S GD&T BC LIU CHNH THC

K THI TH THPT QUC GIA NM 2015Mn: TON

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

:

Cu 1: (2 im) Cho hm s .13

xxy

1. Kho st s bin thin v v th (C) ca hm s cho. 2. Vit phng trnh tip tuyn ca (C) ti im c tung bng -1.

Cu 2: (1 im)

1. Cho gc ;2

c 1sin 3

. Tnh gi tr ca biu thc: sin 2 cos 2A .

2. Gii phng trnh: 3 1 33

log log ( 2) 1 log (4 )x x x

Cu 3: (0.5 im) Cho s phc z tha: (1 ) 2 5 3i z iz i . Tm phn thc, phn o ca s phc 2w z z .

Cu 4: (1 im) Tnh tch phn sau: 21

2 (2 ln )e

I x x x dx .

Cu 5: (1 im Cho hnh chp tam gic u S.ABC c cnh y bng a, cnh bn to vi mt y mt gc 060 . Tnh theo a th tch hnh chp S.ABC v khong cch t A n mt (SBC).

Cu 6: (1 im) Trong khng gian vi h ta Oxyz, cho im A(1; 1; 1), B(2; 2; 2), C(2; 0; 5), D(0; 2; 1). Vit phng trnh mt phng cha A v B

v i qua trung im ca on CD. Cu 7: (1 im) Trong mt phng Oxy cho tam gic ABC c nh A(3;5), trc tm

H(3;3), tm ng trn ngoi tip l I(4;2). Tm ta cc nh B v C bit nh B c honh nh hn 3.

Cu 8: (1 im) Gii h phng trnh:

2 2 2 210 4 2 2 4 10 4( )

1 2 4 2 18 5( 3)

x xy y x xy y x y

x y xy x

Cu 9: (0.5 im) C 20 th ng trong 2 hp khc nhau, mi hp ng 10 th nh s th t t 1 n 10. Ly ngu nhin 2 th t 2 hp (mi hp mt th). Tnh xc sut ly c 2 th c tch hai s ghi trn hai th l mt s chn.

Cu 10: (1 im) Cho ba s thc a, b, c tha 0 a b c . Tm gi tr nh nht ca biu

thc: 2 2 2

2 2 2 22 2

( )( ) ( )a b c a b cP a b c

a b a c a b c

.

------- HT ------- Th sinh khng c s dng ti liu. Cn b coi thi khng gii thch g thm.

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

S GD&T BC LIU CHNH THC

P N - THANG IM THI TH THPT QUC GIA NM 2015

Mn: TON (Gm c 5 trang)

Cu p n im 1. (1im) a. Tp xc nh: }1{ \ D . b. S bin thin:

* Chiu bin thin: Ta c .1,0 )1(

4' 2 x

xy

Suy ra hm s ng bin trn mi khong )1;( v );1( , hm s khng c cc tr. * Gii hn: 1lim

y

x ; 1lim

y

x ;

y

x )1(lim ;

y

x )1(lim

Suy ra th c tim cn ngang l 1y v tim cn ng l 1x . * Bng bin thin

x 1 'y

y

1

1

0,25

0,25

0,25

* th: th ct Ox ti (3 ; 0); ct Oy ti 3; 0 . th nhn giao im )1; 1(I ca hai tim cn lm tm i xng.

0,25

2. (1 im) Gi s ( ; 1) ( )M a C , ta c: 3 1

1a a

1a

Suy ra 2

4'(1) 1(1 1)

y

.

0,25

0,25

Cu 1 (2 )

Vy phng trnh tip tuyn ti M l: '(1)( 1) ( 1)y y x hay 2y x . 0,5

O 1 1 I

y

3

3

x

1. (0.5 im)

V ;2

nn cos 0 , suy ra 2 2 2cos 1 sin 3

Do : 2 1 2 2 2 7 4 2sin 2 cos 2 2sin cos 1 2sin 2. . 1

3 3 9 9A

0,25

0,25

Cu 2 (1 )

2. (0.5 im)

iu kin: 0 2 0 2 4

4 0

x x x

x

, ta c :

3 1 3 3 3 33

log log ( 2) 1 log (4 ) log log ( 2) log [3(4 )]x x x x x x

23 3log [ ( 2)] log [3(4 )] ( 2) 3(4 ) 12x x x x x x x x

3

4 ( )xx loai

Vy phuong trnh c 1 nghim 3 x .

0,25

0,25

Cu 3 (0.5 )

t z a bi vi ,a b R . Ta c: (1 ) 2 5 3i z iz i tr thnh: (1 )( ) 2 ( ) 5 3 3 ( ) 5 3i a bi i a bi i a b a b i i

3 5 2

3 1a b aa b b

Suy ra 2 2 4 2 6w z z i i i . Vy s phc w c phn thc bng 6, phn o bng -1.

0,25

0,25

Cu 4 (1 )

2 31 1 1

2 (2 ln ) 4 2 .lne e e

I x x x dx x dx x xdx

3 4 41

1

4 1e

ex dx x e

t 2

1ln2

du dxu xx

dv xdx v x

, ta c:

2 22 2

11 1 1

12 .ln ln 2 2

ee ee x ex xdx x x xdx e

Vy 2 4 2

4 1 2 11 2 2

e e eI e

0,25

0,25

0,25

0,25

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S

M C

B

A H

Cu 5 (1 )

Theo gi thit 2 34ABC

aS

Gi H l hnh chiu ca S ln (ABC), suy ra 0 0

2 3

.

