thi thu dai hoc lan 3 mon toan chuyen lqd dien bien
TRANSCRIPT
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8/9/2019 Thi Thu Dai Hoc Lan 3 Mon Toan Chuyen Lqd Dien Bien
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TRNG THPT CHUYNL QU NT : Ton - Tin
THI TH I HC LN III NM HC 2009 -2010Mn : Ton Khi: A+B
Thi gian lm bi : 180 pht (khng k thi gian pht )
BI
Cu 1: (2 im)Cho hm s 4 22 1 (1) y x mx m= - + + ( m l tham s)
1. Kho st s bin thin v v th hm s (1) khi m = 1.2. Xc nh m hm s (1) c 3 im cc tr ng thi cc im cc tr ca th hm s to thnh mt tam gic c bn knh ng trn ngoi tip bng 1.
Cu 2: (2 im)
1. Gii phng trnh: 8 8 21 1sin os cos 2 os22 2
x c x x c x- = -
2. Gii h phng trnh:4 2
4 3 0
0
log log
x y
x y
- + = - =
Cu 3: (3 im)1. Trong mt phng vi h trc ta Oxy cho A(4;3), ng thng (d) :xy2 = 0 v (d): x + y 4 = 0 ct nhau ti M. Tm ( ) ( ')B d v C d sao choA l tm ng trn ngoi tip tam gic MBC.
2. Trong khng gian cho hai ng thng :2
1 2 2
1
21
: 1 : 2
3 0
x tx
d y v d y t
z t z
= += = = = + =
a. Chng minh rng d1, d2cho nhau v vung gc vi nhau.b. Lp phng trnh ngvung gc chung gia d1 v d2.3. Cho hnh chp S.ABCD c y ABCD l hnh ch nht vi AB=2; AD= 2 2 v
SA =2 vung gc vi mt phng (ABCD). Gi M, N ln lt l trung im ca AD vSC, I l giao im ca BM v AC. Chng minh rng mt phng (SAC) vung gc vimt phng (SMB). Tnh th tch ca khi t din ANIB.
Cu 4: (3 im)
1. Tnh tch phn:8
3
ln
1
x I dx
x=
+
2. Tm s hng khng cha x trong khai trin 3 2n
xx
+ bit rng: n +
tha mn : 6 7 8 9 8 23 3 2n n n n nC C C C C ++ + + = 3. Cho cc s thc x,y dng thay i tha mn: x2 + y2 = 1.Tm gi tr nh nht ca
biu thc:1 1
(1 )(1 ) (1 )(1 )P x yy x
= + + + + +
--------------Ht-------------
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8/9/2019 Thi Thu Dai Hoc Lan 3 Mon Toan Chuyen Lqd Dien Bien
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TRNG THPT CHUYNL QU NT : Ton - Tin
THI TH I HC LNIII NM HC 2009 -2010Mn : Ton Khi: D
Thi gian lm bi : 180 pht (khng k thi gian pht )
BI
Cu 1: (2 im)Cho hm s 4 22 1 (1) y x mx m= - + + ( m l tham s)
1. Kho st s bin thin v v th hm s (1) khi m = 1.2. Xc nh m hm s (1) c 3 im cc tr ng thi cc im cc tr ca th hm s to thnh mt tam gic c bn knh ng trn ngoi tip bng 1.
Cu 2: (2 im)
1. Gii phng trnh: 8 8 21 1sin os cos 2 os22 2
x c x x c x- = -
2. Gii h phng trnh:4 2
4 3 0
0log logx y
x y
- + = - =
Cu 3: (3 im)1. Trong mt phng vi h trc ta Oxy cho A(4;3), ng thng (d) :xy2 = 0 v (d): x + y 4 = 0 ct nhau ti M. Tm ( ) ( ')B d v C d sao choA l tm ng trn ngoi tip tam gic MBC.
2. Trong khng gian cho hai ng thng :2
1 2 2
1
21
: 1 : 2
3 0
x tx
d y v d y t
z t z
= += = = = + =
a. Chng minh rng d1, d2cho nhau v vung gc vi nhau.b. Lp phng trnh ngvung gc chung gia d1 v d2.
3. Cho hnh chp S.ABCD c y ABCD l hnh ch nht vi AB=2; AD= 2 2 vSA =2 vung gc vi mt phng (ABCD). Gi M, N ln lt l trung im ca AD vSC, I l giao im ca BM v AC. Chng minh rng mt phng (SAC) vung gc vimt phng (SMB). Tnh th tch ca khi t din ANIB.
