≤2在[，2]恒成立，可得﹣2≤x+m≤2，则（﹣2﹣x）max≤m≤（2﹣x）min，由y＝﹣2﹣x在[，2]上递减，可得y＝﹣2﹣x的最大值为﹣；由y＝2﹣x在[，2]上递减，可得y＝2﹣x的最小值为0，故m的取值范围是[﹣，0]．。

2019-2020学年长沙市长郡中学高三英语期末试卷及答案

2019-2020学年长沙市长郡中学高三英语期末试卷及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AWashingtonDCBusToursDC Highlights TourThis is their base tour. It begins at 10 am daily and lasts for 5 hours. This part-bus, part-walking tour includes guided stops in theUS. All of their tours include a driver as well as a tour guide, and yourtour guide will get off the bus and give you walking tours of each stop, while your bus and driver wait for you. Capitol Building, the White House, Washington Monument as well as the Lincoln and MLK Memorials and the Vietnam War and Korean War Veterans Memorials.$54—Adult I $44—Child (3—12)Discover DC TourIf you want pretty much to explore every famous monument and landmark in DC and take a 1 -hour cruise on thePotomac River, then consider the Discover DC Tour. This 6-hour tour will take you to all locations (景点) listed on their DC Highlights Tour as well as the World WarⅡMemorial, the Franklin Delano Roosevelt Memorial, and the Thomas Jefferson Memorial.TIP: If you are planning on visiting NYC, you will get 30% off the Discover NYC Tour (normally $100) if you buy it at the same time as your Discover DC Tour.$74—Adult I $54—Child (3—12)VIP ExclusiveWashingtonDCCity TourThis 8-hour tour is actually the Discover DC Tour above with a VIP add-on at the beginning and the end. You * 11 meet your guide early for reserved (预留)tickets to tour inside the US Capitol Building. The 45 -minute tour and film have reserved tickets so you don't have to worry about it being sold out. After your day of sightseeing, you 'll be dropped off at the National Archives, again with reserved time tickets so you don't have to worry about waiting in line.$125 for Adults and ChildrenSkyview Changeable Bus TourOn this changeable mini bus, you get to experience a guided tour with panoramic views (全景)without theglare of a window in the photos! A guided bus tour takes you not only through all locations listed on the DC Highlights Tour, but also Old Town Alexandria andNationalHarbor.$69—Adult I $59—Child (3—12)Time: 9 am—4 pm1. What's special about the DC Highlights Tour?A. It uses mini buses.B. It has the fewest locations.C. It has the most tour guides.D. It provides reserved tickets.2. How much should a man pay if he buys the Discover DC Tour and the Discover NYC Tour at the same time?A. $128.B. $174.C. $104.D. $144.3. Which tour lasts the longest?A. DC Highlights Tour.B. Discover DC Tour.C. VIP ExclusiveWashingtonDCCity Tour.D. Skyview Changeable Bus Tour.BA company called Neuralink has shared a video where a monkey is playing a video game. That' s fairly unusual, but what