【文档说明】山东省德州市第一中学2022-2023学年高一下学期3月月考数学试题答案.docx,共(5)页,264.867 KB,由小赞的店铺上传
转载请保留链接:https://www.doc5u.com/view-875cd7c64845bef461a6e38aaff8d858.html
以下为本文档部分文字说明:
高一年级3月份阶段性测试数学试题答案一.单选题:1—5BAAAB6—8CBD二.多选题:9.ABD10.ABD11.AC12.ABD三.填空题:13.()0,2014.315.π3cos4x(答案不唯一)16.
1140,,939四、解答题(共70分)17.解:(1)因为ABab=+uuurrr,28BCab=+uuurrr,()3CDab=−,,所以()283BDBCCDabab=+=++−()283355abababAB=++−=
+=..............................................................................................3所以AB,BD共线,又因为它们有公共点B,所以,,ABD三点共线....
..........................................................................................................5(2)因为kab+和akb+rr共线,所
以存在实数,使()kabakb+=+,所以kabakb+=+,即()()1kakb−=−.又a,b是两个不共线的非零向量,所以10kk−=−=..............................................................
.....................................................................8所以210k−=,所以1k=或1k=−.........................
............................................................................1018.解:(1)3π3πππ33sin3cos42332f=−=−=−........................
.....................................................2(2)由πππ2π22π,Z232kxkk−+−+得,π5πππ,Z1212kxkk−++,()fx\的单调增区间为π5ππ,π,Z1212kkk−
++...................................................................................6(3)当ππ,44
x−时,π5ππ2,366x−−,...................................................................................
......8π1sin21,32x−−,.................................................................................
.........................................10π3sin223,33x−−,故()fx在区间ππ,44−上的值域为33,
2−........................................1219.(1)解:()()()()sin()costanπ==tanπsin2πsin2f−−+++...............
......................................................6(2)解法1:()tan2f==,则22222222sinsincoscostantan1sinsincoscos=sincostan1
+−+−+−=++.........................................104+2-1=14+1=..........................................................
.......................................................................................12解法2:sintan2cos==22sin2cossincos1=+=.
........................................................................8由于是锐角,解得255sin,cos55==......
...........................................................................10所以2222255255sinsincoscos=()-()15555+−+=..............
..............................................1220.【详解】(1)解:()3,2a=,()1,2b=−r-3=(3,2)-(-3,6)=(6,-4)ab..........................
................23=36+16=52=213ab−.........................................................................
........................................4(2)解法1:设(),dxy=,则()4,1dcxy−=−−,()2,4ab+=,................................................
......6因为()()//dcab−+,5dc−=,所以()()()()2244210415xyxy−−−=−+−=,...................................
................................................................................8解得31xy==−或53xy==,所以()3,1
d=−或()5,3d=.............................................................................................................
..........12解法2:设()dcab−=+,()2,4ab+=.................................................................................................65(
)255dcabab−=+=+==11=22=,即...................................................................
...............................................................1011=()(4,1)(1,2)(5,3)22dcab=++=+=时,11=--()(4,1)-(1,2)(3
,-1)22dcab=+==时,所以()3,1d=−或()5,3d=................................................................................................................
......1221.解:(1)以圆环的圆心为坐标原点,过圆心且平行于地面的直线为x轴,过圆心且垂直于地面的直线为y轴建立平面直角坐标系xOy.以x轴非负半轴为始边,1OP为终边的角为π6−;........
................................................................................2点P时刻t所转过的圆心角为:2ππ84tt=.若t时刻时蚂
蚁爬到圆环P点处,那么以x轴非负半轴为始边,OP为终边的角为ππ46t−,则P点纵坐标为ππ4sin46t−,所以()ππ4sin646htt=−+........................6(2)解法1:令()ππ4sin6846htt=−+
,即ππ1sin,462t−.............................................................8所以πππ5π2π<2π,Z6466ktkk+−+,解得4848,Z3ktkk
++,..........................................................10所以在一周范围内,P距离地面超过8m持续时间为:48(48)(8)33kk+−+=分钟.不超过8m的时间是816
8-=33分钟。...........................................................................................................................
......................12(3)解法二:()ππ4sin6846htt=−+即ππ1sin,462t−.....................................................
...............8所以5πππ13π2π<2π,Z6466ktkk+−+,解得28488,Z3ktkk++,....................................................
10在一周内,P距离地面超过8m持续时间为:2816(48)(8)33kk+−+=分钟.不超过8m时间是163分钟......1222.解:()1由三角函数的周期公式可得22==,()()sin2fxx=+,令()22xkkZ+=+,得()422kxkZ=−+,由于直线2
x=−为函数()yfx=的一条对称轴,所以()2422kkZ−=−+,得()32kkZ=+,由于0,1k=−,则2=,因此,()sin2cos22fxxx=+=......................
...............................................................................................3()2将函数()yfx=的图象向
右平移4个单位,得到函数cos2cos2sin242yxxx=−=−=,再将所得的图象上每一点的纵坐标不变,横坐标伸长为原来的2倍后所得到的图象对应的函数为()singxx=...........
.............................................................................................................................
...................4()22sinsin1Fxxx=−++即.令()0Fx=,可得22sinsin10xx−−=,令sin1,1tx=−,得2210tt−−=,280=+,则关于t的二次方
程2210tt−−=必有两不等实根1t、2t,则1212tt=−,1t,2t异号.①当101t且201t时,则方程1sinxt=和2sinxt=在区间()()*0,nnN均有偶数个根,从而方程22sinsin10xx−−=在()()*0,nnN也有偶
数个根,不合题意;.............................................6②当11t=,则212t=−,此时1=,当()0,2x时,1sinxt=只有一根,2sinxt=有两根,所以,关于的方程22sinsin10xx−−=在(
)0,2p上有三个根,由于202136732=+,则方程22sinsin10xx−−=在()0,1346上有36732019=个根,由于方程1sinxt=在区间()1346,1347上只有一个根,在区间(
)1347,1348上无实解,方程2sinxt=在区间()1346,1347上无实数解,在区间()1347,1348上有两个根,因此,关于x的方程22sinsin10xx−−=在区间()0,1347上有2020个根,在区间()0,1348上有2022个根,不合题意;.
........................................................................................8③当11t=−时,则212t=,此时1=−,当()0,2x时,1sinxt
=只有一根,2sinxt=有两根,所以,关于x的方程22sinsin10xx−−=在()0,2p上有三个根,由于202136732=+,则方程22sinsin10xx−−=在()0,1346上有36732019=个根,由于方程1sinxt=在区间
()1346,1347上无实数根,在区间()1347,1348上只有一个实数根,方程2sinxt=在区间()1346,1347上有两个实数解,在区间()1347,1348上无实数解,因此,关于x的方程22sinsin1
0xx−−=在区间()0,1347上有2021个根,满足题意................11④若有一根绝对值大于1,则另一根绝对值大于0且小于1,有偶数个根,不合题意;综上所述:1=−,1347n=..............................
...................................................................................12