TRNG THPT CHUYN H TNH

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

THI TH I HC LN II - NM 2013 Mn: TON ; Khi: A v A1

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

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

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

12

x

xy .

1. Kho st s bin thin v v th (C) ca hm s cho. 2. Vit phng trnh ng thng i qua im I( -1 ; 2) ct th (C) ti hai im phn bit A, B sao

cho tam gic AOB c din tch bng 3 ( O l gc ta ).

Cu 2 ( 1,0 im) Gii phng trnh: .sincos32cos2

1cossin32cos2 2xx

x

xxx

Cu 3 ( 1,0 im) Gii bt phng trnh: .31

2

9

8

x

x

x

x

Cu 4 ( 1,0 im) Tnh tch phn : I = .)sin23)(1cos32(cos

2sinsin33

0

2

dxxxx

xx

Cu 5 ( 1,0 im) Cho hnh lng tr tam gic ABC.ABC c AA = 2a, AB = AC = a (a > 0) v gc gia cnh bn AA v mt phng (ABC) bng 600. Tnh th tch ca khi lng tr ABC.ABC v khong cch t im A n mp(ABC) theo a bit rng hnh chiu ca im A trn mt phng (ABC) trng vi trc tm H ca tam gic ABC.

Cu 6 ( 1,0 im) Cho x, y, z l cc s thc dng v tha mn: 1z z x y x y .

Tm gi tr ln nht ca biu thc T = 3

44

)).().(( xyzzxyyzx

yx

.

II. PHN RING ( 3,0 im): Th sinh ch c lm mt trong hai phn (phn A hoc B) A. Theo chng trnh Chun Cu 7a ( 1,0 im) Trong mt phng vi h ta Oxy, cho tam gic ABC c nh A nm trn ng thng

01: yx , ng thng BC song song vi v ng cao k t B c phng trnh: 2x y 2 = 0.

Tnh din tch tam gic ABC bit im M )4

5;

2

5( nm trn cnh AC v tha mn AM = 3 MC.

Cu 8a ( 1,0 im) Trong khng gian vi h to Oxyz , cho mt phng (P): 2x + y + z 2 = 0 v hai

ng thng d1: 12

2

1

zyx

v d2:

2

3

3

3

1

1

zyx. Vit phng trnh ng thng song

song vi ( P) ng thi ct hai ng thng d1 v d2 ln lt ti M, N sao cho on MN ngn nht.

Cu 9a ( 1,0 im) Tnh mun ca s phc z 2i bit 04)2).(2( iziziz .

B. Theo chng trnh Nng cao

Cu 7b ( 1,0 im) Trong mt phng vi h to Oxy, cho hai ng trn ( C1 ): 0422 yyx v

(C2): 03618422 yyxx . Vit phng trnh ng trn (C) c tm I nm trn ng thng

d: 2x + y 7 = 0 ng thi tip xc ngoi vi c hai ng trn ( C1 ) v ( C2).

Cu 8b (1,0 im) Trong khng gian vi h ta Oxyz, cho tam gic ABC c A(1; 4; 3) v hai ng thng

1 :2

9

11

1

zyx, 2 :

1 3 4

2 1 1

x y z

ln lt cha ng trung tuyn k t nh B v ng

cao k t nh C. Tm ta tm v tnh bn knh ng trn ngoi tip tam gic ABC.

Cu 9b ( 1,0 im) Gii h phng trnh:

1)(log

1loglog

22

2

2

yx

yy

xxxy , ( yx, R )

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

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TRNG THPT CHUYN H TNH

HNG DN CHM THI TH I HC LN II NM 2013 Mn: TON ; Khi: A, A1

Cu

1

Ni dung im

1. * Tp xc nh: R \{-1} * S bin thin:

Chiu bin thin: y =

01

32

x vi mi x -1 nn hm s ng bin trn cc

khong ),1(),1,( .

Cc tr: Hm s khng c cc tr.

0,25

Gii hn v tim cn:

2lim

yx

, 2lim

yx

tim cn ngang y = 2,

yx 1lim ,

y

x 1lim tim cn ng x = -1.

0,25

Bng bin thin:

0,25

th:

th i qua cc im (0, 1); (2, 1) v nhn I(-1, 2) lm tm i xng.

V th

0,25

2. Gi k l h s gc ca t suy ra PT : y = k( x+1) + 2.

PT honh giao im ca v (C) :

1

12

x

x k( x+1) +2 0322 kkxkx (*)

ng thng ct (C) ti hai im phn bit A, B

PT (*) c 2 nghim phn bit

0'

0k k < 0 .

0,25

Vi k < 0 gi A( x1 ; k(x1 + 1) + 2), B( x2 ; k(x2 + 1) + 2) l cc giao im ca vi

( C ) th x1, x2 l cc nghim ca PT (*) . Theo Viet ta c

k

kxx

xx

3.

