25 de Thi Thu Dai Hoc Sp Hn co dap an chi tiet

8/8/2019 25 de Thi Thu Dai Hoc Sp Hn co dap an chi tiet

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I HC SPHM H NI THI THI HC CAO NG 2011KHOA TON-TIN MN: TON- KHI A

------------- Thi gian lm bi: 180 pht( khng kthi gian giao )---------------------------------------------------------------------------------------------------------------------------------------------

A. PHN CHUNG CHO MI TH SINHCu I (2 im).

1. Kho st v v th hm s y = x4 4x2 + 32.

Tm m phng trnh

4 2

24 3 log x x m + =

c ng 4 nghim.Cu II (2 im).

1. Gii bt phng trnh: ( ) ( )325 1 5 1 2 0

x x x+

+ +

2. Gii phng trnh: 2 ( 2) 1 2 x x x x + = Cu III (2 im)

1. Tnh gii hn sau: 1 231

tan( 1) 1lim

1

x

x

e x

x

+

2. Cho hnh chp S.ABCD c y l hnh thoi , BAD = . Hai mt bn (SAB) v (SAD) cng vunggc vi mt y, hai mt bn cn li hp vi y mt gc . Cnh SA = a. Tnh din tch xung quanh

v th tch khi chp S.ABCD.

Cu IV (1 im). Cho tam gic ABC vi cc cnh l a, b, c. Chng minh rng:3 3 3 2 2 2 2 2 23 ( ) ( ) ( )a b c abc a b c b c a c a b+ + + + + + + +

B. PHN TCHN:Mi th sinh ch chn cu Va hoc VbCu Va (3 im). Chng trnh cbn

1. Trong mt phng ta Oxy cho ng thng : 2 3 0x y + = v hai im A(1; 0), B(3; - 4).Hy tm trn ng thng mt im M sao cho 3 MA MB+

nh nht.

2. Trong khng gian Oxyz cho hai ng thng: 1 1: 22

x td y t

z t

= =

= +

v 2 : 1 3

1

x td y t

z t

== +

=

.

Lp phng trnh ng thng i qua M(1; 0; 1) v ct c d1 v d2.

3. Tm s phc z tha mn: 2 2 0z z+ = Cu Vb. (3 im).Chng trnh nng cao

1. Trong mt phng ta cho hai ng trn (C1): x2 + y2 = 13 v (C2): (x - 6)2 + y2 = 25 ct nhau tiA(2; 3). Vit phng trnh ng thng i qua A v ct (C1), (C2) theo hai dy cung c di bng nhau.

2. Trong khng gian Oxyz cho hai ng thng: 11

: 2

2

x t

d y t

z t

=

= = +

v 2 : 1 3

1

x t

d y t

z t

=

= + =

.

Lp phng trnh mt cu c ng knh l on vung gc chung ca d1 v d2.3. Trong cc s phc z tha mn iu kin 1 2 1z i+ + = , tm s phc z c modun nh nht.

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

http://www.VNMATH.com

http://www.VNMATH.com 1 http://www.VNMATH.com


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Gv: Trn Quang Thu n Tel: 0912.676.613 091.565

1

P N THANG IM THI THI HC, CAO NG NM 2011

MN:TON, Khi A

Cu Ni dung im2

1 1TX D = Gii hn : lim

xy

= +

S bin thin : y = 4x3 - 8x

y = 0 0, 2x x = =

Bng bin thin

x 2 0 2+

y - 0 + 0 - 0 +y + +

3-1 -1

Hm sng bin trn cc khong ( ) ( )2;0 , 2; + v nghch bin trn cc khong

( ) ( ); 2 , 0; 2 Hm st cc i ti x = 0, yCD = 3. Hm st cc tiu ti x = 2 , yCT= -1

th

025

025

025

025

2 1

I

th hm s

4 2

4 3 y x x= +

S nghim ca phng trnh 4 2 24 3 log x x m + = bng s giao im ca th hm s

4 24 3 y x x= + v ng thng y = log2m.

Vy phng trnh c 4 nghim khi v ch khi log2m = 0 hoc 21 log m 3< <

hay m = 1 hoc 2


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2

21 1

Vit li bt phng trnh di dng5 1 5 1

2 2 02 2

x x

++

t t =5 1

, 0.2

x

t +

>

khi 5 1 1

2

x

t

=

Bt phng trnh c dng

t +1

2 2 0t

2 2 2 1 0t t +

2 1 2 1t +

5 1 5 1

2 2

5 12 1 2 1

2

log ( 2 1) log ( 2 1)

x

x+ +

+ +

+

025

025

025

025

2 1

II

iu kin : 1x

Phng trnh tng ng vi 2 ( 1 1) 2 1 2( 1) 0 x x x x x = (*)

t 1, 0 y x y= . Khi (*) c dng : x2 x(y - 1) 2y 2y2 = 0( 2 )( 1) 0

2 0( 1 0)

x y x y

x y do x y

+ + =

= + +

2

2 1

4 4 02

x x

x x

x

=

+ =

=

025025

05

21 1

1 2 1 23 2 3

31 1

1 23 2 3 23 3

21 1

3 2 3 23 3

1 1

tan( 1) 1 1 tan( 1)lim lim .( 1)

11

1 tan( 1)lim .( 1) lim .( 1)( 1)

1 1

lim( 1) lim( 1)( 1) 9

x x

x x

x

x x

x x

e x e xx x

xx

e x x x x x x

x x

x x x x x

+ + = + +

= + + + + + +

= + + + + + + =

025

05

025

2 1

III

Kng cao SI ca tam gic SBC. Khi AI BC




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3

(nh l 3 ng vung gc) do SIA = S

AI = a.cot , AB = AD =cot

sin

a

, SI =

sin

a

2 2cot. .sin

sinABCDa

S AB AD

= =

A3 2

.

cot

3sinS ABCDa

V

=

Sxq = SSAB + SSAD SSBC + SSCD B I C

=2 cot 1

.(1 )sin sin

a

+

025

025

025

025

1IV

Ta c 3 3 3 2 2 2 2 2 23 ( ) ( ) ( )a b c abc a b c b c a c a b+ + + + + + + + 2 2 2 2 2 2 2 2 2 3

2 2 2 23

cos cos cos2

a b c b c a c a b

ab bc ca

A B C

+ + + + +

+ +

Mt khc

2 2 2 2

cos cos cos (cos cos ).1 (cos cos sin sin )

1 1 3[(cos cos ) 1 ]+ [sin A+sin B]-cos cos

2 2 2

A B C A B A B A B

A B A sB

+ + = +

+ + =

Do 3

cos cos cos2

A B C + +

025

025

05

31 1

Gi I l trung im ca AB, J l trung im ca IB. Khi I(1 ; -2), J(5

; 32

)

Ta c : 3 ( ) 2 2 2 4 MA MB MA MB MB MI MB MJ + = + + = + =

V vy 3 MA MB+

nh nht khi M l hnh chiu vung gc ca J trn ng thng ng thng JM qua J v vung gc vi c phng trnh : 2x y 8 = 0.

Ta im M l nghim ca h

22 3 0 5

2 8 0 19

5

xx y

x yy

=+ =

= =

vy M(19 2

;5 5

)

025

025025

025

Va

2 1




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5

==============================

Do M(2 14 3

; ;5 5 5

), N(

3 14 2; ;

5 5 5).

Mt cu ng knh MN c bn knh R =2

2 2

MN= v tm I(

1 14 1; ;

10 5 10

) c phng trnh

2 2 21 14 1 1

( ) ( ) ( )10 5 10 2 x y z + + + =

025

0253 1

Gi z = x + yi, M(x ; y ) l im biu din s phc z.2 21 2 1 ( 1) ( 2) 1 z i x y+ + = + + + =

ng trn (C) : 2 2( 1) ( 2) 1x y+ + + = c tm (-1;-2)ng thng OI c phng trnh y = 2xS phc z tha mn iu kin v c mdun nh nht khi v ch khi imBiu din n thuc (C) v gn gc ta O nht, chnh l mt trong haigiao im ca ng thng OI v (C)

Khi ta ca n tha

mn h2 2

1 11 1

2 5 5,

2 2( 1) ( 2) 12 2

5 5

x xy x

x yy y

= = + =

+ + + = = = +

Chon z =1 2

1 ( 2 )5 5

i + + +

025

025

025

025




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I HC SPHM H NI THI THI HC CAO NG 2011

KHOA TON-TIN MN: TON- KHI A------------- Thi gian lm bi: 180 pht( khng kthi gian giao )

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

PHN CHUNG CHO TT C TH SINH (7 im).Cu I ( 2 im)

Cho hm s 2)2()21( 23 ++++= mxmxmxy (1) m l tham s.

1. Kho st s bin thin v v th (C) ca hm s (1) vi m=2.2. Tm tham s m th ca hm s (1) c tip tuyn to vi ng thng d: 07 =++yx gc ,bit

26

1cos = .

Cu II(2 im)

1. Gii bt phng trnh: 544

2log2

2

1

x

x.

2. Gii phng trnh: ( ) .cos32cos3cos21cos2.2sin3 xxxxx +=++ Cu III(1 im)

Tnh tch phn: I( ) ++

+=

4

02

2111 dx

xx .

Cu IV(1 im)

Cho hnh chp S.ABC c y ABC l tam gic vung cn nh A, AB 2a= . Gi I l trung im ca BC, hnh

chiu vung gc H ca S ln mt y (ABC) tha mn: IHIA 2= , gc gia SC v mt y (ABC) bng 060 .Hy tnh th tch khi chp S.ABC v khong cch t trung im K ca SB ti (SAH).Cu V(1 im)

Cho x, y, z l ba s thc dng thay i v tha mn: xyzzyx ++ 222 . Hy tm gi tr ln nht ca biu thc:

xyz

z

zxy

y

yzx

xP

++

++

+=

222.

PHN TCHN (3 im): Th sinhchchn lm mt trong hai phn ( phn A hoc phn B ).A.Theo chng trnh chun:Cu VI.a(2 im)

1. Trong mt phng Oxy, cho tam gic ABC bit A(3;0), ng cao tnh B c phng trnh 01 =++yx ,trung tuyn tnh C c phng trnh: 2x-y-2=0. Vit phng trnh ng trn ngoi tip tam gic ABC.

2. Trong khng gian vi h trc ta Oxyz, cho cc im A(-1;1;0), B(0;0;-2) v C(1;1;1). Hy vit phng

trnh mt phng (P) qua hai im A v B, ng thi khong cch t C ti mt phng (P) bng 3 .Cu VII.a (1 im)

Cho khai trin: ( ) ( ) 14142

210

2210 ...121 xaxaxaaxxx ++++=+++ . Hy tm gi tr ca 6a .B. Theo chng trnh nng cao:

Cu VI.b (2 im)1. Trong mt phng ta Oxy, cho tam gic ABC bit A(1;-1), B(2;1), din tch bng

11

2v trng tm G

thuc ng thng d: 043 =+yx . Tm ta nh C.