makes the video even stranger is that the monkey is playing the video game with just his mind.The monkey in the video is called Pager who has two of Neuralink's special "Link" devices(装置)inside his brain. The devices planted in Pager's brain are connected to 2,048 wires which lead to the parts of Pager's brain that control movements of the arms and hands.Scientists taught Pager to play a video game. At first, Pager controlled the video game using a joystick it, which is a normal gaming controller. But as Pager played, his Link devices wirelessly sent out information about the signals his brain was using to control his arms and hands. Neuralink's scientists recorded all of these signals.Then they used computers to match the signals from Pager's brain to the movements that his hands were actually doing. This was the most difficult work and the scientists counted on artificial intelligence ( AI) to help them decode(解码)Pager's brain signals.The final step was to have a computer make moves inthe video game as if Pager had actually moved the joystick. If Pager thought about moving the joystick up, the computer would send an “up” signal to the video game.At first, the researchers let Pager keep moving the joystick with his hand, even though it was no longer connected to the computer. But soon Pager was able to play the video game using just his brain.Even though Neuralink's work right now focuses on animals and video games, there's a very serious purposebehind it. Neuralink wants to make it possible for humans who have lost the ability to make physical movements to interact with the world around them.4. What are “Link” devices used to do?A. To pick up the arms' and hands' signals.B. To link the computer to the monkey's brain.C. To send out information about the brain's signals.D. To control movements of the arms and hands.5. What challenged scientists most in the study?A. Recording and sending out body signals.B. Training Pager to use the joystick correctly.C. Planting "Link" devices into Pager's brain.D. Matching brain signals to body movements.6. What is Neuralink's real purpose of the study?A. To test artificial intelligence.B. To help those without arms or legs.C. To study how animals play video games.D. To develop more complex video games.7. What can be the best title for the text?A. Video Games for Animals Are Developed.B. Science Proves the Intelligence of Monkeys.C. Monkey Plays Video Games Using His Mind.D. Neuralink Is Leading the World in Technology.CIf there is no difference in general intelligence between boys and girls, what can explain girls’ poor performance in science and math?It hasbeen suggested that girls do not take math courses, not because they are difficult, but for social reasons. Girls