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S GD&T BC NINH TRNG THPT THUN THNH S 1

NGY 05/01/2014

THI TH I HC LN 1 NM 2013 2013 Mn : TON, Khi A, B

Thi gian lm bi: 180 pht (khng k thi gian giao )

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

Cu I (2,0 im). Cho hm s: 2x 1yx 1

1. Kho st s bin thin v v th (C) ca hm s.

2. Tm m ng thng y= 12

x m ct th (C) ti hai im A,B sao cho KA=KB vi K(2;0).

Cu II (2,0 im).

1. Gii phng trnh:

42

cos)sin2(2

cos)2

cos2

(sin22 33 xxxxx .

2. Gii phng trnh : 2 227 21 28

x x x x x

Cu III (1,0 im). Tnh: I=.2 2 3 1

1

x x x

x

x e xe e dxxe

Cu IV (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh thoi,hai ng cho AC = 2 3a , BD = 2a v ct nhau

ti O, hai mt phng (SAC) v (SBD) cng vung gc vi mt phng (ABCD). Bit khong cch t im O

n mt phng (SAB) bng 3

4a , tnh th tch khi chp S.ABCD theo a, v gc gia 2 mt phng (SAB)

vi (SBD). Cu V:(1,0 im). Cho x,y,z > 0 tha mn: 2 2 2x y xz yz xy .

Tm gi tr nh nht ca 4 4 4 4 4 41 1 1

4 4P x y z

x y z

PHN RING (3,0 im). Th sinh ch c lm mt trong hai phn (phn A hoc B) A. Theo chng trnh Chun Cu VI.a (2,0 im). 1. Trong mt phng ta Oxy cho 2 ng thng c phng trnh ln lt l d1: 3x-4y-24=0,

d2: 2x-y-6=0. Vit phng trnh ng trn(C ) tip xc vi d1 ti A v ct d2 ti B, C sao cho BC = 4 5 v sinA = 2

5. Bit tm I ca ng trn (C ) c cc ta u dng.

2. Gii h phng trnh:

2 4

29 3

log log 2

log log 1

y xy

x x y

Cu VII.a (1,0 im). T cc ch s 1,2,3,4,5,6 lp cc s c 4 ch s khc nhau. Ly ngu nhin mt s trong cc s c

lp, tnh xc sut s c ly c 2 ch s chn, 2 ch s l. B. Theo chng trnh Nng cao

Cu VI.b (2,0 im). 1. Trong mt phng ta Oxy cho ng trn 2 2: 2 C x y .Vit phng trnh tip tuyn ca

ng trn (C) bit tip tuyn ct cc tia Ox, Oy ln lt ti A v B sao cho tam gic OAB c din tch nh nht.

2. Trong khng gian vi h ta Oxyz, cho tam gic ABC c A(0;0;2), B(0;1;0), C(-2;0;0). Gi H l trc tm ca tam gic ABC. Vit phng trnh mt cu tm H tip xc vi Oy.

Cu VII.b (1,0 im)Gii bt phng trnh 222log log2 4 20 0x x 2

..Ht H v tn th sinh...................................................................., S bo danh.....................................................

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P N V THANG IM Cu -

Ni dung im

I.1 *Tp xc nh : \ 1D

Tnh 21' 0

( 1)y x D

x

Hm s nghch bin trn cc khong ( ;1) v (1; ) *Hm s khng c cc tr Gii hn

1

xlim y

1

xlim y

2

xlim y 2

xlim y

th c tim cn ng :x=1 , tim cn ngang y=2 *Bng bin thin x 1 y - - y

2

2

*V th (Hc sinh t v)

0.25 0.25 0.25 0.25

I.2 * PT honh giao im ca dm: y =

12

x m vi (C) l :

2 1 11 2

x x mx

2

15 2 2 2 0 1

x

x m x m

dm ct ti hai im khi (1) nghim phn bit khc 24 12 17 0

1 2 5 2 2 0m m

m m

m

* Gi x1, x2 l cc nghim ca PT(1): 1 2 5 2x x m . To giao im ca dm vi (C):

1 1 2 21 1; , ;2 2

A x x m B x x m

.Gi I l trung im ca AB th 5 2 5 2;2 4

m mI

* KA=KB 32m

KI d m

0.25 0.25 0.25 0.25

II.1

Pt(1)

2

sin2

cossin22

cos2

cos2

sin12

cos2

sin4 xxxxxxxx

2

sin2

cossin22

cossin211

2cos

2sin4 xxxxxxx

012

cos2

0sin2

02

sin2

cos

012

cos2)sin2(2

sin2

cosx

x

xx

xxxx

+) x x x xsin cos 0 sin 0 k x k2 (k )2 2 2 4 2 4 2

+) 2xsin0xsin2 (v nghim)

0.25 0.25 0.25

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+) 2cos 1 41 0 cos 42 2 2 3x x x k (t/mk)

Vy nghim ca phng trnh l: 4x k2 , x k4 k2 3

0.25

II.2 III

K: x 0 , Nhn xt x = 0 khng l nghim ca phng trnh Nhn hai v ca phng trnh vi 2 ta c:

* 2 227 272 2 2 2 24 4

x x x x x x x x x

22 271 (*)4

x xx

VT(*) = f(x) c f(x) = 2

1 0, 02

xxx

x

, f(x) l hm nghch bin trn khong 0;

VP(*) = g(x) c g(x) = 27 0, 0 ( )2

x x g x l hm ng bin trn khong 0; .

phng trnh (*) c khng qu mt nghim.

Mt khc x = 23

l nghim ca (*).Vy phng trnh cho c nghim duy nht x = 23

.