60 , SH=AH.tan60

1 1 3 3. .3 3 4 12

S ABC ABC

SAH a

a aV SH S a

Gi M l trung im ca BC, suy ra 21 1 39 39. .

2 2 6 12SBC aS SM BC a a

3 3 13, 13SBC

V ad A SBC S

0,25

0,25

0,25

0,25

Cu 6 (1 )

Gi I l trung im ca on CD, suy ra I(1;1;3) 0;0;2AI

suy ra (P) nhn 2; 2;0AB AI

lm vect php tuyn

Do (P) i qua A(1;1;1) nn phng trnh mp(P) l: 1(x-1)-1(y-1) = 0 Hay x-y=0

0,250,25 0,250,25

Cu 7 (1 )

Cch 1: Gi G l trng tm ABC , M l trung im BC. Ta c 3IH IG

(ng thng -le), suy ra

11 7;3 3

G

V 3AM GM

nn (4;1)M . ng thng BC qua M nhn (0; 2)AH

lm VTPT nn c phng trnh: 1y . ng trn ngoi tip ABC c tm l I, c bn knh 10IA nn c phng trnh 2 2( 4) ( 2) 10x y .

Ta im B v C l nghim ca h 2 2( 4) ( 2) 10

1x y

y

.

Gii h vi ch 3Bx , ta thu c (1;1)B v (7;1)C

0,25

0,25

0,25

0,25

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Cch 2: ng trn ngoi tip ABC c tm l I, c bn knh 10IA nn c phng trnh 2 2( 4) ( 2) 10x y . Phng trnh ng cao AH: 3x nn phng trnh ng thng BC c dng y b .

Ta im B v C l nghim ca h 2 2( 4) ( 2) 10x y

y b

.

v 3Bx nn gii h ta c: 24 10 ( 2) ;B b b , 24 10 ( 2) ;C b b suy ra 21 10 ( 2) ; 5AC b b , 21 10 ( 2) ;3BH b b V BH AC nn . 0BH AC

210 ( 2) 1 ( 5)(3 ) 0b b b

1 5

b b

. * Vi 1b ta c (1;1)B v (7;1)C nhn.

* Vi 5b ta c (3;5)B nn loi.

Ta c 2 2 2 210 4 2 (3 ) ( ) 3x xy y x y x y x y , du bng xy ra khi x y v 3 0x y .

Tng t 2 2 2 22 4 10 ( 3 ) ( ) 3x xy y x y x y x y , du bng xy ra khi x y v 3 0x y .

Do 2 2 2 210 4 2 2 4 10 4( )x xy y x xy y x y khi x y v 0x y 0,25

Cu 8 (1 )

Thay y x vo phng trnh th 2 ta c: 21 2 4 2 18 5( 3)x x x x (iu kin 0 4x ) 25 15 2 18 5( 3) 1 2 4x x x x x 25 15 2 18 1 2 4 0x x x x

2

3

2 18 1 2 4 (1)

x

x x x

Ta c 2(1) 2 18 17 3 4 ( 1)(4 )x x x x

( 1)(2 1) 4 ( 1)(4 ) 0x x x x 1

1(2 1) 4 4 0 (2)

x

x x x

3 2 2 3(2) 4 8 21 63 0 (2 3)(4 14 42) 0

2x x x x x x x

Tm li h c 3 nghim: (-1;-1), 3 3(3;3), ;2 2

.

0,25

0,25

Cu 9 (0.5 )

Rt 2 th t hai hp (mi hp mt th), khng gian mu c s phn t l: 10.10=100

Gi A l bin c nhn c 2 th c tch hai s ghi trn 2 th l s l, ta c A l bin c nhn c 2 th c tch hai s ghi trn 2 th l s chn. S phn t ca bin c A l 5.5=25 (v mi hp c 5 th l).

Suy ra xc sut cn tm l: 25 3( ) 1 1 100 4p A p A

0,25

0,25

Cu 10 (1 ) Ta c:

2 2 2

2 2 2 22 2

( )( ) ( )a b c a b cP a b c

a b a c a b c

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2 2 2 21 1 1 1 2 a b c

a b a c a b c

V 0 a b c nn: 2

2 2 2

2aa b ab b b

du bng xy ra khi 0a .

Tng t: 2

2 2

2aa c c

du bng xy ra khi 0a .

Nn: 2 2

1 1 1 1 2

2 2

P a b ca b ca ab c

du bng xy ra khi 0a 0,25

p dng cc bt ng thc: vi 0, 0x y ta c:

2 2 21 1 8

( )x y x y

du bng xy ra khi x y . (phi chng minh)

1 1 4x y x y

du bng xy ra khi x y .

Ta c: 2

8 4 2P a b ca b ca b c

0,25

t t a b c vi 0t .

Xt hm s 4 28 4( ) 2f t tt t

vi 0t .

Ta c: 5 2

5 3 5

32 8 2 8 32'( ) 2 t tf t t t t

5 2 4 2'( ) 0 2 8 32 0 2( 2)( 2 4 8) 0f t t t t t t t 2t

0,25

Bng bin thin:

Suy ra 112

P , du bng xy ra khi: 2

0,t a b ca b ca b c

02

ab c

Vy gi tr nh nht ca P l 112

.

0,25

--- HT---

t f(t)

f(t)

0 2

0

112

_ +