[vnmath.com]-De Thi Hoc Ki 1 Toan 11 co loi giai chi tiet

8/8/2019 [vnmath.com]-De Thi Hoc Ki 1 Toan 11 co loi giai chi tiet

1/15

1

s 1

THI H C K 1 Nm h cMn TON L p 11 Nng cao

Th i gian lm bi 90 pht

Cu I: (3) Gi i cc ph ng trnh sau :

1) (1) ( ) x x23tan 1 3 tan 1 0 + + = 2) (1) x x2 32 cos 3 cos2 04

+ =

3) (1) x x x2

1 cos21 cot 2

sin 2

+ =

Cu II: (2)

1) (1) Tm s h ng khng cha x trong khai trin c an

x x

24

1 +

, bi t: n n nC C A0 1 22 109 + = .

2) (1) T cc ch s 1, 2, 3, 4, 5, 6 c thl p c bao nhiu s t nhin chn c su ch s v tho mni u ki n: su ch s c a m i s l khc nhau v trong mi s t ng c a ba ch s u l n h nt ng c a ba ch s cu i m t n v .

Cu III: (2) Trn m t gi sch c cc quyn sch v ba mn hc l ton, vt l v ho hc, g m 4quy n sch ton, 5 quyn sch vt l v 3 quyn sch ho hc. L y ng u nhin ra 3 quyn sch. Tnh

xc su t 1) (1) Trong 3 quyn sch ly ra, c t nht m t quy n sch ton.2) (1) Trong 3 quyn sch ly ra, ch c hai loi sch v hai mn hc.Cu IV: (1) Trong mt ph ng to Oxy, cho ng trn ( ) ( )C x y2 2( ) : 1 2 4 + = . G i f l php bin

hnh c c b ng cch sau: thc hi n php tnh ti n theo vect v 1 3;2 2

=

, r i n php v t tm

M 4 1

;3 3

, t s k 2= . Vi t ph ng trnh nh c a ng trn (C ) qua php bin hnh f .

Cu V: (2) Cho hnh chpS.ABCD c y ABCD l hnh bnh hnh. Gi M v N l n l t l tr ng tmc a tam gicSAB v SAD.1) (1) Ch ng minh: MN // ( ABCD ).

2) (1) G i

E l trung

i m c a

CB. Xc

nh thi t di n c a hnh chp

S.ABCDkhi c t b

i m tph ng ( MNE ).

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

s 1

P N THI H C K 1 Nm h cMn TON L p 11 Nng cao


Cu N i dung i m I (3) 1

( ) x x x hoac x2 1

3 tan 1 3 tan 1 0 tan 1 tan 3 + + = = =

0,50

x x k tan 14

= = + 0,25

x x k 1

tan63

= = + 0,25

2 pt x x x x x x

31 cos 2 3 cos2 0 1 sin 2 3 cos2 0 sin 2 3 cos2 1

2

+ + = + = =

0,25

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2/15

2

xsin 2 sin3 6

=

0,25

x k x k x

x k x k

2 23 6 4sin 2 sin

5 73 6 2 23 6 12

= + = +

= = + = +

0,25

0,25

3 K: x x k sin2 0 2

( )( )

x x pt x x x x

x x xx x x x x

22

cos 2 1 cos 21 sin 2 cos2 sin 2 1 cos2

sin2 sin 2sin2 1sin2 1 sin 2 cos2 1 0sin2 cos2 1

+ = + =

= + + =

+ =

0,50

x x k x k sin2 1 2 22 4

= = + = + (tho i u ki n) 0,25

x k (loi) x x x x k

x k sin2 cos2 1 sin 2 sin

4 4 44

= + = + = = + = +

(tho k) 0,25

II (2) 1 K: n n2; ; ( )n n nC C A n n n n0 1 22 109 1 2 1 109 12 + = + = = 0,25

( )k

k k k k

k k x C x x C x

x

12 12 12122 2 4 24 6

12 1240 0

1

= =

+ = =

0,25

k k 24 6 0 4 = = 0,25 V y s h ng khng cha x l C 412 495= 0,25

2 G i s c n tm la a a a a a1 2 3 4 5 6 .Theo ra, ta c:

( )( )

