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THI TH I H C S 01

PH N I. PH N CHUNG (Dnh cho t t c cc th sinh)

Cu I. Cho hm s : 3 2 22 11 4 33 2

y x m x m m x .

1. Kh o st v v th c a hm s khi m = -3.

2. V i gi tr no c a m hm s c c i, c c ti u? G i x1, x2 m c i, c c ti uc a th hm s , hy tm gi tr l n nh t c a bi u th c 1 2 1 2. 2 x x x x .

Cu II.

1. Gi nh 4 421 cot 2 cot 2 sin cos 3cosx x x x

x

2. Tm cc gi tr c a tham s b nh 24 4 5 2 0 x x m x x nghi i

m i gi tr x thu n 2; 2 3

Cu III.

1. Hnh chp S.ABCD c nh ch nh t, 2 AD a , CD = 2a. C nh SA vung gc v i3 2 0SA a a . G i K l trung m c a c nh CD. Ch ng minh m t ph ng (SBK) vung gc

v i m t ph ng (SAC) v tnh th tch kh i chp SBCK theo a.

2. Trong khng gian v i h t ng OAB.O1A1B1 v i A(2; 0; 0), B(0; 4; 0) vO1 nh t m M trn AB m N trn OA1 ng th ng MN song song v im t ph ng ( ): 2 5 0 x y z di MN = 5 .

Cu IV.

1. Tnh t ng:2 2 2 20 1 2

...1 2 3 1

nn n n nC C C C S

n, knC l s t h p

ch p k c a n ph n t .

2. Trong m t ph ng v i h tr c t ng trn (C): 2 2 6 2 6 0 x y x y mB(2; - nh t m A thu ng trn (C) sao cho tam gic ABC cn t m Av c di n tch nh nh t.

PH N II. PH N T CH N (Th sinh ch lm m t trong hai ph n)

Cu Va.

1. Tnh tch phn:ln 5

ln 2 10 1 1x xdxI

e e.


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2. Gi i h nh:

2

21

22 2

32 2 42

2 2 4 1 0 5

x

yx xy

x y x x y x

Cu Vb.

1. Tnh tch phn:4

30

sincosx x I dx

x.

2. Gi nh 22 7 7 2log log 3 2 log 3 log2x x x x x x.

THI TH I H C S 02


Cu I. Cho hm s 3 22 3 1 2 y x mx m x (1) (m l tham s th c)

1. Kh o st s bi n thin v v th hm s (1) khi m = 0.

2. ng th ng : 2y x . Tm cc gi tr c ng th ng c th

hm s (1) t m A(0; 2); B, C sao cho tam gic MBC c di n tch b ng 2 6 .

Cu II.1. Gi nh 2 22sin sin 2 cos sin 2 1 2cos

4 x x x x x

2. Tm cc gi tr c a tham s h nh sau c nghi m th c duy nh t

2 2

1 1 x y x y

x y m.

Cu III.

1. Cho hnh chp S.ABCD c nh thoi c nh a (a > 0). Gc ABC b ng 120o, c nh SA

vung gc v i m t ph ng (ABCD) v SA = a. G i C m c a c nh SC. M t ph ng (AC v song song v i BD c t cc c nh SB, SD l t t i B , D . Tnh th tch kh i c a chpS.AB C D .

2. Trong khng gian v i h tr c t m A(-1; 0; 2), m t ph ng (P): 2 3 0 x y z v

ng th ng (d):23 6

2 4 1yx z . Vi nh ng th ng (d m A, c t (d) t i B

v c t (P) t i C sao cho 2 0 AC ABuuur uuur r

.

Cu IV.

1. Cho s ph c ; ,z x yi x y Z th a mn 3 18 26z i . Tnh 2009 20092 4T z z


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2. Cho cc s th c khng m x, y, z th a mn 3 z y z . Tm gi tr nh nh t c a bi u th c:

1 1 14 2 ln 1 4 2 ln 1 4 2ln 1P x y y z z x


Cu Va.

1. Tnh di n tch hnh ph ng gi i h n b ng 2 3x y , 1 0x y .

2. Trong m t ph ng v i h tr c t Oxy cho tam gic ABC c nh A n ng th ng ( ):2 3 14 0x y , c nh BC song song v i nh: 2 1 0x y . Bi t trung

m c a c nh AB l M(- nh t nh A, B, C.

Cu Vb.

