1. S GIO DC V O TO HNG YN THI TH I HC LN 1 NM HC 2011-2012 TRNG
THPT MINH CHU Mn: TON; Khi: A, B Thi gian lm bi: 180 pht, khng k
thi gian pht A. PHN CHUNG CHO TT C CC TH SINH (7,0 im) 3x 2Cu I
(2,0 im). Cho hm s y , c th (C). x2 1. Kho st s bin thin v v th (C)
ca hm s. 2. Gi M l im bt k trn (C). Tip tuyn ca (C) ti M ct cc ng
tim cn ca (C) ti A v B. Gi I l giao im ca cc ng tim cn. Tm ta M sao
cho ng trn ngoi tip tam gic IAB c din tch nh nht.Cu II (2,0 im). x
x x 1. Gii phng trnh 1 sin sin x cos sin 2 x 2 cos 2 . 2 2 4 2 3x 2
y 4 x y 5 2. Gii h phng trnh 2 y2 x; y . 2 x 5 y 0 x 2 3 x ( x sin
x)sin xCu III (1,0 im). Tnh tch phn I dx . (1 sin x )sin 2 x 3Cu IV
(1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh vung tm O cnh a. Hnh
chiu ca nh S trn mt phng(ABCD) l trung im H ca cnh AD, gc gia hai
mt phng (SAC) v (ABCD) bng 600. Tnh th tch ca khi chp S.HABCv khong
cch t H n mt phng (SBC).Cu V (1,0 im). Tm m sao cho h phng trnh sau
c bn nghim thc phn bit x 3 6 x 3x 2 y 3 3 y 4 2 2 m x 4 y 2 y 3 5 x
8 y 32 B. PHN RING (3,0 im). Th sinh ch c lm mt trong hai phn (phn
1 hoc 2)1. Theo chng trnh ChunCu VI.a (2,0 im). 1. Trong mt phng vi
h ta Oxy , cho tam gic ABC bit A(5; 2). Phng trnh ng trung trc cnh
BC, ng trung tuyn CC ln lt l x + y 6 = 0 v 2x y + 3 = 0. Tm ta cc
nh ca tam gic ABC. 2. Trong khng gian vi h ta Oxyz, hy xc nh to tm
v bn knh ng trn ngoi tip tam gic ABC, bit ta cc nh A 1; 0;1 , B 1;
2; 1 , C 1; 2;3 . 1 8 3 x 1 log 2 3x 1 1 Cu VII.a (1,0 im). Cho
khai trin 2 log 2 9 7 2 5 . Hy tm cc gi tr ca x bit rng s hng th 6
trong khai trin ny l 224.2. Theo chng trnh Nng cao Cu VI.b (2,0
im). 1. Trong mt phng Oxy cho tam gic ABC cn ti nh C. Bit phng trnh
ng thng AB l x y 2 0 ; trng tm 14 5 65 ca tam gic l G ; v din tch
ca tam gic bng (vdt). Vit phng trnh ng trn ngoi tip tam gic 3 3 2
ABC. 2. Trong khng gian vi h ta Oxyz, cho ba im A(0;1;2),
B(2;-2;1), C(-2;0;1). Vit phng trnh mt phng (ABC) v tm im M thuc mt
phng 2x + 2y + z 3 = 0 sao cho MA = MB = MC. x x y2 y Cu VII.b (1,0
im). Gii h phng trnh x; y . log 2 y x y 1
------------------Ht-----------------Th sinh khng c s dng ti liu.
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danh.......................................................
