二、填空题:
13. 14. 15. 16.
三、解答题:
17. 解:
(1)
...............5分
(2)∵为第三象限角,且 ....................................2分
. ...........................................................2分
则 ............................................................1分
18. 解(Ⅰ)甲班学生成绩的中位数为 (154+160)=157.....................................2分
乙班学生成绩的中位数正好是150+x=157,故x=7;........................................2分
(Ⅱ)用A表示事件甲班至多有1人入选.
设甲班两位优生为A,B,乙班三位优生为1,2,3.
则从5人中选出3人的所有方法种数为:(A,B,1),(A,B,2),
(A,B,3),(A,1,2),(A,1,3),(A,2,3),(B,1,2),
(B,1,3),(B,2,3),(1,2,3)共10种情况,..........................3分
其中至多1名甲班同学的情况共(A,1,2),(A,1,3),(A,2,3),
(B,1,2),(B,1,3),(B,2,3),(1,2,3)7种......................3分
由古典概型概率计算公式可得P(A)= .............................................................2分
19. .......................................................2分
f(x)的最小正周期T= =4 ......................................................................1分
当 时,f(x)取得最小值-2;..............................................................1分
当 时,f(x)取得最大值2...................................................................1分高一数学下册期末考试卷答案答案-2
(2)g(x)是偶函数.理由如下:.................................................................................1分
由(1)知
又g(x)
g(x)= ...........................................3..分
∵g(-x)= =g(x),....................................................................2分
函数g(x)是偶函数. ......................................................................................... ...1分
20. 解:(1)证明:∵M是BC的中点,
OM=12(OB+OC).....................................................................................................3分
代入OA=-2OM,得OA=-OB-OC,.................................................................2分
即OA+OB+OC=0........................................................................................................1分
(2)设|AP|=x,则|PM|=2-x(02).....................................................................1分
∵M是BC的.中点,
PB+PC=2PM................................................................................................................2分
PA(PB+PC)=2PAAM=-2|PA||PM|
=-2x(2-x)=2(x2-2x)=2(x-1)2-2,...................................................................2分
当x=1时,取最小值-2.................................................................................................1分高一数学下册期末考试卷答案答案-3
21. (Ⅰ)设 =2R
则a=2RsinA,b=2RsinB,c=2RsinC................................................................................2分
∵2asinA=(2b+c)sinB+(2c+b)sinC
方程两边同乘以2R
2a2=(2b+c)b+(2c+b)c...........................................................................................2分
整理得a2=b2+c2+bc............................................................................................................1分
∵由余弦定理得a2=b2+c2-2bccosA..................................................................................1分
故cosA=- ,A=120........................................................................................................2分
(Ⅱ)由(Ⅰ)得:sinB+sinC=sinB+sin(60-B)....................................................1分
= ...................................................................................2分
故当B=30时,sinB+sinC取得最大值1......................................................................1分
22. 解:(1)因为x= 是函数y=f(x)的图象的对称轴,
所以sin(2 +)=1,即 ++ ,kZ...............................2分
因为-0,所以= .............................................................2分
(2)由(1)知= ,因此y=sin(2x ).
由题意得2k 2x2k+ ,kZ,.......................................2分
所以函数y=sin(2x )的单调区间为[k+ ,k+ ],kZ........2分
(3)由y=sin(2x )知: .................................................................2分高一数学下册期末考试卷答案答案-4
x 0
8
3
8
5
8
7
8
.y
-1 0 1 0
故函数y=f(x)在区间[0,]上的图象是.................................................2分