archive-ThiPDF-KiemTraMonHoc-201506-20150603-thumb11_660370845
description
Transcript of archive-ThiPDF-KiemTraMonHoc-201506-20150603-thumb11_660370845
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Trang 1/5
S GIO DC V O TO H NI TRNG THPT CHU VN AN
P N THANG IM
THI TH K THI THPT QUC GIA NM 2015 Mn: TON
CHNH THC
(p n thang im c 5 trang)
Cu Ni dung im
1
(2,0 )
a) (1,0 im)
* Tp xc nh : D = IR\{-2}.
* S bin thin ca hm s
- Chiu bin thin: 2
3y' 0, x
(x 2)
D.
- Hm s ng bin trn cc khong ( ; 2),( 2; ) .
0,25
- Gii hn v tim cn: limx
y
, limx
y
, ( )
lim
x
y , ( )
lim
x
y .
th )(C nhn ng thng y = 2 lm ng tim cn ngang
v nhn ng thng x = -2 lm ng tim cn ng.
- Cc tr: Hm s khng c cc tr.
0,25
- Bng bin thin:
x - -2
y + +
y
+ 2
2 -
0,25
* th )(C :
0,25
b) (1,0 im)
Gi 0 0M(x ;y ) l tip im ca tip tuyn d vi th (C). Khi : 0y'(x ) 3 0,25
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Ta c phng trnh 020200
x 133 (x 2) 1
x 3(x 2)
0,25
Phng trnh tip tuyn d ca th (C) ti cc im (-1; -1) v (-3; 5) ln lt l: y = 3x +2; y = 3x +14
0,25
T gi thit ta c: y = 3x +2 0,25
2
(1,0 )
a) (0,5 im)
Ta c:
3 1 2 4
A . cos2x cos 2x cos 2x2 2 3 3
0,25
3 1 3 1 3 cos2x 2cos( 2x)cos cos2x cos2x2 2 3 2 2 2
0,25
b) (1,0 im)
K: 1
x ;x 32
vi iu kin , phng trnh tng ng vi:
2 2 2x 3
4log x 3 4log 2x 1 4 log 12x 1
0,25
x 3 x 3 4x 22 x 3 4x 2 x 1
x 3 4x 22x 1
Vy phng trnh c nghim x =1
0,25
3
(1,0 )
I = (x x.sinx) sin sin
xdx x xdx x xdx .
0,25
Tnh 10
I xsin xdx
t u x du dx
dv sin xdx v cosx
0,25
1 0
0
I x cos x cos xdx sinx0
0,25
Vy 3
I3
. 0,25
4
(1,0 )
a) (0,5 im)
Gi biaz ( Rba , ).
Ta c: 2(z 1) 3z i(5 i) 2(a bi 1) 3(a bi) 1 5i a 1 5(1 b)i 0 0,25
a 1
b 1
. Vy z 2 0,25
b) (0,5 im)
Gi X l bin c: Chia 20 bn thnh 4 nhm A, B, C, D, mi nhm c 5 bn sao cho 5 bn n thuc cng mt nhm
0,25
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Ta c: 5 5 5 520 15 10 5C .C .C .C cch chia 20 bn thnh 4 nhm A, B, C, D
Xt 5 bn n thuc nhm A, c 5 5 515 10 5C .C .C cch chia cc bn nam vo 3 nhm cn li
Do vai tr ca cc nhm nh nhau, c 4 5 5 515 10 5C .C .C cch chia cc bn vo cc nhm a,
B,C ,D trong c 5 bn n thuc mt nhm.
Xc sut cn tm l 520
4 1P
3876C .
