2015-2016学年度上学期新安二高期末统考热身练试卷

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2015—2016学年广东省深圳市南山区高二（上)期末数学试卷（文科）一、选择题(本题共12小题，每小题5分，共60分，每小题只有一个选项符合题意）1．“x2＞1”是“x＞1"的（）条件．A．充分不必要B．必要不充分C．充要D．既不充分也不必要2．在△ABC中，若sin2A+sin2B＜sin2C，则△ABC的形状是（）A．钝角三角形B．直角三角形C．锐角三角形D．不能确定3．下列双曲线中，渐近线方程为y=±2x的是（）A．x2﹣=1 B．﹣y2=1 C．x2﹣=1 D．﹣y2=14．设等比数列｛an }的前n项和为Sn，满足an＞0,q＞1，且a3+a5=20，a2a6=64，则S5=（）A．31 B．36 C．42 D．485．若焦点在x轴上的椭圆的离心率为，则m=（）A．B．C．D．6．函数f（x）的定义域为（a，b）,导函数f′（x）在（a，b）内的图象如图所示,则函数f （x）在开区间（a,b）内有极值点（)A．1个B．2个C．3个D．4个7．已知命题p：｜x﹣1|≥2，命题q：x∈Z；如果“p且q”与“非q"同时为假命题，则满足条件的x为（)A．{x｜x≥3}或{x|x≤﹣1，x∉Z} B．｛x｜﹣1≤x≤3，x∈Z}C．{﹣1,0，1,2,3} D．{0，1，2｝8．在△ABC中,三个内角A，B，C所对的边为a，b，c，若S△ABC=2，a+b=6， =2cosC，则c=（）A．2B．4 C．2D．39．已知数列{an ｝中a1=1，a2=，a3=,a4=，…an=…,则数列{an}的前n项的和sn=（）A． B． C． D．10．若x,y满足，则z=x+2y的最大值为（）A．0 B．1 C．D．211．函数y=x3﹣3x2﹣9x（﹣2＜x＜2）有( ）A．极大值5，无极小值B．极小值﹣27,无极大值C．极大值5，极小值﹣27 D．极大值5，极小值﹣1112．如图，F 1、F 2是双曲线=1(a ＞0，b ＞0）的左、右焦点，过F 1的直线l 与双曲线的左右两支分别交于点A 、B ．若△ABF 2为等边三角形，则双曲线的离心率为（ ）A ．4B ．C ．D ．二、填空题（本题共4小题,每小题5分,共20分）13．抛物线y=8x 2的焦点坐标为 ．14．在三角形△ABC 中，角A 、B 、C 所对的边分别为a 、b 、c ，已知A=60°，b=1，其面积为，则a= ．15．设f(x ）=xlnx ，若f′（x 0）=2，则x 0= ．16．递减等差数列｛a n ｝的前n 项和S n 满足S 5=S 10，则欲使S n 最大,则n= ．三、解答题(本题共6小题，共70分）17．已知p ：方程x 2+mx+1=0有两个不等的负实根,q ：方程4x 2+4（m ﹣2）x+1=0无实根．若“p 或q”为真,“p 且q"为假．求实数m 的取值范围．18．△ABC 的内角A ，B,C 所对的边分别为a ，b,c,acosC+ccosA=2bcosA ．（1）求A；（2）若a=，b=2，求△ABC的面积．19．设｛an ｝是等差数列，{bn｝是各项都为正数的等比数列，且a1=1，b1=2，a2+b3=10，a3+b2=7．（1）求数列｛an ｝，{bn}的通项公式;(2）设数列{bn ｝的前n项和为Sn，记，求数列｛cn｝的前n项和Tn．20．解关于x的不等式ax2﹣2（a+1)x+4＞0(a∈R）21．如图，椭圆C： +=1（a＞b＞0）的右焦点为F，右顶点、上顶点分别为点A、B，且｜AB|=|BF｜．（Ⅰ）求椭圆C的离心率；（Ⅱ）若斜率为2的直线l过点(0,2）,且l交椭圆C于P、Q两点,OP⊥OQ．求直线l的方程及椭圆C的方程．22．已知函数．（Ⅰ）若曲线y=f(x）在x=1和x=3处的切线互相平行，求a的值；（Ⅱ）求f（x）的单调区间；（Ⅲ）设g（x）=x2﹣2x，若对任意x1∈（0,2］,均存在x2∈(0，2]，使得f(x1)＜g（x2），求a的取值范围．2015-2016学年广东省深圳市南山区高二（上)期末数学试卷（文科）参考答案与试题解析一、选择题（本题共12小题,每小题5分，共60分，每小题只有一个选项符合题意）1．“x2＞1”是“x＞1”的（）条件．A．充分不必要B．必要不充分C．充要D．既不充分也不必要【考点】必要条件、充分条件与充要条件的判断．【专题】转化思想；不等式的解法及应用；简易逻辑．【分析】由x2＞1，解得:x＞1或x＜﹣1．进而判断出结论．【解答】解：由x2＞1，解得：x＞1或x＜﹣1．∴“x2＞1"是“x＞1”的必要不充分条件．故选：B．【点评】本题考查了不等式的解法、充要条件的判定,考查了推理能力与计算能力，属于基础题．2．在△ABC中，若sin2A+sin2B＜sin2C，则△ABC的形状是( )A．钝角三角形B．直角三角形C．锐角三角形D．不能确定【考点】三角形的形状判断．【专题】三角函数的图像与性质．【分析】利用正弦定理将sin2A+sin2B＜sin2C，转化为a2+b2＜c2，再结合余弦定理作出判断即可．【解答】解：∵在△ABC中,sin2A+sin2B＜sin2C，由正弦定理===2R得，a2+b2＜c2，又由余弦定理得：cosC=＜0,0＜C＜π,∴＜C＜π．故△ABC为钝角三角形．故选A．【点评】本题考查三角形的形状判断,着重考查正弦定理与余弦定理的应用，属于基础题．3．下列双曲线中，渐近线方程为y=±2x的是（）A．x2﹣=1 B．﹣y2=1 C．x2﹣=1 D．﹣y2=1【考点】双曲线的简单性质．【专题】圆锥曲线的定义、性质与方程．