do not want to be in open competition with boys because they are afraid to appear less feminine (女性的) and attractive (有魅力的).However, there are still more high-achieving boys than girls when taking math exams. This difference appears to be worldwide. Biological explanations have been offered for this, but there are other explanations too.Perhaps the difference which comes out during the teenage years has its roots in much earlier experiences. From their first days in kindergarten, boys are encouraged to work on their own and to complete tasks. Facts show that outstanding mathematicians and scientists have not had teachers who gave answers.Besides, there can be little doubt that teachers of math and science expect their boy students to do better at these subjects than their girl students. They even appear to encourage the difference between boys and girls. They spend more time with the boy students, giving them more time to answer questions and working harder to get correct answers from them. They are more likely to call on boys for answers and to allow them to take the lead in classroom discussion. They also praise boys more frequently. All of this seems to encourage boys to work harder in science and math and to give them confidence that they are able to succeed.Such a way of teaching is not likely to encourage girls to take many math and science courses, nor is it likely to support girls who do. When it comes to these subjects it seems certain that school widens the difference between boys and girls.8. Why does the author raise a question in paragraph 1?A. To find differences between boys and girls.B. To explain the poor performance of girls.C. To ask readers a question.D. To introduce the topic.9. The text mainly discusses ________ reasons for the difference between boys and girls in scientific achievements.A. biologicalB. personalC. socialD. historical10. What does the text say about great mathematicians?A. Their teachers did not offer answers to them.B. They started learning math at an earlier age.C. They showed mathematical abilities in their teenage years.D. Their success resulted from their strong interest in math.11. The author would probably agree that ________.A. boys and girls learn in the same wayB. boys and girls are equal in general intelligenceC. girls are more confident in themselves than beforeD. girls should take fewer science courses than boysDYou've probably heard it suggested that you need to move more throughout the day, and as a general rule of thumb, that "more" is often defined as around 10,000 steps. With many Americans tracking their stepsvia new fitness-tracking wearables, or even just by carrying their phone, more and more people use the 10,000-step rule as their marker for healthy living. Dr. Dreg Hager, professor of computer science at Johns Hopkins, decided to take a closer look at that 10,000-step rule, and he found that usingitas a standard may be doing more harm than good for many.