2021-2022学年第一学期三明市期末质量检测高一数学参考答案

1 / 9 2021-2022 学年第一学期三明市期末质量检测 高一数学参考答案及评分细则 评分说明： 1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题 的主要考查内容比照评分标准制定相应的评分细则。 2．对计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的 内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数 的一半；如果后继部分的解答有较严重的错误，就不再给分。 3．解答右端所注分数，表示考生正确做到这一步应得的累加分数。 4．只给整数分数。选择题和填空题不给中间分。 一、选择题：本题共8 小题，每小题5 分，共60 分。 1．A 2．D 3．D 4．B 5．A 6．C 7．B 8．D 二、选择题：本题共4 小题，每小题5 分，共20 分。 全部选对的得5 分，有选错的得0 分， 部分选对的得2 分。 9．AD 10．ABD 11．ACD 12．BD 三、填空题：本题共4 小题，每小题5 分，满分20 分． 13． 3 2 14．2 3 m ≤≤ 15．2 16．3 2 四、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. （1）     1 2 3 2 4 0 2 7 8 3 π 2 8                2 1 3 4 3 2 2 3 2 1        2 2 1 2 3 2      4 3 4 1    ·········································································· 4 分 4  ·························································································· 5 分 （2） 1 lg2 lg 3lg5 lg0.1 4    lg2 2lg2 3lg5 1    ····································································7 分   3 lg 2 lg5 1    3lg10 1  ·····················································································9 分 3 1 1  2  ······························································································10 分 18．解法一 2 / 9 （1）        sin cos 2 sin cos 2 f                      sin cos sin cos 2                   sin cos cos cos       ······································································· 4 分 tan  ·······················································································5 分 则 3 f        tan 3         tan 3         ······························································································6 分 tan 3         3  ．·································································································7 分 （2）由（1）知，tan 3  .········································································8 分 即sin 3 cos   ，所以sin 3cos    ． 因为 2 2 sin cos 1    ， 所以  2 2 3cos cos 1     ，即 2 10cos 1 ，·············································· 9 分 解得 10 cos 10  ．···············································································10 分 当 10 cos 10  时， 3 10 sin 10  ； 当 10 cos 10  时， 3 10 sin 10  ．······················································ 11 分 所以 2 9 sin 10  ， 3 sin cos 10   ， 3 / 9 所以 2 9 3 9 2sin 3sin cos 2 3 10 10 10        ．······································· 12 分 解法二 （1）同解法一．·······················································································7 分 （2）由（1）知，tan 3  . ······································································ 8 分 则 2 2sin 3sin cos     2 2 2 2sin 3sin cos sin cos         ········································································· 9 分 2 2 2 2 2 2sin 3sin cos cos sin cos cos           ······································································· 10 分 2 2 2tan 3tan tan 1       ···············································································11 分 2 2 2 3 3 3 3 1     9 10  ．·······························································································12 分 19． （1） 1 2 , ( 2,2) x x   ，且 1 2 x x  ， 则     1 2 f x f x  1 2 1 2 2 2 2 2 x x x x       ······················································································ 1 分         1 2 2 1 1 2 2 2 2 2 2 2 x x x x x x              1 2 1 2 4 2 2 x x x x     ．················································································· 2 分 因为 1 2 0 x x   ， 1 2 0 x   ， 2 2 0 x   ， 所以    1 2 0 f x f x   ，·········································································· 3 分 即     1 2 f x f x  ，所以  f x 在  2,2  上单调递增.······································4 分 （2）由  0 f x  ，得2 2 x   ，即 g x 的定义域为  2,2  ．····················· 5 分 4 / 9 对于任意的 ( 2,2) x ， 2 2 log 2 x g x x    ,       2 2 log 2 x g x x     ··············································································6 分 2 2 log 2 x x    1 2 2 log 2 x x           2 2 log 2 x x     g x  ························································································7 分 所以 g x 是奇函数．················································································ 8 分 （3）由（1）知， 2 2 x y x    在( 2,2)  上单调递增， 又因为 2 log y x  是增函数，所以 g x 是( 2,2)  上的增函数． 由 2 1 2, 2 2 2 x x         得1 3 x   ．································································ 9 分 由  1 0 2 x g x g          ， 得   1 2 x g g x        因为 g x 是奇函数，所以     1 1 g x g x     ． 所以原不等式可化为   1 2 x g g x        ，······················································10 分 则 1 2 x x  ， 解得 2 x  ．··························································································· 11 分 所以原不等式的解集为  1 2 x x   ．······················································ 12 分 20． （1）该地区2000 年底的恩格尔系数为2000 60% r  ， 则2010 年底的恩格尔系数为 10 2010 2000 10 1.04 1.06 r r   ···········································1 分 5 / 9 1.480 0.6 1.791   ，······································· 2 分 因为1.480 0.6 0.8880   ，1.791 0.5 0.8955   ， 所以1.480 0.6 1.791 0.5    ， 则1.480 0.6 0.5 1.791  ， 所以2010 50% r  ．················································································· 4 分 所以该地区在2010 年底已经达到小康水平．···············································5 分 （2）从2000 年底算起，设经过n 年，该地区达到富裕水平 则 2000 1.04 40% 1.06 n n r  ≤ ，·············································································· 6 分 故1.04 2 1.06 3 n       ≤ ，即52 2 53 3 n       ≤ ． 化为 52 2 ln ln 53 3 n ≤ ．·················································································7 分 因为 52 0 1 53  ，则 52 ln 0 53  ，所以 2 ln 3 52 ln 53 n≥ ．············································8 分 因为 2 ln ln 2 ln3 3 52 ln52 ln53 ln 53    ···········································································9 分 ln3 ln 2 ln53 ln52    1.099 0.693 3.970 3.951    21.37  ．···············································································11 分 所以 22 n≥ ． 所以，最快到2022 年底，该地区达到富裕水平．·········································· 12 分 21． （1）  2 3sin 2 2cos f x x x m    3sin 2 cos2 1 x x m    ······················································· 1 分 3 1 2 sin 2 cos2 1 2 2 x x m            2sin 2 1 6 x m           ···························································· 2 分 因为sin 2 6 x         的最大值为1，所以  f x 的最大值为3 m  ，····················· 3 分 6 / 9 依题意，3 2 m   ， 解得 1 m ．······················································································· 4 分 （2）由（1）知，  2sin 2 6 f x x          ··················································· 5 分 由  1 f x ≥， 得 1 sin 2 6 2 x         ≥ ．·············································································6 分 所以 5 2 2 2 6 6 6 k x k         ≤ ≤ ，k Z ．·············································7 分 解得 3 k x k    ≤≤ ，k Z ． 所以，使  1 f x ≥成立的x 取值集合为 3 x k x k k          Z ≤≤ ．·············· 8 分 （3）依题意， 2 sin 6 g x x t          ，······················································ 9 分 因为4 是 g x 的一个零点，所以sin 0 2 6 t           ， 所以2 6 k t    ，k Z ．··································································· 10 分 所以 3 6 1 t k   ，·················································································· 11 分 因为 0 t  ，所以 1 k≥， 所以t 的最大值为3 5 ．············································································12 分 22．解法一 （1）由 2 1 0, 3 0, 3 2 0, x x x x       ≥ ≥ ≥ ········································································· 2 分 解得3 1 x ≤≤． 所以  f x 的定义域为