1. Cho hnh ph ng H gi i h n b ng 2y x ; 22y x . Tnh th tch c a kh i trn xoay t othnh khi quay hnh H quanh tr c Ox.

2. Trong m t ph ng v i h tr c t m I(-1; 3). Vi nh ng trn c tm I vc ng th ng 3 4 10 0x y t i ha m A, B sao cho AIB b ng 120o.

THI TH I H C S 03


Cu I. m) Cho hm s y =x

x-1(C)

1. Kh o st s bi n thin v v th hm s (C)

2. Vi nh ti p tuy n v th (C), bi t r ng kho ng cch t i x ng c thn ti p tuy n l l n nh t.

Cu II. m)

1. Tm nghi m c nh 2cos4x - ( 3 - 2)cos2x = sin2x + 3 bi t x [ 0 ; ].

2. Gi i h nh3 2 3 2

2

3 5.6 4.2 0

( 2 )( 2 )

x y x x y

x y y y x y x

Cu III. m) Tnh tch phn3

1 42

0

( )1

x x x e dxx

Cu IV m) Cho x, y, z l cc s th n 1 v tho mn u ki n xy + yz + zx2xyz. Tm gi tr l n nh t c a bi u th c A = (x - 1)(y - 1)(z - 1).


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Cu V m) Cho t di n ABCD bi t AB = CD = a, AD = BC = b, AC = BD = c. Tnh th tch c at di n ABCD.


nh nng cao

Cu VIa. m)

1. Trong m t ph ng to ng th ng (d1) : 4x - 3y - 12 = 0 v (d2): 4x + 3y - 12 =0. Tm to ng trn n i ti p tam gic c 3 c nh n m trn (d1), (d2), tr c Oy.

2. Cho hnh l c c nh b ng 2. G m c n AD, Nl tm hnh vung CCDD. Tnh bn knh m t c m B, C, M, N.

Cu VIIa m) Gi i b nh

2 3

3 42log ( 1) log ( 1) 05 6

x xx x

.

nh chu n

Cu VIb. m)

1. Cho elip (E) : 4x2 + 16y2 = 64.G i F1, F2 m b t k trn (E).Ch ng tr ng

t s kho ng cch t M t m F2 v t ng th ng x =8

3c gi tr i.

2. Trong khng gian v i h tr c to m A(1 ;0 ; 1), B(2 ; 1 ; 2) v m t ph ng (Q):

x + 2y + 3z + 3 = 0. L nh m t ph i (Q).

Cu VIIb. m)

Gi i b nh 2 2 321 6

102 x x x

A A C x

( knC ,

k

nA l t h p, ch nh h p ch p k c a n ph n t ).

THI TH I H C S 04

Cu I m). Cho hm s2 1

1

xy

x(1).

1) Kh o st v v th (C) c a hm s (1).

2) Tm m M thu th ti p tuy n c a (C) t i M v ng thng ti m c n c tch h s gc b ng - 9.

Cu II. m)


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1) Gi i ph nh sau: 2

1 12

2x

x

.

2) Gi nh l ng gic:

4 44sin 2 os 2 os 4

tan( ). tan( )4 4

x c xc x

x x.

Cu III m) Tnh gi i h n sau:

3 2

20

ln(2 . os2 ) 1lim

x

e e c x xL

x

Cu IV. m)Cho hnh nn ng trn r. G i I l tm m t

c u n i ti p hnh nn (m t c u bn trong hnh nn, ti p xc v i t t c ng trnc a nn g i l m t c u n i ti p hnh nn).

1. Tnh theo r, l di n tch m t c u tm I;2. Gi s ng sinh c i. V i u ki n no c di n

tch m t c t gi tr l n nh t?

Cu V m) Cho cc s th c x, y, z th a mn: x2 + y2 + z2 = 2.

Tm gi tr l n nh t, gi tr nh nh t c a bi u th c: P = x3 + y3 + z3 3xyz.

Cu VI. m) Trong m t ph ng t Oxy cho hnh ch nh t ABCD c tm1

( ; 0)2

I . ng th ng

nh: x m A m. Tm t nh c a hnhch nh

Cu VII. m) Gi i h nh :

2 2 2

2

3 2

20102009

2010

3 log ( 2 6) 2 log ( 2) 1

y x x

y

x y x y


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V

NG D N GI Iwww.mathvn.com


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NG D N GI THI TH 01

PH N I.