【分析】由双曲线方程﹣=1（a＞0，b＞0）的渐近线方程为y=±x，对选项一一判断即可得到答案．【解答】解：由双曲线方程﹣=1(a＞0，b＞0)的渐近线方程为y=±x,由A可得渐近线方程为y=±2x，由B可得渐近线方程为y=±x，由C可得渐近线方程为y=x，由D可得渐近线方程为y=x．故选:A．【点评】本题考查双曲线的方程和性质，主要考查双曲线的渐近线方程的求法，属于基础题．4．设等比数列{a n ｝的前n 项和为S n ，满足a n ＞0,q ＞1,且a 3+a 5=20，a 2a 6=64，则S 5=（ )A ．31B ．36C ．42D ．48【考点】等比数列的性质．【专题】等差数列与等比数列．【分析】利用等比中项的性质求得a 3a 5=a 2a 6，进而根据a 3+a 5=20，构造出一元二次方程求得a 3和a 5，则a 1和q 可求得，最后利用等比数列的求和公式求得答案．【解答】解：a 3a 5=a 2a 6=64,∵a 3+a 5=20,∴a 3和a 5为方程x 2﹣20x+64=0的两根,∵a n ＞0,q ＞1，∴a 3＜a 5，∴a 5=16，a 3=4, ∴q===2， ∴a 1===1，∴S 5==31．故选A ． 【点评】本题主要考查了等比数列的求和公式，等比数列的等比中项的性质的应用．解题过程中巧妙的构造出一元二次方程，较快的求得a 3和a 5，进而求得a 1和q ．5．若焦点在x 轴上的椭圆的离心率为，则m=( ）A．B．C．D．【考点】椭圆的简单性质．【专题】计算题．【分析】先根据椭圆的标准方程求得a,b，c,再结合椭圆的离心率公式列出关于m的方程，解之即得答案．【解答】解：由题意，则,化简后得m=1.5，故选A【点评】本题考查椭圆的性质与其性质的应用，注意根据椭圆的标准方程求得a,b，c,进而根据题意、结合有关性质，化简、转化、计算，最后得到结论．6．函数f（x）的定义域为（a，b），导函数f′（x）在（a，b）内的图象如图所示，则函数f（x）在开区间（a，b)内有极值点（)A．1个B．2个C．3个D．4个【考点】函数在某点取得极值的条件．【专题】导数的综合应用．【分析】根据当f’（x）＞0时函数f（x）单调递增，f'（x）＜0时f（x）单调递减，可从f′（x）的图象可知f（x）在（a,b)内从左到右的单调性依次为增→减→增→减,然后得到答案．【解答】解：从f′（x）的图象可知f(x)在（a,b)内从左到右的单调性依次为增→减→增→减,根据极值点的定义可知，导函数在某点处值为0，左右两侧异号的点为极值点，由图可知，在（a，b）内只有3个极值点．故答案为 C．【点评】本题主要考查函数的极值点和导数正负的关系．属基础题．7．已知命题p：｜x﹣1｜≥2，命题q：x∈Z；如果“p且q”与“非q"同时为假命题,则满足条件的x为（)A．｛x|x≥3}或｛x｜x≤﹣1，x∉Z｝B．{x｜﹣1≤x≤3，x∈Z}C．{﹣1，0，1，2，3} D．{0，1，2｝【考点】复合命题的真假．【专题】计算题．【分析】由题设条件先求出命题P：x≥4或x≤0．由“p且q”与“¬q”同时为假命题知0＜x＜4,x∈Z．由此能得到满足条件的x的集合．【解答】解：由命题p：｜x﹣1｜≥2，得到命题P：x﹣1≥2或x﹣1≤﹣2，即命题P：x≥3或x≤﹣1;∵¬q为假命题，∴命题q：x∈Z为真翕题．再由“p且q"为假命题，知命题P:x≥4或x≤0是假命题．故﹣1＜x＜3，x∈Z．∴满足条件的x的值为:0,1，2．故选D．【点评】本题考查命题的真假判断和应用，解题时要认真审题,仔细解答，注意公式的灵活运用．8．在△ABC中，三个内角A，B，C所对的边为a，b，c，若S△ABC=2，a+b=6， =2cosC,则c=（）A．2B．4 C．2D．3【考点】正弦定理;余弦定理．【专题】三角函数的求值；解三角形．【分析】运用正弦定理和两角和的正弦公式和诱导公式,化简可得角C，再由面积公式和余弦定理，计算即可得到c的值．【解答】解： ===1，即有2cosC=1，可得C=60°，若S△ABC=2,则absinC=2，即为ab=8，又a+b=6，由c2=a2+b2﹣2abcosC=（a+b)2﹣2ab﹣ab=(a+b）2﹣3ab=62﹣3×8=12,解得c=2．故选C．【点评】本题考查正弦定理、余弦定理和面积公式的运用，同时考查两角和的正弦公式和诱导公式的运用，考查运算能力，属于中档题．9．已知数列｛an }中a1=1，a2=，a3=，a4=，…an=…,则数列｛an}的前n项的和sn=（）A．B．C．D．【考点】数列的求和．【专题】转化思想；等差数列与等比数列．【分析】an===2．，利用“裂项求和”即可得出．【解答】解：∵an===2．∴数列{an ｝的前n项的和sn=2++…+==．故选：A．【点评】本题考查了等差数列的前n项和公式、“裂项求和”方法，考查了推理能力与计算能力，属于中档题．10．若x，y满足,则z=x+2y的最大值为（)A．0 B．1 C．D．2【考点】简单线性规划．【专题】不等式的解法及应用．【分析】作出题中不等式组表示的平面区域，再将目标函数z=x+2y对应的直线进行平移，即可求出z取得最大值．【解答】解：作出不等式组表示的平面区域，当l经过点B时，目标函数z达到最大值∴z最大值=0+2×1=2．故选：D．【点评】本题给出二元一次不等式组,求目标函数z=x+2y的最大值，着重考查了二元一次不等式组表示的平面区域和简单的线性规划等知识,属于基础题．11．函数y=x3﹣3x2﹣9x（﹣2＜x＜2）有( )A．极大值5，无极小值B．极小值﹣27，无极大值C．极大值5,极小值﹣27 D．极大值5，极小值﹣11【考点】利用导数研究函数的极值．【专题】导数的综合应用．