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S GD&T BNH NH THI TH I HC LN IV, NM 2013

TRNG THPT CHUYN L QU N Mn thi: TON, Khi: D.Thi gian lm bi:180 pht. Ngy thi: 28/04/2013

PHN CHUNG CHO TT C TH SINH (7,0 im)

Cu 1 (2,0 im) Cho hm s 3 21 3 9

( 1)

8 4 2

x m x x m= + + (1), vi m l tham s thc.

1. Kho st s bin thin v v th ca hm s (1) ng vi 1=m .2. Xc nh m hm s (1) t cc tr ti 21 ,xx sao cho 1 2 4x x .

Cu 2 (1,0 im) Gii phng trnh: s in2x sin 4 sin 6 2 os os2x c xc x+ + = .

Cu 3 (1,0 im) Gii h phng trnh:2 2

4 1 1

x xy x y

xy y

+ + =

+ + =

Cu 4 (1,0 im) Tnh din tch hnh phng (H) gii hn bi hai ng: 3 23 2x x v y x= = .Cu 5 (1,0 im) Cho hnh chp S.ABCD c y ABCD l hnh thang vung ti A v B , bit mt bnSAB l tam gic u c di cnh bng 2a v nm trong mt phng vung gc vi mt y (ABCD),

cnh AD= 3a v cnh SC 2a 2= . Gi H l trung im ca AB v M l hnh chiu vung gc ca imH trn ng SC. Tnh th tch ca khi chp S.HCD v khong cch t M n mp(SAD) theo a.

Cu 6 (1,0 im) Chox, y, z l ba s thc dng tha mn: 3xy yz zx+ + = . Tm gi tr nh nht ca biu

thc 2 2 25

A x y zx y z

= + + ++ +

.

PHN RING (3,0 im) Th sinh ch c lm mt trong hai phn (Phn A hoc B)A. Theo chng trnh Chun

Cu 7.a (1,0 im) Trong mt phng vi h ta Oxy, cho tam gic ABCc nh A( 6 ; 5), cnh BC=

2 10 , im E(-5; 2) nm trn ng thngBCv im I( 2; 3) l tm ng trn ngoi tip tam gic ABC. Tnhdin tch ca tam gicABC.

Cu 8.a (1,0 im) Trong khng gian vi h ta Oxyz, cho ng thng d: 2 2 12 2 3

x y z = =

v

mt phng (P): 2 2 1 0x y z + = . Vit phng trnh mt phng (Q) vung gc vi mp(P) ng thi songsong v cch ng thng d mt khong bng 1.

Cu 9.a (1,0 im) Cho s phc z tha mn (3 ) (2 )( 1) 5i z i z + + = . Tnh m un ca s phc2w 1 z z= + .

B. Theo chng trnh Nng caoCu 7.b (1,0 im) Trong mt phng vi h ta Oxy, vit phng trnh chnh tc ca elip (E), bitphng trnh mt ng chun: 4x = v ng trn i qua hai nh nm trn trc nh ca (E) cng iqua hai tiu im ca (E)

Cu 8.b (1,0 im) Trong khng gian vi h trc ta Oxyz, cho cc im(1; 0; 0), (2; 2; 3), (4;1; 2)A B C v mt phng ( ) : 3 0.x z + = Tm to ca im bit rng cch

u cc im CBA ,, v mt phng ).(

Cu 9.b (1,0 im) Gi z1 v z2 l nghim ca phng trnh n z trn tp s phc : ( )2z.z z 8 1 2i+ = .

Tnh4 4

1 2

1 1A

z z= +

HtH v tn th sinh:S bo danh:

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S GD&T BNH NH P N THI TH I HC LN IV, NM 2013TRNG THPT CHUYN L QU N Mn thi: TON, Khi: D.

Ngy thi: 28/04/2013

Cu Ni dung im

Khi m= 1: 3 21 3 9

18 2 2

y x x x= + 1,00

TX : , Tnh: lim , limx xy y += = + 0,25

23 9' 3 ,8 2x x x R= + /

6 (6) 10

2 (2) 3

x yy

x y

= = = = =

0,25

Lp BBT 0,25

1.1

th 0,25

. 3 21 3 9

( 1)8 4 2

y x m x x m= + + 1,00

. / 2 23 3 9

( 1) 4( 1) 12 08 2 2

y x m x x m x= + + + + = (*) 0,25

Pt (*) c hai nghim phn bit: / 21 3

4( 1) 12 01 3

mm

m

> + = + >

< (**)

Khi ( ) ( )1 1 2 2; , ;x y N x y , vi 1 2,x l 2 nghim ca (*).

