2017-2018学年度必修五综合测试卷

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2017-2018学年(新课标)北师大版高中数学必修五模块检测试题及答案解析

2017-2018学年(新课标)北师大版高中数学必修五模块检测试题及答案解析

(新课标)2017-2018学年北师大版高中数学必修五模块检测(时间:120分钟 满分:150分)一、选择题(本大题共10小题,每小题5分,共50分) 1.已知数列{a n }的前n 项和S n =n 3,则a 5+a 6的值为( ). A .91 B .152 C .218 D .279 解析 a 5+a 6=S 6-S 4=63-43=152. 答案 B2.在△ABC 中,sin A ∶sin B ∶sin C =4∶3∶2,则cos A 的值是( ). A .-14B.14C .-23 D.23解析 由正弦定理得a ∶b ∶c =4∶3∶2,设a =4k ,b =3k ,c =2k ,则cos A = 9k 2+4k 2-16k 22×3k ×2k =-14.答案 A3.在正项等比数列{a n }中,a 1和a 19为方程x 2-10x +16=0的两根,则a 8·a 10·a 12等于( ).A .16B .32C .64D .256解析 ∵{a n }是等比数列且由题意得a 1·a 19=16=a 102(a n >0),∴a 8·a 10·a 12=a 103=64. 答案 C4.等差数列{a n }满足a 42+a 72+2a 4a 7=9,则其前10项之和为( ). A .-9 B .-15 C .15 D .±15解析 a 42+a 72+2a 4a 7=(a 4+a 7)2=9.∴a 4+a 7=±3, ∴a 1+a 10=±3,∴S 10=10(a 1+a 10)2=±15.答案 D5.在坐标平面上,不等式组⎩⎪⎨⎪⎧y ≥x -1,y ≤-3|x|+1所表示的平面区域的面积为( ).A. 2B.32C.322D .2解析 |CD|=1+1=2,⎩⎨⎧y =x -1,y =-3x +1,∴x A =12.⎩⎪⎨⎪⎧y =x -1,y =3x +1,∴x B =-1,∴S △CDA =12×2×12=12,S △CDB =12×2×1=1.故所求区域面积为32.答案 B6.如果不等式2x 2+2mx +m4x 2+6x +3<1对一切实数x 均成立,则实数m 的取值范围是( ).A .(1,3)B .(-∞,3)C .(-∞,1)∪(2,+∞)D .(-∞,+∞)解析 ∵4x 2+6x +3=⎝ ⎛⎭⎪⎫2x +322+34>0,∴原不等式⇔2x 2+2mx +m <4x 2+6x +3⇔2x 2+(6-2m)x +(3-m)>0,x ∈R 恒成立⇔Δ=(6-2m)2-8(3-m)<0,∴1<m <3. 答案 A7.△ABC 中,a ,b ,c 分别是内角A ,B ,C 的对边,且cos 2B +3cos(A +C)+2=0,b =3,则c ∶sin C 等于( ). A .3∶1 B.3∶1 C.2∶1 D .2∶1解析 cos 2B +3cos(A +C)+2=2cos 2B -3cos B +1=0, ∴cos B =12或cos B =1(舍).∴B =π3.∴c sin C =b sin B =332=2. 答案 D8.已知各项都为正数的等比数列{a n }的公比不为1,则a n +a n +3与a n +1+a n +2的大小关系是( ).A .a n +a n +3<a n +1+a n +2B .a n +a n +3=a n +1+a n +2C .a n +a n +3>a n +1+a n +2D .不确定的,与公比有关 解析 因为a n +a n +3=a n (1+q 3), a n +1+a n +2=a n (q +q 2),a n +a n +3-(a n +1+a n +2)=a n (1+q 3-q -q 2)= a n (1-q)(1-q 2)=a n (1-q)2(1+q)>0. 答案 C9.已知公差不为0的等差数列的第4,7,16项恰好分别是某等比数列的第4,6,8项,则该等比数列的公比是( ). A. 3 B.2C .±3D .± 2解析 等差数列记作{a n },等比数列记作{b n }, 则q 2=b 8b 6=b 6b 4=b 8-b 6b 6-b 4=a 16-a 7a 7-a 4=9d3d =3,∴q =± 3.答案 C10.若实数x ,y 满足不等式组⎩⎪⎨⎪⎧x +3y -3≥0,2x -y -3≤0,x -my +1≥0,且x +y 的最大值为9,则实数m 等于( ).A .-2B .-1C .1D .2 解析 如图,作出可行域,由⎩⎪⎨⎪⎧x -my +1=0,2x -y -3=0,得A ⎝⎛⎭⎪⎫1+3m -1+2m ,5-1+2m ,平移y =-x ,当其经过点A 时,x +y 取得最大值,即1+3m -1+2m +5-1+2m=9,解得m= 1. 答案 C二、填空题(本大题共5小题,每小题5分,共25分)11.正项等比数列{a n }满足a 2a 4=1,S 3=13,b n =log 3a n ,则数列{b n }的前10项和是________.解析 ∵{a n }成等比数列,a n >0,∴a 2a 4=a 32=1. ∴a 3=1,∴a 1q 2=1.①∵S 3=a 1+a 2+1=13,∴a 1(1+q)+1=13.② 由①②得,a 1=9,q =13,a n =33-n .∴b n =3-n.∴S 10=-25. 答案 -2512.如图所示,为测一树的高度,在地面上选取A 、B 两点,从A 、B 两点分别测得树尖的仰角为30°,45°,且A 、B 两点之间的距离为60 m ,则树高的高度为________.解析 ∵∠A =30°,∠ABP =45°,∴∠APB =15°,AB sin ∠APB =PA sin ∠PBA ,60sin 15°=PAsin 135°,∴PA =60(3+1),PQ =PA ·sin ∠A =60(3+1)·sin 30°=30(3+1).答案 (30+303)m13.设,x ,y 满足约束条件⎩⎪⎨⎪⎧2x -y +2≥0,8x -y -4≤0,x ≥0,y ≥0,若目标函数z =abx +y(a >0,b >0)的最大值为8,则a +b 的最小值为________.解析 如图所示,线性约束条件表示的区域为图中的阴影部分,A(0,2),B ⎝ ⎛⎭⎪⎫12,0,C(1,4),当直线l :y =-abx+z 过点C 时,z 取最大值8,即8=ab +4, ∴ab =4.又∵a >0,b >0,∴a +b ≥2ab =24=4(a =b =2时取等号).答案 414.在△ABC 中,D 为BC 边上一点,BC =3BD ,AD =2,∠ADB =135°,若AC =2AB ,则BD =________. 解析 如图,设AB =k , 则AC =2k ,再设BD =x , 则DC =2x.在△ABD 中,由余弦定理得 k 2=x 2+2-2·x ·2·⎝ ⎛⎭⎪⎫-22=x 2+2+2x ,① 在△ADC 中,由余弦定理得 2k 2=4x 2+2-2·2x ·2·22=4x 2+2-4x , ∴k 2=2x 2+1-2x.② 由①②得x 2-4x -1=0, 解得x =2+5(负值舍去). 答案 2+ 515.设x ,y ∈R ,a >1,b >1,若a x =b y =3,a +b =23,则1x +1y 的最大值为________.解析 因为a >1,b >1,a x =b y =3,a +b =23, 所以x =log a 3,y =log b 3.1x +1y =1log a 3+1log b 3=log 3a +log 3b =log 3ab ≤ log 3⎝⎛⎭⎪⎫a +b 22=log 3⎝ ⎛⎭⎪⎫2322=1,当且仅当a =b 时,等号成立.答案 1三、解答题(本大题共6小题,共75分)16.(12分)已知{a n }是首项为19,公差为-2的等差数列,S n 为{a n }的前n 项和. (1)求通项a n 及S n ;(2)设{b n -a n }是首项为1,公比为3的等比数列,求数列{b n }的通项公式及前n 项和T n .解 (1)∵{a n }是首项为a 1=19,公差为d =-2的等差数列,∴a n =19-2(n -1)=21-2n ,S n =19n +12n(n -1)×(-2)=20n -n 2.(2)由题意得b n -a n =3n -1,即b n =a n +3n -1,∴b n =3n -1-2n +21, ∴T n =S n +(1+3+…+3n -1)=-n 2+20n +3n -12.17.(12分)已知不等式ax 2-3x +6>4的解集为{x|x <1或x>b}, (1)求a ,b ;(2)解不等式ax 2-(ac +b)x +bc <0.解 (1)因为不等式ax 2-3x +6>4的解集为{x|x <1或x >b},所以x 1=1与x 2=b 是方程ax 2-3x +2=0的两个实数根,且b >1.由根与系数的关系,得⎩⎪⎨⎪⎧1+b =3a,1×b =2a.解得⎩⎪⎨⎪⎧a =1,b =2.所以a =1,b =2.(2)所以不等式ax 2-(ac +b)x +bc <0, 即x 2-(2+c)x +2c <0,即(x -2)(x -c)<0.当c >2时,不等式(x -2)(x -c)<0的解集为{x|2<x <c}; 当c <2时,不等式(x -2)(x -c)<0的解集为{x|c <x <2}; 当c =2时,不等式(x -2)(x -c)<0的解集为∅,综上,当c >2时,不等式ax 2-(ac +b)x +bc <0的解集为{x|2<x <c}; 当c <2时,不等式ax 2-(ac +b)x +bc <0的解集为{x|c <x <2}; 当c =2时,不等式ax 2-(ac +b)x +bc <0的解集为∅.18.(12分)在△ABC 中,a 比b 长2,b 比c 长2,且最大角的正弦值是32,求△ABC 的面积.解 据题意知a -b =2,b -c =2,∴边长a 最大,∴sin A =32, ∴cos A =±1-sin 2A =±12.∵a 最大,∴cos A =-12.又a =b +2,c =b -2,∴cos A =b 2+c 2-a 22bc =b 2+(b -2)2-(b +2)22b (b -2)=-12,解得b =5,∴a =7,c =3,∴S △ABC =12bcsin A =12×5×3×32=1534.19.(12分)已知某地今年年初拥有居民住房的总面积为a(单位:m 2),其中有部分旧住房需要拆除.当地有关部门决定每年以当年年初住房面积的10%建设新住房,同时也拆除面积为b(单位:m 2)的旧住房.(1)分别写出第一年末和第二年末的实际住房面积的表达式.(2)如果第五年末该地的住房面积正好比今年年初的住房面积增加了30%,则每年拆除的旧住房面积b 是多少?(计算时取1.15=1.6) 解 (1)第一年末的住房面积为 a ·1110-b =(1.1a -b)(m 2). 第二年末的住房面积为⎝ ⎛⎭⎪⎫a ·1110-b ·1110-b=a ·⎝⎛⎭⎪⎫11102-b ⎝ ⎛⎭⎪⎫1+1110=(1.21a -2.1b)(m 2).(2)第三年末的住房面积为⎣⎢⎡⎦⎥⎤a ·⎝ ⎛⎭⎪⎫11102-b ⎝ ⎛⎭⎪⎫1+1110·1110-b =a ·⎝⎛⎭⎪⎫11103-b ⎣⎢⎡⎦⎥⎤1+1110+⎝ ⎛⎭⎪⎫11102,第四年末的住房面积为 a ·⎝⎛⎭⎪⎫11104-b ⎣⎢⎡⎦⎥⎤1+1110+⎝ ⎛⎭⎪⎫11102+⎝ ⎛⎭⎪⎫11103,第五年末的住房面积为 a ·⎝⎛⎭⎪⎫11105-b ⎣⎢⎡⎦⎥⎤1+1110+⎝ ⎛⎭⎪⎫11102+⎝ ⎛⎭⎪⎫11103+⎝ ⎛⎭⎪⎫11104=1.15a -1-1.151-1.1b =1.6a -6b.依题意可知1.6a -6b =1.3a ,解得b =a 20,所以每年拆除的旧住房面积为a20 m 2.20.(13分)已知1≤x +y ≤5,-1≤x -y ≤3,求2x -3y 的取值范围.解 法一 作出一元二次方程组⎩⎪⎨⎪⎧1≤x +y ≤5-1≤x -y ≤3所表示的平面区域(如图)即可行域.考虑 z =2x -3y ,把它变形为y =23x -13z ,得到斜率为23,且随z 变化的一组平行直线,-13z 是直线在y 轴上的截距,当直线截距最大且满足约束条件时目标函数z =2x -3y 取得最小值;当直线截距最小且满足约束条件时目标函数z =2x -3y 取得最大值.由图可知,当直线z =2x -3y 经过可行域上的点A 时,截距最大,即z 最小.解方程组⎩⎪⎨⎪⎧x -y =-1,x +y =5,得A 的坐标为(2,3).所以z min =2x -3y =2×2-3×3=-5.解方程组⎩⎪⎨⎪⎧x -y =3,x +y =1,得B 的坐标为(2,-1),所以z max =2x -3y =2×2-3×(-1)=7. ∴2x -3y 的取值范围是[-5,7].法二 设2x -3y =m(x +y)+n(x -y)=mx +my +nx -ny =(m +n)x +(m -n)y则⎩⎪⎨⎪⎧m +n =2,m -n =-3,⇒⎩⎪⎨⎪⎧m =-12,n =52.则2x -3y =-12(x +y)+52(x -y)∵1≤x +y ≤5,-1≤x -y ≤3,∴-52≤-12(x +y)≤-12,-52≤52(x -y)≤152,∴-5≤2x -3y ≤7. 即2x -3y 的取值范围为[-5,7].21.(14分)某港口O 要将一件重要物品用小艇送到一艘正在航行的轮船上,在小艇出发时,轮船位于港口O 北偏西30°且与该港口相距20海里的A 处,并正以30海里/时的航行速度沿正东方向匀速行驶.假设该小艇沿直线方向以v 海里/时的航行速度匀速行驶,经过t 小时与轮船相遇.(1)若希望相遇时小艇的航行距离最小,则小艇航行速度的大小应为多少?(2)为保证小艇在30分钟内(含30分钟)能与轮船相遇,试确定小艇航行速度的最小值.解 (1)若相遇时小艇的航行距离最小,又轮船沿正东方向匀速行驶,则小艇航行方向为正北方向.如图所示,设小艇与轮船在C 处相遇.在Rt △OAC 中,OC =20cos 30°=103,AC =20sin 30°=10.又AC =30t ,OC =vt.此时,轮船航行时间t =1030=13,v =10313=303,即小艇以303海里/时的速度航行,相遇时小艇的航行距离最小.(2)如图所示,设小艇与轮船在B 处相遇.由题意,可得(vt)2=202+(30t)2-2·20·30t ·cos(90°-30°),化简,得v 2=400t 2-600t+900=400⎝ ⎛⎭⎪⎫1t -342+675.由于0<t ≤12,即1t≥2,所以当1t=2时,v 取得最小值1013,即小艇航行速度的最小值为1013海里/时.。

2017-2018学年高二英语综合检测练习:专项练习1(新人教版必修5) Word版含答案

2017-2018学年高二英语综合检测练习:专项练习1(新人教版必修5)   Word版含答案

专项练习(一)Ⅰ.语法填空阅读下面材料,在空白处填入适当内容(1个单词)或括号内单词的正确形式。

(ⅰ)It was a nice Monday morning and I waswaiting for the bus at the station. “Here__1__ (come) the bus,” a boy nearby cried out. Butit was not the bus I was waiting for.__2__hadthe bus stopped when crowds of people gathered in front of the door, waiting__3__ (anxious)to get on the bus. Out__4__(rush)ayoung man with a cell -phone in his hand assoon as the door opened. Chasing him__5__atall and strong man. He was a policeman! The young man was caught stealing a cell-phone.Caught, the young man had something to say for himself. He said his father was ill in bed and couldn't take care of himself.__6__could his mother, because she had been blind ever since she was born. Two younger brothers and a younger sister were waiting for money to go to school. Try__7__he might, the young man said,he couldn't find a job. He said only in this way__8__he provide for his big family and__9__should he have another chance, he would not do this again.So moving was his story that some ladies present insisted on his being set free. Some passengers even took out some money and wanted to give it to the young man. The policeman stopped them and said they would go and check whether he was telling the truth.But the young man had to be responsible__10__what he had done.1.______ 2.______ 3.______ 4.______ 5.______6.______ 7.______ 8.______ 9.______ 10.______答案:1.comes 根据语境可知是车来了,here放句首句子要用完全倒装。