Cu 4: (3 im)
1. Tnh tch phn:
2
ln ln(ln )e
e
x x
I dxx
+=
2. Tm snguyn dng n nh nht sao cho khai trin (1+x)nc t s hai h s lin tip
bng7
15.
3. Cho cc s thc x,y dng thay i tha mn: x2 + y2 = 1.Tm gi tr nh nht cabiu thc:
1 1(1 )(1 ) (1 )(1 )P x y
y x= + + + + +
--------------Ht-------------
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P N V THANG IM THI TH I HC LN III
Mn: Ton A,B- Nm hc: 2009 2010
Cu Ni dung im1 1 m=1 ta c y = x
4-2x
2+ 2
+ TX: D =
+ limx y = +
+ y=4x34x0
' 01
xy
x
== =
BBT
x - -1 0 1 + y - 0 + 0 - 0 +y +
1
2
1
+
Hm s nghch bin trn mi khong ( - ;-1) v (0;1)Hm s ng bin trn mi khong (-1;0) v ( 1; + )
0.5
Hm s t cc tiu ti cc im 1x = gi tr cc tiu ca hm s l ( 1) 1y = Hm s t cc i ti im 0x = gi tr cc i ca hm s l (0) 2y = 0.25
10
8
6
4
2
-2
-4
-15 -10 -5 5 10 15x 1-1-2 2
0.25
2 Ta c y = 4x34mx = 4x(x2m)
y = 02
0x
x m
= =
iu kin hm s c 3 cc tr : y=0 c 3 nghim phn bit
m > 0.
0.25
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Khi th hm s c 3 im cc tr2
2
(0; 1)
( ; 1)
( ; 1)
A m
B m m m
C m m m
- - + - - - + -
Ta thy
4
2 4 3
4 32
( ; ) 1os
. 1( ; )
AB AC m m
AB m m AB AC m m mc BAC
AB AC m m m AC m m
= = +
- - + - = = = + +- -
uuur uuur uuur
uuur
0.25
2
3
2 2
2sin 1 cos
1
( ) 0 2
m m BAC BAC
m
BC m m m
= - =+
= + + =
3
3
3
3 2
2 12
sin 2
1
112
1
2 1 0 ( 1)( 1) 0 1 5
2
BC m mR
A mm m
m
mR m
m
m m m m mm
+= = =
+
+
= =
= - + = - + - = - + =
0.5
2
2im
1Gii phng trnh:
8 8 21 1sin os cos 2 os22 2
x c x x c x- = -
PT
4 4 4 4 2
2
2
2
1 1(sin os )(sin os ) cos 2 os2
2 2
1 1os2 (1 sin 2 ) os2 ( os2 1)2 2
os2 (1 os 2 ) os2 ( os2 1)
os2 ( os 2 os2 ) 0
os2 0 4 2
os2 1
2
x c x x c x x c x
c x x c x c x
c x c x c x c x
c x c x c x
kx
c xk
c xx k
p p
pp
+ - = -
- - = -
- + = -
+ =
= + = = -
= +
0.25
0.25
0.5
2
Gii h phng trnh:4 2
4 3 0
0log log
x y
x y
- + = - =
iu kin : 4
2
0 1
10
log
log
x x
yy
0.25
Vi iu kin trn h cho tng ng vi:
2
4 2 44 2
4 3 0 4 3 0
log log loglog log
x y x y
x y yx y
- + = - + = = ==
0.25
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2
2
2 2
4 31
4 33
x yx y x y
yx y y y
y
= = - = = = = - =
0.25
Tp nghim ca h phng trnh l: ( ) ( ){ }1;1 ; 9;3S = 0.25
Cu3
1 Trong mt phng vi h trc ta Oxy cho A(4;3), ng thng (d) :xy2 = 0 v (d): x + y 4 = 0 ct nhau ti M. Tm ( ) ( ')B d v C d sao
cho A l tm ng trn ngoi tip tam gic MBC.M(3;1):
( ) ( ; 2) ( ') ( ';4 ')B d B t t C d C t t - -
A l tm ng trn ngoi tip2 2
2 2
6
' 2
MA AB t
t MA AC
= = ==
B(6;4) v C(2;2)
0.25
0.5
0.25
2Trong khng gian cho hai ng thng :
2
1 2 2
1
21: 1 : 2
3 0
x txd y v d y t
z t z
= += = = = + =
a. Chng minh rng d1, d2cho nhau v vung gc vi nhau.b. Lp phng trnh on vung gc chung gia d1 v d2.