“It turns out that in 1960 in Japan they figured out that the average Japanese man, when he walked 10,000 steps a day burned something like 3,000 calories and that is what they thought the average person should consume so they picked 10,000 steps as a number” Hager said.According to Hager, asking everyone to shoot for 10,000 steps each day could be harmful to the elderly or those with medical conditions, making it unwise for them to jump into that level of exercise, even if it's walking. The bottom line is that 10,000 steps may be too many for some and too few for others. He also noted that those with shorter legs have an easier time hitting the 10,000-step goal because they have to take more steps than people with longer legs to cover the distance. It seems that 10,000 steps may be suitable for the latter.A more recent study focused on older women and how many steps can help maintain good health and promote longevity (长寿).The study included nearly 17,000 women with an average age of 72. Researchers found that women who took 4,400 steps per day were about 40% less likely to die during a follow-up period of just over four years: Interestingly, women in the study who walked more than 7,500 steps each day got no extra boost in longevity.12. What does the underlined word "it' in Paragraph 1 refer to?A. The phone recording.B. The 10,000-step rule.C. The healthy living.D. The fitness-tracking method.13. What does Paragraph 2 mainly talk about?A. How many steps a Japanese walks.B. How we calculate the number of steps.C. If burning 3,000 calories daily is scientific.D. Where 10,000 steps a day came from.14. Who will probably benefit from 10,000 steps each day according toHager?A. Senior citizens.B. Young short-legged people.C. Healthy long-legged peopleD. Weak individuals.15. How many steps may the researchers suggest senior citizens take each day?A. 4,400 steps.B. 10,000 steps.C. 2,700 steps.D. 7,500 steps.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

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2019-2020学年高三第二学期段考数学试卷（理科）（3月份）一、选择题1．若i为虚数单位，复数z满足z（1+i）＝|1﹣i|+i，则z的虚部为（）A．B．C．D．2．设集合A＝{x|＜0}，B＝{x|x≤﹣3}，则集合{x/x≥1}＝（）A．A∩B B．A∪B C．（∁R A）∪（∁R B}D．（∁R A）∩（∁R B} 3．中国古代数学著作《九章算术》中有这样一个问题：“某贾人擅营，月入益功疾（注：从第2月开始，每月比前一月多入相同量的铜钱），3月入25贯，全年（按12个月计）共入510贯”，则该人12月营收贯数为（）A．35B．65C．70D．604．“石头、剪刀、布”，又称“猜丁壳”，是一种流传多年的猜拳游戏，起源于中国，然后传到日本、朝鲜等地，随着亚欧贸易的不断发展，它传到了欧洲，到了近代逐渐风靡世界．其游戏规则是：出拳之前双方齐喊口令，然后在话音刚落时同时出拳，握紧的拳头代表“石头”，食指和中指伸出代表“剪刀”，五指伸开代表“布”．“石头”胜“剪刀”、“剪刀”胜“布”、而“布”又胜过“石头”．若所出的拳相同，则为和局．小千和大年两位同学进行“五局三胜制”的“石头、剪刀、布”游戏比赛，则小千和大年比赛至第四局小千胜出的概率是（）A．