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THI HC K I MN TON LP 11 NM HC 2009-2010 Thi gian 90(khng k thi gian giao )

I . Phn chung (Gm 5 bi, bt buc cho mi hc sinh): Bi 1: (2 im)

a. Gii phng trnh : cos 2x sin x 1+ = b. Gii phng trnh : ( ) ( )2 2 22sin x 1 tan 2x 3 2cos x 1 0 + =

Bi 2: (1,5 im) Cho tp { }10...,,3,2,1X = .Chn ty ba s khc nhau , khng k th t t X a. Tnh xc sut tng 3 s c chn l 12. b. Tnh xc sut tng 3 s c chn l s l.

Bi 3: (2 im)

a. Tm hng t khng cha x trong khai trin nh thc 121

xx

+

; x 0 .

b. Gii bt phng trnh 2 2 32x x x1 6A A C 102 x

+ .

( y k kn nA ; C ln lt l s chnh hp , t hp chp k ca n ). Bi 4:( 1 im) . Trong mt phng oxy, tm nh ca ng thng (d) c phng trnh 3x 2y 4 0 = qua php v t tm S (-1; 4) v t s k = -2 .

Bi 5 : (1,5 im) Cho hnh chp S.ABCD vi ABCD l t gic li . Ly M, N l hai im ln lt trn cc cnh AB, CD );;;( DNCNBMAM .

Gi ( P ) l mt phng qua MN v song song vi SA 1.Xc nh thit din ca hnh chp vi mt phng ( P ) . 2. Chng minh thit din ny l hnh thang khi v ch khi MN song song vi BC II. Phn t chn (Hc sinh chn mt trong hai phn sau):

Phn dnh cho ban c bn ( 6A) Bi 6A: (2 im)

Ba s hng lin tip ca mt cp s cng c tng bng 27, cn tch ca chng bng 693. Tm cc s hng . Phn dnh cho ban nng cao (6B)

Bi 6B: (2 im). Cho ng trn ng knh AB v C l mt im trn on AB( )C B;C A . Mt ng knh PQ thay i ca ng trn khng trng vi AB. ng thng CQ ct cc ng thng PA v PB theo th t ti M v N. Tm qu tch cc im M v N khi PQ thay i./. ==========================================================

====

Trng QH Hu T Ton chnh thc

HNG DN CHM THI HC K I

Mn Ton lp 11 Bi Ni dung im 1 a. cos 2x sin x 1+ = 1,0

22sin x sin x 0 + =

1sin x 0,sin x

2 = =

* sin x 0 x k (k )= = pi

* 1 5

sin x x k2 , x k2 (k )2 6 6

pi pi= = + pi = + pi .

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b. iu kin: cos 2x 0 ( ) ( )2 2 22sin x 1 tan 2x 3 2cos x 1 0 + =

2 2 cos 2x tan 2x 3cos 2x 0 tan 2x 3 + = =

tan 2x 3 x k , k Z6 2pi pi

= = + (tha iu kin)

0,5 0,5

2

a.

b.

Cc kh nng c th 310C 120=

Xc xut tng 3 s c chn l ( ) 7P A120

=

3 s c chn l l khi v ch khi tng 3 s l 10C35 = hoc tng gm 2 s chn v 1 s l: 1 25 5C C 5 .10 50= = .

( ) 10 50 1P B120 2+

= = .

1.5 0,25 0,5

0,5

0,25

3 a. Vit ng cng thc khai trin Tm c hng t khng cha x

k k12 12 k

1C x . k 12 k k 6x

= = .

612C 924=

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b. iu kin x Nx N, 2x 2, x 2, x 3

x 3

Bin i a v bpt : x 4. Kt lun : x = 3, x = 4.

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0,5 0,25

4 * ( )M x;y d , gi ( )M' x';y ' l nh ca M qua php v t tm S t s k , ta c

( )( )

=

=

0 0

0 0

x ' x k x x

y' y k y y , trong k = -2 ,

= =0 0x 1;y 4 .

