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TRNG I HC VINH TRNG THPT CHUYN
KHO ST CHT LNG LP 12, LN 1 - NM 2014 Mn: TON; Khi: A v A1; Thi
gian lm bi: 180 pht
I. PHN CHUNG CHO TT C TH SINH (7,0 im) Cu 1 (2,0 im). Cho hm s 2
3.
1xyx
=
a) Kho st s bin thin v v th (H) ca hm s cho. b) Tm m ng thng : 3
0d x y m+ + = ct (H) ti hai im M, N sao cho tam gic AMN vung ti im
(1; 0).A
Cu 2 (1,0 im). Gii phng trnh sin3 2cos2 3 4sin cos (1 sin ).x x
x x x+ = + + + Cu 3 (1,0 im). Gii bt phng trnh 24 1 2 2 3 ( 1)(
2).x x x x+ + +
Cu 4 (1,0 im). Tnh tch phn 1
20
3 2ln(3 1) d .( 1)x xI x
x
+ +=
+
Cu 5 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh ch nht, mt bn
SAD l tam gic vung ti S, hnh chiu vung gc ca S ln mt phng (ABCD) l
im H thuc cnh AD sao cho 3 .HA HD= Gi M l trung im ca AB. Bit rng 2
3SA a= v ng thng SC to vi y mt gc 030 . Tnh theo a th tch khi chp
S.ABCD v khong cch t M n mt phng (SBC).
Cu 6 (1,0 im). Gi s x, y, z l cc s thc khng m tha mn 2 2 25( )
6( ).x y z xy yz zx+ + = + + Tm gi tr ln nht ca biu thc 2 22( ) (
).P x y z y z= + + +
II. PHN RING (3,0 im) Th sinh ch c 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 tam gic ABC c (2; 1)M l trung im cnh AC, im
(0; 3)H l chn ng cao k t A, im (23; 2)E thuc ng thng cha trung
tuyn k t C. Tm ta im B bit im A thuc ng thng : 2 3 5 0d x y+ = v im
C c honh dng.
Cu 8.a (1,0 im). Trong khng gian vi h ta ,Oxyz cho ng thng 2 1
2:1 1 2
x y zd + = =
v hai mt
phng ( ) : 2 2 3 0, ( ) : 2 2 7 0.P x y z Q x y z+ + + = + = Vit
phng trnh mt cu c tm thuc d, ng thi tip xc vi hai mt phng (P) v
(Q).
Cu 9.a (1,0 im). Cho tp hp { }1, 2, 3, 4, 5 .E = Gi M l tp hp tt
c cc s t nhin c t nht 3 ch s, cc ch s i mt khc nhau thuc E. Ly ngu
nhin mt s thuc M. Tnh xc sut tng cc ch s ca s bng 10.
b. Theo chng trnh Nng cao Cu 7.b (1,0 im). Trong mt phng vi h ta
,Oxy cho hai im (1; 2), (4; 1)A B v ng thng
: 3 4 5 0.x y + = Vit phng trnh ng trn i qua A, B v ct ti C, D
sao cho 6.CD =
Cu 8.b (1,0 im). Trong khng gian vi h ta ,Oxyz cho im (1; 1; 0)M
v hai ng thng
1 21 3 1 1 3 2
: , : .1 1 1 1 2 3
x y z x y zd d + = = = =
Vit phng trnh mt phng (P) song song vi 1d v 2d ng
thi cch M mt khong bng 6.
Cu 9.b (1,0 im). Tm s nguyn dng n tha mn 0 1 2 31 1 1 1 ( 1)
1
. . . .
2 3 4 5 2 156
nn
n n n n nC C C C Cn
+ + + =+
------------------ Ht ------------------
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TRNG I HC VINH TRNG THPT CHUYN
P N KHO ST CHT LNG LP 12, LN 1 - NM 2014 Mn: TON Khi A, A1; Thi
gian lm bi: 180 pht
Cu p n im a) (1,0 im) 10. Tp xc nh: \{1}.R 20. S bin thin: * Gii
hn ti v cc: Ta c lim 2
xy
= v lim 2.
xy
+=
Gii hn v cc: 1
limx
y+
= v 1
lim .x
y
= +
Suy ra th (H) c tim cn ngang l ng thng 2,y = tim cn ng l ng thng
1.x = * Chiu bin thin: Ta c 2
1' 0, 1.( 1)y xx= >
Suy ra hm s ng bin trn mi khong ( ); 1 v ( )1; .+
0,5
* Bng bin thin:
30. th:
th ct Ox ti 3 ; 0 ,2
ct Oy ti (0;3).
Nhn giao im (1; 2)I ca hai tim cn lm tm i xng.
0,5
b) (1,0 im) Ta c 1: .