2

21

21

. 0,25

Ta c AB = 21222

12

2

12 1)( xxkxxkxx =

= k

kxxxxk12

.1.41 2212

12

2 , d( O ; ) = 1

2

2

k

k

0,25

Theo bi ra din tch tam gic ABC bng 3 nn ta c :

.2

1AB d( O ; ) = 3 4,1 kk tha mn k < 0.

Vy c 2 PT ng thng l y = - x + 1; y = - 4x -2.

0,25

Cu

2

iu kin: cos2x 0 (*) Pt cho

x

xxxx

2cos2

sincossin32cos3 22 xx sincos3

.2cos2)sincos3( 2 xxx ( xx sincos3 )

0,5

x -1 + y + || +

y 2

+2

2

xxx

xx

sincos32cos2

0sincos3

)6

(cos2cos

3tan

xx

x

3

2

18,2

6

3

kxkx

kx

Cc nghim u TMK ( *) nn phng trnh cho c 3 h nghim:

)(3

2

18,2

6,

3Zkkxkxkx

.

0,5

Cu

3 K : 1< x< 9 ( * ). Vi k ( * ) ta c : ( 1 ) 31

1

9

119

xxxx

31.9

1919

xx

xxxx ( 2 ) t t = 19 xx , t > 0.

Ta c: 16198)1).(9(288 2 xxxxt .

0,5

422 t ( ** ) v 2

8)1).(9(

2

txx . Khi BPT ( 2 ) tr thnh: 3

8

22

t

tt

024103 23 ttt ( do ( **) ).

40)4).(2).(3( tttt . Kt hp vi ( ** ) ta suy ra t = 4 hay

19 xx = 4 5 x . Vy BPT cho c tp nghim T = 5 .

0,5

Cu

4 Ta c I = .

)sin23)(cos3cos2(

2sinsin33

0

22

dxxxx

xx=

3

0

2 )cos21(.cos).3cos2(

)cos23(sin

dxxxx

xx

= 3

0

2 )cos21(.cos).3cos2(

)cos23(sin

dxxxx

xx

3

0

2 )cos21(.cos

sin

dxxx

x

0,25

= 3

0

22 )cos21(.cos

).sin.(cos

xx

dxxx. t t = x2cos2 dxxxdt )sin.(cos4 . 0,25

i cn: Khi x = 0 2 t ; khi2

1

3 tx

. Khi I =

2

1

2)1.(2

1

tt

dt=

= dttt

2

1

21

11

2

1 =

1ln

2

1

2

.2

1

t

t= )

3

2ln

3

1ln.(

2

1 =

2

1ln.

2

1. Vy I = 2ln

2

1 .

0,5

Cu

5

Theo bi ra gc gia cnh bn AA v mt

phng (ABC) bng 600 nn gc 060' AHA

v do AA = 2a nn AH = 3a l mt ng

cao ca khi lng tr ABC.ABC v AH = a. Mt khc tam gic ABC cn ti A nn nu gi M l trung im ca cnh BC th on AM l mt ng cao ca tam gic ABC v AM < AC = AB = AH = a nn H nm ngoi tam gic ABC v nm trn tia i ca tia AM suy ra A l trng tm ca tam gic HBC.

60

2a

a

a

M

C

A

A'

B'

C'

B

H

K

Khi ta c AM = 322

aMCBCa

4

3..

2

1 2aAMBCS ABC .

Th tch khi lng tr cho l : V = 4

3.'

3aSHA ABC .

0,25

0,25

Ni AM, ta c mp(AHM) BC khi k HK MAKMA ',' th HK )'( BCA nn di

on HK l d( H ; (ABC)) = HK. Ta c : 7

3....

1

'

11222

aHK

HMHAHK .

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suy ra khong cch d( H ; (ABC)) = 7

3a.

Ta li c : 3 BC))A' ( ;A d(

BC))A' ( ; H d(

AM

HM. Vy khong cch d( A ; (ABC)) =

7

a.

0,25

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

0,25

Cu

6 V 1z z x y x y (z + 1)( x + y) = z2 - 1 v do z > 0 nn ta c: zyx 1 .

Khi : T = 3

44

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

yx =

4244

)1)(1(.)( yxyx

yx

0,25

p dng BT Csi cho cc s dng x, y ta c :

27

.427

41333

13

4

4

4

34

4 xxxxxx

,

27

.427

41333

13

4

4

4

34

4 yyyyyy

, xyyx 42 .

0,25

Do 42 )1)(1(.)( yxyx 446

9

6

338 ..

3

4

3

..4.4 yx

yxxy suy ra

9

6

4

3T ( * ) 0,25

Du = ( * ) xy ra 7,3,3

1

133

zyx

yxz

yx

.Vy Max9

6

4

3T khi x= 3, y =3, z = 7 0,25

Cu

7a.

