B GIO DC V O TO

THI MINH HA - K THI THPT QUC GIA NM 2015 Mn: TON

Thi gian lm bi: 180 pht.

Cu 1.(2,0 im) Cho hm s 2 1.1

xyx

=

+

a) Kho st s bin thin v v th (C) ca hm s cho. b) Vit phng trnh tip tuyn ca th (C), bit tip im c honh 1.x =

Cu 2.(1,0 im) a) Cho gc tha mn:

2< < v 3sin .

5= Tnh 2

tan .

1 tan A =

+

b) Cho s phc z tha mn h thc: (1 ) (3 ) 2 6 .i z i z i+ + = Tnh mun ca z. Cu 3.(0,5 im) Gii phng trnh: 3 3log ( 2) 1 log .x x+ =

Cu 4.(1,0 im) Gii bt phng trnh: 2 22 3( 2 2).x x x x x+ +

Cu 5.(1,0 im) Tnh tch phn: 2

3

1

(2 ln )d .I x x x= +

Cu 6.(1,0 im) Cho hnh chp S.ABC c y ABC l tam gic vung ti B, AC = 2a, o30 ,ACB = Hnh chiu vung gc H ca nh S trn mt y l trung im ca cnh AC v 2 .SH a= Tnh theo a th tch khi chp S.ABC v khong cch t im C n mt phng (SAB). Cu 7.(1,0 im) Trong mt phng vi h ta Oxy , cho tam gic OAB c cc nh A v B thuc ng thng : 4 3 12 0x y + = v im (6; 6)K l tm ng trn bng tip gc O. Gi C l im nm trn sao cho AC AO= v cc im C, B nm khc pha nhau so vi im A. Bit im C c

honh bng 24 ,5

tm ta ca cc nh A, B.

Cu 8.(1,0 im) Trong khng gian vi h ta Oxyz, cho hai im (2; 0; 0)A v (1; 1; 1).B Vit phng trnh mt phng trung trc (P) ca on thng AB v phng trnh mt cu tm O, tip xc vi (P). Cu 9.(0,5 im) Hai th sinh A v B tham gia mt bui thi vn p. Cn b hi thi a cho mi th sinh mt b cu hi thi gm 10 cu hi khc nhau, c ng trong 10 phong b dn kn, c hnh thc ging ht nhau, mi phong b ng 1 cu hi; th sinh chn 3 phong b trong s xc nh cu hi thi ca mnh. Bit rng b 10 cu hi thi dnh cho cc th sinh l nh nhau, tnh xc sut 3 cu hi A chn v 3 cu hi B chn l ging nhau.

Cu 10.(1,0 im) Xt s thc x. Tm gi tr nh nht ca biu thc sau: 2

2 2

3 2 2 1 1 13 2 3 3 3 2 3 3 3

+ += + +

+ + + + +

( ).

( ) ( )

x xP

x x x x

----------- HT -----------

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B GIO DC V O TO P N - THANG IM THI MINH HA - K THI THPT QUC GIA NM 2015

Mn: TON

CU P N IM Cu 1

(2,0 im)

a) (1,0 im) Tp xc nh: { }\ 1 .D = Gii hn v tim cn:

( 1)lim

xy

+ = ,

( 1)lim

xy

= + ; lim lim 2.

x xy y

+= =

Suy ra, th hm s c mt tim cn ng l ng thng 1x = v mt tim cn ngang l ng thng 2.y =

0,25

S bin thin:

- Chiu bin thin: y' = 23

( 1)x + > 0 x D.

Suy ra, hm s ng bin trn mi khong ( ); 1 v ( )1; + . - Cc tr: Hm s cho khng c cc tr.

0,25

Lu : Cho php th sinh khng nu kt lun v cc tr ca hm s.

- Bng bin thin:

x 1 +

y' + +

y + 2 2

0,25

th (C):

0,25

O x

y

1 1

2

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b) (1,0 im) Tung 0y ca tip im l: 0

1(1) .2

y y= = 0,25

Suy ra h s gc k ca tip tuyn l: 3'(1) .4

k y= = 0,25

Do , phng trnh ca tip tuyn l: 3 1( 1) ;4 2

y x= + 0,25

hay 3 1 .4 4

y x= 0,25

Cu 2 (1,0 im)

a) (0,5 im) Ta c: 22

tan 3tan .cos sin .cos cos.

1 tan 5A = = = =

+ (1) 0,25

22 2 3 16cos 1 sin 1 .

