së gi¸o dôc vµ ®µo t¹o hµ tÜnh Tr­êng THPT NguyÔn Trung Thiªn

§Ò THi thö ®¹i Häc LÇN I n¨m 2014 Môn thi: To¸n - KHỐI A, A1, B Thời gian làm bài: 180 phút

I. PhÇn chung cho tÊt c¶ thÝ sinh(7,0 ®iÓm)

=

y

-

C©u I (2,0 ®iÓm) Cho hµm sè

3 + cã ®å thÞ (C) 1

x x

1. Kh¶o s¸t sù biÕn thiªn vµ vÏ ®å thÞ (C) cña hµm sè. 2. ViÕt ph­¬ng tr×nh tiÕp tuyÕn cña (C) biÕt kho¶ng c¸ch tõ giao ®iÓm I cña 2 tiÖm cËn cña (C) ®Õn

tiÕp tuyÕn b»ng 2 2 .

C©u II (2,0 ®iÓm)

.

1. Gi¶i ph­¬ng tr×nh 1

dx

2. TÝnh: I =

+ + = + 2 sin(2 x ) cos x cos 3 x p 4

t anx + 2 c os

x

1

2

2

(cid:242)

+

+

=

x

y

4

y

1

(cid:236) - (cid:239) (cid:237)

C©u III (1,0 ®iÓm) Gi¶i hÖ ph­¬ng tr×nh:

+

+ = y

x

2

2

xy y +

x

1

2

2

(cid:239) (cid:238)

.

+ + + + + ) ( ) ( ( ) = P x y z

C©u IV (1,0 ®iÓm) Cho h×nh chãp S.ABCD cã ®¸y ABCD lµ h×nh thang. §¸y lín AB = 2a ; BC = CD = DA = a; SA vu«ng gãc víi ®¸y, mÆt ph¼ng(SBC) t¹o víi ®¸y mét gãc 60o . TÝnh thÓ tÝch khèi chãp S.ABCD theo a. C©u V (1,0 ®iÓm) Cho 3 sè thùc d­¬ng x, y, z. T×m gi¸ trÞ nhá nhÊt cña biÓu thøc : 2 z 3

x 3 y 3 2 xy 2 yz 2 xz

. §­êng th¼ng AB cã ph­¬ng tr×nh

- = -

y+ 3 1 0 x

II. PhÇn riªng (3,0 ®iÓm): ThÝ sinh chØ ®­îc lµm mét trong hai phÇn (phÇn A hoÆc phÇn B) A. Theo ch­¬ng tr×nh chuÈn C©u VI. a. (1,0 ®iÓm) Trong mÆt ph¼ng víi hÖ täa ®é Oxy cho tam gi¸c ABC cã träng t©m G (2;-1). §­êng trung trùc cña c¹nh BC cã ph­¬ng tr×nh d : 3x y 4 0 + = . T×m täa ®é c¸c ®Ønh A, B, C. d 1 :10 C©u VII. a. (1,0 ®iÓm) Trong mÆt ph¼ng víi hÖ täa ®é Oxy cho hai ®iÓm A(2;0), B(6;4). ViÕt ph­¬ng tr×nh ®­êng trßn (C) tiÕp xóc víi trôc hoµnh t¹i ®iÓm A vµ kho¶ng c¸ch tõ t©m (C) ®Õn B b»ng 5.

n

3

(

.

( ) P x

(cid:230) (cid:246) = + x > 0) x

C©u VIII. a. (1,0 ®iÓm ) T×m sè h¹ng kh«ng chøa x trong khai triÓn

.

9 n

7 n

6 n

8 C + 2 n

2

(cid:231) (cid:247) Ł ł 2 x + + = C C 3 + 8 C C 3 n

tiÕp xóc víi

- = -

5 0 x y+ + = ; d 1 : 3

1 0 x

BiÕt r»ng n tháa m·n: B. Theo ch­¬ng tr×nh n©ng cao C©u VI. b. (1,0 ®iÓm) Trong mÆt ph¼ng víi hÖ täa ®é Oxy cho tam gi¸c ABC vu«ng c©n t¹i A(1;2). ViÕt ph­¬ng tr×nh ®­êng trßn (T) ngo¹i tiÕp tam gi¸c ABC biÕt ®­êng th¼ng d : x y 1 0 (T) t¹i B. C©u VII. b. (1,0 ®iÓm) Trong mÆt ph¼ng víi hÖ täa ®é Oxy cho hai ®­êng th¼ng y+ + = vµ ®iÓm I(1;-2). ViÕt ph­¬ng tr×nh ®­êng th¼ng ®i qua I c¾t d

2

.