2.Trong khng gian vi h trc Oxyz, cho mt phng (P) 01 =++ zyx ,ng thng d:3

1

1

1

1

2

=

=

zyx

Gi I l giao im ca d v (P). Vit phng trnh ca ng thng nm trong (P), vung gc vi d v cch

I mt khong bng 23 .

Cu VII.b (1 im) Gii phng trnh: .13

=

+

zi

iz

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




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---------------------------------------------------------------------------------------------------------------------Gv: Trn Quang Thu n Tel: 0912.676.613 091.5657.95

1

P N THANG IM THI THI HC, CAO NG NM 2011

MN:TON, Khi A

PHN CHUNG CHO TT C TH SINH.

Cu Ni dung im1(1) Kho st hm skhi m = 2Khi m = 2, hm s trthnh: y = x3 3x 2 + 4a) TX: Rb) SBTGii hn: lim ; lim

x xy y

+

= = + 0,25

Chiu bin thin:C y = 3x2 6x; y=0 x =0, x =2

x 0 2 +y + 0 0 +

y

4

0

+

Hm sB trn cc khong ( ; 0) v (2 ; +), nghch bin trn (0 ; 2).

0,25

Hm st cc i ti x = 0, yC = y(0) = 4;Hm st cc tiu ti x = 2, yCT = y(2) = 0.

0,25

c) th:Qua (-1 ;0)Tm i xng:I(1 ; 2)

0,25

2(1) Tm m ...

Gi k l h s gc ca tip tuyn tip tuyn c vctphp )1;(1 = kn

d: c vctphp )1;1(2 =n

Ta c

=

=

=+

+

==

32

2

3

0122612

12

1

26

1.cos

2

12

221

21

k

k

kk

k

k

nn

nn

0,5

I(2)

Yu cu ca bi ton tha mn t nht mt trong hai phng trnh: 1/ ky = (1)

v 2/ ky = (2) c nghim x

=++

=++

3

22)21(23

2

32)21(23

2

2

mxmx

mxmx

0

0

2/

1/

0,25c nghim

1

I

2

2

-1

4

0 x

y

c nghim




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2

034

01282

2

mm

mm

1;4

32

1;

4

1

mm

mm

4

1m hoc

2

1m 0,25

II(2) 1(1) Gii bt phng trnh ...

Bpt

)2(34

2log2

)1(24

2log3

94

2log

044

2log

2

1

2

1

2

2

1

2

2

1

x

x

x

x

x

x

x

x

0,25

. Gii (1): (1)5

16

3

8

04

165

04

83

84

24

x

x

x

x

x

x

x 0,25

. Gii (2): (2) 9

4

17

4

04

49

04

417

4

1

4

2

8

1

x

x

x

x

x

x

x

0,25

Vy bt phng trnh c tp nghim4 4 8 16

; ;17 9 3 5

. 0,25

2(1) Gii PT lng gic

Pt )1cos2()12(cos)cos3(cos)1cos2(2sin3 ++=+ xxxxxx

)1cos2(sin2cossin4)1cos2(2sin3 22 +=+ xxxxxx

0)1sin22sin3)(1cos2( 2 =+++ xxx

0,5

1)6

2sin(22cos2sin301sin22sin3 2 ===++

xxxxx

kx +=6

0,25

)(

23

2

23

2

01cos2 Zk

kx

kx

x

+=

+=

=+

Vy phng trnh c nghim: 23

2 kx += ; 23

2 kx += v kx +=6

0,25

III(1) 1(1) Tnh tch phn.




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3

I( ) ++

+=

4

02

211

1dx

x

x.

t dttdxx

dxdtxt )1(

21211 =

+=++= v

2

22 ttx

=

i cnx 0 4

t 2 4

0,25

Ta c I =

dttt

tdtt

tttdt

t

ttt

+=

+=

+4

22

4

2

4

22

23

2

2 243

2

1243

2

1)1)(22(

2

1

=

++

ttt

t 2ln43

22

1 2

0,5

=4

12ln2 0,25

(1) Tnh thtch v khong cch

Ta c = IHIA 2 H thuc tia i ca tia IA v IA = 2IH

BC = AB 2 a2= ; AI= a ; IH=2

IA=

2

a

AH = AI + IH =2

3a

0,25

Ta c2

545cos.2 0222

aHCAHACAHACHC =+=

V )(ABCSH 060))(;( ==

SCHABCSC

21560tan 0 aHCSH ==

0,25

IV

6

15

2

15)2(

2

1.

3

1.

3

1 32.

aaaSHSV ABCABCS === 0,25

H

K

I

BA

S

C




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4

)(SAHBISHBI

AHBI

Ta c22

1)(;(

2

1))(;(

2

1

))(;(

))(;( aBISAHBdSAHKd

SB

SK

SAHBd

SAHKd=====

0,25

V (1) Tim gi tr ln nht ca P

xyz

z

zxy

y

xyx

xP

++

++

+=

222.

V 0;; >zyx , p dng BT Csi ta c:xyz

z

zxy

y

yzx

xP

222 222++ =

++=

xyzxyz

222

4

1

0,25

++

++=

+++++

xyz

zyx

xyz

xyzxyz

yxxzzy

222

2

1

2

1111111

4

1

2

1

2

1=

xyz

xyz

0,5

Du bng xy ra 3=== zyx . Vy MaxP =2

1

0,25

PHN TCHN:

Cu Ni dung imVIa(2) 1(1) Vit phng trnh ng trn

KH: 022:;01: 21 ==++ yxdyxd

1d c vctphp tuyn )1;1(1 =n v 2d c vctphp tuyn )1;1(2 =n

AC qua im A( 3;0) v c vct ch phng )1;1(1 =n phng trnhAC: 03 =yx .

= 2dACC Ta C l nghim h: )4;1(022

03

=

=C

yx

yx.

0,25

Gi );( BB yxB )2;2

3

(BB yx

M

+

( M l trung im AB)

Ta c B thuc 1d v M thuc 2d nn ta c: )0;1(02

23

01

=+

=++

Byx

yx

B

B

BB

0,25

Gi phng trnh ng trn qua A, B, C c dng:02222 =++++ cbyaxyx .

Thay ta ba im A, B, C vo pt ng trn ta c:




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5

=

=

=

=+

=+

=+

3

2

1

1782

12

96

c

b

a

cba

ca

ca

Pt ng trn qua A, B, C l:

034222 =++ yxyx . Tm I(1;-2) bn knh R = 22

0,5

2(1) Vit phng trnh mt phng (P)

G

i Ocban=

);;( l vct

php tuyn c

a (P)

V (P) qua A(-1 ;1 ;0) pt (P):a(x+1)+b(y-1)+cz=0

M (P) qua B(0;0;-2) a-b-2c=0 b = a-2c

Ta c PT (P):ax+(a-2c)y+cz+2c =0

0,25

d(C;(P)) = 0141623)2(

23 22

222=+=

++

+ caca

ccaa

ca

=

=

ca

ca

7

0,5

TH1: ca = ta chn 1== ca Pt ca (P): x-y+z+2=0

TH2: ca 7= ta chn a =7; c = 1 Pt ca (P):7x+5y+z+2=00,25

VII.a (1 ) Tm h sca khai trin

Ta c 4

3

)12(4

1

1

22++=++

xxx nn

( ) 1012142210 )21(16

9)21(

8

3)21(

16

1)1(21 xxxxxx +++++=+++

0,25

Trong khai trin ( )1421 x+ h s ca 6x l: 61462 C



0,5

Vy h s .4174821692

832

161 6106612661466 =++= CCCa 0,25

Tm ta ca im CVI.b(2) 1(1)

Gi ta ca im )3

;3

1();( CCCCyx

GyxC + . V G thuc d

)33;(330433

13 ++==+

+ CCCC

CC xxCxyyx

ng thng AB qua A v c vctch phng )2;1(=AB

0,25




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6

032: = yxptAB

5

11

5

3332

5

11);(

2

11);(.

2

1=

+===

CC

ABC

xxABCdABCdABS

=

=

=

5

17

11165

C

C

C

x

x

x 0,5

TH1: )6;1(1 = CxC

TH2: )5

36;

5

17(

5

17= CxC .

0,25

2(1) Vit phng trnh ca ng thng

(P) c vc tphp tuyn )1;1;1()( =Pn v d c vc tch phng 3;1;1(. =u

)4;2;1()( IPdI =

v dP);( c vc tch phng )2;2;4(;)(

==

unuP

)1;1;2(2 =

0,25

Gi H l hnh chiu ca I trn )(QmpH qua I v vung gc Phng trnh (Q): 0420)4()2()1(2 =+=+ zyxzyx

Gi 11 )()( dQPd = c vcto ch phng

)1;1;0(3)3;3;0(; )()( ==QP nn v 1d qua I

+=

+=

=

tz

ty

x

ptd

4

2

1

:1

Ta c );;0()4;2;1(1 ttIHttHdH =++

=

===

3

323223 2

t

ttIH

0,5

TH1:1

7

1

5

2

1:)7;5;1(3

=

=

=

zyxptHt

TH2:1

1

1

1

2

1:)1;1;1(3

=

+=

=

zyxptHt

0,25

VII.b 1 Gii phng trnh trn tp sphc.K: iz

tzi

izw

+= ta c phng trnh: 0)1)(1(1 23 =++= wwww

0,5




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7

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

=

+=

=

=++

=

2

31

2

31

1

01

12

iw

iw

w

ww

w

Vi 011 ==

+= zzi

izw

Vi 333)31(2

31

2

31==+

+=

+

+= zizi

i

zi

iziw

Vi 333)31(2

31

2

31==

=

+

= zizi

i

zi

iziw

Vy pt c ba nghim 3;0 == zz v 3=z .

0,5




16/147Gv: Trn Quang Thu n 0912.6

I HC SPHM H NIKHOA TON TIN

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

THI THI HC CAO NG 2010MN: TON

Thi gian lm bi: 180 pht (khng kthi gian giao )

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

I. PHN CHUNG CHO TT C TH SINH

Cu I( 2,0 im): Cho hm s: (C)

1.

Kho st v v th (C) hm s2. Cho im A( 0; a) Tm a t A kc 2 tip tuyn ti th (C) sao cho 2 tip im tng ng nm 2 pha ca trc honh.

Cu II (2,0 im):1. Gii phng trnh lng gic.

2. Gii h phng trnh.

Cu III(1,0 im): Tnh tch phn sau.

=3

4

42 cos.sin

xx

dxI

Cu IV(1,0 im): Cho ba s thc tha mn ,Chng minh rng:

Cu V(1,0 im): Cho t din ABCD c AC = AD = , BC = BD = a, khong cch t B n mt phng

(ACD) bng . Tnh gc gia hai mt phng (ACD) v (BCD). Bit th ca khi t din ABCD bng .