a a a a a a a a a a a a a a a

a a a a a a

1 2 3 4 5 6 1 2 3 1 2 3 4 5 6

1 2 3 1 2 3

1 2 1

2 21 1 11

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

+ + = + + + =

0,25

+TH 1:{ } { }a a a1 2 3; ; 2;4;5= th { } { }a a a4 5 6; ; 1;3;6= nn c (1.2!).(3!) = 12 (s)+TH 2:{ } { }a a a1 2 3; ; 2;3;6= th { } { }a a a4 5 6; ; 1;4;5= nn c (1.2!).(3!) = 12 (s)+TH 1:{ } { }a a a1 2 3; ; 1;4;6= th { } { }a a a4 5 6; ; 2;3;5= nn c (1.2!).(3!) = 12 (s)

0,50

Theo quy tc c ng, ta c: 12 + 12 + 12 = 36 (s) 0,25 III (2) 1 A l bi n c Trong 3 quy n sch ly ra, c t nht m t quy n sch ton.

A l bi n c Trong 3 quyn sch ly ra, khng c quyn sch ton no.

( ) C P AC

38312

14

55

= = 0,50

( ) ( )P A P A 14 411 1 55 55= = = 0,50 2 B l bi n c Trong 3 quyn sch ly ra, cng hai loi sch v hai mn hc

B C C C C C C C C C C C C 1 2 2 1 1 2 2 1 2 1 1 24 5 4 5 4 3 4 3 5 3 5 3 145 = + + + + + =

0,50

( )P BC 312

145 2944

= = 0,50

IV (1) G i I l tm c a (C ) th I (1 ; 2) v R l bn knh ca (C ) th R = 2.



3/15

3

G i A l nh c a I qua php tnh ti n theo vect 3v ;2

1

2

=

, suy ra 7 A ;

2

3

2

0,25

G i B l tm c a (C ) th B l nh c a A qua php vt tm 1 M ;3

4

3

t s k 2=

nn : B A M

B A M

x x x MB MA

y y y

52

321423

= ==

= =

. V y 20 B ;

3

5

3

0,25

G i R l bn knh ca (C ) th R = 2 R = 4 0,25

V y C x y2 2

5 20( ') : 16

3 3

+ =

0,25

V (2)

O F

Q

P

G

K

E

N M

J

ID

A

B C

S

0,50

1 G i I , J l n l t l trungi m c a AB v AD , ta c:SM SN

MN IJ SI SJ

2 / /

3= = 0,50

M IJ ABCD( ) nn suy ra MN // ( ABCD ). 0,50 2 + Qua E v ng th ng song song v i BD c t CD t i F , c t AD t i K .

+ KN c t SD t i Q, KN c t SA t i G; GM c t SB t i P .Suy ra ng gic EFQGP l thi t di n c n d ng.

0,50

H T

s 2



Cu I: (3) Gi i cc ph ng trnh sau :1) (1) x xsin3 3 cos3 1 = 2) (1) x x x3cos 3 2 sin2 8cos+ =

3) (1)( ) x x

x

22 3 cos 2sin2 4

12 cos 1

=

Cu II: (2)

1) (1) Tm h s c a x 31 trong khai trin c an

x x2

1 +

, bi t r ng n nn n nC C A1 21 821

2+ + = .

2) (1) T cc ch s 0; 1; 2; 3; 4; 5; 6; 7; 8; 9 c thl p c t t c bao nhiu s t nhin chn cnm ch s khc nhau v trong nm ch s cng hai ch s l v hai ch s l ny khng ngc nh nhau.