B．C．D．5．已知a＝log0.62，b＝log20.6，c＝0.62，则（）A．a＞b＞c B．b＞c＞a C．c＞b＞a D．c＞a＞b6．椭圆C：+＝1，F1，F2是其焦点，点P是椭圆C上一点，若△F1PF2是直角三角形，则点P到x轴的距离为（）A．B．C．D．27．若α为锐角，且（4cos50°﹣tan40°）tanα＝1，则α＝（）A．60°B．50°C．40°D．30°8．设等比数列{a n}的前n项和为S n，公比为q，且S3，S9，S6成等差数列，则8q3等于（）A．﹣4B．﹣2C．2D．49．在平面直角坐标系xOy中，圆C的方程为x2+y2﹣8x+15＝0，若直线y＝kx+2上至少存在一点，使得以该点为圆心，半径为1的圆与圆C有公共点，则k的最小值是（）A．B．C．D．10．已知函数的图象关于直线对称，若f（x1）f（x2）＝﹣4，则|x1﹣x2|的最小值为（）A．B．C．D．11．如图，在梯形ABCD中已知|AB|＝2|CD|，＝，双曲线过C，D，E三点，且以A，B为焦点，则双曲线的离心率为（）A．B．2C．3D．12．如图，棱长为4的正方体ABCD﹣A1B1C1D1，点A在平面α内，平面ABCD与平面α所成的二面角为30°，则顶点C1到平面α的距离的最大值是（）A．2（2+）B．2（+）C．2（+1）D．2（+1）二、填空题13．已知n＝（﹣2x）dx，则x（1﹣）n的展开式中的常数项为．14．某封闭几何体的三视图如图所示，则该几何体的表面积为15．对于数列{a n}，若∀m，n∈N*（m≠n），都有成立，则称数列{a n}具有性质P（t）．若数列{a n}的通项公式为，且具有性质P（10），则实数a的取值范围是．16．若∀x∈[e，+∞），满足恒成立，则实数m的取值范围为．三.解答题17．已知在△ABC中，a，b，c分别为角A，B，C的对应边，点D为BC边的中点，△ABC 的面积为．（1）求sin∠BAD•sin∠BDA的值；（2）若BC＝6AB，AD＝2，求b．18．如图，矩形ABCD中，AB＝6，，点F是AC上的动点．现将矩形ABCD沿着对角线AC折成二面角D'﹣AC﹣B，使得．（Ⅰ）求证：当时，D'F⊥BC；（Ⅱ）试求CF的长，使得二面角A﹣D'F﹣B的大小为．19．已知F为抛物线C：y2＝2px（p＞0）的焦点，过F的动直线交抛物线C于A，B两点．当直线与x轴垂直时，|AB|＝4．（1）求抛物线C的方程；（2）设直线AB的斜率为1且与抛物线的准线l相交于点M，抛物线C上存在点P使得直线PA，PM，PB的斜率成等差数列，求点P的坐标．20．已知函数f（x）＝e﹣x﹣ax（x∈R）．（1）当a＝﹣1时，求函数f（x）的最小值；（2）若x≥0时，f（﹣x）+ln（x+1）≥1，求实数a的取值范围．21．如图，直角坐标系中，圆的方程为x2+y2＝1，A（1，0），B（﹣，），C（﹣，﹣）为圆上三个定点，某同学从A点开始，用掷骰子的方法移动棋子．规定：①每掷一次骰子，把一枚棋子从一个定点沿圆弧移动到相邻下一个定点；②棋子移动的方向由掷骰子决定，若掷出骰子的点数为偶数，则按图中箭头方向移动；若掷出骰子的点数为奇数，则按图中箭头相反的方向移动．设掷骰子n次时，棋子移动到A，B，C处的概率分别为P n（A），P n（B），P n（C）．例如：掷骰子一次时，棋子移动到A，B，C处的概率分别为P1（A）＝0，P1（B）＝，P1（C）＝（1）分别掷骰子二次，三次时，求棋子分别移动到A，B，C处的概率；（2）掷骰子n次时，若以x轴非负半轴为始边，以射线OA，OB，OC为终边的角的余弦值记为随机变量X n，求X4的分布列和数学期望；（3）记P n（A）＝a n，P n（B）＝b n，P n（C）＝c n．，其中a n+b n+c n＝1．证明：数列{b n ﹣}是等比数列，并求a2020．[选修4-4：坐标系与参数方程]22．在平面直角坐标系中，曲线C1：（a为参数）经过伸缩变换后的曲线为C2，以坐标原点为极点，x轴正半轴为极轴建立极坐标系．（Ⅰ）求C2的极坐标方程；（Ⅱ）设曲线C3的极坐标方程为ρsin（﹣θ）＝1，且曲线C3与曲线C2相交于P，Q 两点，求|PQ|的值．[选修4-5：不等式选讲]23．已知函数f（x）＝|x+b2|﹣|﹣x+1|，g（x）＝|x+a2+c2|+|x﹣2b2|，其中a，b，c均为正实数，且ab+bc+ac＝1．（Ⅰ）当b＝1时，求不等式f（x）≥1的解集；（Ⅱ）当x∈R时，求证f（x）≤g（x）．参考答案一.选择题1．若i为虚数单位，复数z满足z（1+i）＝|1﹣i|+i，则z的虚部为（）A．B．C．D．【分析】把已知等式变形，再由复数代数形式的乘除运算化简得答案．解：由z（1+i）＝|1﹣i|+i＝，得z＝．∴z的虚部为．故选：D．2．设集合A＝{x|＜0}，B＝{x|x≤﹣3}，则集合{x/x≥1}＝（）A．A∩B B．A∪B C．（∁R A）∪（∁R B}D．（∁R A）∩（∁R B}【分析】解不等式得集合A，根据补集的定义写出∁R A、∁R B，即可得出结论解：集合A＝{x|＜0}＝{x|﹣3＜x＜1}，B＝{x|x≤﹣3}，则∁R A＝{x|x≤﹣3或x≥1}，∁R B＝{x|x＞﹣3}；∴（∁R A）∩（∁R B}＝{x|x≥1}．故选：D．3．中国古代数学著作《九章算术》中有这样一个问题：“某贾人擅营，月入益功疾（注：从第2月开始，每月比前一月多入相同量的铜钱），3月入25贯，全年（按12个月计）共入510贯”，则该人12月营收贯数为（）A．35B．65C．70D．60【分析】设每个月的收入为等差数列{a n}．公差为d．由a3＝25，S12＝510．可得a1+2d ＝25，12a1+d＝510，联立解出即可得出．解：设每个月的收入为等差数列{a n}．公差为d．则a3＝25，S12＝510．∴a1+2d＝25，12a1+d＝510，解得a1＝15，d＝5，∴a12＝15+11×5＝70．故选：C．4．“石头、剪刀、布”，又称“猜丁壳”，是一种流传多年的猜拳游戏，起源于中国，然后传到日本、朝鲜等地，随着亚欧贸易的不断发展，它传到了欧洲，到了近代逐渐风靡世界．其游戏规则是：出拳之前双方齐喊口令，然后在话音刚落时同时出拳，握紧的拳头代表“石头”，食指和中指伸出代表“剪刀”，五指伸开代表“布”．“石头”胜“剪刀”、“剪刀”胜“布”、而“布”又胜过“石头”．若所出的拳相同，则为和局．