2. S GD V T HNG YN HNG DN CHM THI TH I HC LN 1 NM TRNG THPT
MINH CHU 2011-2012 Mn: TON-khi A-BPhn im p nchungCu I 1.(1.5 im)(2
im) *Tp xc nh: R{-2} *S bin thin 4 -Chiu bin thin: y 0 x-2 0,25 ( x
2) 2 Hm s ng bin trn cc khong (-;-2) v (-2;+) -Cc tr: hm s khng c
cc tr -Gii hn v tim cn: lim y lim y 3 y=3 l tim cn ngang ca th 0,25
x x lim y ; lim y x=2 l tim cn ng ca th x 2 x 2 Bng bin thin x - -2
+ y + + 0,25 + 3 y - 3 * th: x=0y=1 2 y=0x=- 3 y f(x)=(3x+2)/(x+2)
8 x=-2 7 y=3 6 5 0,25 4 3 2 1 x -7 -6 -5 -4 -3 -2 -1 1 O2 3 4 -1 -2
-3 2. (1 im) 3a 2 Gi M (a; ) (C ), a 2 a2 Phng trnh tip tuyn ca (C)
ti M l: 0,25 4 3a 2 y 2 ( x a) () (a 2) a2 ng thng d1:x+2=0 v
d2:y-3=0 l hai tim cn ca th 3a 2 0,25 d1=A(-2; ) , d2=B(2a+2;3)
a2
3. Tam gic IAB vung ti I AB l ng knh ca ng trn ngoi tip tam gic
IAB AB 2 64 0,25 din tch hnh trn S= 4( a 2) 2 8 4 4 (a 2) 2 16 a 0
Du bng xy ra khi v chi khi (a 2)2 2 ( a 2) a 4 0,25 Vy c hai im M
tha mn bi ton M(0;1) v M(-4;5) 1.(1 im) 1 cos( x) x x 2 2 Phng trnh
1 sin . sin x cos . sin x 2 2 2 2 0,25 x x 1 sin . sin x cos . sin
2 x 1 sin x 2 2 x x sin x.(sin cos .sin x 1) 0 2 2 sin x 0 x k , k
Z 0,25 x sin cos x .sin x 1 0 (*) 2 2 x x x x x x (*) sin 2 sin .
cos 2 1 0 sin 2 sin .(1 sin 2 ) 1 0 2 2 2 2 2 2 0,25 x x 2 sin 3
sin 1 0 2 2 x t sin t ,1 t 1 2 Cu II x 0,25 Ta c phng trnh:
2t2-t-1=0t=1 sin 1 x k 4 , k Z (2 im) 2 Vy phng trnh cho c nghim
x=k,kZ 2.(1 im) iu kin x>0 2 x 2 5 xy 2 y 2 0 2 x y x 2 y 0 0,5
3x 2 y 4 x y 5 3 x 2 y 4 x y 5 y 2x 3x 2 y 4 x y 5 0.25 x 2 y 3x 2
y 4 x y 5 y 2 x x 6 x 5(l ) y 1 Vy h c nghim duy nht (2;1) x 2 y y
0 x 2 0,5 4 y 9 y 5 Cu 2 x ( x sin x )sin x 2 x (1 sin x ) sin 2
xIII I 3 dx 3 dx(1,0 3 (1 sin x)sin 2 x 3 (1 sin x)sin 2 x) 2 2 3 x
dx 0,25 2 dx 3 3 sin x 3 1 sin x
4. u x du dx + t dx dv sin 2 x v cot x 2 2 2 2 0,25 x 3 3 sin x
2 dx x cot x| 3 3 cot xdx x cot x ln sin x | 3 3 3 3 3 2 2 2 dx dx
dx 3 3 3 1 sin x 2 x 3 3 1 cos x 3 2cos 4 2 0,25 2 2 x tan | 3 3 2
4 2 3 0,25 Vy I 32 3Cu IV: Hnh khng gian S K A I B H C O J Dng HI
AC => SI AC (nh l 3 ng vung gc) SIH 600 0,25 D SH a 2 a 6 Xt SHI