4/15

4

Cu III: (2) C hai ci hp ch a cc qu c u, h p th nh t g m 3 qu c u mu trng v 2 qu c u mu ; h p th hai g m 3 qu c u mu trng v 4 qu c u mu vng. Ly ng u nhin t m i h p ra 2 qu c u. Tnh xc sut :1) (1) Trong 4 qu c u l y ra, c t nht m t qu c u mu trng.2) (1) Trong 4 qu c u l y ra, c c ba mu: trng, v vng.

Cu IV: (1) Trong mt ph ng to Oxy, cho ng trn ( ) ( )C x y2 2( ) : 2 1 9 + = . G i f l php bin

hnh c c b ng cch sau: thc hi n php i x ng tm M 4 1;3 3 , r i n php v t tm N 1 3;

2 2 ,

t s k 2= . Vi t ph ng trnh nh c a ng trn (C ) qua php bin hnh f .Cu V: (2) Cho hnh chpS.ABCD c y ABCD l hnh thang ( AD // BC, AD > BC ). G i M l m t

i m b t k trn c nh AB ( M khc A v M khc B). G i ( ) l m t ph ng qua M v song song v iSB v AD .1) (1) Xc nh thi t di n c a hnh chp khi ct b i m t ph ng ( ). Thi t di n ny l hnh g ?2) (1) Ch ng minhSC // ( ).

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

s 2

p n THI H C K 1 Nm h c

Mn TON L p 11 Nng caoTh i gian lm bi 90 pht

Cu N i dung i mI (3)1

x x x1 3 1

sin3 cos3 sin 3 sin2 2 2 3 6

= =

0,50

x k x k

x k x k

23 2

3 6 6 35 7 2

3 23 6 18 3

= + = +

= + = +

0,25

0,25

2 ( ) pt x x x x x x x x

x x (*)

3 2

2

4cos 6 2 sin cos 8cos cos 2 cos 3 2 sin 4 0

cos 0

2sin 3 2 sin 2 0

+ = + =

=

+ =

0,25

x x k cos 02

= = + 0,25

x k x x

x k x (loi)

2 22sin 4(*) sin2 32 2sin 24

= +=

=

= +=

0,25

0,25

3 i u ki n: x x k 1cos 22 3

+

( ) pt x x x x x x2 3 cos 1 cos 2cos 1 sin 3 cos 0 tan 32

+ = = =

0,50

x x k tan 33

= = + 0,25

i chi u i u ki n, ta c nghim c a pt l: x k 43

= + 0,25

II (2)1 K: n n2;

0,25



5/15

5

( )n nn n n

n nC C A n n n n1 2 2

11821 1 821 1640 0 40

2 2

+ + = + + = + = =

k k k k k

k k x C x x C x

x

40 40 4040 2 40 3

40 4020 0

1

= =

+ = =

0,25

k k 40 3 31 3 = = 0,25V y h s c a x31 l C 3

409880= 0,25

3 + S t nhin chn g m 5 ch s khc nhau v cng hai ch s l c:C C C C 2 2 2 15 4 5 35 4! 4 3! 6480 = (s ) 0,25

+ S t nhin chn g m 5 ch s khc nhau v cng hai ch s l ng c nh nhau c A A A2 2 25 4 55 3 4 2 3 3120 = (s )

0,50

Suy ra c: 6480 3120 = 3360 (s) 0,25III (2)1 C C 2 25 7 210 = = 0,25

G i A l bi n c Trong 4 qu c u l y ra, c t nht m t qu c u mu trng. A l bi n c Trong 4 qu c u l y ra, khng c quc u mu trng.

