Tuyển Tập Các Đề Thi Đại Học Cao Đẳng Môn Toán ( Có Đáp Án )

7/26/2019 Tuyn Tp Cc Thi i Hc Cao ng Mn Ton ( C p n )

1/148

b gio dc v o to K thi tuyn sinh i hc, cao nG nm 2002------------------------------ Mn thi : ton

chnh thc (Thi gian lm bi: 180 pht)_____________________________________________

Cu I (H : 2,5 im; C : 3,0 im)

Cho hm s : (1) ( l tham s).23223 )1(33 mmxmmxxy +++= m

1. Kho st s bin thin v v th hm s (1) khi .1=m 2. Tm k phng trnh: c ba nghim phn bit.033 2323 =++ kkxx3. Vit phng trnh ng thng i qua hai im cc tr ca th hm s (1).Cu II.(H : 1,5 im; C: 2,0 im)

Cho phng trnh : 0121loglog 232

3 =++ mxx (2) ( l tham s).m

1 Gii phng trnh (2) khi .2=m

2. Tm phng trnh (2) c t nht mt nghim thuc on [m 33;1 ].Cu III. (H : 2,0 im; C : 2,0 im )

1. Tm nghim thuc khong )2;0( ca phng trnh: .32cos

2sin21

3sin3cossin +=

+

++ x

x

xxx5

2. Tnh din tch hnh phng gii hn bi cc ng: .3,|34| 2 +=+= xyxxyCu IV.( H : 2,0 im; C : 3,0 im)1. Cho hnh chp tam gic u nh c di cnh y bng a. GiABCS. ,S v ln ltN

l cc trung im ca cc cnh v Tnh theo din tch tam gic , bit rngSB .SC a AMN mt phng ( vung gc vi mt phng .)AMN )(SBC

2. Trong khng gian vi h to cac vung gc Oxyzcho hai ng thng:

v .

=++

=+

0422

042:1

zyx

zyx

+=

+=

+=

tz

ty

tx

21

2

1

:2

a) Vit phng trnh mt phng cha ng thng)(P 1 v song song vi ng thng .2

b) Cho im . Tm to im)4;1;2(M Hthuc ng thng 2 sao cho on thng H c di nh nht.

Cu V.( H : 2,0 im)1. Trong mt phng vi h to cac vung gc Oxy , xt tam gic vung ti ,ABC A

phng trnh ng thng lBC ,033 =yx cc nh vA B thuc trc honh v

bn knh ng trn ni tip bng 2. Tm ta trng tm ca tam gic .G ABC 2. Cho khai trin nh thc:

nx

n

n

nxx

n

n

xn

x

n

nx

n

nxx

CCCC

+

++

+

=

+

3

1

32

1

13

1

2

1

12

1

032

1

22222222 L

( n l s nguyn dng). Bit rng trong khai trin C v s hng th t13 5 nn C=

bng , tm vn20 n x .----------------------------------------Ht---------------------------------------------

Ghi ch: 1)Th sinhch thi cao ngkhng lm Cu V.

2) Cn b coi thi khng gii thch g thm.

H v tn th sinh:.................................................... S bo danh:.....................


2/148

1

b gio dc v o to Kthi tuyn sinh i hc, cao ng nm 2002------------------------------------- p n v thang im mn ton khi A

Cu Ni dung H CI 1

23 31 xxym +==

Tp xc nh Rx . )2(363' 2 =+= xxxxy ,

=

==

2

00'

2

1

x

xy

10",066" ===+= xyxy

Bng bin thin

+ 210x

'y +0 0

+ 0"y

y + lm U 4

CT 2 C 0 li

=

== 300xxy , 4)1( =y

th:

( Th sinh c th lp 2 bng bin thin)

1 ,0

0,25

0,5

0,25

1 ,5

0,5

0,5

0,5

-1 1 2 3 x0

2

4

y


3/148

2

I 2

Cch I. Ta c 2332323 33033 kkxxkkxx +=+=++ .

t 23 3kka += Da vo th ta thy phng trnh axx =+ 23 3

c 3 nghim phn bit 43040 23 =+= ymm c 2 nghim 21 xx

v 'y i du khi qua 1x v 2x hm s t cc tr ti 1x v 2x .

Ta c 23223 )1(33 mmxmmxxy +++=

( ) .2336333

1 222 mmxmmxxmx ++++

=

T y ta c mmxy += 211 2 v mmxy += 2

22 2 .

Vy phng trnh ng thng i qua 2 im cc tr l mmxy += 2

2 .

1 ,0

0,25

0,25

0,25

0,25

----------

0,25

0,25

0,25 0,25

1 ,0

0,25

0,25

0,25

0,25

-----------

0,25

0,25

0,25 0,25

II 1.

Vi 2=m ta c 051loglog 232

3 =++ xx

iu kin 0>x . t 11log23 += xt ta c

06051 22 =+=+ tttt .2

3

2

1

=

=t

t

5,0

0,25

0,1

0,5


4/148

3

31 =t (loi) ,3

3

2

32 33log3log2 ==== xxxt

33=x tha mn iu kin 0>x .(Th sinh c th gii trc tip hoc t n ph kiu khc)

0,25 0,5

2.

0121loglog 232

3 =++ mxx 2)

iu kin 0>x . t 11log23 += xt ta c

0220121 22 =+=+ mttmtt 3)

.21log13log0]3,1[ 2333 += xtxx

Vy (2) c nghim ]3,1[ 3 khi v ch khi (3) c

nghim [ ]2,1 . t tttf += 2)(

Cch 1.

Hm s )(tf l hm tng trn on ][ 2;1 . Ta c 2)1( =f v 6)2( =f .Phng trnh 22)(222 +=+=+ mtfmtt c nghim [ ]2;1

.20622

222

22)2(

22)1(

+

+

+

+ m

m

m

mf

mf

Cch 2.

TH1. Phng trnh (3) c 2 nghim 21, tt tha mn 21 21


5/148

4

2.

V (0x ; )2 nn ly3

1

=x v

3

52

=x . Ta thy 21,xx tha mn iu

kin2

12sin x . Vy cc nghim cn tm l:

31

=x v

3

52

=x .

(Th sinh c th s dng cc php bin i khc)

Ta thy phng trnh 3|34| 2 +=+ xxx c 2 nghim 01 =x v .52 =x

Mt khc ++ 3|34| 2 xxx [ ]5;0x . Vy

( ) ( ) ( )dxxxxdxxxxdxxxxS ++++++=++=1

0

3

1

22

5

0

2 343343|34|3

( )dxxxx +++5

3

2 343

( ) ( ) ( )dxxxdxxxdxxxS +++++=5

3

2

3

1

2

1

0

2 5635

5

3

233

1

231

0

23

2

5

3

16

2

3

3

1

2

5

3

1

++

++

+= xxxxxxxS

6

109

3

22

3

26

6

13=++=S (.v.d.t)

(Nu th sinh v hnh th khng nht thit phi nu bt ng thc ++ 3|34| 2 xxx [ ]5;0x )

0,25

1 ,0

0,25

0,25

0,25

0,25

0,25

1 ,0

0,25

0,25

0,25

0,25

IV 1. 1 1

x510-1

y

3

32

1

8

-1


6/148

5

S

N

I

M C

A K

B

Gi Kl trung im ca BC v MNSKI =

. T gi thitMNa

BCMN ,22

1== //BC I l trung im ca SKv MN.

Ta c = SACSAB hai trung tuyn tng ng ANAM = AMN cn ti A MNAI .

Mt khc

( ) ( )( ) ( )

( ) ( ) SKAISBCAI

MNAI

AMNAI

MNAMNSBC

AMNSBC

=

.

Suy ra SAK cn ti 2

3a

AKSAA == .

244

3 222222 aaaBKSBSK ===

4

10

84

3

2

222

222 aaaSKSASISAAI ==

== .

Ta c16

10.

2

1 2aAIMNS AMN == (vdt)

ch 1) C th chng minh MNAI nhsau:

( ) ( ) AIMNSAKMNSAKBC .2) C th lm theo phng php ta :Chng hn chn h ta cac vung gc Oxyz sao cho

h

aS

aA

aC

aBK ;

6

3;0,0;

2

3;0,0;0;

2,0;0;

2),0;0;0(

trong h l di ng cao SH ca hnh chp ABCS. .

0,25

0,25

0,25

0,25

0,25

0,25

0,25

0,25


7/148

6

2a)

Cch I. Phng trnh mt phng )(P cha ng thng 1 c dng:

( ) ( ) 042242 =++++ zyxzyx ( 022 + )( ) ( ) ( ) 044222 =+++ zyx

Vy ( ) 2;22; ++=Pnr

.Ta c ( )2;1;12 =ur

// 2 v ( ) 22 1;2;1 M

( )P //( ) ( ) ( )

=

=

PMPM

unP

22

2

2

0

1;2;1

0. rr

Vy ( ) 02: =zxP

Cch II Ta c th chuyn phng trnh 1 sang dng tham s nhsau:

T phng trnh 1 suy ra .02 =zx t

=

=

=

=

'4

2'3

'2

:'2 1

tz

ty

tx

tx

( ) )4;3;2(,0;2;0 111 = uM r // 1 .(Ta c th tm ta im 11 M bng cch cho 020 === zyx

v tnh ( )4;3;221

21;12

11;22

121 =

=u

r).

Ta c ( )2;1;12 =ur

// 2 . T ta c vc t php ca mt phng )(P l :

[ ] ( )1;0;2, 21 == uunPrrr

. Vy phng trnh mt phng )(P i qua ( )0;2;01 Mv ( )1;0;2 =Pn

rl: 02 =zx .

Mt khc ( ) ( ) PM 1;2;12 phng trnh mt phng cn tm l: 02 =zx

5,0

0,25

0,25 -----------

0,25

0,25

0,1

0,5

0,5 -----------

0,5

0,5

2b)

b)Cch I. ( ) MHtttHH +++ 21,2,12 = ( )32;1;1 + ttt

( ) ( ) ( ) 5)1(6111263211 22222 +=+=+++= ttttttMHt gi tr nh nht khi v ch khi ( )3;3;21 Ht =Cch II. ( )tttHH 21;2;12 +++ .MHnh nht ( )4;3;210. 22 HtuMHMH ==

r

5,0 0,25

0,25 -----------

0,25 0,25

0,1 0,5

0,5 -----------

0,5 0,5

V 1.

Ta c ( )0;1BOxBC =I . t axA = ta c );( oaA v

.33 == ayax CC Vy 33; aaC .

T cng thc( )

( )

++=

++=

CBAG

CBAG

yyyy

xxxx

3

13

1

ta c

+

3

)1(3;3

12 aaG .

Cch I.

Ta c :

|1|2|,1|3|,1| === aBCaACaAB . Do

1

0,25


8/148

7

( )212

3.

2

1== aACABS ABC .

Ta c( )

|1|3|1|3

132 2

+

=

++=

aa

a

BCACAB

Sr = .2

13

|1|=

+

a

Vy .232|1| +=a

TH1.

+++=

3

326;

3

347332 11 Ga

TH2

=

3

326;

3

134132 22 Ga .

Cch II.

y C

I

O B A x

GiI l tm ng trn ni tip ABC . V 22 == Iyr .

Phng trnh ( ) 3213

11.30: 0 =

== Ixx

xtgyBI .

TH1 Nu A v O khc pha i vi .321+= IxB T 2),( =ACId

.3232 +=+= Ixa

++

3

326;

3

3471G

TH 2.Nu A v O cng pha i vi .321= IxB Tng t

ta c .3212 == Ixa

3

326;

3

1342G

0,25

0,25

0,25 -----------

0,25

0,25

0,25

2.

T 13 5 nn CC = ta c 3n v1


9/148

8

( ) ( ) 02835

6

)2)(1(

!1

!5!3!3

! 2 ==

=

nnnnnn

n

n

n

n

41 =n (loi) hoc .72 =n

Vi 7=n ta c

.4421402.2.3514022 2223

34

21

3

7 ====

xC xxxxx

0,25

0,25

0,5


10/148

b gio dc v o to k thi tuyn sinh i hc, cao ng nm 2002 chnh thc Mn thi : ton, Khi B.

(Thi gian lm bi : 180 pht)_____________________________________________

Cu I (H : 2,0 im; C : 2,5 im) Cho hm s : ( ) 109 224 ++= xmmxy (1) (m l tham s).1. Kho st s bin thin v v th ca hm s (1) khi 1=m .2. Tm m hm s (1) c ba im cc tr.