2 : 3 sao cho

,d d lÇn l­ît t¹i A vµ B 1

3

.

AB = 2 2 (cid:230) (cid:246) (cid:230) (cid:246) + = log log 2 (cid:231) (cid:247)

C©u VIII. b. (1,0 ®iÓm) Gi¶i ph­¬ng tr×nh:

x

2

2

(cid:231) (cid:247) Ł ł Ł ł 2 x

x 2 --------------------- HÕt --------------------

§¸p ¸n K.A gåm cã 6 trang. L­u ý : Mäi c¸ch gi¶i ®óng ®Òu cho ®iÓm tèi ®a.

§¸p ¸n vµ h­íng dÉn chÊm §iÓm C©u

{ } 1

- D R= \ 1 (1,0 ®iÓm) ________________________________________________________________________ + TËp x¸c ®Þnh:

C©u I. 2,0 ®iÓm

2

)

) ; 1

= > „ - y ' 0 + Sù biÕn thiªn: , , suy ra hµm sè ®ång biÕn trªn c¸c kho¶ng x" 1 4 + 1) +¥ - ¥ - - 1; ( x . vµ (

fi+¥ x ; lim fi+¥ x

x lim x 1

y y 1 1 ( ________________________________________________________________________ + Giíi h¹n: lim fi - ¥ = ; lim = +¥ = => TiÖm cËn ngang: y=1 = - ¥ y => TiÖm cËn ®øng: x=-1. y- fi -

________________________________________________________________________ + B¶ng biÕn thiªn: ¥+ ¥ - -1 x y’ +

y 1

¥ +¥ 1 -

1

x

0

-1

3

-3

________________________________________________________________________ + §å thÞ : Giao víi Ox: (3;0), giao víi Oy: (0;-3). 0,25 0,25 0,25 0.25

)

0

( ;M x y 0

0

) +

2

2

- = - §å thÞ nhËn I(-1;1) lµm t©m ®èi xøng. 2 (1,0 ®iÓm) ________________________________________________________________________ x „ 1 Gi¶ sö thuéc (C), . , y 0 + 3 1 x 0 x 0 t¹i M lµ: - = - y x x 0 + 4 + Khi ®ã ph­¬ng tr×nh tiÕp tuyÕn D 3 ( 1 x 0 x 0

( x 0 (

+ + (cid:219) - - -

(

) =

) 1 ) 1

2 x 0

2

2

4 x y 6 3 0 x 0 x 0

0

+ + - - - - ________________________________________________________________________ Theo ®Ò : ) ( d I D = , ( ( 2 2 ) 1 4 6 x x 0 x 0

) 3 =

4

(

2

4

(cid:219) 2 2 16

(

) + 1 ) 1

) 1

x 0 + + + (cid:219) - + ( 8 16 = 0 x 0 x 0 0.25 0.25

= Ø (cid:219) Œ 1 = - 3 º x 0 x 0

0

D : y 0,5

0

D : x y 3 = - x 2 ; = + . 6 x = , ph­¬ng tr×nh 1 x = - , ph­¬ng tr×nh

2

2

+ + = (cid:219) x x x + (cid:219) - x 2cos cos 2 x x 2cos cos 2 2 cos 2 x 2sin cos + (cid:219) - -

(

________________________________________________________________________ Víi Víi 1 (1,0 ®iÓm) ________________________________________________________________________ PT 2 cos cos cos sin sin 0 x x

C©u II. 2,0 ®iÓm

+ + (cid:219) - x )( x ( 1 sin 2 x 2 cos ( cos 1 cos cos sin sin x x x x x

= 0 x ) ) = x ) = 0 ________________________________________________________________________ Ø = + Œ x k p Œ Ø 0 cos x (cid:219) Œ Œ tan 1 p 2 = - x = + = (cid:219) cos x 0 sin Œ Œ (cid:230) (cid:246) 1 Œ Œ - + = cos x sin x = x 1 º cos x (cid:231) (cid:247) Œ Ł ł º p 4 2