2017-2018学年高一数学必修5第1章 章末综合测评1 含答

2017-2018学年高一数学必修5第1章 章末综合测评1 含答

章末综合测评(一)(时间120分钟,满分160分)一、填空题(本大题共14小题,每小题5分,共70分.请把答案填在题中的横线上)1.在△ABC 中,AB =6,A =75°,B =45°,则AC = .【导学号:92862026】【解析】 C =180°-75°-45°=60°,由正弦定理得AB sin C =AC sin B ,即6sin 60°=AC sin 45°,解得AC =2.【答案】 22.在△ABC 中,已知c =6,a =4,B =120°,则b = . 【解析】 由b 2=16+36-2×4×6cos 120°, 得b =219. 【答案】 2193.在△ABC 中,a =4,b =43,A =30°,则B = . 【解析】 sin B =b sin A a =43sin 30°4=32.又a <b ,故B >A ,∴B =60°或120°. 【答案】 60°或120°4.在△ABC 中,化简b cos C +c cos B = .【解析】 利用余弦定理,得b cos C +c cos B =b ·a 2+b 2-c 22ab +c ·a 2+c 2-b 22ac =a .【答案】 a5.在△ABC 中,若sin A ∶sin B ∶sin C =2∶3∶4,则cos C = . 【解析】 ∵sin A ∶sin B ∶sin C =a ∶b ∶c , ∴a ∶b ∶c =2∶3∶4. 设a =2k ,b =3k ,c =4k ,则 cos C =4k 2+9k 2-16k 22×2k ×3k=-14.【答案】 -146.在△ABC 中,若A =60°,b =16,S △ABC =2203,则a = . 【解析】 由12bc sin A =2203,可知c =55. 又a 2=b 2+c 2-2bc cos A =2 401, ∴a =49. 【答案】 497.在△ABC 中,若sin A =34,a =10,则边长c 的取值范围是 . 【解析】 ∵c sin C =a sin A =403, ∴c =403sin C ,∴0<c ≤403. 【答案】 ⎝ ⎛⎦⎥⎤0,4038.根据下列情况,判断三角形解的情况,其中正确的是 .(填序号)【导学号:92862027】(1)a =8,b =16,A =30°,有两解; (2)b =18,c =20,B =60°,有一解; (3)a =5,c =2,A =90°,无解; (4)a =30,b =25,A =150°,有一解. 【解析】 (1)中,∵a sin A =bsin B , ∴sin B =16×sin 30°8=1,∴B =90°,即只有一解; (2)中,sin C =20sin 60°18=539,且c >b ,∴C >B ,故有两解;(3)中,∵A =90°,a =5,c =2,∴b =a 2-c 2=25-4=21,即有解,故(1)(2)(3)都不正确.所以答案为(4).【答案】 (4)9.已知锐角△ABC 的内角A ,B ,C 的对边分别为a ,b ,c,23cos 2A +cos 2A =0,a =7,c =6,则b = .【解析】 化简23cos 2A +cos 2A =0,得23cos 2A +2cos 2A -1=0,解得cos A =15.由余弦定理,知a 2=b 2+c 2-2bc cos A ,代入数据解方程,得b =5.【答案】 5 10.在△ABC 中,若acos A 2=b cos B 2=c cos C 2,那么△ABC 是 三角形. 【解析】 由正弦定理得,sin A cos A 2=sin B cos B 2=sin Ccos C 2, ∴sin A 2=sin B 2=sin C 2.∵0<A 2,B 2,C 2<π2,∴A 2=B 2=C2,即A =B =C ,∴△ABC 是等边三角形. 【答案】 等边11.如图1所示,在△ABC 中,∠ACB 的平分线CD 交AB 于D ,AC →的模为2,BC →的模为3,AD →的模为1,那么DB →的模为 .图1【解析】 由三角形内角平分线的性质得 |AC →|∶|BC →|=|AD →|∶|DB →|, 故|DB →|=32. 【答案】 3212.如图2所示,在山底测得山顶仰角∠CAB =45°,沿倾斜角为30°的斜坡走1 000 m 至S 点,又测得山顶仰角∠DSB =75°,则山高BC 为 m.图2【解析】 由题可知,∠SAB =45°-30°=15°,又∠SBD =15°,∴∠ABS =45°-15°=30°,AS =1 000.由正弦定理可知BS sin 15°=1 000sin 30°,∴BS =2 000sin 15°,∴BD =BS ·sin 75°=2 000sin 15°cos 15°=1 000sin 30°=500,且DC =1 000sin 30°=500,∴BC =DC +BD =1 000 m.【答案】 1 00013.已知角A ,B ,C 是三角形ABC 的内角,a ,b ,c 分别是其对边长,向量m =⎝ ⎛⎭⎪⎫23sin A 2,cos 2A 2,n =⎝ ⎛⎭⎪⎫cos A 2,-2,m ⊥n ,且a =2,cos B =33,则b= .【解析】 ∵m·n =0,∴23sin A 2cos A 2-2cos 2A 2=0,∵cos A2≠0,∴tan A 2=33,∴A2=30°,∴A =60°, ∵a sin A =bsin B ,sin B =1-⎝ ⎛⎭⎪⎫332=63, ∴b =a sin B sin A =2×6332=43 2.【答案】 43 214.在锐角△ABC 中,角A ,B ,C 的对边分别为a ,b ,c ,若b a +ab =6cos C ,则tan C tan A +tan Ctan B 的值是 .【解析】 ∵b a +ab =6cos C ,∴a 2+b 2ab =6·a 2+b 2-c 22ab , 即a 2+b 2=32c 2, ∴tan C tan A +tan C tan B =tan C ⎝ ⎛⎭⎪⎫cos A sin A +cos B sin B=sin 2Ccos C sin A sin B =2c 2a 2+b 2-c 2=4. 【答案】 4二、解答题(本大题共6小题,共90分.解答时应写出文字说明、证明过程或演算步骤)15.(本小题满分14分)在△ABC 中,a ,b ,c 分别是角A ,B ,C 所对的边,已知b 2=ac ,且a 2-c 2=ac -bc ,求角A 的大小及b sin Bc .【解】 由b 2=ac 及a 2-c 2=ac -bc ,得b 2+c 2-a 2=bc . 在△ABC 中,cos A =b 2+c 2-a 22bc =12. ∵0°<A <180°,∴A =60°. 在△ABC 中,由正弦定理得sin B =b sin A a .又∵b 2=ac ,A =60°, ∴b sin B c =b 2sin A ac =sin 60°=32.16.(本小题满分14分)已知△ABC 的内角A ,B ,C 所对的边分别为a ,b ,c ,且a =2,cos B =35.(1)若b =4,求sin A 的值;(2)若△ABC 的面积S △ABC =4,求b ,c 的值. 【解】 (1)∵cos B =35>0,且0<B <π,∴sin B =1-cos 2B =45.由正弦定理得a sin A =bsin B ,sin A =a sin Bb =2×454=25.(2)∵S △ABC =12ac sin B =4, ∴12×2×c ×45=4,∴c =5.由余弦定理得b 2=a 2+c 2-2ac cos B =22+52-2×2×5×35=17,∴b =17. 17.(本小题满分14分)在△ABC 中,内角A ,B ,C 的对边分别为a ,b ,c .已知cos A -2cos C cos B=2c -a b .(1)求sin Csin A 的值;(2)若cos B =14,△ABC 的周长为5,求b 的长. 【解】 (1)由正弦定理,设a sin A =b sin B =csin C =k , 则2c -a b =2k sin C -k sin A k sin B=2sin C -sin Asin B,所以cos A -2cos C cos B =2sin C -sin Asin B,即(cos A -2cos C )sin B =(2sin C -sin A )cos B ,化简可得sin(A +B )=2sin(B +C ). 又A +B +C =π,所以sin C =2sin A . 因此sin Csin A =2.(2)由sin Csin A =2,得c =2a ,由余弦定理及cos B =14得b 2=a 2+c 2-2ac cos B =a 2+4a 2-4a 2×14=4a 2, 所以b =2a .又a +b +c =5,从而a =1, 因此b =2.18.(本小题满分16分)在△ABC 中a ,b ,c 分别为内角A ,B ,C 的对边,且2a sin A =(2b +c )sin B +(2c +b )sin C .(1)求A 的大小;(2)若sin B +sin C =1,试判断△ABC 的形状.【导学号:92862028】【解】 (1)由2a sin A =(2b +c )sin B +(2c +b )sin C ,得2a 2=(2b +c )b +(2c +b )c ,即a 2=b 2+c 2+bc , ∴b 2+c 2-a 2=-bc , ∴2bc cos A =-bc ,∴cos A =-12,又A ∈(0,π), ∴A =2π3.(2)由(1)得sin 2A =sin 2B +sin 2C +sin B sin C , 又sin B +sin C =1,得sin B =sin C =12. 又B ,C ∈⎝ ⎛⎭⎪⎫0,π2,故B =C .所以△ABC 是等腰三角形.19.(本小题满分16分)在△ABC 中,角A ,B ,C 的对边分别为a ,b ,c ,且cos(A -B )cos B -sin(A -B )sin(A +C )=-35.(1)求sin A 的值;(2)若a =42,b =5,求向量BA→在BC →方向上的投影.【解】 (1)由cos(A -B )cos B -sin(A -B )sin(A +C )=-35,得cos(A -B )cos B-sin(A -B )sin B =-35,则cos(A -B +B )=-35,即cos A =-35. 又0<A <π,则sin A =45.(2)由正弦定理,有a sin A =bsin B , 所以sin B =b sin A a =22.由题知a >b ,则A >B ,故B =π4.根据余弦定理,有(42)2=52+c 2-2×5c ×⎝ ⎛⎭⎪⎫-35,解得c =1或c =-7(负值舍去).故向量BA →在BC →方向上的投影为|BA →|cos B =22.20.(本小题满分16分)如图3,游客从某旅游景区的景点A 处下山至C 处有两种路径.一种是从A 沿直线步行到C ,另一种是先从A 沿索道乘缆车到B ,然后从B 沿直线步行到C .现有甲、乙两位游客从A 处下山,甲沿AC 匀速步行,速度为50 m/min.在甲出发2 min 后,乙从A 乘缆车到B ,在B 处停留1 min 后,再从B 匀速步行到C .假设缆车匀速直线运行的速度为130 m/min ,山路AC 长为1 260 m ,经测量,cos A =1213,cos C =35.图3(1)求索道AB 的长;(2)问乙出发多少min 后,乙在缆车上与甲的距离最短?(3)为使两位游客在C 处互相等待的时间不超过3 min ,乙步行的速度应控制在什么范围内?【解】 (1)在△ABC 中,因为cos A =1213,cos C =35, 所以sin A =513,sin C =45. 从而sin B =sin []π-(A +C ) =sin(A +C )=sin A cos C +cos A sin C =513×35+1213×45=6365. 由AB sin C =ACsin B ,得AB =AC sin B ×sin C =1 2606365×45=1 040(m).所以索道AB 的长为1 040 m.(2)设乙出发t min 后,甲、乙两游客距离为d m ,此时,甲行走了(100+50t )m ,乙距离A 处130t m ,所以由余弦定理得d 2=(100+50t )2+(130t )2-2×130t ×(100+50t )×1213=200(37t 2-70t +50), 因0≤t ≤1 040130,即0≤t ≤8,故当t =3537 min 时,甲、乙两游客距离最短.(3)由BC sin A =AC sin B ,得BC =AC sin B ×sin A =1 2606365×513=500(m).乙从B 出发时,甲已走了50×(2+8+1)=550(m),还需走710 m 才能到达C .设乙步行的速度为v m/min ,由题意得-3≤500v -71050≤3,解得1 25043≤v ≤62514,所以为使两位游客在C 处互相等待的时间不超过3 min ,乙步行的速度应控制在⎣⎢⎡⎦⎥⎤1 25043,62514(单位:m/min)范围内.。

2017-2018学年高二英语必修五综合能力评估试卷含答案

2017-2018学年高二英语必修五综合能力评估试卷含答案

2017-2018学年高二英语必修五综合能力评估试卷含答案必修5综合能力评估(Units1—5)时间:120分钟满分:150分第Ⅰ卷(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.How often does the man play golf?A.Once a week.B.Once a month.C.Every day.2.What did the woman win in the contest?A.A free trip.B.A good camera.C.A medal.3.What is the boy going to do next?A.Have dinner.B.Do his reading.C.Watch TV.4.When does the man leave home?A.At6:00.B.At7:00.C.At about6:40.5.How is the girl going to spend the weekend?A.Staying with her parents.B.Boating in the Water Park.C.Preparing for the competition.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,回答第6~8题。

6.Why does the girl want to change her coat?A.Because the coat doesn't fit her.B.Because the coat is too expensive.C.Because she doesn't like the colour.7.Who bought the coat for the girl?A.Her brother.B.Her sister.C.Her mother.8.Which coat will the girl probably take?A.A red one,Size11.B.A yellow one,Size9.C.A blue one,Size11.听第7段材料,回答第9~11题。

2017人教版高中英语必修五综合测试题

2017人教版高中英语必修五综合测试题

必修5学业质量标准检测时间:120分钟,满分:150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)错误!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,您都有10秒钟的时间来回答有关小题与阅读下一小题。

每段对话仅读一遍。

1、What’s the woman going to do this evening?_C__A、Go to a movie、B、Go to a concert、C、Study in the library、2、How will the man go to the office?_B__A、By train、B、By taxi、C、By bus、3、What’s the probable relationship between the two speakers?_B__A、Classmates、B、Workmates、C、Friends、4、What’s the woman doing?_A__A、Ordering some food、B、Selling some food、C、Cooking some food、5、Where does the conversation probably take place?_A__A、In a store、B、In an office、C、In a bank、第二节(共15小题;每题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,您将有时间阅读各个小题,每小题5秒钟;听完后,每小题将给出5秒钟的作答时间.每段对话或独白读两遍。

听第6段材料,回答第6至7题。

导学号 491525386、Where does the man work?_B__A、In a factory、B、In a hotel、C、In a school、7、What does the man do?_C__A、A manager、B、A cleaner、C、An electrician、听第7段材料,回答第8至9题。