Cu II (H : 3,0 im; C : 3,0 im)

1. Gii phng trnh: xxxx 6cos5sin4cos3sin 2222 = .

2. Gii bt phng trnh: ( ) 1)729(loglog 3 xx .

3. Gii h phng trnh:

++=+

=

.2

3

yxyx

yxyx

Cu III ( H : 1,0 im; C : 1,5 im) Tnh din tch ca hnh phng gii hn bi cc ng :

4

42x

y = v24

2xy= .

Cu IV (H : 3,0 im ; C : 3,0 im)1. Trong mt phng vi h ta cac vung gc Oxy cho hnh ch nht ABCD c tm

0;

2

1I , phng trnh ng thng AB l 022 =+ yx v ADAB 2= . Tm ta cc nh

DCBA ,,, bit rng nh A c honh m.2. Cho hnh lp phng 1111 DCBABCDA c cnh bng a .

a) Tnh theo a khong cch gia hai ng thng BA1 v DB1 .

b) Gi PNM ,, ln lt l cc trung im ca cc cnh CDBB ,1 , 11DA . Tnh gc gia

hai ng thng MPv NC1 .

Cu V (H : 1,0 im) Cho a gic u nAAA 221 L ,2( n n nguyn ) ni tip ng trn ( )O . Bit rng s tam gic c cc nh l 3 trong n2 im nAAA 221 ,,, L nhiu gp 20 ln s hnh ch nht

c cc nh l 4 trong n2 imn

AAA221

,,, L , tm n .

--------------------------------------Ht-------------------------------------------Ghi ch : 1)Th sinhch thicao ngkhng lm Cu IV 2. b) v Cu V.

2) Cn b coi thi khng gii thch g thm.

H v tn th sinh:................................................................... S bo danh:...............................


11/148

1

B gio dc v o to k thi tuyn sinh i hc, cao ng nm 2002------------------------- p n v thang im thi chnh thc

Mn ton, khi b

Cu Ni dung H CI 1 Vi 1=m ta c 108 24 += xxy l hm chn th i xng qua Oy .

Tp xc nh Rx , ( )44164' 23 == xxxxy , 0'=y

=

=

2

0

x

x

,3

4121612" 22

== xxy

3

20" == xy .

Bng bin thin:

+

23

20

3

22x

'y 0 + 0 0 +

"y + 0 0 + + 10 +

y lm U C U lmCT li CT6 6

Hai im cc tiu : ( )6;21 A v ( )6;22 A .Mt im cc i: ( )10;0B .

Hai im un:

9

10;

3

21U v

9

10;

3

22U .

Giao im ca th vi trc tung l ( )10;0B . th ct trc honh ti 4 im c honh :

64 +=x v 64 =x .

(Th sinh c th lp 2 bng bin thin)

0,1

0,25

0,5

0,25

5,1

0,5

0,5

0,5

x0

10

y

-6

-2 2

A2A1

B

U1 U2


12/148

2

I 2 ( ) ( )922924' 2223 +=+= mmxxxmmxy ,

=+

==

092

00'

22 mmx

xy

Hm s c ba im cc tr phng trnh 0'=y c 3 nghim

phn bit (khi 'y i du khi qua cc nghim) ph

ng trnh092 22 =+mmx c 2 nghim phn bit khc 0.

092 22 =+mmx

=

m

mx

m

2

90

22 . Phng trnh 092

22 =+mmx

c 2 nghim khc 0

>

>

>

x

xxx

x

x (2).

Do 173log9 >>x nn ( ) xx 729log)1( 3 ( ) 072333729 2 xxxx (3).

t xt 3= th (3) tr thnh

2938980722 xttt x .Kt hp vi iu kin (2) ta c nghim ca bt phng trnh l:

273log9 < x .

0,1

0,25

0,25

0,25

0,25

0,1

0,25

0,25

0,25

0,25


13/148

3

3

++=+

=

).2(2

)1(3

yxyx

yxyx iu kin: )3(

.0

0

+

yx

yx

( )

+=

==

.101)1( 63

yx

yxyxyx

Thay y= vo (2), gii ra ta c .1== yx

Thay 1+= yx vo (2), gii ra ta c:2

1,

2

3== yx .

Kt hp vi iu kin (3) h phng trnh c 2 nghim:

1,1 == yx v2

1,

2

3== yx

Ch :Th sinh c th nng hai v ca (1) ln lu tha bc 6 di n kt qu:

+=

=

.1yx

yx

0,1 0,25

0,25

0,25

0,25

0,1 0,25

0,25

0,25

0,25

III

Tm giao im ca hai ng cong4

42x

y = v24

2xy= :

44

2x =

24

2x8804

432

224

===+ xxxx

.

Trn 8;8 ta c24

2x

44

2x v do hnh i xng qua trc tung

nn dxxxS

=

8

0

22

24442 21

8

0

2

8

0

2

22116 SSdxxdxx == .

tnh 1S ta dng php i bin tx sin4= , khi4

0

t th 80 x .

tdtdx cos4= v

>

4;00cos

tt . Do

0,1

0,25

0,25

5,1

0,5

0,25

x0-4 4

2

y

-2 2 2 2

2 A2A1

4

x4y

2

=24

xy

2

=


14/148

4

( ) 422cos18cos16164

0

4

0

2

8

0

2

1 +=+===

dtttdtdxxS .

3

8

26

1

22

1 8

0

3

8

0

2

2 === xdxxS . Vy 34

221 +== SSS .

Ch : Th sinh c th tnh din tch dxxx

S

=8

8

22

2444 .

0,25

0,25

0,5

0,25

IV 1

Khong cch t I n ng thngAB bng2

55= AD v

2

5

== IBIA .

Do BA, l cc giao im ca ng thng AB vi ng trn tm I v bn

knh2

5=R . Vy ta BA, l nghim ca h :

=+

=+2

2

2

2

5

2

1

022

yx

yx

Gii h ta c ( ) ( )2;2,0;2 BA (v 0


15/148

5

IV 2a) Tm khong cch gia BA1 v DB1 .

Cch I. Chn h ta cac vung gc Oxyzsao cho

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )aaDaaaCaaBaaCaAaDaBA ;;0,;;;;0;;0;;;0;0,0;;0,0;0;,0;0;0 1111

( ) ( ) ( )0;0;,;;,;0; 1111 aBAaaaDBaaBA === v ( )22211 ;2;, aaaDBBA = .

Vy ( )[ ] 66,

.,,

2

3

11

1111

11

a

a

a

DBBA

BADBBADBBAd === .

Cch II. ( ) DBBADCABBAADBA

ABBA11111

1

11

.

Tng t DBCA 111 ( )111 BCADB .

Gi ( )111 BCADBG = . Do aCBBBAB === 11111 nn

GGCGBGA == 11 l tm tam gic u 11BCA c cnh bng 2a .

GiI l trung im ca BA1 th IG l ng vung gc chung ca BA1 v

DB1 , nn ( )62

3

3

1

3

1, 1111

aBAICIGDBBAd ==== .

Ch :

Th sinh c th vit phng trnh mt phng ( )P cha BA1 v song song vi

DB1 l: 02 =++ azyx v tnh khong cch t 1B (hoc t D ) ti ( )P ,

hoc vit phng trnh mt phng ( )Q cha DB1 v song song vi BA1 l:

022 =++ azyx v tnh khong cch t 1A (hoc t B) ti ( )Q .

0,1

0,25

0,25

0,25

0,25

0,25

0,25

0,25

5,1

0,25

0,5

0,25

0,5

0,25

0,5

0,5

x

D

D

CB1

A1

z

y

x

A

CB

I


16/148

6

2b)Cch I.

T Cch Ica 2a) ta tm c

a

aPa

aN

aaM ;

2;0,0;;

2,

2;0;

0.;0;2

,2

;2

; 11 =

=

= NCMPa

aNC

aaaMP .

Vy NCMP 1 .

Cch II.

Gi El trung im ca 1CC th ( ) 11CCDDME hnh chiu vung gc caPtrn ( )11CCDD l 1ED . Ta c

NCEDNCDNCCEDCECDCNC 11110

111111 90 === . T y

theo nh l ba ng vung gc ta c NCMP 1 .

0,1

0,25

0,5

0,25

0,25

0,25

0,25 0,25

V

S tam gic c cc nh l 3 trong n2 im nAAA 221 ,,, L l3

2nC .

Gi ng cho ca a gic u nAAA 221 L i qua tm ng trn ( )O lng cho ln th a gic cho c n ng cho ln.

Mi hnh ch nht c cc nh l 4 trong n2 im nAAA 221 ,,, L c cc ng

cho l hai ng cho ln. Ngc li, vi mi cp ng cho ln ta c cc umt ca chng l 4 nh ca mt hnh ch nht. Vy s hnh ch nht ni trnbng s cp ng cho ln ca a gic nAAA 221 L tc

2

nC .

Theo gi thit th:

0,1

0,25

0,25

D1A

B1 C1

C

B

A

ME

N

P

y

x

z


17/148

7

( )( ) ( )

( )( ) ( )2

120

6

2212.2

!2!2

!20

!32!3

!220 232

=

=

=

nnnnn

n

n

n

nCC nn

81512 == nn .

Ch :

Th sinh c th tm s hnh ch nht bng cc cch khc. Nu l lun ng i

n kt qu s hnh ch nht l2

)1( nnth cho im ti a phn ny.

0,5


18/148

B gio dc v o to K thi Tuyn sinh i hc ,cao ng nm 2002 chnh thc Mn thi : Ton Khi D

(Thi gian lm bi : 180 pht) _________________________________________

CuI ( H : 3 im ; C : 4 im).

Cho hm s :( )

1x

mx1m2y

2

= (1) ( m l tham s ).

1. Kho st s bin thin v v th (C) ca hm s (1) ng vi m = -1.

2. Tnh din tch hnh phng gii hn bi ng cong (C) v hai trc ta .3. Tm m thca hm s (1) tip xc vi ng thng xy= .

Cu II ( H : 2 im ; C : 3 im).

1. Gii bt phng trnh : ( )x3x2

. 02x3x22

.

2. Gii h phng trnh :

=+

+

=+

.y22

24

y4y52

x

1xx

2x3

Cu III

( H : 1 im ; C : 1 im ). Tm x thuc on [0 ; 14 ] nghim ng phng trnh : 04xcos3x2cos4x3cos =+ .

Cu IV ( H : 2 im ; C : 2 im).1. Cho hnh t din ABCD c cnh AD vung gc vi mt phng (ABC); AC = AD = 4 cm ;

AB = 3 cm ; BC = 5 cm . Tnh khong cch t im A ti mt phng (BCD).2. Trong khng gian vi h ta cac vung gc Oxyz, cho mt phng (P) : 02yx2 =+

v ng thng md :( ) ( )

( )

=++++

=+++

02m4z1m2mx

01mym1x1m2( m l tham s ).

Xc nh m ng thng md song song vi mt phng (P).

Cu V (H : 2 im ).

1. Tm s nguyn dng n sao cho 243C2....C4C2C nnn2

n1n

0n =++++ .

2. Trong mt phng vi h ta cac vung gc Oxy , cho elip (E) c phng trnh

19

y

16

x 22=+ . Xt im M chuyn ng trn tia Ox v im N chuyn ng trn tia Oy sao cho

ng thng MN lun tip xc vi (E). Xc nh ta ca M , N on MN c di nhnht . Tnh gi tr nh nht .

-------------------------Ht-------------------------

Ch :

1. Th sinh ch thi cao ng khng lm cu V 2. Cn b coi thi khng gii thch g thm.

H v tn th sinh : ................................................................ S bo danh.............................


19/148

1

B gio dc v o to K thi tuyn sinh i hc , cao ng nm 2002Mn Ton khi D

p n v thang im thi chnh thc

Cu Ni dung im

H CI

3 41. 1 1,5

Khi m = -1 ,ta c1x

43

1x

1x3y

=

=

-TX : 1x

- CBT :( )

>

= 1x,01x

4y2

, hm s khng c cc tr.1/4 1/4

3ylimx

=

; =+=+ 1x1x

ylim;ylim .

- BBT :

x - 1 +

y/ + ++

y -3 -3

- 1/4 1/4- TC: x=1 l tim cn ng v =

ylim

1x.

y=-3 l tim cn ngang v 3ylimx

=

1/4 1/4

- Giao vi cc trc : x = 0 y = 1; y = 0 x = - 1/3. 1/4- th :

x

y

1/4 1/2


20/148

2

2. 1

1,5

Din tch cn tnh l :

dx1x

1x3S

0

3/1

=

1/4 1/2

=

0

3/1

0

3/1 1x

dx

4dx3 1/4 1/4

3/1

01xln4

3

1.3

=

1/4 1/2

3

4ln41 += ( vdt).

