[blogtoanli.net]Đề thi thử toán khối d chuyên Đh vinh lần 2 2014

www.MATHVN.com Ton hc Vit Nam

www.DeThiThuDaiHoc.com Thi Th i Hc 1

TRNG I HC VINH TRNG THPT CHUYN

KHO ST CHT LNG LP 12, LN 2 - NM 2014 Mn: TON; Khi: B v D; Thi gian lm bi: 180 pht

I. PHN CHUNG CHO TT C TH SINH (7,0 im) Cu 1 (2,0 im). Cho hm s 3 26 3( 2) 4 5y x x m x m= + + + c th ( ),mC vi m l tham s thc.

a) Kho st s bin thin v v th ca hm s cho khi 1.m = b) Tm m trn ( )mC tn ti ng hai im c honh ln hn 1 sao cho cc tip tuyn ti mi im ca ( )mC vung gc vi ng thng : 2 3 0.d x y+ + =

Cu 2 (1,0 im). Gii phng trnh sin 1 cot 2.1 cos 1 cos

xx

x x+ + =

+

Cu 3 (1,0 im). Gii h phng trnh ( )

( )1 1 2 2 0

( , ).1 4 0

x y yx y

y y x x

+ =

+ + =

Cu 4 (1,0 im). Tnh din tch hnh phng c gii hn bi cc ng 3 1 ; 0; 1.(3 1) 3 1

x

x xy y x

= = =

+ +

Cu 5 (1,0 im). Cho t din ABCD c 2, 3, 2 ,AB AC a BD CD a BC a= = = = = gc to bi hai mt phng (ABC) v (BCD) bng 045 . Tnh theo a th tch khi t din ABCD v khong cch t B n mt phng (ACD).

Cu 6 (1,0 im). Gi s ,x y l cc s thc dng tha mn ( )2 2 23( ) 4 1 .x y x y+ = + + Tm gi tr ln nht ca biu thc 2 2 2 2

2 2.

2 2x y x yP

x y x y+ +

= ++ +

II. PHN RING (3,0 im) Th sinh ch c lm mt trong hai phn (phn a hoc phn b) a. Theo chng trnh Chun Cu 7.a (1,0 im). Trong mt phng vi h ta ,Oxy cho tam gic ABC c nh (3; 3),A tm ng trn ngoi tip

(2; 1),I phng trnh ng phn gic trong gc BAC l 0.x y = Tm ta cc nh B, C bit rng 8 55

BC = v

gc BAC nhn. Cu 8.a (1,0 im). Trong khng gian vi h ta ,Oxyz cho mt phng ( ) : 2 1 0P x y z + = v cc ng thng

1 23 7 2 1 1 3

: ; : ; : .2 1 2 1 2 1 1 1 2

x y z x y z x y zd d d+ = = = = = =

Tm 1 2,M d N d sao cho ng thng MN

song song vi (P) ng thi to vi d mt gc c 1cos .3

=

Cu 9.a (1,0 im). Cho phng trnh 28 4( 1) 4 1 0 (1),z a z a + + + = vi a l tham s. Tm a (1) c hai nghim 1 2,z z tha mn 1

2

z

z l s o, trong 2z l s phc c phn o dng.

b. Theo chng trnh Nng cao Cu 7.b (1,0 im). Trong mt phng vi h ta ,Oxy cho tam gic ABC c phng trnh ng thng cha ng cao k t B l 3 18 0,x y+ = phng trnh ng thng trung trc ca on thng BC l 3 19 279 0,x y+ = nh C thuc ng thng : 2 5 0.d x y + = Tm ta nh A bit rng 0135 .BAC =

Cu 8.b (1,0 im). Trong khng gian vi h ta ,Oxyz cho im (4; 4; 5), (2; 0; 1)A B v mt phng ( ) : 3 0.P x y z+ + + = Tm ta im M thuc mt phng (P) sao cho mt phng (MAB) vung gc vi (P) v

2 22 36.MA MB =

Cu 9.b (1,0 im). Cho th 2 2( ) :

