# 2014年大连市高三双基测试 数学理试题_图文

2014 年大连市高三双基考试 数学（理科）参考答案及评分标准

3．C

4．D

5．A

6．B

7．D

8．B

9．A

10．A

11．C

12．D

1 2

14． 80

15． 13?

16． ( ?8, ?7)

2

?

sin 2 x 3 cos 2 x 3 ? ? 2 2 2

? sin( 2 x ?

?
3

)?

3 .···················································································································· 4 分 2

?
3

? 2k? ?

?
2

（ k ? Z) ，

5? 3 .················································ 6 分 , k ? Z } 时， f ( x) 取最大值 1 ? 12 2

(Ⅱ) f ( ) ? ?
2

A 2

3 ? ? ,可得 sin( A ? ) ? 0 ，因为 A 为△ ABC 内角，所以 A ? .········· 8 分 2 3 3
2 2 2 2

1 3 .································································································ 12 分 bc sin A ? 2 4

18. （Ⅰ） 2 ? 2 列联表如下 甲 厂 优质品 非优质品 合计 400 100 500 乙 厂 300 200 500 合计 700 300 1000 ········································· 4 分

?2 ?

1000(400 ? 200 ? 100 ? 300) 2 ? 47.619 ? 10.828 500 ? 500 ? 700 ? 300

X 取值为 0,1,2 . P ( x ? 0) ?
1 1 2 C2 C3 3 C 32 C2 1 3 ? , P ( x ? 1 ) ? ? , P ( x ? 2 ) ? ? , 2 2 2 5 C 5 10 C5 C 5 10

X

0

1

2

P

1 10

3 5

3 10

························································································································································ 10 分

3 3 6 所以 EX ? 1 ? ? 2 ? ? . 5 10 5 (或者直接利用超几何分布的期望公式求得) ································································ 12 分

19. 解：(Ⅰ)取 BC 中点 O ，因为三角形 ABC 是等边三角形，所以 AO ? BC ， 又因为面 BCC ' B ' ? 底面 ABC ， AO ? 面 ABC ，面 BCC ' B '? 面 ABC = BC ，

(Ⅱ) 取 B ' C ' 中 点 O ' ， 所 以 OO' ? 底 面 ABC . 分 别 以

OC , OA, OO' 为 x, y, z 轴建立空间直角坐标系如图所示.

M (a, 3 (1 ? a ),3)

?

? ? ? ? BE ? n ? 0 ? x ? 3 y ? 2 z ? 0 ?? 则? ，不妨令 x ? 1 ，则 n ? (1, 3 ,?2) . ? ? ? BF ? n ? 0 ? 2 x ? z ? 0

········· 8 分

BM 和 面 BEF 所 成 角 的 余 弦 值 为 ? | cos ? n , BM ?|? | 2a ? 2 | 8 4a ? 4a ? 13
2

58 ， 则 8

6 . 8 ? 6 1 23 ，解得 a ? 或 a ? ? （舍）. 8 2 2

（I）设点 20.解： 得 即 因为点 M,N 在椭圆 所以 故

，则由 , ·····································································································2 分 上， ······················································································· 4 分

， ，···················································································································· 6 分

(II)将曲线 C 与直线 l 联立: 消 y 得： 直线 l 与曲线 C 交于 A、B 两点,设

?

····························································································································· 7 分 ································································································· 8 分

,

AB ? (1 ? k 2 ) x3 ? x4 ? (1 ? k 2 )[( x3 ? x4 ) 2 ? 4 x3 x4 ]

··························································································· 10 分 . 当且仅当 m ? 30 ? m ，即 m ? 15 时取等号.
2 2 2

. ················································································ 12 分

x ? ln( x ? 1) 21．解：(Ⅰ) f ' ( x) ? x ? 1 2 ，·············································································· 1 分 x x 设 g ( x) ? ? ln( x ? 1) ，不妨令 x ? ?1 ， x ?1

1 1 ?x ，当 x ? ( ?1,0) 时， g ' ( x) ? 0 ， g ( x) 为增函数；当 x ? (0,??) 时， ? ? 2 x ? 1 ( x ? 1) 2 ( x ? 1)

g ' ( x) ? 0 ， g ( x) 为减函数.

1 1 x ? 1 等价于 ln( x ? 1) ? kx 3 ? x 2 ? x ? 0 ， 2 2 1 3 2 设函数 h( x) ? ln( x ? 1) ? kx ? x ? x ，对于函数 h( x) ，不妨令 x ? 0 . 2
（Ⅱ） f ( x) ? kx ?
2

h' ( x ) ?

1 ? 3kx 3 ? x 2 ? 3kx 2 x 2 (?3kx ? 1 ? 3k ) ··························· 8 分 ? 3kx 2 ? x ? 1 ? ? x ?1 x ?1 x ?1

1 1 ? 3k 1 ? 3k ， 在 x ? [0, ] 时 ， h' ( x) ? 0 ， 所 以 h( x) 在 x ? [0, ] 为增函数，所以 3 3k 3k

h( x) ? h(0) ? 0 ，不符合题意； 1 时，在 x ? [0,??) 时， h' ( x) ? 0 ，所以 h( x) 在 x ? [0,??) 为减函数，所以 h( x) ? h(0) ? 0 ， 即 3 1 ln( x ? 1) ? kx 3 ? x 2 ? x ? 0 在 x ? 0 上成立，符合题意； 2 1 综上，实数 k 的最小值为 .····································································································· 12 分 3

0

A H O K D

∵ CD ? AB ，∴ ?OAG ? ?HKA ? 90 ，
0

B

F G E ∵ OA ? OG ，∴ ?OGA ? ?OAG , ∴ ?KGE ? ?HKA ? ?GKE ，∴ KE ? GE . ······································································· 5 分

KG GE KE , ? ? KD KG KG ∵ ?DKG ? ?GKE ，∴△ KDG ∽△ KGE ∴ ?AGD ? ?E ，又∵ ?AGD ? ?ACD ，∴ ?ACD ? ?E .
（Ⅱ）连接 DG , ,∵ KG ? KD gGE ，∴
2

∴ AC P EF .

····························································································································· 10 分

23.解： （I）圆 C1 的普通方程为： ( x ? 4) ? y ? 16 ，则 C1 的极坐标方程为： ? ? 8cos ?
2 2

2 2

··········· 5 分

(II) 设 P (?, ?) ， 则 有 8cos ? ? 4sin ? ， 解 得 tan ? ? 2 ， sin ? ?

2 5 ，所以 P 点的极坐标为 5

(

8 5 2 5 , arcsin ) 5 5

························································································································································ 10 分 24.解： （ I）

5 1 ? 3 ?1 ?? x ? ? 2 ? x ? ? 2 原不等式等价于 ? 2 或 ?2 或 2 2 ? ? x ?1 ? ? 1? x ? 3

5 ?3 ? x? ?2 2 ?2 ? ? x?3

1 3

5 ? 3 ? x ? , x ?1 ? 2 2 ? 1 1 ?1 (II) f ( x) ? x ? 1 ? | x ? 3 |? ? x ? ,1 ? x ? 3 2 2 ?2 5 ? 3 ? 2 x? 2,x ?3 ? f ( x) 图象如图所示，其中 A(1,1)， B (3, 2) ，

············································· 7 分

y

1 , 0) 旋转， 2 x O 1 3 1 由图可得不等式 f ( x) ? a ( x ? ) 的解集非空时， a 的范围为 2 3 4 ············································································································ 10 分 （ -?， - ） U[ ， +?） 2 7

1 2

A

B

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