1/4 1/43. 1 1

K hiu( )

1x

mx1m2)x(f

2

= . Yu cu bi ton tng ng vi tm

m h phng trnh sau c nghim:

(H)( )

=

=

.x)x(f

x)x(f//

1/4 1/4

Ta c (H)

( )

( )

=

=

01x

mx

01x

mx

/2

2

1/4 1/4

( )

( )( ) ( )( )

=

+

=

0

1x

mx1xmx2

0

1x

mx

2

2

2

1/4 1/4Ta thy vi 1m ; x = m lun tho mn h ( H ) . V vy 1m , (H)lun c nghim , ng thi khi m = 1 th h ( H ) v nghim. Do th hm s (1) tip xc vi ng thng y = x khi v ch khi 1m .

S : 1m . 1/4 1/4II

2 31. 1 1,5

Bt phng trnh

>

=

0x3x

02x3x2

02x3x2

2

2

2

1/4 1/2

TH 1: .2

1x2x02x3x202x3x2 22 ====

1/4 1/4

TH 2:

>

>

0x3x

02x3x2

0x3x

02x3x22

2

2

2

>


21/148

3

3x2

1x <

1/4 1/4

T hai trng hp trn suy ra S: 3x2x2

1x =

1/4 1/4

2.1 1,5

H phng trnh

=

=

y2

y4y52x

2x3

1/4 1/2

=+

>=

0y4y5y

0y223

x

1/4 1/4

===

>=

4y1y0y

0y2x

1/4 1/4

=

=

=

=

4y

2x

1y

0x

1/4 1/2

III1 1

Phng trnh ( ) ( ) 01x2cos4xcos3x3cos =++ 0xcos8xcos4 23 =

( ) 02xcosxcos4 2 = 0xcos = 1/4 1/2

+

= k2

x .1/4 1/4

[ ] 3k2k1k0k14;0x ==== 1/4

S : ;2

x

= 2

3x

= ;

2

5x

= ;

2

7x

= .

1/4 1/4IV

2 21. 1 1Cch 1T gi thit suy ra tam gic ABC vung ti A , do .ACAB 1/4 1/4Li c ( )ABCmpAD ABAD v ACAD , nn AB, AC, AD imt vung gc vi nhau. 1/4 1/4Do c th chn h to cac vung gc, gc A sao cho B(3;0;0) ,

C(0;4;0), D( 0;0;4). Mt phng (BCD) c phng trnh :

014

z

4

y

3

x=++ .

1/4 1/4

Khong cch cn tnh l :17

346

16

1

16

1

9

1

1=

++

(cm).

1/4 1/4


22/148

4

Cch 2T gi thit suy ra tam gic ABC vung ti A , do .ACAB 1/4 1/4Li c ( )ABCmpAD ABAD v ACAD , nn AB, AC, AD imt vung gc vi nhau. 1/4 1/4

D

H C

A E

BGi AE l ng cao ca tam gic ABC; AH l ng cao ca tam gicADE th AH chnh l khong cch cn tnh.

D dng chng minh c h thc:2222 AC

1

AB

1

AD

1

AH

1++= .

1/4 1/4Thay AC=AD=4 cm; AB = 3 cm vo h thc trn ta tnh c:

cm17

346AH=

1/4 1/4Cch 3:T gi thit suy ra tam gic ABC vung ti A , do .ACAB 1/4 1/4

Li c ( )ABCmpAD ABAD v ACAD , nn AB, AC, AD imt vung gc vi nhau. 1/4 1/4

Gi V l th tch t din ABCD, ta c V= 8ADACAB6

1= .

p dng cng thc)BCD(dt

V3AH

= vi V = 8 v dt( BCD) =2 34

ta tnh c cm17

346AH= .

1/2 1/22 1 1

Cch 1:Mt phng (P) c vect php tuyn ( )0;1;2n

. ng thng md c vec

t ch phng ( )( ) ( ) ( )( )m1m;1m2;1m2m1u 2 ++

. 1/4 1/4

Suy ra

u .

n =3(2m+1).

md song song vi (P)

)P(d

nu

m

1/4 1/4


23/148

5

( )

=

PA,dA

0n.u

m

Ta c : iu kin 0n.u =

2

1m =

1/4 1/4

Mt khc khi m = - 1/2 th md c phng trnh :

=

=

0x

01y, mi im

A( 0;1;a) ca ng thng ny u khng nm trong (P), nn iu kin( )PA,dA m c tho mn. S : m = - 1/2 1/4 1/4

Cch 2:Vit phng trnh dmdi dng tham s ta c

=

+=

+=

m)t.m(12z

t1)(2m1y

1)tm)(2m(1x2

1/4 1/4

md // (P) h phng trnh n t sau

=+

=

+=

+=

02yx2

t)m1(m2z

t)1m2(1y

t)1m2)(m1(x2

v nghim

1/4 1/4 phng trnh n t sau 3(2m+1)t+1 = 0 v nghim 1/4 1/4 m=-1/2 1/4 1/4Cch 3:

md // (P) h phng trnh n x, y, z sau

(H) ( ) ( )

=++++

=+++=+

02m4z)1m2(mx

01myx1x1m2

02yx2

v nghim 1/4 1/4

T 2 phng trnh u ca h phng trnh trn suy ra

+=

=

3

4m2y

3

1mx

1/4 1/4

Th x , y tm c vo phng trnh th ba ta c :

)6m11m(3

1z)1m2( 2 ++=+

1/4 1/4

H (H) v nghim2

1m =

1/4 1/4V

2

1. 1

Ta c : ( ) =

=+n

0k

kkn

nxC1x ,

1/4

Cho x = 2 ta c ==n

0k

kknn 2C31/4

5n32433 5n === . 1/2


24/148

6

2. 1Cch 1Gi s M(m;0) v N(0;n) vi m > 0 , n > 0 l hai im chuyn ng trnhai tia Ox v Oy.

ng thng MN c phng trnh : 01n

y

m

x=+

1/4ng thng ny tip xc vi (E) khi v ch khi :

1n

19

m

116

22

=

+

.

1/4Theo BT Csi ta c :

( )2

2

2

2

22

22222

n

m9

m

n1625

n

9

m

16nmnmMN ++=

++=+=

499.16225 =+ 7MN 1/4

ng thc xy ra

>>

=+=

0n,0m

49nm

n

m9

m

n16

22

2

2

2

2

21n,72m == .

KL: Vi 21;0N,0;72M th MN t GTNN v GTNN (MN) = 7. 1/4Cch 2Gi s M(m;0) v N(0;n) vi m > 0 , n > 0 l hai im chuyn ng trnhai tia Ox v Oy.

ng thng MN c ph

ng trnh : 01n

y

m

x=+

1/4ng thng ny tip xc vi (E) khi v ch khi :

1n

19

m

116

22

=

+

.

1/4Theo bt ng thc Bunhiacpski ta c

( ) 49n

3.n

m

4.m

n

9

m

16nmnmMN

2

22

22222 =

+

++=+= .

7MN 1/4

- ng thc xy ra

>>

=+

=

0n,0m

7nmn

3

:nm

4

:m22 21n,72m == .

KL: Vi 21;0N,0;72M th MN t GTNN v GTNN (MN) = 7. 1/4Cch 3:

Phng trnh tip tuyn ti im (x0; y0) thuc (E) : 19

yy

16

xx 00 =+

1/4


25/148

7

Suy ra to ca M v N l

0;

x

16M

0

v

0y

9;0N

+

+=+=

20

2

20

220

20

20

2

20

22

y

9

x

16

9

y

16

x

y

9

x

16MN

1/4S dng bt ng thc Csi hoc Bunhiacpski (nhcch 1 hoc cch 2)ta c : 22 7MN

1/4

- ng thc xy ra7

213y;

7

78x 00 == .

- Khi 21;0N,0;72M v GTNN (MN) = 7 1/4

-----------------------Ht----------------------


26/148

8

B gio dc v o to K thi tuyn sinh i hc ,cao ng nm 2002------------------------ ---------------------------------------------

H

ng dn chm thi mn ton khi D

Cu I:

1. -Nu TS lm sai bc no th k t tr i s khng c im.-Nu TS xc nh ng hm s v ch tm ng 2 tim cn th c 1/4 im.

2. Nu TS lm sai bc no th k t tr i s khng c im.3. -Nu TS dng iu kin nghim kp th khng c im.

-Nu TS khng loi gi tr m = 1 th b tr 1/4 im.

Cu II:

1. -Nu TS lm sai bc no th k t tr i s khng c im.-Nu TS kt lun nghim sai b tr 1/4 im .

-Nu TS s dng iu kin sai:

> l trung im cnh CC .'

a) Tnh th tch khi t din 'BDA M theo a v b .

b) Xc nh t sa

b hai mt phng v( ' )A BD ( )BD vung gc vi nhau.

Cu 4 ( 2 im).

1) Tm h s ca s hng chax8trong khai trin nh thc Niutn ca

n

xx

+ 5

3

1 , bit rng

)3(73

14

+= +++ nCC

nn

nn

( n l s nguyn dng,x> 0, l s t hp chp k ca n phn t).knC

2) Tnh tch phn

+=

32

52 4xx

dxI .

Cu 5 (1 im).Cho x,y, zl ba s dng v x+y+ z1. Chng minh rng

.821

1

1

2

2

2

2

2

2 +++++z

zy

yx

x

HT

Ghi ch: Cn b coi thi khng gii thch g thm.

H v tn th sinh: .. . S bo danh: .


28/148

1

B gio dc v o to k thi tuyn sinh i hc, cao ng nm 2003

p n thang im thi chnh thc Mn thi : ton Khi A

Ni dung imCu 1. 2im

1)

Khi2 1 1

1 .1 1

x xm y x

x x

+ = = =

+ Tp xc nh: \{ 1 }.R

+2

2 2

01 2' 1 . ' 0

2.( 1) ( 1)

xx xy y

xx x

= += + = = =

+ [ ] =

=

01

1lim)(lim

xxy

xxtim cn xin ca th l: y = .

=

yx 1lim tim cn ng ca th l: 1=x .

Bng bin thin:

th khng ct trc honh. th ct trc tung ti im (0; 1).

1 im

0,25

0,5

0, 25

x 0 1 2 + y 0 + + 0

+ + 3y CT C

1

y

xO 1 2

3

1

1


29/148

2

2)

th hm s1

2

++=

x

mxmxy ct trc honh ti 2 im phn bit c honh

dng phng trnh 2( ) 0f x mx x m= + + = c 2 nghim dng phn bit khc 1

2

0

1 4 0

(1) 2 1 0

10, 0

m

m

f m

mS P

m m

= >

= + = > = >

0

1

12 01 2

2

0

m

m

m

m

m

x x x x x .

Cch 2. t 43

1( ) 2 ( ) min ( ) 0

4

= + + = >

xf x x x f x f x f

R

.

Trng hp ny h v nghim.Vy nghim ca h phng trnh l:

1 5 1 5 1 5 1 5( ; ) (1;1), ; , ;

2 2 2 2x y

+ + =

.

0, 25

Cu 3. 3im

1)

Cch 1.t AB a= . Gi Hl hnh chiu vung gc ca B trn AC, suy ra BH AC,m BD (AAC) BD AC, do AC(BHD) ACDH.Vy gc

phng nh din [ ], ' ,B A C D l gc BHD .Xt 'A DC vung ti D c DH l ng cao, ta c . ' . 'DH A C CD A D=

. '

'

CD A DDH

A C =

. 2 2

3 3

a a a

a= = . Tng t, 'A BC vung tiB cBHl ng

cao v2

3

aBH= .

Mt khc:

2 2 2

2 2 2 2 2 2 22 2 . cos 2. cos3 3 3a a aa BD BH DH BH DH BHD BHD= = + = + ,

do 1

cos2

BHD= o120BHD = .

Cch 2. Ta cBDACBD AC (nh l ba ng vung gc).Tng t,BCAC(BCD) AC. Gi H l giao im ca 'A Cv ( ' )BC D

BHD l gc phng ca [ ]; ' ;B A C D .

Cc tam gic vungHAB,HAD, HAC bng nhau HB=HC=HD

H l tm BCD u o120BHD = .

1 im

0, 25

0, 25

0, 25

0, 25

hoc

0, 25

0,25 0,5

A

A

B C

D

D

CB

H

I


31/148

4

2)a) T gi thit ta c

)2

;;();;('0);;;( b

aaMbaaCaaC .

Vy ( ; ; 0), (0; ; )2

bBD a a BM a= =

2, ; ;

2 2

ab abBD BM a

=

.

