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闵行区 2010 年高考模拟试卷数学试卷(文理科)答案
一、填空题: (每题 4 分) 1.-1;
2 2 2

2.0.35; 8.3;

3.2; 9.8;

4. 60;

5.17.64;

6. 36π ; 7. 1 + 3 + 5 + L + 2009 ; 11.

5π ; 12. 理 k ? 5 ;文(-2,4) ; 13.理符合 a + b < 0 且 a ≤ ?2 的一个特例均可;文符 6 2 合 a ≥ ?4 的一个特例均可; 14.理 k + k ? 1 ;文 29. 二、选择题: (每题 4 分)15. A ; 16. D ; 17. B; 18. C 4 π 3 三、解答题:19.(本题满分 14 分)理:(1)Q sin x = , x ∈ ? , π ? ,∴ cos x = ? (2 分) ? ?
5 ?2 ? 5

10.理 2 2 ;文(0,2)

? 3 ? 1 f ( x) = 2? sin x + cos x ? ? 2 cos x (4 分) = 3 sin x ? cos x = 4 3 + 3 (8 分) ? 2 ? 2 5 5 ? ? π? π π 5π π ? ≤ x ≤ π ,∴ ≤x? ≤ (2) f ( x ) = 2 sin ? x ? ? (10 分)Q (12 分) 6? 2 3 6 6 ? 1 π? ? ≤ sin? x ? ? ≤ 1 , ∴ 函数 f (x) 的值域为 [ 1, 2 ] . (14 分) 2 6? ? 文:设 z = a + bi ( a , b ∈ R ) (2 分)因为 (2 + i ) z = (2a ? b) + ( a + 2b)i 为纯虚数 (5 分)
?2 a ? b = 0 4 ? ? ?a = 5 4 8 ? a + 2b ≠ 0 (9 分)解得 ? (12 分) 故复数 z = + i (14 分) 所以 ? ? 2 2 5 5 ?b = 8 ?(a ? 2) + b = 4 ? 5 ? 20.(本题满分 14 分)理:解法一: (1)以 A 点为坐标原点建立空间直角坐标系 A ? xyz (图略), 由 PA = AD = AB = 2 BC = 2 得 1 A(0, 0, 0) , P (0, 0, 2) , B (2, 0, 0) , M (1, ,1) D(0, 2, 0) (2 分) 2 uuu uuuu r r 1 因为 PB ? AM = (2, 0, ?2) ? (1, ,1) = 0 (5 分) 所以 PB ⊥ AM . (7 分) 2 uuu uuur r (2)因为 PB ? AD = (2, 0, ?2) ? (0, 2, 0) = 0 ,所以 PB ⊥ AD ,又 PB ⊥ AM , uuu r 故 PB ⊥ 平面 ADMN ,即 PB = (2, 0, ? 2) 是平面 ADMN 的法向量.(9 分) uuu r uuu uuu r r 设 BD 与平面 ADMN 所成的角为 θ ,又 BD = ( ?2, 2, 0) ,设 BD 与 PB 夹角为 α , uuu uuu r r BD ? PB ?4 1 则 sin θ = cos α = uuu uuu = (12 分) = , r r 8? 8 2 BD ? PB

6 6 解法二: (1)证明:因为 N 是 PB 的中点, PA = AB , 所以 AN ⊥ PB (2 分) 由 PA ⊥ 底面 ABCD ,得 PA ⊥ AD ,又 ∠BAD = 90° ,即 BA ⊥ AD ,∴ AD ⊥ 平面 PAB , ∴ AD ⊥ PB (4 分) ∴ PB ⊥ 面 ADMN ,∴ PB ⊥ AM (7 分) (2)联结 DN ,Q BP ⊥ 平面 ADMN ,故 ∠BDN 为 BD 与面 ADMN 所成角(9 分)

又 θ ∈ [0 ,

π
2

] ,故 θ =

π

,故 BD 与平面 ADMN 所成的角是

π

.

(14 分)

第 1 页 共 5 页

在 Rt ?ABD 中, BD = 在 Rt ?PAB 中, PB =

BA2 + AD 2 = 2 2 , PA2 + AB 2 = 2 2 ,故 BN =

1 PB = 2 , 2
(12 分) (14 分)

在 Rt ?BDN 中, sin ∠BDN =

BN 1 = ,又 0 ≤ ∠BDN ≤ π , BD 2

故 BD 与平面 ADMN 所成的角是 文(同理 19 题)

π

6

21.(本题满分 16 分) (1)设第 n 区每平方米的重量为 an 千克,则

an = 1000(1 ? 2%) n ?1 = 1000 × 0.98n ?1 (2 分)

第 1225 米位于第 25 区, (4 分)

∴ a25 = 1000 × 0.9824 = 616 (千克)故第 1225 米处每平方米火山灰约重 616 千克(6 分)
(2)设第 n 区内的面积为 bn 平方米,则 bn = π 50 n ? π 50 ( n ? 1) = 2500π (2n ? 1) 则第 n 区
2 2 2 2

内火山灰的总重量为 Cn = an bn = 25 × 10 π (2n ? 1) × 0.98
5

n ?1

(千克) 分) (9

设第 n 区火山灰总重量最大, 则?