* ( )( )

+= + = +

= =

x ' 3xx ' 1 2 x 1 2

y' 12y' 4 2 y 4y

2

* x ' 3 y ' 123 2 4 0 3x ' 2y ' 41 0

2 2+

= + =

Pt cn tm 3x 2y 41 0 + = .

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5 . 1.

2.

V hnh ng Xc nh c thit din l MPQN Ch c hai kh nng MP QN hoc MN QP

Nu MP QN do MP SA SA QN suy ra SA song song vi mp (SCD) v l . Nu MN QP th MN song song vi BC. o li v kt lun

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Phn dnh ring cho tng ban

6.A.

Gi ba s cho l a, b, c ta c:a b c 27 (1)a.b.c 693 (2)

+ + =

=

Do a c 2b+ = nn 3b 27 b 9= = . T (2) suy ra ( ) ( )b d .b. b d 693 + =

2 2 2693 9 d 77 d 81 77 4 d 2

9 = = = = =

Vy ba s cn tm l: 7; 9; 11 hoc 11; 9; 7.

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0,5

6.B.

V C nm trn AB nn : ( )CA kCB; k 0= . BQ // AP CM kCQ =

M l nh ca Q qua php v t kCV do Q chy trn (O) nn qu tch ca M l ng trn ( ) ( )k1 CO V O= AQ // BP CQ kCN =

hay 1CN CQk

=

. Vy qu tch ca N l

ng trn ( ) ( )1k

2 CO V O= .

Ch : Do Q khc A v B nn tp hp im M khng phi ton b ng trn ( 10 ) . Tng t tp hp im N khng phi ton b ng trn )0( 2

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S GD & T THA THIN HU KIM TRA HC K 1 TRNG THPT CHUYN QUC HC MN: TON LP 11 - NM HC: 2010 - 2011 Thi gian: 90 pht (khng k thi gian pht ) --------------------------------------------------------------------------------------------------------------------------------

A. PHN CHUNG CHO TT C CC HC SINH Cu 1 (3 im). Gii cc phng trnh lng gic sau: a) cos 2 5sin 2 0x x+ + = .

b) sin (2sin 3) cos2cos 1

x xx

x

+=

.

c) 21 3sin (tan 1) sin (sin cos )x x x x x+ = + . Cu 2 (1 im). T tp hp { }0;1;2;3;4;5;6A = , c th lp c bao nhiu s t nhin chn c 4 ch s khc nhau v ln hn 3000.

Cu 3 (2 im). Mt hp c cha 4 qu cu mu , 5 qu cu mu xanh v 7 qu cu mu vng. Ly ngu nhin cng lc 4 qu cu t hp . Tnh xc sut sao cho:

a) 4 qu cu chn c khng cng mu. b) 4 qu cu chn c c ng mt qu cu mu v khng qu hai qu cu mu vng.

Cu 4 (1 im). Trong mt phng vi h ta Oxy cho ng thng : 2 0d x y+ = v ng trn 2 2( ) : 2 4 20 0.C x y x y+ + = Tm trn ng thng d im M v trn ng trn ( )C im N sao cho N l

nh ca M qua php tnh tin theo vect (3; 1).v =

Cu 5 (2 im). Cho t din ABCD. Gi M, N ln lt l trung im ca AB, AC v G l im trn on thng DN sao cho 4DN NG= . Trn on thng BG ly im I (I khc vi B v G).

a) Dng thit din ca t din ct bi mt phng (IMN), thit din l hnh g?

b) Xc nh v tr im I trn on thng BG thit din l hnh bnh hnh. Khi hy tnh t s BIBG

.

B. PHN RING (Hc sinh ch c lm mt trong hai phn) Cu 6a (1 im) (Theo chng trnh chun). Cho dy s ( )nu bit 1 12; 3 n nu u u n+= = + vi 1.n Lp cng thc s hng tng qut nu ca dy s trn.

