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7/25/2019 [DeThiThu.net]20 de Thi Thu Toan 2016 Co Dap An

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TUYNTP

20THI THTHPT QUCGIA NM 2016MN : TON ( Phn 1)

(Ca cc trng THPT, S GD & T trong nc)

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MC LC THISTT Tn trng/ S GD STT Tn trng/ S GD

1 THPT chuyn H Vinhln 1

11 THPT Kim Lin ln 1

2 THPT chuyn H Vinh

ln 2

12 THPT L Li

3 S GD Vnh Phc (L1) 13 THPT H Huy Tp (L1)

4 THPT chuyn Vnh Phcln 3

14 THPT L Thi T (L1)

5 S GD Qung Ninh 15 THPT Vit Tr - Ph Th

6 THPT chuyn Nguyn Huln 2

16 Chuyn Nguyn Tt Thnh

7 THPT chuyn Bin Ha

ln 1

17 THPT Hm Nghi

8 THPT Ng S Lin Ln 3 18 THPT Nghn- H Tnh

9 THPT chuyn H Long ln2

19 THPT Hng Kh - H nh

10 THPT TrnPh ln 1 20 THPT Phan Thc Trc-NA

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TRNG I HC VINHTRNG THPT CHUYN

THI TH THPT QUC GIANM 2016 LN 1Mn: TON

Thi gian lm bi: 180 pht, khng k thi gian pht

b)Cho = Tnh theo

Cu 4(1,0 ).Tnh tch phn+ +

=+

Cu 5 (1,0 ).Trong khng gian vi h ta cho mt phng + + = v

ng thng+

= =

Tm ta giao im ca vi v lp phng trnh mt

phng cha ng thi vung gc vi

Cu 6(1,0 ).a) Gii phng trnh + = +

b) Nhn dp knim ngy Nh gio Vit Nam, trng THPT tuyn chn c tit mc vnngh tiu biu, trong s lp c tit mc cng din trong ton trng. Ban t chc cho

bc thm ngu nhin chia thnh hai bui cngdin, mi bui tit mc. Tnh xc sut tit mc ca lp c biu din trong cng mt bui.

Cu 7(1,0 ). Cho hnh chp c y l hnh ch nht tm vung

gc vi mt phng = = gc gia hai mt phng v

bng Tnh theo th tch khi chp v khong cch gia hai ng thng

Cu 8(1,0 ).Trong mt phng vi h ta cho tam gic c phng trnh cc ngthng cha trung tuyn v ng cao k t ln lt l + = v + = ng thng

cha trung tuyn k t i qua Tnh bit rng im c tung m v thuc

ng thng + + =

Cu 9 (1,0 ).Gii bt phng trnh + + + + +

Cu 10(1,0 ). Gi s l cc s thc khng m tha mn + + = Tm gi tr

ln nht ca biu thc = + ++ + +

------------------ Ht ------------------

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Cu 1 (1,0 ). Kho st s bin thin v v th ca hm s = +

Cu 2 (1,0 ). Vit phng trnh tip tuyn ca th hm s = + bit rng tip tuyn

song song vi ng thng + =

Cu 3 (1,0 ).

a) Gii bt phng trnh + + ++ < <

Suy ra hm sngbintrn mikhong ( ; 1)- v (3; );+ hm snghch bintrn

khong(1; 3).

* Cc tr: Hm s t cc i ti 1,x = y (1) 3y= = ;

hm s t cc tiu ti 3, (3) 1.CT

x y y= = = -

* Gii hn ti v cc:

32 3

6 9 1lim lim 1 ;x x

y xx x x- -

= - + - = -

32 3

6 9 1lim lim 1 .x x

y xx x x+ +

= - + - = +

0

Cu 1.(1,0

im)

* Bng bin thin:

3o. th:

0

H s gc ca dl3

.4

k= - Suy ra h s gc ca tip tuyn cng l3

.4

-

Ta c

( )2

3 , 1.1

y x

x

= - -

Honh tip im ca tip tuyn vi th l nghim ca phng trnh

2

2

13 3 3 ( 1) 4

34 4( 1)

xy x

xx

= -= - - = - - =

=-

0

Cu 2.(1,0

im)

* Vi 1x = - ta c1

.2

y= Suy ra tip tuyn l3 1

( 1) ,4 2

y x= - + + hay3 1

.4 4

y x= - -

* Vi 3x =

ta c

7

.2y=

Suy ra tip tuyn l

3 7

( 3) ,4 2y x= - - +

hay

3 23

.4 4y x= - +

Vy c hai tip tuyn cn tm l3 1

y x= - - v3 23

y x= - +

0

x

y

y

1- + 3

3

-

+

1-

+ 0 0 +

x

O

3

y

1-

1 3

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a) iu kin: 3.x -

t 32 0,x t+ = > bt phng trnh cho tr thnh

222 5 2 5 2 0,t t tt

+ < - + < (v 0t> )1

22

t < <

1 32 2 2 1 3 1 3 2.x x x- + < < - < + < - < - Vy bt phng trnh cho c nghim 3 2.x- < -

0

Cu 3.(1,0

im)

b) Ta c3

45453

log 75

log 75 2 log 75 2 log 45= =

23 3

233

log (3.5 ) 1 2 log 5 2 4

2 2 .2 log 5 2log (3 .5)

a

a

+ +

= = =+ + 0

t2

dln(2 1), d .

