S GD&T QUNG NAM K THI TH THPT QUC GIA NM 2015TRNG THPT CHUYN MN TON NGUYN BNH KHIM Thi gian lm bi : 180 pht

CHNH THC:Cu 1) (2,0 im) Cho hm s 3 23 2y x x= + - (1)

a) Kho st s bin thin v v th (C) hm s

b) Vit phng trnh tip tuyn vi th (C), bit tip tuyn vung gc vi ng thng y = 19x-

Cu 2) (1,0 im)

a) Gii phng trnh: 2cos 2cos 3 03xx + - =

b) Tm s phc z tha mn iu kin 6z z+ = v 2 2 8z z i+ - l mt s thc.

Cu 3) (0,5 im) Gii phng trnh: 2

4 4 14

log ( 7 10) log ( 2) log ( 5)x x x x- + - - = +

Cu 4) (1,0 im) Gii h phng trnh:

2

2

( 6 4) 3 (3 4) 8 2( ) ( ) 4(1 ) 2

3 22 1 2 3

x x y y y x y x y xy

x xy y x y

+ - + - + + + = + + - +

- + - - = - +

Cu 5) (1,0 im) Tnh tch phn I = 4

2

0

( 2 tan )sinx x xdx

p

+ +Cu 6) (1,0 im) Cho hnh lng tr ABC.ABC, y ABC c AC = 3a , BC = 3a , 030ACB = . Cnh bn hp vi mt phng y gc 060 v mt phng (ABC) vung gc vi mt phng (ABC). im H trn cnh BC sao cho BC = 3BH v mt phng (AAH) vung gc vi mt phng (ABC). Tnh th tch khi lng tr ABC.ABC ' v khong cch t B n mt phng (AAC).

Cu 7) (1,0 im) Trong mt phng ta Oxy cho tam gic ABC vi A( 3; 4), tm ng trn ni tip

I(2; 1) v tm ng trn ngoi tip J(1 ;12

- ). Vit phng trnh ng thng BC.

Cu 8) (1,0 im) Trong khng gian ta Oxyz, cho hai im A(4; 2; 11), B( 2; 10; 3) v mt phng (P): x + y z 4 = 0 . Vit phng trnh mt phng trung trc on AB v tm im M trn mt phng (P) sao cho MA = MB = 13.

Cu 9) (0,5 im) Mt hp ng 3 xanh , 4 bi v 5 bi vng . Ly ngu nhin 5 bi t hp. Tnh xc sut trong 5 bi ly ra c 3 mu v s bi xanh v s bi bng nhau.

Cu 10) (1,0 im) Cho hai s thc a, b thuc khong (0, 1) tha mn 3 3( )( ) ( 1)( 1) 0a b a b ab a b+ + - - - = .Tm gi tr ln nht ca biu thc sau:

P = 4 4

2 2

12 336 (1 9 )(1 9 )

a bababa b+

+ -+ + +

1

WWW.VNMATH.COM

HNG DN CHM MN TON THI TH THPT QUC GIA NM 2015

Cu p n im

Cu 1(2,0)

Cu1) a) 3 23 2y x x= + -

+ TX D = R , limx y- = - , limx y+ = +

+ 2' 3 6y x x= + , 0 2

' 02 2

x yy

x y= = -

= = - =------------------------------------------------------------------------------------------------------------+ BBT x - 2- 0 + y + 0 - 0 + y

- 2----------------------------------------------------------------------------------------------------------------+ Hm B trn cc khong (- ; 2- ), (0; + ) v NB trn khong ( 2- ; 0). im cc i th ( 2- ; 2); im cc tiu th (0; 2- )--------------------------------------------------------------------------------------------------------------+ th

4

2

-2

-4

-10 -5 5 10

---------------------------------------------------------------------------------------------------------------

b)Tip tuyn vung gc vi ng thng y = 19x- nn tip tuyn c h s gc bng 9.