Cu 6b (1 im) (Theo chng trnh nng cao).

Tm h s ca s hng cha 9x trong khai trin 21 2n

xx

bit rng : 3 2 2 18 3( 1).n nA n C = +

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S GD & T THA THIN HU P N KIM TRA HC K 1 TRNG THPT CHUYN QUC HC MN: TON LP 11 - NM HC 2010 - 2011 -------------------------------------------------------------------------------------------------------------------------------

CU NI DUNG IM

1a)

2 2cos 2 5sin 2 0 1 2sin 5sin 2 0 2sin 5sin 3 0sin 3

1sin

2

26 ( ).

7 26

(loi)

x x x x x x

x

x

x kk

x k

pipi

pipi

+ + = + + = =

=

=

= +

= +

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0,5

1b)

iu kin: 1cos 2 ( ).2 3

x x k kpi pi +

Vi iu kin , phng trnh tng ng vi 2 22sin 3 sin 2cos cos cos 3 sin 2cos 2

1 3cos sin cos 2 cos cos 2

2 2 3

2 2 23 3

22 23 9 3

(loi)

(tha iu kin).

x x x x x x x

x x x x x

x x k x k

x x k x k

pi

pi pipi pi

pi pi pipi

+ = + =

+ = =

= + = +

= + + = +

Vy phng trnh c nghim l 2 , ( ).9 3

x k kpi pi= + Z

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1c)

iu kin: ( ).2

x k kpi pi + Z Vi iu kin , phng trnh tng ng vi

2 2 2

2 2 2 2

2 2 2

sin sin cos3sin 1 1 sin sin cos 0 3sin cos (cos sin ) 0cos cos

3sin (sin cos ) cos (sin cos ) 0 (3sin cos )(sin cos ) 0ta

sin cos 0 tan 13sin cos 0 3tan 1

x x xx x x x x x x x

x x

x x x x x x x x x x

x x x

x x x

+ = + =

= =

= =

= =

n 11

tan3

4 ( ).

6

x

x

x kk

x k

pipi

pipi

= =

= +

= +

Z

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Cu 2 Gi abcd l s t nhin chn c 4 ch s khc nhau v ln hn 3000 c lp t A, khi

{3;4;5;6}a v {0;2;4;6}d . C 2 trng hp:

Nu {3;5}a : C 2 cch chn a, 4 cch chn d v 25A cch chn bc . Do trng

hp ny c 252.4. 160A = s.

Nu {4;6}a : C 2 cch chn a, 3 cch chn d v 25A cch chn bc . Do trng

hp ny c 252.3. 120A = s.

Tm li c 160+120=280 s tha yu cu.

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Cu 3 S phn t ca khng gian mu l 416 1820C = = . 0,25

3a)

Gi A l bin c 4 qu chn c khng cng mu. Khi A l bin c 4 qu ly c c cng mu.

Ta c: 4 4 44 5 7 41.A C C C = + + =

Do xc sut ca bin c A l: 41( )1820

AP A

= =

.

Vy xc sut ca bin c A l 41 1779( ) 1 ( ) 1 0,98.1820 1820

P A P A= = =

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3b)

Gi B l bin c 4 qu ly c c ng mt qu cu mu v khng qu 2 qu cu mu vng. Khi

1 3 1 1 2 1 2 14 5 4 7 5 4 7 5. . . . . 740.B C C C C C C C C = + + =

Xc sut ca bin c B l 740 37( ) 0,41.1820 91

BP B

= = =

0,5

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Cu 4

Gi ( ; 2 )M x x d . V ( )v

N T M= nn ta ca N l ( 3; 2 1).N x x+ 2 2

2

( ) ( 3) ( 2 1) 2( 3) 4( 2 1) 20 05 20 2.

