DaToan B ct_DH_K10

8/9/2019 DaToan B ct_DH_K10

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B GIO DC V O TO

CHNH THC

THI TUYN SINH I HC NM 2010

Mn: TON; Khi: BThi gian lm bi: 180 pht, khng kthi gian pht

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

Cu I (2,0 im) Cho hm s 2 11

xyx

+=+

.

1. Kho st s bin thin v v th (C) ca hm s cho.2. Tm mng thngy=2x+m ct th (C) ti hai im phn bitA,B sao cho tam gic OAB

c din tch bng 3 (O l gc ta ).Cu II (2,0 im)1. Gii phng trnh (sin .2 cos 2 ) cos 2cos 2 sin 0x x x x x+ + =2. Gii phng trnh 23 1 6 3 14 8x x x x+ + = 0 (xR).Cu III (1,0 im) Tnh tch phn

( )21

lnd

2 ln

ex

I xx x

=

+.

Cu IV (1,0 im) Cho hnh lng tr tam gic u ' cAB=a, gc gia hai mt phng. ' 'ABC A B C ( ' )A BC v ( )ABC bng . Gi G l trng tm tam gic . Tnh th tch khi lng tr chov tnh bn knh mt cu ngoi tip t din GABCtheo a.

60o 'A BC

Cu V (1,0 im) Cho cc s thc khng m a, b, c tha mn: a+b+c = 1. Tm gi tr nh nht

ca biu thc 2 2 2 2 2 2 2 2 23( ) 3( ) 2a b b c c a ab bc ca a b c= + + + + + + + + .

PHN RING (3,0 im)Th sinh chc lm mt trong hai phn (phn A hoc B)A. Theo chng trnh Chun

Cu VI.a (2,0 im)1. Trong mt phng toOxy, cho tam gicABCvung tiA, c nh C( 4; 1), phn gic trong gcA cphng trnh x+y 5 = 0. Vit phng trnh ng thngBC, bit din tch tam gicABCbng 24 vnhA c honh dng.

2. Trong khng gian toOxyz, cho cc imA(1; 0; 0),B(0; b; 0), C(0; 0; c), trong b, c dngv mt phng (P):yz+ 1 = 0. Xc nh b v c, bit mt phng (ABC) vung gc vi mt phng

(P) v khong cch tim On mt phng (ABC) bng1

3.

Cu VII.a (1,0 im) Trong mt phng ta Oxy, tm tp hp im biu din cc s phcztha mn:(1 )z i i z = + .

B. Theo chng trnh Nng caoCu VI.b (2,0 im)

1. Trong mt phng toOxy, cho imA(2; 3 ) v elip (E): 2 2 13 2

x y+ = . GiF1 vF2 l cc

tiu im ca (E) (F1 c honh m); Ml giao im c tung dng ca ng thngAF1 vi(E);Nl im i xng caF2 qua M. Vit phng trnh ng trn ngoi tip tam gicANF2.

2. Trong khng gian toOxyz, cho ng thng : 12 1 2

y z= = . Xc nh ta im Mtrn

trc honh sao cho khong cch tMn bng OM.

Cu VII.b (1,0 im) Gii h phng trnh2

2

log (3 1)

4 2 3x xy x

y

=

+ = (x,yR).

---------- Ht ----------Th sinh khng c sdng ti liu. Cn b coi thi khng gii thch g thm.

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


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Trang 1/4

B GIO DC V O TO CHNH THC

P N THANG IM THI TUYN SINH I HC NM 2010

Mn: TON; Khi B(p n - thang im gm 04 trang)

P N THANG IM

Cu p n im

1. (1,0 im)

Tp xc nh: R \ {1}.

S bin thin:

- Chiu bin thin:2

1'

( 1)y

x=

+> 0, x1.

0,25

Hm sng bin trn cc khong (; 1) v (1; +).

- Gii hn v tim cn: lim lim 2x x

y y +

= = ; tim cn ngang: y= 2.

( 1)

limx

y

= + v( 1)

limx

y+

= ; tim cn ng: x=1.