1ax axC y

x

+ =

v ng thng : 2 1.d y x= + Tm cc s thc a d ct

( )aC ti hai im phn bit ,A B tha mn ,IA IB= vi ( 1; 2).I ------------------ Ht ------------------



TRNG I HC VINH TRNG THPT CHUYN

P N KHO ST CHT LNG LP 12, LN 2 - NM 2014 Mn: TON Khi B, D; Thi gian lm bi: 180 pht

Cu p n im a) (1,0 im) Khi 1m = hm s tr thnh 3 26 9 1.y x x x= + a) Tp xc nh: .R b) S bin thin: * Gii hn ti v cc: Ta c lim

xy

= v lim .

xy

+= +

* Chiu bin thin: Ta c 2' 3 12 9;y x x= + 1 1

' 0 ; ' 0 ; ' 0 1 3.3 3

x xy y y x

x x

= < <

Suy ra hm s ng bin trn mi khong ( ) ( ); 1 , 3; ; + nghch bin trn khong ( )1; 3 . * Cc tr: Hm s t cc i ti 1, 3,Cx y= = hm s t cc tiu ti 3, 1.CTx y= =

0,5

* Bng bin thin:

c) th:

0,5

b) (1,0 im) ng thng d c h s gc 1 .

2k = Do tip tuyn ca ( )mC vung gc vi d c h s gc

' 2.k = Ta c 2' ' 3 12 3( 2) 2y k x x m= + + = 23 12 4 3 .x x m + = (1) Yu cu bi ton tng ng vi phng trnh (1) c hai nghim phn bit ln hn 1.

0,5

Cu 1. (2,0 im)

Xt hm s 2( ) 3 12 4f x x x= + trn (1; ).+ Ta c bng bin thin:

Da vo bng bin thin ta suy ra phng trnh ( ) 3f x m= c hai nghim phn bit ln hn 1 khi v ch khi 5 88 3 5 .

3 3m m < < < < Vy 5 8.

3 3m< <

0,5

Cu 2. (1,0 im)

iu kin: cos 1, sin 0 , .x x x k kpi Z Phng trnh cho tng ng vi 0,5

x

'y

y

1 + 3

3

+

1

+ 0 0 +

x O

3

y

1

1

3

x

( )f x

1 + +

8

2

5

+



2sin sin cos 1 cos cos 2

sinsinx x x x x

xx

+ ++ =

2sin cos 1 2sinsin cos cos2 0(sin cos )(1 cos sin ) 0.

x x x

x x x

x x x x

+ + =

+ + =

+ + =

*) sin cos 0 ,4

x x x kpi pi+ = = + .k Z

*) 211 cos sin 0 sin 24 2 2 , .

x kx x x

x k k

pipipi

pi pi

= + + = = = + Z

i chiu iu kin, ta c nghim ca phng trnh l , 2 , .4 2

x k x k kpi pipi pi= + = + Z

0,5

iu kin: 1.x t 1, 0.t x t= Khi 2 1x t= + v h tr thnh

2 2 2 2

(1 2 ) 2 0 2 2 0 ( ) 2 2 0( ) 3 0 3 0 ( ) 3 3 0

t y y t y ty t y ty

y y t t y ty t t y ty

+ = + = + =

+ + = + + = + =

Suy ra 20

2( ) 3( ) 0 3 3.

2 2

t y y tt y t y

t y y t

= =

+ =

= = +

0,5

Cu 3. (1,0 im)