【分析】求出y的导函数得到x=﹣1，x=3(因为﹣2＜x＜2,舍去）,讨论当﹣2＜x＜﹣1时,y′＞0；当﹣1＜x＜2时，y′＜0,得到函数极值即可．【解答】解：y′=3x2﹣6x﹣9=0，得x=﹣1,x=3，由于﹣2＜x＜2，则当﹣2＜x＜﹣1时，y′＞0；当﹣1＜x＜2时，y′＜0，=5；x取不到3，无极小值．当x=﹣1时,y极大值故选：A【点评】本题考查学生利用导数研究函数极值的能力,属于基础题12．如图，F1、F2是双曲线=1（a＞0，b＞0)的左、右焦点，过F1的直线l与双曲线的左右两支分别交于点A、B．若△ABF2为等边三角形，则双曲线的离心率为( )A．4 B．C．D．【考点】双曲线的简单性质．【专题】解三角形；圆锥曲线的定义、性质与方程．【分析】由双曲线的定义，可得F1A﹣F2A=F1A﹣AB=F1B=2a，BF2﹣BF1=2a,BF2=4a,F1F2=2c,再在△F1BF2中应用余弦定理得,a，c的关系，由离心率公式，计算即可得到所求．【解答】解：因为△ABF2为等边三角形，不妨设AB=BF2=AF2=m，A为双曲线上一点,F1A﹣F2A=F1A﹣AB=F1B=2a，B为双曲线上一点，则BF2﹣BF1=2a，BF2=4a，F1F2=2c，由,则，在△F1BF2中应用余弦定理得：4c2=4a2+16a2﹣2•2a•4a•cos120°,得c2=7a2,则．故选：B．【点评】本题考查双曲线的定义、方程和性质，考查余弦定理的运用，考查运算能力，属于中档题．二、填空题(本题共4小题，每小题5分，共20分）13．抛物线y=8x2的焦点坐标为．【考点】抛物线的简单性质．【专题】计算题；圆锥曲线的定义、性质与方程．【分析】化抛物线方程为标准方程,即可求得焦点坐标．【解答】解：抛物线y=8x2可化为，焦点在y轴上∵，∴∴抛物线y=8x2的焦点坐标为故答案为:【点评】本题考查抛物线的性质，化抛物线方程为标准方程是关键．14．在三角形△ABC中，角A、B、C所对的边分别为a、b、c，已知A=60°,b=1，其面积为,则a= ．【考点】余弦定理的应用．【专题】计算题；方程思想；分析法；解三角形．【分析】根据三角形的面积公式，求出c,然后利用余弦定理即可得到a的值【解答】解：∵A=60°,b=1,△ABC的面积为,∴S=bcsinA=csin60°=，△即c=,解得c=4，由余弦定理得a2=b2+c2﹣2bccos60°=1+16﹣2×1×4×=13，解得a=，故答案为：．【点评】本题主要考查余弦定理的应用，以及三角形的面积公式的运用，考查运算能力，属于基础题．15．设f （x ）=xlnx ，若f′（x 0）=2,则x 0= e ．【考点】导数的运算．【专题】计算题．【分析】先根据乘积函数的导数公式求出函数f(x)的导数,然后将x 0代入建立方程，解之即可．【解答】解：f(x ）=xlnx∴f’（x ）=lnx+1则f′（x 0）=lnx 0+1=2解得：x 0=e故答案为：e【点评】本题主要考查了导数的运算，以及乘积函数的导数公式的运用，属于基础题之列．16．递减等差数列｛a n ｝的前n 项和S n 满足S 5=S 10，则欲使S n 最大,则n= 7或8 ．【考点】等差数列的前n 项和．【专题】计算题．【分析】根据题意,由S 5=S 10，可得S 10﹣S 5=a 6+a 7+a 8+a 9+a 10=0，结合等差数列的性质，可得a 8=0,又由数列｛a n }是递减等差数列，则可得a 1＞a 2＞…a 7＞a 8=0＞a 9…,分析可得答案．【解答】解:根据题意，数列{a n ｝满足S 5=S 10，则S 10﹣S 5=a 6+a 7+a 8+a 9+a 10=0，由等差数列性质得：5a 8=0,可得a 8=0，又由数列{a n }是递减的等差数列，则由a 1＞a 2＞…a 7＞a 8=0＞a 9…，则当n=7或8时,s n 取最大值,故答案为7或8．【点评】本题考查等差数列前n项和的性质，要牢记其前n项和s取最大或最小值的条件以及n判断方法．三、解答题（本题共6小题，共70分）17．已知p：方程x2+mx+1=0有两个不等的负实根，q：方程4x2+4(m﹣2）x+1=0无实根．若“p 或q"为真,“p且q”为假．求实数m的取值范围．【考点】复合命题的真假；一元二次方程的根的分布与系数的关系．【专题】分类讨论;简易逻辑．【分析】根据题意，首先求得p、q为真时m的取值范围，再由题意p，q中有且仅有一为真，一为假，分p假q真与p真q假两种情况分别讨论,最后综合可得答案．【解答】解：由题意p，q中有且仅有一为真，一为假，若p为真，则其等价于，解可得,m＞2；若q为真，则其等价于△＜0，即可得1＜m＜3，若p假q真，则,解可得1＜m≤2；若p真q假，则，解可得m≥3；综上所述：m∈（1，2]∪[3，+∞）．【点评】本题考查命题复合真假的判断与运用，难点在于正确分析题意,转化为集合间的包含关系，综合可得答案．18．△ABC的内角A，B，C所对的边分别为a，b，c，acosC+ccosA=2bcosA．（1）求A；（2)若a=，b=2，求△ABC的面积．【考点】余弦定理；正弦定理．【专题】转化思想；解三角形．【分析】(1）利用正弦定理、和差公式即可得出；（2）利用余弦定理可得c ，再利用三角形面积计算公式即可得出．【解答】解:（1）∵acosC+ccosA=2bcosA,由正弦定理可得：sinAcosC+sinCcosA=2sinBcosA ， 化为：sin （A+C ）=sinB=2sinBcosA ，sinB≠0,可得cosA=，A∈（0，π）， ∴A=．（2）由余弦定理，得a 2=b 2+c 2﹣2bccosA ，∴7=22+c 2﹣4ccos ，化为c 2﹣2c ﹣3=0，解得c=3．