[blogtoanli.net]Đề thi thử Toán lần 3 Bình Xuyên Vĩnh Phúc 2014

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S GD&T Vnh Phc Trng THPT Bnh Xuyn

THI KHO ST I HC LN 3 NM 2014 Mn thi: TON 12 - KHI A, A1

Thi gian lm bi 180 pht, khng k thi gian giao I. PHN CHUNG CHO TT C TH SINH (7 im) Cu 1.( 2 im) Cho hm s mx 1y

x 1+

=

c th (C) 1/Kho st v v th ca hm s khi m =1.

2/Vit phng trnh tip tuyn d vi (C) ti im c honh x =2, tm m khong cch t im A(3;5) ti tip tuyn d l ln nht. Cu 2.(1 im)Gii phng trnh 24sinx.sin x .sin x 4 3.cosx.cos x .cos x 2

3 3 3 3pi pi pi pi

+ + + = .

Cu 3.(1 im) Gii h phng trnh: 2 2 2

3 2

x y 2x y 02x 3x 4y 12x 11 0

+ =

+ + + =

Cu 4.(1 im) Tnh tch phn 1

x

0

2I x e dxx 1

= + +

.

Cu 5.(1 im) Cho hnh hp ng ABCDABCDc y l hnh thoi cnh a, gc ABC bng 600, gc gia mt phng (ABD) v mt phng y bng 600. Tnh theo a th tch ca hnh hp v khong cch gia CD v mt phng (ABD). Cu 6.(1 im) Cho a, b, c dng, a +b +c =3. Chng minh rng:

2 2 2a 4a 2b b 4b 2c c 4c 2a 7b 2c c 2a a 2b+ + + + + +

+ + + + +

.

II. PHN RING (3 im): Th sinh ch c lm mt trong 2 phn (Phn A hoc phn B) A.Theo chng trnh chun Cu 7.a (1 im). Trong mt phng vi h ta Oxy cho ng trn (C) ( ) ( )2 2x 3 y 2 1 + = Tm M thuc Oy sao cho qua M k c hai tip tuyn MA, MB vi ng trn, A, B l tip im sao cho ng thng AB qua N(4;4). Cu 8.a (1 im). Trong khng gian vi h ta Oxyz cho ba im A(1;1;1) , B(3;5;2) v C(3;1; 3) . Chng minh 3 im A, B, C l 3 nh ca mt tam gic vung. Tnh bn knh ng trn ngoi tip ABC . Cu 9.a (1 im). C 5 bng hoa hng bch, 7 bng hoa hng nhung v 4 bng hoa cc vng. Chn ngu nhin 3 bng hoa. Tnh xc sut 3 bng hoa c chn khng cng mt loi. B.Theo chng trnh nng cao Cu 7.b (1 im) Trong mt phng vi h ta Oxy cho e lp 2 2x y 1

4 3+ = v ng thng

:3x 4y 12 0 + = . T im M bt k trn k ti (E) cc tip tuyn MA, MB. Chng minh ng thng AB lun i qua mt im c nh. Cu 8.b (1 im). Trong khng gian vi h ta Oxyz cho ba im A(1;4;2) , B(2;5;0) v C(0;0;7) . Tm im M thuc (Oxy) sao cho 2 2 2MA MB MC+ + t gi tr nh nht. Cu 9.b (1 im). Gii phng trnh 22 3log x (x 5) log x 6 2x 0+ + =

-------Ht------ Th sinh khng s dng ti liu, cn b coi thi khng gii thch g thm. H tn th sinh:..........................................................................SBD:................................



K THI KS I HC NM HC 2014 LN 3 P N MN: TON 12 - KHI A,A1

Cu p n im 1.(1,0 im) Kho st hm s mx 1y

x 1+

=

khi m=1.

Khi m=1 x 1yx 1

+=

. Tp xc nh: { }R \ 1 S bin thin: ( )2

2y' 0 x 1x 1

= <

Do hm s nghch bin trn mi khong

( ;1) v (1;+ ). Hm s khng c cc tr.

0,25

Tim cn: + Tim cn ng x =1 v x 1lim f (x)

+= + ,

x 1lim f (x)

= .

+ Tim cn ngang y =1 v xlim f (x) 1

=

0,25

Bng bin thin:

0,25

f(x)=(x+1)/(x-1)

-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

x

y

th nhn I(1;1) lm tm i xng, ct Oy ti (0;-1), ct Ox ti (-1;0).

0,25

2.(1,0 im). Vit phng trnh tip tuyn d vi (C) x =2 y =2m +1 v f '(2) m 1= Phng trnh tip tuyn d vi (C) ti im (2;2m+1) l y (m 1)x 4m 3= + + + 0,5 Phng trnh (d) m(x-4) = -x y+3 tip tuyn d qua im c nh H(4;-1) 0,25

(2im)

khong cch t im A(3;5) ti tip tuyn (d) l ln nht (d) AH du .AH 0=

1.1 +6(m+1) =0 7m6

=

0,25

2 (1,0 im) Gii phng trnh.

