扫描全能王 创建
{#{QQABKY6QogioAIIAAQgCAQVaCkIQkBEAAQgOxEAAIAAAwQFABAA=}#}
扫描全能王 创建
{#{QQABKY6QogioAIIAAQgCAQVaCkIQkBEAAQgOxEAAIAAAwQFABAA=}#}
扫描全能王 创建
{#{QQABKY6QogioAIIAAQgCAQVaCkIQkBEAAQgOxEAAIAAAwQFABAA=}#}
扫描全能王 创建
{#{QQABKY6QogioAIIAAQgCAQVaCkIQkBEAAQgOxEAAIAAAwQFABAA=}#}
扫描全能王 创建
{#{QQABKY6QogioAIIAAQgCAQVaCkIQkBEAAQgOxEAAIAAAwQFABAA=}#}
扫描全能王 创建
{#{QQABKY6QogioAIIAAQgCAQVaCkIQkBEAAQgOxEAAIAAAwQFABAA=}#}
扫描全能王 创建
{#{QQABKY6QogioAIIAAQgCAQVaCkIQkBEAAQgOxEAAIAAAwQFABAA=}#}
扫描全能王 创建
{#{QQABKY6QogioAIIAAQgCAQVaCkIQkBEAAQgOxEAAIAAAwQFABAA=}#}高一数学试卷参考答案及评分标准
一 单选题：本大题共 8小题，每小题 5分，共 40 分.在每小题给出的四个选项中，只有一项是符合题目要求的.
题目 1 2 3 4 5 6 7 8
答案 D C A C A B C B
二、多项选择题:本大题共 3 小题,每小题 6 分,共 18 分.在每小题给出的四个选项中,有多项符合题目要求.全部
选对的得 6 分,部分选对的得部分分（9 题每个选项 3 分，10 题，11 题每个选项 2 分）,有选错的得 0分.
题号 9 10 11
答案 B C AB D AC D
三 填空题：本大题共 3 小题，每小题 5 分，共 15 分. 把正确答案写在答题卡相应题的横线上.
3
12、 1 13、-2（填 a 范围内任意一个实数即可） 14 33、
2 6
四. 解答题：本大题共 5 小题，共 77 分，解答应写出文字说明、解答过程或演算步骤.
15.（13 分）
解：（1）取 AB的中点H，连接EH ,HG, 1∵G为 BC的中点，∴HG // AC，············································································2 分
2
E AC EC // 1∵ 为 1 1的中点，∴ 1 AC，·····································································································································3 分2
∴HG //EC1，∴四边形EHGC1为平行四边形，·······································································································5 分
∴C1G∥ EH . ···················································································································································································································································7 分
又 C1G 平面 ABE , EH 平面 ABE ,∴C1G∥平面 ABE .···············································································································································8 分
（2） AB BC ,AC 2,BC 1, AB 3, ··················································································································································································································9 分
1 3
∴ SΔABC 1 3 .····································································································································································································································································································11 分2 2
∵ A1C1∥平面 ABC ,∴点E到面 ABC的距离等于点 A1到面 ABC的距离，
∴VE ABC VA1 ABC ,·····································································································································································································································································································································12分
又∵ AA 1 31 平面 ABC，∴VC ABE VE ABC VA ABC S ABC AA1 ,. ··························································································13 分1 3 3
备注：本题第一问若建系解决，第一问 8分分值不变，建系读点 2分，线向量 1分与法向量正交列式 1分，
结果 1分，线与向量内积 2分，结论 1分，对于线在面内和不在面内的语言不占采分点，没写不扣分.