1 2
1 2
1 2
1 2
1 2
(1;1;3) (2;0;0): d
(0;0;1) (1;2;0)
, ( 2;1;0) , . 3 0
(1; 1;3)
& o .
qua A qua Bta c d
VTCP u VTCP u
u u u u AB
AB
d d ch nhau
= - = - -
ur uur
ur uur
ur uur uuuruuur
0.25
1 1 2 2 2
2 2 1
11 1 1 1
2 2 2 22 2
(1;1;3 ) (2 ;2 ;0)
( 1;2 1; 3)
3. 0 3 0
11 4 2 0. 0
5
M t d N t t d
MN t t t
t MN d MN u MN u t
Ta c MN d t t t MN u MN u
+ +
+ - - -
= - ^ ^ = - - = ^ + + - = =^ =
uuur
uuur ur uuur ur
uuur uur uuur uur
0.25
(1;1;0)6 3
( ; ;0)11 25 5( ; ; 0)
5 5
M
MNN
-
uuur
0.25
ng vung gc chung MN c phng trnh:
61
5
31
5
0
x t
y t
z
= += -
=
0.25
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3 Cho hnh chp S.ABCD c y ABCD l hnh ch nht vi AB=2; AD= 2 2 vSA =2 vung gc vi mt phng (ABCD). Gi M, N ln lt l trung im ca ADv SC, I l giao im ca BM v AC. Chng minh rng mt phng (SAC) vunggc vi mt phng (SMB). Tnh th tch ca khit din ANIB.
j
H
I
S
A
B C
DM
N
0
0
1( . . )
2
90
90
( ) ( ) ( )
AM AB ABM BCA c g c do
AB BC
ABM BCA
ABM BAC BCA BAC
AIB MB AC
MB SA MB SAC SBM SAC
= =
=
+ = + =
= ^
^ ^ ^
V : V
0.5
Gi H l trung im ca AC. Ta c HN l ng trung bnh ca SACV
( ) ( ) 12
SA HN ABCD NH ABI NH ^ ^ = = 0.25
1 1 2 2. . . ( )3 6 9
NABI ABI V NH S NH AI BI dvtt = = =V 0.25
1Tnh tch phn:
8
3
ln
1
x I dx
x=
+
t8
8
3
3
ln
2 11
1(2 1ln ) 2 6ln8 4ln 3 2
x u dxdu
xdxdv
v xx
x I x x dx J
x
= = = = + +
+ = + - = - -
0.5
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Tnh
8
3
1x J dx
x
+= t
3 3 2
2 2
2 2
21 2
1 1
t tt x J tdt dt
t t= + = =
- - 0.253
3
2
2
1 1 1(2 ) (2 ln ) 2 ln 3 ln 2
1 1 1
20 ln2 6 ln3 4
t J dt t
t t t
I
-= + + = + = + -
- + +
= - -
0.25
2
Tm s hng khng cha x trong khai trin 3 2
n
xx
+ bit rng:
n +
tha mn : 6 7 8 9 8 23 3 2n n n n nC C C C C ++ + + =
k , 9n n+ 6 7 7 8 8 9 8
2
7 8 8 9 8
1 1 1 1 2
8 9 8
2 2 2
9 8 9 62 2 2 2
( ) 2( ) ( ) 2
2
2
6 9 15
n n n n n n n
n n n n n
n n n
nn n n n
gt C C C C C C C
C C C C C
C C C
C C C C n n
+
+ + + + +
+ + +
-+ + + +
+ + + + + =
+ + + =
+ =
= = - = =
0,25
Khi
( )15 30 515 15
3 3 615 15
0 0
2 22
n kk k
k k k
k k
x C x C xx x
- -
= =
+ = =
0,25
S hng khng cha x tng ng vi:30 5
0 66
kk
-= = 0,25
S hng khng cha x phi tm l:
6 6
122 320320C =
0,25
3 Cho cc s thc x,y dng thay i tha mn: x2 + y2 = 1.Tm gi tr nh nht cabiu thc:
1 1(1 )(1 ) (1 )(1 )P x y
y x= + + + + +
D
2 2
1 1 4 2 22 2 2 4
2 24 2 ( ) 4 3 2
( ) 2( )
B T c si
x yP x y x y x y
x y y x x y x y x y
x y
x y x y
= + + + + + + + + + + = + + + ++ + +
+ + + = +