N C x x x xx x

+ + + + =

= =

Vi 2x = ta c (2; 4)M v (5; 5).N Vi 2x = ta c ( 2;4)M v (1;3).N

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5a

V hnh thit din ng: 0,25

P

Q

GN

M

B

A

C

D I

Gi Q l giao im ca NI v BD. Ta c ( ) ( )Q MNI BCD ,

( ), ( )MN MNI BC BCD v //MN BC nn giao tuyn ca (MNI) v (BCD) l ng thng d i qua Q song song vi BC, ct CD ti P. Khi t gic MNPQ l thit din ca hnh chp ct bi (IMN). V MN//PQ nn thit din l hnh thang.

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CU NI DUNG IM

5b

Q H

P

I G

N

M

D

C

A

B

Thit din MNPQ l hnh bnh hnh khi

2BCMN PQ= = . Do , gi Q l trung im BD

v I l giao im ca BG v NQ. Khi vi im I xc nh nh vy th thit din thu c khi ct t din ABCD bi mt phng (MNI) l hnh bnh hnh. Trong (BDN), k GH//NQ ( )H BD . Ta c:

1 4 .4

HQ HQ NG QB HQQD QB ND= = = =

4 4.

4 5BI BQ BQ QHBG BH BQ QH QH QH= = = =+ +

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6a)

Ta c 1 3n nu u n+ = vi mi 1n , do :

2 1

3 2

4 3

1

369

.............

3( 1)n n

u u

u u

u u

u u n

=

=

=

=

Suy ra 1 13 6 9 ... 3( 1)n nu u n S = + + + + = trong 1nS l tng ca 1n s hng lin tip ca cp s cng c s hng u bng 3 v cng sai d=3. Do

2

1( 2)( 1).3 3( )3 6 9 ... 3( 1) ( 1).3 .

2 2nn n n nS n n

= + + + + = + =

Vy 2 2

1 13 3 3 3 42 .

2 2n nn n n n

u u S

= + = + =

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6b)

iu kin: 3,n n N . 3 2 2 2

1

2 3 2 2 2

3 2 2

! ( 1)!8 3( 1) 8 3. 3( 3)! 2!( 3)!3( 2)( 1)( 2)( 1) 8 3 2( 3 2 ) 16 3 9 12

22 25 13 12 0 ( 12)(2 1) 0

12.

n n

n nA n C nn n

n nn n n n n n n n n n

n n n n n n

n

= + = +

= + + = +

+ = + =

=

Khi 2 2121 12 2 .

n

x xx x

=

S hng tng qut

12 22

1 12 12 121

.( 2 ) .( 2) .k k

k k k kk k

xT C x Cx x

+

= =

1kT + cha 9x khi 2 (12 ) 9 3 21 7.k k k k = = = Vy s h s ca s hng cha 9x l: 7 712.( 2) 101376.C =

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Ghi ch: Cc cch gii khc nu ng vn c im ti a v im thnh phn cng c cho mt cch tng ng.

TRNG THPT CHUYN QUC HC KIM TRA HC K 1 T Ton MN: TON LP 11 - NM HC: 2011 - 2012 Thi gian: 90 pht (khng k thi gian pht ) --------------------------------------------------------------------------------------------------------------------------------

Cu 1 (3 im). Gii cc phng trnh lng gic sau: a) 24sin 4cos 1 0.x x+ = .

b) ( )(2cos 1)(cos 1) 3 2cos 1sin

x xx

x

+ = .

c) ( )tan sin 2 cos 2 tan 6x x x x = + . Cu 2 (1 im). C bao nhiu s t nhin c 6 ch s trong ch s 9 xut hin 3 ln, cc ch s cn li c mt mt ln.