高一数学试卷参考答案及评分标准 第 1 页（共 5页）
{#{QQABKY6QogioAIIAAQgCAQVaCkIQkBEAAQgOxEAAIAAAwQFABAA=}#}
16.（15 分）
b
解：（1）∵ c a cosB sin B，由正弦定理可知， sinC sin Acos B .············································································2 分
2 2
又 A+B+C= π， sinC sin(A B) sin Acos B cos Asin B, ···················································································································3 分
sin B
∴ cos Asin B, ·················································································································································································································································································································4 分
2
1
又 sin B 0, cos A ,又 A (0, ) A ,∴ .·················································································································································································6 分
2 3
（2）∵ AD为 BC边上的中线，∴ 2AD AB AC , ······································································································································································8 分
2 2 2
∴ 4AD AB AC 2AB AC ,∴36 b2 c2 bc ,·································································································································································10 分
由余弦定理可得 a2 b2 c2 2bc cosA，即 24 b2 c2 bc , ·························································································································12 分
∴bc 6, ·······························································································································································································································································································································································14 分
∴ S 1 ABC bc sin A
1 3 3 3
6 .·······················································································································································································································15 分
2 2 2 2
备注：本题第一问可以采用余弦定理解决，得出b2 c2 a2 bc给 4分，角 A值给 2分.
17.（15 分）
解：（1） AE
1
AB BE a b ，······················································································································································································································································2 分
2
BF BC CF b 2 a ，························································································································································································································································································4 分
3
过 E作 EH //CF交 BF于H 1 1 1， EH FC CD AB，
H 2 3 3
AG 3 AE，
4
3 3
∴ AG a b . ·············································································································································································································································································································6 分
4 8
（2）解法 1：
1 2 2 1 AE BF (a b ) ( a b ) a 2 2 b 2 a b 2，·································································································································10 分
2 3 3 2 3
AE (a 1b )2 a 2 1 b 2 a b 13, ·······················································································································································································12 分
2 4
且在 BCF 中， BC CF 2, BCF 60
高一数学试卷参考答案及评分标准 第 2 页（共 5页）
{#{QQABKY6QogioAIIAAQgCAQVaCkIQkBEAAQgOxEAAIAAAwQFABAA=}#}
∴ΔBCF为等边三角形，∴ BF 2, ·································································································································································································································13 分
cos EGF A E B F 2 13∴ . ···············································································································································································································15 分AE BF 2 13 13
解法 2： 如图所示建系
y 7 3
D F C 则 A(0,0) ,B(3,0),E( , ),F (2, 3), ································7 分2 2
∴ AE (7 , 3 ),BF ( 1, 3), ···························································9 分
G E 2 2
∴ AE BF
7 3
2, ················································································10 分
A B 2 2x
7 AE ( )2 ( 3 )2 13, BF ( 1)2 ( 3)2 2, ·······························································································································································13分
2 2
∴cos EGF
A E B F 2 13 ·······················································································································································································································15 分
AE BF 2 13 13
备注：计算题主要以计算结果为采分点，计算正确就可以得分，关于解法 2的坐标点，写对一个就给 1分.
第 2 问的两个解法的关键步骤得分点一致，内积 4分，AE 模 2分，BF 模 1 分，结论 2分.
18.（17 分）
2 1
解： f x 2cos x 2 3 sin x cos x cos 2x 1 3 sin 2x 2 cos 2x
3
sin 2x 1
2 2
2sin 2x
π
1. ··························································································································································································································································································································4 分
6
（1）∴T ， ··························································································································································································································································································································5 分
2x k x k 令 （ k Z），∴ （ k Z）.
