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S GIO DC V O TO BNH NH THI TH I HC LN 4 NM 2013TRNG THPT CHUYN L QU N Mn: TON; Khi A, A1, B

---------------------- Thi gian: 180 pht, khng k thi gian pht Ngy thi: 28 - 04 - 2013

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

Cu 1 (2,0 im). Cho hm s 4 22 2 (1)y x mx m= + (m l tham s).a. Kho st s bin thin v v th ca hm s (1) khi m = 1.

b. Tm m hm s (1) c ba cc tr v ng trn i qua ba im cc tr ca th hm s cbn knh bng 1.

Cu 2 (1,0 im). Gii phng trnh: sin 2 2 3 cos 2 2sin .6 6 6

x x x

+ + = + + +

Cu 3 (1,0 im). Gii bt phng trnh: 2 22 3 1 2 2 4.x x x x x+ + + + + +

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

322

1 2

x xI d

=

2 x

x x.

Cu 5 (1,0 im). Cho hnh chp S.ABCD c yABCD l hnh ch nht v c 3 , 2AB a AD a= = , tam

gic SAB c gcSAB nhn v nm trong mt phng vung gc vi mt phng (ABCD). Bit hai mt

phng (SAD) v (SCD)cng to vi mt phng (ABCD) cc gc bng nhau v khong cch giaACvSB bng

3a

2. Tnh th tch khi chp S.ABCD theo a.

Cu 6 (1,0 im). Cho a, b, c l cc s thc dng tha mn 2 2 2 2 1.a b c abc+ + + = Tm gi tr nhnht ca biu thc

( )2 2 22 2 21 1 1

2 .2 2 2

F a b ca b c

= + + + + +

II. PHN RING (3,0 im) (Th sinh ch 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 elp (E) c phng trnh:

2 2

4 4.x y+ = GiF1, F2 l cc tiu im ca (E). TmM(E) sao cho s o ca gc1 2F MF t gi tr ln nht.

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz cho mt phng (P):x + y + z 5 = 0 v haiimB(1; 4; 5), C(0; 0; 6). Tm imA thuc mt phng (P) sao cho tam gicABCvung cn tiA.

Cu 9.a (1,0 im). Cho s phcz tha:19

1 .6

zz

z

=

Tm .

i

z i

+

B. Theo chng trnh nng cao

Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy cho im M(2;1) v ng thng

( ) : 1 0.x y + = Vit phng trnh ng trn i qua M v ct () hai imA, B phn bit sao cho

MAB vung tiMv c din tch bng 2.

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho cc im ( ) ( )1;1;1 , 0;0; 3A B v ng

thng (d) c phng trnh:4 1 5

.1 1 1

x y z + = = Vit phng trnh ng thng () bit () vung gc

vi ng thng (d), () i quaA v khong cch tB ti () t gi tr nh nht.Cu 9.b (1,0 im). Cho s phc ( ) ( )3sin 1 4cos , .z t t i t= + Tm tp hp im Mbiu din chos phc . 1.w i z=

------ HT ------

Th sinh khng c s dng ti liu.Cn b coi thi khng gii thch g thm. H v tn th sinh: ; S bo danh: ...

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2

S GIO DC V O TO BNH NH P N - THI TH I HC LN 4 NM2013TRNG THPT CHUYN L QU N Mn: TON; Khi A, A1, B

Bi p n im

1a

Khi m = 1, ta c: 4 22 1y x x= - TX:D = - S bin thin:

+ Chiu bin thin: ' 4 4y = 3x x ;

( )3 21

' 0 4 4 0 4 1 00

xy x x x x

x

= = = = =

`

0,25

+ Hm s ng bin trn mi khong: ( )1;0 v ( )1;+ , nghch bin trn mi

khong ( ); 1 v ( )0;1

+ Cc tr: Hm s t cc i ti im ( )0, 0 1x y= =

Hm s t cc tiu ti cc im ( )1, 1 2x y= = v ( )1, 1 2x y= = + Gii hn v tim cn: lim ; lim .

x x

y y +

= + = +

th hm s khng c tim cn.

0,25

Bng bin thinx - -1 0 1 +y' - 0 + 0 - 0 +

y+ -1 +

-2 -2

0,25

- th:

0,25

1b

Tp xc nh: D =

o hm: ( )3 2' 4 4 4x mx x x m= =

( )2 20

' 0 4 0(2)

xy x x m

x m=

= = =

0,25

Hm s c ba cc tr khi v ch khi phng trnh ' 0y = c ba nghim phn bit, hayphng trnh (2) c hai nghim phn bit khc 0 hay m > 0(*).Vi m tha iu kin (*) phng trnh ' 0y = c ba nghim phn bit

0, , .x x m x m= = =

0,25

Gi s cc im cc tr ca th hm s l :

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( ) ( ) ( )2 20; 2 , ; 2 , ; 2 .A m B m m m C m m m + + Ta c:

4 , 2 ,AB AC m m BC m= = + = phng trnh cnh2( ) : 2 0BC y m m+ + =

GiR l bn knh ng trn ngoi tip tam gicABC. Khi :

( ) ( )

( )

( )

4

2 2 4

1 . .. ;

2 4

.21.2 . 2 2 2 .