故△ABC 的面积为bcsinA=×3×=． 【点评】本题考查了正弦定理余弦定理、和差公式、三角形面积计算公式,考查了推理能力与计算能力，属于中档题．19．设｛a n }是等差数列，{b n }是各项都为正数的等比数列，且a 1=1，b 1=2，a 2+b 3=10,a 3+b 2=7．（1)求数列｛a n ｝，｛b n ｝的通项公式；(2）设数列｛b n ｝的前n 项和为S n ，记，求数列｛c n }的前n 项和T n ． 【考点】数列的求和；数列递推式．【专题】综合题；方程思想;转化思想;数学模型法；等差数列与等比数列．【分析】（1）利用等差数列与等比数列的通项公式即可得出；（2）利用“错位相减法"、等比数列的通项公式与前n 项和公式即可得出．【解答】解：（1）设等差数列｛a n }的公差为d ，等比数列｛b n ｝的公比为q ，且a 1=1，b 1=2，a 2+b 3=10，a 3+b 2=7． ∴，即， 消去d 得2q 2﹣q ﹣6=0,（2q+3)（q ﹣2）=0，∵{bn}是各项都为正数的等比数列，∴q=2，d=1,∴an =n，bn=2n．（2）Sn=2n+1﹣2，…c n =an•(+1）=n•2n，设Tn=1•21+2•22+3•23+…+n•2n,2Tn=1•22+2•23+…+（n﹣1)•2n+n•2n+1，相减，可得Tn=(n﹣1）•2n+1+2．【点评】本题考查了等差数列与等比数列的通项公式与前n项和公式、“错位相减法"，考查了推理能力与计算能力，属于中档题．20．解关于x的不等式ax2﹣2（a+1）x+4＞0(a∈R）【考点】一元二次不等式的解法．【专题】不等式的解法及应用．【分析】对a分类:a=0，a＜0，0＜a＜1，a=1，a＞1，分别解不等式即可．【解答】解：ax2﹣2（a+1）x+4＞0⇔（ax﹣2）（x﹣2）＞0…(ⅰ）a=0时，x﹣2＜0⇔x∈（﹣∞，2）…(ⅱ）0＜a＜1时,…（ⅲ）a=1时，(x﹣2)2＞0⇔x∈(﹣∞，2)∪（2,+∞）…（ⅳ）a＞1时，…(ⅴ）a＜0时，…【点评】本题考查不等式的解法，考查分类讨论思想,是中档题．21．如图，椭圆C: +=1（a＞b＞0）的右焦点为F,右顶点、上顶点分别为点A、B，且|AB|=｜BF|．（Ⅰ)求椭圆C的离心率；（Ⅱ）若斜率为2的直线l过点（0，2)，且l交椭圆C于P、Q两点，OP⊥OQ．求直线l的方程及椭圆C的方程．【考点】直线与圆锥曲线的综合问题．【专题】综合题；圆锥曲线的定义、性质与方程．【分析】(Ⅰ)利用|AB|=｜BF｜，求出a，c的关系，即可求椭圆C的离心率;（Ⅱ）直线l的方程为y﹣2=2（x﹣0），即2x﹣y+2=0与椭圆C：联立,OP⊥OQ，可得,利用韦达定理，即可求出椭圆C的方程．【解答】解:(Ⅰ）由已知,即,4a2+4b2=5a2，4a2+4（a2﹣c2）=5a2,∴．…(Ⅱ）由（Ⅰ）知a2=4b2，∴椭圆C：．设P(x1，y1)，Q(x2,y2)，直线l的方程为y﹣2=2（x﹣0)，即2x﹣y+2=0．由，即17x2+32x+16﹣4b2=0．．，．…∵OP⊥OQ，∴，即x1x2+y1y2=0,x1x2+（2x1+2）（2x2+2)=0，5x1x2+4（x1+x2)+4=0．从而，解得b=1，∴椭圆C的方程为．…【点评】本题考查椭圆方程，考查直线与椭圆的位置关系，考查韦达定理的运用，属于中档题．22．已知函数．（Ⅰ）若曲线y=f（x）在x=1和x=3处的切线互相平行，求a的值；（Ⅱ）求f（x）的单调区间；（Ⅲ）设g（x）=x2﹣2x，若对任意x1∈(0，2]，均存在x2∈（0,2］，使得f（x1）＜g（x2)，求a的取值范围．【考点】导数在最大值、最小值问题中的应用；利用导数研究曲线上某点切线方程．【专题】综合题；压轴题．【分析】(Ⅰ）由函数，知（x＞0）．由曲线y=f（x）在x=1和x=3处的切线互相平行，能求出a的值．（Ⅱ）（x＞0）．根据a的取值范围进行分类讨论能求出f(x）的单调区间．（Ⅲ）对任意x1∈（0，2］，均存在x2∈（0，2］，使得f（x1）＜g（x2），等价于在（0，2］上有f（x）max ＜g（x)max．由此能求出a的取值范围．【解答】解：（Ⅰ)∵函数,∴（x＞0)．∵曲线y=f(x)在x=1和x=3处的切线互相平行,∴f'（1）=f'(3)，即，解得．（Ⅱ）（x＞0）．①当a≤0时,x＞0,ax﹣1＜0，在区间（0，2)上，f'（x）＞0；在区间（2,+∞)上f’(x）＜0,故f（x）的单调递增区间是(0，2）,单调递减区间是(2，+∞）．②当时，，在区间（0，2）和上，f'（x)＞0;在区间上f’（x）＜0,故f（x）的单调递增区间是（0，2)和,单调递减区间是③当时，，故f(x）的单调递增区间是(0，+∞）．④当时，，在区间和（2,+∞）上,f'（x）＞0；在区间上f’（x）＜0，故f（x)的单调递增区间是和（2，+∞）,单调递减区间是．(Ⅲ)由已知，在（0，2]上有f（x）max ＜g（x）max．由已知,g（x）max=0，由（Ⅱ）可知，①当时，f（x）在（0，2]上单调递增，故f（x）max=f（2）=2a﹣2（2a+1）+2ln2=﹣2a﹣2+2ln2，所以,﹣2a﹣2+2ln2＜0，解得a＞ln2﹣1，故．②当时,f（x）在上单调递增，在上单调递减，故．由可知，2lna＞﹣2，﹣2lna＜2，＜0，所以，﹣2﹣2lna＜0，f（x）max综上所述，a＞ln2﹣1．【点评】本题考查导数在求函数的最大值与最小值问题中的综合运用，考查运算求解能力，推理论证能力;考查化归与转化思想．对数学思维的要求比较高，有一定的探索性．综合性强，难度大，是高考的重点．易错点是分类不清导致致出错,解题时要认真审题，仔细解答,注意分类讨论思想的合理运用．。