Cu 3 (2 im). a) Cn chn ngu nhin 5 hc sinh trong mt lp hc c 15 nam v 20 n tham gia ng din. Tnh

xc sut sao cho 5 hc sinh c chn c c nam ln n v s hc sinh n t hn s hc sinh nam. b) Mt ng xu do ch to khng cn i nn xc sut xut hin mt nga ch bng 80% xc sut xut

hin mt sp. Tnh xc sut khi gieo 4 ln c lp th c t nht mt ln xut hin mt nga.

Cu 4 (1 im). Tm s hng khng cha x trong khai trin 3 213

n

xx

bit rng:

( ) 222 4 5 . 3 .nn n nP n P A + = Cu 5 (1 im). Trong mt phng vi h ta Oxy cho ng trn 2 2( ) : 2 10 0.C x y x y+ + = Tm trn ng trn ( )C cc im ,M N sao cho N l nh ca M qua php v t tm O t s 2k = (vi O l gc ta ). Cu 6 (2 im). Cho hnh chp S.ABCD c y ABCD l hnh thang v // .AD BC Gi E, F ln lt l trung im ca AB, CD; H, K ln lt l trung im ca SE v SF; G l trng tm ca tam gic ABD. Trn

on SG ly im I sao cho 3 .SI IG=

a) Xc nh thit din ca hnh chp khi ct bi mt phng (HIK). Thit din l hnh g? b) Bit rng SA BC a= = v 2 .SD AD a= = Hy tnh theo a chu vi ca thit din va tm c.

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TRNG THPT CHUYN QUC HC P N KIM TRA HC K 1 T Ton MN: TON LP 11 - NM HC 2011 - 2012 -------------------------------------------------------------------------------------------------------------------------------

CU NI DUNG IM

1a)

2 2 24sin 4cos 1 0 4 4cos 4cos 1 0 4cos 4cos 3 03

cos2

1cos

22 23 ( ) .2 23

x x x x x x

x

x

x kk

x k

pipi

pipi

+ = + = =

=

=

= +

= +

(loi)

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0,5

1b)

iu kin: sin 0 ( ).x x k kpi Vi iu kin , phng trnh tng ng vi

22cos cos 1 3(2cos 1)sin cos 2 cos 3(sin 2 sin )3 sin cos 3 sin 2 cos 23 1 3 1

sin cos sin 2 cos 2 sin sin 22 2 2 2 6 6

22 26 6

4 22 2 9 36 6

x x x x x x x x

x x x x

x x x x x x

x kx x k

x kx x k

pi pi

pi pipipi

pi pipi pi

pi pi

= =

=

= =

= = +

= +

= + +

(loi)

(t

ha iu kin).

Vy phng trnh c nghim l 4 2 , ( ).9 3

x k kpi pi= + Z

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1c)

iu kin: ( ).2

x k kpi pi + Z

Vi iu kin , phng trnh tng ng vi ( ) ( )

( ) ( )2 2

2 2

3 3 2 3 2

tan 2sin cos cos sin . tan 6

tan (1 tan ) 2 tan 1 tan . tan 6tan tan tan 6 tan tan 6 2 tan 6 tan 2 tan 6 0

tan 1 ( ).4tan 3

arctan( 3)

x x x x x x

x x x x x

x x x x x x x x

x x kk

xx k

pipi

pi

= +

+ = +

= + + + =

= = + =

= +

Z

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Cu 2

C 2 trng hp: Ch s hng u tin (hng trm ngn) bng 9: Xp 2 ch s 9 vo 5 v tr: c 25C cch. Chn 3 ch s trong 9 ch s (khc vi 9) v sp chng vo 3 v tr cn li: c 39A cch. Do trng hp ny c 2 35 9. 5040C A = s. Ch s hng u tin (hng trm ngn) khc 9:

Chn ch s cho hng u tin: c 8 cch. Xp 3 ch s 9 vo 5 v tr: c 35C cch. Chn 2 ch s trong 8 ch s (khc vi ch s chn hng u tin v khc 9) v

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sp th t chng vo 2 v tr cn li: c 28A cch. Vy trng hp ny c 3 25 88. . 4480C A = s

Tm li c 5040+4480=9520 s tha yu cu.