( ) C C P A 2 22 4 1210 35= = 0,50

Suy ra: ( ) ( )P A P A 1 341 1 35 35= = = 0,25

2 G i B l bi n c Trong 4 qu c u l y ra, c c ba mu: trng, v vng.+Tr ng h p 1: 1 trng, 1 h p m t; 2 vng h p hai c( )C C C 1 1 22 3 4 (cch)+Tr ng h p 2: 2 h p m t; 1 vng, 1 trng h p hai c ( )C C C 2 1 12 3 4 (cch)+Tr ng h p 3: 1 , 1 tr ng h p m t; 1 vng, 1 trng h p hai c( )( )C C C C 1 1 1 13 2 4 3 (cch)

Suy ra: ( ) ( ) ( )( ) B C C C C C C C C C C 1 1 2 2 1 1 1 1 1 12 3 4 2 3 4 3 2 4 3 120 = + + =

0,75

Suy ra: ( )P B 120 4210 7

= = 0,25

IV (1)G i I l tm ca (C ) th I (2 ; 1) v R l bn knh ca (C ) th R = 3.

G i A l nh c a I qua php i x ng tm 1 M ;3

4

3

, suy ra 1 A ;3

2

3

0,25

G i B l tm ca (C ) th B l nh c a A qua php vt tm 3 N ;2

1

2

t s k 2= nn :

B A N

B A N

x x x NB NA

y y y

52

6213

26

= ==

= =

. V y 13 B ;6

56

0,25

G i R l bn knh ca (C ) th R = 2 R = 6 0,25

V y C x y2 2

5 13( ') : 36

6 6

+ + =

0,25

V (2)



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6

PN

C

A D

B

S

M

0,50

1( )SB SAB MN SB N SASB SAB

( ) / / ( ) ( ) / / ,( )

=

( ) AD SAD NP AD P SD AD SAD( ) / / ( ) ( ) / / ,

( )

=

( ) AD ABCD MQ AD Q CD AD ABCD( ) / / ( ) ( ) / / ,

( )

=

V y thi t di n l hnh thang MNPQ ( MQ // NP ).

0,50

2 Ta c: DP AN AN AM AM DQ DP DQ SC PQ DS AS AS AB AB DC DS DC

; ; / / = = = =

M ( )PQ nn suy ra ( )SC / / (pcm).1,00

H T

s 3

THI H C K 1 Nm h cMn TON L p 11 C b n


Bi 1(2i m). Gi i cc ph ng trnh sau:

a) x 0 2cos 102 2

=

b) x xsin 3 cos 1 = c) x x23tan 8tan 5 0 + =

Bi 2(2 i m). Trong mt h p ng 5 vin bi xanh v 4 vin bi . L y ng u nhin ng th i 3 vin bi.Tnh xc sut trong 3 vin bi ly ra:a) C 2 vin bi mu xanh b) C t nht m t vin bi mu xanh.

Bi 3(2i m).

a) Xt tnh tng gi m c a dy s ( )nu , bi t nn

un

1

2 1

+=

+

b) Cho cp s c ng ( )nu c u1 8= v cng said 20= . Tnhu101 v S 101 .Bi 4(3,5i m). Cho hnh chp S.ABCD cy ABCD l hnh bnh hnh. Gi M, N, P ln l t l trung

i m c a cc cnh AB, AD v SB.a) Ch ng minh rng: BD//(MNP).

b) Tm giaoi m c a m t ph ng (MNP) v

i BC.c) Tm giao tuyn c a hai mt ph ng (MNP) v (SBD).

d) Tm thit di n c a hnh chp v i m t ph ng (MNP).

Bi 5(0,5i m). Tm s h ng khng cha x trong khai trin x x

15

4

12

.