2017-2018学年高中数学北师大版必修5:模块综合测评 含解析 精品

2017-2018学年高中数学北师大版必修5:模块综合测评 含解析 精品

模块综合测评(时间120分钟,满分150分)一、选择题(本大题共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.已知等差数列{a n}满足a2+a4=4,a3+a5=10,则它的前10项的和S10=()A.138B.135C.95 D.23【解析】由a2+a4=4,a3+a5=10得a1=-4,d=3,所以S10=10×(2a1+9d)2=10×(-8+27)2=5×19=95.【答案】 C2.在△ABC中,已知a、b和锐角A,要使三角形有两解,则应该满足的条件是()A.a=b sin A B.b sin A>aC.b sin A<b<a D.b sin A<a<b【解析】当a=b sin A时,有一解,当b sin A<a<b时,有两解,当a>b 时有一解.【答案】 D3.已知不等式x2+ax+4<0的解集为空集,则a的取值范围是()A.-4≤a≤4 B.-4<a<4C.a≤-4或a≥4 D.a<-4或a>4【解析】欲使不等式x2+ax+4<0的解集为空集,则Δ=a2-16≤0,∴-4≤a≤4.【答案】 A4.已知等差数列的前n项和为18,若S3=1,a n+a n-1+a n-2=3,则n的值为()A.9 B.21C.27 D.36【解析】 ∵S 3=a 1+a 2+a 3=1, 又a 1+a n =a 2+a n -1=a 3+a n -2, ∴3(a 1+a n )=1+3,∴a 1+a n =43.又S n =n (a 1+a n )2=23n =18,∴n =27,故选C.【答案】 C5.关于x 的不等式ax -b >0的解集是(1,+∞),则关于x 的不等式(ax +b )(x -3)>0的解集是( )A .(-∞,-1)∪(3,+∞)B .(-1,3)C .(1,3)D .(-∞,1)∪(3,+∞) 【解析】 (ax +b )(x -3)>0等价于 ⎩⎨⎧ ax +b >0,x -3>0或⎩⎨⎧ax +b <0,x -3<0, ∴⎩⎨⎧x >-1,x >3或⎩⎨⎧x <-1,x <3. ∴x ∈(-∞,-1)∪(3,+∞). 【答案】 A6.“神七”飞天,举国欢庆,据科学计算,运载“神舟七号”飞船的“长征2号”系列火箭,点火1分钟内通过的路程为2 km ,以后每分钟通过的路程比前一分钟增加2 km ,在到达离地面240 km 的高度时,火箭与飞船分离,则这一过程大约需要的时间是( )A .10分钟B .13分钟C .15分钟D .20分钟【解析】 由题设条件知,火箭每分钟通过的路程构成以a 1=2为首项,公差d =2的等差数列,∴n 分钟内通过的路程为S n =2n +n (n -1)2×2=n 2+n =n (n +1).检验选项知,n =15时,S 15=240 km.故选C.【答案】 C7.在△ABC 中,内角A 、B 、C 满足6sin A =4sin B =3sin C ,则cos B =( ) A.154 B.34 C.31510D.1116【解析】 由6sin A =4sin B =3sin C 得sin A ∶sin B ∶sin C =2∶3∶4,设△ABC 中角A 、B 、C 的对边分别为a ,b ,c ,则由正弦定理知a ∶b ∶c =2∶3∶4.不妨设a =2k ,b =3k ,c =4k (k >0), 则cos B =a 2+c 2-b 22ac =(22+42-32)k 22×2k ×4k =1116.【答案】 D8.设变量x ,y 满足约束条件⎩⎨⎧x -y +2≥0,2x +3y -6≥0,3x +2y -9≤0,则目标函数z =2x +5y 的最小值为( )【导学号:47172142】A .-4B .6C .10D .17【解析】 由约束条件作出可行域如图所示,目标函数可化为y =-25x +15z ,在图中画出直线y =-25x ,平移该直线,易知经过点A 时z 最小. 又知点A 的坐标为(3,0), ∴z min =2×3+5×0=6.故选B. 【答案】 B9.y =3+x +x 21+x(x >0)的最小值是( )A .2 3B .-1+2 3C .1+2 3D .-2+2 3【解析】 y =3+x +x 21+x =31+x +x =31+x +x +1-1≥23-1,当且仅当31+x =1+x ,即x =3-1时取等号,故y 有最小值23-1.【答案】 B10.对于每个自然数n ,抛物线y =(n 2+n )x 2-(2n +1)x +1与x 轴交于A n ,B n 两点,以|A n B n |表示该两点间的距离,则|A 1B 1|+|A 2B 2|+…+|A 2 015B 2 015|的值是( )A.2 0142 015 B.2 0162 015 C.2 0152 014D.2 0152 016【解析】 |A n B n |=|x 1-x 2|=⎝ ⎛⎭⎪⎫2n +1n 2+n 2-4n 2+n =1n 2+n =1n ·(n +1)=1n -1n +1, ∴|A 1B 1|+|A 2B 2|+…+|A 2 015B 2 015|=⎝ ⎛⎭⎪⎫1-12+⎝ ⎛⎭⎪⎫12-13+…+⎝ ⎛⎭⎪⎫12 015-12 016=2 0152 016.【答案】 D11.设f (x )=3ax -2a +1,若存在x 0∈(-1,1)使f (x 0)=0,则实数a 的取值范围是( )A .-1<a <15 B .a <-1 C .a <-1或a >15D .a >15 【解析】 由于f (x )=3ax -2a +1,故f (x )一定是一条直线,又由题意,存在x 0∈(-1,1),使得f (x 0)=0,故直线y =3ax -2a +1在x =-1和x =1时的函数值异号,即f (-1)f (1)<0,得(1-5a )(a +1)<0,解得a <-1或a >15.【答案】 C12.已知圆C :(x -a )2+(y -b )2=1,平面区域Ω:⎩⎨⎧x +y -7≤0,x -y +3≥0,y ≥0.若圆心C ∈Ω,且圆C 与x 轴相切,则a 2+b 2的最大值为( ) 【导学号:47172143】A .5B .29C .37D .49【解析】 作出可行域,如图,由题意知,圆心为C (a ,b ),半径r =1,且圆C 与x 轴相切,所以b =1.而直线y =1与可行域的交点为A (6,1),B (-2,1),目标函数z =a 2+b 2表示点C 到原点距离的平方,所以当点C 与点A 重合时,z 取到最大值,z max =37.【答案】 C二、填空题(本大题共4个小题,每小题5分,共20分.将答案填在题中的横线上)13.已知二次函数f (x )=ax 2-3x +2,不等式f (x )>0的解集为{x |x <1或x >b },则b =________.【解析】 由题意知1,b 是方程ax 2-3x +2=0的两根, 由根与系数的关系得⎩⎪⎨⎪⎧1+b =3a ,1×b =2a ,∴⎩⎨⎧a =1,b =2.【答案】 214.在△ABC 中,B =120°,AC =7,AB =5,则△ABC 的面积为________.【导学号:47172144】【解析】 设AB =c ,BC =a ,AC =b ,由余弦定理b 2=a 2+c 2-2ac cos B ,得49=a 2+25-2×5a ⎝ ⎛⎭⎪⎫-12,解得a =3,∴S △ABC =12ac sin B =12×3×5×sin120°=1534. 【答案】 153415.若x ,y 满足约束条件⎩⎨⎧x -y +1≥0,x -2y ≤0,x +2y -2≤0,则z =x +y 的最大值为________.【解析】 画出可行域如图所示.由⎩⎨⎧x -2y =0,x +2y -2=0, 得⎩⎪⎨⎪⎧x =1,y =12.∴A ⎝ ⎛⎭⎪⎫1,12. 由z =x +y ,得y =-x +z ,平移直线l 0:x +y =0. 当直线过点A 时,z 最大,z max =1+12=32. 【答案】 3216.若a >0,b >0,且a 2+14b 2=1,则a 1+b 2的最大值为________.【解析】 a 1+b 2=12·2a 1+b 2≤4a 2+1+b 24=54,当且仅当⎩⎨⎧4a 2=1+b 2,4a 2+b 2=4时等号成立, 即a =104,b =62时成立.【答案】 54三、解答题(本大题共6小题,共70分,解答应写出文字说明,证明过程或演算步骤)17.(本小题满分10分)已知△ABC 的周长为2+1,且sin A +sin B =2sin C . (1)求边AB 的长;(2)若△ABC 的面积为16sin C ,求C 的度数. 【解】 (1)由题意△ABC 的周长为2+1, ∴AB +BC +AC =2+1.由正弦定理,得 BC +AC =2AB ,∴AB =1.(2)由△ABC 的面积为12BC ·AC ·sin C =16sin C ,得BC ·AC =13.由(1)知BC +AC =2,由余弦定理,得cos C =AC 2+BC 2-AB 22AC ·BC =12,∴C =60°.18.(本小题满分12分)已知等比数列{a n }中,a 2=2,a 5=128,若b n =log 2a n ,数列{b n }前n 项的和为S n .(1)若S n =35,求n 的值; (2)求不等式S n <2b n 的解集.【导学号:47172145】【解】 (1)由a 2=a 1q =2,a 5=a 1q 4=128得q 3=64, ∴q =4,a 1=12,∴a n =a 1q n -1=12·4n -1=22n -3, ∴b n =log 2a n =log 222n -3=2n -3. ∵b n +1-b n =[2(n +1)-3]-(2n -3)=2,∴{b 1}是以b 1=-1为首项,2为公差的等差数列, ∴S n =(-1+2n -3)n 2=35,n 2-2n -35=0,(n -7)(n +5)=0,即n =7.(2)∵S n -2b n =n 2-2n -2(2n -3)=n 2-6n +6<0, ∴3-3<n <3+3, ∵n ∈N +,∴n =2,3,4,即所求不等式的解集为{2,3,4}.19.(本小题满分12分)如图1,矩形ABCD 是机器人踢球的场地,AB =170 cm ,AD =80 cm ,机器人先从AD 中点E 进入场地到点F 处,EF =40 cm ,EF ⊥AD .场地内有一小球从点B 向点A 运动,机器人从点F 出发去截小球.现机器人和小球同时出发,它们均作匀速直线运动,并且小球运动的速度是机器人行走速度的2倍.若忽略机器人原地旋转所需的时间,则机器人最快可在何处截住小球?图1【解】 设该机器人最快可在点G 处截住小球,点G 在线段AB 上.连接FG .设FG =x cm.根据题意,得BG =2x cm.则AG =AB -BG =(170-2x )cm.连接AF ,在△AEF 中,EF =AE =40 cm ,EF ⊥AD , 所以∠EAF =45°,AF =402cm , 于是∠F AG =45°.在△AFG 中,由余弦定理,得 FG 2=AF 2+AG 2-2AF ·AG cos ∠F AG ,所以x 2=(402)2+(170-2x )2-2×402×(170-2x )×cos 45°, 解得x 1=50,x 2=3703.所以AG =170-2x =70 cm 或AG =-2303cm(不合题意,舍去). 即该机器人最快可在线段AB 上离A 点70 cm 处截住小球.20.(本小题满分12分)解关于x 的不等式ax 2-2≥2x -ax (a ∈R ).【导学号:47172146】【解】 原不等式可化为ax 2+(a -2)x -2≥0⇒(ax -2)(x +1)≥0. (1)当a =0时,原不等式化为x +1≤0⇒x ≤-1. (2)当a >0时,原不等式化为⎝ ⎛⎭⎪⎫x -2a (x +1)≥0⇒x ≥2a 或x ≤-1; (3)当a <0时,原不等式化为⎝ ⎛⎭⎪⎫x -2a (x +1)≤0.①当2a >-1,即a <-2时,原不等式的解集为-1≤x ≤2a ; ②当2a =-1,即a =-2时,原不等式的解集为x =-1; ③当2a <-1,即-2<a <0时,原不等式的解集为2a ≤x ≤-1. 综上所述,当a <-2时,原不等式的解集为⎣⎢⎡⎦⎥⎤-1,2a ;当a =-2时,原不等式的解集为{-1}; 当-2<a <0时,原不等式的解集为⎣⎢⎡⎦⎥⎤2a ,-1;当a =0时,原不等式的解集为(-∞,-1];当a >0时,原不等式的解集为(-∞,-1]∪⎣⎢⎡⎭⎪⎫2a ,+∞.21.(本小题满分12分)某汽车运输公司购买了一批豪华大客车投入运营.据市场分析,每辆客车运营的总利润y (单位:十万元)与运营年数x 满足二次函数的关系:y =-a (x -6)2+11,且该二次函数图像过点(4,7).问每辆客车运营多少年,运营的年平均利润最大?最大值为多少?(年平均利润=总利润年数) 【解】 设年平均利润为z 十万元,依题意, ∵二次函数y =-a (x -6)2+11的图像过点(4,7), ∴7=-a (4-6)2+11, ∴a =1,∴y =-(x -6)2+11,z =y x =-(x -6)2+11x=-x 2+12x -25x =-x -25x +12=-⎝ ⎛⎭⎪⎫x +25x +12.∵x >0,∴x +25x ≥10, ∴-⎝ ⎛⎭⎪⎫x +25x ≤-10,∴-⎝ ⎛⎭⎪⎫x +25x +12≤2,∴z ≤2,当且仅当x =25x 即x =5时,z 有最大值为2十万元.即每辆客车运营5年,运营的年平均利润最大,最大值为2十万元. 22.(本小题满分12分)已知数列{a n }是首项为a 1=14,公比q =14的等比数列,设b n +2=3log 14a n (n ∈N +),数列{c n }满足c n =a n ·b n .(1)求证:{b n }是等差数列; (2)求数列{c n }的前n 项和S n ;(3)若c n ≤14m 2+m -1对一切正整数n 恒成立,求实数m 的取值范围.【导学号:47172147】【解】 (1)证明:由题意知,a n =⎝ ⎛⎭⎪⎫14n (n ∈N +),∵b n =3log 14a n -2,b 1=3log 14a 1-2=1,∴b n +1-b n =3log 14a n +1-3log 14a n =3log 14a n +1a n =3log 14q =3,∴数列{b n }是首项b 1=1,公差d =3的等差数列. (2)由(1)知,a n =⎝ ⎛⎭⎪⎫14n ,b n =3n -2(n ∈N +),∴c n =(3n -2)×⎝ ⎛⎭⎪⎫14n (n ∈N +),∴S n =1×14+4×⎝ ⎛⎭⎪⎫142+7×⎝ ⎛⎭⎪⎫143+…+(3n -5)×⎝ ⎛⎭⎪⎫14n -1+(3n -2)×⎝ ⎛⎭⎪⎫14n ;于是14S n =1×⎝ ⎛⎭⎪⎫142+4×⎝ ⎛⎭⎪⎫143+7×⎝ ⎛⎭⎪⎫144+…+(3n -5)×⎝ ⎛⎭⎪⎫14n +(3n -2)×⎝ ⎛⎭⎪⎫14n +1,两式相减得34S n =14+3⎣⎢⎡⎦⎥⎤⎝ ⎛⎭⎪⎫142+⎝ ⎛⎭⎪⎫143+…+⎝ ⎛⎭⎪⎫14n -(3n -2)×⎝ ⎛⎭⎪⎫14n +1=12-(3n +2)×⎝ ⎛⎭⎪⎫14n +1. ∴S n =23-12n +83×⎝ ⎛⎭⎪⎫14n +1(n ∈N +). (3)∵c n +1-c n =(3n +1)·⎝ ⎛⎭⎪⎫14n +1-(3n -2)·⎝ ⎛⎭⎪⎫14n =9(1-n )·⎝ ⎛⎭⎪⎫14n +1(n ∈N +), ∴当n =1时,c 2=c 1=14, 当n ≥2时,c n +1<c n ,即c 1=c 2>c 3>c 4>…>c n ,∴当n =1或2时,c n 取得最大值是14.又c n ≤14m 2+m -1对一切正整数n 恒成立,∴14m 2+m -1≥14,即m 2+4m -5≥0,解得m ≥1或m ≤-5.故实数m 的取值范围为{m |m ≥1或m ≤-5}.。

2017-2018学年高中数学北师大版必修5课时作业:模块综合测试卷

2017-2018学年高中数学北师大版必修5课时作业:模块综合测试卷
模块综合测试卷
班级__________ 姓名__________ 考号__________ 分数__________
本试卷满分 100 分,考试时间 90 分钟.
一、选择题:本大题共 10 题,每题 4 分,共 40 分.在下列各题的四个选项中,只有
( ) a+b b+c c+a ·· ∴lg 2 2 2 >lg(abc), a+b b+c c+a ∴lg 2 +lg 2 +lg 2 >lga+lgb+lgc.
18.解:(1)方案:①在未破损的圆周上任取三点 M,N,P.
②连结三点 M,N,P 得圆内接三角形(△MNP).
③用直尺量得 MN=a,用量角器量得∠MPN=α.
a ④由正弦定理得:sinα=2R,即为所求圆块的直径.
abc (2)由sinA=sinB=sinC=2R,在△ABC 中,等式(sin∠CBA)2+(sin∠ACB)
2-(sin∠CAB)2=sin∠CBA·sin∠ACB,可化为 b2+c2-a2=bc,
b2+c2-a2 1 即:cos∠CAB= 2bc =2.
2 2=2a2b2,∴cosC=± 2 ,∴C=45°或 135°.
10.D 取等比数列 1,2,4,令 n=1 得 X=1,Y=3,Z=7 代入验算,只有选项 D 满
足.
二、填空题
1 11.56-210
111
1
1
1
1
1
解析:12+24+38+…+10210=(1+2)+(2+4)+(3+8)+…+(10+210)
8.已知 0<x<1,则 x(3-3x)取最大值时 x 的值为( )
1132 A.3 B.2 C.4 D.3 9.在△ABC 中,已知 a4+b4+c4=2c2(a2+b2),则 C 等于( )

2017-2018学年度第一学期必修5 Modules1-3综合测试题

2017-2018学年度第一学期必修5 Modules1-3综合测试题

必修5 Modules1-3综合测试题第一部分阅读理解(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和D)中,选出最佳选项。

AAre you planning for your first ever international trip? No worries! We have a list of places that are perfect for first-time getaways.BaliBali is ever charming. Photographs cannot describe the beauty of its beaches, coral reefs, temples and forest areas. The temples, especially Tanah Lot, are quite a sight. The Kit. Batur is another amazing experience.Top attractions: Tanah Lot, Kuta Beach, Nusa Dua and Jimbaran Beach.MaldivesMaldives is an amazing beach destination. Here you can spend quality time away from the city life. Its bright blue shoreline, clean air and white sand can make you start feeling as if you had landed in heaven, then you get on a seaplane to the hotel that you have booked. The seaplane journey is an adventure that helps you see some of the most beautiful islands.Top attractions: water sports, beach and outdoor activities, local markets and more.MauritiusWhether you are into scenic beauty, adventure, or food, this island will not disappoint you. Mark Twain said, "Mauritius was made first and then heaven.” You can spend an evening in Port Louis and have dinner in Namaste Restaurant. Spend all the other days visiting only the beaches. If you are visiting lie Aux Cerfs, ensure you reach the island as early as possible as it closes at 4 pm.Top attractions: Grand Bay, Blue Bay, skydiving and bird watching.SingaporeClassical buildings stand tall with elegant skyscrapers (摩天大楼) all around the island, city, mixing perfectly with fancy shopping malls. This small country has good enough places for your 5-6 days‟ stay.Top attractions: shopping, street foods, museums and theme parks.1. If you are interested in water skiing, which place is the best choice?A. Bali.B. Maldives.C. Mauritius.D. Singapore.2. What do we know about Mauritius?A. Mark Twain loved Mauritius more than his country.B. Port Louis is not open at daytime.C. Only beaches are worth visiting in Mauritius.D. You can visit lie Aux Cerfs at night.3. What can we learn from this text?A. You‟re not allowed to take pictures in Bali.B. People take seaplanes rather than ships in Maldives.C. Singapore combines the classical and modem culture.D. Summer is the best time to visit the four places.BAs a mom who had already lost her son, a Houston nurse became concerned when she heard about the popular Netflix show 13 Reasons Why. By then her 15-year-old daughter had already been attracted to it. In the show, the main character feels hopeless and kills herself.Mental health experts believe that the show's messages on suicide (自杀) are dangerous. Educators have warned parents that it glamorizes the idea of suicide and could lead to an increase in imitation (模仿) among student s. They‟re worried about children watching it alone without parents beside them. Employees from some elementary and middle schools have seen an increase in self-harm among students since the show began.Netflix believed that the show has helped bring up important conversations in many families. Though some parents loved the show, for the nurse‟s family, it was a disaster. Her daughter has been suffering from depression since her brother‟s death. With treatment, things got better until she watched the series. “If I‟d known about the show beforehand I would have monitored (监视) her Netflix account,” said the nurse. “If kids have a history of depressio n or suicidal thoughts, I don ‟t think they need to watch it. ”The show‟s creators said they did make part of t he show painful to watch because they wanted to make it clear that there is nothing worthwhile about suicide. Before its release, a producer sought the support of the JED Foundation, a teen suicide prevention group. “I think they were looking for us to say it was a great educational tool, or that they handled the issues in a psychologically helpfu l way,” said a medical officer. The group issued a guide about the show, advising people who choose to watch it to view it with someone else and to take breaks between episodes (剧集).4. What does the underlined word “glamorizes” in Paragraph 2 mean?A. Makes something become more strange.B. Makes something seem more attractive.C. Makes something become more fearful.D. Makes something seem more touching.5. What do we know about the nurse?A. She‟d known about the TV show before her daughter watched it.B. The show has helped improve her family relationship.C. She didn‟t suggest teens with self-harm thoughts watch the show.D. She has been monitoring her daughter‟s Netflix account.6. The creators may agree that the show ________.A. includes dangerous messagesB. encourages self-harmC. is pleasant to watchD. is educational and helpful7. What is the passage mainly about?A. The story of a mom and her daughter.B. A TV show that caused discussion.C. An educational TV show.D. Some common teen problems.CAn “exercise pill” that could reduce the need for visits to the gym is a step closer. A study found that preventing production of the protein myostatin (肌肉抑制素) increases muscle. Although the research was conducted in mice, the team hopes it will lead to human treatments (治疗)Lead researcher Dr Joshua Butcher, from Augusta University in Georgia, the US, said, "Finally, the goal of our research would be to create a pill that mimics the effect of exercise and protects against obesity. A pill that stops myostatin could also be used for muscle-wasting diseases, such as cancer, muscle dystrophy and AIDS. Myostatin is known to be a powerfu l …brake‟ that holds back muscle growth.”Studies suggest that obese people produce more of the protein myostatin, making it harder for them to exercise and build up muscle. “Given that exercise is one of the most effective ways to deal with obesity, this creates a cycle by which a perso n becomes trapped in obesity,”said Dr Butcher.The researchers got four groups of thin and obesemice that were genetically programmed either to produce unlimited levels of myostatin or to be completely lacking in the protein. As expected, mice with no myostatin became markedly more muscular. Obese mice unable to produce myostatin remained fat, but they were also stronger and showed signs of heart and metabolic (新陈代谢) health compared with the thin mice. They were in greatly better shape than obese mice with unlimited myostatin production.Dr Butcher added, “In our muscular obese mice, despite full presentation of obesity, it appears that several of these key points are prevented. While much more research is needed, at this point, myostatin appears to be a very promising way for protection against some heart problems caused by obesity.” The research was presented at the American Physiological Society‟s annual meeting, Experimental Biology 2017, in Chicago, the US.8. What b enefit will the ''exercise pill” bring to people?A. Consumers will look much younger than before.B. Users can keep in good shape without going to the gym.C. Patients don‟t have to be treated in hospital at all.D. It offers more opportunities to take exercise outdoors.9. The underlined word “mimics” in Paragraph 2 means ________.A. lacksB. reducesC. is similar toD. is limited to10. The cycle mentioned in the third paragraph should be ________.①It‟s harder to exercise.②Obesity appears.③People tend to be fatter and fatter.④More myostatin is produced.A. ②④①③B. ①④③②C. ④③②①D. ③②①④11. How did researchers reach the conclusion?A. By analyzing causes.B. By referring to documents.C. By turning to scientists.D. By making comparisons.DSometimes “you” doesn‟t mean “you”, a new study finds.People often say “you” to make it easier to talk about a negative personal experience. For example, people may say “You win some, you lose some.” when they have just failed at a task. By using “you” instead of “I”, they explain that failure can happen to anyone. The “you” here used to refer to people in general rather than a specific (特定的) audience is called “generic-you”. Researchers conducted a series of experiments, asking 2,500 people to write about personal experiences and answer questions. In one experiment, a group of participants were asked to write about a negative (消极的) personal experience and another group of participants were asked to write about a neutral (中立的) personal experience. Researchers noted that people in the negative group used the ……generic-you” more in their responses.When they use “you”to describe the event, it allows them to move beyond the negative experience and helps them feel better.In another experiment, all participants were asked to write about a negative experience. One group focused on lessons they had learned, while another group wrote about the emotions they felt. Researchers found that peop le who focused on lessons used “you” more than those who focused on emotions. Besides, the people who used “you” said they felt further away from the event than those who used “I”. This suggests that “generic-you” is a way for people to give meaning t o an experience.Researchers concluded that the “generic-you” provides a way to normalize one‟s negative experiences and find meaning from personal experiences.12. When people say “You win some, you lose some.”, failure actually happened to __________.A. the listenerB. the speakerC. anyoneD. nobody13. Which of the following is an example of “generic-you”?A. Do you believe it?B. When will you be ready to go?C. You don‟t know how long I‟ve waited for you.D. Treat others as you hope they will treat you.14. In the first experiment, the “generic-you” _________.A. is only used in a neutral situationB. is only used in a negative situationC. helps leave good impressions on othersD. helps overcome a negative situation15. What do we know about the second experiment?A. “Lessons” group used “you” less than “emotions” group.B. “Emotions” group was less affected by negative experiences.C. People tend to use “you” to explain lessons they've learned.D. People never use “you” to express personal feelin gs.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项。