Cu I. Cho hm s : 3 2 22 11 4 33 2

y x m x m m x .

1. Kh o st v v th c a hm s khi m = -3.

2. V i gi tr no c a m hm s c c i, c c ti u? G i x1, x2 m c i, c c ti u

c a hm s , hy tm gi tr l n nh t c a bi u th c 1 2 1 2. 2 x x x x .

Ta c 2 22 2 1 4 3 y x m x m m .

Hm s c c i, c c ti u khi v ch khi y = 0 c hai nghi m phn bi t x1, x2 hay

2 2 21 2 4 3 0 6 5 0 5 1m m m m m m

nh l Vi-t, ta c 1 2 1 x x m ,2

1 21. 4 32

x x m m

Suy ra 2 21 14 3 2 1 8 72 2

m m m m m

Ta nh n th y, v i 5; 1m th 229 8 7 4 9 0m m m

Do n nh t b ng 92

khi m = -4.

Cu II.

1. Gi nh 4 42

1 cot 2 cot 2 sin cos 3cos

x x x xx

u ki n: sin2x 0.

nh 2 4 22

2 12 1 sin 2 3 sin 2 sin 2 2 02sin

x x xx

22

2

sin 2 2sin 2 1 cos 2 0

4 4sin 2 1

x k x x x k

x

2. Tm cc gi tr c a tham s b t nh 24 4 5 2 2 x x m x x nghi i

m i gi tr x thu n 2; 2 3

t 2 4 5t x x . T 2; 2 3 1; 2x t . B nh cho t i:2

2 55 2 02

tt m t m g t t

(do 2 0t )

B nh nghi 2; 2 3 max , 1; 2 x m g t t .


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ng bi n 11; 2 max 2 , 1; 24

t m g t m t

Cu III. 1. Hnh chp S.ABCD c nh ch nh t, 2 AD a , CD = 2a. C nh SA vunggc v 3 2 0SA a a . G m c a c nh AC. Ch ng minh m t ph ng (SBK)

vung gc v i m t ph ng (SAC) v tnh th tch kh i chp SBCK theo a.

1. G i H l giao c a AC v BK th

BH = 23

BK 2 33

a v CH = 13

; CA = 63

a

2 2 2 22 BH CH a BC BK AC

T BK AC v BK SA BK (SAC) (SBK)(SAC)

VSBCK = 13SA.SBCK = 13

2323 2

2a

a a

2. Trong khng gian v i h t ng OAB.O1A1B1 v i A(2; 0; 0), B(0; 4; 0) vO1 nh t m N trn OA1 ng th ng MN song song v i

m t ph ng ( ): 2 5 0 x y z di MN = 5 .

C A1(2; 0; 4) 1 2; 0; 4OAuuuur

nh OA1:20 2 ; 0; 4

4

x n y N n n

z n

C 2; 4; 0ABuuur

nh AB:2 2

4 2 2 ; 4 ; 0

0

x m

y m N m m

z

V y 2 2 2; 4 ; 4 MN n m m muuuur

T

1

// . 0 2 2 2 2 4 4 0 1; 0; 22 MN MN n n m m n n N

uuuur uuuur

.

2 12 2

2

8 41 ; ; 05 552 1 16 4 5

0 2; 0; 0

Mm MN m m

m M A

Cu IV. 1. Tnh t ng:2 2 2 20 1 2

...1 2 3 1

nn n n nC C C C S

n, knC l

s t h p ch p k c a n ph n t .


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Ta c:111 !!1 1 , 0,1,...,

1 1 1! ! 1 1 ! !

k k

n nC Cnn k nk k nk n k n k n k

V y:2 2 2 21 2 3 1

1 1 1 121 ...1

nn n n nS C C C C

n

T 1 1 2 21 . 1 1n n n x x x , cn b ng h s 1nx hai v ta c:

2 2 2 2 20 1 2 3 1 11 1 1 1 1 2 2...

n nn n n n n nC C C C C C

V y:1

2 22

1

1

nnCS

n

2. Trong m t ph ng v i h tr c t ng trn (C):2 2

6 2 6 0 x y x y mB(2; - nh t m A thu ng trn (C) sao cho tam gic ABC cn t m Av c di n tch nh nh t.