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

P N THI H C K 1 Nm h cMn TON L p 11 C b n



7/15

7

s 3 Th i gian lm bi 90 pht

Bi Ni dung i m1 2.0

a) xk x

x k

0 0 0

0

0 0 0

10 60 .3601 2cos 102 2 10 60 .360

2

+ = + + =

+ = +

( ) x k k x k

0 0

0 0100 .720

140 .720

= +

= +

V y nghi m c a pt l: x k x k k 0 0 0 0100 .720 ; 140 .720 ,= + = +

0,25

0,25

0,25b)

x x x3 sin cos 3 2sin 36

= =

( ) x k

k

x k

.225

.26

= +

= +

V y nghi m c a pt l: x k x k k 5.2 ; .2 ,2 6

= + = +

0,25

0,25

0,25

c) x x x

x2

tan 13tan 5tan 8 0 8

tan3

=+ =

=

x k

x k k

48

arctan ,3

= +

= +

V y nghi m c a pt l: x k x k k 8; arctan ,4 3

= + = +

0,25

0,252 2.0

a) V l y ng u nhin 3 vin bi trong ti c 9 vin bi nn spt c a khng gian mu l:( )n C 39 84 = =

K hi u: A: 3 vin ly ra c hai vin bi mu xanhTa c: ( )n A C C 2 15 4. 40= =

V y xc sut c a bi n c A l: ( ) ( )( )n A

P An

40 10

84 21 = = =

0,25

0,5

0,25

b) K hiu: B: 3 vin ly ra c t nht 1 vin bi mu xanhTa c: B : C 3 vin bi ly ra u mu

( )n B C 34= ( ) ( )( )n A

P Bn

1

21 = =

V y xc sut c a bi n c B l: ( ) ( )P B P B 1 201 1 21 21= = =

0,5

0,5

3 2.0a)

Ta c: ( )( )n nn n

u unn1

1 1 1

2 12 1 1+

+ =

++ + 0,25



8/15

8

Q

R

I

P

N

M

C

A B

D

S

( )( )n n3

02 3 2 1

= >+ +

V y dy s nu( ) l dy tng.

0,5

0,25b) u u d 100 1 99 2008= + =

( )S u u100 1 10050 101800= + = 0,50,5

4 1,5a) Hnh v

Do BD//MN (t/c ng trung bnh)M: MN (MNP) nn BD//(MNP)

0,5

0,75

b) G i I MN BC =

Ta c: ( ) I BC I MNP BC I MN

0,75c) V ( ) ( )P MNP SBD v MN//BD nn (MNP) (SBD) l ng th ng d qua P v

song song v i BD.0,5

d) G i R SD d = . N i IP c t SC t i Q, n i RQ.Ta c: ( ) ( ) MNP ABCD MN =

( ) ( )( ) ( )( ) ( )( ) ( )

MNP SAB MP

MNP SBC PQ MNP SCD QR

MNP SDA RN

=

=

=

=

V y thi t di n c a hnh chp S.ABCD v i mp(MNP) l ng gic MPQRN 1,0

5 0.5

( ) ( )k

k k k k k k k T C x C x

x

12 12 12 41 12 123

12 . 1 .2 . .

+

= =

S h ng khng cha x c: k k 12 4 0 3 = = V y s h ng khng cha x trong khai trin trn l:( ) C 3 9 3121 .2 . 112640 =

0,25

0,25

s 4

THI H C K 1 Nm h cMn TON L p 11


I. PH N CHUNG CHO TT C H C SINH(7 i m):Cu I: (2,0 i m)



9/15

9

1) Tm tp xc nh c a hm s x y x

1 sin5

1 cos2

=

+.

2) C bao nhiu st nhin l c ba ch s khc nhau, trong ch s hng trm l ch s ch n?

Cu II: (1,5 i m) Gi i ph ng trnh: x x23sin2 2cos 2+ = .

Cu III: (1,5 i m) M t h p ng 5 vin bi xanh, 3 vin bi v 4 vin bi vng (chng chkhc nhau v mu). Chn ng u nhin 3 vin bi th p . Tnh xc sut c:1) Ba vin bi ly ra 3 mu khc nhau.2) Ba vin bi ly ra c t nht m t vin bi mu xanh.