湖南省衡阳市市石鼓区新安中学高二语文上学期期末试题含解析一、现代文阅读（35分，共3题）1. 阅读下面一篇文章，然后回答问题。

（15分）读书断想钟长江①书和太阳一样，能够给人带来光明和温暖，是我们须臾不可或缺的精神食粮。

塞缪尔说过，好书可以引为诤友。

好书与我们亲善相处，年轻时从中汲取乐趣与教诲，到鬓发染霜，则带给我们以亲抚和安慰。

②一窗明月的夜晚，远离了纷繁喧嚣的尘世、是是非非的人群，一书在握，热茶一杯，静心读了下去，渐渐便觉茶香满口、书香满纸，尽可以刘伶醉酒，放浪形骸——开心处开怀大笑，伤心处黯然神伤，激奋处拍案而起……仰观苍天，俯察大地，一种腋生双翼的感觉油然而生。

心骛八极，思接千载，神游古今，可以大荒无限，咫尺千里，豆芥巨观，缩龙成寸。

自由自在晶莹剔透的灵魂啊，分明已化作奔月的嫦娥。

③很静很美的冬夜，就着火炉，温一壶酒，读那些又静又美的书：可以是《红楼梦》，可以是《瓦尔登湖》、《追忆似水年华》，可以是张岱的《西湖梦寻》、袁枚的《小仓山房尺牍》、蒲松龄的《聊斋志异》，也可以是周作人的小品、唐弢的书话。

当然，日本古代作家的随笔也可以，美国作家包斯威尔的《约翰生传》也可以，《板桥杂记》也没什么不可以。

④古人刘向云：书犹药也，善读之可以医愚。

列夫·托尔斯泰说：理想的书籍是智慧的钥匙。

忧愁烦恼袭来时，不妨读唐诗，吟宋词，看艾芜的《南行记》。

如此，春天的风秋天的雨，会渐渐冷却你急躁的情绪；弯弯的小桥潺潺的流水，可慢慢松弛你绷紧的神经。

不知不觉中，你悠悠南行，那溪头卧剥莲蓬的小儿，那岭南见客羞遮面的村姑，无一不使你怡然自得，超然物外。

此时烦恼何在!忧愁何存!⑤一个生活得丰富圆满的人，是不会没有寂寞的。

对于每个人来说，他人永远是水，自身永远是鱼。

即或能在水里自由地游弋，也不能够溶解于水，更不能与水合一。

固然有挚友的安慰，情人的蜜语，但是所有的思想与情感，只能被理解，无法被分享。

无论一人独处，还是朋友聚会，那一丝半缕的寂寞之感，都会不召自来，涌上心头。

新安中学2016—2017学年度第一学期高二年级期末考试英语试卷第Ⅰ卷第一部分听力（共两节，满分30分）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，并标在试卷的相应位置。

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。

每段对话仅读一遍。

1. What is the man going to do?A.Repair the car.B. Make lunch.C. Go fishing.2. Who is ill?A. Mary.B. Bill.C. John.3. What does the man want to do?A. Stay in bed.B. Take a journey.C. Breathe fresh air.4. What can we learn from the conversation?A. T he woman’s mother was ill.B. The woman did n’t go home last weekend.C. The woman advised the man to put on more clothes.5. What are the two speakers mainly talking about?A. A cupboard.B. A bookshelf.C. A fridge.第二节 (共15小题,每小题1.5分,满分22.5分)听下面5段对话或独白。

每段对话后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟，听完每段对话后，各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6段材料，回答6-8题6. How does the man feel?A. Angry.B. Excited.C. Disappointed.7. Whom has the woman told the man’s salary to?A. Her brothers.B. Mrs Wallace.C. Her neighbors.8. What is the woman doing?A. Asking for help.B. Making an apology.C. Offering information.听第7段材料，回答9-11题9. What did the man think of the football match?A. Boring.B. Disappointing.C. Wonderful.10. What happened to the woman yesterday afternoon?A. She lost her bike.B. She fell off her bike.C. She hurt her right leg.11. When did the football match begin?A. At 3:10.B. At 3:20.C.At 3:30.听第8段材料，回答12-14题12. What could the man speaker most probably be?A. A policeman.B. A taxi driver.C. A shop assistant.13. Where are the speakers?A. On Frost Street.B. On Main Street.C. On Wall Street.14. How far away is the department st ore?A. One block.B. Two blocks.C. Three blocks.听第9段材料，回答15-17题15. Why has the woman come to school so early?A. To find a parking lot.B. To avoid traffic jams.C.To review her lessons.16. Why has the woman been working hard?A. To get high marks.B. To win support from the man.C.To get a job from the government.17. How does the man feel?A. Happy.B. Worried.C. Confident.听第10段材料，回答18-20题18. Why did the speaker go to Egypt?A. To keep warm.B. To visit pyramids.C. To see camels.19. Where is the speaker now?A. In France.B. In China.C. In the USA.20. What can students learn about in the speaker’s classes?A.How they can find jobs abroad.B. How Americans live and work.C.How they can learn English.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，从每题所给的四个选项(A、B、C、D)中，选出最佳选项，并在答题卡上将该项涂黑。

2015-2016学年第一学期宝安区期末调研测试卷高二理科数学2016.1本试卷共6页，22小题，满分150分•考试用时120分钟.注意事项：1 •答卷前，考生首先检查答题卡是否整洁无缺损，监考教师分发的考生信息条形码是否正确；之后务必用 0.5毫米黑色字迹的签字笔在答题卡指定位置填写自 己的学校、姓名和考生号，同时，将监考教师发放的条形码正向准确粘贴在答 题卡的贴条形码区，请保持条形码整洁、不污损2 •选择题每小题选出答案后，用 2B 铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其它答案，答案不能答在试卷上.不按要求 填涂的，答案无效.3 .非选择题必须用 0.5毫米黑色字迹的签字笔作答，答案必须写在答题卡各题目指定区域内相应位置上，请注意每题答题空间，预先合理安排；如需改动，先 划掉原来的答案，然后再写上新的答案； 不准使用铅笔和涂改液.不按以上要求 作答无效. 4 •作答选做题时，请先用2B 铅笔填涂选做题的题号对应的信息点，再做答.漏涂、错涂、多涂的答案无效.一、选择题：本大题共 12小题，每小题5分，满分 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 .不等式X 2-2x -5 - 2x 的解集是（)A .| x 亠 5或 x _ -1 匚B .^x | x 5或 x ：：： -1C . ：x|-1 ：： x ：：5；—&—¥■—FD—►.| - 仁 x 二 5』 2.已知向量a =（-1,0,2）,b = （1,1,0），且a kb 与2b -a 相互垂直，则k 值为（ ）2 24.若方程E ：-上 y 1表示焦点在y 轴上的双曲线，则实数m 的取值范围为1 -m m -2（） A . 1,2 B .：,1） （2, ：： C . （-：：,2） D . （1,：：）5.在=ABC 中,a = 2、3,b= 2、2,B = 45，则角 A 等于（）7 3 A .B .-553.“ x 2 = y 2”是“ x = y ”的()A .充分不必要条件C .必要不充分条件C .丄D . 15B .充分必要条件D .既不充分也不必要条件A. 30 B . 60 C . 60 或120 D . 30 或1506•已知-14盘,8成等差数列，—1,b ib ,b 3,-4成等比数列，那么 岂空 的值为( )b 255A • 5B • -5C •D •-227.若动点M(x, y)始终满足关系式.x 2 (y 2)^ . x 2 (y-2)2=8，则动点M 的轨迹方程为()2 2 2 2 2 2 2 2xy, xy, xy, xy,A •1 B •1 C •1 D • 116 12 12 16 12 16 16 128 •已知等差数列:a n [的前n 项和S n ,且满足S n 1 =n 2 -n -2,则a ^：()A • 4B • 2C • 0D • -2x - y _ 09•已知x, y 满足约束条件《x + yE2，若z = x + ay 的最大值为4,则a=()、y 兰0A • 3B • 2C • -2D • -310 •在 ABC 中，a =2,c =1，则角C 的取值范围是()(八31A •陀丿B • —,—<6 3 .丿C •—,— 丨 <6 2丿D • (0,611 •已知直线l :^kx 2k 1与抛物线C : y 2 = 4x ，若I 与C 有且仅有一个公共点，则实数k 的取值集合为()尸r f1 IA • J -1,- >B • {-1,。

