1 (C)-1 (D)-1004.5 查看更多

 

題目列表(包括答案和解析)

“|-1|<2成立”是“<0成立”的(      )

(A)充要條件;                (B)必要不充分條件;

(C)充分不必要條件;          (D)既不充分也不必要條件.

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(1+i)20-(1-i)20的值為

A.0                  B.1024              C.-1024              D.-1024i

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(6)設函數(shù)y=f(x)的反函數(shù)為x=f-1(x),且y=f(2x-1)的圖象過點(,1),則y=f--1(x)的圖象必過點

(A)(,1)                          (B)(1,

(C)(1,0)                           (D)(0,1)

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(7)已知f(x)為R上的減函數(shù),則滿足f(||)<f(1)的實數(shù)x的取值范圍是

A (-1,1)  B(0,1)  C (-1,0)(0,1)    D(-,-1)(1,+

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(理)設偶函數(shù)f (x)=loga|xb|在(-∞,0)上遞增,則f (a+1)與f (b+2)的大小關系是(    )

         A.f(a+1)=f (b+2)                             B.f (a+1)>f (b+2)  

         C.f(a+1)<f (b+2)                             D.不確定

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1-10.CDBBA   CACBD

11. 12. ①③④   13.-2或1  14. 、  15.2  16.  17..

18.

解:(1)由已知            7分

(2)由                                                                   10分

由余弦定理得                          14分

 

19.(1)證明:∵PA⊥底面ABCD,BC平面AC,∴PA⊥BC,                                  3分

∵∠ACB=90°,∴BC⊥AC,又PA∩AC=A,∴BC⊥平面PAC.                             5分

(2)解:過C作CE⊥AB于E,連接PE,

∵PA⊥底面ABCD,∴CE⊥面PAB,

∴直線PC與平面PAB所成的角為,                                                    10分

∵AD=CD=1,∠ADC=60°,∴AC=1,PC=2,

中求得CE=,∴.                                                  14分

 

20.解:(1)由①,得②,

②-①得:.                              4分

(2)由求得.          7分

,   11分

.                                                                 14分

 

21.解:

(1)由得c=1                                                                                     1分

,                                                         4分

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        市一次模文數(shù)參答―1(共2頁)

                                                                                                5分

        (2)時取得極值.由,.                                                                                          8分

        ,,∴當時,

        上遞減.                                                                                       12分

        ∴函數(shù)的零點有且僅有1個     15分

         

        22.解:(1) 設,由已知,

        ,                                        2分

        設直線PB與圓M切于點A,

        ,

                                                         6分

        (2) 點 B(0,t),點,                                                                  7分

        進一步可得兩條切線方程為:

        ,                                   9分

        ,

        ,                                          13分

        ,又時,,

        面積的最小值為                                                                            15分

         

         

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