2 4.1

ABC

AB AC BCBC d A BC S

R

m m mm m m m m m m m m

= =

+ + + = = +

0,25

( ) ( )3 21

2 1 0 1 1 0 1 5

2

m

m m m m mm

= + = + = =

i chiu iu kin (*) ta c1 5

1, .2

m m +

= =

0,25

2

Phng trnh cho tng ng vi:

1 3sin 2 3 cos 2 2sin 0

6 2 2 6 6x x x

+ + + + + =

0,25

( )

2sin .cos 2 3 sin .sin 2sin 06 6 6

2sin . cos 3.sin 1 06

x x x x x

x x x

+ + + + =

+ + =

0,25

sin 0sin 0 6

6

cos coscos 3.sin 13 3

xx

xx x

+ = + = =+ =

0,25

62

232

k

k

x k

= + = +

=

. Vy, ( )2

, 2 , 2 ,6 3

x k x k x k k

= + = + = 0,25

3

iu kin 1x .Vi 1x , bnh phng hai v bt phng trnh cho v thu gn ta c

2 22 2 3. 1 0x x x x x + + + 0,25

2 22 3 2 2 3. 1 3( 1) 0 (*)x x x x x x + + + + + +

R rng 1x = khng l nghim ca bt phng trnh (*) nn chia 2 v bt pt chox + 1 > 0 v t

2 2 3, 0

1

x xt t

x

+ += >

+ta c

2 2 3 0 0 3t t t < (do t> 0).

0,25

222 3 7 73 7 733 7 6 0

1 2 2

x xx x x

x

+ + +

+. 0,25

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4

4

i chiu iu kin 1x ta c nghim bpt l7 73 7 73

2 2x

+ . 0,25

( )

322

21

.1 -1

x xI d

x

=

x

t 1 sin ,

2 2x t t

=

0,25

Suy ra cos .dx t dt = i cn:

1 03

2 6

x t

x t

= =

= =

0,25

Do

( ) ( )( ) ( )

26 6 62

20 0 0

sin 1 sin 1 1.cos . sin sin 1 2sin cos 2

21 sin

t tI t dt t t dt t t dt

t

+ += = + = +

0,25

( )6

0

1 1 1 2 3 1 3 1 5 32cos sin 2 . 2 1

2 2 2 6 2 2 2 2 12 8t t t

= = = +

0,25

5

Ta c ( ) ( ) ( ) ( ),ABC ABC AB =SAB D SAB D . Nn nu k ( ),SH AB H AB

th ( )SH ABC D .

K ( )HK C D, K CD th gc gia hai mt phng (SCD) v (ABCD) l gcSKH.

Gc gia hai mt phng (SAD) v (ABCD) l gcSAB .

0,25

Theo gi thit =SAB SKH nn hai tam gic SAH v SKHbng nhau. Suy raAHKD l hnh vung cnh 2a.KBI//AC(ICD) thAC//(SBI)nn:

( ) ( )( ) ( )( ) ( )( ); ; ; 3d AC SB d AC SBI d A SBI = = = d H; SBI

0,25

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(vAHct (SBI) tiB mAB=3BH).KHMBI(MBI) thBI(SHM) do cBISH. Suy ra (SBI)(SHM).Li c (SBI)(SHM)=SMnn khi kHJSM (J SM)thHJ(SBI)v

( )( );d H SBI HJ =3

.2 2

aa =

1=

3.

Ta c HBMng dng vi tam gic BICSuy ra

2 2 13 2 13.

13 1313

BC a aHM HB

BI a

= = = =HM

HB

.

0,25

Trong tam gic vung SHMta c:

2 2 2

1 1 1 2SH

HJ SH HM= + =

a 3

3.

Vy, th tch khi chp S.ABCD:

( ) 31 4 3

. . .3 3

V SH dt ABC a= =D (vtt).

0,25

6

t 2 2 2t a b c= + + , t gi thit ta c t< 1 v1

.2

tabc

=

p dng bt ng thc Cauchy ta c:

( )

( ) ( )

( ) ( )

22 2 2 3

3 23 3

23 3

3

1 2 3 2

2 3 1 0

1 2 1 0

1 1

2 8

a b c abc abc abc

abc abc

abc abc

abc abc

= + + + +

+

+

Do 1 1 3

.2 8 4

tt

Nh vy ,

31.

4t <

0,25

p dng bt ng thc Cauchy ta li c

( )( )

( )2 2 2 2 2 22 2 2 2 2 21 1 1 9 92 2 2 .

2 2 2 66F a b c a b c t

a b c t a b c= + + + + + + + + = +

+ + 0,25

Xt hm s ( )9 3

2 , 16 4

f t t tt

= +