河南省洛阳市新安县第二高级中学2016届毕业班考前化学适应性训练（5月月考）（解析版）1．已知X、Y是主族元素，I为电离能，单位是KJ/mol．根据下表所列数据判断错误的是（）A．元素X的常见化合价是+1价B．元素Y是ⅢA族的元素C．元素X与氯形成化合物时，化学式可能是XClD．若元素Y处于第3周期，它可与冷水剧烈反应【答案】D【解析】试题分析：A．元素X的第二电离能比第一电离能大很多，说明X较容易失去一个电子，第二个电子就不再容易失去，所以X原子核外最外层只有一个电子，常见化合价是+1价，A 正确；B．元素Y失去三个电子后要是去第四个电子很困难，说明Y的最外层有3个电子，是ⅢA族的元素，B正确；C．X原子核外最外层只有一个电子，与氯形成化合物时，化学式可能是XCl，C正确；D．若元素Y处于第3周期，那么Y是Al，Al与冷水不反应，D错误，答案选D。

考点：考查电离能与原子结构2．下列说法正确的是（）①合金的硬度一般比它的各成分金属的小，熔点一般比它的各成分金属的高；②SiC俗称金刚砂，可用作砂纸、砂轮的磨料；③漂白粉和漂白精的有效成分都是Ca(ClO)2；④NO在人体的血管系统内具有传送信号的功能，也是大气污染物之一；⑤氧化铜可作为制造玻璃、搪瓷的红色颜料。

A．①③⑤ B．②③④C．①②⑤D．②③⑤【答案】B【解析】合金的硬度一般比它的各成分金属的大，熔点一般比它的各成分金属的低，①错；氧化铜是黑色固体，⑤错。

3．制取一氯乙烷最好采用的办法是（）A．乙烷和氯气反应 B．乙烯和氯气反应C．乙烯和氯化氢反应 D．乙烯和氯气、氢气反应【答案】C【解析】试题分析：A、乙烷和氯气发生取代反应，产物为多种产物的混合物，难以分离，错误；B、乙烯和氯气发生加成反应，产物为1,2—二氯乙烷，错误；C、乙烯和氯化氢发生加成反应，生成氯乙烷，正确；D、乙烯和氯气、氢气反应的产物为多种物质的混合物，正确。

考点：考查有机物的制备及相关物质的性质。

2015年新安二高高三考前热身练英语试题二英语第一部分听力（略）第二部分阅读理解（共两节，满分60分）第一节（共15小题；每小题3分，满分45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，并在答题卡上将该项涂黑。

