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TRNG I HC VINH TRNG THPT CHUYN

KHO ST CHT LNG LP 12, LN 1 - NM 2014 Mn: TON; Khi: A v A1; 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 2 3.

1xyx

=

a) Kho st s bin thin v v th (H) ca hm s cho. b) Tm m ng thng : 3 0d x y m+ + = ct (H) ti hai im M, N sao cho tam gic AMN vung ti im (1; 0).A

Cu 2 (1,0 im). Gii phng trnh sin3 2cos2 3 4sin cos (1 sin ).x x x x x+ = + + + Cu 3 (1,0 im). Gii bt phng trnh 24 1 2 2 3 ( 1)( 2).x x x x+ + +

Cu 4 (1,0 im). Tnh tch phn 1

20

3 2ln(3 1) d .( 1)x xI x

x

+ +=

+

Cu 5 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh ch nht, mt bn SAD l tam gic vung ti S, hnh chiu vung gc ca S ln mt phng (ABCD) l im H thuc cnh AD sao cho 3 .HA HD= Gi M l trung im ca AB. Bit rng 2 3SA a= v ng thng SC to vi y mt gc 030 . Tnh theo a th tch khi chp S.ABCD v khong cch t M n mt phng (SBC).

Cu 6 (1,0 im). Gi s x, y, z l cc s thc khng m tha mn 2 2 25( ) 6( ).x y z xy yz zx+ + = + + Tm gi tr ln nht ca biu thc 2 22( ) ( ).P x y z y z= + + +

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 (2; 1)M l trung im cnh AC, im

(0; 3)H l chn ng cao k t A, im (23; 2)E thuc ng thng cha trung tuyn k t C. Tm ta im B bit im A thuc ng thng : 2 3 5 0d x y+ = v im C c honh dng.

Cu 8.a (1,0 im). Trong khng gian vi h ta ,Oxyz cho ng thng 2 1 2:1 1 2

x y zd + = =

v hai mt

phng ( ) : 2 2 3 0, ( ) : 2 2 7 0.P x y z Q x y z+ + + = + = Vit phng trnh mt cu c tm thuc d, ng thi tip xc vi hai mt phng (P) v (Q).

Cu 9.a (1,0 im). Cho tp hp { }1, 2, 3, 4, 5 .E = Gi M l tp hp tt c cc s t nhin c t nht 3 ch s, cc ch s i mt khc nhau thuc E. Ly ngu nhin mt s thuc M. Tnh xc sut tng cc ch s ca s bng 10.

b. Theo chng trnh Nng cao Cu 7.b (1,0 im). Trong mt phng vi h ta ,Oxy cho hai im (1; 2), (4; 1)A B v ng thng

: 3 4 5 0.x y + = Vit phng trnh ng trn i qua A, B v ct ti C, D sao cho 6.CD =

Cu 8.b (1,0 im). Trong khng gian vi h ta ,Oxyz cho im (1; 1; 0)M v hai ng thng

1 21 3 1 1 3 2

: , : .1 1 1 1 2 3

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

Vit phng trnh mt phng (P) song song vi 1d v 2d ng

thi cch M mt khong bng 6.

Cu 9.b (1,0 im). Tm s nguyn dng n tha mn 0 1 2 31 1 1 1 ( 1) 1

. . . .

2 3 4 5 2 156

nn

n n n n nC C C C Cn

+ + + =+

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

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TRNG I HC VINH TRNG THPT CHUYN

P N KHO ST CHT LNG LP 12, LN 1 - NM 2014 Mn: TON Khi A, A1; Thi gian lm bi: 180 pht

Cu p n im a) (1,0 im) 10. Tp xc nh: \{1}.R 20. S bin thin: * Gii hn ti v cc: Ta c lim 2

xy

= v lim 2.

xy

+=

Gii hn v cc: 1

limx

y+

= v 1

lim .x

y

= +

Suy ra th (H) c tim cn ngang l ng thng 2,y = tim cn ng l ng thng 1.x = * Chiu bin thin: Ta c 2

1' 0, 1.( 1)y xx= >

Suy ra hm s ng bin trn mi khong ( ); 1 v ( )1; .+

0,5

* Bng bin thin:

30. th:

th ct Ox ti 3 ; 0 ,2

ct Oy ti (0;3).

Nhn giao im (1; 2)I ca hai tim cn lm tm i xng.

0,5

b) (1,0 im) Ta c 1: .