( 1)

xu x x v

x= + + =

+Suy ra

2 1d 1 d , .

2 1 1u x v

x x

= + = -

+ +

Theo cng thc tch phn tng phn ta c1 1

0 0

ln(2 1) 1 2d

1 1 (2 1)( 1)

x xI x

x x x x

+ += - + +

+ + + +

0

Cu 4.(1,0

im)

1 1

0 0

1 1 4 2 1 4 1(1 ln 3) d (1 ln 3) d

2 1 2 1 1 2 2 1 1x x

x x x x x

= - + + + - = - + + -

+ + + + +

( )1

0

1(1 ln 3) 2 ln(2 1) ln( 1)

2x x= - + + + - +

1(1 ln 3) 2 ln 3 ln 2

2= - + + -

( )3 1 1

ln 3 ln 2 3 ln 3 2 ln 2 1 .2 2 2

= - - = - -

0

Gi ( ).M d P= V M d nn ( 2 3; 4 8; ).M t t t - + - -

Suy ra ( ) ( 2 3) (4 8) ( ) 7 0 12,M P t t t t - + + - + - - = = hay ( 21; 40; 12).M - - 0Cu 5.

(1,0im)

Mt phng ( )Qcha dv vung gc vi ( )Pnn ( )Qc cp vtcp( 2; 4; 1)

(1; 1; 1)

d

P

u

n

= - -

=

Suy ra , (5; 1; 6).Q d Pn u n = = -

Ly (3; 8; 0)N d- nn ( ).N Q

Suy ra phng trnh ( ) : 5 6 7 0.Q x y z + - - =

0

a) iu kin: sin 0.x Khi phng trnh cho tng ng vi

( ) ( )

( ) ( )

cos sin sin 2 1 cot 0 cos sin 2 cos sin cos 0

cos sin4cos sin 1 2 cos 0 ( ).1

cos 22 3

x x x x x x x x x

x x x kx x x k

x x k

p

p

p

p

- + - = - + - =

= = + - - = = = +

0

Cu 6.(1,0

im)

b) Gi hai bui cng din l , .I IIS cch chia 24 tit mc thnh hai bui cng din

chnh l s cch chn 12 tit mc cho bui ,I l 1224

.C

Gi l bin c 2 tit mc ca lp 11 c biu din trong cng mt bui.Nu 2 tit mc ca lp 11 cng biu din trong bui Ith s cch chn 10 tit mc cn

li chobui Il 1022

.C Hai tit mc ca lp 11 cng c th cng biu din trong bui .II

V vy, s cch chia bin cAxy ra l 1022

2. .C

Do 1022

1224

2. 11( ) 0, 4783.

23

CP

C= =

2.C2 11 ( ) 12

0

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V( )SD BCD

DC BC

^

^ nn .SC BC^

Suy ra ( ) 0( ), ( ) 45SCD SBC BCD= = (do SCDD vung ti Dnn 090 ).SCD< V BCD l hnh ch nht nn ,O OD=

kt hp vi0 0

180 60 .OD OB = - = Suyra O DD u.

Do 0, 60 .O OD a DO= = =

Suy ra 0. tan 60 3.B D a= =

Suy ra 2. 3BCDS B D a = = v 0. tan 45 3.SD CD a = =

Suy ra 3.

1. .

3S BCD BCD V SD S a = =

0

Cu 7.(1,0

im)

K Bx// C mp( , )S Bx//C

( ) ( )

1

( , ) , ( , ) , ( , ) .2d C SB d O S Bx d D S Bx = =

(1)H , .DK Bx DH SK^ ^ V ( )Bx SDK^ nn ( , ).Bx DH DH S Bx^ ^ (2)

V 2 2BD DO a = = v 060DBK DO= = (ng v) nn 0sin 60 3.DK BD a = =

Suy ra SDKD vung cn ti2 6

.2 2 2

SK SD a D DH = = = (3)

Kt hp (1),(2) v (3) ta suy ra1 6

( , ) .2 4

ad C SB DH= =

0

T h2 0

( 4; 2).3 2 8 0

y

Cx y

+ = - - - + =

Gi ,M Nl trung im , .B BC

Ta c: 2 2 0 ( 2 2; ) ( 0)d x y a a a + + = - - <

: 2 0 ( ; 2).M CM y M m + = -

M Ml trung im Bnn6

(2 2 2; 4) 1; .2

aB a m a N a m

- -+ + - - + -

V CH B^ nn . 0 2(2 2) 3( 2) 0 2 2.CHu M a m a a m = + + + - - = = - + (1)

0

Cu 8.(1,0im)