2017-2018学年高中数学模块综合检测新人教B版必修5

2017-2018学年高中数学模块综合检测新人教B版必修5

模块综合检测(时间120分钟 满分150分)一、选择题(本大题共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f (x )=3x 2-x +1,g (x )=2x 2+x -1,则f (x )与g (x )的大小关系为( ) A .f (x )>g (x ) B .f (x )=g (x ) C .f (x )<g (x )D .随x 值变化而变化解析:选A 因为f (x )-g (x )=(3x 2-x +1)-(2x 2+x -1)=x 2-2x +2=(x -1)2+1>0,所以f (x )>g (x ).2.在△ABC 中,角A ,B ,C 所对的边分别为a ,b ,c ,若a =2,b =3,B =60°,那么角A 等于( )A .135°B .90°C .45°D .30°解析:选C 由正弦定理知a sin A =bsin B, ∴sin A =a sin Bb =2sin 60°3=22. 又a <b ,B =60°,∴A <60°,∴A =45°.3.若a 1=1,a n +1=a n3a n +1,则给出的数列{a n }的第4项是( )A.116B.117C.110D.125解析:选C a 2=a 13a 1+1=13+1=14,a 3=a 23a 2+1=1434+1=17,a 4=a 33a 3+1=1737+1=110. 4.若关于x 的不等式x 2-3ax +2>0的解集为(-∞,1)∪(m ,+∞),则a +m =( ) A .-1 B .1 C .2D .3解析:选D 由题意,知1,m 是方程x 2-3ax +2=0的两个根,则由根与系数的关系,得⎩⎪⎨⎪⎧1+m =3a ,1×m =2,解得⎩⎪⎨⎪⎧a =1,m =2,所以a +m =3,故选D.5.已知x >0,y >0,且x +y =8,则(1+x )(1+y )的最大值为( ) A .16 B .25 C .9D .36解析:选B (1+x )(1+y )≤⎣⎢⎡⎦⎥⎤+x ++y 22=⎣⎢⎡⎦⎥⎤2+x +y 22=⎝ ⎛⎭⎪⎫2+822=25,因此当且仅当1+x =1+y 即x =y =4时,(1+x )(1+y )取最大值25,故选B.6.已知数列{a n }为等差数列,且a 1=2,a 2+a 3=13,则a 4+a 5+a 6等于( ) A .40 B .42 C .43D .45解析:选B 设等差数列{a n }的公差为d , 则2a 1+3d =13,∴d =3,故a 4+a 5+a 6=3a 1+12d =3×2+12×3=42.7.钝角三角形ABC 的面积是12,AB =1,BC =2,则AC =( )A .5 B. 5 C .2D .1 解析:选 B ∵S △ABC =12AB ·BC sin B =12×1×2sin B =12,∴sin B =22,∴B =45°或135°,若B =45°,则由余弦定理得AC =1,∴△ABC 为直角三角形,不符合题意,因此B =135°,由余弦定理得AC 2=AB 2+BC 2-2AB ·BC cos B =1+2-2×1×2×⎝ ⎛⎭⎪⎫-22=5,∴AC =5,此时△ABC 为钝角三角形,符合题意.故选B.8.已知S n 为正项等比数列{a n }的前n 项和,S 3=3a 1+2a 2,且a 2-12,a 4,a 5成等差数列,则a 1=( )A .2 B.12 C.14D .4解析:选C 设数列{a n }的公比为q (q >0),则由S 3=3a 1+2a 2可得q 2-q -2=0,解得q =2或q =-1(舍去),又a 2-12,a 4,a 5成等差数列,所以2a 4=a 2-12+a 5,即a 2=12,所以a 1=14.9.在△ABC 中,AC =7,BC =2,B =60°,则BC 边上的高等于( ) A.32B.332C.3+62D.3+394解析:选B 由余弦定理得AB 2+4-2·AB ×2×cos 60°=7,解得AB =3或AB =-1(舍去),设BC 边上的高为x ,由三角形面积关系得12·BC ·x =12AB ·BC ·sin 60°,解得x =332,故选B.10.某汽车公司有两家装配厂,生产甲、乙两种不同型的汽车,若A 厂每小时可完成1辆甲型车和2辆乙型车;B 厂每小时可完成3辆甲型车和1辆乙型车.今欲制造40辆甲型车和40辆乙型车,若要使所费的总工作时数最少,那么这两家工厂工作的时间分别为( )A .16,8B .15,9C .17,7D .14,10解析:选A 设A 工厂工作x 小时,B 工厂工作y 小时,总工作时数为z ,则目标函数为z =x +y ,约束条件为⎩⎪⎨⎪⎧x +3y ≥40,2x +y ≥40,x ≥0,y ≥0作出可行域如图所示,由图知当直线l :y =-x +z过Q 点时,z 最小,解方程组⎩⎪⎨⎪⎧x +3y =40,2x +y =40,得Q (16,8),故A 厂工作16小时,B 厂工作8小时,可使所费的总工作时数最少.11.若log 4(3x +4b )=log 2ab ,则a +b 的最小值是( ) A .6+2 3 B .7+2 3 C .6+4 3D .7+4 3解析:选D 由log 4(3a +4b )=log 2ab ,得12log 2(3a +4b )=12log 2(ab ),所以3a +4b =ab ,即3b +4a=1.所以a +b =(a +b )⎝ ⎛⎭⎪⎫3b +4a=3a b +4ba +7≥43+7,当且仅当3ab =4b a,即a =23+4,b =3+23时取等号,故选D.12.已知x ,y 满足约束条件⎩⎪⎨⎪⎧x -y ≥0,x +y ≤2,y ≥0.若z =ax +y 的最大值为4,则a =( )A .3B .2C .-2D .-3解析:选B 画出不等式组表示的平面区域如图阴影部分所示,若z =ax +y 的最大值为4,则最优解为x =1,y =1或x =2,y =0,经检验x =2,y =0符合题意,∴2a +0=4,此时a =2,故选B.二、填空题(本大题共4小题,每小题5分,共20分.请把正确答案填在题中的横线上)13.若实数x ,y 满足xy =1,则x 2+2y 2的最小值为________. 解析:因为实数x ,y 满足xy =1,所以x 2+2y 2≥2x 2·2y 2=2xy2=22,并且仅当x 2=2y 2且xy =1,即x 2=2y 2=2时等号成立,故x 2+2y 2的最小值为2 2.答案:2 214.已知△ABC 的一个内角为120°,并且三边长构成公差为4的等差数列,则△ABC 的面积为________.解析:由于三边长构成公差为4的等差数列, 故可设三边长分别为x -4,x ,x +4.由一个内角为120°,知其必是最长边x +4所对的角. 由余弦定理得,(x +4)2=x 2+(x -4)2-2x (x -4)·cos 120°, ∴2x 2-20x =0,∴x =0(舍去)或x =10, ∴S △ABC =12×(10-4)×10×sin 120°=15 3.答案:15 315.设S n 是数列{a n }的前n 项和,且a 1=-1,a n +1=S n S n +1,则S n =________. 解析:∵a n +1=S n +1-S n ,a n +1=S n S n +1, ∴S n +1-S n =S n S n +1.∵S n ≠0,∴1S n -1S n +1=1,即1S n +1-1S n=-1.又1S 1=-1,∴⎩⎨⎧⎭⎬⎫1S n 是首项为-1,公差为-1的等差数列.∴1S n=-1+(n -1)×(-1)=-n ,∴S n =-1n.答案:-1n16.若a >0,b >0,a +b =2,则下列不等式①ab ≤1;②a +b ≤2;③a 2+b 2≥2;④1a+1b≥2,对满足条件的a ,b 恒成立的是________.(填序号)解析:因为ab ≤⎝⎛⎭⎪⎫a +b 22=1,所以①正确;因为(a +b )2=a +b +2ab =2+2ab ≤2+a +b =4,故②不正确;a 2+b 2≥a +b22=2,所以③正确;1a +1b =a +b ab =2ab≥2,所以④正确.答案:①③④三、解答题(本大题共6小题,共70分,解答时写出必要的文字说明、证明过程或演算步骤)17.(10分)等差数列{a n }的前n 项和记为S n ,已知a 10=30,a 20=50. (1)求通项a n ; (2)若S n =242,求n .解:(1)设{a n }的首项为a 1,公差为d ,则⎩⎪⎨⎪⎧a 1+9d =30,a 1+19d =50.解得⎩⎪⎨⎪⎧a 1=12,d =2.∴通项a n =a 1+(n -1)d =10+2n . (2)由S n =na 1+n n -2d =242,得12n +n n -2×2=242,解得n =11,或n =-22(舍去).故n =11.18.(12分)已知f (x )=2x 2+bx +c ,不等式f (x )<0的解集是(0,5). (1)求f (x )的解析式;(2)若对于任意的x ∈[-1,1],不等式f (x )+t ≤2恒成立,求t 的取值范围. 解:(1)因为f (x )=2x 2+bx +c ,不等式f (x )<0的解集是(0,5), 所以2x 2+bx +c <0的解集是(0,5), 所以0和5是方程2x 2+bx +c =0的两个根, 由根与系数的关系,知-b 2=5,c2=0,所以b =-10,c =0,所以f (x )=2x 2-10x .(2)对任意的x ∈[-1,1],f (x )+t ≤2恒成立等价于对任意的x ∈[-1,1],2x 2-10x +t -2≤0恒成立.设g (x )=2x 2-10x +t -2,则由二次函数的图象可知g (x )=2x 2-10x +t -2在区间[-1,1]上为减函数,所以g (x )max =g (-1)=10+t ,所以10+t ≤0,即t ≤-10,所以t 的取值范围为(-∞,-10].19.(12分)已知等差数列{a n }的前n 项和S n 满足S 3=0,S 5=-5. (1)求{a n }的通项公式; (2)求数列⎩⎨⎧⎭⎬⎫1a 2n -1a 2n +1的前n 项和.解:(1)设{a n }的公差为d ,则S n =na 1+n n -2d .由已知可得⎩⎪⎨⎪⎧3a 1+3d =0,5a 1+10d =-5.解得a 1=1,d =-1.故{a n }的通项公式为a n =2-n . (2)由(1)知1a 2n -1a 2n +1=1-2n-2n=12⎝ ⎛⎭⎪⎫12n -3-12n -1,从而数列⎩⎨⎧⎭⎬⎫1a 2n -1a 2n +1的前n 项和为12( 1-1-11+11-13+…+12n -3-12n -1 )=n1-2n. 20.(12分)某气象仪器研究所按以下方案测试一种“弹射型”气象观测仪器的垂直弹射高度:在C 处进行该仪器的垂直弹射,观察点A ,B 两地相距100 m ,∠BAC =60°,在A 地听到弹射声音的时间比B 地晚217s .A 地测得该仪器在C 处时的俯角为15°,A 地测得最高点H 的仰角为30°,求该仪器的垂直弹射高度CH .(声音的传播速度为340 m/s)解:由题意,设AC =x m , 则BC =x -217×340=(x -40)m ,在△ABC 内,由余弦定理:BC 2=BA 2+CA 2-2·BA ·CA ·cos∠BAC , 即(x -40)2=1002+x 2-100x ,解得x =420.在△ACH 中,AC =420 m ,∠CAH =30°+15°=45°,∠CHA =90°-30°=60°, 由正弦定理:CH sin ∠CAH =ACsin ∠AHC ,可得CH =AC ·sin∠CAHsin ∠AHC=1406(m).即该仪器的垂直弹射高度CH 为140 6 m.21.(12分)在△ABC 中,BC =6,点D 在BC 边上,且(2AC -AB )cos A =BC cos C . (1)求角A 的大小;(2)若AD 为△ABC 的中线,且AC =23,求AD 的长;(3)若AD 为△ABC 的高,且AD =33,求证:△ABC 为等边三角形.解:(1)由(2AC -AB )cos A =BC cos C 及正弦定理,有(2sin B -sin C )cos A =sin A cos C , 得2sin B cos A =sin C cos A +sin A cos C =sin(A +C )=sin B ,所以cos A =12.因为0°<A <180°,所以A =60°. (2)由正弦定理BC sin A =ACsin B,得sin B =AC sin A BC =12. 因为A +B <180°,所以B =30°,所以C =90°. 因为D 是BC 的中点,所以DC =3, 由勾股定理,得AD =AC 2+DC 2=21.(3)证明:因为12AD ·BC =12AB ·AC sin A ,且AD =33,BC =6,sin A =32,所以AB ·AC=36.因为BC 2=AB 2+AC 2-2AB ·AC cos A , 所以AB 2+AC 2=72,所以AB =AC =6=BC , 所以△ABC 为等边三角形.22.(12分)已知数列{a n }的前n 项和S n 和通项a n 满足2S n +a n =1,数列{b n }中,b 1=1,b 2=12,2b n +1=1b n +1b n +2(n ∈N +).(1)求数列{a n },{b n }的通项公式;(2)数列{c n }满足c n =a n b n ,求证:c 1+c 2+c 3+…+c n <34.解:(1)由2S n +a n =1,得S n =12(1-a n ).当n ≥2时,a n =S n -S n -1=12(1-a n )-12(1-a n -1)=-12a n +12a n -1,即2a n =-a n +a n -1,∴a n a n -1=13(由题意可知a n -1≠0). ∴{a n }是公比为13的等比数列,而S 1=a 1=12(1-a 1),∴a 1=13,∴a n =13×⎝ ⎛⎭⎪⎫13n -1=⎝ ⎛⎭⎪⎫13n.由2b n +1=1b n +1b n +2,1b 1=1,1b 2=2,得d =1b 2-1b 1=1⎝ ⎛⎭⎪⎫d 为等差数列⎩⎨⎧⎭⎬⎫1b n 的公差,∴1b n =n ,∴b n =1n.(2)证明:c n =a n b n =n ⎝ ⎛⎭⎪⎫13n,设T n =c 1+c 2+…+c n ,则T n =1×⎝ ⎛⎭⎪⎫131+2×⎝ ⎛⎭⎪⎫132+3×⎝ ⎛⎭⎪⎫133+…+n ×⎝ ⎛⎭⎪⎫13n ,13T n =1×⎝ ⎛⎭⎪⎫132+2×⎝ ⎛⎭⎪⎫133+…+(n -1)×⎝ ⎛⎭⎪⎫13n +n ×⎝ ⎛⎭⎪⎫13n +1,由错位相减,得23T n =13+⎝ ⎛⎭⎪⎫132+…+⎝ ⎛⎭⎪⎫13n -n ×⎝ ⎛⎭⎪⎫13n +1=13⎣⎢⎡⎦⎥⎤1-⎝ ⎛⎭⎪⎫13n 1-13-n ×⎝ ⎛⎭⎪⎫13n +1=12-12×⎝ ⎛⎭⎪⎫13n-n ×⎝ ⎛⎭⎪⎫13n +1,所以T n =34-34×⎝ ⎛⎭⎪⎫13n -12n ×⎝ ⎛⎭⎪⎫13n =34-2n +34×13n <34.。