Cu IV: (2,0 i m) Trong mt ph ng t a Oxy cho vect v (1; 5)=

, ng th ng d : 3x + 4y 4 = 0 v ng trn (C) c ph ng trnh (x + 1)2 + (y 3)2 = 25.1) Vi t ph ng trnh ng th ng d l nh c a d qua php tnh ti n theo vect v

.

2) Vi t ph ng trnh ng trn (C) lnh c a (C) qua php vt tm O t s k = 3.

II. PH N DNH RING CHO HC SINH T NG BAN(3 i m):Th sinh ch c ch n m t trong hai ph n: Theo ch ng trnh Chu n ho c Nng cao1. Theo ch ng trnh Chu n

Cu V.a: (1,0 i m) Tm c p s c ng (un) c 5 s h ng bi t: u u uu u2 3 5

1 5

410

+ =

+ = .

Cu VI.a: (2,0 i m) Cho hnh chp S.ABCD cy ABCD l hnh bnh hnh. Gi M l trungi m c ac nh SA.1) Xc nh giao tuyn d c a hai mt ph ng (MBD) v (SAC). Chng t d song song v i m t ph ng(SCD).2) Xc nh thi t di n c a hnh chp ct b i m t ph ng (MBC). Thit di n l hnh g ?

2. Theo ch ng trnh Nng caoCu V.b: (2,0 i m) Cho t di n ABCD. Gi M, N ln l t l trungi m c a cc c nh AB, AD; P l mt

i m trn cnh BC (P khng trng v i i m B v C) v R li m trn cnh CD sao cho BP DR BC DC .

1) Xc nh giaoi m c a ng th ng PR v mt ph ng (ABD).2) nhi m P trn cnh BC thi t di n c a t di n v i m t ph ng (MNP) l hnh bnh hnh.

Cu VI.b: (1,0 i m) Tm s nguyn d ng n bi t: n 0 n 1 1 n 2 2 n 1n n n nC C C C 203 3 3 3 2 1 + + + + = .

(trong k nC l s t h p ch p k c a n ph n t )

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

s 4

P N THI H C K 1 Nm h cMn TON L p 11


Cu Ni dung i mI (2,0 i m)

1 Tm TX c a hm s x

y x

1 sin5

1 cos2

=

+. 1,0i m

Ta c: sin5x 1 1 sin5x 0 x (do x1 sin5 c ngh a) 0,25Hm s xc nh x1 cos2 0 + xcos2 1 0,25



10/15


11/15

11

pt c a d: 3x + 4y + C = 0 (C 4) (0,25)L y i m M(0; 1) d, g i M l nh c a M qua

vT . Ta c: M(1;4)

d. Thay ta i m M vo pt ca d, ta c C = 13. (0,50)V y pt d: 3x + 4y + 13 = 0. (0,25)

(1,0 i m)

2 Vi t ph ng trnh ng trn (C') l nh c a (C) qua V (O, 3) 1,0i m(C) c tm I(1; 3), bn knh R = 5. 0,25

G i I'(x; y) l tm v R' l bn knh ca (C'). Ta c: R' = |k|R = 3.5 = 15; 0,25 OI OI ' 3=

, I '(3; 9) 0,25V y (C') c pt: (x 3)2 + (y + 9)2 = 225. 0,25

V.a Tm c p s c ng (u n) c 5 s h ng bi t:u u uu u

2 3 5

1 5

410

+ =

+ = (*) 1,0i m

G i d l cng sai ca CSC (un). Ta c:(u d u d u d u u d

1 1 1

1 1

) ( 2 ) ( 4 ) 4(*)

( 4 ) 10

+ + + + =

+ + =

0,25

u d 2u d

1

1

44 10

=

+ = u d u d

1

1

42 5

=

+ = ud

1 13

=

= 0,50

V y c p s c ng l: 1; 2; 5; 8; 11. 0,25 VI.a (2,0 i m)