II. PHN RING (Th sinh chc lm 1 trong 2 phn A hoc B)A. Theo chng trnh chun.

Cu VIa(2,0 im):1. Trong khng gian vi h ta Oxyz cho 4 im : A(1;2; 2) B(-1;2;-1) C(1;6;-1) D(-1;6;2). Tm ta

hnh chiu vung gc ca im A trn mt phng (BCD)2. Trong mp vi h ta Oxy cho ng trn : x2 +y2 -2x +6y -15=0 (C ).

Vit PT ng thng () vung gc vi ng thng : 4x-3y+2 =0 v ct ng trn (C) ti A; Bsao cho AB = 6

Cu VIIa(1,0 im): Xc nh h s ca x5 trong khai trin (2+x +3x2 )15B. Theo chng trnh nng cao.

Cu VIb(2,0 im):1. Trong khng gian vi h ta Oxyz cho 4 im : A(1;2; 2) B(-1;2;-1) C(1;6;-1) D(-1;6;2). Tm ta

hnh chiu vung gc ca im A trn mt phng (BCD)2. Trong mp vi h ta Oxy cho ng trn : x2 +y2 -2x +6y -15=0 (C ).

Vit PT ng thng ( ) vung gc vi ng thng : 4x-3y+2 =0 v ct ng trn (C) ti A; Bsao cho AB = 6

Cu VIIb(1,0 im):Gii phng trnh:





P N

CU NI DUNG I

I 2,0

1 1,0

TX: D= R\{1}

y=Hm s lung nghch bin trn D v khng c cc tr

0,25

Gii hn:

PT ng TC: x=1; PT ng TCN: y=1

0,25

Bng bin thin:t - 1 +

f(t) - +f(t)

1 +

- 1

0,25

th: 0,25

2 1,0

Gi k l h s gc ca t i qua A(0;a). PT t d c dng y= kx+a (d) 0,25

x

y f x( ) =x+2

x-1

1

4

-2

-2 O 1 2 3

5/2




18/147



H cho trthnh

Vy h d cho c mt nghim

0,25

III 1,0

=3

4

42 cos.sin

xx

dxI =

3

4

22 cos.2sin.4

xx

dx

t : t = tanx

i cn: x =

x =

0,5

Khi 3

438)

32

1()2

1(

)1( 3

1

33

1

22

3

12

22

=++=++=+

= t

tt

dtttt

dttI

0,5

IV 1,0

BT cn chng minh tng ng vi

Nhn xt: Do nn l cc s thc dng

0,25

Xt : A = vi x,y > 0 Chia t v mu cho v t t = ta c A = vi t > 0

Xt hm s f(t) = trn (0;+ ) Ta c : f(t) = Bng bin thin:

t 0 1 +

f(t) - 0 +f(t)

1 1

0,5





Da vao bng bin thin ta c f(t) vi mi t > 0 T A = vi x,y > 0; du bng xy ra khi t = 1 nn x = y.

Do vai tr l nh nhau nn BT cn chng minh tng ng

p dng BT c si ta c Thay vo ta suy BT c chng minh, du ng thc xy ra khi a = b = c =

0,25

V 1,0

Gi E l trung im ca CD, k BH AE

Ta c ACD cn ti A nn CD AE

Tng t BCD cn ti B nn CD BE

Suy ra CD (ABE) CD BH

M BH AE suy ra BH (ACD)

Do BH = v gc gia hai mt phng

(ACD) v (BCD) l

0,25

Th tch ca khi t din ABCD l

M

Khi : l 2 nghim ca pt: x2 - x + = 0

trng hp v DE



VIa 2,0

1 1,0

Ta c ;[ , ] = (12; -6;8)

Mp (BCD) i qua B v c VTPT =(6;-3;4) nn c PT: 6x-3y+4z+16=0Gi d l t i qua A v vung gc vi mp(BCD) th d c PT:

0,5

Hnh chiu vung gc H ca A ln mp(BCD) l giao im ca d vi mp(BCD)Ta ca H l nghim ca h :

Vy H( -2; -4; -4)

0,5

2 1,0

ng trn ( C) c tm I(1;-3); bn knh R=5Gi H l trung im AB th AH=3 v IH AB suy ra IH =4Mt khc IH= d( I; )V || d: 4x-3y+2=0 nn PT ca c dng3x+4y+c=0

0,5

d(I; )=

vy c 2 t tha mn bi ton: 3x+4y+29=0 v 3x+4y-11=0

0,5

VIIa 1,0

Ta c (2+x+3x2 )15 =M =Vy (2+x+3x2 )15 =

0,5

Theo gt vi x5 ta c cc cp s : (k=3; i=2) ( k=4; i=1) (k=5; i=0)Vy h s ca x5 trong khai trin trn l :a=

0,5

VIb 1,0

K: x > 1 Vi K trn phng trnh cho tng ng

0,25

0,5

I

A H B





Vy phng trnh cho c mt nghim :

0,25http://www.VNMATH.com



23/147------------------------------------------------------------------------------------------------------------------------------Gv: Trn Quang Thu n 0912.676.613 09

I HC SPHM H NI

KHOA TON TIN

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

THI THI HC -- CAO NG 2010

MN: TON


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

Cu I: (2 im) Cho hm s: ( )3 23 1 9 2y x m x x m= + + + (1) c th l (Cm)

1) Kho st v v th hm s (1) vi m=1.2) Xc nh m (Cm) c cc i, cc tiu v hai im cc i cc tiu i xng vi nhau qua

ng thng1

2y x= .

Cu II: (2,5 im)1) Gii phng trnh:

( ) ( )3sin 2 cos 3 2 3 os 3 3 os2 8 3 cos sinx 3 3 0 x x c x c x x+ + = .2) Gii bt phng trnh :

( )22 12

1 1log 4 5 log

2 7x x

x

+ >

+ .

3) Tnh din tch hnh ph

ng gi

i h

n b

i cc

ng: y = x.sin2x, y = 2x, x = 2

.Cu III: (2 im)

1) Cho hnh lng tr ABC.ABC c y ABC l tam gic u cnh a, cnh bn hp vi y mtgc l 45

0. Gi P l trung im BC, chn ng vung gc h t A xung (ABC) l H sao cho

1

2 AP AH =

. gi K l trung im AA, ( ) l mt phng cha HK v song song vi BC ct BB v CC

ti M, N. Tnh t s th tch' ' '

ABCKMN

A B C KMN

V

V.

2) Gii h phng trnh sau trong tp s phc:( )

2

2

2 2 2 2

65

6 0

a a

a aa b ab b a a

+ =

+ + + + =

Cu IV: (2,5 im)1) Cho m bng hng trng v n bng hng nhung khc nhau. Tnh xc sut lyc 5 bng hng trong c t nht 3 bng hng nhung? Bit m, n l nghim ca h sau:

2 2 1

3

1

9 19

2 2

720

m

m n m

n

C C A

P

+

+ +


24/147


25/147------------------------------------------------------------------------------------------------------------------------------Gv: Trn Quang Thu n 0912.676.613 09

P N THI THI HC

Bi1

1

Khi m = 1 ta c hm s: 3 26 9 1 y x x x= +

BBT:x - 1 3 +

y/

+ 0 - 0 +

3 + y

- 1

1

2 9)1(63' 2 ++= xmxy

hm s c cc i, cc tiu:

09.3)1(9' 2 >+= m );31()31;( ++ m

Ta c ( ) 14)22(29)1(633

1

3

1 22+++++

+= mxmmxmx

mxy

Vy ng thng i qua hai im cc i v cc tiu l

14)22(2 2 +++= mxmmy

V hai im cc i v cc tiu i xng qua t xy2

1= ta c iu kin cn l

[ ] 12

1.)22(2 2 =+ mm

=

==+

3

10322

m

mmm

Khi m = 1 ptt i qua hai im C v CT l:y = - 2x + 5. Ta trung im C v

CT l:

=++

=+

==+

12

10)(2

2

2

2

4

22121

21

xxyy

xx

Ta trung im C v CT l (2; 1) thuc ng thng xy2

1= 1= m tm .

Khi m = -3 ptt i qua hai im C v CT l: y = -2x 11.3= m khng tha mn.

Vy m = 1 tha mn iu kin bi.

1

Bi2

1 phng trnh a v:

=

=

=

=+

=

=+

)(4cos

1cos

3tan

04cos3cos

0sincos3

0)8cos6cos2)(sincos3(

2

2

loaix

x

x

xx

xx

xxxx

=

+= k

kx

kx,

2

3

1

2




26/147------------------------------------------------------------------------------------------------------------------------------Gv: Trn Quang Thu n 0912.676.613 09

k:

>

+

>+

>+

7

);1()5;(

07

0542

x

x

x

xx)1()5;7( + x

T pt7

1log2)54(log 2

2

2+

>+x

xx 2 22 2

27log ( 4 5) log ( 7)

5 x x x x

+ > + <

Kt hp iu kin: Vy BPT c nghim: )5

27;7(

x

0.75

3 Ta c: x.sin2x = 2x x.sin2x 2x = 0 x(sin2x 2) =0 x = 0Din tch hnh phng l:

==2

0

2

0)22(sin)22sin.(

dxxxdxxxxS

t

=

=

=

=

xx

v

dxdu

dxxdv

xu

22

2cos)22(sin 44424

222

=+= S (vdt)

0.75

Bi3

1 Gi Q, I, J ln lt l trung im BC, BB, CC

ta c:

2

3aAP = 3aAH =

V ''AHA vung cn ti H.

Vy 3' aHA = Ta c

4

3

2

3.

2

12

aaaSABC == (vdt)

4

3

4

3.3

32

'''

aaaV CBABCA == (

vtt) (1)V ''AHA vung cn

( )CCBBHKAAHK ''' G i E = MN KH BM = PE= CN (2)

m AA = 22' AHHA + = 633 22 aaa =+

4

6

2

6 aCNPEBM

aAK ====

Ta c th tch K.MNJI l:

1.

31 1 6

'2 4 4

MNJIV S KE

aKE KH AA

=

= = =

26 6

. . ( )4 4

MNJI

a aS MN MI a dvdt = = =

2 31 6 6

( )3 4 4 8

KMNJI

a a aV dvtt = =

3 3

2 3

' ' '

318 8

3 2

8 8

ABCKMN

A B C KMN

a a

V

a aV

= =

+

1

45

E

K

J

I

A

B

C

C'

B'

A'

P

H

Q

N

M




27/147------------------------------------------------------------------------------------------------------------------------------Gv: Trn Quang Thu n 0912.676.613 09

2 K: 02 + aa

T (1) 06)(5)( 222 =++ aaaa

=+

=+

6

1

2

2

aa

aa

Khi 12 =+ aa thay vo (2)

2

1 23.