PT ng thng d i qua M v vung gc vi ng cao k t B l: x + 2y - 5 = 0.

Ta nh A l nghim ca h: )2;1(2

1

01

052A

y

x

yx

yx

)

4

3;

2

3( AM

Do im M )4

5;

2

5( nm trn cnh AC v tha mn AM = 3 MC nn ta c AMAC

3

4 suy

ra ta im C l C(3 ; 1).

0, 5

ng thng BC song song vi 01: yx v i qua C(3 ; 1) nn c PT: x y - 2 = 0.

Ta nh B l nghim ca h: )2;0(2

0

022

02

B

y

x

yx

yx

Ta c: 23)3;3( BCBC , d( A; BC) = 2

3

2

221

Vy din tch tam gic ABC l: );(..2

1BCAdBCS

2

9

0,5

Cu

8a. Do M 1d , N 2d nn ta cc im M, N c dng: M( t ; 2 2t; t), N( 1 + s; 3 - 3s; - 3+2s)

suy ra )23;321;1( ststtsMN .Do song song vi ( P ) nn ta c: 0. PnMN

023321)1(2 ststts ts . Khi

22 )3()1(1)3;1;1( ttMNttMN = 1182 2 tt 33)2(2 2 t

vi mi t. Du = xy ra khi t = 2 M( 2 ; -2 ; 2) )( P ( tha mn MN song song vi (P)).

0,5

0,25

on MN ngn nht khi v ch khi M( 2 ; -2 ; 2), ).1;1;1( MN

Vy PT ng thng cn tm l: 1

2

1

2

1

2

zyx.

0,25

Cu

9a.

t z = a + bi ( a, b R ). Khi :

04)2).(2( iziziz ( a + ( b- 2)i).( a ( b + 2)i) + 4i ( a + bi ) = 0

( a2 + b2 4 4b) + [a( b 2) a( b + 2) + 4a] i = 0 a2 + b

2 4b 4 = 0

0,5

Ta li c: 44)2(2 22 bbaibaiz = 22884422 bba

Vy mun ca z 2i bng 22 . 0,5

Cu

7b.

ng trn ( C1 ) c tm I1 ( 0 ; 2 ) v bk R1 = 2, ng trn ( C2 ) c tm I2 ( -2 ; -9) v bk R2 = 7. Gi R l bk ng trn ( C ). Do I thuc d nn ta I c dng ( a; 7 -2a). Do ( C ) tip xc ngoi vi hai ng trn ( C1 ) , ( C2 ) nn ta c:

22

11

RRII

RRII51212 RRIIII Ta c: )52;(1 aaII ; )162;2(2 aaII

0,5

55216225 222212 aaaaIIII

525205260605 22 aaaa aaa 42125205 2

22 1616844125205

4

21

aaaa

a.4

11

1044

4

21

a

ahoaca

a

35)1;4( 1 RIII .Vy PT ng trn ( C ) cn tm l: 91422 yx .

0,5

Cu

8b. V B 1 B( -1- t ; - t ; 9 + 2t), )26;4;2( tttAB , C 2 C ( 1 + 2k ; 3 k

; 4 k ) . ng thng 2 c mt vtcp )1;1;2(2 u . Theo bi ra ta c : AB 2 nn

0. 2 uAB 2( - t 2) + 4 + t 6 2t = 0 t = -2 suy ra B( 1 ; 2; 5).

0,25

Gi M l trung im ca on AC suy ra M(

2

7;

2

7;1

kkk

).

Do M 1 14

11

2

7

1

2

k

kkk C( 3 ; 2; 3).

Ta c ),2;2;0( AB )0;2;2( AC , )2;0;2( BC suy ra AB = BC = AC = 22 nn

tam gic ABC l tam gic u c cnh a = 22 .

0,25

0,25

Vy ng trn ngoi tip tam gic ABC c tm I

3

11;

3

8;

3

5 v bk

3

62

2

3..

3

2 aIAR . 0,25

Cu

9b. K:

22,0

10,10

yxy

xxy( * ) + Vi y = 1 thay vo h cho ta c 332 xx (Do ( *)) 0,25

+ Vi 0 < y 1 v x, y tha mn K ( * ) ta c PT: 1loglog 2 yy

xxxy

1log)(log

1

)(log

1 2 yxyxy

x

yx

1loglog1

1

log1

1 2

yxy

x

yx

0,25

t t = yxlog khi ta c PT: 111

1 2

tt

t

t 0223 ttt t = 0 y = 1

( Loi)

0,25

Vy HPT cho c nghim duy nht 1;3; yx 0,25

Lu : Mi cch gii ng v khc vi hng dn chm ny gim kho cho im tng ng.