ABC lm tam gic cn t i A th A ph i n ng trung tr c ( ) qua m BC l

M(3; 1) v nh n 2; 4BCuuur

n nn ( nh:

2 3 4 1 0 2 1 0 x y x y

V A (C) nn t A l nghi m c a h :2 2 6 2 6 0

2 1 0

x y x y

x y

Gi i h tm ra hai m A1(-1; 1) v A2( 215; 13

5)

Do 1 21820

5 A M A M nn

1 2 A BC A BC S S . V m c n tm l A(-1; 1)

PH N II.

Cu Va. 1. Tnh tch phn:ln 5

ln 2 10 1 1x xdxI

e e.

t2

1 1 2 x x x

t e t e tdt e dx . Khi x = ln2 th t = 1; khi x = ln5 th t = 2.2ln 5 2 2 2

22ln 2 1 1 1 1

2 3 51 1 1 1 12 ln ln3 3 3 3 3 3 29910 1x x

dx tdt dt t I dt t t ttt te e

2. Gi i h nh:

2

21

22 2

32 2 42

2 2 4 1 0 5

x

yx xy

x y x x y x

u ki n: x 0


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2

21 25 2 2 2 1 0 2 1 x x xy x xy x xy y

x

Thay vo (4) nh c:

2

2 21 1 2

2

2 22 1 3 1 2 11 12 2

2 2

x x

x x x x xx x x x

2

2 21 1 2

2 2

2 2 2 21 1 2 1 2 12 2

x x

x x x x x xf f x x x x

22

t tf t ng bi n v i m i t.

T2

2 21 2 1 32

4x x x y

x x

V y nghi m c a h nh l 324

x y .

Cu Vb. 1. Tnh tch phn:4

30

sincosx x I dx

x.

t u = x v3

sincos

xdv dx du dxx

v2

12cos

vx

.

T4

4 42 2

00 01 1 1tan2 4 2 4 22 cos cos

x dxI xx x

2. Gi nh 22 7 7 2log log 3 2 log 3 log2x x x x x x (6)

u ki n: x > 0

2 2 76 log log 2 log 3 02x x x x

Xt 22ln ln 2log 2

2 2xx xx x

x(7). t: ln 1 lnx x f x f x

x x; 0 f x x e.

V nh f(x) = 0 c nhi u nh t hai nghi m. D th y x = 2 v x = 4 l nghi m c a (7).

Xt 2 7log 2 log 3x x (8). t: 2log 2t

x t x

2 4 2 18 7 2 3 6 9 17 7 7

t t tt t c nghi m duy nh t t = 2.

V y p nh c nghi m x = 2 v x = 4.

H t


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NG D N GI THI TH 02

PH N I.

Cu I. Cho hm s 3 22 3 1 2 y x mx m x (1) (m l tham s th c)

1. Kh o st s bi n thin v v th hm s (1) khi m = 0.

2. ng th ng : 2y x . Tm cc gi tr c ng th ng c th

hm s (1) t m A(0; 2); B, C sao cho tam gic MBC c di n tch b ng 2 6 .

nh honh m c th hm s v ng th ng l:

2 22

0 22 3 1 2 32 3 2 0

x y x mx m x xg x x mx m

ng th ng c th hm s (1) t m A(0; 2), B, C

nh g(x) = 0 c hai nghi m phn bi t x 0

2 20 3 2 021;0 3 2 03

mm m

m mg x m

Chi u cao MBC: h = d(M; ( )) = 3 1 2 22

.

V y2

4 3MBCS

BCh

.

V xB, xC l hai nghi nh g(x) = 0 v B, C nn:

2 2 2 22

2 2 2

2 2 4

2 4 12 8 8 3 2 48 3 4 0

B C B C B C B C B C BC x x y y x x x x x x

m m m m m m

1m (lo i) ho c m = 4 (th a mn).

Cu II.

1. Gi nh 2 22sin sin 2 cos sin 2 1 2cos4

x x x x x

nh cho t i

2sin sin 2 cos sin 2 1 1 cos 2 1 sin 22

sin 2 sin cos sin 2 1 0

x x x x x x

x x x x


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* sin 2 02

k x x k

* 2 2sin cos sin 2 1 0 sin 1 2 cos sin 0 sin 1 1 2sin 2sin 0 x x x x x x x x x

21 2sin 2sin 0x x (v nghi m) ho c sinx = 1

22

x k k

2. Tm cc gi tr c a tham s h nh sau c nghi m th c duy nh t.

2 2

1 1 x y x y

x y m

Do h i x ng nn n u (x; y) l m t nghi m c a h th (y; x) c m t nghi m c a h . Doh nh c nghi m duy nh t th x = y.

nh (2) m = 2.