A

B C

D

S

M

O

N

0,25

1 Xc nh giao tuy n d c a hai m t ph ng (MBD) v (SAC). Ch ng t d // mp(SCD). 1,0i mTa c M mp(MBD); M SA M mp(SAC)Suy ra M l mt i m chung ca hai mp trn. 0,25Trong mp(ABCD), gi O l giaoi m c a AC v BD, ta c O li m chung

th hai c a hai mp trn. 0,25V y giao tuyn l ng th ng MO. 0,25Ta c d chnh l ng th ng MO, m MO // SC nn MO // mp(SCD). 0,25

2 Xc nh thi t di n c a hnh chp c t b i m t ph ng (MBC). Thi t di n lhnh g ? 0,75i m

Ta c M li m chung ca hai mp (MBC) v (SAD) 0,25BC (MBC); AD (SAD) v BC // AD nn giao tuyn c a hai mp ny l

ng th ngi qua M v song song v i AD c t SD t i N.0,25

V MN // BC nn thit di n c n tm l hnh thang BCNM(haiy l MN v BC). 0,25

V.b (2,0 i m)1 Xc nh giao i m c a ng th ng PR v mp(ABD). 1,0i m

Ch : Hnh v ct 02 li tr ln thkhng cho im phn hnh v.



12/15

12

C

B

D

A

M N

PQ

R

I

0,25

V BP DR BC DC

nn PR/ / BD. Trong mp (BCD), gi I = BD PR. 0,50Ta c: I PR v I BD, suy ra I mp(ABD). Vy PR mp(BCD) I = . 0,25

2 nh i m P trn c nh BC thi t di n c a t di n v i m t ph ng (MNP) lhnh bnh hnh. 1,0i m

Ta c MN (MNP); BD (BCD) v MN // BD. Do giao tuyn c amp(MNP) v mp(BCD) l ng th ngi qua P song song v i MN c t CDt i Q.

0,25

Thi t di n l hnh thang MNQP (MN // PQ). 0,25 thi t di n trn l hnh bnh hnh th PQ = MN = ( ) BD 0,25Suy ra PQ l ng trung bnh ca tam gic BCD, hay P l trungi m c a BC.

V y khi P l trungi m c a BC th thit di n l hnh bnh hnh.[ Ch : N u h c sinh ch ra trung i m sau c/m hnh bnh hnh th ch cho

2/: 0,75 i m.]

0,25

VI.bTm s nguyn d ng n bi t:

n 0 n 1 1 n 2 2 n 1n n n nC C C C

203 3 3 3 2 1 + + + + = (*) 1,0i m

Ta cn 0 n 1 1 n 2 2 n 1 n

n n n n nC C C C C 20

(*) 3 3 3 3 2

+ + + + + = 0,25

n n20 20(3 1) 2 4 2 + = = 2n 202 2 = 0,50n 10 = . V y n = 10 l gi trc n tm. 0,25

s 5



Bi 1(2,5i m) Gi i cc ph ng trnh :1) 2sin( 2x + 150 ).cos( 2x + 150 ) = 1 2) cos2x 3cosx + 2 = 0

3) x x x x

2 2sin 2sin 2 5cos 02sin 2

=+

Bi 2(0,75i m) Tm gi trl n nh t v gi trnh nh t c a hm s : y x x3sin 3 4 cos 36 6

= + + +

Bi 3(1,5i m)1) Tm h s c a s h ng ch a x31 trong khai trin bi u th c x x3 15(3 ) .2) T cc ch s 1, 2, 3, 4, 5, 6, 7 c thl p c bao nhiu s ch n c b n ch s khc nhau.

Bi 4(1,5i m) M t h p ch a 10 qu c u tr ng v 8 qu c u , cc qu c u ch khc nhau v mu. Lyng u nhin 5 qu c u.

Ch : Hnh v ct 02 li tr ln thkhng cho im phn hnh v.



13/15

13

1) C bao nhiu cch ly ng 3 qu c u .2) Tm xc sut l y c t nh t 3 qu c u .