26 0

1 23.

2

ib

b bi

b

=

= +

=

;

+=

=

=++

2

31

2

31

012

ia

ia

aa

Khi 62 =+ aa

=

=

2

3

a

aThay vo (2) 2

1 5

26 6 6 0

1 5

2

b

b b

b

+=

+ =

=

Vy h pt c nghim (a, b) l:

+

2

31;

2

231,

2

31;

2

231 iiii

+

+

2

31;

2

231,

2

31;

2

231 iiii;

+

+

2

51;2,

2

51;2,

2

51;3,

2

51;3

Bi4 1)

=


28/147------------------------------------------------------------------------------------------------------------------------------Gv: Trn Quang Thu n 0912.676.613 09

=

2255

6;0 aAB ; 2 2 2

10 100 100 12525 25 25

3 9 9 9a a a = = = =

3

55= a Vy phng trnh ng thng:

3

55,

3

55=

= xx

3)ng thng d2 c PTTS l:

+=

+=

+=

'51

'2

'21

tz

ty

tx

vectCP ca d1 v d2 l:1 2

(1;1; 1), (2;1;5)d d

u u= =

VTPT ca mp() l1 2. (6; 7; 1)

d dn u u

= =

pt mp() c dng 6x 7y z + D = 0ng thng d1 v d2 ln lt i qua 2 M(2; 2; 3) v N(1; 2; 1)

( , ( )) ( , ( ))

|12 14 3 | | 6 14 1 |

| 5 | | 9 | 7

d M d N

D D

D D D

=

+ = +

+ = + =

Vy PT mp() l: 3x y 4z + 7 0=

Bi 5

Ta c: P + 3 = 22

32

2

32

2

3

111a

a

cc

c

bb

b

a+

+

++

+

++

+

24

1

121224

6 2

2

2

2

3b

b

a

b

aP

++

+

+

+

=+ 24

1

1212

2

2

2

2

3c

c

b

c

b ++

+

+

+

+

24

1

1212

2

2

2

2

3 a

a

c

a

c ++

+

+

+

+ 3

6

3

6

3

6

2163

2163

2163

cba++

6

222

3 82

9)(222

3

22

3=+++ cbaP

2

3

22

3

22

9

22

3

22

96 3

== P

PMin khi a = b = c = 1




29/147..Gv: Trn Quang Thu n 0912.676.613 091.565

1


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

THI THI HC CAO NG 2010MN: TON


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

( thi gm 2 trang )

I. PHN CHUNG DNH CHO TT C TH SINH (7,0 im)

Cu I: (2,0 im) Cho hm s 4 2 2 42 2y x m x m m= + + (1), vi m l tham s.1. Kho st s bin thin v v th ca hm s (1)khi 1m = .2. Chng minh th hm s (1) lun ct trc Ox ti t nht hai im phn bit,

vi mi 0m < .

Cu II: (2,0 im)1. Gii phng trnh :

2sin 2 4sin 16

x x

+ + =

.

2. Tm cc gi tr ca tham sm sao cho h phng trnh2

1

y x m

y xy

=

+ =

c nghim

duy nht.Cu III: (2,0 im)

1. Tm nguyn hm ca hm s ( )( )

( )

2

4

1

2 1

xf x

x

=

+.

2. Vi mi s thc dng ; ;x y z tha iu kin 1x y z+ + . Tm gi tr nh nht

ca biu thc: 1 1 12P x y zx y z

= + + + + +

.

Cu IV: (1,0 im) Cho khi t din ABCD. Trn cc cnhBC,BD,ACln lt lycc im M, N, P sao cho 4 , 2BC BM BD BN = = v 3AC AP= . Mt phng (MNP) chiakhi t dinABCD lm hai phn. Tnh t s th tch gia hai phn .

II. PHN RING (3,0 im)

Tt c th sinh chc lm mt trong hai phn: A hoc B.

A. Theo chng trnh ChunCu Va: (1,0 im )Trong mt phng ta (Oxy), chong thng ( ) : 2 4 0d x y = .

Lp phng trnh ng trn tip xc vi cc trc ta v c tm trn ng thng (d).Cu VIa: (2,0 im)

1. Gii phng trnh : log log4 22 8x xx = .

2. Vit phng trnh cc ng thng ct th hm s 12

xy

x

=

ti hai im

phn bit sao cho honh v tung ca mi im l cc s nguyn.




30/147..Gv: Trn Quang Thu n 0912.676.613 091.565

2

B. Theo chng trnh Nng caoCu Vb: (1,0 im) Trong khng gian Oxyz , cho cc im

( ) ( ) ( )1;3;5 , 4;3;2 , 0;2;1A B C . Tm ta tm ng trn ngoi tip tam gicABC.

Cu VIb: (2,0 im)1. Gii bt phng trnh :

( )2 4 82 1 log log log 0x x x+ + < .2. Tm m th hm s ( )3 25 5y x m x mx= + c im un trn th

hm s 3y x= .

........HT.......................................




31/147..Gv: Trn Quang Thu n 0912.676.613 091.565

3

I HC SPHM H NIKHOA TON - TIN

P N THI THI HC CAO NG 2010

Mn thi: TON

CU NI DUNG IM

Khi 4 21 2 3m y x x= = + .Tp xc nh D=R .

0,25

Gii hn: lim ; limx x

y y +

= + = + .

( )3 2' 4 4 4 1y x x x x= = . ' 0 0, 1y x x= = = .0,25

Bng bin thin:

Hm sng bin trn khong ( ) ( )1; 0 , 1; + v nghch bin

trn khong ( ) ( ); 1 , 0;1 .

Hm st C ti 0, 3CDx y= = v t CT ti 1, 2CTx y = .

0,25

1(1,0)

th ct Oy ti (0;3). thi xng qua Oy. 0,25 Phng trnh HG ca th (1) v Ox:

4 2 2 42 2 0x m x m m + + = ().0,25

t ( )2 0t x t= , ta c : 2 2 42 2 0t m t m m + + = (). 0,25

Ta c : ' 2 0m = > v 22 0S m= > vi mi 0m > .Nn PT () c nghim dng.

0,25

Cu I(2,0)

2(1,0)

KL: PT () c t nht 2 nghim phn bit (pcm). 0,25

PT 3 sin 2 cos 2 4sin 1 0x x x + + = 22 3 sin cos 2sin 4sin 0x x x x + = .

0,25

( )2 3 cos sin 2 sin 0x x x + = . 0,25

Khi : 5sin 3 cos 2 sin 1 23 6

x x x x k

= = = +

. 0,25

1(1,0)

Khi: sin 0x x k = = .

KL: nghim PT l 5, 26

x k x k

= = + . 0,25

Ta c : 2x y m= , nn : 22 1y my y = . 0,25

PT1

12

y

m yy

= +

( v y = 0 PTVN). 0,25

Cu II(2,0)

2(1,0)

Xt ( ) ( ) 21 1

2 ' 1 0f y y f yy y

= + = + > 0,25




32/147


33/147..Gv: Trn Quang Thu n 0912.676.613 091.565

5

Khi: 2t= th 2log 2 4( )x x th= = . KL: Nghim PT2, 4x x= = .

0,25

Ta c: 112

yx

= +

0,25

Suy ra: ; 2 1 3, 1x y Z x x x = = = 0,25

Ta cc im trn th c honh v tung lnhng s

nguyn l ( ) ( )1;0 , 3;2A B

0,25

2(1,0)

KL: PT ng thng cn tm l 1 0x y = . 0,25

Ta c: ( )3;0; 3 3 2AB AB= =

. 0,25 Tng t: 3 2BC CA= = . 0,25 Do : ABC u, suy ra tm I ng trn ngoi tipABC ltrng tm ca n.

0,25

Cu Vb(1,0)

KL:5 8 8

; ;3 3 3I

. 0,25

K : 0x > . t 2logt x= , ta c : ( )1 03t

t t+ + < 0,25

BPT 2 43 4 0 03

t t t + < < < . 0,25

1(1,0)

KL: 2 34 1

log 0 13 2 2

x x < < < < . 0,50

Ta c: ( )2' 3 2 5 5 ; " 6 2 10y x m x m y x m= + = + . 0,25

5" 03

my x = = ; yi du qua 53

mx = .

Suy ra: ( ) ( )3

2 5 5 55;

3 27 3

m m mmU

+

l im un0,50

Cu VIb(2,0)

2(1,0)

KL: 5m = . 0,25

..HT...




34/147

.

Gv: Trn Quang Thu n 0912.676.613 091.5

1

I HC SPHM H NI

KHOA TON TIN

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

THI THI HC CAO NG 2010

MN: TON


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

I. Phn chung cho tt c th sinh (7 im)

Bi 1: Cho hm s 3 21 5

3

3 3

y x x x= + +

1. Kho st v v th hm s.2. Gi A v B l giao im ca (C) v trc Ox. Chng minh rng trn th (C) tn ti hai im cngnhn on AB di mt gc vung.Bi 2 : Gii phng trnh sau

1. 2 2 2sin cos 02 4 2

x xtg x

=

2. 32222 2

=+ xxxx

Bi 3 : 1. Tm gi tr ln nht v gi tr nh nht ca hm s y =1

12

+

+

x

xtrn on [1; 2].

2. Tnh tch phn : I = dxxx 2

0

2

Bi 4: Cho hai mp(P) v (Q) vung vi nhau, c giao tuyn l ng thng (). Trn () ly hai im Av B vi AB = a. Trong mp(P) ly im C, trong mp(Q) ly im D sao cho AC, BD cng vung gc vi() v AC = BD = AB.Tnh bn knh mt cu ngoi tip t din ABCD v tnh khong cch t A n mp (BCD) theo a.II) Phn ring (3 im)- Th sinh chc lm 1 trong 2 phn.

A. PHN IBi 5a1. Trong mp Oxy cho ng trn (C) :

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

Vit phng trnh ng trn (C) i xng vi ng trn (C) qua (d). Tm ta cc giao im ca(C) v (C).2. Trong khng gian Oxyz cho ng thng (dk): l giao tuyn ca hai mt phng(P): x +3ky z +2 = 0; (Q): kx y +z +1 = 0Tm k ng thng (dk) vung vi(R): x y 2z +5 = 0Bi 6a. T mt t gm 6 bn nam v 5 bn n, chn ngu nhin 5 bn xp vo bn u theo nhng tht khc nhau. Tnh xc sut sao cho trong cch xp trn c ng 3 bn nam.B. PHN IIBi 5b. Cho hai ng thng

( )

x 2 t

d : y 1 t

z 2t

= + = =

v ( )

x 0 2t '

d' : y 3

z 1 t '

= = = +

1) Chng minh (d) v (d) cho nhau. Hy vit pt ng vung gc chung ca (d) v (d).2) Vit pt mp song song cch u (d) v (d)

Bi 6b: Cho hm s2 2 4

2

x xy

x

+=

c th (C), chng minh (C) c tm i xng

________________HT ________________




35/147

.