0,25

Gi thit: 21 2 1 2 1 24 ( ) 4 16x x x x x x + v 1 2 1 24( 1) & 12x x m x x+ = + = .

Ta c: 216( 1) 48 16 3 1m m+ (***)0,25

1.2

T (**) v (***) ta c:1 3 1

3 1 3

m

m

+ <

< 0,25

Gii: s in2x sin 4 sin 6 2 os os2x x c xc x+ + = 1,002s in3x.cos 2sin 3 os3 os3 ospt x xc x c x c x + = + 0,25

( os3 cos )(2sin 3 1) 0c x x x + = 0,25

Gii pt: os3 osx=0 2cos2x.cos 0 ;4 2 2

c x c x x k x k

+ = = + = + 0,25

2

Gii pt:2 5 2

2sin 3 1=0 ;18 3 18 3

x k x k

= + = + 0,25

Gii h:2 2

4 1 1

x xy x y

xy y

+ + =

+ + =

1,00

KXD: 4 1xy v pt u: ( 1)( 2) 0x x y + + = 0,25

3

Gii h:1

1 4 1

x

y y

=

+ =

11 1

;22 0

0

xx x

yy y

y

= = =

= = = =

0,25

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Gii h2

4 (2 ) 1 1

x y

y y

=

+ + = 2

2 22

1 0 1 00

05 10 0

2

x y x yx

y yy

yy y

y

= = =

= = = =

0,25

Tp nghim ca h: { }( 1; 0); ( 1; 2); (2; 0)S= 0,25

Din tch (H) gii hn bi hai ng: 3 23 2x x v y x= = . 1,00

Pt honh giao im: 3 23 2 0, 1, 2x x x x x x = = = = 0,252

3 2

0

3 2S x x x dx= + 0,25

1 23 2 3 2

0 1

( 3 2 ) ( 3 2 )S x x x dx x x x dx= + + 0,254

*1 2

4 3 2 4 3 2

0 1

1 1 1

4 4 2

= + + =

S x x x x x x 0,25

Cho hnh chp S.ABCD c y ABCD l hnh thang vung ti A v B , bit mt bnSAB l tam gic u c di cnh bng 2a v nm trong mt phng vung gc vi mt

y (ABCD), cnh AD= 3a v cnh SC 2a 2= . Gi H l trung im ca AB v M lhnh chiu vung gc ca im H trn ng SC. Tnh th tch ca khi chp S.HCD vkhong cch t M n mp(SAD) theo a.

1,00

H

D

BC

S

M

V (SAB) (ABCD) v AH AB nn SH (ABCD) suy ra: SH a 3=

Trong tam gic vung SBC c: 2 2BC SC SB 2a= = .

Tnh 5, 10. , 5CH a HD a CD a HCD C = = =

0,25

Tnh: 31 1 5 3

. . .3 2 6SHCD

V SH HC CD a= = 0.25

Xc nh:2

2( ,( ) . ( , ( )) . ( , ( )) . ( , )

SM SM SH d M SAD d C SAD d B SAD d B SA

SC SC SC = = = 0.25

5

Tnh3 3

( , ( )8

d M SAD a= 0.25

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Cho cc s thc zy,, > 0 tha: 3xy yz zx+ + = . Tm gi tr nh nht ca biu thc

2 2 2 5A x y zx y z

= + + ++ +

.

1,00

zxt ++= 2 2 2 2 2 2 2 26 6t x y z x y z t = + + + + + = .

Ta c 29 3( ) ( ) 3xy yz zx x y z x y z= + + + + + + , suy ra 3t

Khi 25

6 .A t

t

= + 0,25

2 5( ) 6, 3.f t t tt

= +

Ta c:

3

2 2

5 2 5'( ) 2 0, 3

tf t t t

t t

= = >

0,25

)(tf ng bin 3t . Do 14

( ) (3)3

f t f = 0,25

6

Vy minA =3

14, khi .1=== zyx 0,25

Cho tam gicABCc nhA( 6 ; 5), cnhBC= 2 10 , im E(-5; 2) nm trn ng thngBCv im I( 2; 3) l tm ng trn ngoi tip tam gic ABC. Tm din tch ca tam gicABC.