小千和大年两位同学进行“五局三胜制”的“石头、剪刀、布”游戏比赛，则小千和大年比赛至第四局小千胜出的概率是（）A．B．C．D．【分析】小千和大年比赛至第四局小千胜出，由指前3局中小千胜2局，有1局不胜，第四局小千胜，由此能求出小千和大年比赛至第四局小千胜出的概率．解：根据“石头”胜“剪刀”，“剪刀”胜“布”，而“布”又胜“石头”，可得每局比赛中小千胜大年、小千与大年和局和小千输给大年的概率都为，∴小千和大年两位同学进行“五局三胜制”的“石头、剪刀、布”游戏比赛，则小千和大年比赛至第四局小千胜出，由指前3局中小千胜2局，有1局不胜，第四局小千胜，∴小千和大年比赛至第四局小千胜出的概率是：p＝＝．故选：B．5．已知a＝log0.62，b＝log20.6，c＝0.62，则（）A．a＞b＞c B．b＞c＞a C．c＞b＞a D．c＞a＞b【分析】a＝log0.62＝﹣1，又ab＝1．可得b＝log20.6∈（﹣1，0），而c ＞0，即可得出大小关系．解：a＝log0.62＝﹣1，又ab＝×＝1．∴b＝log20.6∈（﹣1，0），c＝0.62＞0，则c＞b＞a．故选：C．6．椭圆C：+＝1，F1，F2是其焦点，点P是椭圆C上一点，若△F1PF2是直角三角形，则点P到x轴的距离为（）A．B．C．D．2【分析】分两种情况讨论，是∠P为90°还是∠F1或∠F2为90°，注意P的纵坐标的取值范围，将P的坐标代入椭圆中，再由角为90°可得P的纵坐标的绝对值，即是P 到x轴的距离．解：设P（m，n），|n|2≤5，由题意可得：+＝1，m2＝9（1﹣），a2＝9，b2＝5，所以c2＝a2﹣b2＝9﹣5＝4，所以c＝2，F1（﹣2，0），F2（2，0），△F1PF2是直角三角形，当∠PF2F1＝90°，或∠PF1F2＝90°结果一样的，则m＝c＝2，代入椭圆可得|n|＝＝；当∠F1PF2＝90°时，而＝（m+2，n），＝（m﹣2，n），所以＝0，即（m+2）（m﹣2）+n2＝0，m2+n2＝4，即9（1﹣）+n2＝4，解得n2＝＞5，不成立，综上所述|n|＝，故选：A．7．若α为锐角，且（4cos50°﹣tan40°）tanα＝1，则α＝（）A．60°B．50°C．40°D．30°【分析】先利用三角函数公式化简4cos50°﹣tan40°＝，则tan，从而求出α的值．解：4cos50°﹣tan40°＝4sin40°﹣tan40°＝＝＝＝＝＝，∴，又∵α为锐角，∴α＝300，故选：D．8．设等比数列{a n}的前n项和为S n，公比为q，且S3，S9，S6成等差数列，则8q3等于（）A．﹣4B．﹣2C．2D．4【分析】利用等差数列的性质、等比数列的通项公式即可得出．解：）∵S3，S9，S6成等差数列，∴2S9＝S3+S6，∴（S9﹣S6）+（S9﹣S3）＝0，即（a7+a8+a9）+（a7+a8+a9）+（a4+a5+a6）＝0，∴2q3（a4+a5+a6）+（a4+a5+a6）＝0，∵，∴q3＝﹣，∴8q3＝﹣4．故选：A．9．在平面直角坐标系xOy中，圆C的方程为x2+y2﹣8x+15＝0，若直线y＝kx+2上至少存在一点，使得以该点为圆心，半径为1的圆与圆C有公共点，则k的最小值是（）A．B．C．D．【分析】化圆C的方程为（x﹣4）2+y2＝1，求出圆心与半径，由题意，只需（x﹣4）2+y2＝4与直线y＝kx+2有公共点即可．解：∵圆C的方程为x2+y2﹣8x+15＝0，整理得：（x﹣4）2+y2＝1，即圆C是以（4，0）为圆心，1为半径的圆；又直线y＝kx+2上至少存在一点，使得以该点为圆心，1为半径的圆与圆C有公共点，∴只需圆C′：（x﹣4）2+y2＝4与直线y＝kx+2有公共点即可．设圆心C（4，0）到直线y＝kx+2的距离为d，则d＝≤2，即3k2≤﹣4k，∴﹣≤k≤0．∴k的最小值是．故选：A．10．已知函数的图象关于直线对称，若f（x1）f（x2）＝﹣4，则|x1﹣x2|的最小值为（）A．B．C．D．【分析】根据函数的对称性，利用f（0）＝f（﹣），建立方程求出a的值，然后利用辅助角公式求出f（x）的解析式，利用最值性质转化为周期关系进行求解即可．解：∵f（x）的图象关于直线对称，∴f（0）＝f（﹣），即﹣＝a sin（﹣）﹣cos（﹣）＝﹣a﹣，得a＝，得a＝1，则f（x）＝sin2x﹣cos2x＝2sin（2x﹣），∵f（x1）f（x2）＝﹣4，∴f（x1）＝2，f（x2）＝﹣2或f（x1）＝﹣2，f（x2）＝4，即f（x1），f（x2）一个为最大值，一个为最小值，则|x1﹣x2|的最小值为，∵T＝＝π，∴＝，即|x1﹣x2|的最小值为，故选：D．11．如图，在梯形ABCD中已知|AB|＝2|CD|，＝，双曲线过C，D，E三点，且以A，B为焦点，则双曲线的离心率为（）A．B．2C．3D．【分析】以AB所在的直线为x轴，以AB的垂直平分线为y轴，建立如图所示的坐标系，求出C的坐标，根据向量的运算求出点E的坐标，代入双曲线方程即可求出解：由|AB|＝2|CD|，以AB所在的直线为x轴，以AB的垂直平分线为y轴，建立如图所示的坐标系，设双曲线的方程为﹣＝1，由双曲线是以A，B为焦点，∴A（﹣c，0），B（c，0），把x＝c，代入﹣＝1，可得y＝b，即有C（c，b），又设A（﹣c，0），∴＝（c，b），设E（x，y），∴＝（x+c，y），∵＝，∴（x+c，y）＝（c，b），解得x＝﹣c，y＝b•），可得E（﹣c，b•），代入双曲线的方程可得﹣（﹣1）＝1，即e2﹣（﹣1）＝，即e2＝7，即e＝，故选：A．12．如图，棱长为4的正方体ABCD﹣A1B1C1D1，点A在平面α内，平面ABCD与平面α所成的二面角为30°，则顶点C1到平面α的距离的最大值是（）A．2（2+）B．2（+）C．2（+1）D．2（+1）【分析】如图所示，O在AC上，C1O⊥α，垂足为E，则C1E为所求，∠OAE＝30°，由题意，设CO＝x，则AO＝4﹣x，由此可得顶点C1到平面α的距离的最大值．解：如图所示，AC的中点为O，C1O⊥α，垂足为E，则C1E为所求，∠AOE＝30°由题意，设CO＝x，则AO＝4﹣x，C1O＝，OE＝OA＝2﹣x，∴C1E＝+2﹣x，令y＝+2﹣x，则y′＝﹣＝0，可得x＝，∴x＝，顶点C1到平面α的距离的最大值是2（+）．故选：B．二、填空题13．已知n＝（﹣2x）dx，则x（1﹣）n的展开式中的常数项为60．【分析】根据题意，由定积分计算公式可得n的值，进而由二项式定理分析（1﹣）6的展开式含x﹣1次方的项，据此分析可得答案．