0,25
5
(1,0 )
Xt tam gic ABC c
0
2ABC
BC AB.tan 60 2a 3
2a 3S
0,25
2 3ABCS.ABCD
1 1SA.S .a 3.2a 3 2a
3 3V 0,25
Gi N l trung im ca SA. Do SB // (CMN) nn d(SB,CM) = d(SB, (CMN)) = d(B, (CMN)) = d(A, (CMN))
K AE MC,E MC v k AH NE,H NE
Chng minh c: AH (CMN) d A,(CMN) AH
0,25
Tnh AE = AMC2S
MC
, trong :
2AMC
1 1 3S AM.AC.sinCAM a.4a. a 3
2 2 2
MC = a 13
Suy ra: AE = 2a 3
13
Tnh c AH = 2a 3
29
2a 3d(SB;CM)
29
0,25
6
(1,0 )
Do 0 0AIB 90 ACB 45 hoc 0 0ACB 135 ACD 45 tam gic ACD vung cn ti D nn DA = DC Hn na, IA = IC
Suy ra, DI AC ng thng AC tha
mn iu kin: AC qua im M v AC vung gc vi ID
0,25
Vit phng trnh ng thng AC: x 2y + 9 = 0 0,25
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Gi A(2a 9; a) AC . Do DA 2d(D,AC) 2 10 nn
2 2 2a 1 A( 7;1)
2a 8 a 1 2 10 a 6a 5 0a 5 A(1;5)
Theo gi thit bi cho ta c A(1; 5)
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Vit phng trnh ng thng DB: x + 3y + 4 = 0 Gi B(-3b -4; b)
Tam gic IAB vung ti I nn IA.IB 0 3( 3b 2) 4(b 1) 0 b 2 B(2; 2)
Vy A(1; 5) v B(2; -2)
0,25
7
(1,0 )
Mt cu (S) cn tm c tm I l trung im ca AB, vi I(2; 3; 0) 0,25
Bn knh ca (S) l: AB
R 32
Phng trnh ca (S) l: 2 2 2x 2 y 3 z 3 0,25
Gi M(0; 0; t) thuc Oz. Do MABCV 5 nn 1
AB,AC AM 5 11 4t 156
0,25
t 1 M(0;0;1)11 4t 15
11 4t 15 13 1311 4t 15 t M(0;0; )
2 2
0,25
8
(1,0 )
K: x 1 Vi iu kin ,
2 2 2
2
2 22 2 2
2
8 2BPT 6(x 2) 2 x x 6 x x 1 0
x x 1
4 23 x 1 x x x 1 2 x x 5 0
x x 1
0,25
Xt hm s 4 2
f (t) t 5t 1
vi t 0 . Ta c:
2 2f '(t) 1
t 1 t 1
f '(t) 0 t 1
Bng xt du: x 0 1 + f(x) - 0 +
Suy ra: f (t) f (1) t 0; ) f (t) 0 t [0; ) Du = xy ra t 1
0,25
Do 2 2
2
4 2x x 0 x [0; ) x x 5 0, x [0; )
x x 1
Du = xy ra khi 21 5
x x 1 x2
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Khi ,
2 22 2 2
2
2
2
2
2
4 23 x 1 x x x 1 2 x x 5 0
x x 1
x 1 x 0
1 5x x 1 0 x
24 2
x x 5 0
x x 1
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Vy tp nghim ca bt phng trnh cho l: 1 5
S [1; ) \{ }2
9
(1,0 )
Ta c: 2(x + y) = z(xy 7). Do x, y, z l cc s dng nn xy 7 > 0
Khi , t gi thit ta c z = 2(x y)
xy 7
Suy ra: 4(x y)
S f (x, y) 2x yxy 7
vi iu kin x > 0, y > 0, xy > 7 (*)
0,25
Vi mi x c nh, ta xt o hm ca hm s f(x, y) theo n y ta c: 2
'y 2 2
4(xy 7) 4x(x y) 28 4xf (x, y) 1 1
(xy 7) (xy 7)
' 2 2 2y 0 2
7 7f (x,y) 0 x y 14xy 21 4x 0 y 2 1
x x
Suy ra: 0 211 7
f (x;y ) 2x 4 1x x
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Xt hm s 2
11 7g(x) 2x 4 1
x x vi x > 0
23
2
11 28g '(x) 2
7xx 1
x
g'(x) 0 x 3
Khi : g(x) g(3) g(x) 15
0,25
Vi iu kin (*), ta c: 0S f (x,y ) g(x) 15 Vy minS= 15 khi x = 3, y = 5, z = 2
0,25
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