1,00

Khong cch t I n ng thng BC: d(I,BC) =2

2 BCIA 102

=

0,25

Phng trnh ng thng BC: ax + by + 5a 2b = 0 ( )2 2a b 0+ >

( )2 2

7a bd I,BC 10

a b

+= =

+( )( )

3a b 03a b 13a 9b 0

13a 9b 0

= + = + =

0,25

Khi 3a b = 0. Chn a = 1 v b = 3. Pt BC: x + 3y 1 = 0 .Din tch tam gic ABC bng 20(vdt)

0,25

7.a

Khi 13a + 9b = 0. Chn a = 9 v b = -13. Pt BC: 9x 13y +71 = 0Din tch tam gic ABC bng 12 (vdt) 0,25

ng thng d:2 2 1

2 2 3

y z = =

v mt phng (P): 2 2 1 0x y z + = . Vit phng

trnh mt phng (Q) vung gc vi mp(P) ng thi song song v cch ng thng dmt khong bng 1.

1,00

T pt (d) : (2; 2;1) , : (2; 2;3)M d vtcp u =

v vtpt ca (P) : (1; 2;2)n =

0,25

Gi vtpt ca (Q): 1 , (2; 1; 2)n u n = =

( ) : 2 2 0mp Q x y z D + = 0,25

3( , ( )) 1

33

DDKC M Q

D

== =

=

0,25

8.a

Hai mp(Q) cn tm: 2 2 3 0 2 2 3 0x y z v x y z + = = 0,25

Cho s phc z tha mn (3 ) (2 )( 1) 5i z i z + + = . Tnh m un ca s phc2w 1 z z= + .

1.0

Gi ; ,z x yi x y R= + . Gi thit (5 2 2) ( 1) 5y y i + + = (1) 0.25

Gii (1): 1, 1 1x y z i= = = 0.25

Xc nh: 2w 1 z z i= + = 0,25

9.a

M un ca w: w 1= 0.25

7.b Vit phng trnh chnh tc ca elip (E), bit phng trnh mt ng chun: 4x=

v 1,00

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ng trn i qua hai nh nm trn trc nh ca (E) cng i qua hai tiu im ca (E)

Pt chnh tc (E):2 2

2 21, 0

x ya b

a b+ = > > 0,25

Pt ng chun:2

2 2 2 24 ,a

c a c a bc

= = = (1) 0,25

Gi thit cho ta: 2 2c b= (2) 0,25

T (1) v (2):

2 22 2 2

8, 4 ( ) : 18 4

x ya b c E = = = + = 0,25

Cc im (1; 0; 0), (2; 2; 3), (4;1; 2)A B C v mt phng ( ) : 3 0.x z + = Tm to caim bit rng cch u cc im CBA ,, v mt phng ).(

1,00

Gi2 2 2 2 2 2

0 0 0 0 0 00 0 0 2 2 2 2 2 2

0 0 0 0 0 0

( 1) ( 2) ( 2) ( 3)( ; ; ) :

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

x y z x y zM x y z AM BM CM

x y z x y z

+ + = + + + = =

+ + = + +

0 0 0 0 00 0 0

0 0 0 0 0

2 4 6 16 4( ;2 ;4 )

6 2 4 20 2

x y z z xx x x

x y z y x

+ = =

+ + = = (1)

0,25

2

2 2 2 0 00 0 0 ( 3 )( , ) ( 1) 10

x zd M AM x y z += + + = (2) 0,25

T (1) v (2), ta c:0

20 0

0

113 46 33 0 33

13

x

x xx

= + = =

0,25

8.b

C hai im tha: 1 233 7 19

(1;1;3); ; ;13 13 13

M M

0,25

Gi z1 v z2 l nghim ca phng trnh n z trn tp s phc : ( )2z.z z 8 1 2i+ = . Tnh

4 41 2

1 1A

z z= +

1

Gi s phc z = x + i.y ( )x , y

( ) ( ) ( )2 2z.z z 8 1 2i 2x 8 16 2xy i 0+ = + + = 0,25

22x 8 0

2xy 16 0

=

+ =. 0,25

Gii h (x;y) = (2;-4), (-2;4). Suy ra: 1 2z 2 4i; z 2 4i= = + 0,25

9.b

4 4

1 2

1 1A

z z= + =

1

200 0,25

Ch : Mi cch gii khc ng u cho im ti a.