0,25

- Bng bin thin:

0,25

th:

0,25

2. (1,0 im)

Phng trnh honh giao im: 2 11

xx

++

=2x+m

2x+ 1 = (x+ 1)(2x+m) (dox=1 khng l nghim phng trnh)

2x2+ (4 m)x+ 1 m= 0 (1).

0,25

=m2+ 8 > 0 vi mi m, suy ra ng thng y=2x+m lun ct th (C) ti hai imphn bitA,B vi mi m. 0,25

Gi A(x1;y1) vB(x2;y2), trong x1 vx2 l cc nghim ca (1); y1=2x1+m v y2 =2x2+m.

Ta c: d(O,AB) =| |

5

mvAB= ( ) ( )

2 2

1 2 1 2x x y y + = ( )2

1 2 1 25 20x x x x+ =25( 8)

2

m +.

0,25

I

(2,0 im)

SOAB =1

2AB. d(O,AB) =

2| | 8

4

m m +, suy ra:

2| | 8

4

m m += 3 m = 2. 0,25

x 1 +

'y + +

y

2

2+

2

1 O x

y

1


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Trang 2/4

Cu p n im

1. (1,0 im)

Phng trnh cho tng ng vi: 22sin cos sin cos2 cos 2cos2 0x x x x x x + + = 0,25

cos2 sin (cos 2)cos2 0x x x x+ + = (sin cos 2)cos2 0x x x+ + = (1). 0,25

Do phng trnh sin cos 2 0x x+ + = v nghim, nn: 0,25

(1) cos2 0x = 4 2x k

= + (kZ). 0,25

2. (1,0 im)

iu kin:1

63

x . 0,25

Phng trnh cho tng ng vi: 2( 3 1 4) (1 6 ) 3 14 5 0x x x x+ + + = 0,25

3( 5) 5

( 5)(3 1) 03 1 4 6 1

x xx x

x x

+ + + =

+ + +

x = 5 hoc

3 1

3 1 03 1 4 6 1 xx x+ + + =+ + + .

0,25

II

(2,0 im)

3 1 13 1 0 ; 6

33 1 4 6 1x x

x x

+ + + > + + +

, do phng trnh cho c nghim:x= 5. 0,25

t 2 lnt x= + , ta c1

d dt xx

= ; x= 1 t= 2; x = et= 3. 0,25

3

22

2d

tI t

t

=

3 3

22 2

1 1d 2 dt t

t t= . 0,25

33

2 2

2ln t

t= +

0,25

III

(1,0 im)

1 3ln

3 2= + . 0,25

Thtch khi lng tr.

GiD l trung imBC, ta c:

BCAD BC 'D, suy ra:' 60ADA = .

0,25

Ta c: 'AA =AD.tan'ADA =3

2

a; SABC =

2 3

4

a.

Do : 3. ' ' ' 3 3V S . '8

ABC A B C ABC aAA= = .

0,25

Bn knh mt cu ngoi tip tdin GABC.

GiHl trng tm tam gicABC, suy ra:

GH// 'A A GH (ABC).

GiIl tm mt cu ngoi tip t din GABC, ta cIl giaoim ca GHvi trung trc caAG trong mt phng (AGH).

GiEl trung imAG, ta c:R=GI=.GE GA

GH=

2

2

GA

GH.

0,25

IV

(1,0 im)

Ta c: GH='

3

A=

2

a;AH=

3

3

a; GA2=GH2+AH2=

27

12

a. Do : R=

27

2.12

a.

2

a=

7

12

a. 0,25

HA

B

C

'A

'B

'C

G

D

A

E

H

G

I


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Trang 3/4

Cu p n im

Ta c: M (ab+bc+ca)2+ 3(ab+bc+ca) + 2 1 2( )ab bc ca + + . 0,25

t t=ab+bc+ca, ta c:2( ) 1

03 3

a b ct

+ + = .

Xt hm 2( ) 3 2 1 2f t t t t = + + trn1

0;2

, ta c:2

'( ) 2 31 2

f t t t

= +

;

32''( ) 2

(1 2 )f t

t=

0, du bng ch xy ra ti t= 0; suy ra '( )f t nghch bin.

0,25

Xt trn on1

0;3

ta c:1 11

'( ) ' 2 3 03 3

f t f

= >

, suy ra f(t) ng bin.

