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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 .
0,25
0,25 0,25
0,25
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=
0,25
0,5 0,25
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.
0,25
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 + = .
0,5
0,25
0,25
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
0,25 0,5 0,25
0,25 0,25
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.
0,5
0,5 0,5
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
0,5 0,5 0,5
0,5
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 = +
-----------------------------------------------------HT-----------------------------------------------------
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
+ + = + + = =
=
=
= +
= +
0,25
0,25
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
0,25
0,25
0,25
0,25
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
0,25 0,25
0,25
0,25
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.
0,5
0,25
0,25
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= = =
0,25 0,25
0,25
0,25
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
0,25
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
0,25
0,25 0,25 0,25
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.
0,25
0,25
0,25 0,25
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= = = =+ +
0,25
0,25
0,25
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
= + = + =
0,25
0,25
0,25
0,25
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 =
0,25
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.
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.
-----------------------------------------------------HT-----------------------------------------------------
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)
0,25
0,25
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
0,25
0,25
0,25
0,25
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
0,25 0,25
0,25
0,25
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
0,25
0,5
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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.
0,25
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 = =
0,25
0,25 0,25
0,25
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 = =
0,25
0,25
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 =
0,25
0,25
0,25
0,25
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
0,25
0,25
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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
0,25
0,25
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.
0,25
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0,25
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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+ + + = + + + =
0,25
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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