2017-2018学年数学人教A版必修五优化练习:综合检测

2017-2018学年数学人教A版必修五优化练习:综合检测

综合检测时间:120分钟 满分:150分一、选择题(本大题共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.已知{a n }是等比数列,a 3=,a 6=2,则公比q =( )14A .- B .-212C .2D.12解析:=q 3=8,∴q =2.a 6a 3答案:C2.若a 、b 为实数,则下面一定成立的是( )A .若a >b ,则a 4>b 4B .若|a |>b ,则a 2>b 2C .若a >|b |,则a 2>b 2D .若a ≠|b |,则a 2≠b 2解析:a >|b |⇔a 2>b 2.答案:C3.下列命题中正确的是( )A .a >b ⇒ac 2>bc 2 B .a >b ⇒a 2>b 2C .a >b ⇒a 3>b 3D .a 2>b 2⇒a >b 解析:选项A 中,当c =0时,ac 2=bc 2,所以A 不正确;选项B 中,当a =0,b =-1时a >b ,但a 2<b 2,所以B 不正确;选项D 中,当a =-2,b =-1时,a 2>b 2,但a <b ,所以D 不正确.很明显C 正确.答案:C4.已知各项均为正数的等比数列{a n },a 1·a 9=16,则a 2·a 5·a 8的值为( )A .16 B .32C .48D .64解析:由等比数列的性质可得,a 1·a 9=a =16.25∵a n >0,∴a 5=4,∴a 2·a 5·a 8=a =64,故选D.35答案:D5.在△ABC 中,角A ,B ,C 所对的边分别为a ,b ,c ,且a 2=b 2-c 2+ac ,则角B 的大2小是( )A .45°B .60°C .90°D .135°解析:由已知得a 2+c 2-b 2=ac ,所以cos B ===.又0°<B <180°,所2a 2+c 2-b 22ac2ac2ac 22以B =45°.答案:A6.设等差数列{a n }的前n 项和为S n ,若a 1=-11,a 4+a 6=-6,则当S n 取最小值时,n 等于( )A .6 B .7C .8D .9解析:∵{a n }是等差数列,∴a 4+a 6=2a 5=-6,即a 5=-3,∴d ===2,故{a n }是首项为-11的递增数列,所有的非正项之和最小.a 5-a 15-1-3+114∵a 6=-1,a 7=1,∴当n =6时,S n 取得最小值.答案:A7.在△ABC 中,AB =3,BC =,AC =4,则AC 边上的高为( )13A. B.322332C.D .3323解析:由BC 2=AB 2+AC 2-2AB ·AC cosA ,可得13=9+16-2×3×4×cosA ,得cosA =.∵A 为△ABC 的内角,∴A =,∴AC 边上的高h =AB sin A =3×=.12π332332答案:B8.如果关于x 的不等式5x 2-a ≤0的正整数解是1,2,3,4,那么实数a 的取值范围是( )A .80≤a <125 B .80<a <125C .a <80D .a >125解析:由5x 2-a ≤0,得-≤x ≤.而5x 2-a ≤0的正整数解是1,2,3,4,所以4≤a5a5<5,所以80≤a <125.a5答案:A9.若实数x ,y 满足不等式组Error!且x +y 的最大值为9,则实数m 等于( )A .-2 B .-1C .1D .2解析:作出可行域.如图中阴影部分所示.由Error!得A .(1+3m-1+2m ,5-1+2m )平移y =-x ,当其经过点A 时,x +y 取得最大值.即+=9,解得m =1.1+3m-1+2m 5-1+2m 答案:C10.在△ABC 中,如果sin A sin B +sin A cos B +cos A sin B +cos A cos B =2,则△ABC 是( )A .等边三角形 B .钝角三角形C .等腰直角三角形D .直角三角形解析:由已知,得cos(A -B )+sin(A +B )=2,又|cos(A -B )|≤1,|sin(A +B )|≤1,故cos(A -B )=1且sin(A +B )=1,即A =B 且A +B =90°,故选C.答案:C11.设x ,y ∈R ,a >1,b >1.若a x =b y =3,a +b =2,则+的最大值为( )31x 1y A .2 B.32C .1D.12解析:∵2=a +b ≥2,∴ab ≤3.3ab 由a x =b y =3得x =log a 3,y =log b 3,∴+=+=log 3a +log 3b =log 3ab ≤log 33=1.故选C.1x 1y 1log a 31log b 3答案:C12.数列{a n }中,a n >0且{a n a n +1}是公比为q (q >0)的等比数列,满足a n a n +1+a n +1a n +2>a n +2a n +3(n ∈N *),则( )A .0<q <B .0<q <1+221+52C .0<q <D .0<q <-1+22-1+52解析:∵{a n a n +1}是公比为q 的等比数列,∴a n a n +1=(a 1a 2)·q n -1,∴(a 1a 2)·q n -1+(a 1a 2)·q n >(a 1a 2)·q n +1,∴1+q >q 2,∴q 2-q -1<0,∴0<q <.1+52答案:B二、填空题(本大题共4小题,每小题4分,共16分,把答案填在题中的横线上)13.不等式≤x 的解集是________.1x 解析:≤x 等价于x -≥0,1x 1x 即≥0,所以不等式的解集为{x |-1≤x <0或x ≥1}.x 2-1x答案:{x |-1≤x <0或x ≥1}14.等比数列{a n }中,a 2=2,a 5=16,那么数列{a n }的前6项和S 6=________.解析:设公比为q ,由题意,得Error!解得a 1=1,q =2,所以S 6===63.a 1(1-q 6)1-q1-261-2答案:6315.如图,△ ABC 中,AB =AC =2,BC =2,点D 在BC 边上,3∠ADC =45°,则AD 的长度等于________.解析:在△ABC 中,由余弦定理得:cos C ===,AC 2+BC 2-AB 22·AC ·BC4+12-42×2×2332∴∠C =30°.在△ADC 中由正弦定理,得=,ADsin C ACsin ∠ADC∴=.故AD =.AD 122222答案:216. 不等式ax 2+4x +a >1-2x 2对一切x ∈R 恒成立,则实数a 的取值范围是________.解析:不等式ax 2+4x +a >1-2x 2对一切x ∈R 恒成立,即(a +2)x 2+4x +a -1>0对一切x ∈R 恒成立.若a +2=0,显然不成立;若a +2≠0,则Error!⇔Error!⇔Error!⇔a >2.答案:(2,+∞)三、解答题(本大题共有6小题,共74分,解答应写出文字说明、证明过程或演算步骤)17.(12分)△ABC 中,BC =7,AB =3,且=.sin Csin B 35(1)求AC ;(2)求角A .解析:(1)由正弦定理,得=,ACsin B ABsin C ∴==.ABAC sin C sin B 35∴AC ==5.5×33(2)由余弦定理,得cos A ===-.AB 2+AC 2-BC 22AB ·AC9+25-492×3×512又0°<A <180°,∴A =120°.18.(12分)已知不等式ax 2-3x +6>4的解集为{x |x <1或x >b }.(1)求实数a ,b 的值;(2)当c >2时,解不等式ax 2-(ac +b )x +bc <0.解析:(1)因为不等式ax 2-3x +6>4的解集为{x |x <1或x >b },所以x 1=1与x 2=b 是方程ax 2-3x +2=0的两个实数根,且b >1,a >0,由根与系数的关系,得Error!解得Error!(2)不等式ax 2-(ac +b )x +bc <0,即x 2-(2+c )x +2c <0,即(x -2)(x -c )<0.当c >2时,不等式(x -2)(x -c )<0的解集为{x |2<x <c }.19.(12分)设二次方程a n x 2-a n +1x +1=0(n ∈N *)有两个实根α和β,且满足6α-2αβ+6β=3.(1)试用a n 表示a n +1;(2)求证:是等比数列;{an -23}(3)当a 1=时,求数列{a n }的通项公式.76解析:(1)由根与系数的关系,得α+β=,αβ=,代入6α-2αβ+6β=3,并化简,an +1an 1an 得a n +1=a n +.1213(2)证明:因为a n +1=a n +,1213所以a n +1-=.2312(an -23)因此,数列是公比为的等比数列.{an -23}12(3)当a 1=时,a 1-=,所以是首项为,公比为的等比数列.762312{an -23}1212所以a n-=·n -1=n,2312(12)(12)故a n=+n.23(12)20.(12分)要设计如图的一张矩形广告,该广告含有大小相等的左、中、右三个矩形栏目,这三栏的面积之和为60 000 cm 2,四周空白的宽度为10 cm ,栏与栏之间的中缝空白的宽度为5 cm ,怎样确定广告矩形栏目高与宽的尺寸(单位:cm),能使整个矩形广告面积最小.解析:设矩形栏目的高为a cm ,宽为b cm ,则ab =20 000,所以b =,20 000a广告的高为(a +20)cm ,宽为(3b +30)cm(其中a >0,b >0),广告的面积S =(a +20)(3b +30)=30(a +2b )+60 600=30+60 600(a +40 000a)≥30×2+60 600a ·40 000a=12 000+60 600=72 600.当且仅当a =,40 000a即a =200时等号成立,此时b =100.故当广告矩形栏目的高为200 cm ,宽为100 cm 时,可使整个矩形广告的面积最小.21.(13分)在△ABC 中,角A 、B 、C 的对边分别为a 、b 、c ,已知sin 22C +sin 2C ·sinC +cos 2C =1,且a +b =5,c =.7(1)求角C 的大小;(2)求△ABC 的面积.解析:(1)∵sin 22C +sin 2C ·sin C +cos 2C =1,∴4sin 2 C ·cos 2 C +2sin 2 C ·cos C +1-2sin 2 C =1,即2sin 2 C (2cos 2 C +cos C -1)=0.∴2sin 2 C (2cos C -1)(cos C +1)=0.∵在△ABC 中,sin C ≠0,cos C >-1,∴cos C =,∴C =.12π3(2)∵cos C ===,b 2+a 2-c 22ab(a +b )2-c 2-2ab2ab12∴=,∴ab =6.25-72ab 32∴S △ABC =ab sin C =×6×=.12123233222.(13分)已知各项均为正数的数列{a n },满足a -a n +1a n -2a =0(n ∈N *),且a 1=2.2n +12n (1)求数列{a n }的通项公式;(2)设b n =a n ·log a n ,若b n 的前n 项和为S n ,求S n ;12(3)在(2)的条件下,求使S n +n ·2n +1>50成立的正整数n 的最小值.解析:(1)∵a -a n +1a n -2a =0,2n +12n ∴(a n +1+a n )(a n +1-2a n )=0,∵数列{a n }的各项均为正数,∴a n +1+a n >0,∴a n +1-2a n =0,即a n +1=2a n (n ∈N *),所以数列{a n }是以2为公比的等比数列.∵a 1=2,∴数列{a n }的通项公式a n =2n .(2)由(1)及b n =a n log a n 得,b n =-n ·2n ,12∵S n =b 1+b 2+…+b n ,∴S n =-2-2·22-3·23-4·24-…-n ·2n ①∴2S n =-22-2·23-3·24-4·25-…-(n -1)·2n -n ·2n +1②②-①得,S n =2+22+23+24+25+…+2n -n ·2n +1=-n ·2n +1=(1-n )·2n +1-2.2(1-2n )1-2(3)要使S n +n ·2n +1>50成立,只需2n +1-2>50成立,即2n +1>52,∴使S n +n ·2n +1>50成立的正整数n 的最小值为5.。

2017-2018学年高中数学必修5模块综合检测题含答案

2017-2018学年高中数学必修5模块综合检测题含答案

2017-2018学年高中数学必修5模块综合检测题2018.1.23本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分.满分150分.考试时间120分钟.第Ⅰ卷(选择题共60分)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)⎛⎤112.定义符号函数sgn x =⎩⎪⎨⎪⎧0,x =0,-1,x <0,则当x ∈R 时,不等式x +2>(2x -1)sgn x的解集是( )A.⎩⎪⎨⎪⎧⎭⎪⎬⎪⎫x ⎪⎪⎪-3+334<x <-3+334 B.⎩⎪⎨⎪⎧⎭⎪⎬⎪⎫x ⎪⎪⎪x >-3+334 C.⎩⎪⎨⎪⎧⎭⎪⎬⎪⎫x ⎪⎪⎪x <-3+334[Z|X D.⎩⎪⎨⎪⎧⎭⎪⎬⎪⎫x ⎪⎪⎪-3+334<x <3第Ⅱ卷(非选择题 共90分)二、填空题(本大题共4个小题,每小题5分,共20分,把正确答案填在题中横线上) 13.若函数f (x )=(2-a 2)x +a 在区间[0,1]上恒为正,则实数a 的取值范围是________. 14.在R 上定义运算⊗,a ⊗b =ab +2a +b ,则满足x ⊗(x -2)<0的实数x 的取值范围为________.必修5模块综合检测题参考答案【第1题解析】由9x 2+6x +1≤0,得(3x +1)2≤0,可求得其解为x =-31.故选D.【第2题解析】利用线性规划知识,求解目标函数的取值范围.如下图.根据题意得C (1+,2).作直线-x +y =0,并平移,过点B (1,3)和C (1+,2)时,z =-x +y 分别取最大值和最小值,则-(1+)+2<z <-1+3,∴z =-x +y 的取值范围是(1-,2).故选A.【第5题解析】∵不等式x +a x +1<2的解集为P ,且1∉P ,∴1+a 1+1≥2,即a +12a≤0,∴-1<a ≤0.故选D.【第6题解析】∵x >1,y >1,且xy =16,∴log 2x >0,log 2y >0且log 2x +log 2y =log 216=4.∴log 2x ·log 2y ≤2log2x +log2y 2=4(当且仅当x =y =4时取等号).故选D.【第7题解析】由图象知抛物线顶点坐标为(6,11),且过点(4, 7).设y =a (x -6)2+11,将点(4,7)代入,得7=a (4-6)2+11,∴a =-1.∴y =-(x -6)2+11=-x 2+12x -25.∴年平均利润为x y =-x -x 25+12=12-x 25.∵x +x 25≥10,即x =5时,取等号25,∴当x =5时,x y有最大值2.故选C.【第8题解析】∵不等式x 2+ax +1≥0对一切x ∈21成立,∴对一切x ∈21,ax ≥-x 2-1,即a ≥-x x2+1成立.令g (x )=-x x2+1=-x 1.易知g (x )=-x 1在21内为增函数.∴当x =21时,g (x )max =-25.∴a 的取值范围是a ≥-25,即a 的最小值是-25.故选C.【第9题解析】“求(1-a i x )2<1(i =1,2,3)都成立的x 的取值范围”实质上是求不等式组2<12<1的解集,由于这几个不等式结构一样,则其中解集“最小”的一个不等式的解集即是不等式组的解集.(1-a i x )2<1即a i 2x2-2a i x <0,a i x (a i x -2)<0.∵a i >0,∴这个不等式可化为x ai 2<0,∴0<x <ai 2.若ai 2取最小值,则a i 应取最大值,因此0<x <a12,故选B.【第11题解析】设租用A 型车x 辆,B 型车y 辆,目标函数为z =1 600x +2 400y ,则约束条件为x ,y ∈N ,y -x≤7,作出可行域,如图中阴影部分所示,可知目标函数过点P (5,12)时,有最小值z min =36 800(元).故选C.【第12题解析】当x >0时,不等式化为x +2>2x -1,解得x <3,即0<x <3;当x =0时,不等式恒成立;当x <0时,不等式化为x +2>(2x -1)-1,即2x 2+3x -3<0,解得-433<x<433,即-433<x <0.综上可知,不等式的解集为<x <333.故选D.【第13题解析】当2-a 2=0时,a =±.由题意知a =时符合题意. 当2-a 2≠0,即a ≠±时,f (x )是一次函数,在[0,1]上是单调的,∴>0,0>0,即-a2+a +2>0.a>0,解得0<a <2且a ≠±.综上可知0<a <2.故填(0,2).【第14题解析】∵x ⊗(x -2)=x ·(x -2)+2x +x -2=x 2+x -2,∴x ⊗(x -2)<0,即x 2+x -2<0,即(x +2)(x -1)<0,∴实数x 的取值范围为-2<x <1.故填(-2,1).【第15题解析】设f (x )=x 2+ax +2b ,由题意可知f (x )的图象如图1所示,则有>0<0,⇔a +b +2>0.a +2b +1<0,图1图2【第16题解析】作出可行域如图所示的阴影部分,平移直线l :ax +by =0,由于a >0,b >0,∴直线l 的斜率为-b a<0,∴当直线l 经过点A 时,z =ax +by 取得最大值6.由x -y +2=0,3x -y -6=0,解得y =6,x =4,∴A (4,6).∴4a +6b =6.∴32a +b =1且a >0,b >0.∴a 1+b 2=b 2a +b 2=38+a b +3b 4a ≥38+23b 4a =33.(当且仅当a b =3b 4a ,即a =23b 时取等号)故填33.【第17题答案】y min =3.【第17题解析】令t =x 2+1,则t ≥1,且x 2=t -1.∴y =x2+1x4+3x2+3=t t -1+3=t t2+t +1=t +t 1+1.∵t ≥1,∴t +t 1≥2t 1=2,当且仅当t =t 1,即t =1时,等号成立.∴当t =1时,t 1min =2,此时x =0,y min =t +t 1+1=3.故当x =0时,函数y 取最小值,y min =3.【第18题答案】x >2或x <-1m ∈,31,问题转化为g (m )在m ∈,31上恒大于0,则>0,>0,解得x >2或x <-1.故填x >2或x <-1.【第19题答案】23【第19题解析】(1)若a =2,则不等式f (x )≥0化为2x 2-5x +3≥0,∴不等式f (x )≥0的解集为或x≤13. (2)∵ax 2-(2a +1)x +a +1=a (x -1)2-(x -1),令g (a )=a (x -1)2-(x -1),则g (a )是关于a 的一次函数,且一次项的系数为(x -1)2≥0,∴当x -1=0时,f (x )=0不合题意;当x ≠1时,g (a )为[-2,2]上的增函数.∵f (x )<0恒成立,∴只要使g (a )的最大值g (2)<0即可,即g (2)=2(x -1)2-(x -1)<0,解得1<x <23. 综上,x 的取值范围是23. 学科*网【第20题答案】(1)f (x )=38-x +10180-5x,x ∈[0,100];(2)分别用20万元和80万元资金投资A ,B 两种金融产品,可以使公司获得最大利润,最大利润为28万元.【第21题答案】存在常数a =41,b =21,c =41满足题意.【第21题解析】由f (-1)=0,得a -b +c =0.①又对x ∈R ,不等式x ≤f (x )≤21(x 2+1)成立,取x =1,有1≤f (1)≤1,∴f (1)=1,故a +b +c =1.②由①②可得b =21,c =21-a ,将其代入x ≤f (x )≤21(x 2+1),得x ≤ax 2+21x +21-a ≤21(x 2+1)对x ∈R 恒成立,即x -a≤0 ④1对x ∈R 恒成立.由③得Δ≤0a>0,⇒a =41.由④得Δ≤0<0,⇒a =41.综合可知,存在常数a =41,b =21,c =41满足题意.【第22题答案】存在实数k ∈[3,+∞)使不等式恒成立.【第22题解析】存在.将不等式4-kx -x 4≤0变形,即-x 4≤kx -4(x >0).可设f 1(x )=-x 4,f 2(x )=kx -4.故f 2(x )中参数k 的几何意义是直线y =kx -4的斜率.由下图知当直线y =kx -4与曲线y =-x 4相切时,关于x 的方程-x 4=kx -4有唯一大于0的解,将方程整理成关于x 的一元二次方程kx 2-4x +4=0.由Δ=(-4)2-4×4×k =0,可得k =3.又直线y =kx -4过定点(0,-4),故要使f 1(x )≤f 2(x )(x >0)恒成立,只需k ≥3即可.综上,存在实数k ∈[3,+∞)使不等式恒成立.。

广东省深圳市2017-2018学年人教版高三英语专题练习(小题):必修五综合练习(word解析版)

广东省深圳市2017-2018学年人教版高三英语专题练习(小题):必修五综合练习(word解析版)

必修五综合练习第一部分阅读理解(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和D)中,选出最佳选项。

