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宁夏高二期中文科数学试卷
参考答案及评分标准
一、 选择题(本大题共 12 小题,每小题 5 分,共 60 分)
1.C 2.B 3.B 4.B 5.B 6.A 7.C 8.D 9.A 10.A 11.C 12.D
二、填空题(本大题共 4 小题,每小题 5 分,共 20 分)
13. 14. 64 15. 10 16. ②③④⑤
三、解答题(共 70 分)
17.解(1)在△ABC 中,角 A,E,C 的对边分别为 a,b,c,由于 tan A=3,
∴ ,························(1 分)
又∵cosC= ,
∴ ,··············································(4 分)
∴cos B=﹣cos(A+C)=﹣cos Acos C+sin Asin C=﹣ ,
由于 0<B<π,
∴B= .·····························································(5 分)
(2)在△ABC 中,c=4,sinA= , ,
由正弦定理,得 = ,解得 ,···············(8 分)
∴ =6.·······························(10 分)
18.解:(1)证明:由题意,得数列{an}满足 an+1=2an+3×2
n,
等式两边除以 2n+1,得 ;····································(3 分)
故数列 是以 为首项, 为公差的等差数列.·····················(4 分)
(2)由 bn+1=bn+2n+1,得 bn+1﹣bn=2n+1,
则 bn﹣bn﹣1=2(n﹣1)+1=2n﹣1,·········································(6 分)
则 bn=(bn﹣bn﹣1)+(bn﹣1﹣bn﹣2)+……+(b2﹣b1)+b1
=(2n﹣1)+(2n﹣3)+……+3+1·········································(10 分)
= =n2···············································(12 分)
19.解:(1)∵不等式 x2+bx+c>0 的解集为{x|x>2 或 x<1},
∴1,2 是方程不等式 x2+bx+c=0 的两个根
由根与系数的关系得
b=﹣(1+2)=﹣3; c=1×2=2·······································(6 分)
(2)cx2+bx+1≤0
将 b= -3,c=2 代入
∴2x2﹣3x+1≤0
∴(2x﹣1)(x﹣1)≤0
∴ .·························································(10分)
∴cx2+bx+1≤0 的解集为 ···································(12 分)
20. 解:(1)解:当 n=1 时,有 ,
解得 a1=4.···························································(2 分)
当 n≥2 时,有 Sn﹣1= (an﹣1﹣1),
则 ,
整理,得 an=4an﹣1,
∴数列{an}是以 q=4 为公比,以 4 为首项的等比数列.·······················(4 分)
∴ ,
即数列{an}的通项公式为: . ························(6 分)
(2)证明:由(1)有 ,则
,·················(8 分)
∴Tn= +……+
= ,得证.············································(12 分)
21.解:(1)函数 ,
当 时,有不等式化为 ,·····························(2 分)
即 ,
∴不等式的解集为 .·······································(6 分)
(2)∵不等式 ,
当 时,有 0<a<1,∴不等式的解集为 ;················(8 分)
当 时,有 a>1,∴不等式的解集为 ;··················(10 分)
当 时,有 a=1,∴不等式的解集为{1}.········