________________________________________________________________________ Ø = + x k p Œ p 2 Œ 0,25 0,5 0,25 Œ (cid:219) x k p Œ k ˛  Œ = k x p = - + 4 2 p Œ Œ º

2 (1,0 ®iÓm) ________________________________________________________________________

2

= = I dx dx Ta cã: (cid:242) (cid:242) ) x = (cid:222) t cos x §Æt x x sin cos + 2 cos (1 cos x xdx x 2sin cos

= - I Suy ra: (cid:242) x = - dt + x tan + 2 1 cos dt 1 2 1) ( t t

________________________________________________________________________ + (cid:230) (cid:246) t 1 = = + - I dt C ln (cid:231) (cid:247) (cid:242) ł 1 +Ł t 1 t t 1 2 1 1 2

2

________________________________________________________________________ (cid:230) (cid:246) x = + KÕt luËn: . ln I C (cid:231) (cid:247) 0,25 0,5 0,25 1 2 + 1 cos 2 cos x Ł ł

NhËn xÐt y=0 kh«ng tháa m·n hÖ ph­¬ng tr×nh.

C©u III. 1,0 ®iÓm

0,25

2

2

(cid:236) + 1 x + + = x y 4 (cid:239) (cid:239) y (cid:237) HÖ t­¬ng ®­¬ng víi (cid:239) + + = y x 2 (cid:239) y + (cid:238) x 1

________________________________________________________________________ + = (cid:236) 4 u v (cid:239) x = (cid:237) u , v = x + y. HÖ trë thµnh: §Æt + 2 1 y v 2 (cid:239) (cid:238) 1 = + u

2

Gi¶i hÖ ta cã: u =1 v = 3 ________________________________________________________________________ Ø 0,25 0,25 0,25 (cid:236) (cid:237) + Œ 1 x = (cid:236) 1 2 =(cid:236) x = y (cid:239) (cid:238) 1 1 u Œ (cid:222) (cid:219) (cid:237) (cid:237) Víi Œ = = = - (cid:236) 3 v (cid:238) 2 x (cid:239) Œ 3 y + = y x (cid:237) (cid:238) = Œ 5 y (cid:238) º

C©u IV. 1,0 ®iÓm

N

B

A

60 0

D

C

(cid:236) Gäi N lµ trung ®iÓm AB. AN // DC (cid:237) Ta cã: nªn ADCN lµ h×nh b×nh hµnh. (cid:238)

D ^ . vu«ng t¹i C suy ra AC BC ^ ^ SA ABCD ( . nªn SA BC ^ . = — — (cid:176) => 60

= D — (cid:176) AN = DC = a Suy ra: NC = AD = a => NA = NB = NC =a hay ACB ) Do ¸p dông ®Þnh lý ba ®­êng vu«ng gãc ta suy ra SC BC Suy ra: Gãc gi÷a (SBC) vµ (ABCD) lµ SCA SCA ________________________________________________________________________ MÆt kh¸c: NBC NBC 60

= = a 3 AC AB

2

(cid:176) = = ®Òu nªn 3 2 = SA AC .tan 60 a 3 . 3 a 3 ________________________________________________________________________

ABCD

= S a 3 3 4 0,25 0,25 0,25

________________________________________________________________________ 3 a TÝnh ®­îc thÓ tÝch chãp S.ABCD b»ng . 0,25 3 3 4

3

3

2

2

2

3

2

2

2

2

2

(cid:230) (cid:246) + + + + x z x z = + Ta cã : P 2 (cid:231) (cid:247) y xyz y 3 Ł ł

+ + + + (cid:222) ‡ + ‡ " xy yz zx x y z a b , ab 2 a b , .