( )23

' ; 0; , . ' .2

a bBA a b BD BM BA

= =

Do 2

'1

, . '6 4

BDA Ma b

V BD BM BA = =

.

b) Mt phng ( )BDM c vct php tuyn l 21 , ; ;2 2

ab abn BD BM a

= =

,

mt phng ( ' )BD c vct php tuyn l 22 , ' ( ; ; )n BD BA ab ab a = =

.

Do 2 2 2 2

41 2( ) ( ' ) . 0 02 2

a b a bBDM A BD n n a a b = + = =

1ab

= .

2 im

0, 25

0, 25

0, 25

0, 25

0, 5

0, 5

Cu 4. 2im

1)

Ta c ( )1 14 3 3 3 37( 3) 7( 3)n n n n nn n n n nC C n C C C n+ ++ + + + + = + + = + ( 2)( 3)

7( 3) 2 7.2! 14 12.2!

n nn n n

+ + = + + = = =

S hng tng qut ca khai trin l ( )12

5 60 113 2 2

12 12.

kk

kk kC x x C x

=

.

Ta c

60 1182

60 118 4.

2

= = =

kk

x x k

Do h s ca s hng cha 8x l .495)!412(!4

!12412 =

=C

2) Tnh tch phn2 3

2 25 4

xdxI

x x=

+ .

t 22

4

4

dxt x dt

x

= + =+

v 2 2 4.x t=

Vi 5x= th 3t= , vi 2 3x= th 4t= .

Khi 2 3 4 4

22 23 35

1 1 1

4 2 244

xdx dtI dt

t ttx x

= = = + +

4

3

1 2 1 5ln ln .

4 2 4 3

t

t

= = +

1 im

0, 5

0, 25

0, 25

1 im

0, 25

0, 25

0,25

0, 25

A

A

BC

D

D

CB

y

x

z


32/148

5

Cu 5. 1im

Vi mi ,u v

ta c | | | | | | (*)u v u v+ +

(v ( )22 22 2 2| | 2 . | | | | 2 | | . | | | | | |u v u v u v u v u v u v+ = + + + + = +

)

t ,1

;

=

xxa

=

yyb

1; ,

=

zzc

1; .

p dng bt ng thc (*) ta c | | | | | | | | | | | | .a b c a b c a b c+ + + + + +

Vy

22 2 2 2

2 2 2

1 1 1 1 1 1( )P x y z x y z

y zx y z

= + + + + + + + + + +

.

Cch 1. Ta c

( )22

22 3 31 1 1 1 9

( ) 3 3 9P x y z xyz tx y z xyz t

+ + + + + + = +

, vi

( )

22

3 1

0 3 9

x y z

t xyz t

+ + = <

.

t2

9 9 1( ) 9 '( ) 9 0, 0; ( )

9Q t t Q t t Q t

t t

= + = <

gim trn1

0;9

1( ) 82.

9Q t Q

=

Vy ( ) 82.P Q t

(Du = xy ra khi 13

x y z= = = ).Cch 2.

Ta c2 2

2 2 21 1 1 1 1 1( ) 81( ) 80( )x y z x y z x y zx y z x y z

+ + + + + = + + + + + + +

21 1 118( ) 80( ) 162 80 82.x y z x y zx y z

+ + + + + + =

Vy 82.P

(Du = xy ra khi 13

x y z= = = ).Ghi ch:Cu ny cn c nhiu c ch gii khc.

0, 25

0, 25

0, 25

0, 25

hoc

0,25

0,5


33/148

B gio dc v o to k thi tuyn sinh i hc, cao ng nm 2003----------------------- Mn thi : ton khi B

chnh thc Thi gian lm bi: 180 pht_______________________________________________

Cu 1(2 im). Cho hm s ( l tham s).3 23 (1)y x x m= + m 1) Tm th hm s (1) c hai im phn bit i xng vi nhau qua gc ta .m 2) Kho st s bin thin v v th hm s (1) khi m =2.Cu 2(2 im).

1) Gii phng trnh2

otg tg 4sin 2sin2

x x xcx

+ = .

2) Gii h phng trnh

2

2

2

2

23

23 .

yy

x

xx

y

+=

+=

Cu 3(3 im).1) Trong mt phng vi h ta cac vung gc Ox cho tam gic cy ABC

0, 90 .AB AC BAC= = Bit (1; 1)M l trung im cnh BCv2

; 03

G l trng

tm tam gic . Tm ta cc nh .

ABC , ,A B C

2) Cho hnh lng tr ng c y l mt hnh thoi cnh ,

gc

. ' ' ' 'ABCD A B C D ABCD a

060BAD= . Gi l trung im cnh v l trung im cnh ' .Chng minh rng bn im

' NAA CC

', , ,B M D N

'

cng thuc mt mt phng. Hy tnh

di cnh ' theo a t gicAA B MDN l hnh vung.

3) Trong khng gian vi h ta cac vung gc Ox cho hai im

v im sao cho . Tnh khong cch t

trung im

yz

0)(2; 0; 0), (0; 0; 8)A B C (0; 6;AC

=

I ca BCn ng thng OA .Cu 4 (2 im).

1) Tm gi tr ln nht v nh nht ca hm s 24 .y x x= +

2) Tnh tch phn

4 2

0

1 2sin

1 sin 2I dxx

= + .Cu 5 (1 im). Cho l s nguyn dng. Tnh tngn

2 3 10 1 22 1 2 1 2 1

2 3 1

nn

n n nC C Cn

+ + + + +

+ nC

( C l s t hp chp kca phn t).kn n

----------------------------------Ht---------------------------------

Ghi ch: Cn b coi thi khng gii thch g thm.

H v tn th sinh.. S bo danh


34/148

1


p n thang im thi chnh thc Mn thi : ton Khi B

Ni dung imCu 1. 2im

1) th hm s (1) c hai im phn bit i xng nhau qua gc ta tn ti 0 0x sao cho 0 0( ) ( )y x y x=

tn ti 0 0x sao cho3 2 3 2

0 0 0 03 ( ) 3( )x x m x x m + = +

tn ti 0 0x sao cho2

03x m=

0m > .2) Kho st s bin thin v v th hm s khi m = 2.

Khi 2m= hm s tr thnh 3 23 2.y x x= +

Tp xc nh : .

2 0' 3 6 , ' 02.

xy x x yx

== = =

" 6 6. '' 0 1.y x y x= = =

"y trit tiu v i du qua 1 (1;0)x= l im un.

Bng bin thin:

th ct trc honh ti cc im (1; 0), (1 3; 0) v ct trc tung ti im (0;2) .

1 im0, 25

0, 25

0,25

0,25 1 im

0,25

0,25

0,25

0,25

x 0 2 + y + 0 0 +

2 + C CT

y 2

x

y

O

2

21

2


35/148

2

Cu 2. 2im

1) Gii phng trnh:2

cotg tg 4sin 2 (1).sin2

x x xx

+ =

iu kin:sin 0

(*).cos 0

x

x

Khi (1)cos sin 2

4sin2sin cos sin 2

x xx

x x + =

2 2cos sin 24sin2

sin cos sin 2

x xx

x x x

+ =

22cos 2 4sin 2 2x x + = 22cos 2 cos 2 1 0x x =

cos 2 1

1cos2

32

x kx

kx

== = +=

( )kZ .

Kt hp vi iu kin (*) ta c nghim ca (1) l

( ).3

x k k= + Z

2) Gii h phng trnh

2

2

2

2

23 (1)

2

3 (2).

yy

x

x

x y

+=

+=

iu kin 0, 0x y .

Khi h cho tng ng vi2 2

2 22 2

( )(3 ) 03 2

3 2.3 2

x y xy x yx y y

xy xxy x

+ + == +

= += +

TH1:2 2

1

1.3 2

x y x

yxy x

= =

== +

TH2:2 2

3 0

3 2

xy x y

xy x

+ + =

= + v nghim, v t (1) v (2) ta c , 0x y> .

Vy nghim ca h phng trnh l: 1.x y= =

1 im

0,25

0,25

0,25

0,25

1 im

0,25

0,5

0,25

Cu 3. 3im1)V G l trng tm BC v l trung im BCnn

3 ( 1;3)MA MG= =

(0;2)A .

Phng trnh BC i qua (1; 1)M v vung gc vi

( 1, 3)MA=

l: 1( 1) 3( 1) 0 3 4 0 (1).x y x y + + = + + =

Ta thy 10MB MC MA= = = ta ,B C tha mn

phng trnh: 2 2( 1) ( 1) 10 (2).x y + + =

Gii h (1),(2) ta

c ta ca ,B Cl (4; 0), ( 2; 2).

2)Ta c ' // 'A M NC A MCN= l hnh bnh hnh,do 'A C v MN ct nhau ti trung im I cami ng. Mt khcADCBl hnh bnh hnh nntrung imI caACcng chnh l trung im ca

BD.Vy MN v BD ct nhau ti trung im Icami ng nnBMDNl hnh bnh hnh. Do B,

M, D, Ncng thuc mt mt phng.Mt khc DM2 = DA2+ AM2= DC2 + CN2= DN2,

hay DM = DN. Vy hnh bnh hnh BMDN l hnh thoi. Do BMDN l hnh

1 im

0,25

0,25

0,250,251 im

0,5

GA

B

C

M.

D

A

D C

B N

M

A B

C


36/148

3

vung MN = BD AC = BD AC2= BD2= BB2+BD23a2 = BB2+ a2

BB= 2a AA= 2a .3)

T (0;6;0)AC=

vA(2; 0; 0) suy ra C(2; 6; 0), do I(1; 3; 4).

Phng trnh mt phng () quaIv vung gc vi OA l : 1 0.x = ta giao im ca () vi OA lK(1; 0; 0).

khong cch tIn OAl 2 2 2(1 1) (0 3) (0 4) 5.IK= + + =

0,5

1 im0,25

0,25

0,25

0,25

Cu 4. 2im

1) Tm gi tr ln nht v nh nht ca hm s 24 .y x x= +

Tp xc nh: [ ]2; 2 .

2' 1

4

xy =

,

22 2

0' 0 4 2

4

xy x x x

x x

= = =

=.

Ta c ( 2) 2, ( 2) 2 2, (2) 2y y y = = = ,

Vy[ 2;2]max ( 2) 2 2y y

= = v[ 2;2]min ( 2) 2y y

= = .

2) Tnh tch phn

4 2

0

1 2sin.

1 sin 2I dx

x

=

+

Ta c

4 42

0 0

1 2sin cos 2

1 sin 2 1 sin 2

x xI dx dx

x x

= =

+ + .

t 1 sin 2 2cos 2t x dt xdx= + = .

Vi 0x= th 1,t= vi

4x= th 2t= .

Khi 2

1

21 1 1ln | | ln 2.

12 2 2

dtI t

t= = =

1 im

0,25

0,25

0,25

0,25

1 im

0,25

0,25

0,25

0,25

Cu 5. 1im

Ta c 0 1 2 2(1 ) ...n n nn n n nx C C x C x C x+ = + + + + .

Suy ra ( )2 2

0 1 2 2

1 1

(1 ) ...n n nn n n nx dx C C x C x C x dx+ = + + + +

22 2 3 1

1 0 1 2

11

1 (1 ) ...1 2 3 1

n

n nn n n nx x xx C x C C Cn n

+

+

+ = + + + + + +

2 3 1 1 10 1 22 1 2 1 2 1 3 2

2 3 1 1

n n nn

n n n nC C C C n n

+ + + + + + + =

+ + .

0,5

0,5


37/148

B gio dc v o to k thi tuyn sinh i hc, cao ng nm 2003---------------------- Mn thi: ton Khi D chnh thc Thi gian lm bi: 180 pht

_______________________________________________

Cu 1 (2 im).1) Kho st s bin thin v v th ca hm s

2 2 4 (1)

2

x xy

x

+=

.

2) Tm ng thng d ym : 2 2m mx m= + ct th ca hm s (1) ti hai im

phn bit.Cu 2 (2 im).

1) Gii phng trnh 2 2 2

sin tg cos 02 4 2

x xx

=

.

2) Gii phng trnh .2 222 2x x x x + = 3

Cu 3 (3 im).1) Trong mt phng vi h ta cac vung gc cho ng trnOxy

4)2()1(:)( 22 =+ yxC v ng thng : 1 0d x y = .

Vit phng trnh ng trn ( i xng vi ng trn qua ng thng

Tm ta cc giao im ca v .

')C

(C

( )C .d

) ( ')C

2) Trong khng gian vi h ta cac vung gc Oxyz cho ng thng

3 2:

1 0.k

x ky zd

kx y z

0+ + =

+ + =

Tm ng thng vung gc vi mt phngk kd ( ) : 2 5 0P x y z + = .

