datoandct_dh_k10

8/8/2019 DaToanDCt_DH_K10

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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 D(p n - thang im gm 04 trang)

P N THANG IM

Cu p n im

1. (1,0 im)

Tp xc nh: R.

S bin thin:

- Chiu bin thin: 'y = 4x3 2x= 2x(2x2+ 1); 'y (x) = 0 x= 0.

0,25

- Hm sng bin trn khong (; 0); nghch bin trn khong (0; +).

- Cc tr: Hm st cc i ti x= 0; yC= 6.

- Gii hn: limx

y

= limx

y +

=.0,25

- Bng bin thin:

0,25

th:

0,25

2. (1,0 im)

Do tip tuyn vung gc vi ng thng y=1

6

x 1, nn tip tuyn c h s gc bng 6. 0,25

Do , honh tip im l nghim ca phng trnh 4x3 2x = 6 0,25

x= 1, suy ra ta tip im l (1; 4). 0,25

I(2,0 im)

Phng trnh tip tuyn: y= 6(x 1) + 4 hay y = 6x+ 10. 0,25

1. (1,0 im)

Phng trnh cho tng ng vi: 2sinxcosx cosx (1 2sin2x) + 3sinx 1 = 0 0,25

(2sinx 1)(cosx+ sinx+ 2) = 0 (1). 0,25

Do phng trnh cosx+ sinx+ 2 = 0 v nghim, nn: 0,25

II(2,0 im)

(1) sinx =1

2x=

6

+k2 hoc x =5

6

+k2 ( kZ). 0,25

'y + 0

y6

x 0 +

y

x

6

2 2

O


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Cu p n im

2. (1,0im)

iu kin: x 2.

Phng trnh cho tng ng vi: ( )( )32 24 4 42 2 2 2 0

xx x+ = .0,25

24x 24 = 0 x = 1. 0,25

2 22 x + 3 42x = 0 2 2x + =x3 4 (1).

Nhn xt:x 3 4 .0,25

Xt hm sf(x) = 2 2x + x3 + 4, trn )3 4 ; + .

'f (x) =1

2x + 3x2 < 0, suy ra f(x) nghch bin trn )3 4 ; + .

Ta c f(2) = 0, nn phng trnh (1) c nghim duy nhtx= 2.

Vy phng trnh cho c hai nghim: x= 1; x= 2.

0,25

I =1

32 ln d

e

x x xx

=

1

2 ln d

e

x x x 1

ln3 d

ex

xx . 0,25

t u= lnx v dv= 2xdx, ta c: du=dx

xv v =x2.

1

2 ln d

e

x x x = ( )21

lne

x x 1

d

e

x x =e2

2

12

ex

=2 1

2

e +.

0,25

1

lnd

ex

xx

= ( )1

ln d ln

e

x x =2

1

1ln

2

e

x =1

2. 0,25

III(1,0 im)

Vy I =2

2

e 1. 0,25

M l trungim SA.

AH=2

4

a, SH= 2 2SA AH =

14

4

a.

0,25

HC=3 2

4

a, SC= 2 2SH HC + = a 2 SC=AC.

Do tam gic SACcn ti C, suy ra Ml trung im SA.

0,25

Thtch khi tdin SBCM.

Ml trung im SA SSCM=1

2SSCA

VSBCM = VB.SCM= 12

VB.SCA= 12

VS.ABC

0,25

IV(1,0 im)

VSBCM =1

6SABC.SH=

3 14

48

a. 0,25

iu kin: 2 x 5.

Ta c (x2+ 4x+ 21) (x2+ 3x+ 10) =x+ 11 > 0, suy ray> 0.0,25

y2= (x+ 3)(7 x) + (x+ 2)(5 x) 2 ( 3)(7 )( 2)(5 )x x x x+ +

= ( )2

( 3)(5 ) ( 2)(7 )x x x x+ + + 2 2, suy ra:0,25

y

2 ; du b

ng x

y ra khi v ch

khi x

=

1

3 . 0,25

V(1,0 im)

Do gi tr nh nht cay l 2 . 0,25

S

CD

BA

M

H


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Cu p n im

1.(1,0im)

ng trn ngoi tip tam gicABCc phng trnh:

(x+ 2)2+y2 = 74.