5 25

= = =

(2)

V ;2pi

pi

nn cos 0.< Do , t (2) suy ra 4cos .5

= (3)

Th (3) vo (1), ta c 12 .25

A =

0,25

b) (0,5 im) t z = a + bi, ( ,a b ); khi z a bi= . Do , k hiu () l h thc cho trong bi, ta c: () (1 )( ) (3 )( ) 2 6i a bi i a bi i+ + + = (4 2 2) (6 2 ) 0a b b i + =

0,25

{4 2 2 06 2 0a bb = = { 23.ab == Do 2 2| | 2 3 13.z = + =

0,25

Cu 3 (0,5 im)

iu kin xc nh: 0.x > (1) Vi iu kin , k hiu (2) l phng trnh cho, ta c: (2) 3 3log ( 2) log 1x x+ + = 3 3log ( ( 2)) log 3x x + =

0,25

2 2 3 0x x+ = 1x = (do (1)). 0,25

Cu 4 (1,0 im)

iu kin xc nh: 1 3.x + (1) Vi iu kin , k hiu (2) l bt phng trnh cho, ta c: (2) 2 22 2 2 ( 1)( 2) 3( 2 2)x x x x x x x+ + +

0,25

( 2)( 1) ( 2) 2( 1)x x x x x x + + ( )( )( 2) 2 ( 1) ( 2) ( 1) 0.x x x x x x + + + (3) Do vi mi x tha mn (1), ta c ( 2) ( 1) 0x x x + + > nn (3) ( 2) 2 ( 1)x x x +

0,50

2 6 4 0x x 3 13 3 13.x + (4) Kt hp (1) v (4), ta c tp nghim ca bt phng trnh cho l:

1 3 ; 3 13 . + +

0,25

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Cu 5 (1,0 im) Ta c:

2 23

1 1

2 d ln d .I x x x x= + (1) 0,25

t 2

31

1

2 dI x x= v 2

21

ln d .I x x= Ta c:

24

11

1 15.

2 2I x= =

0,25

2 22 2

2 1 11 1

.ln d(ln ) 2ln 2 d 2ln 2 2ln 2 1.I x x x x x x= = = =

Vy 1 213 2 ln 2.2

I I I= + = + 0,50

Cu 6 (1,0 im)

Theo gi thit, 12

HA HC AC a= = = v SH mp(ABC).

Xt v. ABC, ta c: o.cos 2 .cos 30 3 .BC AC ACB a a= = = 0,25

Do o 21 1 3. .sin .2 . 3 .sin 30 .2 2 2ABC

S AC BC ACB a a a= = =

Vy 3

2.

1 1 3 6. . 2 . .

3 3 2 6S ABC ABCaV SH S a a= = =

0,25

V CA = 2HA nn d(C, (SAB)) = 2d(H, (SAB)). (1) Gi N l trung im ca AB, ta c HN l ng trung bnh ca ABC. Do HN // BC. Suy ra AB HN. Li c AB SH nn AB mp(SHN). Do mp(SAB) mp(SHN). M SN l giao tuyn ca hai mt phng va nu, nn trong mp(SHN), h HK SN, ta c HK mp(SAB). V vy d(H, (SAB)) = HK. Kt hp vi (1), suy ra d(C, (SAB)) = 2HK. (2)

0,25

V SH mp(ABC) nn SH HN. Xt v. SHN, ta c:

2 2 2 2 21 1 1 1 1

.

2HK SH HN a HN= + = +

V HN l ng trung bnh ca ABC nn 1 3 .2 2

aHN BC= =

Do 2 2 2 21 1 4 11

.

2 3 6HK a a a= + = Suy ra 66 .

11aHK = (3)

Th (3) vo (2), ta c ( ) 2 66, ( ) .11

ad C SAB =

0,25

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Cu 7 (1,0 im)

Trn , ly im D sao cho BD = BO v D, A nm khc pha nhau so vi B. Gi E l giao im ca cc ng thng KA v OC; gi F l giao im ca cc ng thng KB v OD. V K l tm ng trn bng tip gc O ca OAB nn KE l phn gic ca gc

.OAC M OAC l tam gic cn ti A (do AO = AC, theo gt) nn suy ra KE cng l ng trung trc ca OC. Do E l trung im ca OC v KC = KO. Xt tng t i vi KF, ta cng c F l trung im ca OD v KD = KO. Suy ra CKD cn ti K. Do , h KH , ta c H l trung im ca CD. Nh vy: + A l giao ca v ng trung trc 1d ca on thng OC; (1) + B l giao ca v ng trung trc 2d ca on thng OD, vi D l im i xng ca C qua H v H l hnh chiu vung gc ca K trn . (2)

0,50

V C v c honh 0245

x = (gt) nn gi 0y l tung ca C, ta c:

0244. 3 12 0.5

y+ = Suy ra 012

.