Khi m = 2 th h tr thnh2 2

1 1

2

x y x y

x y

2

2

00

11

2 2

x yx y

x y xy x yxy

x y xy

ho c2

1

x y

xy

D th y h c ba nghi m (1; -1); (-1; 1) v (1; 1). V y khng t n t i gi tr m th a mn.

Cu III. 1. Cho hnh chp S.ABCD c nh thoi c nh a (a > 0). Gc ABC b ng 120o,c nh SA vung gc v i m t ph ng (ABCD) v SA = a. G i C m c a c nh SC. M t ph ng

( v song song v i BD c t cc c nh

SB, SD l t t i B , D . Tnh th tch kh i c a

chp S.AB C D .

G i m c a AC v BD;

m c a SO v AC .

Trong m t ph ng (SBD), qua I k ng th ng

song song BD c t SB, SD l t t i B v D .

T BD (SAC) B D (SAC) B D AC .

Ta c: 13 22

AC a SC a AC SC a.

S

A

a D

D

I

B

C

C

B

a

O

2a


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Do I l tr ng tm c a SAC 223 3

a B D BD . V y21 .

2 3AB C DaS AC N D

T B D (SAC) (AB C D ) (SAC ). V ng cao h c a hnh chp S.AB C D ng

cao c u SAC 32

ah .

V y3

.31 .

3 18S AB C D AB C Da

V h S

2. Trong khng gian v i h tr c t m A(-1; 0; 2), m t ph ng (P): 2 3 0 x y z v

ng th ng (d):23 6

2 4 1yx z . Vi nh ng th ng (d m A, c t (d) t i B

v c t (P) t i C sao cho 2 0 AC AB

uuur uuur r

.G m c a (d) v (P).

nh tham s c a (d) l:3

2 4

6

x m

y m

z m

.

Thay vo (P) ta c: 6 4 2 4 6 3 0 1m m m m

V y M(5; 6; 7).

K ng th ng (d1 m c a (d1) v (P) ta c:

1

1 2

: 4

2

x t

d y t

z t

c 2 4 4 2 3 0 1t t t t

V y N(-3, -4, 1).

G m trn (P) sao cho 2 0 19; 24; 11 NC NM C uuur uuuur r

ng CA c t (d) t i B th a mn yu c u. V y (d ng th nh:

1 2

18 24 13

yx z .

Cu IV.

1. Cho s ph c ; , z x yi x y th a mn 3 18 26z i . Tnh 2009 20092 4T z z

ta c3 2

3 3 2 2 3

2 3

3 183 3 18 26

3 26

x xy z x xy x y y i i

x y y

Do x = y = 0 khng l nghi m h t y = tx

MN

C

A

d1

dd

B

P


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3 22

3 3

1 3 183 1 3 12 13 0

3 26

x tt t t

x t t

Khi 13

t th x = 3 v y = 1, th a mn x, y Z.

Khi 23 12 13 0t t th x, y . V y s ph c c n tm l: z = 3 + i

V y 2009 2009 2009 2009 1004 1004 10052 4 1 1 2 1 2 1 2T z z i i i i

2. Cho cc s th c khng m x, y, z th a mn 3 z y z . Tm gi tr nh nh t c a bi u th c:

1 1 14 2 ln 1 4 2ln 1 4 2 ln 1

P x y y z z x

T gi thi t 0 , , 3 x y z suy ra 4 2 ln 1 0; 4 2 ln 1 0 x y y z v

4 2 ln 1 0z x . Theo b ng th c C-si ta c:

94 2 ln 1 4 2 ln 1 4 2 ln 1

P x y y z z x

Xt hm s 2 ln 1 , 0;3 f t t t t , c 11

tf tt

.

L p b ng bi n thin hm f(t), v i 0; 3t suy ra 0 2 ln 2 1f t .

9 33 2 ln 212

Pf x f y f z

.

V y 3min3 2 ln 2

P , khi x = y = z = 1.

PH N 2 (th sinh lm m t trong hai cu)

Cu Va. 1. Tnh di n tch hnh ph ng gi i h n b ng 2 3x y , 1 0x y .

nh honh m c ng 23x y v 1x y l:

2 23 1 2 0 1 y y y y y ho c y = 2.