3) Cho hai mt phng v vung gc vi nhau, c giao tuyn l ng thng( )P ( )Q .Trn ly hai im vi ,A B AB a= . Trong mt phng ly im , trong

mt phng ( ly im sao cho ,

( )P C

)Q D AC BD cng vung gc vi v

. Tnh bn knh mt cu ngoi tip t din v tnh khongcch t n mt phngAC BD

A

AB== ABCD( )BCD theo .a

Cu 4 ( 2 im).

1) Tm gi tr ln nht v gi tr nh nht ca hm s2

1

1

xy

x

+=

+ trn on [ ]1; 2 .

2) Tnh tch phn2

2

0

I x x d= x .

Cu 5 (1 im).

Vi l s nguyn dng, gin 3 3na l h s ca3 3nx trong khai trin thnh a

thc ca ( 1 . Tm n 2 ) ( 2)nx x+ + n 3 3 26na n= .

------------------------------------------------ Ht ------------------------------------------------Ghi ch: Cn b coi thi khng gii thch g thm.

H v tn th sinh:.. . S bo danh:


38/148


p n thang im thi chnh thc Mn thi : ton Khi D

Ni dung im

Cu 1. 2im

1) Kho st s bin thin v v th ca hm s2 2 4

2

x xy

x

+=

. 1 im

Tp xc nh :R \{ 2 }.

Ta c2 2 4 4

.2 2

x xy x

x x

+= = +

2

2 2

04 4' 1 . ' 0

4.( 2) ( 2)

xx xy y

xx x

== = = =

[ ] 4

lim lim 02x xy x x = = tim cn xin ca th l: y x= ,

tim cn ng ca th l:2

limx

y

= 2x= .

Bng bin thin:

th khng ct trc honh. th ct trc tung ti im (0; 2).

0,25

0,5

0,25

2) 1 im

ng thng ct th hm s (1) ti 2 im phn bitmd

phng trnh4

2 22

x mx mx

+ = +

c hai nghim phn bit khc 2

2( 1)( 2) 4m x = c hai nghim phn bit khc 2 1 0m > 1.m >

Vy gi tr cn tm lm 1.m>

0,5

0,5

x

2

6

2

2 4O

y

x 0 2 4 + y + 0 0 +

2 + + y C CT

6

1


39/148

Cu 2. 2im

1) Gii phng trnh 2 2 2

tg cos 02 4 2

x xx

sin (1)= 1 im

iu kin: (*). Khi cos 0x

( )2

2

1 sin 1(1) 1 cos 1 cos

2 2 2cos

xx x

x

= +

( ) ( )2 21 sin sin 1 cos cosx x = + x

( ) ( )1 sin (1 cos )(1 cos ) 1 cos (1 sin )(1 sin )x x x x + = + + x

( )1 sin (1 cos )(sin cos ) 0x x x x + + =

2sin 1 2

cos 1 2

tg 1

4

x kx

x x k

xk

= +=

= = + = = +

( )kZ .

Kt hp iu kin (*) ta c nghim ca phng trnh l:

2

4

k

k

= + = +

( ) .kZ

0,5

0,25

0,25

2) Gii phng trnh (1).2 222 2x x x x + 3= 1 im

t .2

2 0x xt t= >

Khi (1) tr thnh 24

3 3 4 0 ( 1)( 4) 0t t t t t t

= = + = =4t (v t )0>

Vy2 22 4x x x x = =2

1

2.

= =

x

x

Do nghim ca phng trnh l1

2.

= =

x

x

0,5

0,5

Cu 3. 3im

1) 1 im

T ( ) suy ra c tm v bn knh2 2: ( 1) ( 2) 4 + =C x y ( )C (1;2)I 2.R=

ng thng c vct php tuyn l nd (1; 1).= uur

Do ng thng i qua

v vung gc vi d c phng trnh:(1;2)I 1 2

1 1

x yx y 3 0

= +

= .

Ta giao im ca v l nghim ca h phng trnh:H d

1 0 2 (2;1).3 0 1

x y x Hx y y

= = + = =

Gi l im i xng vi qua . Khi J (1;2)I d

2 3(3;0)

2 0J H I

J H I

x x xJ

y x x

= =

= =.

V i xng vi ( qua nn c tm l v bn knh

Do c phng trnh l:

( ')C

(C

)C d ( ')C

2 2

(3;0)J 2.R=

') ( 3) 4 +x y = .

Ta cc giao im ca ( v l nghim ca h phng trnh:)C ( ')C

2 2

2 2 22 2

1 0 1( 1) ( 2) 4 1, 0

3, 2.( 3) 4 2 8 6 0( 3) 4

x y y xx y x y

x yx y x xx y

= = + = = = = = + = + = + =

Vy ta giao im ca v ( l v( )C ')C (1;0)A (3;2).B

0,5

0,25

0,25

2


40/148

2) 1 im

Ta c cp vect php tuyn ca hai mt phng xc nh lkd 1 (1;3 ; 1)= uur

n k

v . Vect php tuyn ca l2 ( ; 1;1)= uur

n k ( )P (1; 1; 2)= r

n .

ng thng c vect ch phng l:kd

21 2, (3 1; 1; 1 3 ) 0 k k k

r

Nn21 1 3

1.1 1 2

k k kk = = =

Vy gi tr cn tm l

0,5

0,5

3) 1 imTa c (P) (Q) v = (P) (Q), mAC AC (Q) AC AD, hay

. Tng t, ta cBDnn

BD(P), do CBD . VyAvB

A,Bnm trn mt cu ng knh CD.

090=CAD 090=

V bn knh ca mt cu l:

2 21

2 2

CDR BC BD= = +

2 2 21 3

2 2

aAB AC BD= + + = .

GiHl trung im caBCAHBC. DoBD(P) nn BD AHAH(BCD).

VyAHl khong cch t A n mt phng (BCD)v1 2

.2 2

aAH BC= =

0,25

0,25

0,5

Cu 4. 2im

1) Tm gi tr ln nht v gi tr nh nht ca hm s2

1

1

xy

x

+=

+trn on [ ]1; 2 . 1 im

2 3

1' .

( 1)

xy

x

=

+

' 0 1y x= = .

Ta c 3

( 1) 0, 2, (2) .5

y(1)y y = = =

Vy [ ]1;2 (1) 2max y y = = v [ ]1;2min ( 1) 0.y y = =

0,5

0,5

2) Tnh tch phn2

2

0

I x x d= x . 1 im

Ta c 2 0 0 1x x , suy ra1 2

2 2

0 1

( ) ( )= + I x x dx x x dx

1 22 3 3 2

0 1

1.2 3 3 2

= + =

x x x x

0,5

0,5

u n n k = =

r uur uur

3 1( ) || kd P u n

r r

k 1.=k

.

A B

C

D

P

Q

H

3


41/148

Cu 5. 1im

Cch 1:Ta c ( 2 0 2 1 2 2 2 2 41) ...n n n nn n nnnC x C x C x C

+ = + + + + ,

0 1 1 2 2 2 3 3 3( 2) 2 2 2 ... 2n n n n n nn n n nnnx C x C x C x C x C

+ = + + + + + .

D dng kim tra 1, 2= =n n khng tha mn iu kin bi ton.

Vi th3n 3 3 2 3 2 2 1.n n n n nx x x x x = =

Do h s ca 3 3nx trong khai trin thnh a thc ca l2( 1) ( 2+ +n nx x )

nC3 0 3 1 1

3 3 2 . . 2. .n n n na C C C = + .

Vy2

3 3

52 (2 3 4)

26 26 73

2

= + = =

=

n

nn n n

a n nn

Vy l gi tr cn tm (v nguyn dng).5=n nCch 2:

Ta c

2 3 2

3 3 2

20 0 0 0

1 2( 1) ( 2) 1 1

1 22 .

n n

n n n

i kn n n nn i k n i i k k k

n n n ni k i k

x x x xx

x C C x C x C xxx

= = = =

+ + = + +

= =

Trong khai trin trn, lu tha ca l 3 3n khi 2 3i k = 3k

, hayTa ch c hai trng hp tha iu kin ny l

2 3i k+ = .0,i= = hoc i 1, 1k= = .

Nn h s ca 3 3nx l .0 3 3 1 13 3 . .2 . .2n n n n na C C C C = +

Do 2

3 3

52 (2 3 4)26 26 7

32

= + = =

=

n

nn n na n n

n

Vy l gi tr cn tm (v nguyn dng).5=n n

0,75

0,25

hoc

0,75

0,25

4


42/148

B gio dc v o to thi tuyn sinh i hc, cao ng nm 2004------------------------------ Mn thi : Ton , Khi A

chnh thc Thi gian lm bi : 180 pht, khng k thi gian pht --------------------------------------------------------------

Cu I (2 im)

Cho hm s2x 3x 3

y2(x 1)

+ =

(1).

1) Kho st hm s (1).2) Tm m ng thng y = m ct th hm s (1) ti hai im A, B sao cho AB = 1.

Cu II (2 im)

1) Gii bt phng trnh22(x 16) 7 x

x 3 >x 3 x 3

+

.

2)

Gii h phng trnh 1 442 2

1

log (y x) log 1y

x y 25.

=

+ =

Cu III (3 im)

1) Trong mt phng vi h ta Oxy cho hai im ( )A 0; 2 v ( )B 3; 1 . Tm ta trctm v ta tm ng trn ngoi tip ca tam gic OAB.

2) Trong khng gian vi h ta Oxyz cho hnh chp S.ABCD c y ABCD l hnh thoi,

AC ct BD ti gc ta O. Bit A(2; 0; 0), B(0; 1; 0), S(0; 0; 2 2 ). Gi M l trung im

ca cnh SC.a) Tnh gc v khong cch gia hai ng thng SA, BM.

b) Gi s mt phng (ABM) ct ng thng SD ti im N. Tnh th tch khi chp S.ABMN.

Cu IV (2 im)

1) Tnh tch phn I =2

1

xdx

1 x 1+ .

2) Tm h s ca x8trong khai trin thnh a thc ca8

21 x (1 x) + .

Cu V (1 im)

Cho tam gic ABC khng t, tha mn iu kin cos2A + 2 2 cosB + 2 2 cosC = 3.Tnh ba gc ca tam gic ABC.

------------------------------------------------------------------------------------------------------------------------

Cn b coi thi khng gii thch g thm.

H v tn th sinh............................................................................S bo danh.................................................


43/148

1

B gio dc v o to p n - Thang im..................... thi tuyn sinh i hc, cao ng nm 2004

........................................... chnh thc Mn:Ton, Khi A

(p n - thang im c 4 trang)

Cu Ni dung imI 2,0

I.1 (1,0 im)

( )12332

+=

x

xxy =

( )

1 1x 1

2 2 x 1 +

.

a) Tp xc nh: { }R \ 1 .

b) S bin thin:

2

x(2 x)y '

2(x 1)

=

; y ' 0 x 0, x 2= = = . 0,25

yC= y(2) =1

2 , yCT= y(0) =

3

2.

ng thng x = 1 l tim cn ng.

ng thng 1

y x 12

= + l tim cn xin. 0,25

Bng bin thin:x 0 1 2 +

y'

0 + + 0

y + + 1

2

3

2

0,25

c) th:

0,25


44/148

2

I.2 (1,0 im)Phng trnh honh giao im ca th hm s vi ng thng y = ml :

( ) m

x

xx=

+

12

332 ( ) 023322 =++ mxmx (*). 0,25

Phng trnh (*) c hai nghim phn bit khi v ch khi:

0> 24m 4m 3 0 > 3m2

> hoc 1m2

< (**). 0,25

Vi iu kin (**), ng thng y = m ct th hm s ti hai im A, B c honh x1, x2 l nghim ca phng trnh (*).

AB = 1 121 = xx 2

1 2x x 1 = ( )1 22

1 2x x 4x x 1+ = 0,25

( ) ( ) 123432 2 = mm

1 5m

2

= (tho mn (**)) 0,25

II 2,0

II.1 (1,0 im)

iu kin : x 4 . 0,25

Bt phng trnh cho tng ng vi bt phng trnh:2 2

2(x 16) x 3 7 x 2(x 16) 10 2x + > > 0,25

+ Nu x > 5 th bt phng trnh c tho mn, v v tri dng, v phi m. 0,25

+ Nu 4 x 5 th hai v ca bt phng trnh khng m. Bnh phng hai v ta

c: ( ) ( )22 22 x 16 10 2x x 20x 66 0 > + < 10 34 x 10 34 < < + .

Kt hp vi iu kin 4 x 5 ta c: 10 34 x 5 < . p s: x 10 34> 0,25

II.2 (1,0 im)

iu kin: y > x v y > 0.