6 2 12
(kπ π∴对称中心为 ,1)（ k Z）； ·····················································································································································································································································7 分
2 12
37
（2）∵ f (x0 ) ，∴ sin(2x
) 12 x π , π0 ，又∵ 0
，∴ 2x
π π , 5π
13 6 13 6 3 0 6
，
2 6
∴ cos(2x
π
0 )
5
， ·········································································································································································································································································································9 分
6 13
∴ sin 2x0 sin[(2x
π π 12 3 1 5 12 3 5
0 ) ] ( ) . ···························································································································11 分6 6 13 2 2 13 26
高一数学试卷参考答案及评分标准 第 3 页（共 5页）
{#{QQABKY6QogioAIIAAQgCAQVaCkIQkBEAAQgOxEAAIAAAwQFABAA=}#}
π
（3）由题意， g x sin 2x
， ···················································································································································································································································13 分
6
x [ π π] 2 当 ， 时， 2x [ ,
]，∴ g(x) [ 1,
1]， ··········································································································································15 分
4 6 6 3 6 2
g(x) m 1 0 m g(x) 1 m g(x) 1 1 1∵ 恒成立 恒成立 max 1. ··········································16 分2 2 2 2 2
∴m 1. ·····················································································································································································································································································································································17 分
备注：化简部分见到 2sin 2x
π
1后才给 4分，如果只写周期，没有化简，给 1分.
6
19.（17 分）
（1）证明：∵ PE DE ,BE DE，∴DE 平面PBE , ·····················································································································································1 分
∵DE / /BC，∴BC 平面 PBE， ··························································································································································································································································2 分
∴ BC ME，·······················································································································································································································································································································································3 分
∵ BE PE ,M 是 PB中点，∴ME PB， ··············································································································································································································4 分
∵ BC PB B,∴BCM,PEB 面BMN . ···························································································································································································································································5分
（2）延长CB至点Q，使得 BQ= BC，由（1）可知，DE⊥平面PBE，又∵DE 平面BCDE，
∴平面 PBE 平面BCDE，∴CH =QH，··························································································································································································································6 分
∴CH FH QH FH QF ，···································································································································································7 分
∴当QF CP,且QF PB H 时，CH +FH 最小， ······························································8 分
∴CF 2BC BC 4 3 cos FCB 2BC . ·································································································10 分
PC 3
（3）假设存在点N 满足题意，∵平面PDE 平面BCDE，PE DE ,平面 PDE 平面
BCDE DE ∴ PE 平面 BCDE， ·········································································································································································································································11 分，
∴如图所示建系，
∴ E(0,0,0), P(0,0, 2),M (1,0,1) ，设 N (2,m,0)(0 m 2) , EM (1,0,1),EN (2,m,0)，· ····························12 分
设平面 EMN 法向量为 n (x, y, z)，
n EM 0
∴ ， ··················································································································································································································13 分
n EN 0
∴ n (m, 2, m)， ··················································································································································································································14 分
又平面 BCDE的法向量为 EP (0,0, 2)， ··················································································································································································································15 分
高一数学试卷参考答案及评分标准 第 4 页（共 5页）
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EP n
∴ cos EP,n m 2 m 1 ， ······························································································································································16 分
EP n 2m2 4 6 2 .
即此时N 为线段 BC上靠近点B的四等分点. ·································································································································································································17 分
备注：第 1问中立体几何证明垂直时候，两条线交于一个点，必须要写，但不占采分点，不写也不扣分；
第 2问 CF 道理说明白，值求对就给 5分，没有道理，只有答案正确，给 2分；
第 3问建立直角坐标系的时候，需要声明 x轴⊥y轴，z 轴⊥面Oxy，或者证明出 x，y，z 轴两两垂
直（教学不建议这样处理），如果没证明 z轴⊥面 Oxy，扣 1 分，但后面的读点和求解角度的步骤（6分）
按正常采分点给分（证明建系和建系解决是两个事件，不合为一谈），坐标写对一个就给 1分，法向量写出
内积为 0，就给 1分，法向量对一个给 1分，m值求对 1分，结论 1分。
（特此说明：请在阅卷前详细阅读各题的评分细则和备注说明。本套试题中解题步骤对应采分
标准参考国标的评分标准，只要框内的答案正确，就给相应的采分点，希望教师在阅卷中，能
够落实高考评价体系，如果遇到其他解法，每问分值不变，其他步骤酌情处理）
高一数学试卷参考答案及评分标准 第 5 页（共 5页）
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