V y22 2 3 2

2 2

11 1

93 1 2 23 2 2

y yS y y dy y y dy y

2. Trong m t ph ng v i h tr c t Oxy cho tam gic ABC c nh A n ng th ng ( ):

2 3 14 0x y , c nh BC song song v i nh: 2 1 0x y . Bi t trung

m c a c nh AB l M(- nh t nh A, B, C.


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V AB nh: 2 0 x y c .

Do M(-3; 0) AB nn c = 6. V rnh AB l: 2 6 0x y .

Do A nn t A l nghi m c a h :2 3 14 0

4;22 6 0

x yA

x y

V M(- m AB nn B(-2; -2)

C nh BC // nh: 2 2 3 2 0 2 3 2 0 x y x y . V y t

C l nghi m c a h2 3 2 0

1;02 1 0

x yC

x y

Cu Vb. 1. Cho hnh ph ng H gi i h n b ng2

y x ;2

2y x . Tnh th tch c a kh i trnxoay t o thnh khi quay hnh H quanh tr c Ox.

nh honh m c ng cong l:

2 2 4 22 2 0 1 x x x x x ho c x = 1.

Khi 1; 1x th 2 22 x x th cc hm 2y x v 22y x cng n m pha trn tr c Ox.

V y11 3 5

2 4

11

442 23 5 5

x xV x x dx x

2. Trong m t ph ng v i h tr c t m I(-1; 3). Vi nh ng trn c tm I vc ng th ng 3 4 10 0x y t m A, B sao cho AIB b ng 120o.

G i H l hnh chi u c ng th ng (d): 3 4 10 0x y

3 12 10, 15

IH d I d

Suy ra R = AI =o

2cos60

IH .

V nh ng trn c n tm l:

22

1 3 4x y H t


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THI TH 03

PH N CHUNG CHO T T C CC TH SINH (7.0 m)

CU N I DUNG m

: D = R\{1} 0.25

Chi u bi n thin

lim ( ) lim ( ) 1x x

f x f x nn y = 1 l ti m c n ngang c th hm s

1 1lim ( ) , lim

x xf x nn x = 1 l ti m c ng c th hm s

y =2

10

( 1)x

0.25

B ng bi n thin

1

+

-

1

- -

y

y'

x - 1 +

Hm s ngh c bi n trn ( ;1) v (1; )

Hm s khng c c c tr

0.25

Cu I.

th .(t v )

m c th v i tr c Ox l (0 ;0)

V th

Nh n xt th nh m c a ng ti m c n I(1 i x ng

0.25

Gi s M(x0 ; y0) thu c (C) m ti p tuy n v th t ng cch t ix n ti p tuy n l l n nh t.

nh ti p tuy n t i M c d ng : 0020 0

1( )

( 1) 1

x y x x

x x

0.25


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-+

f(t)

f'(t)

x

2

0

10 +

20

2 20 0

10

( 1) ( 1)

xx y

x x

Ta c d(I ;tt) = 0

40

21

11

( 1)

x

x

Xt hm s f(t) =4

2( 0)

1

tt

tta c f(t) =

2

4 4

(1 )(1 )(1 )

(1 ) 1

t t t

t t

0.25

f(t) = 0 khi t = 1

B ng bi n thin

t b ng bi n thin ta c

d(I ;tt) l n nh t khi vch khi t = 1 hay

00

0

21 1

0

xx

x

0.25

+ V i x0 = 0 ta c ti p tuy n l y = -x

+ V i x0 = 2 ta c ti p tuy n l y = -x+4

0.25

Cu II

nh cho t i

2(cos4x + cos2x) = 3 (cos2x + 1) + sin2x

0.25


18/25


2cosx=0

4 os3xcosx=2 3 os 2s inxcosx

2cos3x= 3 osx+sinx

c c x

c

0.25

+ osx=0 x=2

c k

+3x=x- 2

62 os3x= 3 osx+sinx cos3x=cos(x- )6

3 26

k

c c

x x k

0.25

12

24 2

x k

kx

v x11 13

0; , , ,2 12 24 24

x x x x 0.25

, 0x y

x y

H nh3 2 3 2 3 2 3 23 5.6 4.2 0 3 5.6 4.2 0

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

x y x x y x y x x y

x y y y x y x x y y x y x x y y

0.25

3 2 3 2 3 2 3 23 5.6 4.2 0 3 5.6 4.2 0

2 0(2 )[( 2 )( ) 1] 0

x y x x y x y x x y

y x y x y x x y y

(do 2 )( ) 1 0 y x x y y )3 2 3 2 2 23 5.6 4.2 0 3 5.6 4.2 0 (1)