+ +
Du =1
2x y = =
Vy1
min 4 3 22
P x y= + = =
0.5
0.25
0.25
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P N V THANG IM THI TH I HC LN IIIMn: Ton D- Nm hc: 2009 2010
Cu Ni dung im1 1 m=1 ta c y = x
4-2x
2+ 2
+ TX: D = + lim
xy
= +
+ y=4x34x0
' 01
xy
x
== =
BBTx - -1 0 1 + y - 0 + 0 - 0 +y +
1
2
1
+
Hm s nghch bin trn mi khong ( - ;-1) v (0;1)Hm s ng bin trn mi khong (-1;0) v ( 1; + )
0.5
Hm s t cc tiu ti cc im 1x = gi tr cc tiu ca hm s l ( 1) 1y = Hm s t cc i ti im 0x = gi tr cc i ca hm s l (0) 2y = 0.25
10
8
6
4
2
-2
-4
-15 -10 -5 5 10 15x 1-1-2 2
0.25
2 Ta c y = 4x34mx = 4x(x2m)
y = 02
0x
x m
= =
iu kin hm s c 3 cc tr : y=0 c 3 nghim phn bit
m > 0.
0.25
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Khi th hm s c 3 im cc tr2
2
(0; 1)
( ; 1)
( ; 1)
A m
B m m m
C m m m
- - + - - - + -
Ta thy
4
2 4 3
4 32
( ; ) 1os
. 1( ; )
AB AC m m
AB m m AB AC m m mc BAC
AB AC m m m AC m m
= = +
- - + - = = = + +- -
uuur uuur uuur
uuur
0.25
2
3
2 2
2sin 1 cos
1
( ) 0 2
m m BAC BAC
m
BC m m m
= - =+
= + + =
3
3
3
3 2
2 12
sin 2
1
112
1
2 1 0 ( 1)( 1) 0 1 5
2
BC m mR
A mm m
m
mR m
m
m m m m mm
+= = =
+
+
= =
= - + = - + - = - + =
0.5
2
2im
1Gii phng trnh:
8 8 21 1sin os cos 2 os22 2
x c x x c x- = -
PT
4 4 4 4 2
2
2
2
1 1(sin os )(sin os ) cos 2 os2
2 2
1 1os2 (1 sin 2 ) os2 ( os2 1)2 2
os2 (1 os 2 ) os2 ( os2 1)
os2 ( os 2 os2 ) 0
os2 0 4 2
os2 1
2
x c x x c x x c x
c x x c x c x
c x c x c x c x
c x c x c x
kx
c xk
c xx k
p p
pp
+ - = -
- - = -
- + = -
+ =
= + = = -
= +
0.25
0.25
0.5
2
Gii h phng trnh:4 2
4 3 0
0log log
x y
x y
- + = - =
iu kin : 4
2
0 1
10
log
log
x x
yy
0.25
Vi iu kin trn h cho tng ng vi:
2
4 2 44 2
4 3 0 4 3 0
log log loglog log
x y x y
x y yx y
- + = - + = = ==
0.25
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2
2
2 2
4 31
4 33
x yx y x y
yx y y y
y
= = - = = = = - =
0.25
Tp nghim ca h phng trnh l: ( ) ( ){ }1;1 ; 9;3S = 0.25
Cu3
1 Trong mt phng vi h trc ta Oxy cho A(4;3), ng thng (d) :xy2 = 0 v (d): x + y 4 = 0 ct nhau ti M. Tm ( ) ( ')B d v C d sao
cho A l tm ng trn ngoi tip tam gic MBC.M(3;1):
( ) ( ; 2) ( ') ( ';4 ')B d B t t C d C t t - -
A l tm ng trn ngoi tip2 2
2 2
6
' 2
MA AB t
t MA AC
= = ==
B(6;4) v C(2;2)
0.25
0.5
0.25
2Trong khng gian cho hai ng thng :
2
1 2 2
1
21: 1 : 2
3 0
x txd y v d y t
z t z
= += = = = + =
a. Chng minh rng d1, d2cho nhau v vung gc vi nhau.b. Lp phng trnh on vung gc chung gia d1 v d2.
