0  275212  275220  275226  275230  275236  275238  275242  275248  275250  275256  275262  275266  275268  275272  275278  275280  275286  275290  275292  275296  275298  275302  275304  275306  275307  275308  275310  275311  275312  275314  275316  275320  275322  275326  275328  275332  275338  275340  275346  275350  275352  275356  275362  275368  275370  275376  275380  275382  275388  275392  275398  275406  447090 

22.  After many years of hard work, his dream _________ at last.

A. come true        B. was come true    C. was realized       D. realized

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第一节:单项填空(共15小题;每小题1分,满分15分)

从A、B、C、D四个选项中,选出可以填入空白处的最佳选项。

21.  Despite the fact that there are a lot of complaints about ________  CCTV’s annual Spring Festival Gala(Chunwan),most people think it was really ________ feast for our eyes this year.

A. /; a            B. the; the        C. the; /           D.  / ; /

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21.解:(1) ,据题意

(2) 由 (1) 知,,则

x
– 1
(– 1,0)
0
(0,1)
1

– 7
-
0
+
1

– 1

– 4

– 3

∴ 对于的最小值为

的对称轴为,且抛物线开口向下,

的最小值为中较小的

∴ 当的最小值为 – 7

的最小值为 – 7

的最小值为 – 11

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20.解:(1) 第r + 1项项系数为,第r项系数为,第r + 2项系数为

∴ 展开式中系数最大的项为

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18.解:(1)

(2)

在R上递增,满足题意;

,   ∴

∴ 综上,a的取值范围是

  19.解法一:

(1) 过O作OF⊥BC于F,连接O1F,

OO1⊥面AC,∴BCO1F,

∴∠O1FO是二面角O1-BC-D的平面角,······················· 3分

OB = 2,∠OBF = 60°,∴OF =

在Rt△O1OF中,tan∠O1FO =

∴∠O1FO=60° 即二面角O1-BC-D的大小为60°·············································· 6分

(2) 在△O1AC中,OE是△O1AC的中位线,∴OEO1C

OEO1BC,∵BC⊥面O1OF,∴面O1BC⊥面O1OF,交线O1F

OOHO1FH,则OH是点O到面O1BC的距离,································· 10分

OH = ∴点E到面O1BC的距离等于····················································· 12分

解法二:

(1) ∵OO1⊥平面AC

OO1OAOO1OB,又OAOB,······················· 2分

建立如图所示的空间直角坐标系(如图)

∵底面ABCD是边长为4,∠DAB = 60°的菱形,

OA = 2OB = 2,

A(2,0,0),B(0,2,0),C(-2,0,0),O1(0,0,3)·········· 3分

设平面O1BC的法向量为=(xyz),则

,则z = 2,则x=-y = 3,

=(-,3,2),而平面AC的法向量=(0,0,3)··························· 5分

∴ cos<>=

O1BCD的平面角为α, ∴cosα=∴α=60°.

故二面角O1BCD为60°.············································································· 6分

(2) 设点E到平面O1BC的距离为d

   ∵E是O1A的中点,∴=(-,0,),············································· 9分

则d=

∴点E到面O1BC的距离等于.··································································· 12分

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17.解:(1) 法一:设两项技术指标达标的概率分别为

由题意得:  ······················································ 3分

解得:,∴

即,一个零件经过检测为合格品的概率为.·············································· 6分

法二:

(2) 任意抽出5个零件进行检查,其中至多3个零件是合格品的概率为

············································································· 13分

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21.(本小题满分12分)

已知函数

(1)    若函数的图象在点P(1,)处的切线的倾斜角为,求实数a的值;

(2)    设的导函数是,在 (1) 的条件下,若,求的最小值.

(3)    若存在,使,求a的取值范围.

(命题人:周  静   审题人:赵文丽)

a的取值范围为

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