解：根据题意，n＝（﹣2x）dx＝（）dx﹣（2x）dx＝××π﹣（x2）＝6，（1﹣）6的展开式通项为T r+1＝C6r（﹣）r，当r＝2时，有T3＝C62（﹣）2＝，则x（1﹣）n的展开式中的常数项为60；故答案为：6014．某封闭几何体的三视图如图所示，则该几何体的表面积为222+6【分析】由已知中的三视图可得该几何体是一个三棱柱切去一个三棱锥所得的组合体，画出直观图，计算各个面的面积，相加可得答案．解：由已知中的三视图可得该几何体是一个三棱柱切去一个三棱锥所得的组合体，其直观图如图所示：底面△ABC的面积为：×8×6＝24；侧面ACDE的面积为：×10＝100，侧面ABFE的面积为：（4+10）×6＝42，侧面CBFD的面积为：（4+10）×8＝56，面EFD中，EF＝6，FD＝10，ED＝10，故面积为：×6×＝6，故几何体的表面积S＝222+6，故答案为：222+615．对于数列{a n}，若∀m，n∈N*（m≠n），都有成立，则称数列{a n}具有性质P（t）．若数列{a n}的通项公式为，且具有性质P（10），则实数a的取值范围是[36，+∞）．【分析】由题意知恒成立，从而可得数列为单调递增数列，从而可得恒成立，即a≥﹣n（n+1）（2n﹣9），从而解得．解：∵数列通项公式且数列具有性质P（10），∴，∴恒成立，∴数列为单调递增数列，∴恒成立，即a≥﹣n（n+1）（2n﹣9），由数轴标根法作图如下，故最大值在n＝1，2，3或4上取得，当n＝1时，﹣n（n+1）（2n﹣9）＝14，当n＝2时，﹣n（n+1）（2n﹣9）＝30，当n＝3时，﹣n（n+1）（2n﹣9）＝36，当n＝4时，﹣n（n+1）（2n﹣9）＝20，故a≥36．故答案为：[36，+∞）．16．若∀x∈[e，+∞），满足恒成立，则实数m的取值范围为（﹣∞，2e]．【分析】通过①m≤0，判断是否满足题意；②m＞0时，由，利用函数的单调性转化求解即可．解：①m≤0，恒成立，所以满足恒成立，显然成立；②m＞0时，由，由f（x）＝xe x在[e，+∞）为增⇒m≤2xlnx在[e，+∞）恒成立，由g（x）＝2xlnx在[e，+∞）为增函数，g（x）min＝2e，0＜m≤2e，综上，m≤2e，故答案为：（﹣∞，2e]．三.解答题17．已知在△ABC中，a，b，c分别为角A，B，C的对应边，点D为BC边的中点，△ABC 的面积为．（1）求sin∠BAD•sin∠BDA的值；（2）若BC＝6AB，AD＝2，求b．【分析】（1）由ABC的面积为且D为BC的中点可得△ABD的面积为，再由三角形的面积公式及正弦定理可求sin∠BAD•sin∠BDA；（2）由（1）可得BC＝6AB，可求sin∠BAD，3sin∠BDA，再由余弦定理可求．解：（1）∵D为BC边的中点，△ABC的面积为，∴△ABD的面积为，∴，∴3AB•BD＝，由正弦定理可得，＝∴3AB•BD＝＝，∴sin∠BAD•sin∠BDA＝（2）∵BC＝6AB，且D为BC的中点，∴BC＝2BD＝6AB，即BD＝3AB，△ABD中，由正弦定理可得，，∴sin∠BAD＝3sin∠BDA，由（1）可知，sin∠BAD•sin∠BDA＝∴sin∠BAD＝1，sin∠BDA＝，∴∠BAD＝90°，Rt△ABD中，AD＝2，∴AB＝1，BD＝3，∴BC＝2BD＝6，△ABC中，由余弦定理可得，b2＝a2+c2﹣2ac cos B＝1+36﹣2×1×6×＝33，∴b＝．18．如图，矩形ABCD中，AB＝6，，点F是AC上的动点．现将矩形ABCD沿着对角线AC折成二面角D'﹣AC﹣B，使得．（Ⅰ）求证：当时，D'F⊥BC；（Ⅱ）试求CF的长，使得二面角A﹣D'F﹣B的大小为．【分析】（Ⅰ）连结DF，BF．通过计算DF2+AF2＝9+3＝DA2，推出DF⊥AC，得到D'F⊥AC，证明BF⊥D'F，然后证明D'F⊥平面ABC．推出D'F⊥BC．（Ⅱ）说明OE，OC，OD'两两垂直，以O为原点，的方向为x轴的正方向建立空间直角坐标系O﹣xyz，求出平面AD'F的一个法向量．平面BD'F的法向量通过向量的数量积求解二面角的平面角的余弦值即可．【解答】满分．（Ⅰ）证明：连结DF，BF．在矩形ABCD中，，∴，∠DAC＝60°．…（1分）在△ADF中，∵，∴DF2＝DA2+AF2﹣2DA•AF•cos∠DAC＝9，．…∵DF2+AF2＝9+3＝DA2，∴DF⊥AC，即D'F⊥AC．…又在△ABF中，BF2＝AB2+AF2﹣2AB•AF•cos∠CAB＝21，…∴在△D'FB中，，∴BF⊥D'F，…又∵AC∩FB＝F，∴D'F⊥平面ABC．∴D'F⊥BC．…（Ⅱ）解：在矩形ABCD中，过D作DE⊥AC于O，并延长交AB于E．沿着对角线AC翻折后，由（Ⅰ）可知，OE，OC，OD'两两垂直，以O为原点，的方向为x轴的正方向建立空间直角坐标系O﹣xyz，则O（0，0，0），E（1，0，0），， (7)）k AB＝﹣1平面AD'F，∴为平面AD'F的一个法向量．…设平面BD'F的法向量为＝（x，y，z），∵F（0，t，0），∴，由得取y＝3，则，∴．…∴，即，∴．∴当时，二面角A﹣D'F﹣B的大小是．…19．已知F为抛物线C：y2＝2px（p＞0）的焦点，过F的动直线交抛物线C于A，B两点．当直线与x轴垂直时，|AB|＝4．（1）求抛物线C的方程；（2）设直线AB的斜率为1且与抛物线的准线l相交于点M，抛物线C上存在点P使得直线PA，PM，PB的斜率成等差数列，求点P的坐标．【分析】（1）由题意可得|AB|＝2p＝4，即可求出抛物线的方程，（2）设直线AB的方程为y＝x﹣1，联立消去x，得y2﹣4y﹣4＝0，根据韦达定理结合直线PA，PM，PB的斜率成等差数列，即可求出点P的坐标解：（1）因为，在抛物线方程y2＝2px中，令，可得y＝±p．于是当直线与x轴垂直时，|AB|＝2p＝4，解得p＝2．所以抛物线的方程为y2＝4x．（2）因为抛物线y2＝4x的准线方程为x＝﹣1，所以M（﹣1，﹣2）．设直线AB的方程为y＝x﹣1，联立消去x，得y2﹣4y﹣4＝0．设A（x1，y1），B（x2，y2），则y1+y2＝4，y1y2＝﹣4．若点P（x0，y0）满足条件，则2k PM＝k PA+k PB，即，因为点P，A，B均在抛物线上，所以．代入化简可得，将y1+y2＝4，y1y2＝﹣4代入，解得y0＝±2．将y0＝±2代入抛物线方程，可得x0＝1．于是点P（1，±2）为满足题意的点．20．已知函数f（x）＝e﹣x﹣ax（x∈R）．