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3a)

S phn t ca khng gian mu l 535| | 324632.C = = Gi A l bin c 5 hc sinh chn c c c nam ln n v s hc sinh n t hn s hc sinh nam. Khi c cc trng hp xy ra l: 1 n v 4 nam; 2 n v 3 nam. S kt qu thun li cho A l 1 4 2 320 15 20 15| | . . 113750.A C C C C = + = Vy xc sut ca bin c A l: | | 113750( ) 0,35.| | 324632

AP A = =

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3b)

Gi x l xc sut xut hin mt sp ca ng xu khi gieo. Khi xc sut xut hin mt

nga l 0,8x . Ta c 1 50,8 1 .1,8 9

x x x+ = = =

Gi A l bin c gieo ng xu 4 ln c lp th c t nht mt ln xut hin mt nga. Lc A l bin c gieo ng xu 4 ln c lp th c khng xut hin mt nga ln no. Ta c 1 2 3 4A A A A A= , trong iA l bin c ln gieo th ( {1, 2,3,4}) i i xut hin mt sp.

V 1 2 3 4, , ,A A A A c lp vi nhau nn 4

1 2 3 45( ) ( ). ( ). ( ). ( ) .9

P A P A P A P A P A = =

Vy 45( ) 1 ( ) 1 0,905.

9P A P A = =

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0,25

0,25

Cu 4

iu kin: 2, n n N . 2

2

2

!2 (4 5) 3 2. ! (4 5).( 2)! 3.2!

3 ( 1)2 ( 1) (4 5) 9 10 02

10.

1

n

n n n

nP n P A n n n

n nn n n n n

n

n

+ = + =

+ = =

=

= (loi)

Khi 3 32 2101 13 3 .

n

x xx x

=

S hng tng qut

( ) 30 3103 101 10 102 213 . .3 ( 1) .k kkk k k k

k kxT C x C

x x

+

= =

1kT + khng cha x khi 30 3 2 0 5 30 6.k k k k = = = Vy s hng khng cha x ca khai trin l: 6 4 610.3 .( 1) 17010.C =

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Cu 5

Gi ( ; ) ( )M x y C . Khi 2 2 2 10 0 (1).x y x y+ + =

Ta c: ( , 2)2( ) 2 ( 2 ; 2 ).2

NO

N

x xN V M ON OM N x y

y y=

= = =

2 2

2 2

( ) ( 2 ) ( 2 ) 2( 2 ) ( 2 ) 10 04 4 4 2 10 0 (2).

N C x y x yx y x y

+ + =

+ + =

T (1) v (2) ta c h

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2 2 2 2

2 2 2 2

2 2 2 2

2

2 10 0 4 4 8 4 40 04 4 4 2 10 0 4 4 4 2 10 012 6 30 0 2 5

2 10 0 (2 5) 2 2 5 10 02 5 2

.

15 20 20 0

x y x y x y x y

x y x y x y x y

x y y x

x y x y x x x x

y x xyx x

+ + = + + =

+ + = + + =

+ = = +

+ + = + + + + =

= + =

=+ + =

Vy ( 2;1)M v (4; 2).N

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6a

L

Q

M

P

NJ

I

K H

E FG

A D

B

S

C

Trong (SED) gi .J HI ED= Khi ( ) ( )J HIK ABCD .

Ta c ( ), ( )EF ABCD HK HIK m //EF HK nn giao tuyn ca (HIK) v

(ABCD) l ng thng qua J song song vi EF, ct AB ti M, ct CD ti N. Trong (SCD), gi .P NK SC= Lc

( ) ( ).P HIK SBC V ( ), ( )HK HIK BC SBC v //BC HK

nn giao tuyn ca (HIK) v (SBC) l ng thng qua P song song vi BC, ct SB ti Q. Khi t gic MNPQ l thit din cn tm. V //MN PQ (do cng song song vi BC) nn thit din l hnh thang.