2 2 (2)

x y x x y x x x

y x y x

Gi i (1): 2 2 2

3( ) 1

3 3 23 5.6 4.2 0 ( ) 5.( ) 4 032 2

( ) 42

x

x x x x x

x3

2

0

log 4

x

x

0.25

0.25


19/25


V i x = 0 thay v c y = 0

V i 32

log 4x c y = 32

1 log 42

K t h p v u ki c nghi m c nh l 32

log 4x ,y = 32

1log 4

2

0.25

t I =3

1 42

0

( )1

x x x e dx

x. Ta c I =

31 1 4

2

0 0 1x x

x e dx dxx

0.25

Ta tnh

31

2

10

x

I x e dx t t = x3

ta c

11

1 00

1 1 1 1

3 3 3 3t t

I e dt e e

0.25

Ta tnh1 4

2

0 1

x I dx

xt t = 4 x 4 34 x t dx t dt

0.25

Cu III.

1 142

2 2 20 0

1 24 4 ( 1 ) 4( )

1 1 3 4

t I dx t dt

t t

V y I = I1+ I21

33

e

0.25

Ta c1 1 1

2 2 xy yz xz xyz x y z

nn0.25

1 1 1 1 1 ( 1)( 1)1 1 2 (1)

y z y z

x y z y z yz

ta c1 1 1 1 1 ( 1)( 1)

1 1 2 (2) x z x z

y x z x z xz

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

x y x y

y x y x y xy

0.25

Cu IV.

Nhn v v i v c c1

( 1)( 1)( 1)8

x y z 0.25


20/25


B'

Y

X

Z

N

D'

C'

A'

C

D A

B

M

BD

A

C

P

M

N

v y Amax =1 3

8 2 x y z

0.25

Cu V. Qua B, C, D l t d ng th ng

Song song v i CD, BD, BC c t nhau t i M, N, P

Ta c MN = 2BD, MP = 2CD, NP = 2BC

t

vung t i A t x = AM, y = AN, AP = z ta c

2 2 2 2 2 2

2 2 2

2( ), 2( )

2( )

x a c b y b c a

z a b c

V y V =1

122 2 2 2 2 2 2 2 22( )( )( )a c b b c a a b c

1.0

G i A l gi m d1 v d2 ta c A(3 ;0)

G m d1 v i tr c Oy ta c B(0 ; - 4)

G m d2 v i Oy ta c C(0 ;4)

0.5CuVIa.

G ng phn gic trong gc B v i I thuI(4/3 ; 0), R = 4/3

0.5

Ch n h tr c to nh nh v

Ta c M(1 ;0 ;0), N(0 ;1 ;1)

B(2 ;0 ;2), C(0 ;2 ;2)

G nh m t c mM,N,B,C c d ng

x

2

+ y

2

+ z

2

+2Ax + 2By+2Cz +D = 0V m t c m nn ta c

1.0


21/25


5

21 2 052 2 2 02

8 4 4 01

8 4 4 0 24

A

A D

B C D B

A C D

C B C D

D

. V y bn knh R = 2 2 2 15 A B C D

- 1 0.25

b nh

33

3

3log ( 1)2log ( 1)

log 4 0( 1)( 6)

xx

x x

3log ( 1) 06

x

x

0.25

0.25

0 6x 0.25

Ta c 1 2( 12;0), ( 12;0)F F Gi s M(x0 ; y0)thu c (E) H l hnh chi u c a M trn

ng th ng 83

x . Ta c MF2 = a - cx0/a = 08 32

x

0.5

CuVIa

Cu VIb

MH = 08 3

3

x. V y 2

MF

MHi

0.5

Ta c (1;1;1), (1;2;3), ; (1; 2;1)Q Q AB n AB n

uuur uur uuur uur

V ; 0Q AB nuuur uur r

nn m t ph ng (P) nh n ; Q AB nuuur uur

n

V nh x - 2y + z - 2 = 0

1.0

CuVIb Nghi m b nh l x = 3 v x = 4 1.0

H t


22/25


THI TH 04

CU N I DUNG i m

I.1 Hm s :2 1 3

21 1

xy

x x

+) Gi i h n, ti m c n:( 1) ( 1)

2; 2; ;lim lim lim limx x x x

y y y y

- ng: x = -1; TCN: y = 2.