（1）当a＝﹣1时，求函数f（x）的最小值；（2）若x≥0时，f（﹣x）+ln（x+1）≥1，求实数a的取值范围．【分析】（1）求出函数的导数，解关于导函数的不等式，求出函数的单调区间，从而求出函数的最小值；（2）得到e x+ax+ln（x+1）﹣1≥0．（*）令g（x）＝e x+ax+ln（x+1）﹣1，通过讨论a 的范围，确定函数的单调性，从而求出满足条件的a的具体范围即可；解：（1）当a＝﹣1时，f（x）＝e﹣x+x，则f′（x）＝﹣+1．令f'（x）＝0，得x＝0．当x＜0时，f'（x）＜0；当x＞0时，f'（x）＞0．∴函数f（x）在区间（﹣∞，0）上单调递减，在区间（0，+∞）上单调递增．∴当x＝0时，函数f（x）取得最小值，其值为f（0）＝1f（x）的最小值为1．（2）若x≥0时，f（﹣x）+ln（x+1）≥1，即e x+ax+ln（x+1）﹣1≥0（*）令g（x）＝e x+ax+ln（x+1）﹣1，则①若a≥﹣2，由（1）知e﹣x+x≥1，即e﹣x≥1﹣x，故e x≥1+x∴函数g（x）在区间[0，+∞）上单调递增，∴g（x）≥g（0）＝0．∴（*）式成立．②若a＜﹣2，令，则∴函数ϕ（x）在区间[0，+∞）上单调递增，由于ϕ（0）＝2+a＜0，．故∃x0∈（0，﹣a），使得ϕ（x0）＝0，则当0＜x＜x0时，ϕ（x）＜ϕ（x0）＝0，即g'（x）＜0．∴函数g（x）在区间（0，x0）上单调递减，∴g（x0）＜g（0）＝0，即（*）式不恒成立．综上所述，实数a的取值范围是[﹣2，+∞）．21．如图，直角坐标系中，圆的方程为x2+y2＝1，A（1，0），B（﹣，），C（﹣，﹣）为圆上三个定点，某同学从A点开始，用掷骰子的方法移动棋子．规定：①每掷一次骰子，把一枚棋子从一个定点沿圆弧移动到相邻下一个定点；②棋子移动的方向由掷骰子决定，若掷出骰子的点数为偶数，则按图中箭头方向移动；若掷出骰子的点数为奇数，则按图中箭头相反的方向移动．设掷骰子n次时，棋子移动到A，B，C处的概率分别为P n（A），P n（B），P n（C）．例如：掷骰子一次时，棋子移动到A，B，C处的概率分别为P1（A）＝0，P1（B）＝，P1（C）＝（1）分别掷骰子二次，三次时，求棋子分别移动到A，B，C处的概率；（2）掷骰子n次时，若以x轴非负半轴为始边，以射线OA，OB，OC为终边的角的余弦值记为随机变量X n，求X4的分布列和数学期望；（3）记P n（A）＝a n，P n（B）＝b n，P n（C）＝c n．，其中a n+b n+c n＝1．证明：数列{b n ﹣}是等比数列，并求a2020．【分析】（1）由概率的乘法公式，可得所求值；（2）随机变量X4的可能取值为1，﹣，结合（1）运用概率乘法公式，可得随机变量的分布列和期望；（3）易得b n＝c n，即b n﹣1＝c n﹣1，n≥2，由条件推得2b n+b n﹣1＝1，由构造等比数列，可得b n＝+•（﹣）n﹣1，即可得到所求值．解：（1）P2（A）＝•+•＝，P2（B）＝•＝，P2（C）＝•＝，P3（A）＝••+••＝，P3（B）＝（+）•＝，P3（C）＝（+）•＝；（2）随机变量X4的可能取值为1，﹣，由（1）可得P（x4＝1）＝（P3（B）+P3（C））•＝（+）•＝，P（x4＝﹣）＝（P3（A）+P3（C））•+（P3（A）+P3（B））•＝，则X4的分布列为x41﹣PE（X4）＝1•+（﹣）•＝；（3）证明：易得b n＝c n，即b n﹣1＝c n﹣1，n≥2，n≥2时，b n＝（a n﹣1+c n﹣1）＝（a n﹣1+b n﹣1），又a n﹣1+b n﹣1+c n﹣1＝1，可得2b n+b n﹣1＝1，由b n﹣＝﹣b n﹣1+﹣＝﹣（b n﹣1﹣），可得数列{b n﹣}是首项为，公比为﹣的等比数列，则b n﹣＝•（﹣）n﹣1，即b n＝+•（﹣）n﹣1，又a n＝1﹣b n＝1﹣2[+•（﹣）n﹣1]＝[1﹣（﹣）n﹣1]，故a2020＝[1+（）2019]．[选修4-4：坐标系与参数方程]22．在平面直角坐标系中，曲线C1：（a为参数）经过伸缩变换后的曲线为C2，以坐标原点为极点，x轴正半轴为极轴建立极坐标系．（Ⅰ）求C2的极坐标方程；（Ⅱ）设曲线C3的极坐标方程为ρsin（﹣θ）＝1，且曲线C3与曲线C2相交于P，Q 两点，求|PQ|的值．【分析】（Ⅰ）求出C2的参数方程，即可求C2的极坐标方程；（Ⅱ）C2是以（1，0）为圆心，1为半径的圆，曲线C3的极坐标方程为ρsin（﹣θ）＝1，直角坐标方程为x﹣y﹣2＝0，求出圆心到直线的距离，即可求|PQ|的值．解：（Ⅰ）C2的参数方程为（α为参数），普通方程为（x′﹣1）2+y′2＝1，∴C2的极坐标方程为ρ＝2cosθ；（Ⅱ）C2是以（1，0）为圆心，1为半径的圆，曲线C3的极坐标方程为ρsin（﹣θ）＝1，直角坐标方程为x﹣y﹣2＝0，∴圆心到直线的距离d＝＝，∴|PQ|＝2＝．[选修4-5：不等式选讲]23．已知函数f（x）＝|x+b2|﹣|﹣x+1|，g（x）＝|x+a2+c2|+|x﹣2b2|，其中a，b，c均为正实数，且ab+bc+ac＝1．（Ⅰ）当b＝1时，求不等式f（x）≥1的解集；（Ⅱ）当x∈R时，求证f（x）≤g（x）．【分析】（Ⅰ）当b＝1时，把f（x）用分段函数来表示，分类讨论，求得f（x）≥1的解集．（Ⅱ）当x∈R时，先求得f（x）的最大值为b2+1，再求得g（x）的最小值，根据g（x）的最小值减去f（x）的最大值大于或等于零，可得f（x）≤g（x）成立．解：（Ⅰ）由题意，当b＝1时，f（x）＝|x+b2|﹣|﹣x+1|＝，当x≤﹣1时，f（x）＝﹣2＜1，不等式f（x）≥1无解，不等式f（x）≥1的解集为∅；当﹣1＜x＜1时，f（x）＝2x，由不等式f（x）≥1，解得x≥，所以≤x＜1；当x≥1时，f（x）＝2≥1恒成立，所以不等式f（x）≥1的解集为[，+∞）．（Ⅱ）（Ⅱ）当x∈R时，f（x）＝|x+b2|﹣|﹣x+1|≤|x+b2 +（﹣x+1）|＝|b2+1|＝b2+1；g（x）＝|x+a2+c2|+|x﹣2b2|＝≥|x+a2+c2﹣（x﹣2b2）|＝|a2+c2+2b2|＝a2+c2+2b2．而a2+c2+2b2﹣（b2+1）＝a2+c2+b2﹣1＝（a2+c2+b2+a2+c2+b2）﹣1≥ab+bc+ac﹣1＝0，当且仅当a＝b＝c＝时，等号成立，即a2+c2+2b2≥b2+1，即f（x）≤g（x）．。

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