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6b

+ Gi L l trung im HE, ta c 3 // 2 .SL SI LI EJ EJ LILE IG

= = =

Mt khc, 3 3 .4 4

LI SI LI EGEG SG

= = =

Do 3 3 3 1 12. .4 2 2 3 2

EJ EG EG ED ED= = = = do J l trung im ED.

Suy ra M, N ln lt l trung im ca AE, DF.

+ Vy //MQ SA v do 3 3 3 .4 4 4

MQ MB aMQ SASA AB

= = = =

Tng t 3 6 .4 4

aNP SD= =

+ Ta cng c 1 1 .4 4 4

PQ SQ AM aPQ BCBC SB AB

= = = = =

3 6 72.

2 2 4 4 4

AD BC ADEF AD BC AD a a aMN

+++ + +

= = = = =

+ Vy chu vi ca thit din MNPQ l

7 6 3 17.

4 4 4 4 4a a a a aMN NP PQ QM+ + + = + + + =

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0,25

0,25

0,25 Ghi ch: Cc cch gii khc nu ng vn c im ti a v im thnh phn cng c cho mt cch tng ng.

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TRNG THPT CHUYN QUC HC T TON

KIM TRA KC K I Mn TON - lp 11 (Thi gian lm bi: 90 pht)

Bi 1 (3 im). Gii cc phng trnh sau

a) 2cos 3sin 3 0+ + =x x .

b) sin2 osx 01-sinx

x c=

c) 1 t anx 1 sin21 t anx

x

= ++

Bi 2 (2 im). Cho tp hp { }1;2;3;4;5;6;7;8;9X = a) C bao nhiu s t nhin l c 6 ch s khc nhau c ly trong tp X. b) C bao nhiu s t nhin l c 6 ch s khc nhau c ly trong tp X, trong

c ng 2 ch s chn v hai ch s chn ny khng ng lin k nhau.

Bi 3 (2 im). Trong mt lp hc c 8 bng n, mi bng c xc sut b chy l 0,025. Lp hc c nh sng nu c t nht 6 bng n sng. Tnh xc sut lp hc khng c nh sng.

Bi 4 (1 im). Trong mt phng ta Oxy, cho ng thng : 2 1 0d x y + = . Gi 1d l

nh ca d qua php tnh tin theo vect ( )2;0v = . Vit phng trnh ca ng thng 1d . Bi 5 ( 2 im). Cho hnh chp S.ABCD c ABCD l hnh bnh hnh, im M thay i trn cnh SD, M khng trng S.

a) Dng giao im N ca SC vi mt phng (ABM); T gic ABNM l hnh g? C th l hnh bnh hnh khng?

b) Gi I l giao im ca AM v BN. Chng minh rng: khi M chy trn cnh SD th I chy trn mt ng thng c nh. Hy ch ra ng thng c nh .

-------------------- Ht -------------------

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P N KIM TRA HC K I MN TON LP 11 (NC) NM HC 2012 2013

Bi cu Bi gii gi im Bi 1 a) ( ) 2sin 3sin 4 0 =PT x x

0.25

sin 1sin 4( )

=

=

x

x l

0.5

sin 1 22

x x kpi

pi= = +

0.25

b) iu kin:1 sinx 0 sinx 1 x 2

2k

pipi +

0,25

(Pt sin2 osx = 0 cosx(2sinx-1) = 0x c osx=0

1sinx=

2

c

0.25

osx=02

c x kpi

pi = +

2 ,1 6

sinx=2 5

26

x k

x k

pipi

pipi

= +

= +

0.25

i chiu vi iu kin, phng trnh c 3 h nghim: 5

2 ; 2 ; 22 6 6

x k x k x kpi pi pi

pi pi pi= + = + = +

0,25 c)