21 1

11 1

ln 1 ln 1 ln 1

x x x xx

x x

x x x x x x x x

xe e xe d xeI dx xe dx

xe xex xd e xe x xe xe e dx x xe xe e C

0.25 0.25 0.25 0.25 0,5 0,5

IV

T gi thit AC = 2 3a ; BD = 2a v AC , BD vung gc vi nhau ti trung im O ca mi ng cho.Ta c tam gic ABO vung ti O v AO = 3a ; BO = a. Gi K l hnh chiu ca O trn AB, gi I l hnh chiu ca O trn SK. T gi thit hai mt phng (SAC) v (SBD) cng vung gc vi mt phng (ABCD) nn giao tuyn ca chng l SO (ABCD). Ta chng minh c khong cch O ti (SAB) l on OI

Ta c trong tam gic vung AOB ta c: 2 2 2 2 21 1 1 1 1 3

3 2aOK

OK OA OD a a

.Tam gic SOK vung ti O, OI l ng cao 2 2 21 1 1

2aSO

OI OK SO .

Din tch y 24 2. . 2 3D SABC ABOS OA OB a ;

ng cao ca hnh chp 2aSO .

Th tch khi chp S.ABCD: 3

.1 3.3 3D DS ABC ABC

aV S SO

Ta c hnh chiu ca tm gic SAB trn mf(SBD) l Tam gic SBO . Gi l gc gia hai mt phng

(SAB) v (SBD) ta c os SBOSAB

scs

Ta c : 2

21 1 1. , os arccos2 4 4 4SBO SAB

as OB SO SK a s a c

0.25 0.25 0.25 0.25

A

B C

O

I D

a

K

S

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V

p dng bt ng thc C-Si cho hai s dng v bt ng thc: 2

2 2

2a b

a b

Ta c:

2 42 24 4

42 2 4 4

1 1 8 12 2 8

x y x yP z z

x y z zx y

t 4

4 0 1x y

tz

Khi ta c: 8 81 1 2

8 8t tP

t t

Xt hm s

28 1 8( ) 2 '( ) 0, 0;1

8 8tf t f t t

t t

Ta c f(x) nghch bin trn 0;1 0;1

81min (1)8t

P f

Khi x = y = 2z

0.25 0.25 0.25 0.25

VIa.1 Gi I(x;y), R ln lt l tm v bn knh ng trn (C ) p dng nh l sin trong tam gic ta c: R = d(I; d1) =5 ( do (C ) tip xc vi d1) Gi M l trung im ca BC theo nh l Pitago ta c MI = d(I;d2) = 2 2 5R MB .

Khi ta c h: 3 4 24 25

2 6 5

x y

x y

Gii h ta c 2 nghim tha mn yu cu

TH1 1;1I ta c phng trnh (x -1)2+(y-1)2=25 TH2 I(9;7) ta c phng trnh (x -9)2+(y-7)2=25

0.25 0.25 0.25 0.25

VIa.2

k: 29 3

00 0 log log2 0

x yy x x xxy

Khi ta c h 2

2

23

y xyx xy

2

22

1( )31

123 3

x y loaixx y

x yyx xy x xy

(t/mk)

0.25 0.25 0.5

VIIa T 6 ch s cho ta lp c 46 360A s c 4 ch s khc nhau S cch chn 2 ch s chn t 3 ch s 2,4,6 l 23 3C S cch chn 2 ch s l t 3 ch s 1,3,5 l 23 3C T 4 ch s c chn ta lp s c 4 ch s khc nhau, mi s lp c ng vi mt hon v ca 4 phn t. theo quy tc nhn ta c s cc s lp c tha mn yu cu l:

2 23 3. .4! 216C C

Xc sut chn c s c 4 ch s khc nhau c chn t cc ch s 1,2,3,4,5,6 trong

c 2 ch s chn 2 ch s l l: 216 3360 5

P

0.25 0.25 0.25 0.25

VIa.1 +

Tm : 0;0

Ban knh : 2

C O

C R

. Gi ta ;0 , 0;A a B b vi 0, 0a b .

0.25

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+ Phng trnh AB: 1 1 0x y x ya b a b

AB tip xc (C) 2 2

2 2

1, 2 2 21 1

abd O ABa b

a b

(***)

2 2 2 2

2 22 2a OABa b a b S

a b b

OABS nh nht khi a b . T a b v (***) suy ra 2a b .

Kt lun: Phng trnh tip tuyn l 1 02 2x y .

0.25 0.25 0.25

VIa.2 *Ta c ( )

AH BCBC AOH BC OH

AO BC

.

Tng t AB OH Suy ra ( )OH ABC .

*Phng trnh mp (ABC): 1 2 2 02 1 2x y z x y z

*mp(ABC) c vtpt 1;2;1n

nn OH c vtcp (1;2; 1)u n

*Phng trnh ng thng OH: 1 2 12 ; ;3 3 3

x ty t Hx t

Khong cch t H ti Oy l 23

R

Phng trnh mt cu tm H tip xc vi Oy l 2 2 21 2 1 2

3 3 3 9x y z

0.5 0.25 0.25

VIIb iu kin: x> 0 ; BPT 22

224log log2 4 20 0x x

t.22log4 xy , y 1

0.25

. BPT tr thnh y2 + y- 20 0 - 5 y 4.Do y 1 nn ta c y 4 0.25

Khi ta c : 22log 2

2 24 4 log 1 1 log 1x x x

1 22

x

0.25 0.25

Lu : Nu th sinh lm cch khc ng th gim kho chm theo cc bc lm ca cch .

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