*) Vi ,y t= ta c 22 2 0 1.t t + = = Suy ra 2, 1.x y= =

*) Vi 3 ,2

y t= + ta c 23 3 3 132 2 0 4 6 1 0 .2 2 4

t t t t t +

+ + = + = =

Suy ra 19 3 13 3 13, .8 4

x y += =

Vy nghim (x; y) ca h l 19 3 13 3 13(2; 1), ; .8 4

+

0,5

Ta c 3 1 0 3 1 0.(3 1) 3 1

xx

x xx

= = =+ +

R rng 3 1 0(3 1) 3 1

x

x x

+ +

vi mi [ ]0; 1 .x Do din tch ca hnh phng l

1 1

0 0

3 1 3 1d .3 d .(3 1) 3 1 (3 1) 3 1

x xx

x x x xS x x

= =

+ + + +

0,5

Cu 4. (1,0 im)

t 3 1,xt = + ta c khi 0x = th 2,t = khi 1x = th 2t = v 23 1.x t=

Suy ra 3 ln3d 2 d ,x x t t= hay 2 d3 d .ln3

x t tx = Khi ta c

( )22 223 2

22 2

2 3 2 22 2 2 2 2 2d 1 d .ln3 ln3 ln3 ln 3

tS t t t ttt t

= = = + =

0,5



Gi M l trung im BC. T cc tam gic cn ABC, DBC , .AM BC DM BC

T gi thit

00

0

45( , ) 45135

AMDAM DM

AMD

= =

=

TH 1. 045AMD =

S dng nh l Pitago , 2.AM a DM a = = K AH MD ti H. V ( ) ( ).BC AMD BC AH AH BCD Khi

0 22 1.sin 45 ; . 2.

2 2BCDaAH AM S DM BC a= = = =

Suy ra 31

. .

3 3ABCD BCDaV AH S= =

0,5

Cu 5. (1,0 im)

S dng nh l cosin cho 2 2 2 23AMD AD a AC AD a CD ACD = + = = vung ti A. Suy ra ( )

2 31 2. , ( ) 2.

2 2ABCD

ACDACD

VaS AC AD d B ACD aS

= = = =

TH 2. 0135AMD =

Tng t ta c ( )3 6

; ,( ) ,3 3ABCDa aV d B ACD= = ( 5AD a= ).

0,5

Ta c 2 2 2 2 2 22 1 3 3 3

. . . .

22 ( ) 2x y xy xy x

x y x y x y x y x yx y x y y xy y+

= =+ + + + ++ + + +

Tng t, ta cng c 2 22 1 3

. .

22x y y

x y x y x yx y+

+ + ++

Mt khc, ta c 2 ,2 2 3

x yx y x y

+ + +

v bt ng thc ny tng ng vi

2 2

2 24 2

32 2 5x y xyx y xy

+ + + +

, hay 2( ) 0.x y

0,5

Cu 6. (1,0 im)

T ta c 2 3 2 3 2. . .2 2 3

x yPx y x y x y x y x y x y

+ =

+ + + + + + Suy ra 4 .P

x y

+ (1)

T gi thit ta li c 2 2 2 23( ) 4( ) 4 2( ) 4.x y x y x y+ = + + + + Suy ra 2( ) 4,x y+ hay 2.x y+ (2) T (1) v (2) ta c 2.P Du ng thc xy ra khi v ch khi 1.x y= = Vy gi tr ln nht ca P bng 2, t c khi 1.x y= =

0,5

Cu 7.a (1,0 im)

V AD l phn gic trong gc A nn AD ct ng trn (ABC) ti E l im chnh gia cung BC .IE BC V E thuc ng thng 0x y = v (0; 0).IE IA R E= = Chn (2; 1)BCn EI= =

pt BC c dng 2 0.x y m+ + =

T gi thit 2 24 5 35 5

HC IH IC HC = = =

0,5

A

B D

M H

C

A

B C E

I

D H



3( , )5

d I BC =2| 5 | 385 5

mm

m

= + =

=

: 2 2 0: 2 8 0.

BC x yBC x y

+ = + =

V BAC nhn nn A v I phi cng pha i vi BC, kim tra thy : 2 2 0BC x y+ = tha mn.

T h 2 22 2 0 8 6(0; 2), ;

5 5( 2) ( 1) 5x y

B Cx y

+ =

+ = hoc 8 6; , (0; 2)

5 5B C

.