( ) 11

loglog 44

1 =y

xy ( ) 11

loglog 44 =y

xy 0,25

4y x

log 1y

=

4

3yx = . 0,25

Th vo phng trnh x2+ y2= 25 ta c:

2

23y y 25 y 4.4

+ = =

0,25

So snh vi iu kin , ta c y = 4, suy ra x= 3 (tha mn y > x).Vy nghim ca h phng trnh l (3; 4). 0,25

III 3,0 III.1 (1,0 im)

+ ng thng qua O, vung gc vi BA( 3 ; 3)

c phng trnh 3x 3y 0+ = .

ng thng qua B, vung gc vi OA(0; 2)

c phng trnh y = 1

( ng thng qua A, vung gc vi BO( 3 ; 1)

c phng trnh 3x y 2 0+ = )0,25

Gii h hai (trong ba) phng trnh trn ta c trc tm H( 3; 1) 0,25

+ ng trung trc cnh OA c phng trnh y = 1.ng trung trc cnh OB c phng trnh 3x y 2 0+ + = .

( ng trung trc cnh AB c phng trnh 3x 3y 0+ = ).0,25


45/148

3

Gii h hai (trong ba) phng trnh trn ta c tm ng trn ngoi tip tam gicOAB l ( )I 3 ; 1 . 0,25

III.2.a (1,0 im)

+ Ta c: ( )C 2; 0; 0 , ( )D 0; 1; 0 , ( )2;0;1M ,

( )22;0;2 =SA , ( )BM 1; 1; 2=

. 0,25 Gi l gc gia SA v BM.

Ta c: ( ) SA.BM 3

cos cos SA, BM2SA . BM

= = =

30 = .0,25

+ Ta c: ( )SA, BM 2 2; 0; 2 =

, ( )AB 2; 1; 0=

. 0,25

Vy:

( )SA, BM AB 2 6

d SA,BM

3SA,BM

= =

0,25

III.2.b (1,0 im)

Ta c MN // AB // CD N l trung im SD

2;

2

1;0N .

0,25

( )SA 2; 0; 2 2=

, ( )2;0;1 =SM , ( )22;1;0 =SB , 1SN 0; ; 22

=

( )SA, SM 0; 4 2; 0 =

. 0,25

S.ABM 1 2 2V SA,SM SB6 3

= =

0,25

S.AMN

1 2V SA,SM SN

6 3 = =

S.ABMN S.ABM S.AMNV V V 2= + = 0,25

IV 2,0 IV.1 (1,0 im)

2

1

xI dx

1 x 1=

+ . t: 1= xt 1

2+=tx tdtdx 2= .

01 == tx , 12 == tx . 0,25


46/148

4

Ta c:

1 1 12 32

0 0 0

t 1 t t 2I 2t dt 2 dt 2 t t 2 dt

1 t 1 t t 1

+ + = = = +

+ + +

0,25

I

1

3 2

0

1 12 t t 2t 2 ln t 1

3 2

= + +

0,25

1 1 11I 2 2 2 ln 2 4ln 23 2 3 = + =

. 0,25

IV.2 (1, 0 im)

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

8 2 3 42 0 1 2 2 4 3 6 4 8

8 8 8 8 8

5 6 7 85 10 6 12 7 14 8 16

8 8 8 8

1 x 1 x C C x 1 x C x 1 x C x 1 x C x 1 x

C x 1 x C x 1 x C x 1 x C x 1 x

+ = + + + +

+ + + +

0,25

Bc ca x trong 3 s hng u nh hn 8, bc ca x trong 4 s hng cui ln hn 8. 0,25

Vy x8 ch c trong cc s hng th t, th nm,vi h s tng ng l:3 2 4 0

8 3 8 4C .C , C .C 0,25

Suy ra a8 168 70 238= + = . 0,25V 1,0

Gi 3cos22cos222cos ++= CBAM

32

cos2

cos2221cos2 2

+

+= CBCB

A . 0,25

Do 0

2sin >

A, 1

2cos

CBnn 2

AM 2cos A 4 2 sin 4

2 + . 0,25

Mt khc tam gic ABC khng t nn 0cos A , AA coscos2 . Suy ra:

42sin24cos2 + A

AM 42sin242sin212 2 +

=

AA

22

sin242

sin4 2 += AA

012

sin222

=

A. Vy 0M . 0,25

Theo gi thit: M = 0

=

=

=

2

1

2sin

12

cos

coscos2

A

CB

AA

A 90

B C 45

=

= =

0,25


47/148

B gio dc v o to------------------------

chnh thc

thi tuyn sinh i hc, cao ng nm 2004Mn: Ton, Khi B

Thi gian lm bi: 180 pht, khng k thi gian pht -------------------------------------------

Cu I(2 im)

Cho hm s y = xxx 3231 23 + (1) c th (C).

1) Kho st hm s (1).2)Vit phng trnh tip tuyn ca (C) ti im un v chng minh rng l tip tuyn ca (C)

c h s gc nh nht.

Cu II(2 im)

1)

Gii phng trnh xtgxx 2)sin1(32sin5 = .

2) Tm gi tr ln nht v gi tr nh nht ca hm s

x

xy

2ln= trn on [1; 3e ].

Cu III(3 im)1) Trong mt phng vi h ta Oxy cho hai im A(1; 1), B(4; 3 ). Tm im C thuc ng

thng 012 = yx sao cho khong cch t C n ng thng AB bng 6.

2) Cho hnh chp t gic u S.ABCD c cnh y bng a, gc gia cnh bn v mt y bng

( o0 < < o90 ). Tnh tang ca gc gia hai mt phng (SAB) v (ABCD) theo . Tnh thtch khi chp S.ABCD theo a v .

3) Trong khng gian vi h ta Oxyz cho im A )4;2;4( v ng thng d:

+=

=

+=

.41

1

23

tz

ty

tx

Vit phng trnh ng thng i qua im A, ct v vung gc vi ng thng d.

Cu IV(2 im)

1) Tnh tch phn I = dxx

xxe

+

1

lnln31.

2) Trong mt mn hc, thy gio c 30 cu hi khc nhau gm 5 cu hi kh, 10 cu hi trungbnh, 15 cu hi d. T 30 cu hi c th lp c bao nhiu kim tra, mi gm 5 cu

hi khc nhau, sao cho trong mi nht thit phi c 3 loi cu hi (kh, trung bnh, d) vs cu hi d khng t hn 2 ?

Cu V(1 im)Xc nh m phng trnh sau c nghim

22422 1112211 xxxxxm ++=

++ .

------------------------------------------------------------------------------------------------------------------------

Cn b coi thi khng gii thch g thm.

H v tn th sinh.................................................................................................S bo danh...........................


48/148

1


........................................... chnh thc Mn:Ton,Khi B

(p n - thang im c 4 trang)

Cu Ni dung imI 2,0

1 Kho st hm s (1,0 im)

3 21y x 2x 3x3

= + (1).

a) Tp xc nh: R.b) S bin thin:y' = x2 4x + 3; 3,10' === xxy . 0,25

yC= y(1) = 43 , yCT= y(3) = 0; y" = 2x 4, y'' = 0 ( ) 2x 2, y 2 3 = = . th

hm s li trn khong ( ; 2), lm trn khong ( 2; + ) v c im un l

2U 2;

3

.

0,25

Bng bin thin:x 1 3 +

y' + 0 0 +

y 43

+

0

0,25

c) th:Giao im ca th vi cc trcOx, Oy l cc im ( ) ( )0;0 , 3;0 .

0,25


49/148

2

2 Vit phng trnh tip tuyn ca (C) ti im un, ...(1,0 im)

Ti im un U 22;

3

, tip tuyn ca (C) c h s gc 1)2(' =y . 0,25

Tip tuyn ti im un ca th (C) c phng trnh:

2 8y 1.(x 2) y x

3 3

= + = + . 0,25

H s gc tip tuyn ca th (C) ti im c honh x bng:

y'(x) = x2 34 + x = 1)2( 2 x 1 y' (x) y' (2), x.0,25

Du " =" xy ra khi v ch khi x = 2 ( l honh im un).Do tip tuyn ca th (C) ti im un c h s gc nh nht.

0,25

II 2,01 Gii phng trnh (1,0 im)

5sinx 2 = 3 tg2x ( 1 sinx ) (1) .

iu kin: cosx 0 x k , k Z2

+ (*). 0,25

Khi (1) 2

2

3sin x5sin x 2 (1 sin x)

1 sin x =

02sin3sin2 2 =+ xx . 0,25

2

1sin = x hoc 2sin =x (v nghim).

0,25

+

== 262

1sin kxx hoc +

= 2

6

5kx , Zk ( tho mn (*)).

0,25

2 Tm gi tr ln nht v gi tr nh nht ca hm s (1,0 im)

y =2ln x

x

2

ln x(2 ln x )y '

x

= 0,25

y'= 03

2 3

ln x 0 x 1 [1; e ]

ln x 2 x e [1; e ].

= = = =

0.25

Khi : y(1) = 0, 2 32 3

4 9y(e ) , y(e )

e e= =

0,25

So snh 3 gi tr trn, ta c:33

2

2[1; e ][1; e ]

4max y khi x e , min y 0 khi x 1

e= = = = .

0,25

III 3,01 Tm im C (1,0 im)

Phng trnh ng thng AB:4

1

3

1

=

yx 4x + 3y 7 = 0. 0,25

Gi s );( yxC . Theo gi thit ta c: 012 = yx (1).

d(C, (AB)) = 62 2

4x 3y 37 0 (2a)4x 3y 76

4x 3y 23 0 (2b).4 3

+ =+ = + + =+

0,25

Gii h (1), (2a) ta c: C1( 7 ; 3). 0,25

Gii h (1), (2b) ta c: 2 43 27C ;11 11 . 0,25

2 Tnh gc v th tch (1,0 im)


50/148

3

Gi giao im ca AC v BD lO th SO (ABCD) , suy ra

SAO= .

Gi trung im ca AB l M thOM AB v ABSM Gcgia hai mt phng (SAB) v(ABCD) lSMO .

0,25

Tam gic OAB vung cn ti O, nn === tg

aSO

aOA

aOM

2

2

2

2,

2.

Do : SO

tgSMO 2 tgOM= = . 0,25

2 3

S.ABCD ABCD

1 1 a 2 2V S .SO a tg a tg .

3 3 2 6= = = 0,50

3 Vit phng trnh ng thng (1,0 im)

ng thng d c vect ch phng )4;1;2( =v . 0,25 B d )41;1;23( tttB ++ (vi mt s thc t no ).

( )AB 1 2t;3 t; 5 4t = + +

. 0,25

AB d 0. =vAB 2(1 2t) (3 t) 4( 5 4t) 0 + + + = t = 1. 0,25

AB (3; 2; 1) =

Phng trnh ca1

4

2

2

3

4:

=

+=

+

zyx. 0,25

IV 2,01 Tnh tch phn (1,0 im)

dxx

xxI

e

+

=1

lnln31.

t: 2 dx

t 1 3ln x t 1 3ln x 2tdt 3x

= + = + = .

x 1 t 1= = , x e t 2= = . 0,25Ta c: ( )

2 222 4 2

1 1

2 t 1 2I t dt t t dt

3 3 9

= = .

0,25

25 3

1

2 1 1I t t

9 5 3

=

.

0,25

I =

135

116.

0,25


51/148

4

2 Xc nh s kim tra lp c ... (1,0 im)Mi kim tra phi c s cu d l 2 hoc 3, nn c cc trng hp sau: c 2 cu d, 2 cu trung bnh, 1 cu kh, th s cch chn l:

23625.. 15210

215 =CCC . 0,25

c 2 cu d, 1 cu trung bnh, 2 cu kh, th s cch chn l:

10500..25

110

215 =CCC .

0,25 c 3 cu d, 1 cu trung bnh, 1 cu kh, th s cch chn l:

22750.. 15110

315 =CCC . 0,25

V cc cch chn trn i mt khc nhau, nn s kim tra c th lp c l:56875227501050023625 =++ . 0,25

V Xc nh m phng trnh c nghim 1,0 iu kin: 1 x 1.t t 2 21 x 1 x= + .

Ta c: 2 21 x 1 x t 0+ , t = 0 khi x = 0.2 4t 2 2 1 x 2 t 2= , t = 2 khi x = 1.

Tp gi tr ca t l [0; 2 ] ( t lin tc trn on [ 1; 1]). 0,25

Phng trnh cho tr thnh: m ( ) 2t 2 t t 2+ = + +

2t t 2m

t 2

+ + =

+(*)

Xt f(t) =2t t 2

t 2

+ +

+vi 0 t 2 . Ta c f(t) lin tc trn on [0; 2 ].