+) 23

' 0,1

y x Dx

+) BBT:

x - - 1 +

y' + || +

y 2

||

2

+

m

I.2 +) Ta c I(- 1; 2). G i 0 20 0

3 3( ) ( ;2 )

1 ( 1)M I

IM

M I

y y M C M x k

x x x x m

8

6

4

2

-2

-4

-6

-10 -5 5 10


23/25


+) H s gc c a ti p tuy n t i M: 0 20

3'( )

1Mk y x

x

+) . 9 M IM ycbt k k

+) Gi c x0 = 0; x0 = - m M th a mn: M(0; - 3), M(- 2; 5)

II.1 ( 2; 2) \{0}x

t 22 , 0 y x y Ta c h : 2 22

2

x y xy

x y

+) Gi i h c x = y = 1 v

1 3 1 3

2 2;1 3 1 3

2 2

x x

y y

+) K t h u ki c: x = 1 v1 3

2x

m

II.2,

4 2 x k k Z

4 4 2 2

4 2

) tan( ) tan( ) tan( ) cot( ) 14 4 4 41 1 1

sin 2 os 2 1 sin 4 os 42 2 2

2cos 4 os 4 1 0

x x x x

x c x x c x

pt x c x

+) Gi c cos24x = 1 cos8x = 14

x k v cos24x = -1/2 (VN)

+) K t h c nghi m c nh l ,2

x k k Z

m

III 3 32 22 20 0

32 2 2

2 2 2 32 2 230 02 22 2

ln(2 . os2 ) 1 ln(1 1 os2 ) 1 1lim lim

ln(1 2 sin 2 ) 1 1 ln(1 2 sin 2 ) 1lim lim

(1 ) 1 12 sin 2 sin2 sin 2 sin

1 52

3 3

x x

x x

e e c x x c x xL

x x

x x x

x x x x xx xx x

m


24/25


IV.1 +) G i Cr l bn knh m t c u n i ti p nn, v c bn

ng trn n i ti p tam gic SAB. Ta c:

2 2

1( ). .

2

.2

2( )

SAB C C

C

S pr l r r SM AB

l r r l r r r

l r l r

+) Sc u =2 24 4C

l rr r

l r

m

IV.2 t :2 3

2 2

2

( ) ,0

5 12 ( ) 2) '( ) 0

( ) 5 1

2

lr r y r r l

l r

r lr r rl l

y rl r

r l

+) BBT:

r 0 5 12

l l

y'(r)

y(r) ymax

+) Ta c max Sc u t t max5 1

2

r l

m

V +) Ta c

2 2 2

2 2 2 22 2 2

2 2

( )( )

( )( )

2

2 ( ) ( )( ) 2 ( ) 3

2 2

P x y z x y z xy yz zx

x y z x y zP x y z x y z

x y z x y zP x y z x y z

r

l

I

M

S

A B


25/25


t x +y + z = t, 6( cov )t Bunhia xki c: 3

1( ) 3

2P t t t

+) '( ) 0 2P t t , P( 6 ) = 0; ( 2) 2 2P ; ( 2) 2 2P

+) KL: ax 2 2; 2 2 M P MinP

m

VI+) 5( , )

2d I A B AD = 5 AB = 2 5 BD = 5.

ng trn - 1/2)2 + y2 = 25/4

+) T A, B l nghi m c a h :2 2

2

1 25 2( )( 2; 0), (2; 2)2 4

22 2 00

x

yx yA B

xx y

y

(3;0), ( 1; 2)C D .

1m

VII 2 2 2

2

3 2

20102009 (1)

2010

3 log ( 2 6) 2 log ( 2) 1(2)

y x x

y

x y x y

+) L pt:2 2 2 2

2009 2009log ( 2010) log ( 2010) x x y y

+) Xt v CM HS 2009( ) log ( 2010), 0 f t t t t ng bi n,

t 2 = y2 x= y, x = - y

+) V i x = y th pt: 3log3(x +2) = 2log2(x + 1) = 6t

d ng1 8 19 9

t t

, cm pt ny c nghi m duy nh t t = 1

x = y =7

+) V i x = - y th c pt: log3(y + 6) = 1 y = - 3 x = 3

m

H t

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