3 3md y x= Honh giao im ca d v (H) l nghim ca phng trnh

2 3 1,

1 3 3x m

xx

=

hay 2 ( 5) 9 0, 1.x m x m x+ + = (1)

Ta c 2( 7) 12 0,m = + + > vi mi m. Suy ra phng trnh (1) c 2 nghim phn bit. Hn na c 2 nghim 1 2,x x u khc 1. Do d lun ct (H) ti 2 im phn bit 1 1 2 2( ; ), ( ; ).M x y N x y

0,5

Cu 1. (2,0 im)

Ta c 1 1 2 2( 1; ), ( 1; ).AM x y AN x y= =

Tam gic AMN vung ti A . 0.AM AN =

Hay 1 2 1 2( 1)( 1) 0x x y y + = 1 2 1 2

1( 1)( 1) ( )( ) 09

x x x m x m + + + =

21 2 1 210 ( 9)( ) 9 0.x x m x x m + + + + = (2)

p dng nh l Viet, ta c 1 2 1 25, 9.x x m x x m+ = = Thay vo (2) ta c

210( 9) ( 9)( 5) 9 0m m m m + + + = 6 36 0 6.m m = = Vy gi tr ca m l 6.m =

0,5

Cu 2. (1,0 im)

Phng trnh cho tng ng vi sin3 sin 2cos2 3(sin 1) cos (sin 1)x x x x x x + = + + +

0,5

x

'y

y

+ 1

2

+ +

+

2

x O

y

I 3

2

1 32

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2

2cos2 sin 2cos2 (sin 1)(cos 3)(sin 1)(2cos2 cos 3) 0(sin 1)(4cos cos 5) 0(sin 1)(cos 1)(4cos 5) 0.

x x x x x

x x x

x x x

x x x

+ = + +

+ =

+ =

+ + =

*) sin 1 2 ,2

x x kpi pi= = + .k Z

*) cos 1 2 ,x x kpi pi= = + .k Z *) 4cos 5 0x = v nghim. Vy phng trnh c nghim 2 , 2 , .

2x k x k kpi pi pi pi= + = + Z

0,5

iu kin: 1.x Nhn thy 1x = l mt nghim ca bt phng trnh. Xt 1.x > Khi bt phng trnh cho tng ng vi ( ) ( ) 3 24 1 2 2 2 3 3 2 12x x x x x+ + +

( )

2

2

4( 3) 4( 3) ( 3)( 2 4)1 2 2 3 3

4 43 ( 1) 3 0. (1)1 2 2 3 3

x xx x x

x x

x xx x

+ + ++ + + +

+ +

+ + + +

0,5

Cu 3. (1,0 im)

V 1x > nn 1 0x + > v 2 3 1.x + > Suy ra 4 4 3,1 2 2 3 3x x

+ V (C) i qua A, B nn IA IB R= =

2 2 2 2

2 2

( 1) ( 2) ( 4) ( 1)3 6 ( ; 3 6)

10 50 65 10 50 65 (1)

a b a b Rb a I a a

R a a R a a

+ = + =

=

= + = +

0,5

Cu 7.b (1,0 im)

K IH CD ti H. Khi 9 29

3, ( , )5

aCH IH d I

+= = =

22 2 (9 29)9

25aR IC CH IH = = + = + (2)

T (1) v (2) suy ra 2

2 2(9 29)10 50 65 9 169 728 559 025

aa a a a

+ = + + =

14313

a

a

= =

(1; 3), 543 51 5 61

; ,13 13 13

I R

I R

=

=

Suy ra 2 2( ) : ( 1) ( 3) 25C x y + + = hoc 2 243 51 1525( ) : .

13 13 169C x y + =

0,5

V ( )P // 1 2,d d nn (P) c cp vtcp 1 1 22

(1; 1; 1), (1; 2; 1)

( 1; 2; 3) Pu

n u uu

= = = =

Suy ra pt (P) c dng 2 0.x y z D+ + + =

( ) 33, ( ) 6 6 96DD

d M PD

=+ = =

=

( ) : 2 3 0 (1)( ) : 2 9 0 (2)P x y zP x y z

+ + + = + + =

0,5

Cu 8.b (1,0 im)

Ly 1(1; 3; 1)K d v 2(1; 3; 2)N d th vo cc phng trnh (1) v (2) ta c ( ) : 2 3 0N P x y z + + + = nn 2 ( ) : 2 3 0d P x y z + + + = . Suy ra phng trnh mt phng (P)

tha mn bi ton l ( ) : 2 9 0.P x y z+ + = 0,5

Vi mi x R v mi s nguyn dng n, theo nh thc Niutn ta c 0,5

I

H

A

B

C D

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( )0 1 2 1 0 1. . . ( 1) . . . ( 1) (1 ) .n n n n n n nn n n n n nC x C x C x C C x C x x x x+ + + = + + = Suy ra ( )1 10 1 2 1

0 0

. . . ( 1) d (1 ) d .n n n nn n nC x C x C x x x x x+ + + =

Cu 9.b (1,0 im)

Hay 1 1

0 1 1

0 0

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

2 3 2

nn n n

n n nC C C x x x xn

+ + + =

+

1 1 11 2 ( 1)( 2)n n n n= =+ + + + , vi mi

*.n N

T ta c 21 1 3 154 0 11( 1)( 2) 156 n n nn n = + = =+ + (v *).n N

0,5