Phng trnhAH:x= 3 vBCAH, suy ra phng trnhBCc dng:y=a (a 7, doBCkhng i quaA).

Do honh B, Ctha mn phng trnh:

(x+ 2)2

+a2

= 74 x2

+ 4x+a2

70 = 0 (1).

0,25

Phng trnh (1) c hai nghim phn bit, trong c t nht

mt nghim dng khi v ch khi: | a | < 70 .

Do Cc honh dng, nnB( 2 274 a ; a) v C( 2 + 274 a ; a).

0,25

ACBH, suy ra: .AC BH

= 0

( )274 5a ( )274 5a + + (a+ 7)( 1 a) = 0 a2+ 4a 21 = 0

0,25

a= 7 (loi) hoc a= 3 (tha mn).

Suy ra C( 2 + 65 ; 3).0,25

2. (1,0 im)

Ta c vectphp tuyn ca (P) v (Q) ln lt l

Pn

= (1; 1; 1) v Qn

= (1; 1; 1), suy ra:

,P Qn n

= (2; 0; 2) l vectphp tuyn ca (R).

0,25

Mt phng (R) c phng trnh dng xz+D= 0. 0,25

Ta c d(O,(R)) = ,

2

Dsuy ra:

2

D= 2 D= 2 2 hocD= 2 2 . 0,25

VI.a(2,0 im)

Vy phng trnh mt phng (R): xz+ 2 2 = 0 hocxz 2 2 = 0. 0,25

Gi z = a+bi, ta c: 2 2z a b= + v z2 =a2 b2 + 2abi. 0,25

Yu cu bi ton tha mn khi v ch khi:2 2

2 2

2

0

a b

a b

+ =

= 0,25

2

2

1

1.

a

b

=

= 0,25

VII.a(1,0 im)

Vy cc s phc cn tm l: 1 +i; 1 i; 1 +i; 1 i. 0,251. (1,0 im)

Gi ta Hl (a; b), ta c: 2 2 2( 2)AH a b= + v khong cch

tHn trc honh l | b |, suy ra: a2 + (b 2)2=b2.0,25

DoHthuc ng trn ng knh OA, nn: a2+ (b 1)2= 1. 0,25

T, ta c:2

2 2

4 4 0

2 0.

a b

a b b

+ =

+ =

Suy ra: (2 5 2; 5 1)H hoc ( 2 5 2; 5 1)H .

0,25

VI.b(2,0 im)

Vy phng trnh ng thng l

( 5 1) 2 5 2 0x y = hoc ( 5 1) 2 5 2 0x y + = .0,25

I

A

B C

H

O

H

y

x

A

P Q

R

O


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Trang 4/4

Cu p n im

2. (1,0 im)

Ta c: + M1, nn M(3 +t; t; t).

+ 2i quaA(2; 1; 0) v c vectch phng v

= (2; 1; 2).0,25

Do : AM

= (t+ 1; t 1; t); ,v AM

= (2 t; 2; t 3). 0,25

Ta c: d(M, 2) =

,v AM

v

=

22 10 17

3

t t +, suy ra:

22 10 17

3

t t += 1 0,25

t2 5t+ 4 = 0 t= 1 hoc t= 4.

Do M(4; 1; 1) hoc M(7; 4; 4).0,25

iu kin: x> 2,y> 0 (1). 0,25

T h cho, ta c:2 4 2 0

2

x x y

x y

+ + =

= 0,25

2 3 0

2

x x

y x

=

=

0

2

x

y

=

=

hoc3

1.

x

y

=

=

0,25

VII.b(1,0 im)

i chiu vi iu kin (1), ta c nghim ca h l (x;y) = (3; 1). 0,25

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

M

2

1

d =1

H

datoandct_dh_k10

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