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江西省重点中学协作体2017届高三下学期第一次联考数学(理)试题及答案
江西省重点中学协作体2017届高三第一次联考
数学(理科)试卷参考答案
一、选择题
1-5: DBCCB 6-10: BACCB 11、12:AD
12.详解:解析:设点则,所以,即,又,即,所以,则,令则,考查函数,由,知时单调递减,时单调递减,所以当时,取得唯一极小值即为最小值,此时,所以
二、填空题
13. 14. 15. 16.
16.详解:由得,则,所以[来源:学科网]
,可化为,
则,又为锐角三角形,所以,又,所以,则,所以,解得
三、解答题
17.解:(1)由,得,即,所以为等差数列,且···································5(分)
(2)因为,·······························8(分)
所以,
则·······12(分)
18.解:(1)众数:8.6;中位数:8.75 ·······································2(分)
(2)由茎叶图可知,满意度为“极满意”的人有4人。
设表示所取3人中有个人是“极满意”,至多有1人是“极满意”记为事件,
································6(分)
(3)从16人的样本数据中任意选取1人,抽到“极满意”的人的概率为,故依题意可知,从该顾客群体中任选1人,抽到“极满意”的人的概率.的可能取值为0,1,2,3高.考.;;
; ·······························9(分)
所以的分布列为
.
另解:由题可知, 所以=.·····················12(分)
19.解:(Ⅰ)连延长交于,
因为点为的重心,所以
又,所以,所以//;···················3(分)
为中点,为中点, //,又//,
所以//,得四点共面
//平面··································6(分)
(Ⅱ)平面平面,平面,连接易得,[来源:学科网ZXXK]
以为原点,为x轴,为y轴,为z轴建立空间直角坐标系,
则,设,
, ,[来源:学+科+网]
因为与所成角为,所以,[来源:Z#xx#k.Com]
得,,,··············8(分)
设平面的法向量,则,取,
平面的法向量,所以二面角的余弦值····················12(分)
20.解:(1)设,则,,,.
则有,解得.·······················3(分)
,,,,
,.
于是,在△中,,[来源:学,科,网]
所以,所以,椭圆的方程为.········6(分)
(2)由条件可知、两点关于轴对称,设,,则,
,,所以,.
直线的方程为,······················9(分)
令得点的横坐标,同理可得点的横坐标.于是
,
所以,为常数.····················12(分)
21.解:(1)证明:令,
.
当时,,故在区间上为减函数,
因此,故.···················2(分)
再令,当时,,
故在区间上为增函数.,所以,故是和在上的一个“严格分界函数”···················5(分)
(2)由(1)知.
又,···················7分)
令
解得,易得在单调递减,在单调递增,则
···················9(分)
又在存在使得,故在上先减后增,则有,则,所以,则····················12(分)
22.解析:(1)由(为参数),得,即,所以···················5(分)
(2)设直线的参数方程是(为参数)(1)
曲线的直角坐标方程是,(2)联立方程可得,所以,且,所以,则或,所以···················10(分)
23.解析:(1) ····················4(分)
(2)
即,化简或或
解得或,即为所求····················10(分)
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