?25 × 105 π (2n ? 1) × 0.98n ?1 ≥ 25 × 105 π (2n ? 3) × 0.98n ? 2 ? , 5 n ?1 5 n ?25 × 10 π (2n ? 1) × 0.98 ≥ 25 × 10 π (2n + 1) × 0.98 ?

(13 分) (16 分)

解得 49.5 ≤ n ≤ 50.5 ,即得第 50 区火山灰的总重量最大. 22.(本题满分 16 分) (理)(1)设 y =

( x ⊕ a) ? ( x ? a) ,则 y 2 = ( x ⊕ a ) ? ( x ? a )
又由 y =

= ( x + a ) 2 ? ( x ? a ) 2 = 4ax

(2 分)

( x ⊕ a) ? ( x ? a) ≥0 可得

P( x , ( x ⊕ a ) ? ( x ? a ) )的轨迹方程为 y 2 = 4ax ( y ≥ 0) , 轨迹 C 为顶点在原点, 焦点为 (a,0) 的 抛物线在 x 轴上及第一象限的内的部分 (4 分)

? y 2 = 4ax (2) 由已知可得 ? ? 1 ? y = x +1 ? 2

, 整理得 x 2 + (4 ? 16a ) x + 4 = 0 ,

由 ? = (4 ? 16a ) 2 ? 16 ≥ 0 ,得 a ≥ ∴ ( x1 ? x2 ) + ( y1 ? y2 ) =

1 1 或a ≤ 0 .∵ a > 0 ,∴ a ≥ 2 2

(6 分)

( x1 ? x2 ) 2 + ( y1 ? y2 ) 2 = ( x1 ? x2 ) 2 + (

x1 ? x2 2 ) 2
(8 分)

=

5 5 ( x1 + x2 ) 2 ? 4 x1 x2 = (4 ? 16a ) 2 ? 16 = 8 15 , 2 2

第 2 页 共 5 页

解得 a = 2 或 a = ? (3)∵ d ( AB ) =

3 (舍) ;∴ a = 2 2

(10 分)

y1 ? y2 =| y1 ? y2 | ∴

| d ( ST ) | | d ( ST ) | | ST | | ST | + = + (12 分) | d ( SP) | | d ( SQ) | | SP | | SQ |

设直线 l 2 : x = my + c ,依题意 m ≠ 0 , c ≠ 0 ,则 T (c, 0) ,分别过 P、Q 作 PP1⊥y 轴,QQ1⊥y 轴, 垂 足 分 别 为

P1



Q1





| ST | | ST | | OT | | OT | | c | | c | + = + = + . | SP | | SQ | | PP | | QQ1 | | xP | | xQ | 1
由?

y

? y2 = 8x 2 2 2 消去 y 得 x ? (2c + 8m ) x + c = 0 x = my + c ? | ST | | ST | 1 1 + =| c | ( + ) | SP | | SQ | | xP | | xQ |

T

Q1

Q



P1 S O

P

x

1 2|c| xP xQ

= 2|c|

1 = 2. c2

(14 分)

∵ xP 、 xQ 取不相等的正数,∴取等的条件不成立

| d ( ST ) | | d ( ST ) | + 的取值范围是(2,+ ∞ ) . | d ( SP ) | | d ( SQ ) | (文)解: (1)设 AB 所在直线的方程为 y = x + m


(16 分)

? x2 + 3 y 2 = 4 2 2 由? 得 4 x + 6mx + 3m ? 4 = 0 . (2 分) ?y = x + m 4 3 4 3 2 因为 A、B 在椭圆上,所以 ? = ?12m + 64 > 0 . ? <m< 3 3 设 A、B 两点坐标分别为 ( x1,y1 )、 2,y2 ) ,中点为 P ( x 0 , y 0 ) (x 3m 4 4 1 则 x1 + x2 = ? , m = ? x0 , y 0 = x0 ? x0 = ? x0 2 3 3 3 1 3 所以中点轨迹方程为 y = ? x ( ? 3 < x < 3,且x ≠ ? ) (4 分) 3 2 (2)Q AB / /l ,且 AB 边通过点 (0, ,故 AB 所在直线的方程为 y = x . 0)
此时 m = 0 ,由(1)可得 x = ±1 ,所以 AB =

2 x1 ? x2 = 2 2

(6 分) (8 分) (10 分)

又因为 AB 边上的高 h 等于原点到直线 l 的距离,所以 h =

2

S△ ABC =

1 AB ? h = 2 . 2

第 3 页 共 5 页

(3)由(1)得 x1 + x2 = ?