ALondon bus toursThe Big Bus Company offers open-top sightseeing tours with live guides, or with a digitally-recorded commentary in eight languages. Tickets are available for 24 hours. Admissions are listed as follows:Adults: £ 25; children: £ 10; family: £ 55(2 adults+up to 3 children).The Big Bus Company has won Visit London sightseeing tour of the year three times. Central departure points include Marble Arch, Green Park, Victoria Station, Baker Street and Trafalgar Square.London walking toursOne of London’s most established walking tour companies offers over 40 walks including many classics, including Along the Thames Pub Walk, Historic City, Hidden London, Historic Westminster, Little Venice, Ghost walks, Shakespeare and Dickens walk, Caters for clubs, schools and other group outings. This content has been supplied by London Walks.Opening Times: Walks take place every day.London running toursWant to explore London and get fit at the same time? Try one of these London running tours and see the sights of London on the run! Whether you’re new to running or an experience marathon runner, City Running Tours has a tour for you. Runner guides lead daily tours past sights such as London Eye, Big Ben, the Houses of Parliament and Buckingham Palace. On a London sightseeing running tour, you’ll get training tips from a qualified fitness trainer, as well as a guided tour of London.London river toursFor a unique view of London, take a river boat along the river Thames and see some of London’s best-known attractions from the water. Beginning at Westminster Pier, the tours take you past the Houses of Parliament, London Eye, Shakespeare’s Globe, Tower of London and Tower Bridge on the way to Greenwich. Take one of City Cruises’ regular sightseeing tours and you’ll also hear colourful local stories about London as you travel along the River Thames.2. What is London Walks according to the passage?A. It is a kind of t ransportation they offered.B. It is a company which offered the tours.C. It is an organization they offered training.D. It is a class which offered suggestion of tours.2. Which way can we choose if we not only tour London but also build up our body?A. London bus tours.B. London walking tours.C. London running tours.D. London river tours.3. What is special about London running tours?A. Some advice about the tours is told in advance and it also has a live guide.B. Opening times of London running tours can take place for 24 hours every day.C. People can see some of London’s best-known attractions from the water.D. People can tour some London’s interests by live guides with eight languages.4. Where can you find the information on the web?A. London tours rules.B. London famous interests.C. London tours guide.D. London popular transportation.BBalancing work and school is not an easy task for me. My first semester in college has come and gone and I’ve had to balance a job in there too. Many times I wanted to cut one loose. Honestly I think I would have dropped out of school for my job because I need the money. For a while I made myself believe that school was disturbing my job and my money. Then not only was I trying to do work and school, I wanted a social life. Many times I found that there were not enough hours in the day, but I realized I wasn’t managing my time right. Hanging out with friends may have been wonderful but it wasn’t putting money in my pocket or knowledge in my brain. Friends would have to wait. I was back on the see-saw(拉锯局面) of balancing work and school.Scheduling and planning became the key to my success. Making deadlines for myself and meeting them were important for my college experience. Planning was going well and I felt ready to try and throw a social life back in there. I tried to make plans with friends which didn’t conflict with my deadline. That was actually harder than I thought. So once again my social life had to be put off. My friends were very understanding and encouraged me to do my work. Some friends would joke that I was all work and no play, but sometimes that’s what has to happen to get work done.As the semester came to an end I can say I learned a lot. Planning is necessary for organization. Writing a to-do list at the beginning of the week and actually completing it make me feel good, and it’s an accomplishment to be proud of. Not only have I learned something new on organization, but also I’ve learned a new tool that I can take every where with me and use effectively.5. Why did the author stop hanging out with his friends?A. He thought it was meaningless for him.B. His friends didn’t like to wait for him.C. He had no money to pay for the meals.D. He had to work and study.6. How did the author succeed in balancing work and school?A. By following his friends’ advice.B. By turning to his friends for help.C. By organizing and planning in advance.D. By making plans with his friends and his teachers.7. Which of the following statements is TRUE?A. The author had some difficulty in balancing his job and school.B. The author had enough hours to balance his job and school.C. The author found it was easy to finish the to-do list on time with the help of his friends.D. None of the aut hor’s friends understood him and they all laughed at his planning and scheduling.8. We may know from the passage that ____________.A. the author has been proud of his college lifeB. planning has become a useful tool for the author’s lifeC. the author didn’t succeed in planning his daily lifeD. planning the author’s task makes him feel goodCIt’s not easy being a teenager ― nor is it easy being a parent of a teenager. You can make your child feel angry, hurt, or misunderstood by what you say without realizing it yourself. It is important to give your child the space he needs to grow while gently letting him know that you’ll still be there for him when he needs you.Expect a lot from your child, just not everything. Except for health and safety problems, such as drug use or careless driving; consider everything else open to discussion. If your child is unwilling to discuss something, don’t insist he tell y ou what’s on his mind. The more you insist, the more likely that he’ll clam up. Instead, let him attempt to solve things by himself. At the same time, remind him that you’re always there for him to seek advice or help. Show respect for your teenager’s privacy. Never read his mail or listen in on personal conversations.Teach your teenager that the family phone is for the whole family. If your child talks on the family’s telephone for too long, tell him he can talk for 15 minutes, but then he must stay off the phone for at least an equal period of time. This not only frees up the line so that other family members can make and receive calls, but teaches your teenager moderation (节制). Or if you are open to the idea, let your teenager have his own phone that he pays for with his own pocket money or a part-time job.9. What does the underlined phrase "clam up"in Paragraph 2 probably mean?A. Become excited.B. Show respect.C. Refuse to talk.D. Seek help.10. The last paragraph is about how to teach a teenager ____________.A. to use the phone in a sensible wayB. to pay for his own telephoneC. to share the phone quicklyD. to answer the phone quickly11. What should parents do in raising a teenager according to the text?A. Not allow him to learn driving or take drugs.B. Give him advice only when necessary.C. Let him have his own telephone.D. Not talk about personal things with him.12. The main purpose of the text is to tell parents ____________.A. how to get along with a teenagerB. how to respect a teenagerC. how to understand a teenagerD. how to help a teenager grow upDIt is widely known that any English conversation begins with the Weather. Such a fixation with the weather finds expression in Dr. Johnson’s famous comment that "When two English meet, their first talk is of weather". Though Johnson’s observation is as accurate now as it was over two hundred years ago, most commentators fail to come up with a convincing explanation for this English weather-speak.Bill Bryson, for example, concludes that, as the English weather is not at all exciting, the obsession with it can hardly be understood. He argues that "To an outsider, the most striking thing about the English weather is that there is not very much of it."Simply, the reason is that the unusual and unpredictable weather is almost unknown in the British Isles.Jeremy Paxman, however, disagrees with Bryson, arguing that the English weather is by nature attractive."Bryson is wrong,"he says, "because the English preference for the weather has nothing to do with the natural phenomena. The interest is less in the phenomena themselves, but in uncertainty."According to him, the weather in England is very changeable and uncertain and it attracts the English as well as the outsider.Bryson and Paxman stand for common misconceptions about the weather-speak among the English. Both commentators, somehow, are missing the point. The English weather conversation is not really about the weather at all. English weather-speak is a system of signs, which is developed to help the speakers overcome the natural reserve andactually talk to each other. Everyone knows conversations starting with weather-speak are not requests for weather data. Rather, they are routine greetings, conversation starters or the blank "fillers". In other words, English weather-speak isa means of social bonding.13. The auth or mentions Dr. Johnson’s comment to show that ____________.A. most commentators agree with Dr. JohnsonB. Dr. Johnson is famous for his weather observationC. the comment was accurate two hundred years agoD. English conversations usually start with the weather14. What does the underlined word"obsession"most probably refer to?A. A social trend.B. An emotional state.C. A historical concept.D. An unknown phenomenon.15. What is the author’s main purpose of writing the passage?A. To explain what English weather-speak is about.B. To analyse misconceptions about the English weather.C. To find fault with both Bill Bryson and Jeremy Paxman.D. To convince people that the English weather is changeable.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。

2017-2018学年人教A版数学必修五模块综合检测三 含答

2017-2018学年人教A版数学必修五模块综合检测三 含答

模块综合检测(三) (时间120分钟,满分150分)一、选择题(本大题共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.不等式x 2<x +6的解集为( ) A .{x |-2<x <3} B .{x |x <-2} C .{x |x <-2或x >3} D .{x |x >3}解析:选A 不等式化为x 2-x -6=(x -3)(x +2)<0,解得-2<x <3. 2.等差数列{a n }中,a 4+a 8=10,a 10=6,则a 18=( ) A .8.5 B .8 C .7.5D .7解析:选B a 4+a 8=2a 6=10,即a 6=5,d =14(a 10-a 6)=14,则a 18=a 10+8d =6+2=8.3.若a >0,b >0,且a +b =4,则下列不等式恒成立的是( ) A.1ab >12 B.1a +1b≤1C.ab ≥2D.1a +b ≤18解析:选D 因为2=a +b2≤a 2+b 22,所以a 2+b 2≥8,所以1a 2+b 2≤18. 4.已知等比数列{a n }的公比q =3,前3项和S 3=133,则a n 等于( )A .3nB .3n -1C .3n -2D .3n +1解析:选C 由q =3,S 3=133得a 1-331-3=133,解得a 1=13.所以a n =13×3n -1=3n -2. 5.在△ABC 中,角A ,B ,C 所对的边分别为a ,b ,c ,若sin 2A +sin 2C -sin 2B =3sinA sin C ,则角B 为( )A.π6 B.π3 C.2π3D.5π6解析:选A 由正弦定理可得a 2+c 2-b 2=3ac ,所以cos B =a 2+c 2-b 22ac =3ac 2ac =32,所以B =π6,故选A. 6.在R 上定义运算⊗:x ⊗y =x (1-y ).若不等式(x -a )⊗(x +a )<1对任意实数x 恒成立,则( )A .-1<a <1B .0<a <2C .-12<a <32D .-32<a <12解析:选C 因为(x -a )⊗(x +a )=(x -a )(1-x -a ),又不等式(x -a )⊗(x +a )<1对任意实数x 恒成立,所以(x -a )(1-x -a )<1对任意实数x 恒成立,即x 2-x -a 2+a +1>0对任意实数x 恒成立,所以Δ=(-1)2-4(-a 2+a +1)<0,解得-12<a <32.7.已知实数x ,y 满足⎩⎪⎨⎪⎧y ≤x ,x +y ≤1,y ≥-1,则目标函数z =2x -y -1的最大值为( )A .5B .4 C.12D .-3解析:选B 作出不等式组表示的平面区域,如图所示,其中A (-1,-1),B (2,-1),C ⎝ ⎛⎭⎪⎫12,12,z =2x -y -1可变形为:y =2x -z -1,表示斜率为2,在y 轴上截距为-z -1的一组平行线,将直线l :z =2x -y -1进行平移,当直线经过点B 时,目标函数z 达到最大值,所以z max =2×2-(-1)-1=4,故选B.8.在△ABC 中,角A ,B ,C 所对应的边分别为a ,b ,c ,若角A ,B ,C 依次成等差数列,且a =1,b =3,则S △ABC 等于( )A. 2B. 3C.32D .2解析:选C ∵A ,B ,C 成等差数列,∴B =60°.又由正弦定理得a sin A =bsin B,∴sin A =a sin Bb=1×323=12, ∴A =30°或A =150°(舍去),∴C =90°,∴S △ABC =12ab =32.9.已知等差数列{a n }中,a 2=6,a 5=15.若b n =a 2n ,则数列{b n }的前5项和等于( ) A .30 B .45 C .90D .186解析:选C 依题意得⎩⎪⎨⎪⎧a 1+d =6,a 1+4d =15,∴⎩⎪⎨⎪⎧a 1=3,d =3,∴a n =a 1+(n -1)d =3+(n -1)·3=3n , ∴b n =a 2n =6n ,∴{b n }的前5项和为S 5=b 1+b 2+b 3+b 4+b 5=6+12+18+24+30=90. 10.在△ABC 中,若cos A cos B =b a =43,则△ABC 是( )A .直角三角形B .等腰三角形C .等腰或直角三角形D .钝角三角形解析:选A 由正弦定理得cos A cos B =b a =sin Bsin A ,即sin A cos A =sin B cos B ,所以sin 2A=sin 2B ,所以2A =2B 或2A +2B =π,即A =B 或A +B =π2.又b a =43,所以a ≠b ,故A =B舍去,所以A +B =π2,即△ABC 为直角三角形.11.已知a >b ,则不等式:①a 2>b 2;②1a <1b ;③1a -b >1a 中不能恒成立的个数是( )A .0B .1C .2D .3解析:选D 对于①,a 2-b 2=(a -b )(a +b ),a -b >0,但a +b 的符号无法确定; 对于②,1a -1b =b -aab,b -a <0,但ab 的符号无法确定;对于③,1a -b -1a =b a -b a ,a -b >0,但ba的符号不确定.所以这三个不等式都不能恒成立.12.设a >0,b >0,且不等式1a +1b +ka +b≥0恒成立,则实数k 的最小值等于( )A .0B .4C .-4D .-2解析:选C 由1a +1b +ka +b ≥0得k ≥-a +b 2ab,而a +b 2ab=b a +a b+2≥4,所以-a +b2ab≤-4,因此要使k ≥-a +b2ab恒成立,应有k ≥-4,即实数k 的最小值等于-4.二、填空题(本大题共4小题,每小题5分,共20分,把正确答案填在题中的横线上) 13.已知数列{a n }中,a 1=12,a n +1=1-1a n ,则a 16=________.解析:由题意可知a 2=-1,a 3=2,a 4=12,a 5=-1,a 6=2,a 7=12,…,所以数列{a n }是以3为周期的周期数列. 又16=3×5+1,所以a 16=a 1=12.答案:1214.如图,为了测量河对岸A ,B 两点之间的距离,观察者找到一个点C ,从点C 可以观察到点A ,B ;找到一个点D ,从点D 可以观察到点A ,C ;找到一个点E ,从点E 可以观察到点B ,C .并测量得到一些数据:CD =2,CE =23,∠D =45°,∠ACD =105°,∠ACB =48.19°,∠BCE =75°,∠E =60°,则A ,B 两点之间的距离为________.⎝⎛⎭⎪⎫其中cos 48.19°取近似值23解析:依题意知,在△ACD 中,∠A =30°, 由正弦定理得AC =CD sin 45°sin 30°=2 2.在△BCE 中,∠CBE =45°, 由正弦定理得BC =CE sin 60°si n 45°=3 2.在△ABC 中,由余弦定理AB 2=AC 2+BC 2-2AC ×BC cos ∠ACB =10,所以AB =10. 答案:1015.已知a ∈,不等式x 2+(a -4)x +4-2a >0恒成立,则x 的取值范围为________. 解析:把不等式的左端看成关于a 的一次函数,记f (a )=(x -2)a +(x 2-4x +4),则f (a )>0对于任意的a ∈恒成立,易知只需⎩⎪⎨⎪⎧f-=x 2-5x +6>0,f =x 2-3x +2>0,解得x <1或x >3.所以x 的取值范围是(-∞,1)∪(3,+∞). 答案:(-∞,1)∪(3,+∞)16.某房地产开发公司用800万元购得一块土地,该土地可以建造每层1 000平方米的楼房,已知第一层每平方米的建筑费用为600元,楼房每升高一层,每平方米的建筑费用增加40元.若把楼房建成n 层后,每平方米的平均综合费用最低(综合费用是建筑费用与购地费用之和),则n =________.解析:易知每层的建筑费用构成等差数列,设为{a n },则n 层的建筑总费用为S n =600×103+(600+40)×103+…+×103=(2n 2+58n )×104,所以每平方米的平均综合费用为 800×104+n 2+58n41 000n=10⎝⎛⎭⎪⎫2n +800n+58≥102 2n ×800n+58=1 380元,当且仅当2n =800n,即n =20时等号成立.答案:20三、解答题(本大题共6小题,共70分,解答时应写出必要的文字说明、证明过程或演算步骤)17.(本小题满分10分)已知函数f (x )=ax 2-4ax -3. (1)当a =-1时,求关于x 的不等式f (x )>0的解集;(2)若对于任意的x ∈R ,均有不等式f (x )≤0成立,求实数a 的取值范围. 解:(1)当a =-1时,不等式ax 2-4ax -3>0,即-x 2+4x -3>0. 可化为x 2-4x +3<0,即(x -1)(x -3)<0,解得1<x <3, 故不等式f (x )>0的解集为(1,3).(2)①当a =0时,不等式ax 2-4ax -3≤0恒成立; ②当a ≠0时,要使得不等式ax 2-4ax -3≤0恒成立;只需⎩⎪⎨⎪⎧a <0,Δ≤0,即⎩⎪⎨⎪⎧a <0,-4a 2-4a -,解得⎩⎪⎨⎪⎧a <0,-34≤a ≤0,即-34≤a <0,综上所述,a 的取值范围为⎣⎢⎡⎦⎥⎤-34,0. 18.(本小题满分12分)在△ABC 中,内角A ,B ,C 的对边分别为a ,b ,c ,且A ,B ,C成等差数列.(1)若b =23,c =2,求△ABC 的面积;(2)若sin A ,sin B ,sin C 成等比数列,试判断△ABC 的形状.解:因为A ,B ,C 成等差数列,所以2B =A +C .又A +B +C =π,所以B =π3.(1)法一:因为b =23,c =2,所以由正弦定理得b sin B =csin C ,即b sin C =c sin B ,即23sin C =2×32,得sin C =12. 因为b >c ,所以B >C ,即C 为锐角,所以C =π6,从而A =π2.所以S △ABC =12bc =2 3.法二:由余弦定理得b 2=a 2+c 2-2ac cos B , 即a 2-2a -8=0,得a =4.所以S △ABC =12ac sin B =12×4×2×32=2 3.(2)因为sin A ,sin B ,sin C 成等比数列,所以sin 2B =sin A ·sinC . 由正弦定理得b 2=ac ;由余弦定理得b 2=a 2+c 2-2ac cos B =a 2+c 2-ac . 所以ac =a 2+c 2-ac ,即(a -c )2=0,即a =c . 又因为B =π3,所以△ABC 为等边三角形.19.(本小题满分12分)货轮在海上自B 点以40 km/h 的速度沿方向角(从指北方向顺时针转到目标方向线的旋转角)为140°的方向航行,为了确定船位,船在B 点观测灯塔A 的方向角为110°,航行半小时后,船到达C 点,观测灯塔A 的方向角是65°,求货轮到达C 点时与灯塔A 的距离.解:在△ABC 中,BC =40×0.5=20 km , ∠ABC =140°-110°=30°,∠ACB =65°+(180°-140°)=105°, ∠BAC =45°.根据正弦定理,得AC sin ∠ABC =BCsin ∠BAC,AC =BC ·sin∠ABC sin ∠BAC =20·sin 30°sin 45°=10 2.故货轮到达C 点时与灯塔的距离为10 2 km.20.(本小题满分12分)已知函数f (x )=ax 2+a 2x +2b -a 3,当x ∈(-2,6)时,其值为正,而当x ∈(-∞,-2)∪(6,+∞)时,其值为负.(1)求实数a ,b 的值及函数f (x )的解析式;(2)设F (x )=-k4f (x )+4(k +1)x +2(6k -1),问k 取何值时,函数F (x )的值恒为负值?解:(1)由题意可知-2和6是方程f (x )=0的两根,∴⎩⎪⎨⎪⎧-a =-2+6=4,2b -a 3a=-2×6=-12.∴⎩⎪⎨⎪⎧a =-4,b =-8.∴f (x )=-4x 2+16x +48.(2)F (x )=-k4(-4x 2+16x +48)+4(k +1)x +2(6k -1)=kx 2+4x -2.当k =0时,F (x )=4x -2不恒为负值; 当k ≠0时,若F (x )的值恒为负值,则有⎩⎪⎨⎪⎧k <0,16+8k <0,解得k <-2.21.(本小题满分12分)已知S n 为数列{a n }的前n 项和,且a 2+S 2=31,a n +1=3a n -2n(n ∈N *).(1)求证:{a n -2n}为等比数列; (2)求数列{a n }的前n 项和S n . 解:(1)由a n +1=3a n -2n可得a n +1-2n +1=3a n -2n -2n +1=3a n -3·2n =3(a n -2n),即a n +1-2n +1a n -2n=3.又a 2=3a 1-2,则S 2=a 1+a 2=4a 1-2, 得a 2+S 2=7a 1-4=31,得a 1=5,∴a 1-21=3≠0,且a n +1-2n +1a n -2n=3.故{a n -2n}为等比数列. (2)由(1)可知a n -2n=3n -1(a 1-21)=3n,故a n =2n+3n, ∴S n =-2n1-2+-3n1-3=2n +1+3n +12-72. 22.(本小题满分12分)已知定义域为的函数f (x )是增函数,且f (1)=1.(1)若对于任意x ∈,总有4f 2(x )-4(2-a )·f (x )+5-4a ≥0,求实数a 的取值范围;(2)证明:f ⎝ ⎛⎭⎪⎫122+223+…+n 2n +1<1.解:(1)f (x )在上是增函数,则f (x )≤f (1)=1,故1-f (x )≥0, 当f (x )=1时,不等式化为0·a +1≥0,显然a ∈R ;当f (x )<1时,不等式化为a ≤4f2x -8f x +54-4f x对于x ∈恒成立.设y =4f2x -8f x +54-4f x=1-f (x )+14[1-f x≥1.当且仅当f (x )=12时取等号,∴y min =1,从而a ≤1, 综上所述,a ∈(-∞,1]. (2)令T n =122+223+…+n2n +1,①则12T n =123+224+…+n -12n +1+n2n +2,② ①-②化简得,T n =12+122+…+12n -n 2n +1=1-12n -n2n +1<1,又由①知T n >0,∵f (x )在上是增函数,∴f ⎝ ⎛⎭⎪⎫122+223+…+n 2n +1<f (1)=1.。