Bi 5 ( 1,5 i m) Trong mt ph ng to Oxy, cho haii m A( 2; 3) , B(1; 4); ng th ng d: x y3 5 8 0 + = ; ng trn (C ): x y2 2( 4) ( 1) 4+ + = . G i B, (C) l n l t l nh c a B, (C) qua

php i x ng tm O. Gi d l nh c a d qua php tnh ti n theo vect AB

.1) Tm to c a i m B, ph ng trnh ca d v (C) .

2) Tm ph ng trnh ng trn (C ) nh c a (C) qua php vtm O t s k = 2.Bi 6(2,25i m) Cho hnh chp S.ABCD cy ABCD l mt hnh bnh hnh. Gi M, N ln l t ltrungi m c a SA, SD v P l mt i m thu c o n th ng AB sao cho AP = 2PB .1) Ch ng minh rng MN song song v i m t ph ng (ABCD).2) Tm giao tuyn c a hai mt ph ng (SBC) v (SAD).3) Tm giaoi m Q c a CD v i m t ph ng (MNP). Mt ph ng (MNP) ct hnh chp S.ABCD theom t thi t di n l hnh g ? .4) G i K l giaoi m c a PQ v BD. Chng minh rng ba ng th ng NK, PM v SB ng qui t im t i m.

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

s 5

P N THI H C K 1 Nm h cMn TON L p 11 Nng cao


Bi Cu H ng d n i m

1

12sin( 2x + 150 ).cos( 2x + 150 ) = 1 sin(4x +300) = 1 x k , k Z 0 0 04 30 90 360+ = + x k , k Z 0 015 .90 = +

0,5

2

cos2x 3cosx + 2 = 0 2cos2x 1 3cosx + 2 = 0 2cos2x 3cosx + 1 = 0

x k x

, k Z x k x

2cos 11 2cos

32

==

= +=

1

3

x x x

x

2 2sin 2sin 2 5cos0

2sin 2

=

+(1)

K : x m

x , m,n Z x n

22 4sin52 24

+

+(*)

V i i u ki n (*) ta c: (1) sin2x 4sinx.cosx 5cos2x = 0 cosx = 0 khng thomn ph ng trnh (1) cosx 0 , chia hai v c a (1) cho x2cos ta c:

(1) tan2x 4tanx 5 = 0 x x

tan 1tan 5

= =

x k x k

4arctan5

= +

= +

K t h p v i i u ki n (*), ta c nghi m c a ph ng trnh cho l:

x k , x k , k Z (2 1) arctan 54

= + + = +

1



14/15


15/15

15

6

1

KQ

I

P

NM

DA

BC

S

MN l ng trung bnh ca tam gic SAD .V MN nm ngoi mt ph ng (ABCD) v MN // AD nn MN // (ABCD).

0,75

2 Giao tuyn c a hai mt ph ng (SBC) v (SAD) l ng th ngi qua S vsong song v i AD . 0,25

3

3/ Tm giaoi m Q c a CD v i m t ph ng (MNP). Mt ph ng (MNP) ct hnhchp S.ABCD theo mt thi t di n l hnh g ? .

Ba m t ph ng (MNP), (SAD) v (ABCD) ct nhau theo ba giao tuyn MN, PQ,AD, ng th i MN //AD nn ba ng th ng PQ, MN, ADi m t songsong .

Trong mt ph ng (ABCD), quai m P k ng th ng song song v i AD, c tCD t i Q. i m Q l giaoi m c n tm.

0,75

4

Trong mt ph ng (SAB), hai ng th ng SB v PM khng song song nnchng ct nhau t i I .

Suy ra I li m chung ca hai mt ph ng (MNP) v (SBD) .L i c (SBD) v (MNP) ct nhau theo giao tuyn KN nni m I ph i thu c

ng th ng NK .V y ba ng th ng SB, MP, NK ng qui t i I .

0,5


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