3m 3m2 ? 4 , x1 x2 = , 2 4
(12 分)

32 ? 6m2 所以 AB = 2 x1 ? x2 = . 2
又因为 BC 的长等于点 (0,m) 到直线 l 的距离,即 BC =
2 2 2 2 2

2?m 2

. (14 分)

所以 AC = AB + BC = ? m ? 2m + 10 = ?( m + 1) + 11 . 所以当 m = ?1 时, AC 边最长, (这时 ? = ?12 + 64 > 0 ) 此时 AB 所在直线的方程为 y = x ? 1 . 23.(本题满分 18 分) (16 分)

r 1 uuuu uuur (OM + ON ) ,所以 P 是 MN 的中点,有 2 3 x1 3 x2 3 x1 x2 x1 + x2 = 1 ,∴ y1 + y2 = log 3 + log 3 = log 3 = 1 (4 分) 1 ? x1 1 ? x2 1 ? ( x1 + x2 ) + x1 x2 (2)由(1)知当 x1 + x2 = 1 时, y1 + y2 = f ( x1 ) + f ( x2 ) = 1. 1 2 n ?1 n ?1 2 1 Sn = f ( ) + f ( )L + f ( ) ① Sn = f ( ) +L + f ( ) + f ( ) ② n n n n n n Sn Sn n ?1 n ?1 n ?1 ①+②得 S n = (6 分) lim 4n+1 ? 9 Sn+1 = lim 2 n ? 3n = ? 1 (10 分) S n →∞ 4 n →∞ 2 + 3 +9 3 2 1 1 1 (3)当 n ≥ 2 时, an = = ? . n +1 n + 2 n +1 n + 2 4× ? 2 2 1 1 1 又当 n = 1 时, a1 = , 所以 an = ? (12 分) 6 n +1 n + 2 1 1 1 1 1 1 n 故 Tn = ( ? ) + ( ? ) + L + ( (14 分) ? )= 2 3 3 4 n +1 n + 2 2(n + 2) Tn n Q Tn < m( S n +1 + 1) 对一切 n ∈ N* 都成立,即 m > = 恒成立(16 分) S n +1 + 1 (n + 2)2 n 1 1 1 又 = ≤ ,所以 m 的取值范围是 ( , +∞) (18 分) 2 4 (n + 2) 8 8 n+ +4 n a+ a n (文) (1) an +1 = 2an + 2 , nn 1 = nn 1 + 1 , (2 分) 2 2? bn +1 = bn + 1 , 故 {bn } 为等差数列, b1 = 1 , bn = n . (4 分)
(理) (1)证明:由已知可得, OP = (2)由(1)可得 an = n 2
0
n ?1

uuu r

(6 分)
2

S n = 1 ? 2 0 + 2 ? 21 + 3 ? 2 2 K + n ? 2 n ?1
n ?1

2 S n = 1 ? 21 + 2 ? 2 2 + 3 ? 2 3 K + (n ? 1) ? 2 n ?1 + n ? 2 n
两式相减,得 ? S n = 2 + 2 + 2 K + 2
1

? n ? 2 n = 2 n ? 1 ? n ? 2 n ,即
(10 分)

S n = (n ? 1)2 n + 1 (8 分)

∴ lim
n →∞

Sn (n ? 1)2n + 1 1 = lim = n ? 2n +1 n →∞ n ? 2n +1 2

第 4 页 共 5 页

Tn 1 = n , 4a ? Tn 4 ? 1 1 (d1 + d 2 + d 3 + L + d n + d n +1 ) ? (d1 + d 2 + d 3 + L + d n ) = d n +1 = n +1 >0 4 ?1 1 ∴ {d1 + d 2 + d 3 + L + d n } 单调递增,即 d1 + d 2 + d3 + L + d n ≥ d1 = , (14 分) 3 要使 d1 + d 2 + d3 + L + d n ≥ log 8 (2m + t ) 对任意正整数 n 成立, 1 必须且只需 ≥ log 8 (2m + t ) ,即 0 < 2m + t ≤ 2 对任意 m ∈ [1, 恒成立. (16 分) 2] 3 ?2 + t > 0 ∴ [2 + t, + t ] ? (0, ,即 ? 4 2] ? ?2 < t ≤ ?2 矛盾. ?4 + t ≤ 2 ∴满足条件的实数 t 不存在. (18 分)
(3)由(1)可得 Tn = n , (12 分) ∴ d n =
2

2 n

闵行区 2009 学年第二学期高三年级质量调研考试

第 5 页 共 5 页


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以上会员名单排名不分前后