Gv: Trn Quang Thu n 0912.676.613 091.5

2

P N

Ni dung Li gii chi titBi 1: Cho hm s

3 21 533 3

y x x x= + +

1) Kho st v v thhm s.

3 21 533 3

y x x x= + + , D = R

y = x2 +2x 3 , y = 0 1

3

x

x

=

=

lim , limx x

y y +

= = +

y tng trn (- ; -3) v (1 ; + )y gim trn (-3; 1),C(-3; -32/2) v CT(1; 0)

+

-

+

0

32

3

BBT_

f(x)

f'(x)

x - +1-3

00 +

2) Gi A v B l giao imca (C) v trc Ox. Chng

minh rng trn th (C)tn ti hai im cng nhnon AB di mt gcvung.

Phng trnh honh 3 211 5

3 053 3

x x x x

x

=+ + =

=

=> A(-5; 0) v B(1; 0)

Gi M thuc (C) => M 3 21 5

; 33 3

a a a a

+ +

khc A v B

3 2 3 21 5 1 55 ; 3 ; 1 ; 33 3 3 3

AM a a a a BM a a a a

+ + + + +

Theo gi thit AM BM . 0AM BM =

(a +5)(a -1) +[1

3(a -1)2(a +5)]2 = 0

Do M khc A v B nn a khc -5 v a khc 1 nn pt trn tng ng

1 +1

9(a -1)3(a +5) = 0 hay a4 +2a3 -12a2 +14a +4 = 0 (*)

t y = a4 +2a3 -12a2 +14a +4 c tp xc nh D = R

y = 4a3 +6a2 -12a +14 ; y = 0 c 1 nghim thc0 0

7 2043a y

2 16

=>

++

9

+

y0


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3

S x = 2k+ v x =

k+4

(k Z)

2) 32222 2

=+ xxxx

t t =xx 22 , vi t > 0 th phng trnh thnh: 34 =

tt t2 3t 4 = 0

t = 1 (loi) hoc t = 4

xx

2

2 = 22 x2 x 2 = 0S x = 1 v x = 2

Bi 3:1) Tm gi tr ln nht vgi tr nh nht ca hm s

y =1

12

+

+

x

xtrn on

[1; 2].

y =1

12

+

+

x

xc min xc nh D = [1; 2]

y =)1(1

122

++

xx

x= 0 khi x = 1

So snh cc gi tr y(1) = 2 ; y(-1) = 0 v y(2) =5

53ta c

max y = y(1) = 2 ; min y = y(-1) = 0

2) Tnh tch phn

I = dxxx 2

0

2 I = ( ) ( )dxxxdxxxdxxx

+=

2

1

21

0

22

0

2

= 12323

2

1

231

0

23

=

+

xxxx

Bi 4: Cho hai mp(P) v(Q) vung vi nhau, cgiao tuyn l ng thng(). Trn () ly hai imA v B vi AB = a. Trongmp(P) ly im C, trongmp(Q) ly im D sao cho

AC, BD cng vung gcvi () v AC = BD = AB.Tnh bn knh mt cungoi tip t din ABCDv tnh khong cch t An mt phng (BCD) theoa.

Cch 1: Cc gc B, C u nhn DC dimt gc vung nn mt cu ngoi tipc tm I l trung im ca BC.V AF// BD => F l trung im ca BC

V ABC vung cn nn AF BC (1)V DB (ABC)=> DB AF (2)(1) v (2) th AF (DBC) Nn d(A;

(BCD)) =AF =2

2a

Cch 2: Chn h trc Oxyz sao cho A(0; 0; 0), B(0; a; 0), C(0; 0; a), I(x; y;

z). Theo gi thit ta c: IA = IB = IC = ID =21

CD = R

x = y = z =2

a R = IA =

2

3a=> n BCD = (0; a

2; a2)

Pt (BCD): y +z a = 0 => d(A; (BCD)) =2

2a

PHN RING

A. Phn 1Bi 5a :1) Trong mp Oxy chong trn (C) :(x 1)2 +(y 2)2 = 4 vng thng(d) : x y 1 = 0.Vit phng trnh ngtrn (C) i xng ving trn (C) qua (d).

(C) c tm I(1 ; 2) v bn knh R = 2.() l ng thng qua I v () (d) c pt : (x 1) +(y 2) = 0

Giao im H ca () vi (d):1 0

3 0

x y

x y

=

+ ==> H(2; 1)

Gi I l im i xng ca I qua (d) th H l trung im ca II.p dng cng thc trung im => I(3; 0)V R = R = 2 nn phng trnh (C): (x 3)2 +y2 = 4.




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4

Tm ta cc giao imca (C) v (C). Gii h

( ) ( )

( )

( ) ( )2 2 2 2

2 2

( ) : 1 2 4 1 2 4

1 0( ') : 3 4

C x y x y

x yC x y

+ = + =

= + =

Ta c giao im A(1; 0) v B(3; 2)2) Trong khng gian Oxyzcho ng thng (dk): lgiao tuyn ca hai mtphng(P): x +3ky z +2 = 0,

(Q): kx y +z +1 = 0Tm k ng thng (dk)vung vi(R): x y 2z +5 = 0

(P) c vtpt ( )1;3 ; 1n k=

, (Q) c vtpt ( )' ; 1;1n k=

=> ng thng dk c vtcp ( )2, ' 3 1; 1; 3 1u n n k k k = =

Ga thit dk vung (R) nn ta c , 0Ru n =

2

2

3 2 1 0

3 6 3 0 1

2 2 0

k k

k k k

k

+ + =

+ = =

+ =

. Vy k = 1 l p s

Chn ngu nhin 5 bn v sp th t (ch ngi)

=> khng gian mu gm 511A (phn t)

K hiu A l bin c: Trong cch xp trn c ng 3 bn namChn 5 bn, trong

- Chn 3 nam t 6 nam, c 36C cch.

- Chn 2 n t 5 n, c 25C cch.- Xp 5 bn chn vo 5 v tr khc nhau th c 5! Cch.-T theo quy tc nhn ta c bin c A chn 5 ngi vo 5 v tr (trong

c ng 3 nam) l: n(A) = 36C .25C .5!

Bi 6a:T mt t gm 6 bn namv 5 bn n, chn ngunhin 5 bn xp vo bnu theo nhng th t khcnhau. Tnh xc sut sao

cho trong cch xp trn cng 3 bn nam.

Vy:3 26 5

511

. .5!( ) 0,433

C CP A

A=

B. Phn 2

Bi 5b. Cho hai ngthng

( )x 2 t

d : y 1 t

2t

= + = =

v

( )

x 0 2t '

d' : y 3

z 1 t '

= = = +

1) Chng minh (d) v (d)cho nhau. Hy vitphng trnh ng vunggc chung ca (d) v (d).

d qua im M(2; 1; 0) c vtcp ( )1; 1; 2u =

d qua im M(0; 3; 1) c vtcp ( )' 2;0;1u=

( )' 2; 2; 1M M

v ( ), ' 1; 5; 2u u =

, ' . ' 2 10 2 10 0u u M M = + + =

nn d v d cho nhau.

ng vung gc chung (D) ct (d) ti I => I(2 +t; 1 t; 2t)(d)v (D) ct (d) ti J =>J(2s; 3; 1 +s) (d)

=> (2 2 ; 2 ; 1 2 ) JI t s t t s= + + +

Theo gi thit ng vung gc chung nn. 0

. ' 0

JI u

JI u

=

=

1

(2 2 ) (2 ) 2 4 2 0 3( 4 2 4 ) 1 2 0

1t s t t s t

t s t ss

+ + + + + =

= + = =

Lc 5 4 2

; ;3 3 3

I

, J(2; 3; 0) v1 5 2

; ;3 3 3

JI

=




38/147


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-------------------------------------------------------------------------------------------------------------------------------------------------Gv: Trn Quang Thu n 0912.

1

THI THI HC, CAO NG NM 2010Mn thi: TON

THI THAM KHO Thi gian: 180 pht ( khng kthi gian giao )--------------------------------------------------------------------------------------------------------------------

Cu 1:( 2,0 im)Cho hm s

2 1

1

xy

x

=

c th (C).

1) Kho st s bin thin v v th (C) ca hm s.2) Tm m ( )m ng thng y x m= + ct (C) ti hai im A, B sao cho 4AB

Cu 2: (2,0 im)1) Gii phng trnh ( )3 cos 2 2cos sin 1 0x x x+ = 2) Gii phng trnh 2 2

2 1 4

2

2log 2log log ( )x x x

x =

Cu 3:(1,0 im)

Tnh tch phn1

0

2I

1

xdx

x=

+

Cu 4:(1,0 im)Cho lng tr ABC.ABC c cnh bn bng a, y ABC l tam gic u, hnh chiu ca A trn

(ABC) trng vi trng tm G ca ABC. Cnh bn to vi y gc 060 . Tnh th tch lng trABC.ABC theo a.

Cu 5:(1,0 im)Trong h to Oxy, cho hai ng thng 1 2: 3 4 20 0, : 1 0d x y d x y+ = + + =

Vit phng trnh ng trn (C) bit rng (C) c bn knh R=5, tip xc vi 1d v c tm nm trn

Cu 6:( 1,0 im)Trong khng gian Oxyz, cho mt cu (S) v mt phng (P) c phng trnh(S): 2 2 2 4 4 2 16 0x y z x y z+ + + = ( ) : 2 2 1 0P x y z+ + =

Vit phng trnh mt phng (Q) song song vi (P) v khong cch t tm mt cu (S) n mt phngbng 3.

Cu 7:( 1,0 im).Cho s phc z tho mn ( )1 3 4i z i+ = . Tnh 2010z .

Cu 8:(1,0 im)

Cho cc s thc khng m x, y, z tho mn 2 2 24

3

x y z+ + = .

Tm gi tr ln nht ca biu thc ( )4

3P x y zx y z

= + + ++ +

.HtTh sinh khng c sdng ti liu. Cn b coi thi khng gii thch g thm.H v tn th sinh:; S bo danh:Ch k gim th:




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2

Hng dn chm TONCu Ni dung iu1

2,0))1,0

1)Kho st s bin thin v v th hm s2 1

1

xy

x

=

1. Tp xc nh: \ {1}D = 2. S bin thin ca hm s* Gii hn ti v cc, gii hn v cc ca hm s. Tim cn ca th hm s.

12

2 1lim lim lim 2

111

x xx

x xyx

x

= = =

=> th hm s nhn ng thng y=2 lm tim cn ngang

1 11 1

2 1 2 1lim lim ;lim lim

1 1x xx x

x xy y

x x+ +

= = + = =

=> th hm s nhn ng thng x=1 lm tim cn ng

0,

* Lp bng bin thin

2

1' 0

( 1)y x D

x

= <

, y khng xc nh x=1

Hm s nghch bin trn tng khong xc nh ca n. Hm s khng c cc tr.