---------------------------------------------------------------------------------------------------------------

Ta c 0 020 0 00 0

1 2'( ) 9 3 6 9

3 2x y

y x x xx y= =

= + = = - = ----------------------------------------------------------------------------------------------------------------+ Phng trnh tip tuyn ti im (1, 2) l 9( 1) 2y x= - +--------------------------------------------------------------------------------------------------------------+Phng trnh tip tuyn ti im ( 3, 2 ) l 9( 3) 2y x= + -

0,25

0,25

0,25

0,25

0,25

0,25

0,25

0,252

WWW.VNMATH.COM

Cu 2(1,0)

-------------------------------------------------------------------------------------------------------------Cu 2)

a) 2cos 2cos 3 03xx + - = 3 24cos 3cos 2cos 3 0

3 3 3x x x- + - =

2(cos 1)(4cos 6cos 3) 03 3 3x x x- + + =

0,25

Cu p n im

Cu 3(0,5)

Cu 4(1,0)

cos 1 2 6 ,3 3x x k x k k Zp p= = =

--------------------------------------------------------------------------------------------------------------b) Gi z x yi= + . Ta c 6 ( ) ( ) 6 3z z x yi x yi x+ = + + - = = (1)

2 2 8z z i+ - = 2 2 2( ) 2( ) 8 ( 2 ) (2 2 8)x yi x yi i x y x xy y i+ + - - = - + + - - l s thc nn2 2 8 0xy y- - = (2). ---------------------------------------------------------------------------------------------------------------T (1) v (2) ta gii c x = 3 v y = 2. Vy z = 3 + 2i--------------------------------------------------------------------------------------------------------------

Cu 3) b)K

2 7 10 0 2 52 0 2 55 0 5

x x x xx x xx x

- + > < > - > > > + > > -

Vi K trn phng trnh tng ng : 24 4 4log ( 7 10) log ( 2) log ( 5)x x x x- + - - = - +2

4 4log ( 7 10)( 5) log ( 2)x x x x - + + = ---------------------------------------------------------------------------------------------------------------

2( 7 10)( 5) 2x x x x - + + = -( 5)( 5) 1x x - + = 26x = (v x > 5)

--------------------------------------------------------------------------------------------------------------

Cu 4) 2

2

( 6 4) 3 (3 4) 8 2( ) ( ) 4(1 ) 2(1)

3 22 1 2 3(2)

x x y y y x y x y xy

x xy y x y

+ - + - + + + = + + - +

- + - - = - +

-----------------------------------------------------------------------------------------------------------+Ta c (1) 2 2( 3 2) 4 ( 3 2) ( ) 4 ( )x y x y y x y x + - + + + - = - + + -

+ Xt hm 2( ) 4f t t t= + + , t R . Ta c 2

2 2

4'( ) 1 0,4 4

t t tf t t Rt t

+ += + = > "

+ +Suy ra f(t) ng bin trn R. ------------------------------------------------------------------------------------------------------------- + Ta c (1) ( 3 2) ( )f x y f y x+ - = - 3 2 1x y y x y x + - = - = ----------------------------------------------------------------------------------------------------------------+ Th y = 1 x vo (2) ta c : 2 22 22 2 1x x x x x+ + - = + + (3) . Vi K x 0. ta c

(3) 2 2( 2 22 5) ( 1) 2 3x x x x x + + - - - = + -

2

2

2 3 1 ( 1)( 3)12 22 5

x x x x xxx x

+ - -- = - +

++ + +

0,25

0,25

0,25

0,25

0,25

0,25

0,25

0,25

3

WWW.VNMATH.COM

2

1 1( 1) ( 3) 1 01 2 22 5

x xx x x

- + + - =

+ + + + x = 1

V vi x 0 th 2

1 1( 3) 1 01 2 22 5

xx x x

+ + - >

+ + + + (phi gii thch)

--------------------------------------------------------------------------------------------------------------x = 1 y = 0 .Vy h c nghim (x ; y) = (1 ; 0) 0,25

Cu p n im

Cu 5(1,0)

Cu 6(1,0)

Cu 5) I = 4

2

0

( 2 tan )sinx x xdx

p

+ + = 4 4

20 0

sin( 1)sincos

xx xdx dxx

p p

+ + --------------------------------------------------------------------------------------------------------------

+ t 1

sin cosu x du dxdv xdx v x= + =

= = - .