-1 x

1

y

y

+ 1 - -

+

1 -



24sinx.sin x .sin x 4 3.cosx.cos x .cos x 23 3 3 3pi pi pi pi

+ + + = .

2PT 2sin x(cos 2x cos ) 2 3.cos x.(cos(2x ) cos ) 23 3pi pi

+ pi + =

2sin x.cos 2x sin x 2 3.cos x.cos 2x 3 cos x 2 + + =

0,25

(sin3x sin x) sin x 3(cos3x cos x) 3 cos x 2 + + + =

0,25 1 3

sin 3x 3 cos3x 2 sin 3x cos3x 12 2

+ = + = 0,25

(1im)

2cos 3x 1 3x k2 3x k2 x k , k Z.

6 6 6 18 3pi pi pi pi pi

= = pi = + pi = + 0,25

(1,0 im) Gii h phng trnh 2 2 2

3 2

x y 2x y 0 (1)2x 3x 4y 12x 11 0 (2)

+ =

+ + + =

T PT (1) ta c x 0 , y 1 0,25 PT (2) 3 24y 2x 3x 12x 11 = + + (3) V phi (3) 4 4y 4 v y 1 . t f(x) = 2x3 +3x2 -12x +11 vi x 0 ta c bng bin thin:

V phi ca (3) 4 3 24y 4 2x 3x 12x 11 + + v x 0 , y 1 Vy nghim ca (2) l (x;y)=(1;-1).

0,5

3 (1im)

Thay (x;y)=(1;-1) vo (1) ta thy tha mn. Vy nghim ca h phng trnh l (x;y) = (1;-1). 0,25

(1,0 im). Tnh tch phn: 1

x

0

2I x e dxx 1

= + +

.

1 1 1x x

1 20 0 0

2 2xI x e + dx xe dx dx I Ix+1 x+1

= = + = +

+1

x

10

I xe dx= . t x xu x du dx

dv e dx v e= =

= =

( )1x x x x10

1 1I xe e dx xe e 1

0 0= = = .

0,5

+ ( )1 1

20 0

12x 2I dx 2 dx 2x 2ln(x 1) 2 2ln 20x+1 x 1

= = = + = +

0,25

4 (1im)

1 2I I I 3 2ln 2= + = 0,25 (1,0 im). Tnh theo a th tch hnh hp... 5

(1im) Gi O l tm hnh thoi ABCD AO BD m AA' (ABCD) A'O BD 0,25

x

f(x) f(x)

0 1 +

4

11 +



A 'OA la goc gia mp(A'BD) vi y oA 'OA 60 = . Do oABC 60 = nen tam giac ABC eu aAO

2= .

Trong tam giac vuong A'AO, ta co o 3AA ' AO. tan 60 a

2= = .

Do o the tch cua hnh hop: 2 3

ABCDa 3 a 3 3aV=S .AA'= . =

2 2 4.

B' C'

A' D'

B

A

C

D

0,25

Theo chng minh tren ta co BD (A ' AO) (A 'BD) (A ' AO) . Trong tam giac vuong A ' AO , dng ng cao AH, ta co AH (A ' BD) hay AH d(A, (A 'BD))= .Do CD '/ /BA ' nen CD '/ /(A 'BD) suy ra d(CD ', (A 'BD)) d(C, (A 'BD))= d(A, (A 'BD))= (v AO CO= ) AH= oAO.sin 60= 3

4a

= .

0,5

(1,0 im). Chng minh rng 2 2 2a 4a 2b b 4b 2c c 4c 2a 7

b 2c c 2a a 2b+ + + + + +

+ + + + +

.

BT 2 2 2a b c

b 2c c 2a a 2b+ +

+ + ++

4a 2b 4b 2c 4c 2a 7b 2c c 2a a 2b

+ + ++ +

+ + +

+ Ta c 2 2a b 2c 2a a b 2c 6a

b 2c 9 3 b 2c 9+ + +

+ + +

(1) (Csi)

Du = khi 2a b 2c

b 2c 9+

=

+

Tng t 2b c 2a 6b

c 2a 9+ +

+ (2)

2c a 2b 6ca 2b 9

+ ++

(3)

Cng (1), (2), (3) v vi v ta c 2 2 2a b c 1

b 2c c 2a a 2b+ +

+ + + (*) du =

khi a=b=c=1.

0,5

6 (1im)

+ Ta c ( )4a 2b 4b 2c 4c 2a 1 1 14 a b c 6b 2c c 2a a 2b b 2c c 2a a 2b+ + +

+ + = + + + + + + + + + +

( ) ( ) ( )4 1 1 1b 2c c 2a a 2b 63 b 2c c 2a a 2b

+ + + + + + + + + +

33

4 13 (b 2c)(c 2a)(a 2b).3 6 63 (b 2c)(c 2a)(a 2b)

+ + + =+ + +

(**)

Cng (*) v (**) ta c iu phi chng minh, du = khi a =b =c =1.