2017-2018学年高二英语综合检测练习:综合测试题(新人教版必修5) Word版含答案

2017-2018学年高二英语综合检测练习:综合测试题(新人教版必修5)   Word版含答案

综合测试题时间:120分钟,满分:150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.What's the woman going to do this evening?A.Go to a movie.B.Go to a concert.C.Study in the library.2.How will the man go to the office?A.By train.B.By taxi.C.By bus.3.What's the probable relationship between the two speakers?A.Classmates.B.Workmates.C.Friends.4.What's the woman doing?A.Ordering some food.B.Selling some food.C.Cooking some food.5.Where does the conversation probably take place?A.In a store.B.In an office.C.In a bank.第二节(共15小题;每题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;听完后,每小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6段材料,回答第6至7题。

6.Where does the man work?A.In a factory.B.In a hotel.C.In a school.7.What does the man do?A.A manager.B.A cleaner.C.An electrician.听第7段材料,回答第8至9题。

2017-2018学年人教A版数学必修五模块综合检测一 含答

2017-2018学年人教A版数学必修五模块综合检测一 含答

模块综合检测(一) (时间120分钟,满分150分)一、选择题(本大题共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)1.一个等差数列的第5项a 5=10,且a 1+a 2+a 3=3,则有( ) A .a 1=-2,d =3 B .a 1=2,d =-3 C .a 1=-3,d =2D .a 1=3,d =-2解析:选A ∵a 1+a 2+a 3=3且2a 2=a 1+a 3, ∴a 2=1.又∵a 5=a 2+3d =1+3d =10, ∴d =3,∴a 1=a 2-d =1-3=-2.2.若a <1,b >1,那么下列命题中正确的是( ) A.1a >1bB.b a>1 C .a 2<b 2D .ab <a +b解析:选D 利用特值法,令a =-2,b =2. 则1a <1b ,A 错;b a<0,B 错;a 2=b 2,C 错.3.已知实数x ,y 满足⎩⎪⎨⎪⎧x -y +2≥0,x +y ≥0,x ≤1,则z =2x +y 的最小值是( )A .-1B .1C .-2D .2解析:选A 由不等式组作出可行域如图所示,由图可知:当直线y =-2x +z 经过点A (-1,1)时,z 取得最小值为-1.4.已知△ABC 的三个内角之比为A ∶B ∶C =3∶2∶1,那么,对应的三边之比a ∶b ∶c 等于( )A .3∶2∶1 B.3∶2∶1 C.3∶2∶1D .2∶3∶1解析:选D ∵A ∶B ∶C =3∶2∶1,A +B +C =180°, ∴A =90°,B =60°,C =30°.∴a ∶b ∶c =sin 90°∶sin 60°∶sin 30° =1∶32∶12=2∶3∶1. 5.已知△ABC 中,三内角A ,B ,C 依次成等差数列,三边a ,b ,c 成等比数列,则△ABC 是( )A .直角三角形B .等腰直角三角形C .钝角三角形D .等边三角形解析:选D 由题意可得B =60°,再由余弦定理可得b 2=a 2+c 2-2ac cos B =a 2+c 2-ac , 又三边a ,b ,c 成等比数列,所以b 2=ac ,上式即为a 2+c 2-2ac =(a -c )2=0, 则a =c ,所以△ABC 是等边三角形.6.等比数列{a n }的前4项和为240,第2项与第4项的和为180,则数列{a n }的首项为( )A .2B .4C .6D .8解析:选C S 4-(a 2+a 4)=60⇒a 1+a 3=60. ∴q =a 2+a 4a 1+a 3=3,a 1=6. 7.在△ABC 中,a ,b ,c 分别是角A ,B ,C 的对边,若A =π3,b =1,△ABC 的面积为32,则a 的值为( ) A .1 B .2 C.32D. 3解析:选D 根据S =12bc sin A =32,可得c =2,由余弦定理得a 2=b 2+c 2-2bc cos A =3,故a = 3.8.关于x 的不等式x 2-ax -6a <0的解集是{x |m <x <n },且n -m ≤5,则实数a 的取值范围是( )A .C .(-25,-24)∪(0,1)D .解析:选D 由题意知,方程x 2-ax -6a =0有两根分别为m 和n , 则有⎩⎪⎨⎪⎧Δ=a 2+24a >0⇒a <-24或a >0,m +n =a ,mn =-6a .又0<n -m ≤5,∴(n -m )2=(n +m )2-4nm =a 2+24a ≤25, 即a 2+24a -25≤0,解得-25≤a ≤1. ∴-25≤a <-24或0<a ≤1. 故实数a 的取值范围是.9.实数x ,y 满足⎩⎪⎨⎪⎧x ≥1,y ≤a a,x -y ≤0,若函数z =x +y 的最大值为4,则实数a 的值为( )A .2B .3C .4D.32解析:选A 由不等式组作出可行域,如图所示的阴影部分,当z =x +y 过y =x 和y =a 的交点A (a ,a )时,z 取得最大值,即z max =a +a =4,所以a =2.10.在△ABC 中,三个内角A ,B ,C 所对的边分别为a ,b ,c ,若内角A ,B ,C 依次成等差数列,且不等式-x 2+6x -8>0的解集为{x |a <x <c },则S △ABC 等于( )A. 3 B .2 3 C .3 3D .4 3解析:选B 由于不等式-x 2+6x -8>0的解集为{x |2<x <4}, ∴a =2,c =4.又角A ,B ,C 依次成等差数列,∴B =π3,∴S △ABC =12×2×4×sin π3=2 3.11.一个等比数列前三项的积为2,最后三项的积为4,且所有项的积为64,则该数列有( )A .13项B .12项C .11项D .10项解析:选B 设该数列的前三项分别为a 1,a 1q ,a 1q 2,后三项分别为a 1qn -3,a 1qn -2,a 1qn-1.所以前三项之积a 31q 3=2,后三项之积a 31q 3n -6=4,两式相乘,得a 61q3(n -1)=8,即a 21qn -1=2.又a 1·a 1q ·a 1q 2·…·a 1q n -1=64,所以a n1·q n n -12=64,即(a 21qn -1)n=642,即2n=642,所以n =12.12.函数y =x 2+2x -1(x >1)的最小值是( )A .23+2B .23-2C .2 3D .2解析:选A ∵x >1, ∴x -1>0.∴y =x 2+2x -1=x 2-2x +2x +2x -1=x 2-2x +1+x -+3x -1=x -2+x -+3x -1=x -1+3x -1+2≥23+2. 二、填空题(本大题共4小题,每小题5分,共20分,把正确答案填在题中的横线上)13.若规定⎪⎪⎪⎪⎪⎪ab cd =ad -bc ,则1<⎪⎪⎪⎪⎪⎪211x <2的解集是________.解析:由已知可得1<2x -1<2,解得1<x <32,所以所求解集为⎩⎨⎧⎭⎬⎫x 1<x <32.答案:⎩⎨⎧⎭⎬⎫x 1<x <32 14.在数列{a n }中,a 1=2,a n +1-2a n =0(n ∈N *),b n 是a n 和a n +1的等差中项,设S n 为数列{b n }的前n 项和,则S 6=________.解析:由a n +1=2a n ,{a n }为等比数列, ∴a n =2n. ∴2b n =2n+2n +1,即b n =3×2n -1,∴S 6=3×1+3×2+…+3×25=189.答案:18915.已知A船在灯塔C北偏东80°处,且A到C的距离为2 km,B船在灯塔C北偏西40°处,A,B两船的距离为3 km,则B到C的距离为________ km.解析:如图所示,在△ABC中,∠ACB=40°+80°=120°,AB=3 km,AC=2 km.设BC=a km.由余弦定理的推论,得cos 120°=a2+4-94a,解得a=6-1或a=-6-1(舍去),即B到C的距离为(6-1) km.答案:(6-1)16.不等式x2+ax+4<0的解集不是空集,则实数a的取值范围是________.解析:∵不等式x2+ax+4<0的解集不是空集,∴Δ=a2-4×4>0,即a2>16.∴a>4或a<-4.答案:(-∞,-4)∪(4,+∞)三、解答题(本大题共6小题,共70分,解答时应写出必要的文字说明、证明过程或演算步骤)17.(本小题满分10分)已知△ABC的内角A,B,C所对的边分别为a,b,c,且a=2,cos B=35 .(1)若b=4,求sin A的值;(2)若S△ABC=4,求b,c的值.解:(1)∵cos B=35>0,且0<B<π,∴sin B=1-cos2B=4 5 .由正弦定理asin A =bsin B,得sin A=a sin Bb=2×454=25.(2)∵S △ABC =12ac sin B =4,∴12·2·c ·45=4. ∴c =5.由余弦定理b 2=a 2+c 2-2ac cos B , 得b =a 2+c 2-2ac cos B =22+52-2×2×5×35=17.18.(本小题满分12分)已知函数f (x )=log 3(x 2-4x +m )的图象过点(0,1). (1)求实数m 的值; (2)解不等式:f (x )≤1.解:(1)由已知有f (0)=log 3m =1, ∴m =3.(2)由(1)知f (x )=log 3(x 2-4x +3). 由x 2-4x +3>0,得x <1或x >3,∴函数的定义域为(-∞,1)∪(3,+∞). ∵log 3(x 2-4x +3)≤1且y =log 3x 为增函数, ∴0<x 2-4x +3≤3, ∴0≤x <1或3<x ≤4,∴不等式的解集为{x |0≤x <1或3<x ≤4}.19.(本小题满分12分)△ABC 中,D 是BC 上的点,AD 平分∠BAC ,△ABD 面积是△ADC 面积的2倍.(1)求sin Bsin C; (2)若AD =1,DC =22,求BD 和AC 的长. 解:(1)S △ABD =12AB ·AD sin ∠BAD ,S △ADC =12AC ·AD sin ∠CAD .因为S △ABD =2S △ADC ,∠BAD =∠CAD , 所以AB =2AC .由正弦定理,得sin B sin C =AC AB =12.(2)因为S △ABD ∶S △ADC =BD ∶DC ,所以BD = 2.在△ABD 和△ADC 中,由余弦定理,知AB 2=AD 2+BD 2-2AD ·BD cos ∠ADB , AC 2=AD 2+DC 2-2AD ·DC cos ∠ADC .故AB 2+2AC 2=3AD 2+BD 2+2DC 2=6. 由(1),知AB =2AC ,所以AC =1.20.(本小题满分12分)设数列{a n }的前n 项和为S n =2n 2,{b n }为等比数列,且a 1=b 1,b 2(a 2-a 1)=b 1.(1)求数列{a n }和{b n }的通项公式; (2)设c n =a n b n,求数列{c n }的前n 项和T n . 解:(1)当n ≥2时,a n =S n -S n -1=2n 2-2(n -1)2=4n -2,当n =1时,a 1=S 1=2满足上式,故{a n }的通项公式为a n =4n -2. 设{b n }的公比为q ,由已知条件a 1=b 1,b 2(a 2-a 1)=b 1知,b 1=2,b 2=12,所以q =14,∴b n =b 1qn -1=2×14n -1,即b n =24n -1.(2)∵c n =a n b n =4n -224n -1=(2n -1)4n -1,∴T n =c 1+c 2+…+c n =1+3×41+5×42+…+ (2n -1)4n -1.4T n =1×4+3×42+5×43+…+(2n -3)4n -1+(2n -1)4n. 两式相减得3T n =-1-2(41+42+43+…+4n -1)+(2n -1)4n=13. ∴T n =19.21.(本小题满分12分)如图所示,将一矩形花坛ABCD 扩建成一个更大的矩形花坛AMPN ,要求B 在AM 上,D 在AN 上,且对角线MN 过C 点,已知AB =3米,AD =2米.(1)要使矩形AMPN 的面积大于32平方米,则AN 的长应在什么范围内? (2)当AN 的长度是多少时,矩形AMPN 的面积最小?请求最小面积.解:(1)设AN =x (x >2)米,则ND =x -2, 因为ND DC =AN AM ,所以x -23=x AM, 所以AM =3xx -2. 所以3xx -2·x >32, 所以3x 2-32x +64>0, 所以(3x -8)(x -8)>0, 所以2<x <83或x >8.即2<AN <83或AN >8.(2)S 矩形AMPN =3x2x -2=x -2+x -+12x -2=3(x -2)+12x -2+12≥236+12=24, 当且仅当x =4时取等号.所以当AN 的长度是4米时,矩形AMPN 的面积最小,最小面积为24平方米.22.(本小题满分12分)设不等式组⎩⎪⎨⎪⎧x >0,y >0,y ≤-nx +3n所表示的平面区域为D n ,记D n 内的整点个数为a n (n ∈N *). (1)求数列{a n }的通项公式;(2)记数列{a n }的前n 项和为S n ,且T n =S n3·2n -1,若对一切的正整数n ,总有T n ≤m ,求实数m 的取值范围.解:(1)由x >0,y >0,3n -nx ≥y ,得0<x <3. 则D n 内的整点在直线x =1和x =2上.记y =-nx +3n 为l ,l 与x =1,x =2的交点的纵坐标分别为y 1,y 2, 则y 1=2n ,y 2=n ,∴a n =3n (n ∈N *).(2)∵S n =3(1+2+…+n )=3n n +2,∴T n =n n +2n.令T n +1T n =n +22n>1,解得n <2, ∴当n ≥3时,T n >T n +1,且T 1=1<T 2=T 3=32,即T n 的最大值为32.所以实数m 的取值范围为⎣⎢⎡⎭⎪⎫32,+∞.。

2017-2018学年高中英语必修五 阶段质量检测一 含答案 精品

2017-2018学年高中英语必修五 阶段质量检测一 含答案 精品

阶段质量检测(一)A卷学业水平达标(时间:120分钟满分:150分)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。

(Text 1)M:How did you like Professor Wang's lecture?W:I shouldn't have attended it. It was a waste of time.1. What does the woman probably think of Professor Wang's lecture?A.Interesting. B.Exciting.C.Disappointing.(Text 2)M:Jim has been sitting there lost in deep thought for half an hour.W:It's typical of him. He is a little different from the other boys.2. What do we know about Jim?A.He is often lost in deep thought.B.He is often laughed at by others.C.He often says something strange.(Text 3)W:Shall we go now? It's already half past ten.M:Don't worry. We still have one and a half hours left to catch the train.W:OK.3. When does the train leave?A.At 10:30. B.At 11:00.C.At 12:00.(Text 4)M:I'm in front of the People's Hospital now. How do I get to the Green Hotel from here?W:Cross the street and walk straight along. You'll see a post office in about ten minutes.M:Is the hotel near the post office?W:Yes. It's only a five­minute walk from there to the hotel.4. Where does the man want to go?A.To a hospital. B.To a hotel.C.To a post office.(Text 5)W:David, look! What a beautiful red dress! And the blue one looks beautiful too.M:Frankly, I like this white dress the best.W:Then I'll take this one.M:OK. Now, Lisa, you put on the dress in the fitting room while I go and pay the money. We've got to hurry, or we'll be late for Jenny's party.5. Which dress does the woman take?A.The red one. B.The blue one.C.The white one.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。

