Thi Thu Dai Hoc Lan 3 Mon Toan Chuyen Lqd Dien Bien

8/9/2019 Thi Thu Dai Hoc Lan 3 Mon Toan Chuyen Lqd Dien Bien

1/12

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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2/12

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-------------


3/12

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


4/12

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


5/12

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


6/12

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


7/12

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


8/12

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


9/12

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


10/12

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


11/12

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


12/12

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

Thi Thu Dai Hoc Lan 3 Mon Toan Chuyen Lqd Dien Bien

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