C©u V. 1,0 ®iÓm

3

3

3

3

3

3

¸p dông bÊt ®¼ng thøc (§¼ng thøc x¶y ra khi x=y=z) (cid:230) (cid:246) (cid:230) (cid:246) (cid:230) (cid:246) (cid:230) (cid:246) + + + + x z xy zx + + + + + (cid:222) ‡ + (cid:222) ‡ P P 2 (cid:231) (cid:247) (cid:231) (cid:247) (cid:231) (cid:247) (cid:231) (cid:247) x 3 y 3 z 3 yz xyz 2 x 2 y 2 z y 3 Ł ł Ł ł Ł ł Ł ł

3

________________________________________________________________________

2

4

= f t ( ) XÐt hµm sè + víi t > 0 ; t 3

= - f t '( ) 0 = (cid:219) = t 2 f t '( ) t . ; 2 t 2 2 t

________________________________________________________________________ B¶ng biÕn thiªn: t ¥+ 0 4 2

y’ - 0 +

y +¥

+¥

8 4 3 2

44 8

44 8

4 2

= = = P = P ‡ x y ________________________________________________________________________ z VËy . §¼ng thøc x¶y ra khi . hay 0,25 0,25 0,25 0,25

A. Theo ch­¬ng tr×nh chuÈn

nªn M (m; 3m-4). = - nªn A (6-2m; 5-6m).  GA  GM 2

(

)

)2; 2M(cid:222) (

- ˛ 2; 7 A , . = 2m(cid:222)

C©u VI. a. 1,0 ®iÓm

)1;3

( B -

)5;1C (

= ˙ . nªn

Gäi M lµ trung ®iÓm BC, v× M d˛ Mµ ________________________________________________________________________ A AB ________________________________________________________________________ BC qua M vµ vu«ng gãc víi d nªn cã ph­¬ng tr×nh x + 3y – 8 = 0. B AB BC ________________________________________________________________________ M lµ trung ®iÓm BC nªn . 0,25 0,25 0,25 0,25

; ) Gäi lµ t©m cña ®­êng trßn (C). I x y ( 0 0

(0;1) lµ vect¬ ®¬n vÞ trªn trôc Ox, ta cã:

) =

)

( 1. 1

( 0. 0

(cid:219) - - (cid:219) 0 = x 0 x 0 y 0

C©u VII. a. 1,0 ®iÓm

2

2

2 =

(

(

)

) =

+ (cid:219) - - 2 6 25 4 IB = i = Khi ®ã, do (C) tiÕp xóc víi Ox t¹i A nªn víi  ^ + 2 . IA i ________________________________________________________________________ Theo gi¶ thiÕt, ta cã: 25 ; R = IB – 5 y 0

Ø 7 = – (cid:219) - (cid:219) 3 Œ y 0 4 = 1 º y 0 y 0

________________________________________________________________________ 0,25 0,25

0

(cid:222) 7 7 I = R . Víi

0

(cid:222) 1 = R I 1 y = th× (2; 7) y = th× (2;1) .

2

2

2 =

)

(

(

) 1

1

k n

0,5 + - - - - Víi VËy ta cã hai ®­êng trßn cÇn t×m: ) ( 2 + = x 2 49 7 y y 1 x

6 n

6 n

9

+ + = + + C C 3 ; ( + k C n C )

8 C n =

) 2 C+ = + 7 C n + C 2

3nC +

2

7 + n 1

+ 1 k , ta cã: + 1 n + 7 C 2( n + 9 C + n 1

8 + n 1

= = C + 8 9 C C n n + 9 8 C+ + n 2 n C ¸p dông c«ng thøc + 8 9 7 C C 3 n n n C Gi¶ thiÕt t­¬ng ®­¬ng víi

C©u VIII. a. 1,0 ®iÓm

9 + n

3

8 + n

2

n

= (cid:219) C C 2 2 . (cid:219) = n 15 n + 3 = 9 ________________________________________________________________________

3

( ) P x

k

k

15

3

(cid:230) (cid:246) 2 0,25 = + x Khi ®ã (cid:231) (cid:247) Ł ł x - (cid:230) (cid:246) 2 =

(

)15

k 15

=

K

0

k

15

k

30 5 6

x C (cid:229) (cid:231) (cid:247) Ł ł x -

.

k 15

=

K

0

___________________________________________________________________

C x 2 = (cid:229)

6

6

- k = (cid:219) = k 0 6 Sè h¹ng kh«ng chøa x t­¬ng øng víi . 30 5 6

15.2

= 320320 C . ________________________________________________________________________ Sè h¹ng ph¶i t×m lµ 0,25 0,25 0,25

B. Theo ch­¬ng tr×nh n©ng cao

B¸n kÝnh cña (T) lµ:

D . D ^ vu«ng c©n t¹i A nªn I lµ trung ®iÓm BC vµ AI BC . = = ^ (cid:222) R d A d ( 2 ) , . (cid:222) / /d AI ^

C©u VI. b. 1,0 ®iÓm

(cid:222) BC d ( ) BC: x + y + c = 0.