2017-2018学年高中语文人教版必修5(测评):模块综合测评

2017-2018学年高中语文人教版必修5(测评):模块综合测评

2017-2018 学年高中语文人教版必修5(测评):模块综合测评模块综合测评本试题分第Ⅰ卷 (阅读题 )和第Ⅱ卷 (表达题 )两部分。

满分150 分。

考试用时150分钟。

第Ⅰ卷阅读题一、现代文阅读 (35 分)(一 )论述类文本阅读 (9 分,每题 3 分)阅读下边的文字,达成1~3 题。

为何汉字是方块字?这个问题固然没有明确的考据,但从先人察看世界的方式中即可窥见一斑。

《淮南子》说:“往古之时,四极废,九州裂于是女娲炼五色石以补苍天,断鳌足以立四极。

” 在先人心目中,“天圆地方”,地是方形的,并且在这四方形地的终点,还有撑着的柱子。

从甲骨文能够看出,汉字基本上是方形。

到了西周末年,汉字一字一方格的特色已经完好成形了。

今后,春秋战国时从篆书到隶书,汉字达成了一次飞腾,表此刻字形上,就是从从前的圆润瘦长变为了方方正正的方块字。

这类方方正正的写法,表现了中国先人所追求的客观美——庄重、庄重、均衡对称,有一种周正之美。

固然此后有了“龙飞凤舞” 的草书,但每个朝代的正体字,仍旧沿袭着工整的书写模样。

最早的金文,马上在青铜器上的铭文,都写得规行矩步,在当时写字是件极其隆重严肃的事。

到了周代后期,各国纷争,文字异姓,但距离周近来的秦国,其正体字仍旧是规正的篆体字。

秦国灭六国,成立秦代以后,正式立小篆为正体字。

但在官方规定的正体字以外,秦简上还出现了一种协助性书体,即此后的隶书,到汉代汉武帝年间,隶书逐渐代替篆书成为正体字。

今后,在魏晋南北朝期间,楷书流行,成为官方认同的正体字,向来流传到现在。

2017-2018 学年高中语文人教版必修5(测评):模块综合测评固然书写方式不一样,但对均衡对称、周正之美的追求却向来没变。

唐代时用科举制选拔人材,开了各样考试科目,书法就是此中一科。

但是,不论是考哪一科,都要求“楷书字体,皆得正样”,文字与书写并重。

因为文字书法的好坏关系到个人的功名和人生出路,所谓“升沉是系”,所以仕进者特别重视。

肥城市2017-2018学年度上学期高二数学必修5试题

肥城市2017-2018学年度上学期高二数学必修5试题

高二数学必修5试题本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分.共6页.共150分.考试用时120分钟. 考试结束后,本试卷和答题卡一并收回. 注意事项:1.答卷前,考生务必将自己的学校、班级、姓名、考号、座号用0.5毫米黑色签字笔填写在答题卡规定的位置上;用2B 铅笔填涂在答题卡上.2.第Ⅰ卷每小题选出答案后,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,再选涂其它答案标号.3.第Ⅱ卷必须用0.5毫米黑色签字笔作答,答案必须写在答题卡各题目指定区域内相应的位置;如需改动,先划掉原来的答案,然后再写上新的答案;不能使用涂改液、胶带纸、修正带.不按以上要求作答的答案无效.4.填空题请直接填写答案,解答题应写出文字说明、证明过程或演算步骤.第Ⅰ卷(选择题 共60分)一、选择题:本大题共12小题,每小题5分,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1,,5a 成等差数列,则实数a 的值为(A )2 (B )3 (C )4 (D (2)任给,ABC ∆设角,,A B C 所对的边分别为,,,a b c 则下列等式成立的是(A )2222cos c a b ab C =++ (B )2222cos c a b ab C =+- (C )2222sin c a b ab C =++ (D )2222sin c a b ab C =+-(3)不等式()()130x x +-<的解集为(A )()1,3- (B )()3,1-(C )()(),13,-∞-+∞(D )()(),31,-∞-+∞(4)若0,x >则4x x+的最小值为 (A )2 (B )3 (C )4 (D )5 (5)已知数列{}n a 是以q 为公比的等比数列.若2,n n b a =-则数列{}n b 是(A )以q 为公比的等比数列 (B )以q -为公比的等比数列 (C )以2q 为公比的等比数列(D )以2q -为公比的等比数列(6)设ABC ∆的内角,,A B C 所对的边分别为,,,a b c若,,34a A B ππ===则b =(A)2(B )1 (C(D )2 (7)若数列{}n a 的前n 项和()1,n S n n =+则3a =(A )5(B )6 (C )7(D )8(8)我国古代著名的思想家庄子在《庄子·天下篇》中说:“一尺之棰,日取其半,万世不竭.”用现代语言叙述为:一尺长的木棒,每日取其一半,永远也取不完. 这样, 每日剩下的部分都是前一日的一半. 如果把“一尺之棰”看成单位“1”,那么从第1次取开始,每次剩下的部分所成的数列111,,248的通项公式为(A )12n a n = (B)n a (C )12nn a ⎛⎫= ⎪⎝⎭(D )2n n a =(9)已知ABC ∆的内角,,A B C 所对的边分别为,,,a b c 若2cos ,b a C =则ABC ∆是(A )等腰三角形 (B )直角三角形 (C )等边三角形 (D )等腰直角三角形(10)若公差为正数的等差数列{}n a 的首项11,a =且248,,a a a 成等比数列,则数列11n n a a +⎧⎫⎨⎬⎩⎭的前n 项和为 (A )n (B )1n + (C )1n n - (D )1nn + (11)建筑工程中,将房屋的窗户面积与地面面积之比称为“窗地面积比”.某办公室的窗户面积为n 平方米,地面面积为m 平方米,窗地面积比为1.λ 将窗户面积和地面 面积同时减少a 平方米后,窗地面积比为2.λ 如果,a n m << 那么 (A )12λλ≤ (B )12λλ≥ (C )12λλ<(D )12λλ>(12)已知实数,x y满足40,20,40.x y y ⎧+-≤⎪⎪-+≤⎨⎪+-≥⎪⎩则z =(A)⎤⎥⎝⎦(B)⎤⎦ (C )[]1,2(D )(]0,2第Ⅱ卷(非选择题 共90分)二、填空题:本大题共4小题,每小题5分,共20分.(13)若55a c =+=- 则,a c 的等比中项b = ▲ .(14)已知12,21,y x <<-<<- 则yx的取值范围是 ▲ . (15)飞机沿水平方向飞行,在A 处测得正前下方地面目标C 的俯角为,α 向前飞行a 千米到达B 处,此时测得正前下方目标C 的俯角为(),βαβ< 这时飞机与地面目标的水平距离为 ▲ 千米.(16)已知ABC ∆中,角,,A B C 所对的边分别是,,,a b c 若,3B b π== 则边c 的取值范围是 ▲ .三、解答题:本大题共6个小题,共70分. (17)(本小题满分10分)已知等差数列{}()n a n *∈N 满足158,0.a a =-=(Ⅰ)求数列{}n a 的公差d 和通项公式;n a(Ⅱ)若数列{}n a 的前n 项和n S 满足14,n S =- 求n 的值.(18)(本小题满分12分)ABC ∆的内角,,A B C 所对的边分别为,,,a b c 且ABC ∆的面积sin sin .sin B CS A=(Ⅰ)求边a 的长; (Ⅱ)若角,4,4C b π== 求边c 的长.(19)(本小题满分12分)已知函数()()222 4.f x x a x =+-+ (Ⅰ)若0,a = 求不等式()0f x >的解集;(Ⅱ)若对任意(),0x f x ∈>R 恒成立,求实数a 的取值范围.(20)(本小题满分12分)ABC ∆的内角,,A B C 所对的边分别为,,,a b c 且)2cos cos cos .a C b C c B =+(Ⅰ)求角C 的值;(Ⅱ)若2,c a b ==求ABC ∆的面积.(21)(本小题满分12分)设实数,x y 满足约束条件360,20,0,0.x y x y x y --≤⎧⎪-+≥⎨⎪≥≥⎩(Ⅰ)在给定的坐标系中画出不等式组表示的平面区域,并求该平面区域的面积; (Ⅱ)若目标函数()0,0z ax by a b =+>>的最大值为4,求123a b+的最小值.(22)(本小题满分12分)已知数列{}n a 的首项11,a = 前n 项和为,n S 且()11;n n na S n n +=++数列{}n b 满足:()()1122131.n n n a b a b a b n n *++⋅⋅⋅+=-⋅+∈N(Ⅰ)求数列{}n a 与{}n b 的通项公式; (Ⅱ)设()1111231111,2,n n n n n S c a c a n b b b b b -⎛⎫==++++⋅⋅⋅+≥ ⎪⎝⎭求数列{}n c 的前n 项 和.n T高二数学(必修5)参考答案及评分标准一、选择题:每小题5分,共60分.二、填空题:每小题5分,共20分. (13)1± (14)122y x -<<- (15)()sin cos sin a αββα- (16)02c <≤ 三、解答题:本大题共6个大题,共70分. (17)(本小题满分10分) 解:(Ⅰ)由题意158,0.a a =-=得:5840a d =-+= …………………………………………………………2分 ∴2d = …………………………………………………………………………3分∴()()11210;n a a n d n n *=+-=-∈N ……………………………………5分(Ⅱ)由(Ⅰ)可得:()129,2n n n a a S n n +==- ………………………………8分 ∵14,n S =-∴2914,n n -=- 即()()2914270,n n n n -+=--= ……………………9分∵,n *∈N ∴2,n =或7.n = ……………………………………………10分 (18)(本小题满分12分) 解:(Ⅰ)∵1sin sin sin ,sin 02sin B CS ab C C A==≠…………………………………3分 ∴1sin ,2sin Bab A=由正弦定理得:sin ,sin B bA a=……………………………………………………… 6分 ∴1,2b ab a=即22,a =∴a = …………………………………………………… 8分 (Ⅱ)由余弦定理得:2222cos c a b ab C =+-……………………………………10分22424cos4π=+-⋅10,= ………………………………………………………11分∴c = …………………………………………………………… 12分 (19)(本小题满分12分)解: (Ⅰ)当0a =时,()()22442,f x x x x =-+=- ………………………… 2分∵()0,f x >∴()220,x ->即2,x ≠ ………………………………………4分 ∴()0f x >的解集是{}|2;x x ≠ ………………………………………6分 (Ⅱ)由题意得:()242160,a ∆=--< ………………………………………9分即()224,a -<∴0 4.a << ………………………………………………………………11分 ∴a 的取值范围是0 4.a << …………………………………………………12分 (20)(本小题满分12分)解:(Ⅰ)∵)2cos cos cos .a C b C c B =+由正弦定理可得:)2sin cos sin cos sin cos .A C B C C B =+……………2分∴()2sin cos .A C B C =+ ………………………………………………4分 ∵,,A B C 为ABC ∆的内角; ∴()sin sin ;B C A +=∴2sin cos ;A C A ∵sin 0;A ≠∴cos 2C =………………………………………………………………5分 ∵0,C π<< ∴;6C π=………………………………………………………………6分(Ⅱ)∵2,c a =∴由余弦定理得:()22222a a b ab =+-即24160,a a +-= ……8分解得2a =-± ∵0,a >∴2,a = …………………………………………………………10分∴()111sin 22.222ABC S ab C ∆==⨯⨯= ……12分(21)(本小题满分12分) 解:(Ⅰ)不等式组表示的平面区域如图所示阴影部分. ……………3分联立36020x y x y --=⎧⎨-+=⎩,得点C 的坐标为()4,6, ∴平面区域的面积11262410;22S =⨯⨯+⨯⨯= ………………………………6分 (Ⅱ)当直线()0,0z ax by a b =+>>过点()4,6C 时,目标函数()0,0z ax by a b =+>> 取得最大值4,即464,a b += …………8分 ∴31,2a b += ∴121233224,33223b a a b a b a b a b⎛⎫⎛⎫+=++=++≥ ⎪⎪⎝⎭⎝⎭ …………………………10分 当且仅当11,23a b ==时,等号成立, ………………………………………11分 ∴123a b+的最小值为4. …………………………………………………12分 (22)(本小题满分12分) 解:(Ⅰ)∵()11,n n na S n n +=++∴当2n ≥时,()()111,n n n a S n n --=+- ……………………………………1分 ∴()()()()11111,n n n n na n a S S n n n n +---=-++-- 即()112,n n n na n a a n +=-++∴12;n n a a +-= ………………………………………………………………2分 ∴()12121;n a n n =+-=- 又当1n =时,适合上式;∴数列{}n a 是以11a =为首项,2d =为公差的等差数列,通项公式为()21;n a n n *=-∈N ………………………………………………3分又∵()1122131,nn n a b a b a b n ++⋅⋅⋅+=-⋅+∴当2n ≥时,()1112211231,n n n a b a b a b n ---++⋅⋅⋅+=-⋅+ …………………4分∴()()()111323213,nn n n n a b n n n --=-⋅--⋅=-⋅即()132;n n b n -=≥ ……………………………………………………………5分又1n =时,1111,1,a b b ==符合上式,∴()13;n n b n -*=∈N ………………………………………………………6分 (Ⅱ)当1n =时,111111,1;a c a T b ==== ……………………………7分 当2n ≥时, 11231111n n n n n S c a b b b b b -⎛⎫=++++⋅⋅⋅+ ⎪⎝⎭1211231111,n n n n a a a a b b b b b -⎛⎫++⋅⋅⋅+=++++⋅⋅⋅+ ⎪⎝⎭ ∴12n n T c c c =++⋅⋅⋅+2231121231112311111a a a b a a a b b b b b b b ⎛⎫⎛⎫++++ ⎪ ⎪⎝⎭⎝⎡⎤⎡⎤⎛⎫⎛⎫=++++⎢⎥⎢⎥ ⎪ ⎪ ⎪ ⎪⎢⎥⎢⎥⎝⎭⎝⎭⎣⎦⎣⎦⎭ 1211231111n n n n a a a a b b b b b -⎡⎤⎛⎫++⋅⋅⋅++⋅⋅⋅+++++⋅⋅⋅+⎢⎥ ⎪⎝⎭⎣⎦ 12123123123111111111111n n n n a a a b b b b b b b b b b b b ⎛⎫⎛⎫⎛⎫=+++⋅⋅⋅+++++⋅⋅⋅++⋅⋅⋅++++⋅⋅⋅+ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭ ()121231111n n a a a b b b b ⎛⎫=+++⋅⋅⋅+++⋅⋅⋅+ ⎪⎝⎭ 1231111,n n S b b b b ⎛⎫=+++⋅⋅⋅+ ⎪⎝⎭ …………………………………………………9分 又由(Ⅰ)可知:2,n S n =13,n n b -= ∴1n b ⎧⎫⎨⎬⎩⎭是以111b =为首项,以13为公比的等比数列,∴1231111113131,12313nn n b b b b ⎛⎫- ⎪⎡⎤⎛⎫⎝⎭+++⋅⋅⋅+==-⎢⎥ ⎪⎝⎭⎢⎥⎣⎦- ……………………10分 ∴当2n ≥时,2311,23n n T n ⎡⎤⎛⎫⋅-⎢⎥ ⎪⎝⎭⎢⎥⎣⎦= 此式对1n =也成立; …………………………………………………………… 11分 ∴()2311.23n n n n T *⎡⎤⎛⎫⋅-∈⎢⎥ ⎪⎝⎭⎢⎥⎦=⎣N …………………………………………… 12分。

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必修五综合测试卷
一、选择题(每题5分,共50分)
1.不等式x -1x +2
>1的解集是( ) A .{x |x <-2} B .{x |-2<x <1} C .{x |x <1}
D .{x |x ∈R } 2.下列说法正确的是( )
A .a >b ⇒ac 2>bc 2
B .a >b ⇒a 2>b 2
C .a >b ⇒a 3>b 3
D .a 2>b 2⇒a >b 3.若,,a b c R ∈,且a b >,则下列不等式中一定成立的是( )
A.a b b c +≥-
B.ac bc ≥
C.2
0c a b
>- D 2()0a b c -≥ 4.对于任意实数,,,a b c d ,命题①若,0,a b c ><则ac bc >;②若a b >,则22ac bc >;
③若22ac bc <,则a b <;④若
a b >,则 11a b <;⑤若0,0a b c d >>>>,则a c b d >。

其中正确的个数是( ) A.1 B.2 C.3 D.4
5.若x +23x -5
<0,化简y =25-30x +9x 2-(x +2)2-3的结果为( ) A .y =-4x B .y =2-x C .y =3x -4 D .y =5-x
6.已知不等式250ax x b -+>的解集为{|32}x x -<<,则不等式250bx x a -+>的解
集为( ) A.11{|}32x x -
<< B.11{|}32
x x x <->或 C.{|32}x x -<< D.{|32}x x x <->或 7.对于任意实数x,不等式04)2(2)2(2<----x a x a 恒成立,则实数a 的取值范围是( )
A.()2,∞-
B.]2,(-∞
C.]2,2(-
D.)2,2(-
8.已知n S 为数列}{n a 的前n 项和,若n n S S a 2,313==+,则=4a ( )
A.6
B.12
C.16
D.24
9.已知}{n a 是递减的等比数列,且5,2312=+=a a a ,则)(13221*+∈+++N n a a a a a a n n 的取值范围是( )
A.[12,16)
B.[8,16)
C.)332,8[
D.)3
32,316[ 10.正项数列}{n a 中,)2(2,2,12121221≥+===-+n a a a a a n n n ,则=6a ( )
A.16
B.8
C.22
D.4
二、填空题(每题5分,共30分)
11.ABC ∆中,o
B b x a 45,2,===,若AB
C ∆有两解,则x 的取值范围是
12.等比数列}{a n 中,===n n n S S S 32,60,48
13.若14,24a b <<-<<,则2a b -的取值范围是
14.若x R ∈,则2x 与1x -的大小关系是
15(A )对于x ∈R ,式子
1kx 2+kx +1
恒有意义,则常数k 的取值范围是_________. (B )不等式log 12(x 2-2x -15)>log 12(x +13)的解集是_________. 16.(A )=-++-+-2
2222212979899100
(B )数列{}n a 的前n 项和为n S ,若)34()1(--=n a n n ,则n S =_________.
三、解答题(13题10分,其余每题12分)
17.在ABC ∆中,已知1)cos(32cos =+-C A A
(1)求角A 的大小
(2)若ABC ∆的面积5,35==b S ,求C B sin sin 的值
18.设数列{}n a 的前n 项和为n S ,已知24,111+==+n n a S a
(1)设n n n a a b 21-=+,证明数列{}n b 是等比数列
(2)求数列{}n a 的通项公式
19.在公差为d 的等差数列{}n a 中,已知101=a ,且3215,22,a a a +成等比数列
(1)求n a (2)若0<d ,求n
a a a a ++++ 321
20.(12分)已知m ∈R ,试解关于x 的不等式:(m +3)x 2-(2m +3)x +m >0.
21.(1)若不等式04
9)1(220822>+++++-m x m mx x x 对任意实数x 都成立,求实数m 的取值范围
(2)已知函数122++=
ax ax y 的定义域为R ,解关于x 的不等式022<+--a a x x
22. 正项数列}{n a 中,)(111,21,12
121*++∈⋅==
=N n a a a a a n n n (1)求n a (2)求数列{}2+⋅n n a a 的前n 项和。

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