Phng trnh cho c nghim x Phng trnh (*) c nghim t [0; 2 ]

]2;0[]2;0[

)(max)(min tfmtf .0,25

Ta c: f '(t) = ( )

2

2

t 4t0, t 0; 2

t 2

+ f(t) nghch bin trn [0;2 ].

0,25

Suy ra:[0; 2 ] [0; 2 ]

min f (t) f ( 2) 2 1 ; max f (t) f (0) 1= = = = .

Vy gi tr ca m cn tm l 2 1 m 1 . 0,25


52/148

B gio dc v o to thi tuyn sinh i hc, cao ng nm 2004------------------------ Mn: Ton, Khi D chnh thc Thi gian lm bi: 180 pht, khng k thi gian pht

-------------------------------------------Cu I (2 im)

Cho hm s 3 2y x 3mx 9x 1= + + (1) vi m l tham s.1) Kho st hm s (1) khi m = 2.2) Tm m im un ca th hm s (1) thuc ng thng y = x + 1.

Cu II(2 im)1) Gii phng trnh .sin2sin)cossin2()1cos2( xxxxx =+

2) Tm m h phng trnh sau c nghim

=+

=+

.31

1

myyxx

yx

Cu III(3 im)1) Trong mt phng vi h ta Oxy cho tam gic ABC c cc nh );0();0;4();0;1( mCBA

vi 0m . Tm ta trng tm G ca tam gic ABC theo m. Xc nh m tam gic GABvung ti G.

2) Trong khng gian vi h ta Oxyz cho hnh lng tr ng 111. CBAABC . Bit ),0;0;(aA

0,0),;0;(),0;1;0(),0;0;( 1 >> babaBCaB .

a) Tnh khong cch gia hai ng thng CB1 v 1AC theo .,ba

b) Cho ba, thay i, nhng lun tha mn 4=+ba . Tm ba, khong cch gia hai ng

thng CB1 v 1AC ln nht.

3) Trong khng gian vi h ta Oxyz cho ba im )1;1;1(),0;0;1(),1;0;2( CBA v mtphng (P): 02 =++ zyx . Vit phng trnh mt cu i qua ba im A, B, C v c tmthuc mt phng (P).

Cu IV(2 im)

1) Tnh tch phn I = 3

2

2 )ln( dxxx .

2) Tm cc s hng khng cha x trong khai trin nh thc Niutn ca

7

43 1

+

x

x vi x > 0.

Cu V(1 im)Chng minh rng phng trnh sau c ng mt nghim

01225 = xxx .

---------------------------------------------------------------------------------------------------------------------Cn b coi thi khng gii thch g thm.

H v tn th sinh.............................................................S bo danh........................................


53/148

1


........................................... chnh thc Mn:Ton, Khi D

(p n - thang im c 4 trang)Cu

Ni dung im

I 2,0

1 Kho st hm s (1,0 im)

1962 23 ++== xxxym .a) Tp xc nh: R .b) S bin thin:

2 2y ' 3x 12x 9 3(x 4x 3)= + = + ; y ' 0 x 1, x 3= = = . 0,25yC= y(1) = 5 , yCT= y(3) =1. y'' = 6x 12 = 0 x = 2 y = 3. th hms li trn khong ( ; 2), lm trn khong );2( + v c im un l

)3;2(U . 0,25Bng bin thin:

x 1 3 +

y' + 0 0 +

y 5 +

10,25

c) th:

th hm s ct trc Oy ti im (0; 1).

0,252 Tm m im un ca th hm s ...(1,0 im)

y = x3 3mx2+ 9x + 1 (1); y' = 3x2 6mx + 9; y'' = 6x 6m .y"= 0 x = m y = 2m3 + 9m + 1. 0,25y" i du t m sang dng khi i qua x = m, nn im un ca th hm s(1) l I( m; 2m3+ 9m +1). 0,25

I thuc

ng thng y = x + 1 2m3

+ 9m + 1 = m + 1 0,252m(4 m2) = 0 m = 0 hoc 2=m . 0,25


54/148

2

II 2,01 Gii phng trnh (1,0 im)

( 2cosx 1) (2sinx + cosx) = sin2x sinx ( 2cosx 1) (sinx + cosx) = 0. 0,25

2cosx 1= 0 cosx =1

x k2 , k

2 3

= + Z .

0,25 sinx + cosx = 0 tgx = 1 x k , k

4

= + Z .

0,25

Vy phng trnh c nghim l: x k23

= + v x k , k

4

= + Z .

0,252 Tm m h phng trnh c nghim (1,0 im)

t: u = x , v y , u 0, v 0.= H cho tr thnh:3 3

u v 1

u v 1 3m

+ =

+ = (*)

0,25u v 1

uv m

+ =

=

u, v l hai nghim ca phng trnh: t2 t + m = 0 (**).

0,25H cho c nghim (x; y) H (*) c nghim u 0, v 0 Phng trnh(**) c hai nghim t khng m. 0,25

1 4m 0

1S 1 0 0 m .

4P m 0

=

=

=

0,25

III 3,01 Tnh to trng tm G ca tam gic ABC v tm m... (1,0 im)

Trng tm G ca tam gic ABC c ta :

A B C A B CG G

x x x y y y mx 1; y3 3 3

+ + + += = = = . Vy G(1; m3

).0,25

Tam gic ABC vung gc ti G GA.GB 0=

. 0,25m m

GA( 2; ), GB(3; )3 3

.0,25

GA.GB 0=

2m6 0

9 + = m 3 6 = .

0,252 Tnh khong cch gia B1C v AC1,... (1,0 im)

a) T gi thit suy ra:

1 1C (0; 1; b), B C (a; 1; b)=

1 1AC ( a; 1; b), AB ( 2a; 0; b)= =

0,25


55/148

3

( )

1 1 1

1 12 2

1 1

B C, AC AB abd B C, AC

a bB C, AC

= = +

.

0,25

b) p dng bt ng thc Csi, ta c:

1 1

2 2

ab ab 1 1 a bd(B C; AC ) ab 2

22ab 2 2a b

+= = =

+

.

0,25 Du "=" xy ra khi v ch khi a = b = 2.

Vy khong cch gia B1C v AC1ln nht bng 2 khi a = b = 2. 0,25

3 Vit phng trnh mt cu (1,0 im)I(x; y; z) l tm mt cu cn tm I (P) v IA = IB = IC .Ta c: IA2= (x 2)2 + y2+ ( z 1)2 ;

IB2= (x 1)2+ y2+ z2;IC2= (x 1)2 + (y 1)2+ ( z 1)2. 0,25

Suy ra h phng trnh:

=

=

=++

22

22

02

ICIB

IBIA

zyx

=+

=+

=++

12

2

zyzx

zyx

0,25

.0;1 === yzx 0,25 == 1IAR Phng trnh mt cu l ( x 1)2+ y2+ ( z 1)2=1. 0,25IV 2,0

1 Tnh tch phn (1,0 im)

I =3

2

2

ln(x x) dx . t2

2

2x 1du dxu ln(x x)

x xdv dx

v x

==

= =

.

0,25 3 332

22 2

2x 1 1I x ln(x x) dx 3ln 6 2 ln 2 2 dx

x 1 x 1

= = +

0,25

( )3

23ln 6 2 ln 2 2x ln x 1= + .

0,25

I = 3ln6 2ln2 2 ln2 = 3ln3 2. 0,25 2 Tm s hng khng cha x... (1, 0 im)

Ta c: ( )7 k7 7 k

k3 374 4

k 0

1 1x C x

x x

=

+ =

0,25

7 k k 28 7k 7 7k k3 4 127 7

k 0 k 0

C x x C x

= =

= = .0,25

S hng khng cha x l s hng tng ng vi k (k Z, 0 k 7) tho mn:

4012

728==

k

k.

0,25

S hng khng cha x cn tm l 47C 35= . 0,25


56/148

4

V Chng minh phng trnh c nghim duy nht 1,0x5 x2 2x 1 = 0 (1) .

(1) x5= ( x + 1)20 x 0 (x + 1)21 x51 x 1. 0,25

Vi x 1: Xt hm s 5 2f (x) x x 2x 1= . Khi f(x) l hm s lin tc

vi mi x 1.Ta c:

f(1) = 3 < 0, f(2) = 23 > 0. Suy ra f(x) = 0 c nghim thuc ( 1; 2). (2) 0,25

f '( x) = 4 4 4 45x 2x 2 (2x 2x) (2x 2) x = + + .3 4 42x(x 1) 2(x 1) x 0, x 1= + + > . 0,25

Suy ra f(x) ng bin trn [ 1; +) (3).T (1), (2), (3) suy ra phng trnh cho c ng mt nghim. 0,25


57/148

BGIO DC V O TO-----------------------

CHNH THC

THI TUYN SINH I HC, CAO NG NM 2005Mn: TON, khi A

Thi gian lm bi: 180 pht, khng kthi gian pht ----------------------------------------

Cu I (2 im)

Gi m(C ) l thca hm s1

y m xx

= + (*) ( m l tham s).

1)

Kho st sbin thin v vthca hm s(*) khi 1m .4

=

2)

Tm m hm s (*) c cc trv khong cch tim cc tiu ca m(C ) n tim

cn xin ca m(C ) bng1

.2

Cu II (2 im)

1) Gii bt phng trnh 5x 1 x 1 2x 4. >

2) Gii phng trnh 2 2cos 3x cos 2x cos x 0. =

Cu III (3 im)1) Trong mt phng vi hta Oxy cho hai ng thng

1d : x y 0 = v 2d : 2x y 1 0.+ =

Tm ta cc nh hnh vung ABCD bit rng nh A thuc 1d , nh C thuc 2d

v cc nh B, D thuc trc honh.

2) Trong khng gian vi hta Oxyz cho ng thngx 1 y 3 z 3

d :1 2 1

+ = =

v mt

phng (P) : 2x y 2z 9 0.+ + =

a)

Tm ta im I thuc d sao cho khong cch t I n mt phng (P) bng 2. b)Tm ta giao im A ca ng thng d v mt phng (P). Vit phng trnh

tham sca ng thng nm trong mt phng (P), bit i qua A v vung

gc vi d.

Cu IV (2 im)

1) Tnh tch phn2

0

sin 2x sin xI dx.

1 3cos x

+=

+

2) Tm snguyn dng n sao cho1 2 2 3 3 4 2n 2n 1

2n 1 2n 1 2n 1 2n 1 2n 1C 2.2C 3.2 C 4.2 C (2n 1).2 C 2005+

+ + + + + + + + + =L

( knC l sthp chp k ca n phn t).

Cu V (1 im)

Cho x, y, z l cc sdng tha mn1 1 1

4.x y z

+ + = Chng minh rng

1 1 11.

2x y z x 2y z x y 2z+ +

+ + + + + +

------------------------------ Ht -----------------------------Cn bcoi thi khng gii thch g thm.

Hv tn th sinh................................................. sbo danh........................................

Mang Giao duc Edunet - http://www.edu.net.vn


58/148

1

BGIO DC V O TO---------------------CHNH THC

P N THANG IMTHI TUYN SINH I HC, CAO NG NM 2005

----------------------------------------Mn: TON, Khi A

(p n thang im gm 4 trang)

Cu Ni dung imI 2,0

I.1 1,01 1 1

m y x4 4 x

= = + .

a) TX: \{0}.

b) Sbin thin:2

2 2

1 1 x 4y '

4 x 4x

= = , y ' 0 x 2,x 2.= = =

0,25

yC ( ) ( )CTy 2 1, y y 2 1.= = = = ng thng x 0= l tim cn ng.

ng thng1

y x4

= l tim cn xin.

0,25

c) Bng bin thin:

x 2 0 2 + y + 0 0 +

y

1 + +

1

0,25

d) th

0,25



59/148

2

I.2 1,0

2

1y ' m , y ' 0

x= = c nghim khi v chkhi m 0> .

Nu m 0> th 1 21 1

y ' 0 x , xm m

= = = .

0,25

Xt du y ' x

1

m 0

1

m +

y ' + 0 || 0 +

Hm slun c cc trvi mi m 0.>

0,25

im cc tiu ca ( )mC l1

M ;2 m .m

Tim cn xin (d) : y mx mx y 0.= =

( )2 2

m 2 m md M,d .

m 1 m 1

= =

+ +

0,25

( ) 22

1 m 1d M;d m 2m 1 0 m 1.