Gäi I lµ t©m cña ®­êng trßn ngo¹i tiÕp ABC V× ABC Theo gi¶ thiÕt BC d ( ) ________________________________________________________________________ = - + + Ø C 1 1 2 = C = (cid:219) (cid:219) d A d ( , ) R= 2 2 Œ = - C 5 º 2

Ø BC x : 1 0 Suy ra Œ BC x : + - = y + - = y 5 0 º

)1; 2A (

(cid:222) - D AI x : y ®i qua d vµ song song víi ( ) + = . 1 0

2

(cid:236) 1 0 §­êng cao AI cña ABC ________________________________________________________________________ + - = y x = (cid:222) ˙ (cid:237) BC x : y+ - = 1 0 NÕu (cid:222) I(0;1). : I BC AI + = - 1 0 x y (cid:238)

( y+

)2 = 1

- 2 T x ( ) : .

2

2 =

(

)

(

2

2

(cid:236) 5 0 x = (cid:222) ˙ (cid:237) y+ - = BC x : 5 0 I(2;3). (cid:222) NÕu I BC AI : Suy ra: ________________________________________________________________________ + - = y + = - x y 1 0 (cid:238) 0,25 0,25 0.25 0,25 + - - T ( ) : x 2 y 3 2 Suy ra:

2 =

)

(

)

) ( y+

+ - - - x . )2 = 1 x 2 2 y 3 2 VËy cã hai ®­êng trßn: vµ ( .

2

1

- - - - B d˛ A a a ( ; 3 5) B b b ( ; 3 1) , V× ; .

- - - nªn gäi täa ®é ) A d˛  ( = AB ) b a b a ; 4 3( .

C©u VII. b. 1,0 ®iÓm

2

(

AB = + - - - Ø ø ________________________________________________________________________ Tõ gi¶ thiÕt ) suy ra: ) 2 = 2 2 ( 4 3 b a b a 2 2 . º ß

2

( + -

)2

2 =Ø t Œ + = - 4 = (cid:219) 8 t 3 t , ta cã: §Æt t b a Œ = t º 2 5

= (cid:222) - (cid:222) (2; 2)  AB ________________________________________________________________________ Víi cÇn t×m. t = 2 - = b a 2 - lµ vect¬ chØ ph­¬ng cña D + y 1 x = (cid:219) 1 0 y x + + = . Suy ra ph­¬ng tr×nh ®­êng th¼ng cña D lµ - 2 2 2 ________________________________________________________________________

(cid:222) - = b a . Víi 0,25 0,25 0,25 0,25 2 t = 5

- = - = x y- y- x x 9 0 9 0 . . 2 5 T­¬ng tù ta cã ph­¬ng tr×nh ®­êng th¼ng cña D lµ 7 y+ + = vµ 7 1 0 VËy cã hai ®­êng th¼ng cÇn t×m lµ

3

2

2

x „ . §k: x > , 0 1 2 (cid:230) (cid:246) log (cid:231) (cid:247) (cid:230) (cid:246) Ł ł 2 + = (cid:219) PT 2 log 2 (cid:231) (cid:247)

C©u VIII. b. 1,0 ®iÓm

Ł ł x x 2 x log 2 2

2

2

2

2

2

2

________________________________________________________________________ - - x 1 1 x (cid:230) (cid:246) + = (cid:219) - (cid:219) - x = x 2 1 log 2 log 0 (cid:231) (cid:247) Ł ł x x 3log + 1 log 1 2

2 2

2

3log + 1 log ________________________________________________________________________ - 1 = + = + = (cid:219) - - (cid:219) - (cid:219) = 0 t log x - = t t 3 1 t t ( 1) 0 t t 1 0 §Æt , ta cã: 1t + 3 t 1 t

(cid:222) (cid:222) = x 1t = = 1x 2

________________________________________________________________________ . log Víi 2 VËy ph­¬ng tr×nh cã nghiÖm x = 2 .

0,25 0,25 0,25 0,25