c tan600 = SH HI .tan 600 . 3 HI 4 4 a a .a ( AH BC ). AB 2 3a 2 S
HABC 2 2 4 0,25 1 1 3a 2 a 6 a 3 6 VS .HABC .S HABC .SH . . 3 3 4 4
16 * Tnh khong cch t H n (SBC) Gi J l trung im ca BC 0,25 Dng HK SJ
=> HK (SBC) => d(H; (SBC)) = HK 1 1 1 1 1 8 1 11 Ta c: 2 2 2
2 2 2 2 2 HK SH HJ a .6 a 3a a 3a 16 a 3 a 33 a 33 0,25 => HK =
. Vy d(H;(SBC)) = 11 11 11V Tm m sao cho h phng trnh sau c 4 nghim
thc phn bit: x 3 6 x 3x 2 y 3 3 y 4 (1) 2 2 m( x 4) y 2 y 3 5 x 8 y
32 (2)
5. (1) ( x 1)3 3( x 1) y 3 3 y 0,25 ( x 1) y ( x 1) 2 ( x 1) y
y 2 3 0 x y 1 (3)Thay (3) vo (2) ta c: m( x 4) x 2 2 5 x 2 8 x 24
m( x 4) x 2 2 ( x 4)2 4( x 2 2) 0,25 x4 x2 2 m (4) do x 4 KTM x2 2
x4 x4 2 4xt y (*) y 0 x 1/ 2 x2 2 ( x 2 2)3lim y 1; lim y 1x x Lp
bng bin thin 0,25 x - 1/2 + y + 0 - y 3 -1 1suy ra 1 y 3 v (*) c 2
nghim phn bit y 1;3 4 PT (4) theo y: m y (5) y 4 4 Xt hm s f ( y )
y y 1;3 => f ( y ) 1 2 0 y 2 y y lim y ; lim y 0,25 x 0 x0Lp bng
bin thin x -1 0 1 2 3 y - - 0 + y -5 + 13/3 5 - 4 13 KL: ycbt PT
(5) c 2 nghim phn bit y 1;3 m 4; 3 B- Theo chng trnh nng caoCu VI.b
1. Vit phng trnh ng trn.... C Gi H l trung im ca AB CH AB CH c phng
trnh: x-y-3=0 0,25 G. 5 1 H CH AB H ; 2 2 A H B CG 2GH C (9; 6) t
A(a;2-a) B( 5-a; a-3) 13 13 AB (5 2a; 2a 5); CH ; 2 2 65 1 65 a 0
Theo gt th S ABC AB.CH 8a 2 40a 0 2 2 2 a 5 0,25 * a = 0 A 0; 2 ; B
5; 3 * a = 5 A 5; 3 ; B 0; 2 .
6. ng trn cn tm c phng trnh dng: x 2 y 2 2ax 2by c 0 (a 2 b 2 c
0) Do ng trn i qua A, B, C nn ta c h: 0,25 4b c 4 a 137 / 26 10a 6b
c 34 b 59 / 26 18a 12b c 117 c 66 / 13 137 59 66 Vy ng trn cn tm c
pt: x 2 y 2 x y 0 0,25 13 13 13VI.b 2 .Vit phng trnh mt phng (ABC)
v tm im M thuc mt phng 2x + 2y + z 3 = 0 sao cho MA = MB = MC. Ta c
AB (2; 3; 1), AC (2; 1; 1) n (2; 4; 8) l 1 vtpt ca (ABC) 0.25 Suy
ra pt (ABC) l (x 0) + 2(y 1) 4(z 2) = 0 hay x + 2y 4z + 6 = 0 0.25
M(x; y; z) MA = MB = MC . 0.25 M thuc mp: 2x + 2y + z 3 = 0 nn ta c
h, gii h c x = 2, y = 3, z = -7 0.25 x 0 0.25VII b K: y x 0 1 1T
phng trnh x x y 2 y ta c ( x )2 ( y )2 2 2 0.25 x y 1 x y * x y 1
thay vo log 2 ( y x ) y 1 ta c y = 1 suy ra x = 0 0.25* x y vy y 0
suy ra y- x