1 2
1 2
1 2
1 2
1 2
(1;1;3) (2;0;0): d
(0;0;1) (1;2;0)
, ( 2;1;0) , . 3 0
(1; 1;3)
& o .
qua A qua Bta c d
VTCP u VTCP u
u u u u AB
AB
d d ch nhau
= - = - -
ur uur
ur uur
ur uur uuuruuur
0.25
1 1 2 2 2
2 2 1
11 1 1 1
2 2 2 22 2
(1;1;3 ) (2 ;2 ;0)
( 1;2 1; 3)
3. 0 3 0
11 4 2 0. 0
5
M t d N t t d
MN t t t
t MN d MN u MN u t
Ta c MN d t t t MN u MN u
+ +
+ - - -
= - ^ ^ = - - = ^ + + - = =^ =
uuur
uuur ur uuur ur
uuur uur uuur uur
0.25
(1;1;0)6 3
( ; ;0)11 25 5( ; ; 0)
5 5
M
MNN
-
uuur
0.25
ng vung gc chung MN c phng trnh:
61
5
31
5
0
x t
y t
z
= += -
=
0.25
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3 Cho hnh chp S.ABCD c y ABCD l hnh ch nht vi AB=2; AD= 2 2 vSA =2 vung gc vi mt phng (ABCD). Gi M, N ln lt l trung im ca ADv SC, I l giao im ca BM v AC. Chng minh rng mt phng (SAC) vunggc vi mt phng (SMB). Tnh th tch ca khit din ANIB.
j
H
I
S
A
B C
DM
N
0
0
1( . . )
2
90
90
( ) ( ) ( )
AM AB ABM BCA c g c do
AB BC
ABM BCA
ABM BAC BCA BAC
AIB MB AC
MB SA MB SAC SBM SAC
= =
=
+ = + =
= ^
^ ^ ^
V : V
0.5
Gi H l trung im ca AC. Ta c HN l ng trung bnh ca SACV
( ) ( ) 12
SA HN ABCD NH ABI NH ^ ^ = = 0.25
1 1 2 2. . . ( )3 6 9
NABI ABI V NH S NH AI BI dvtt = = =V 0.25
1
Tnh tch phn:
2
ln ln(ln )e
e
x x I dx
x
+=
t
2
ln
1
2
dxt x dt
x
x e t
x e t
= =
= = = =
0.25
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2 2 2 22
1 1 1
1 1 1
3( ln ) ln
2 2
t I t t dt tdt tdt I I = + = + = + = + 0.25
Tnh I1:
t2
2 2
1 1 1
1
ln
( ln ) 2ln 2 2ln 2 1
dtu t du
tdv dt
v t
I t t dt t
= = = =
= - = - = -
Vy1
2ln 2 1 2ln 22 2
I3
= + - = +
0.25
2 Tm snguyn dng n nh nht sao cho khai trin (1+x)nc t s hai h s lin
tip bng7
15.
Ta c (1 )n
n k k
n
k o
x C x
=
+ = H s ca hai s hng lin tip l: 1& (0 1, )k kn nC C k n k
+ - 0,5
Theo yu cu ca bi ton:
1
1
7 1 1
15 15
77
1 1515
k
n
k
n
k
n
k
n
C k
C n k
n kC
kC
+
+
+ = = - - == +
Phn s7
15ti gin, nn n nh nht th:
1 7 6
15 21
7 14
1 15 21
k k
n k n
n k k
k n
+ = = - = = - = = + = =
Vy n = 21 tha mn yu cu bi ton.
0,5
3 Cho cc s thc x,y dng thay i tha mn: x2 + y2 = 1.Tm gi tr nh nht cabiu thc:
1 1(1 )(1 ) (1 )(1 )P x y
y x
= + + + + +
D
2 2
1 1 4 2 22 2 2 4
2 24 2 ( ) 4 3 2
( ) 2( )
B T c si
x yP x y x y x y
x y y x x y x y x y
x yx y x y
= + + + + + + + + + + = + + + ++ + +
+ + + = ++ +
Du =1
2x y = =
Vy1
min 4 3 22
P x y= + = =
0.5
0.25
0.25