0,5

1 2( ; 2 2; 1); ( 1; ; 2 3).M d M m m m N d N n n n + + + + Suy ra ( 1; 2 2; 2 2).MN m n m n m n= + + + + +

V MN // (P) nn 2 0 2. 00 0( )

P m n m nn MNn nN P

+ = = =

Suy ra (3; 2; 4)MNu n n= + +

v (2; 1; 2).du =

0,5

Cu 8.a (1,0 im)

Suy ra 2 2

| 3 12 | | 4 | 1cos( , ) cos

33 2 4 29 2 4 29n nMN d

n n n n

+ += = = =

+ + + +

2 2 23( 4) 2 4 29 20 19 0 1n n n n n n + = + + + + = = hoc 19.n = *) 1 3 ( 3; 4; 2), (0; 1; 1).n m M N= = *) 19 21 ( 21; 40; 20), ( 18; 19; 35).n m M N= =

0,5

T gi thit suy ra 1 2,z z khng phi l s thc. Do ' 0, < hay 24( 1) 8(4 1) 0a a+ + <

2 24( 6 1) 0 6 1 0.a a a a < < (*)

Suy ra 2 2

1 2 11 ( 6 1) 1 ( 6 1)

, .

4 4a a a i a a a i

z z z+ + +

= = =

0,5

Cu 9.a (1,0 im)

Ta c 12

z

z l s o 21z l s o ( )2 2 2 0( 1) ( 6 1) 0 2 0 2.

aa a a a a

a

= + = =

=

i chiu vi iu kin (*) ta c gi tr ca a l 0, 2.a a= = 0,5

: 3 18 ( 3 18; ),: 2 5 ( ; 2 5).

B BH x y B b bC d y x C c c

= + +

= + +

T gi thit suy ra B i xng C qua ng trung trc . 0

: 3 19 279 0 l

60 13 357 4 (6; 4)10 41 409 9 (9; 23).

u BCx y

BC Mb c b Bb c c C

= + =

+ = =

+ = =

AC BH chn ( 3; 1) pt : 3 4 0 ( ; 3 4)AC BHn u AC x y A a a= = + + =

(6 ; 8 3 ), (9 ; 27 3 ).AB a a AC a a = =

0,5

Cu 7.b (1,0 im)

Ta c 02 2 2 2

1 (6 )(9 ) (8 3 )(27 3 ) 1135 cos( , )2 2(6 ) (8 3 ) . (9 ) (27 3 )

a a a aA AB ACa a a a

+ = = =

+ +

2 22

3 9(9 )(3 ) 12(3 ) 6 102| 9 | 6 10

aa a

a a aa a a

<



im) Gi

1 0 2 1( ) ( ) pt :3 0 2 1 1

y z x y zd P Q dx y z

= + += = =

+ + + = ( 2 2; ; 1) .M t t t d

Ta c 2 2 2( 2; 0; 1)0

2 36 6 8 0 14 4 14 ; ; .3 3 33

MtMA MB t t

Mt

=

= + = =

0,5

Honh giao im ca d v ( )aC l nghim ca phng trnh 2 2 2 1,

1x ax

xx

+ = +

hay

2 ( 1) 1 0, 1.x a x x + + = (1)

Phng trnh (1) c 2 nghim phn bit khc 1 2 1( 1) 4 0

3.1aa

aa

> = + > <

(2)

Khi gi 1 2,x x l hai nghim phn bit ca (1), ta c 1 1 2 2( ; 2 1), ( ; 2 1).A x x B x x+ +

0,5

Cu 9.b (1,0 im)

Do 2 2 2 21 1 2 2( 1) (2 3) ( 1) (2 3)IA IB x x x x= + + + = + + + ( )2 21 1 2 2 1 2 1 25 14 5 14 ( ) 5( ) 14 0x x x x x x x x + = + + + = 1 25( ) 14 0,x x + + = v 1 2.x x (3) Theo nh l Viet ta c 1 2 1.x x a+ = + Thay vo (3) ta c 195( 1) 14 0 ,5a a+ + = = tha mn

iu kin (2). Vy 19 .5

a =

0,5

[blogtoanli.net]Đề thi thử toán khối d chuyên Đh vinh lần 2 2014

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