AWhether you are already planning your next vacation, or just daydreaming about wonderful beaches, we give you the best places to visit in Mexico.Mexico CityMexico City was founded by Amerindians and is one of the oldest capital cities in the Americas. It was originally built by the Aztecs in 1325 as Tenochtitlan, so the history behind this city is one of the best attractions it has to offer. Around every corner of the Mexican political, cultural and financial center, a secret waits to be revealed by curious tourists. Don‟t let rumors and myths about this city fool you, Mexico City cleaned up nicely and now has a crime rate about a third of Washington, DC‟s. MazatlanA great destination if what you like is relaxing at the beach. Once famous in the 1970s and 1980s, Mazatlan is re-attracting tourists to its historic sites, renovated museums, and, of course, gorgeous beaches.CozumelThis island‟s economy is based mainly on tourism. And, no wonder, for it is extremely beautiful and a great place to learn about Mayan culture. Located in the Caribbean Sea, tourists like to scuba dive or snorkel in the crystal clear warm waters of Cozumel. Other water-related attractions are the cruise boats and the coral reef.TulumTulum is the site of a Mayan walled city and the ruins are located along the coast of the Yucatan Peninsula. But this port is not only of historical relevance, it is also one of the new beach hotspots. A tranquil and rural retreat, Tulum is chosen by tourists who want to void the mega-resorts that have taken over other Mexican beaches.Playa del Carmen“Playa” is now a lively resort town with breathtaking beaches, entertainment venues and commercial buildings, but it once was only a small fishing village. Bars, restaurants, nightclubs, souvenir shops: everything is right by the beach and available at all hours.21. According to the passage Mexico City ________.A. is the oldest city in the history of MexicoB. has a history of more than over 600 yearsC. witnesses the highest crime rate on earthD. is not away from Washington, DC22. According to the passage the feature of Tulum is _______.A. the Yucatan Peninsula is the ruins of a cityB. the wonderful beach there attracts visitorsC. it is regarded as the only city without wallsD. it is a convenient place to go to other beaches23. If you take interest in night life you can go to ________.A. Mexico CityB. Playa del CarmenC. CozumelD. MazatlanBBack when the telegraph was the fastest method of long-distance communication, a young man applied for a job as a Morse Code operator.Answering an ad in the newspaper, he went to the office address that was listed.When he arrived, he entered a large, busy office filled with noise and clatter（喧闹声）, including the sound of the telegraph in the background. A sign on the receptioni st‟s counter instructed job applicants to fill out a form and wait until they were called to enter the inner office.The young man filled out his form and sat down with the seven other applicants in the waiting area. After a few minutes, the young man stood up, crossed the room to the door of the inner office, and walked right in.Naturally the other applicants came to life, wondering what was going on. They whispered among themselves that they hadn‟t heard any calls yet. They assumed that the young man who went into the office made a mistake and would be disqualified.Within a few minutes, however, the employer accompanied the young man out of the office and said to the other applicants, “Gentlemen, thank you very much for coming, but the job has just been filled.”The other applicants began grumbling about the employer, and one spoke up saying, “Wait a minute, I don‟t understand. He was the last to come in, and we never even got a chance to be interviewed. Yet he got the job. That‟s not fair!”The employer said, “I‟m sorry, but the last several minutes while you‟ve been sitting here, the telegraph has been ticking out the following message in Morse Code: …If you understand this message, then come fi ght in. The job is yours.‟ None of you heard it or understood it. This young man did. The job is his.”24. How did the young man get the job according to the passage?A. By jumping the queue.B. By listening attentively.C. By using good relation with the employer.D. By being the first to arrive.25. Why did the other applicants come to life?A. Because it seemed impossible for them to get the job.B. Because employment result was being announced then.C. Because they found a young man in without permission.D. Because they had known what had happened.26. What does the underlined part in the last but one paragraph mean?A. To keep complaining in an unhappy way.B. To shout in honor of something.C. To discuss something in a loud voice.D. To reason with somebody for some reason.27. What would be the best title of the passage?A. Are you waiting to be called?B. How to get employed?C. Are you the first to arrive?D. Tips for you to get employedCTime spent with families can be much stress during the holidays, sometimes even more than the hurried schedules, gifts to buy, food to cook and the money to prepare that can come with this time of year. Relationships differ from family to family. However, every family has its own unique relationship or situations that could cause some family members to be afraid of the holidays.To make the best of your time together, it might be wise to let go of past conflicts and the expectations unfulfilled and simply enjoy your visit, no matter how short.We might have family members going through something challenging in their life that they really do not feel like discussing over the holidays. For some questions like: “Where are you going to college next year?” or “When will there be babies?” can cause great anxiety to the holidays.You might feel that these are harmless questions to ask, yet to your loved one, they could be their worst fear. For someone who‟s been trying to have a baby f or months or even years, that question is just another pain in the heart. They, too, may wish to be celebrating the season with a child. There might be a medical reason that your family member has not had children or the timing is not right for them.A similar situation could be true for single family members. By asking about a significant other, it might appear that you are thinking little of the life they are living because they are not with someone. They might be perfectly happy on their own.For some, the holidays can highlight a difficult year. The loss of a job or loved one, depression or any variety of life events may have caused a challenging year. Don‟t treat them as if they‟re broken or fragile. Try to treat them like everyone else, so they do not feel singled out.Respect the choices that loved ones have made in their lives —and celebrate where they are today. Be generous and merciful with each person, even the family members who tend to rub you the wrong way at family gatherings. They could be experiencing their own challenges.28. What does the author want to tell in the passage?A. Holidays are not always the most wonderful time.B. Every one is seeking for their pleasant holidays.C. Family members should have a good relationship.D. We should take full advantage of our holidays.29. In order to have a happy holiday ________.A. you should prepare enough food and giftsB. you should be informed of all the situationsC. you should avoid referring to unpleasant topicsD. you should have the ability to deal with conflicts30. While talking with a single mother ________.A. you should show concern over her former husbandB. you mustn‟t make her feel embarrassedC. you can think little of her and her childrenD. you need show much enthusiasm for her life31. If your friends makes you uncomfortable _________.A. you should take his challenge at onceB. you should be kind and generousC. you should make him cut off from othersD. you should leave the conflicts risingDOver the past few years, organic products have increasingly found their way into consumers‟ shopping baskets. But what makes them choose organic over highly processed products?There are reasons to be organic and reasons against being organic. It is not necessarily an either-or decision. Rather, the decision to consume or not to consume organic food is dependent on the economic, educational and social circumstances of the consumer.According to many researchers, there is a connection between organic food and protection of the environment. Many consumers who know about this connection take it as an incentive to go organic. To reduce their carbon footprint, many even rely only on local organic products.However, depending on educational level and social status, not every consumer actually knows and cares about the connection between the environment and organic food. That is, some consumers choose not to be organic because they prefer sticking with their old habits. To them, one reason not to go organic is that organic food differs from the food they are used to. For some, organic food tastes different compared with the food they normally consume.And as organic products generally don’t use preservatives（防腐剂）, they are easier to go bad. Consumers who choose not to go organic therefore feel organic food doesn‟t bring them any benefits, no matter what the commercials say.Organic products are advertised in related with a healthier way of life, even though there is no scientific proof for this. Moreover, organic food is generally more expensive than common food. For these two reasons, organic products have become a symbol for a somewhat classy lifestyle. Therefore, some consumers who choose to buy organic products don‟t necessarily have to know about the differences between organic and processed food. Rather, they need to be able to afford it. At the same time, there are consumers who actually care about the production methods behind their food, but their economic situation simply does not allow them to go organic.32. What is the passage mainly talking about?A. The true cost of organic food products.B. The real value of organic agriculture.C. The economic situation on the earth.D. The ideas about how to consume.33. The underlined word “incentive” in the passage means _________.A. prideB. motiveC. resultD. gift34. Some people dislike organic food in that _________.A. organic products don‟t use preservativesB. they can‟t make money from organic foodC. they can‟t get rid of their old habitsD. organic food will change their tastes35. It can be inferred from the last paragraph that _________.A. most rich people will prefer to purchase organic foodB. some advertisements about Organic products are falseC. income is the base as to whether to buy organic foodD. traditional food will be replaced in the soon future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。

河南省新安县第二高级中学2016—2017学年度高二开学考化学试题（解析版） 1．设N A 为阿佛加德罗常数，下列说法正确的是（ ）①标准状况下，11.2L 以任意比例混合的氮气和氧气所含的原子数为N A ②同温同压下，体积相同的氢气和氩气所含的分子数相等 ③1L 2mol/L 的MgCl 2溶液中含氯离子为4N A ④标准状况下22.4LH 2O 中分子数为N A ⑤32g O 2和O 3混合气体中含有原子数为2N AA ．①②③④ B．③④ C．①③④ D．①②③⑤ 【答案】D 【解析】试题分析：①标准状况下，11.2L 以任意比例混合的氮气和氧气所含的原子数为11.2222.4/A A LN N L mol⨯=,故正确。