KX: cos 0 2

, , 't anx 1

'4

x kx

k k Z

x k

pipi

pipi

+

+

0,25

( ) ( )2 3cos sinx sinx cos cos sinx sinx coscos sinx

xpt x x x

x

= + = ++

0.25

Chia 2 v ca pt cho 3os 0c x , c ( ) ( ) ( )32 2 1 t an x t anx 1 t an x t anx 1+ + = +

0.25

( )( ) ( )321 t an x 1 t anx t anx 1 + = + ( )2t anx t an x t anx 2 0 + + =

t anx 0 ,x k k Zpi = = (Tha /k)

0.25

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Bi2

a) S l c 6 ch s c dng { }, 1;3;5;7;9abcdef f Chn f c 5 cch

0.25

Chn abcde c 58 6720A = cch 0.5

Vy, s cc s l cn tm c 585. 33600A = (s) 0.25 b) S l c 6 ch s c dng { }, 1;3;5;7;9abcdef f

Chn f c 5 cch Chn 3 ch s l trong 4 ch s l cn li ca tp X ri xp th t cho chng, c 34 24A = cch

0.25

Chn 2 ch s chn trong 4 ch s chn ca tp X, c 24 6C = cch 0.25

a 2 ch s chn vo 2 trong 4 v tr (gia hai ch s l hoc ch s hng cao nht ca s cn tm), c 24 12A = cch (Minh ha: C C C CL L L L )

0.25

Vy, c 5. 34A .2

4C .2

4A =8640 (s) 0.25 Bi 3 Xc sut mi bng sng l: 1 391

40 40 =

0,25

Bin c A: Lp hc c nh sng, A : Lp hc khng c nh sng B: 6 bng n sng, 2 bng n b chy. C: 7 bng n sng, 1 bng n b chy. D: 8 bng n sang.

0.25

( ) 286 2

39 1. . 0.01540 40

P B C

=

( ) 187

39 1. . 0.1675;40 40

P C C

=

( )8

390.8167

40P D

=

(ng P(B) v P(D) hoc P(C) v P(D) th cho ti a)

0,5

; , ,A B C D B C D= i mt xung khc. 0,25 ( ) ( ) ( ) ( ) P A P B P C P D= + +

2 1

8 8

6 2 7 839 1 39 1 39

. . . . 0.999240 40 40 40 40

C C

+ +

0,25

0.25

( ) ( ) 1 0,0008P A P A= 0,25 Bi 4 Phng trnh 1 : 2 0d x y m + = . 0.25

Ly (1;1)A d v gi ( )'v

A T A= th ( )' 3;1A . 0,25 V ( ) 1' 3;1A d nn 3 2 0 1m m + = = 0.25 Vy 1 : 2 1 0d x y = 0,25

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Bi 5 a)

(V ng thit din l cho im)

0.25

C

/ /

( )

( )

( ); ( )

CD AB

CD Mp SCD

AB Mp ABM

M Mp SCD M Mp ABM

nn giao tuyn ca hai mp (SBC) v (ABM) i qua M v song song vi CD.

0.25

0.25

Trong mp(SCD), v MN//CD, N trn SC. Suy ra N l giao im ca SC vi mp(ABM) 0.25

C / /

/ // /

MN CDMN AB

AB CD

nn ABNM l hnh thang.

Khi M trng D th ABNM l hnh bnh hnh. 0.25

b) C

( )( ) ( )

( )

I AM SADI SAD ABC

I BN SBC

= d 0.25

Do hai mp (SAD) v (SBC) c nh nn giao tuyn d ca chng c nh. Vy, I chy trn ng thng c nh. 0.25

C

/ /

( )

( )

CB AD

CB mp SCB

AD mp SAD

S chung

nn nn giao tuyn d ca hai mp (SBC) v (SAD) i qua S v song song vi CB, AD.

0.25

d

I

N

D

A B

C

S

M

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