3 3md y x= Honh giao im ca d v (H) l nghim ca phng trnh
2 3 1,
1 3 3x m
xx
=
hay 2 ( 5) 9 0, 1.x m x m x+ + = (1)
Ta c 2( 7) 12 0,m = + + > vi mi m. Suy ra phng trnh (1) c 2
nghim phn bit. Hn na c 2 nghim 1 2,x x u khc 1. Do d lun ct (H) ti
2 im phn bit 1 1 2 2( ; ), ( ; ).M x y N x y
0,5
Cu 1. (2,0 im)
Ta c 1 1 2 2( 1; ), ( 1; ).AM x y AN x y= =
Tam gic AMN vung ti A . 0.AM AN =
Hay 1 2 1 2( 1)( 1) 0x x y y + = 1 2 1 2
1( 1)( 1) ( )( ) 09
x x x m x m + + + =
21 2 1 210 ( 9)( ) 9 0.x x m x x m + + + + = (2)
p dng nh l Viet, ta c 1 2 1 25, 9.x x m x x m+ = = Thay vo (2)
ta c
210( 9) ( 9)( 5) 9 0m m m m + + + = 6 36 0 6.m m = = Vy gi tr ca
m l 6.m =
0,5
Cu 2. (1,0 im)
Phng trnh cho tng ng vi sin3 sin 2cos2 3(sin 1) cos (sin 1)x x x
x x x + = + + +
0,5
x
'y
y
+ 1
2
+ +
+
2
x O
y
I 3
2
1 32
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2
2cos2 sin 2cos2 (sin 1)(cos 3)(sin 1)(2cos2 cos 3) 0(sin 1)(4cos
cos 5) 0(sin 1)(cos 1)(4cos 5) 0.
x x x x x
x x x
x x x
x x x
+ = + +
+ =
+ =
+ + =
*) sin 1 2 ,2
x x kpi pi= = + .k Z
*) cos 1 2 ,x x kpi pi= = + .k Z *) 4cos 5 0x = v nghim. Vy phng
trnh c nghim 2 , 2 , .
2x k x k kpi pi pi pi= + = + Z
0,5
iu kin: 1.x Nhn thy 1x = l mt nghim ca bt phng trnh. Xt 1.x >
Khi bt phng trnh cho tng ng vi ( ) ( ) 3 24 1 2 2 2 3 3 2 12x x x x
x+ + +
( )
2
2
4( 3) 4( 3) ( 3)( 2 4)1 2 2 3 3
4 43 ( 1) 3 0. (1)1 2 2 3 3
x xx x x
x x
x xx x
+ + ++ + + +
+ +
+ + + +
0,5
Cu 3. (1,0 im)
V 1x > nn 1 0x + > v 2 3 1.x + > Suy ra 4 4 3,1 2 2 3
3x x
+ V (C) i qua A, B nn IA IB R= =
2 2 2 2
2 2
( 1) ( 2) ( 4) ( 1)3 6 ( ; 3 6)
10 50 65 10 50 65 (1)
a b a b Rb a I a a
R a a R a a
+ = + =
=
= + = +
0,5
Cu 7.b (1,0 im)
K IH CD ti H. Khi 9 29
3, ( , )5
aCH IH d I
+= = =
22 2 (9 29)9
25aR IC CH IH = = + = + (2)
T (1) v (2) suy ra 2
2 2(9 29)10 50 65 9 169 728 559 025
aa a a a
+ = + + =
14313
a
a
= =
(1; 3), 543 51 5 61
; ,13 13 13
I R
I R
=
=
Suy ra 2 2( ) : ( 1) ( 3) 25C x y + + = hoc 2 243 51 1525( ) :
.
13 13 169C x y + =
0,5
V ( )P // 1 2,d d nn (P) c cp vtcp 1 1 22
(1; 1; 1), (1; 2; 1)
( 1; 2; 3) Pu
n u uu
= = = =
Suy ra pt (P) c dng 2 0.x y z D+ + + =
( ) 33, ( ) 6 6 96DD
d M PD
=+ = =
=
( ) : 2 3 0 (1)( ) : 2 9 0 (2)P x y zP x y z
+ + + = + + =
0,5
Cu 8.b (1,0 im)
Ly 1(1; 3; 1)K d v 2(1; 3; 2)N d th vo cc phng trnh (1) v (2) ta
c ( ) : 2 3 0N P x y z + + + = nn 2 ( ) : 2 3 0d P x y z + + + = .
Suy ra phng trnh mt phng (P)
tha mn bi ton l ( ) : 2 9 0.P x y z+ + = 0,5
Vi mi x R v mi s nguyn dng n, theo nh thc Niutn ta c 0,5
I
H
A
B
C D
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( )0 1 2 1 0 1. . . ( 1) . . . ( 1) (1 ) .n n n n n n nn n n n n
nC x C x C x C C x C x x x x+ + + = + + = Suy ra ( )1 10 1 2 1
0 0
. . . ( 1) d (1 ) d .n n n nn n nC x C x C x x x x x+ + + =
Cu 9.b (1,0 im)
Hay 1 1
0 1 1
0 0
1 1 ( 1). . . (1 ) d (1 ) d
2 3 2
nn n n
n n nC C C x x x xn
+ + + =
+
1 1 11 2 ( 1)( 2)n n n n= =+ + + + , vi mi
*.n N
T ta c 21 1 3 154 0 11( 1)( 2) 156 n n nn n = + = =+ + (v *).n
N
0,5
