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8/8/2019 DaToanDCt_DH_K10
1/4
Trang 1/4
B GIO DC V O TO
CHNH THC
P N THANG IM THI TUYN SINH I HC NM 2010
Mn: TON; Khi D(p n - thang im gm 04 trang)
P N THANG IM
Cu p n im
1. (1,0 im)
Tp xc nh: R.
S bin thin:
- Chiu bin thin: 'y = 4x3 2x= 2x(2x2+ 1); 'y (x) = 0 x= 0.
0,25
- Hm sng bin trn khong (; 0); nghch bin trn khong (0; +).
- Cc tr: Hm st cc i ti x= 0; yC= 6.
- Gii hn: limx
y
= limx
y +
=.0,25
- Bng bin thin:
0,25
th:
0,25
2. (1,0 im)
Do tip tuyn vung gc vi ng thng y=1
6
x 1, nn tip tuyn c h s gc bng 6. 0,25
Do , honh tip im l nghim ca phng trnh 4x3 2x = 6 0,25
x= 1, suy ra ta tip im l (1; 4). 0,25
I(2,0 im)
Phng trnh tip tuyn: y= 6(x 1) + 4 hay y = 6x+ 10. 0,25
1. (1,0 im)
Phng trnh cho tng ng vi: 2sinxcosx cosx (1 2sin2x) + 3sinx 1 = 0 0,25
(2sinx 1)(cosx+ sinx+ 2) = 0 (1). 0,25
Do phng trnh cosx+ sinx+ 2 = 0 v nghim, nn: 0,25
II(2,0 im)
(1) sinx =1
2x=
6
+k2 hoc x =5
6
+k2 ( kZ). 0,25
'y + 0
y6
x 0 +
y
x
6
2 2
O
8/8/2019 DaToanDCt_DH_K10
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Trang 2/4
Cu p n im
2. (1,0im)
iu kin: x 2.
Phng trnh cho tng ng vi: ( )( )32 24 4 42 2 2 2 0
xx x+ = .0,25
24x 24 = 0 x = 1. 0,25
2 22 x + 3 42x = 0 2 2x + =x3 4 (1).
Nhn xt:x 3 4 .0,25
Xt hm sf(x) = 2 2x + x3 + 4, trn )3 4 ; + .
'f (x) =1
2x + 3x2 < 0, suy ra f(x) nghch bin trn )3 4 ; + .
Ta c f(2) = 0, nn phng trnh (1) c nghim duy nhtx= 2.
Vy phng trnh cho c hai nghim: x= 1; x= 2.
0,25
I =1
32 ln d
e
x x xx
=
1
2 ln d
e
x x x 1
ln3 d
ex
xx . 0,25
t u= lnx v dv= 2xdx, ta c: du=dx
xv v =x2.
1
2 ln d
e
x x x = ( )21
lne
x x 1
d
e
x x =e2
2
12
ex
=2 1
2
e +.
0,25
1
lnd
ex
xx
= ( )1
ln d ln
e
x x =2
1
1ln
2
e
x =1
2. 0,25
III(1,0 im)
Vy I =2
2
e 1. 0,25
M l trungim SA.
AH=2
4
a, SH= 2 2SA AH =
14
4
a.
0,25
HC=3 2
4
a, SC= 2 2SH HC + = a 2 SC=AC.
Do tam gic SACcn ti C, suy ra Ml trung im SA.
0,25
Thtch khi tdin SBCM.
Ml trung im SA SSCM=1
2SSCA
VSBCM = VB.SCM= 12
VB.SCA= 12
VS.ABC
0,25
IV(1,0 im)
VSBCM =1
6SABC.SH=
3 14
48
a. 0,25
iu kin: 2 x 5.
Ta c (x2+ 4x+ 21) (x2+ 3x+ 10) =x+ 11 > 0, suy ray> 0.0,25
y2= (x+ 3)(7 x) + (x+ 2)(5 x) 2 ( 3)(7 )( 2)(5 )x x x x+ +
= ( )2
( 3)(5 ) ( 2)(7 )x x x x+ + + 2 2, suy ra:0,25
y
2 ; du b
ng x
y ra khi v ch
khi x
=
1
3 . 0,25
V(1,0 im)
Do gi tr nh nht cay l 2 . 0,25
S
CD
BA
M
H
8/8/2019 DaToanDCt_DH_K10
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Trang 3/4
Cu p n im
1.(1,0im)
ng trn ngoi tip tam gicABCc phng trnh:
(x+ 2)2+y2 = 74.