Ta c 4 4

40

0 0

( 1)sin ( 1)cos cosx xdx x x xdx

p pp

+ = - + + = 402 2( 1) 1 sin 1

4 2 8x

pp p- + + + = - +

-------------------------------------------------------------------------------------------------------------

+4 4 4

2 200 0

sin (cos ) 1 2 1cos cos cos

x d xdxx x x

p p p

-= = = -

---------------------------------------------------------------------------------------------------------------

+ Vy I = 2 28p- +

---------------------------------------------------------------------------------------------------------------Cu 6)

B C

A

A'

C'B'

H

( ' ) ( )( ' ) ( )

' ( ' ) ( ' )

A BC ABCA AH ABCA H A BC A AH

^ ^ =

' ( )A H ABC ^

Suy ra 0' 60A AH =----------------------------------------------------------

2 2 2 02 . .cos30AH AC HC AC HC= + - = 2a AH = a

0' tan 60 3A H AH a = =2

. ' ' '3 3. ' . 3

4ABC A B C ABCaV S A H a= = =

394a

--------------------------------------------------------V 2 2 2AH AC HC+ = HA AC^ 'AA AC^

2'

1 1. . ' . 3.2 32 2A AC

S AC AA a a a= = =

0,25

0,25

0,25

0,25

0,25

0,25

0,25

4

WWW.VNMATH.COM

Cu 7(1,0)

-------------------------------------------------------------------------------------------------------------

3

'2

'

93. 3 34( , ( ' ))

43A ABC

A AC

aV ad B A ACS a

= = =

--------------------------------------------------------------------------------------------------------------Cu 7)

+ Phng trnh ng trn ngoi tip tam gic ABC : 2 21 125( ) ( 1)2 4

x y+ + - = (1)

+ Phng trnh ng thng AI : 3 4

2 3 1 4x y+ +

=+ +

1 0x y - - =

--------------------------------------------------------------------------------------------------------------

0,25

0,25

Cu p n im

Cu 8(1,0)

+ ng thng AI ct ng trn ngoi tip ti im th hai l D, trung im cung BC. Honh im D l nghim khc 3 ca phng trnh :

2 23

1 125( ) ( 2) 92 42

xx x

x

= -+ + - = =

. Suy ra D(9 7;2 2

)

--------------------------------------------------------------------------------------------------------------

+ Ta c BID = 2 2A B+ v

2 2B AIBD IBC CBD= + = + suy ra BID IBD= DI = DB = DC

B, C nm trn ng trn tm D bn knh DI c phng trnh :

2 29 7 50( ) ( )2 2 4

x y- + - = (2)

---------------------------------------------------------------------------------------------------------------+ Ta im B v C l nghim h phng trnh (1) v (2)

2 2

2 2

1 125( ) ( 1)2 49 7 50( ) ( )2 2 4

x y

x y

+ + - = - + - =

2 2

2 2

2 30 09 7 20 0

x y x yx y x y

+ + - - = + - - + =

2 2

10 5 50 09 7 10 0

x yx y x y

+ - =

+ - - + =

Suy ra phng trnh ng thng BC : 10 5 50 0x y+ - = hay 2 10 0x y+ - =------------------------------------------------------------------------------------------------------------ Cu 8)+ Mp trung trc (Q) ca on AB qua trung im I(1; 6; 7) ca AB nhn ( 6; 8; 8)AB = - - -

lm VTPT -------------------------------------------------------------------------------------------------------------Suy ra phng trnh mp(Q): 6( 1) 8( 6) 8( 7) 0x y z- - - + - - = 3 4 4 7 0x y z + + - =-------------------------------------------------------------------------------------------------------------+ Gi D= (Q) (P). ng thng D l tp hp cc im tha h phng trnh:

3 4 4 7 0

4 0x y zx y z+ + - =

+ - - =(1)

+ (P) c VTPT (1;1; 1)Pn = -

, (Q) c VTPT (3;4;4)Qn =

suy ra D c VTCP [ , ] (8; 7;1)P Qu n n= = -

. Trong (1) cho x = 1 gii c y = 2; z = 1 suy

0,25

0,25

0,25

0,25

0,25

0,25

5

WWW.VNMATH.COM

Cu 9(0,5)

Cu 10(1,0)

ra D i qua im I(1; 2; 1). Vy phng trnh tham s ng thng D

1 82 71

x ty tz t

= + = - = - +

-------------------------------------------------------------------------------------------------------------+M D th M(P) v MA = MB. Ta c M(1 + 8t ; 2 7t ; 1 + t)MA = 13 2 2 2(8 3) (4 7 ) ( 12) 169t t t - + - + - = 2114 128 0t t - = 0t = hoc 64 / 27t =

Vy c hai im M tha bi ton : 1(1;2; 1)M - , 2569 334 7( ; ; )57 57 57

M -

---------------------------------------------------------------------------------------------------------------Cu 9) + C 512 792C = cch chn 5 bi t hp 12 bi W = 792---------------------------------------------------------------------------------------------------------------+ Gi X l bin c : 5 bi ly ra c 3 mu v s bi xanh v s bi bng nhau TH1 : 1X, 1, 3V c 1 1 33 4 5 120C C C = cch chn TH2 : 2X, 2, 1V c 2 2 13 4 5 90C C C = cch chnSuy ra XW = 120 + 90 = 210

Vy P(X) = 210 35792 132

XW = =W

---------------------------------------------------------------------------------------------------------------

Cu 10) P = 4 4

2 2

12 336 (1 9 )(1 9 )

a bababa b+

+ -+ + +

----------------------------------------------------------------------------------------------------------

GT : 3 3( )( ) ( 1)( 1) 0a b a b ab a b+ + - - - =3 3( )( ) (1 )(1 )a b a b a b

ab+ +

= - - (*)

V 3 3 2 2( )( ) ( ) 2 .2 4a b a b a b a b ab ab ab

ab b a + +

= + + =

v (1 )(1 ) 1 ( ) 1 2a b a b ab ab ab- - = - + + - + , khi t (*) suy ra 4 1 2ab ab ab - +,

t t = ab (t > 0) ta c 2

10 12 1 3 0394 (1 3 )

tt t t

t t

< - < -

-------------------------------------------------------------------------------------------------------

Ta c 2 2(1 9 )(1 9 ) 36a b ab+ + 2 212 2

136 (1 9 )(1 9 ) aba b

++ + +

v 4 4

3 3 2a bab ab ab abab+

- - = .

Suy ra 2

1P ab

ab +

+. Du ng thc xy ra

13

a b = = .

-------------------------------------------------------------------------------------------------------------

0,25

0,25

0,25

0,25

0,25

6

WWW.VNMATH.COM

. Xt hm 2( )

1f t t

t= +

+ vi 0 < t

19

,

ta c 1 1'( ) 1 0, (0, ]

9(1 ) 1f t t

t t= - > "

+ + f(t) ng bin trn (0,

1 ]9

-------------------------------------------------------------------------------------------------------------

f(t)1 6 1( )9 910

f = + , du ng thc xy ra 1

1 39

a ba b

t ab

= = == =

Vy MaxP = 6 1

910+ t c ti a = b =

13

0,25

0,25

7

WWW.VNMATH.COM