0,

bng bin thin

x - 1 + y - || -

y 2 +

- 2

0.




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3

3. th- Giao ca th hm s v Ox: y=0=>x=1/2

- Giao ca th hm s v Oy: x=0=>y=1- th hm s nhn im I(1;2) lm tm i xng.

0,

)1,0 2)Honh giao im ca ng thng y=x+m (d) v th (C) l nghim ca phng trnh

( ) ( )

2 1

1

2 1 1 (*)

xx m

x

x x x m

= +

= +

( x=1 khng phi l nghim ca (*))2 ( 3) 1 0x m x m + + = (1)

0,

2 2( 3) 4(1 ) 2 5 0m m m m m = = + >

Do (d) lun ct (C) ti hai im phn bit1 1 2 2

( ; ), ( ; )A x y B x y vi1 2,x x l hai nghim ca (1)

0,

Theo vit1 2 1 2

3 ; 1x x m x x m+ = = . V , ( )A B d nn 1 1 2 2;y x m y x m= + = +

( )22 2 2

1 2 1 2 1 22( ) 2 4 2( 2 5)AB x x x x x x m m = = + = +

0,

2 2 21

4 16 2( 2 5) 16 2 3 03

mAB AB m m m m

m

= = = + = =

=

0,

Cu 2:2,0)

1)Gii phng trnh ( )3 cos 2 2cos sin 1 0x x x+ =

3 cos 2 sin 2 2cosx x x + =

3 1

cos 2 sin 2 cos2 2x x x + =

0,

cos 2 cos sin 2 sin cos6 6

x x x

+ =

cos(2 ) cos6

x x

=

0,

I(1;2)

2

1 y

x

O




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I HC SPHM H NI====================================================================

-------------------------------------------------------------------------------------------------------------------------------------------------Gv: Trn Quang Thu n 0912.

4

2 26

( )

2 26

x x k

k

x x k

= +

= +

0,

26( )

2

18 3

x kk

kx

= +

= +

KL

0,

)1,02)Gii phng trnh 2 2

2 1 4

2


x = (1)

KX:x>0

( ) 22 2 22

1 log 2log logx x

x

+ =

0,

2

2 2log 3log 2 0(*)x x + = 0,

t t=log2xThay vo (*) ta c

2 3 2 0

1

2

t t

t

t

+ =

=

=

0,

t=1 ta c log2x=1 x=2


kt hp vi KX phng trnh cho c 2 nghim l x=2 v x=4

0,

Cu 3:1,0) Tnh tch phn

1

0

2I

1

xdx

x=

+

t 2 2t x x t dx tdt = = =

32 4

11

xdx t dt

tx=

++

Nu0 0

1 1

x t

x t

= =

= =

0,

1 132

0 0

4 14 ( 1 )

1 1

tI dt t t dt

t t= = +

+ +

0,




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I HC SPHM H NI====================================================================

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5

3 2 1

0

1 14( ln 1 ) )

3 2t t t t = + +

0,

104ln2

3=

0,

Cu 4:

1,0)

A'

G

M'

C'

B'

C

B

A

Hnh chiu ca AA trn (ABC) l AG nn gc to bi AAv (ABC) l 0' 60AA G = gi Ml trung im BC A,G, M thng hng

0,

t x=AB

ABC u cnh x c AM l ng cao 3 2 3

' ' , ' ' '2 3 3

x xA M A G A M = = =

Trong AAG vung c AG=AAsin600=

3

2

a; 0

3 3' ' os60

2 3 2

a x aA G AA c x= = = =

0,

din tch ABC l2 2

0 21 3 3 3 3 3. .sin 60 ( )

2 4 4 2 16ABC

x a aS AB AC

= = = =

0,

th tch khi lng tr l2 3

. ' ' '

3 3 3 9.

2 16 32ABC A B C ABC

a a aV AG S

= = =

0,

Cu 5:1,0)

Gi s l 2( ; 1 )I t t d tm ca ng trn (C)

V (C) tip xc vi1

d nn

12 2

3 4( 1 ) 20( , ) 5

3 4

t td I d R

+ = =

+

0,

24 25 124 25

24 25 49

t tt

t t

+ = = + =

+ = =

0,

Vi1

1 (1; 2)t I= ta c phng trnh ng trn

( ) ( ) ( )2 2

11 2 25C x y + + =

0,




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6

Vi1

49 ( 49;48)t I= ta c phng trnh ng trn

( ) ( ) ( )2 2

249 48 25C x y+ + =

0,

Cu 6:1,0)

(S): 2 2 2 4 4 2 16 0x y z x y z+ + + =

(S) c tm I(2;2;-1)

phng trnh mt phng (Q) c dng: 2 2 0x y z D+ + =

iu kin 1(*)D

0,

( , ( )) 3d I P =2 2 2

| 2.2 1.2 2( 1) |3

2 1 ( 2)

D+ + =

+ +

0,

1| 8 | 9

17

DD

D

= + =

=

Kt hp vi iu kin (*) ta c D = -17

0,

Vy phng trnh ca (Q) 2 2 17 0x y z+ = 0,Cu 7:1,0)

( )

( )2 2

1 3 4

4 1 343

1 3 1 ( 3)

i z i

i iiz i

i

+ =

= = = ++ +

0,

3 12( ) 2 cos sin

2 2 6 6i i

= + = +

0,

Theo cng thc Moa-vr

2010 2010 2010 20102 cos sin6 6

z i

= +

0,

( )2010 20102 1 2= = 0,

Cu 8:1,0)

t t=x+y+zTa c

( )22 2 2 2 2 2 24 2 33( ) 4 2

3 3

43

x y z x y z x y z t t

A tt

+ + + + + +

= +

0,

Xt hm s

4( ) 3f t t

t= +

trn

2 3;2

3

2

2 2

4 3 4 2 3'( ) 3 0

3

tf t t

t t

= =

2 3'( ) 0

3f t t = =

Hm s f(t) ng bin trn2 3

;23

do ( ) (2) 8f t f =

0




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7

Du ng thc xy ra khi t=2

Do 8A Du = xy ra khi v ch khi ( )2 2 2 2 23( )

32

x y z x y zx y z

x y z

+ + = + + = = =

+ + =

Vy gi tr ln nht ca A l 8

0,




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I HC SPHM H NI================================================================================

THI THI HC - CAO NG NM 2010Mn Thi: TON

THI THAM KHO Thi gian: 180 pht ( khng kthi gian giao )---------------------------------------------------------------------------------------------------------------------

I.Phn chung cho tt c th sinh(7 im)

Cu I(2 im). Cho hm s212

+

+=x

xy c th l (C)

1.Kho st s bin thin v v th ca hm s2.Chng minh ng thng: mxy += lun lun ct th (C) ti hai im phn bit A, B. Tm m

on AB c di nh nht.Cu II(2 im)

1.Gii phng trnh 9sinx + 6cosx - 3sin2x + cos2x = 8

2.Gii bt phng trnh )3(log53loglog 242

222 > xxx

Cu III(1 im). Tm nguyn hm =xx

dxI

53

cos.sin

Cu IV(1 im). Cho lng tr tam gic ABC.A1B1C1 c tt c cc cnh bng a, gc to bi cnh bn v mt phy bng 300. Hnh chiu H ca im A trn mt phng (A1B1C1) thuc ng thng B1C1. Tnh khong cch hai ng thng AA1 v B1C1 theo a.Cu V(1 im).Cho cc s dng x, y, z tha mn 2 2 2x y z 1+ + = . Tm gi tr nh nht ca biu thc

2 2 2 2 2 2

x y zP

y z z x x y= + +

+ + +

II.Phn ring(3 im)1.Theo chng trnh chunCu Via:

1.Trong mt phng vi h ta Oxy cho ng trn (C) c phng trnh (x-1) 2 + (y+2)2 = 9 v ng th

d: x + y + m = 0. Tm m trn ng thng d c duy nht mt im A m t k c hai tip tuyn AB, ACng trn (C) (B, C l hai tip im) sao cho tam gic ABC vung.

2.Cho im A(10; 2; -1) v ng thng d c phng trnh

+=

=

+=

tz

ty

tx

31

21

. Lp phng trnh mt phng (P

qua A, song song vi d v khong cch t d ti (P) l ln nht.Cu VIIa:1). C bao nhiu s t nhin c 4 ch s khc nhau v khc 0 m trong mi s lun lun c mt hai s chn v hai ch s l.

2) Gii phng trnh: )(,14

Cziz

iz=

+

2.Theo chng trnh nng cao (3 im)Cu VIb(2 im)

1.Trong mt phng vi h ta Oxy cho ng trn (C): x2 + y2 - 2x + 4y - 4 = 0 v ng thng dphng trnh x + y + m = 0. Tm m trn ng thng d c duy nht mt im A m t k c hai tip tuAB, AC ti ng trn (C) (B, C l hai tip im) sao cho tam gic ABC vung.

2.Cho im A(10; 2; -1) v ng thng d c phng trnh3

1

12

1 ==

zyx. Lp phng trnh mt ph

(P) i qua A, song song vi d v khong cch t d ti (P) l ln nht.Cu VIIb(1 im)C bao nhiu s t nhin c 5 ch s khc nhau m trong mi s lun lun c mt hai chchn v ba ch s l.

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



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P N

I.Phn dnh cho tt c cc th snhCuI p n im

2. (0,75 im)Honh giao im ca th (C ) v ng thng d l nghim ca phng

trnh

=++

+=

+

+

)1(021)4(2

212

2mxmx

xmx

x

x

Do (1) c mmmvam =++>+= 0321)2).(4()2(01 22 nn ngthng d lun lun ct th (C ) ti hai im phn bit A, B

0,25

Ta c yA = m xA; yB = m xB nn AB2 = (xA xB)

2 + (yA yB)2 = 2(m2 + 12)

suy ra AB ngn nht AB2 nh nht m = 0. Khi 24=AB

0,5

1. (1 im)Phng trnh cho tng ng vi9sinx + 6cosx 6sinx.cosx + 1 2sin2x = 8 6cosx(1 sinx) (2sin2x 9sinx + 7) = 0 6cosx(1 sinx) (sinx 1)(2sinx 7) = 0

0,5

(1-sinx)(6cosx + 2sinx 7) = 0

=+

=

)(07sin2cos6

0sin1

VNxx

x

0,25

22

kx += 0,25

2. (1 im)

K:

>

03loglog

02

222 xx

x

Bt phng trnh cho tng ng vi)1()3(log53loglog 2

22

22 > xxx

t t = log2x,

BPT (1) )3(5)1)(3()3(5322 >+> tttttt

0,5

4log3

1log

43

1

)3(5)3)(1(

3

1

2

2

2x

x

t

t

ttt

t

t

0,25

II(2im)


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I HC SPHM H NI================================================================================

Cx

xxxdttt

tt

dtt

ttt

+++=+++=

+++=

22433

3

246

tan2

1tanln3tan

2

3tan

4

1)

33(

133

0,5

Do )( 111 CBAAH nn gc 1AA H l gc gia AA1 v (A1B1C1), theo gi thit

th gc HAA1 bng 300. Xt tam gic vung AHA1 c AA1 = a, gc

HAA1 =300

2

31

aHA = . Do tam gic A1B1C1 l tam gic u cnh a, H

thuc B1C1 v2

31

aHA = nn A1H vung gc vi B1C1. Mt khc 11CBAH

nn )( 111 HAACB 0,5

K ng cao HK ca tam gic AA1H th HK chnh l khong cch gia AA1v B1C10,25

Cu IV

1 im

Ta c AA1.HK = A1H.AH4

3.