5y =

T , trung im E ca OC c ta l 12 6;5 5

v ng thng OC c

phng trnh: 2 0.x y+ = Suy ra phng trnh ca 1d l: 2 6 0.x y = Do , theo (1), ta ca A l nghim ca h phng trnh:

{4 3 12 02 6 0.x yx y+ = = Gii h trn, ta c A = (3; 0).

0,25

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Gi d l ng thng i qua K(6; 6) v vung gc vi , ta c phng trnh ca d l: 3 4 6 0.x y + = T y, do H l giao im ca v d nn ta ca H l nghim ca h phng trnh:

{4 3 12 03 4 6 0.x yx y+ = + = Gii h trn, ta c 6 12; .

5 5H =

Suy ra 12 36; .

5 5D =

Do , trung im F ca OD c ta l 6 18;5 5

v ng thng OD c

phng trnh: 3 0.x y+ = Suy ra phng trnh ca 2d l: 3 12 0.x y + = Do , theo (2), ta ca B l nghim ca h phng trnh:

{4 3 12 03 12 0.x yx y+ = + = Gii h trn, ta c B = (0; 4).

0,25

Cu 8 (1,0 im) Gi M l trung im ca AB, ta c

3 1 1; ; .

2 2 2M =

V (P) l mt phng trung trc ca AB nn (P) i qua M v ( 1; 1; 1)AB =

l mt vect php tuyn ca (P).

0,25

Suy ra, phng trnh ca (P) l: 3 1 1( 1) ( 1) 02 2 2

x y z + + + =

hay: 2 2 2 1 0.x y z + = 0,25

Ta c 2 2 2

| 1| 1( , ( )) .2 32 ( 2) 2

d O P = =+ +

0,25

Do , phng trnh mt cu tm O, tip xc vi (P) l: 2 2 2 112

x y z+ + =

hay 2 2 212 12 12 1 0.x y z+ + = 0,25

Cu 9 (0,5 im)

Khng gian mu l tp hp gm tt c cc cp hai b 3 cu hi, m v tr th nht ca cp l b 3 cu hi th sinh A chn v v tr th hai ca cp l b 3 cu hi th sinh B chn. V A cng nh B u c 310C cch chn 3 cu hi t 10 cu hi thi nn theo quy

tc nhn, ta c ( )2310( ) C .n = 0,25

K hiu X l bin c b 3 cu hi A chn v b 3 cu hi B chn l ging nhau. V vi mi cch chn 3 cu hi ca A, B ch c duy nht cch chn 3 cu hi ging nh A nn ( ) 3 310 10C .1 C .Xn = = V vy ( ) ( )

310

2 331010

C 1 1( ) .( ) C 120CXnP X

n

= = = =

0,25 ww

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Cu 10 (1,0 im)

Trong mt phng vi h ta Oxy, vi mi s thc x, xt cc im ( ; 1)A x x + , 3 1

;2 2

B

v 3 1; .2 2

C

Khi , ta c ,OA OB OCPa b c

= + + trong a = BC, b = CA v c = AB.

0,25

Gi G l trng tm ABC, ta c: . . . 3 . . .

. . . 2 . . .a b cOA GA OB GB OC GC OA GA OB GB OC GCPa GA b GB c GC a m b m c m

= + + = + +

,

trong ,a bm m v cm tng ng l di ng trung tuyn xut pht t A, B, C ca ABC.

0,25

Theo bt ng thc C si cho hai s thc khng m, ta c

( )( )

2 2 2 2

2 2 2 2 2 2 2

1. . 3 2 2

2 33 2 21

. .

22 3 2 3

aa m a b c a

a b c a a b c

= +

+ + + + =

Bng cch tng t, ta cng c: 2 2 2

.

2 3ba b cb m + + v

2 2 2

. .

2 3ca b c

c m+ +

Suy ra ( )2 2 23 3 . . . .P OAGA OB GB OC GCa b c + ++ + (1)

0,25

Ta c: . . . . . . .OAGA OB GB OC GC OA GA OB GB OC GC+ + + +

(2)

( ) ( ) ( )( )

( )2 2 2

2 2 22 2 2

. . .

. . .

.

4. (3)

9 3a b c

OAGA OB GB OC GCOG GA GA OG GB GB OG GC GC

OG GA GB GC GA GB GCa b c

m m m

+ +

= + + + + +

= + + + + +

+ += + + =

T (1), (2) v (3), suy ra 3.P Hn na, bng kim tra trc tip ta thy 3P = khi x = 0. Vy min 3.P =

0,25

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