Do : f(t) f(0) = 2 t1

0;3

.

0,25

V(1,0 im)

V th: M f(t) 2 t1

0;3

; M= 2, khi: ab=bc=ca, ab+bc+ca= 0 v a+b+c = 1

(a; b; c) l mt trong cc b s: (1; 0; 0), (0; 1; 0), (0; 0; 1).Do gi tr nh nht ca Ml 2.

0,25

1. (1,0 im)

GiD l im i xng ca C( 4; 1) qua d:x+y 5 = 0, suy ra ta D(x;y) tha mn:

( 4) ( 1) 0

4 15 0

2 2

x y

x y

+ = +

+ =

D(4; 9).0,25

imA thuc ng trn ng knh CD, nn ta A(x;y)

tha mn:2 2

5 0

( 5) 32

x y

x y

+ =

+ =vi x> 0, suy ra A(4; 1). 0,25

AC= 8 AB=2S

ABCAC

= 6.

B thuc ng thngAD:x= 4, suy ra ta B(4;y) tha mn: (y 1)2= 36

B(4; 7) hocB(4; 5).

0,25

Do dl phn gic trong ca gcA, nn AB

v D

cng hng, suy ra B(4; 7).

Do , ng thngBCc phng trnh: 3x 4y+ 16 = 0.0,25

2. (1,0 im)

Mt phng (ABC) c phng trnh: 11

x y z

b c+ + = . 0,25

Mt phng (ABC) vung gc vi mt phng (P):yz+ 1 = 0, suy ra: 1b

1c

= 0 (1). 0,25

Ta c: d(O, (ABC)) =1

3

2 2

1

1 11

b c+ +

=1

3

2

1

b+

2

1

c= 8 (2).

0,25

VI.a

(2,0 im)

T (1) v (2), do b, c> 0 suy ra b = c =1

2. 0,25

Biu din s phc z=x+yi bi im M(x;y) trong mt phng ta Oxy, ta c:

|zi | = | (1 +i)z| |x+ (y 1)i | = | (xy) + (x+y)i |0,25

x2+ (y 1)2= (xy)2+ (x+y)2 0,25

x2+y2+ 2y 1 = 0. 0,25

VII.a

(1,0 im)

Tp hp im Mbiu din cc s phczl ng trn c phng trnh: x2+ (y+ 1)2= 2. 0,25

d

A

B

D

C


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Trang 4/4

Cu p n im

1. (1,0 im)

Nhn thy:F1(1; 0) vF2(1; 0).

ng thng AF1 c phng trnh:1

3 3

x y+= .

0,25

Ml giao im c tung dng caAF1 vi (E), suy ra:

2 31;

3M =

MA = MF2=

2 3

3.

0,25

DoNl im i xng caF2 qua Mnn MF2=MN, suy ra: MA=MF2=MN. 0,25

Do ng trn (T) ngoi tip tam gicANF2 l ng trn tm M, bn knh MF2.

Phng trnh (T): ( )2

2 2 3 41

3 3x y

+ =

.

0,25

2. (1,0 im)

ng thng i qua imA(0; 1; 0) v c vectch phng v

= (2; 1; 2).

Do Mthuc trc honh, nn Mc ta (t; 0; 0), suy ra: AM

= (t; 1; 0)

,v AM

= (2; 2t; t 2)

0,25

d(M, ) =,v AM

v

=25 4 8

3

t t+ +. 0,25

Ta c: d(M, ) =OM 25 4 8

3

t t+ += | t| 0,25

VI.b

(2,0 im)

t2t 2 = 0 t= 1 hoc t= 2.

Suy ra: M(1; 0; 0) hoc M(2; 0; 0).0,25

iu kin y>1

3, phng trnh th nht ca h cho ta: 3y 1 = 2x. 0,25

Do , h cho tng ng vi:2 2

3 1 2

(3 1) 3 1 3

xy

y y y

=

+ =

2

3 1 2

6 3 0

xy

y y

=

= 0,25

12

2

1

2

x

y

=

=

0,25

VII.b

(1,0 im)

1

1.

2

x

y

=

=

0,25

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

M

y

x

A

F1 F2

O

N

DaToan B ct_DH_K10

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