1

1 a

AA

AHHAHK ==

0,25

Cu V1 im

T gi thit 2 2 2x y z 1+ + = suy ra 0 x,y,z 1< <

p dng bt ng thc C-si cho ba s dng 2 2 22x ,1 x ,1 x ta c

( ) ( )( ) ( )

( )

2 2 22 2

2 2 2 23 3

2

2

2

2x 1 x 1 x 22x 1 x 2x 1 x

3 3

2x 1 x3 3

x 3 3x

1 x 2

+ +

2

2 2

x 3 3x (1)

y z 2

+

0,5

A1

A B

C

C1

B1

K

H



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I HC SPHM H NI================================================================================

Tng t ta c

2

2 2

2

2 2

y 3 3y (2)

z x 2z 3 3

z (3)x y 2

+

+

Cng tng v (1), (2), (3) ta c

( )2 2 22 2 2 2 2 2x y z 3 3 3 3

x y z (*)y z z x x y 2 2

+ + + + =+ + +

Du bng (*) xy ra khi3

x y z3

= = =

Vy 3 3minP2

=

0,5

1.T phng trnh chnh tc ca ng trn ta c tm I(1;-2), R = 3, t A kc 2 tip tuyn AB, AC ti ng trn v ACAB => t gic ABIC l hnh

vung cnh bng 3 23=IA

0,5

=

===

7

56123

2

1

m

mm

m

0,52. (1 im)

Gi H l hnh chiu ca A trn d, mt phng (P) i qua A v (P)//d, khi khong cch gia d v (P) l khong cch t H n (P).G.s im I l hnh chiu ca H ln (P), ta c HIAH => HI ln nht khi IA

Vy (P) cn tm l mt phng i qua A v nhn AH lm vc t php tuyn.

0,5

CuVIa2im

)31;;21( tttHdH ++ v H l hnh chiu ca A trn d nn

)3;1;2((0. == uuAHdAH l vc t ch phng ca d)

)5;1;7()4;1;3( AHH Vy (P): 7(x 10) + (y 2) 5(z + 1) = 0 7x + y -5z -77 = 0

0,5

1) T gi thit bi ton ta thy c 624 =C cch chn 2 ch s chn (v khng c

s 0)v 1025 =C cch chn 2 ch s l => c25C .

25C = 60 b 4 s tha mn bi

ton

Mi b 4 s nh th c 4! s c thnh lp. Vy c tt c 24C .25C .4! = 1440 s

0,5Cu

VIIa1im

2) pt z(z2 1) = 0 z = 0 ; x = 1; z = -1 0,5

2.Ban nng cao.1.( 1 im)Cu

VIa2im

T phng trnh chnh tc ca ng trn ta c tm I(1;-2), R = 3, t A k c 2tip tuyn AB, AC ti ng trn v ACAB => t gic ABIC l hnh vung cnh

bng 3 23=IA

0,5



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=

===

7

56123

2

1

m

mm

m

0,52.Gi H l hnh chiu ca A trn d, mt phng (P) i qua A v (P)//d, khi khong cch gia d v (P) l khong cch t H n (P).Gi s im I l hnh chiu ca H ln (P), ta c HIAH => HI ln nht khi IA

Vy (P) cn tm l mt phng i qua A v nhn AH lm vc t php tuyn.

0,5

)31;;21( tttHdH ++ v H l hnh chiu ca A trn d nn

)3;1;2((0. == uuAHdAH l vc t ch phng ca d)

)5;1;7()4;1;3( AHH Vy (P): 7(x 10) + (y 2) 5(z + 1) = 0 7x + y -5z -77 = 0

0,5

T gi thit bi ton ta thy c 1025 =C cch chn 2 ch s chn (k c s c ch s

0 ng u) v 35C =10 cch chn 2 ch s l => c25C .

35C = 100 b 5 s c chn.

0,5CuVIIa1im Mi b 5 s nh th c 5! s c thnh lp => c tt c 25C .

35C .5! = 12000 s.

Mt khc s cc s c lp nh trn m c ch s 0 ng u l 960!4..

3

5

1

4=

CC .Vy c tt c 12000 960 = 11040 s tha mn bi ton

0,5



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1

THI THI HC - CAO NG NM 2010Mn thi: TON

THI THAM KHO Thi gian: 180 pht ( khng kthi gian giao )--------------------------------------------------------------------------------------------------------------------

Cu 1:( 2,0 im)Cho hm s

2 1

1

xy

x

=

c th (C).

1) Kho st s bin thin v v th (C) ca hm s.2) Tm m ( )m ng thng y x m= + ct (C) ti hai im A, B sao cho 4AB

Cu 2: (2,0 im)1) Gii phng trnh ( )3 cos 2 2cos sin 1 0x x x+ = 2) Gii phng trnh 2 2

2 1 4

2


x =

Cu 3:(1,0 im)

Tnh tch phn1

0

2I

1

xdx

x=

+

Cu 4:(1,0 im)Cho lng tr ABC.ABC c cnh bn bng a, y ABC l tam gic u, hnh chiu ca A trn

(ABC) trng vi trng tm G ca ABC. Cnh bn to vi y gc 060 . Tnh th tch lng trABC.ABC theo a.

Cu 5:(1,0 im)Trong h to Oxy, cho hai ng thng 1 2: 3 4 20 0, : 1 0d x y d x y+ = + + =

Vit phng trnh ng trn (C) bit rng (C) c bn knh R=5, tip xc vi 1d v c tm nm trn

Cu 6:( 1,0 im)Trong khng gian Oxyz, cho mt cu (S) v mt phng (P) c phng trnh(S): 2 2 2 4 4 2 16 0x y z x y z+ + + = ( ) : 2 2 1 0P x y z+ + =

Vit phng trnh mt phng (Q) song song vi (P) v khong cch t tm mt cu (S) n mt phngbng 3.

Cu 7:( 1,0 im).Cho s phc z tho mn ( )1 3 4i z i+ = . Tnh 2010z .

Cu 8:(1,0 im)

Cho cc s thc khng m x, y, z tho mn 2 2 24

3

x y z+ + = .

Tm gi tr ln nht ca biu thc ( )4

3P x y zx y z

= + + ++ +

.HtTh sinh khng c sdng ti liu. Cn b coi thi khng gii thch g thm.H v tn th sinh:; S bo danh:Ch k gim th:




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2

Hng dn chm TONCu Ni dung iu12,0))1,0

1)Kho st s bin thin v v th hm s2 1

1

xy

x

=

1. Tp xc nh: \ {1}D = 2. S bin thin ca hm s* Gii hn ti v cc, gii hn v cc ca hm s. Tim cn ca th hm s.

12

2 1lim lim lim 2

111

x xx

x xyx

x

= = =

=> th hm s nhn ng thng y=2 lm tim cn ngang

1 11 1

2 1 2 1

lim lim ;lim lim1 1x xx x

x x

y yx x+ +

= = + = =

=> th hm s nhn ng thng x=1 lm tim cn ng

0,

* Lp bng bin thin

2

1' 0

( 1)y x D

x

= <

, y khng xc nh x=1

Hm s nghch bin trn tng khong xc nh ca n. Hm s khng c cc tr.

0,

bng bin thinx - 1 +

y - || -

y 2 +

- 2

0.

3. th- Giao ca th hm s v Ox: y=0=>x=1/2

- Giao ca th hm s v Oy: x=0=>y=1- th hm s nhn im I(1;2) lm tm i xng.

0,

I(1;2)

2

1 y

x

O




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3

)1,0 2)Honh giao im ca ng thng y=x+m (d) v th (C) l nghim ca phng trnh

( ) ( )

2 1

1

2 1 1 (*)

xx m

x

x x x m

= +

= +

( x=1 khng phi l nghim ca (*))2 ( 3) 1 0x m x m + + = (1)

0,

2 2( 3) 4(1 ) 2 5 0m m m m m = = + >

Do (d) lun ct (C) ti hai im phn bit1 1 2 2

( ; ), ( ; )A x y B x y vi1 2,x x l hai nghim ca (1)

0,

Theo vit1 2 1 2

3 ; 1x x m x x m+ = = . V , ( )A B d nn 1 1 2 2;y x m y x m= + = +

( )22 2 2

1 2 1 2 1 22( ) 2 4 2( 2 5)AB x x x x x x m m = = + = +

0,

2 2 21

4 16 2( 2 5) 16 2 3 03

mAB AB m m m m

m

= = = + = =

=

0,

Cu 2:2,0) 1)Gii phng trnh ( )3 cos 2 2cos sin 1 0x x x+ = 3 cos 2 sin 2 2cosx x x + =

3 1cos 2 sin 2 cos

2 2x x x + =

0,

cos 2 cos sin 2 sin cos6 6

x x x

+ =

cos(2 ) cos6x x

=

0,

2 26

( )

2 26

x x k

k

x x k

= +

= +

0,

26

( )2

18 3

x k

kk

x

= +

= +

KL

0,

)1,02)Gii phng trnh 2 2

2 1 4

2


x = (1)

KX:x>0

( ) 22 2 22

1 log 2log logx xx

+ =

0,




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4

2

2 2log 3log 2 0(*)x x + = 0,

t t=log2xThay vo (*) ta c

2 3 2 0

1

2

t t

t

t

+ =

=

=

0,



kt hp vi KX phng trnh cho c 2 nghim l x=2 v x=4

0,

Cu 3:1,0) Tnh tch phn

1

0

2I

1

xdx

x=

+

t 2 2t x x t dx tdt = = =

32 4

11

xdx t dt

tx=

++

Nu0 0

1 1

x t

x t

= =

= =

0,

1 132

0 0

4 14 ( 1 )

1 1

tI dt t t dt

t t= = +

+ +

0,

3 2 1

01 14( ln 1 ) )3 2

t t t t = + + 0,

104ln2

3=

0,

Cu 4:1,0)

A'

GM'

C'

B'

C

B

A

0,




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5

Hnh chiu ca AA trn (ABC) l AG nn gc to bi AAv (ABC) l 0' 60AA G = gi Ml trung im BC A,G, M thng hngt x=AB

ABC u cnh x c AM l ng cao 3 2 3

' ' , ' ' '2 3 3

x xA M A G A M = = =

Trong AAG vung c AG=AAsin600=

3

2

a; 0

3 3' ' os60

2 3 2

a x aA G AA c x= = = =

0,

din tch ABC l2 2

0 21 3 3 3 3 3. .sin 60 ( )

2 4 4 2 16ABC

x a aS AB AC

= = = =

0,

th tch khi lng tr l2 3

. ' ' '

3 3 3 9.