0,5

(1,0 im). Tm M thuc Oy Gi s ( )A AA x ;y , ( )B BB x ;y v 0M Oy M(0; y ) , (C) c tm I(3;2) 0,25

7a (1im)

+ Ta c 2 2A A A AA (C) x y 6x 4y 12 0 (1) + + = + Ta c I A.MA 0=

( ) ( )( )A A A A 0x 3 x y 2 y y 0 + =

0,5



2 2A A A 0 A 0x y 3x (y 2)y 2y 0+ + + = (2) Ly (2) tr (1) v vi v ta c A 0 A 03x (y 2)y 2y 12 0 + = (3) Tng t ta c B 0 B 03x (y 2)y 2y 12 0 + = (4) T (3) v (4) phng trinh AB l 0 03x (y 2)y 2y 12 0 + = AB qua N(4;4) 0 0 03.4 (y 2).4 2y 12 0 y 4 + = = . Vy M(0;4) 0,25 (1,0 im). Chng minh 3 im A, B, C l 3 nh ca mt tam gic vung Ta c AB (2;4;1)=

, AC (2;0; 4)=

khng cng phng .

Ta li c AB.AC 0=

vy ABC vung ti A 0,5

8a (1im)

ABC vung ti A theo cm trn c bn knh ng trn ngoi tip

1 41R BC2 2

= = 0,5

(1,0 im). Tnh xc sut Gi A, B, C tng ng l 3 bin c Chn c ba bng hoa hng bch Chn c ba bng hoa hng nhung Chn c ba bng hoa cc vng H l bin c Chn c ba bng hoa cng loi A, B, C i mt xung khc

v H A B C= P(H) =P(A) +P(B) +P(C) vi 35316

C 10P(A)560C

= = ,

37316

C 35P(B)560C

= = , 34316

C 4P(C)560C

= = , 49 7P(H)

560 80= = .

0,5

9a (1im)

Bin c chn ba bng hoa khng cng loi l H , 7 73P(H) 1 P(H) 180 80

= = = . 0,5

(1,0 im). Chng minh ng thng AB lun i qua mt im c nh. Gi M(x0 ;y0 ), A(x1;y1), B(x2;y2) Tip tuyn ti A c dng 1 1

xx yy 14 3

+ =

Tip tuyn i qua M nn 0 1 0 1x x y y 1

4 3+ =

(1)

0,25

7b (1im)

Ta thy ta ca A v B u tha mn (1) nn ng thng AB c pt 0 0xx yy 1

4 3+ =

do M thuc nn 3x0 + 4y0 =12 4y0 =12-3x0

0 04xx 4yy 4

4 3+ = 0 0

4xx y(12 3x ) 44 3

+ =

0,25

Gi F(x;y) l im c nh m AB i qua vi mi M th (x- y)x0 + 4y 4 = 0 { {x y 0 y 14y 4 0 x 1 = = = = Vy AB lun i qua im c nh F(1;1)

0,5

(1,0 im). 8b (1im) Gi G l trng tm ABC . Ta c:

( ) ( ) ( )2 2 22 2 22 2 2MA MB MC MA MB MC MG GA MG GB MG GC+ + = + + = + + + + +



Ch : p n c 5 trang Hc sinh lm theo cch khc m ng cho im ti a theo thang im mi cu cho. Cu 5 nu khng v hnh hay v khng ng khng cho im.

(Lu : Kin thc ra theo tin dy( khng s phc, khng phng trnh mt phng))

( )2 2 2 2 2 2 2 23MG GA GB GC 2MG GA GB GC 3MG GA GB GC= + + + + + + = + + + Do ( )GA GB GC 0 MG GA GB GC 0+ + = + + = . V 2 2 2GA GB GC+ + khng i nn 2 2 2MA MB MC+ + nh nht 2MG nh nht M l hnh chiu ca G trn (Oxy)

0,5

G l trng tm ( )ABC G 1;3;3 . Hnh chiu ca G trn (Oxy) c ta (1; 3;0). Vy ( )M 1;3;0 0,5 (1,0 im). Gii phng trnh 22 2log x (x 5) log x 6 2x 0+ + = iu kin x>0(*) t 2t log x= phng trnh

2 t 2t (x 5)t 6 2x 0t 3 x

=+ + =

=

0,25

+ Vi 2t 2 log x 2 x 4= = = tha mn (*) 0,25 + Vi 2 2t 3 x log x 3 x x log x 3 0= = + = Xt 2f (x) x log x 3, x >0= + . Ta c 1f '(x) 1 >0, x >0

x ln 2= + hm s lun ng bin x >0

f(2) =0 x =2 tha mn (*) l nghim

0,25

9b (1im)

Vy nghim ca phng trnh l x =2; x =4 0,25

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[blogtoanli.net]Đề thi thử Toán lần 3 Bình Xuyên Vĩnh Phúc 2014