②同温同压下，体积相同的氢气和氩气所含的分子数相等，也正确。

③1L 2mol/L 的MgCl 2溶液中含氯离子为4N A ，故正确。

④标准状况下，H 2O 为液态，不能计算出其分子数，故错误。

⑤32g O 2和O 3混合气体中含有原子数为32216A A N N =,故正确。

考点：阿佛加德罗常数N A点评：考查阿佛加德罗常数N A 的应用，难度不大，考生要特别注意标准状况下的具体条件，水为液态，不是气态。

2．下列物质中，不能通过单质之间反应直接制取的是A ．FeCl 2B ．FeCl 3C ．Na 2O 2D ．Cu 2S 【答案】A 【解析】试题分析：A 、铁和氯气反应生成氯化铁，不是氯化亚铁，选A ；B 、氯气和铁反应生成氯化铁，不选B ；C 、钠和氧气在加热条件下反应生成过氧化钠，不选C ；D 、硫和铜反应生成硫化亚铜，不选D 。

考点：金属和非金属的反应。

3．下列各项中的两种物质在溶液中的反应可用同一离子方程式表示的是（ ） A.醋酸与氢氧化钠、盐酸与氢氧化钡溶液 B.碳酸钠与足量盐酸、碳酸氢钾与足量稀硫酸C.硫酸与氢氧化钡溶液、NaHSO 4溶液中滴入少量Ba （OH ）2D.碳酸钠与澄清石灰水、碳酸氢钠与澄清石灰水【答案】C【解析】NaHSO4与少量Ba（OH）2溶液作用，其化学方程式为：NaHSO4+Ba（OH）Ba（OH）2电离出的Ba2+、OH-完全反应，其离子方程式表示为：SO+Ba2++2OH-====BaSO4↓+2H2O2H++ 242====BaSO4↓+2H2O+Na2SO44．现有碳酸钠、碳酸氢钠、氢氧化钠、氯化铁和苯酚五种物质的溶液，它们之间有如图所示转化关系，图中每条线两端的物质可以发生反应，下列判断不合理的是( )A.X一定是碳酸钠溶液B.Y一定是氯化铁溶液C.Z可能为碳酸氢钠溶液D.M、N必定各为苯酚、氢氧化钠溶液中的某一种【答案】A【解析】上述五种物质相互反应：FeCl3溶液能与四种物质反应、碳酸钠两种、碳酸氢钠两种、氢氧化钠三种、苯酚三种，Y一定是氯化铁溶液，X或Z为碳酸钠。

浙江省严州中学高三年级第二次模拟考试英语学科试题第I卷第一部分：英语知识运用（共两节，满分30分）第一节：单项填空（共20小题；每小题0. 5分，满分10分）从A、B、C和D四个选项中，选出可以填入空白处的最佳选项，并在答题纸上将该选项标号涂黑。

1. — I’ve got something weighing on my mind at the moment. Could you give me some advice?— ______. Tell me all about it and I’ll do what I can.A. Never mindB. No wayC. No problemD. Forget it2. The company needs to improve its communication system to show that it ______ allits staff.A. consultsB. valuesC. educatesD. attaches3. We’ve got a gen eral idea of what we want, but nothing ______ at the moment.A. concreteB. ordinaryC. standardD. abstract4. Linda acts as if she were the boss, ______ people around.A. orderedB. orderingC. to orderD. having ordered5. We have our bad times but on the whole we’re ______ happy.A. rarelyB. merelyC. fairlyD. chiefly6. Two of my friends are having their weddings on the same day, and I’m in a real______ about which one to go to.A. situationB. choiceC. trickD. dilemma7. Cynthia is not ashamed of what she does, ______ she ends up doing something wrong.A. as thoughB. in caseC. even ifD. for fear8. When you reach the other end of the bridge, I ______ right there to show you the way.A. waitB. have waitedC. was waitingD. will be waiting9. — Hello, I’m calling to complain about one of your sales staff.— Oh dear, ______A. you must be kidding!B. what a shock!C. why not?D. what happened?10. Most companies’ marketing effort is focused on getting customers, with littleattention ______ to keeping them.A. payingB. to payC. paidD. having paid11. ______ is surprising is that there are so many Chinese restaurants in Toronto.A. ItB. ThatC. SuchD. What12. Human history is rich ______ examples of our ability to overcome and rise abovedifficulties.A. withB. ofC. onD. by13. She would always be an outsider here, no matter how much she ______ try.A. couldB. mustC. mightD. need14. I’ve had enough of my neighbor. I’m going to ______ their noisy parties right now.A. put an end toB. give way toC. look up toD. come up to15. We should ______ for possible delays on the motorway and leave an hour earlier.A. apologizeB. allowC. accountD. ask16. ______ how mirrors produce image, we need to know what light does.A. UnderstandingB. UnderstoodC. To understandD. Having understood17. The rise in consumer spending is an encouraging ______ that the economy may be recovering.A. symptomB. solutionC. subjectD. sign18. — Have you read the newly-published novel by J. K. Rowling?— Yes. I ______ recommend this book to anyone who is sick of the same old stories.A. brieflyB. instantlyC. fullyD. awfully19. Not ______ with the quality of your goods, I will certainly not advise othersto buy them.A. to impressB. impressingC. having impressedD. impressed20. There are a couple of matters which I want to ______ today for discussion.A. bring upB. set upC. make upD. put up第二节：完形填空（共20小题，每小题1分，满分20分）阅读下面短文，掌握其大意，然后从 21～40 各题所给的四个选项（A、B、C和D）中，选出最佳选项，并在答题纸上将该选项标号涂黑。

新安二高15-16学年下学期高一第一次月考（英语）本试卷满分150分考试时间120分钟第I卷（三部分共115分）第一部分听力（共两节，满分30分）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，并标在试卷的相应位置。

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。

每段对话仅读一遍。

1. Where does the conversation take place?A. At the airport.B. In the post office.C. At the railway station.2. How many people died in the accident?A. Five.B. Fifty.C. Fifteen.3. What is Susan doing?A. Helping her friends.B. Cleaning the house.C. Putting up the paper cuts.4. What is the weather like now?A. Hot.B. Warm.C. Cold.5. What class will the speakers have?A. Math.B. English.C. Chemistry.第二节（共15小题；每小题1.5分，满分22.5分）听下面5段对话或独白，每段对话或独白后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟；听完后，各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

请听第6段材料,回答第6、7题。

6. What are the speakers going to do this Sunday?A. Watch a game.B. Have a dinner.C. Have tea.7. When are they going to meet?A. At 9:30.B. At 10:30.C. At 11:30.请听第7段材料,回答第8至10题。