2 16 32ABC A B C ABC

a a aV AG S

= = =

0,

Cu 5:

1,0)

Gi s l 2( ; 1 )I t t d tm ca ng trn (C)

V (C) tip xc vi 1d nn

12 2

3 4( 1 ) 20( , ) 5

3 4

t td I d R

+ = =

+

0,

24 25 124 25

24 25 49

t tt

t t

+ = = + =

+ = =

0,

Vi 11 (1; 2)t I= ta c phng trnh ng trn

( ) ( ) ( )2 2

1 1 2 25C x y + + =

0,

Vi1

49 ( 49;48)t I= ta c phng trnh ng trn

( ) ( ) ( )2 2

2 49 48 25C x y+ + =

0,

Cu 6:1,0)

(S): 2 2 2 4 4 2 16 0x y z x y z+ + + =

(S) c tm I(2;2;-1)

phng trnh mt phng (Q) c dng: 2 2 0x y z D+ + = iu kin 1(*)D

0,

( , ( )) 3d I P =2 2 2

| 2.2 1.2 2( 1) |3

2 1 ( 2)

D+ + =

+ +

0,

1| 8 | 9

17

DD

D

= + =

=

Kt hp vi iu kin (*) ta c D = -17

0,

Vy phng trnh ca (Q) 2 2 17 0x y z+ = 0,Cu 7:1,0)

( )

( )2 2

1 3 4

4 1 343

1 3 1 ( 3)

i z i

i iiz i

i

+ =

= = = ++ +

0,




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6

3 12( ) 2 cos sin

2 2 6 6i i

= + = +

0,

Theo cng thc Moa-vr

2010 2010 2010 2010

2 cos sin6 6z i

= +

0,

( )2010 20102 1 2= = 0,

Cu 8:1,0)

t t=x+y+zTa c

( )22 2 2 2 2 2 24 2 3

3( ) 4 23 3

43

x y z x y z x y z t t

A t

t

+ + + + + +

= +

0,

Xt hm s4

( ) 3f t t t

= +

trn

2 3;2

3

2

2 2

4 3 4 2 3'( ) 3 0

3

tf t t

t t

= =

2 3'( ) 0

3f t t = =

Hm s f(t) ng bin trn2 3

;23

do ( ) (2) 8f t f =

Du ng thc xy ra khi t=2

0

Do 8A Du = xy ra khi v ch khi ( )2 2 2 2 23( )

32

x y z x y zx y z

x y z

+ + = + + = = =

+ + =

Vy gi tr ln nht ca A l 8

0,




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Gv: Trn Quang Thu n 0912.676.613 09


P N THI THmn : Ton

hng dn chm v biu im

Ni dung imCu I : (3,0 im)

Cho hm s y = x3

- 3x2

+2 c th (C) trong h ta Oxy1. Kho st v v th hm s2. Gi E l tm i xng ca th (C).Vit phng trnh ng thng qua E v ct (C) ti

ba im E, A, B phn bit sao cho din tch tam gic OAB bng 2

a) Tp xc nh : R 0,25b) S bin thin

* Gii hnx -

, limyx

Limy+

= + = 0,25

1.

(2,0)

* Bng bin thin

y = 3x

2

-6x , y= 0

0

2

x

x

=

=

x - 0 2 +y + 0 - 0 +

y2 +

- -2

Hm sng bin trn cc khong (- ;0) v ( 2 ; +)

Nghch bi

n trn (0; 2)Hm st cc i ti x = 0, yc = 2

t cc tiu ti x =2, yct = -2

0,25

0,5

0,25

c. th+ im cc i, cc tiu :(0;2), (2;-2)+ Giao vi Oy : (0;2)+ Giao vi Ox :

NX :0,5

1

2E

O x

y




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+E (1;0)0,25

2

(1,0)

+ PT ng thng qua E, tha mn yu cu bi ton phi c dng y = k(x-1)( Do trng hp x =1 khng tha mn)

Hong giao im ca (C ) v l nghim ca PT: (x-1)(x2-2x-2-k)=0

+ ct (C ) ti 3 im phn bit th PT x2

-2x-2-k = 0 phi c hai nghim phnbit khc 1 k>-3 0,25

+ Tnh c dtOAB =1

( , ).2

d O AB = 3k k+

0,25

+ T gi thit suy ra k c 3 gi tr -1; -1 3 .

KL : C 3 ng thng tha mn yu cu l y = -x +1 ; y = ( ) ( )1 3 1x 0,25

Cu II : (2,0 im)1. Tm gi tr ln nht v nh nht ca hm s y = 2cosx + sin2x trn [0; 2]

2. Tnh tch phn

( ) ( )

3 2

2 2

0 1 1 2 1

x I dx

x x

=

+ + + +

+ H m s lin tc trn [0;2]

+ Tnh y = 2cos2x - 2sinx, [ ]0;2x

y= 0 5 3

; ;6 6 2

x

0,5

1.

(1,0)+) y(0)=2, 3 3 5 3 3 3( ) ; ( ) ; ( ) 0; (2 ) 2

6 2 6 2 2 y y y y = = = =

0,25

Suy ra[ ] [ ]0;20;2

3 3 3 3ax , min

2 2m y y

= = 0,25

+ t 2 1 x t+ + = x =(t-2)2 -1, dx = 2(t-2)dt ; x =0 t =3, x = 3 t = 4 0,25

2.

(1,0) + a v4

2

3

42 362 16 I t dt

t t

= +

0,25

+ Tnh ra c I = -12+ 42ln4

3 0,5

Cu III : (1,0 im)Cho hnh chp t gic u S.ABCD, bit khong cch gia AB v mt phng(SCD) bng 2. Gc gia mt bn v mt y bng 600 .Tnh th tch hnh chpS.ABCD




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+ Goij I, J ln lt l trung im caAB v CD, H l hnh chiu ca I trnSJ. Chng tc IH = 2 v

gc 060SJI =

+ Gi O l tm y, chng minh c

SO = 2,4

IJ=3

+ Tnh c VS.ABCD =32

9( vtt)

0,5

0,25

0,25

Cu IV : (1,0 im)Tm cc cp s thc (x ; y) tha mn phng trnh sau:

4 3 2 2 3 21 1 4 2 2 2 2 x x y x y x y x xye e x x y xy x + + ++ = + + +

+ t 4 3 2 2 3 21 , x 1x x y x y u y x xy v + = + + =

PT trthnh 2u ve e u v+ = + + (2)

+ Xt f(t)=et - t - 1. Chng tc( ) 0,

( ) 0 0

f t t

f t t

= =

T PT (2) u = v = 0

0,25

0,25

+ Gii h4 3 2 2

3 2

1 0

1 0

x x y x y

x y x xy

+ =

+ + =

( )2

2 3

2 3

1

1

x xy x y

x xy x y

=

= +.

t2

3

x xy a

x y b

=

= , gii ra ta c

1

0

a

b

=

=hoc

2

3

a

b

=

=

+ Thay trli tm c hai cp (x;y) l (1;0) v (-1;0) . Kt lun

0,25

0,25

Cu V : (2,0 im)Trong khng gian vi h ta Oxyz, cho im M( 1; -1; 1) v hai ng

thng 11

:1 2 3

x y zd

+= =

v 2

1 4:

1 2 5

x y zd

= =

1. Chng minh rng im M v cc ng thng d1 v d2 cng nm trn mt mtphng. Vit phng trnh mt phng

2. Gi A, B, C ln lt l hnh chiu ca im M trn Ox, Oy, Oz . Vit phngtrnh ng thng nm trong mt phng (ABC) sao cho ct ng thng

(d2) ng thi vung gc vi (d1)

d1qua M1(0;-1;0), vc tch phng 1 (1; 2; 3)u

d2 qua M2(0;1;-4), 2 (1;2;5)u

0,25

1.

(1,0)

+ Chng t d1 v d2ng phng v vit c PT mp(d1,d2) : - x - 2y + z -2 = 0

+ Chng t Mmp(d1,d2). Kt lun

0,5

0,25

S

A

BC

D

IJ

60

0

O

H




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2.

(1,0)

+ A(1;0;0), B(0; -1;0), C(0;0;1); mp(ABC): x - y + z -1 = 0

+ d2ct (ABC) ti H(1 3

;0;2 2

+ ng thng cn tm c vc tch phng ( )1, ABCu u n =

=(-5;-4;1) , ng

thi i qua H

Suy ra PT :

15

2

4

3

2

x t

y t

z t

=

=

= +

0,25

0,25

0,25

0,25

Cu VI : (1,0 im)Gii phng trnh sau trn tp cc s phc bit n c mt nghim thc:

3 2(5 ) 4( 1) 12 12 0 z i z i z i + + + =

+ Gi nghim thc l a thay vo pt suy ra h3 2

2

5 4 12 06

4 12 0

a a aa

a a

= =

+ + = 0,25

+ Khi PT cho tng ng vi

( )( )2

2

6 (1 ) 2 2 0

6

(1 ) 2 2 0

z z i z i

z

z i z i

+ + =

=

+ + =

0,25

+ Gii ra c cc nghim l 6, 2i v -1-i . Kt lun 0,5

- Trn y ch l hng dn lm bi; phi l lun hp l mi cho im- Nhng cch gii khc ng vn c im ti a- im ton bi c lm trn n 0,5




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2

B. Theo chng trnh Nng caoCu Vb:1. Trong mpOxy, cho ng trn (C): x2 + y2 6x + 5 = 0. Tm M thuc trc tung sao cho qua

M kc hai tip tuyn ca (C) m gc gia hai tip tuyn bng 600.2.Trong khng gian vi h ta Oxyz, cho im M(2 ; 1 ; 0) v ng thng d vi

d :x 1 y 1 z

2 1 1

+= =

.Vit phng trnh chnh tc ca ng thng i qua im M,

ct v vung gc vi ng thng d v tm to ca im M i xng vi M qua d

Cu VIb: Gii h phng trnh

3 3log log 2

2 2

4 4 4

4 2 ( )

log ( ) 1 log 2 log ( 3 )

xyxy

x y x x y

= +

+ + = + +

....HT.

(Cn b coi thi khng gii thch g thm)




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