2 2m 1= = + = =

+

Kt lun: m 1= .0,25

II. 2,0II.1 1,0

Bt phng trnh: 5x 1 x 1 2x 4 > . K:

5x 1 0

x 1 0 x 2.2x 4 0

0,25

Khi bt phng trnh cho tng ng vi

5x 1 2x 4 x 1 5x 1 2x 4 x 1 2 (2x 4)(x 1) > + > + + 0,25

2 2x 2 (2x 4)(x 1) x 4x 4 2x 6x 4 + > + + > + 2x 10x 0 0 x 10. < < <

0,25

Kt hp vi iu kin ta c : 2 x 10 < l nghim ca bt phng trnh cho. 0,25II.2 1,0

Phng trnh cho tng ng vi( ) ( )1 cos6x cos 2x 1 cos 2x 0+ + = cos6x cos 2x 1 0 =

0,25

cos8x cos 4x 2 0 + = 22cos 4x cos 4x 3 0 + = 0,25

( )

= =

cos 4x 1

3cos 4x loi .

2

Vy ( )

= = cos 4x 1 x k k .2

0,5



60/148

3

III. 3,0III.1 1,0

V ( )1A d A t; t .

V A v C i xng nhau qua BD v B,D Ox nn ( )C t; t . 0,25

V 2C d nn 2t t 1 0 t 1. = = Vy ( ) ( )A 1;1 , C 1; 1 . 0,25

Trung im ca AC l ( )I 1;0 . V I l tm ca hnh vung nn

IB IA 1

ID IA 1

= =

= =

0,25

b 1 1B Ox B(b;0) b 0,b 2

D Ox D(d;0) d 0,d 2d 1 1

= = =

= = =

Suy ra, ( )B 0;0 v ( )D 2;0 hoc ( )B 2;0 v ( )D 0;0 .

Vy bn nh ca hnh vung l( ) ( ) ( ) ( )A 1;1 , B 0;0 , C 1; 1 , D 2;0 ,

hoc

( ) ( ) ( ) ( )A 1;1 , B 2;0 , C 1; 1 , D 0;0 .

0,25

III.2a 1,0

Phng trnh ca tham sca

x 1 t

d : y 3 2t

z 3 t.

=

= +

= +

0,25

( )I d I 1 t; 3 2t;3 t + + , ( )( )2t 2

d I, P .3

+= 0,25

( )( )t 4

d I, P 2 1 t 3t 2.

== =

= 0,25

Vy c hai im ( ) ( )1 2I 3;5;7 , I 3; 7;1 . 0,25

III.2b 1,0

V A d nn( )

A 1 t; 3 2t;3 t + + .

Ta c ( )A P ( ) ( ) ( )2 1 t 3 2t 2 3 t 9 0 t 1 + + + + = = .

Vy ( )A 0; 1;4 .

0,25

Mt phng ( )P c vectphp tuyn ( )n 2;1; 2 .=

ng thng d c vectchphng ( )u 1;2;1=

.

V ( )P v d nn c vectchphng ( )u n,u 5;0;5

= =

.

0,5

Phng trnh tham sca :

x t

y 1

z 4 t.

=

= = +

0,25



61/148

4

IV 2,0IV.1 1,0

2

0

(2cosx 1) sinxI dx

1 3cos x

+=

+ .

0,25

t

2t 1cosx 3

t 1 3cos x3sinx

dt dx.2 1 3cos x

=

= + = +

x 0 t 2, x t 1.2

= = = =

0,25

( )1 22

2

2 1

t 1 2 2I 2 1 dt 2t 1 dt.

3 3 9

= + = +

0,25

23

1

2 2t 2 16 2 34t 2 1 .

9 3 9 3 3 27

= + = + + =

0,25

IV.2 1,0

Ta c ( )2n 1 0 1 2 2 3 3 2n 1 2n 1

2n 1 2n 1 2n 1 2n 1 2n 11 x C C x C x C x ... C x+ + +

+ + + + ++ = + + + + + x . 0,25

o hm hai vta c

( )( ) ( )2n 1 2 3 2 2n 1 2n

2n 1 2n 1 2n 1 2n 12n 1 1 x C 2C x 3C x ... 2n 1 C x+

+ + + ++ + = + + + + + x . 0,25

Thay x 2= ta c:( )1 2 2 3 3 4 2n 2n 12n 1 2n 1 2n 1 2n 1 2n 1C 2.2C 3.2 C 4.2 C ... 2n 1 .2 C 2n 1.

+

+ + + + + + + + + = + 0,25

Theo githit ta c 2n 1 2005 n 1002+ = = . 0,25

V 1,0

Vi a,b 0> ta c : 21 a b 1 1 1 1

4ab (a b) .a b 4ab a b 4 a b

+ + +

+ +

Du " "= xy ra khi v chkhi a b= .

0,25

p dng kt qutrn ta c:

1 1 1 1 1 1 1 1 1 1 1 1 1(1).

2x y z 4 2x y z 4 2x 4 y z 8 x 2y 2z

+ + + = + +

+ + +

Tng t

1 1 1 1 1 1 1 1 1 1 1 1 1

(2).x 2y z 4 2y x z 4 2y 4 x z 8 y 2z 2x

+ + + = + + + + +

1 1 1 1 1 1 1 1 1 1 1 1 1(3).

x y 2z 4 2z x y 4 2z 4 x y 8 z 2x 2y

+ + + = + +

+ + +

0,5

Vy1 1 1 1 1 1 1

1.2x y z x 2y z x y 2z 4 x y z

+ + + + =

+ + + + + +

Ta thy trong cc bt ng thc (1), (2), (3) th du " "= xy ra khi v chkhi

x y z.= = Vy ng thc xy ra khi v chkhi3

x y z .4= = =

0,25

-------------------------------Ht-------------------------------



62/148

BGIO DC V O TO-------------------------

CHNH THC

THI TUYN SINH I HC, CAO NG NM 2005Mn: TON, khi B

Thi gian lm bi: 180 pht, khng kthi gian pht --------------------------------------------------

Cu I (2 im)

Gim

(C ) l thca hm s( )2x m 1 x m 1

y

x 1

+ + + +=

+

(*) ( m l tham s).

1) Kho st sbin thin v vthca hm s(*) khi m 1.= 2) Chng minh rng vi mbt k, th

m(C ) lun lun c im cc i, im cc tiu

v khong cch gia hai im bng 20.

Cu II (2 im)

1)

Gii hphng trnh( )2 39 3

x 1 2 y 1

3log 9x log y 3.

+ =

=

2) Gii phng trnh 1 sin x cos x sin 2x cos2x 0.+ + + + =

Cu III(3 im)1) Trong mt phng vi hta Oxycho hai im A(2;0) v B(6;4) . Vit phng trnh

ng trn (C) tip xc vi trc honh ti im A v khong cch ttm ca (C) nim B bng 5.

2) Trong khng gian vi h ta Oxyz cho hnh lng tr ng 1 1 1ABC.A B C vi

1A(0; 3;0), B(4;0;0), C(0;3;0), B (4;0;4).

a)Tm ta cc nh 1 1A , C . Vit phng trnh mt cu c tm l A v tip xc vi

mt phng 1 1(BCC B ).

b)

Gi M l trung im ca 1 1A B . Vit phng trnh mt phng (P) i qua hai im

A, M v song song vi 1BC . Mt phng (P) ct ng thng 1 1A C ti im N.

Tnh di on MN.

Cu IV(2 im)

1) Tnh tch phn2

0

sin2x cosxI dx

1 cosx

=+

.

2)

Mt i thanh nin tnh nguyn c 15 ngi, gm 12 nam v 3 n. Hi c bao nhiu

cch phn cng i thanh nin tnh nguyn vgip 3 tnh min ni, sao cho mitnh c 4 nam v 1 n?

Cu V(1 im)

Chng minh rng vi mi x , ta c:x x x

x x x12 15 20 3 4 55 4 3

+ + + +

.

Khi no ng thc xy ra?

--------------------------------Ht--------------------------------

Cn bcoi thi khng gii thch g thm.

Hv tn th sinh .................................................. Sbo danh ...............................



63/148

BGIO DC V O TO---------------------CHNH THC

P N THANG IMTHI TUYN SINH I HC, CAO NG NM 2005

----------------------------------------Mn: TON, Khi B

(p n thang im gm 4 trang)

Cu Ni dung imI 2,0

I.1 1,02x 2x 2 1

m 1 y x 1 .x 1 x 1

+ += = = + +

+ +

a) TX: \{ }.1

b) Sbin thin:( ) ( )

2

2 2

1 x 2xy ' 1

x 1 x 1

+= =

+ +y ' 0 x 2, x 0., = = =

0,25

yC ( ) ( )CTy 2 2, y y 0 2.= = =1

=

ng thng l tim cn ng.x= ng thng l tim cn xin.y x 1= +

0,25

Bng bin thin:

x 2 1 0 + y + 0 0 +

y

2 + +

2

0,25

c) th

0,25

1



64/148

I.2 1,0

Ta c:1

y x mx 1

= + ++

.

TX: \{ }.1

( ) ( )( )2 2

x x 21y ' 1 , y ' 0 x 2, x 0.x 1 x 1

+= = = = =+ +

0,25

Xt du y '

x 2 1 0 + y + 0 || 0 +

thca hm s(*) lun c im cc i l ( )M 2;m 3 v im cc tiu l

.( )N 0;m 1+

0,50

( )( ) ( ) ( )( )2 2

MN 0 2 m 1 m 3 20.= + + = 0,25

II. 2,0II.1 1,0

( )

2 39 3

x 1 2 y 1 (1)

3log 9x log y 3 (2)

+ =

=

K:x 1

0 y 2.

<

0,25

( ) ( )3 3 3 32 3 1 log x 3log y 3 log x log y x y. + = = = 0,25

Thay vo (1) ta cy x=

( )( )x 1 2 x 1 x 1 2 x 2 x 1 2 x 1 + = + + =

( )( )x 1 2 x 0 x 1, x 2. = = =

Vy hc hai nghim l ( ) ( )x; y 1;1= v ( ) ( )x; y 2;2 .=

0,50

II.2 1,0

Phng trnh cho tng ng vi2sin x cos x 2sin x cos x 2cos x 0+ + + =

( )sin x cos x 2cos x sin x cos x 0 + + + =

( )( )sin x cos x 2cos x 1 0. + + =

0,50

sin x cos x 0 tgx 1 x k 4

+ = = = + ( )k . 0,25

1 22cos x 1 0 cos x x k22 3

+ = = = + ( )k . 0,25

2



65/148

III. 3,0III.1 1,0

Gi tm ca (C) l ( )I a;b v bn knh ca (C) l R.

(C) tip xc vi Ox ti A va 2 = b R.= 0,25

( ) ( )2 2 2IB 5 6 2 4 b 25 b 8b 7 0 b 1,b 7.= + = + = = = 0,25

Vi ta c ng trna 2,b 1= =

( ) ( ) ( )2 2

1C : x 2 y 1 1. + = 0,25

Vi ta c ng trna 2,b 7= =

( ) ( ) ( )2 2

2C : x 2 y 7 49. + = 0,25

III.2a 1,0

( ) ( )1 1A 0; 3;4 ,C 0;3;4 . 0,25

( ) ( )1BC 4;3;0 ,BB 0;0;4= =

Vectphp tuyn ca ( )1 1mp BCC B l ( )1n BC,BB 12;16;0 = = .

Phng trnh mt phng ( )1 1BCC B :

( )12 x 4 16y 0 3x 4y 12 0. + = + =

0,25

Bn knh mt cu:

( )( )1 1 2 212 12 24

R d A, BCC B53 4

= =

+.=

0,25

Phng trnh mt cu:

( )2

2 2

576x y 3 z 25+ + + = .

0,25

III.2b 1,0

Ta c ( )13 3

M 2; ;4 , AM 2; ;4 , BC 4;3;4 .2 2

= =

0,25

Vectphp tuyn ca (P) l ( )P 1n AM,BC 6; 24;12 = =

.

Phng trnh (P): ( )6x 24 y 3 12z 0 x 4y 2z 12 0. + + = + + =

Ta thy Do i qua v song song viB(4;0;0) (P). (P) A, M 1BC .

0,25

Ta c ( )1 1A C 0;6;0=

. Phng trnh tham sca ng thng l1 1A C

x 0

y 3

z 4.

=

t= + =

( )1 1N A C N 0; 3 t; 4 . +

V ( )N P nn ( )0 4 3 t 8 12 0 t 2+ + + = = .

Vy ( )N 0; 1;4 .

( ) ( )2

2 23 1MN 2 0 1 4 4 .

2 2

Documents

Tuyển Tập Các Đề Thi Đại Học Cao Đẳng Môn Toán ( Có Đáp Án )