Phng trnhAH:x= 3 vBCAH, suy ra phng trnhBCc dng:y=a (a 7, doBCkhng i quaA).
Do honh B, Ctha mn phng trnh:
(x+ 2)2
+a2
= 74 x2
+ 4x+a2
70 = 0 (1).
0,25
Phng trnh (1) c hai nghim phn bit, trong c t nht
mt nghim dng khi v ch khi: | a | < 70 .
Do Cc honh dng, nnB( 2 274 a ; a) v C( 2 + 274 a ; a).
0,25
ACBH, suy ra: .AC BH
= 0
( )274 5a ( )274 5a + + (a+ 7)( 1 a) = 0 a2+ 4a 21 = 0
0,25
a= 7 (loi) hoc a= 3 (tha mn).
Suy ra C( 2 + 65 ; 3).0,25
2. (1,0 im)
Ta c vectphp tuyn ca (P) v (Q) ln lt l
Pn
= (1; 1; 1) v Qn
= (1; 1; 1), suy ra:
,P Qn n
= (2; 0; 2) l vectphp tuyn ca (R).
0,25
Mt phng (R) c phng trnh dng xz+D= 0. 0,25
Ta c d(O,(R)) = ,
2
Dsuy ra:
2
D= 2 D= 2 2 hocD= 2 2 . 0,25
VI.a(2,0 im)
Vy phng trnh mt phng (R): xz+ 2 2 = 0 hocxz 2 2 = 0. 0,25
Gi z = a+bi, ta c: 2 2z a b= + v z2 =a2 b2 + 2abi. 0,25
Yu cu bi ton tha mn khi v ch khi:2 2
2 2
2
0
a b
a b
+ =
= 0,25
2
2
1
1.
a
b
=
= 0,25
VII.a(1,0 im)
Vy cc s phc cn tm l: 1 +i; 1 i; 1 +i; 1 i. 0,251. (1,0 im)
Gi ta Hl (a; b), ta c: 2 2 2( 2)AH a b= + v khong cch
tHn trc honh l | b |, suy ra: a2 + (b 2)2=b2.0,25
DoHthuc ng trn ng knh OA, nn: a2+ (b 1)2= 1. 0,25
T, ta c:2
2 2
4 4 0
2 0.
a b
a b b
+ =
+ =
Suy ra: (2 5 2; 5 1)H hoc ( 2 5 2; 5 1)H .
0,25
VI.b(2,0 im)
Vy phng trnh ng thng l
( 5 1) 2 5 2 0x y = hoc ( 5 1) 2 5 2 0x y + = .0,25
I
A
B C
H
O
H
y
x
A
P Q
R
O
8/8/2019 DaToanDCt_DH_K10
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Trang 4/4
Cu p n im
2. (1,0 im)
Ta c: + M1, nn M(3 +t; t; t).
+ 2i quaA(2; 1; 0) v c vectch phng v
= (2; 1; 2).0,25
Do : AM
= (t+ 1; t 1; t); ,v AM
= (2 t; 2; t 3). 0,25
Ta c: d(M, 2) =
,v AM
v
=
22 10 17
3
t t +, suy ra:
22 10 17
3
t t += 1 0,25
t2 5t+ 4 = 0 t= 1 hoc t= 4.
Do M(4; 1; 1) hoc M(7; 4; 4).0,25
iu kin: x> 2,y> 0 (1). 0,25
T h cho, ta c:2 4 2 0
2
x x y
x y
+ + =
= 0,25
2 3 0
2
x x
y x
=
=
0
2
x
y
=
=
hoc3
1.
x
y
=
=
0,25
VII.b(1,0 im)
i chiu vi iu kin (1), ta c nghim ca h l (x;y) = (3; 1). 0,25
------------- Ht -------------
M
2
1
d =1
H