Ta c ( 2 16; 3)K a a= - + - v12

17; .2

aKN a m

- -= + +

V , ,N Kthng hng nn K cng phng KN. Do

(2)( 2 16)( 12) 2( 3)( 17).- + - - = - + +a a a a m

5Thay (1) vo (2) ta c 2 21 65 0m m2 + - =

m a= = -

23 ( )tm

m a= - =13 28 ( )ktm

Suy ra B(4; 3), (1; 1).- -

Ta c B BC B BC = - = - - = = -(3; 2), ( 5; 1) cos , .( ) 3( 5) ( 2)( 1) 1- + - -9 4 25 1 2+ +

0

S

KB

O

D

H

a

x

045 C

C

H B M

N

K

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iu kin: 2.x -

t 2 3 , 2 ,x u x v + = + = bt phng trnh cho tr thnh

( )2 2 2 23 4 2 3 1 0u v v u u v u v u v - + + - + + - - + ( 1)( 3) 0u v u v - + + -

2 23 2 1 3 2 3 0.x x x x + - + + + + + -

(1)

Ta c2

2

2

13 2 1 1 0.

3 2

x xx x

x x

- ++ - + + = + >+ + +

Do (1) tng ng vi 2 3 2 3 0x x+ + + -

0

Cu 9.(1,0

im)

2

2

3 2 03 3 2

3 9 6 2 2

xx x

x x x

- + + - +

+ - + + +

( ) ( )

2

22 2

2 7, 8 07

6 2 8 36 2 8

x x xx

x x x x x x

- + -

+ + - + + -

( )( )2 2

1 332 2 2 2 3

21

1 4 8 0

x x

xx x x

+ - - -

= - + - -

Vy bt phng trnh cho c nghim 1x = - v 2 2 2 3.x- -

0

t 2 tan , 2 tan , 2 tan ,2 2 2

B Cx y z= = = vi 0 , , .B C p < (1)

T gi thit ta c tan tan tan tan tan tan 1.2 2 2 2 2 2

B B C C+ + =

Khi 1 tan tan

2 2tan cot tan .2 2 2 2

tan tan2 2

B C B C B C

B C

p- + +

= = = - +

Suy ra , .2 2 2

B Ck k

p

p

+= - + Hay 2 .B C kp p+ + = +

T (1) suy ra 0.k= Do .B C p+ + = Khi

0

Cu 10.(1,0

im)

21 1sin sin sin

22 2

CP B= + + 2

1.2 sin cos 1 cos

2 2 22

B B C+ -= + -

22 cos cos 12 2

C C - +

23 1 3

cos .2 2 22

C = - -

Du ng thc xy ra khi

12 2cos 22 2

2.4

C C x y

z B B

p

p

= = = -=

= = = =

Vy gi tr ln nht ca Pbng3

.2

0

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THI TH THPT QUC GIA NM 2016 LN 2Mn: TON


Cu 1(1,0 im).Kho st s bin thin v v th ( )H ca hm s1

.2

xy

x

Cu 2 (1,0 im).Tm cc im cc i, cc tiu ca hm s 4 3 2( ) 3 4 12 .f x x x x

Cu 3 (1,0 im).

a) Cho hm s 2( ) .x xf x e e Tm x '( ) 2 ( ) 3.f x f x

b)Cho s phc ztha mn 2(1 ) 2 4 .i z i Tm phn thc v phn o ca .z

Cu 4(1,0 im).Tnh tch phn1

0

3 1sin d .

5

xI x x

x

Cu 5 (1,0 im).Trong khng gian vi h ta ,Oxyz cho mt phng ( ) : 3 0P x y z v

im (1; 2; 3).I Vit phng trnh mt cu ( )S tm ,I tip xc vi mt phng ( ).P Tm ta tip

im ca ( )S v ( ).P

Cu 6(1,0 im).

a) Cho1

cos .3

a Tnh gi tr biu thcsin 3 sin

.sin 2

a aP

a

b) Nam v Hng chi bng qua li, ai thnh cng nhiu hn l ngi thng cuc. Nu

bng v tr A th xc sut thnh cng ca Nam l 0, 9 cn ca Hng l 0,7; nu bng v

tr Bth xc sut thnh cng ca Nam l 0,7 cn ca Hng l 0, 8. Nam v Hng mi ngi

u 1 qu v tr A v 1 qu v tr .B Tnh xc sut Nam thng cuc.

Cu 7(1,0im

). Cho hnh lng tr . ' ' 'ABC A B C c y ABC l tam gic u cnh ,a gc giacnh bn v mt phng y bng 045 , hnh chiu ca A ln mt phng ( ' ' ')A B C l trung im

ca ' ' .A B Gi M l trung im ca ' ' .B C Tnh th tch khi lng tr . ' ' 'ABC A B C theo a vcsin ca gc gia hai ng thng ' , ' .A M AB

Cu 8(1,0 im).Trong mt phng vi h ta ,Oxy cho hnh thang ABCDvung ti A v ,D

1.

3AB AD CD Giao im ca AC v BDl (3; 3),E im (5; 9)F thuc cnh AB sao

cho 5 .AF FB Tm ta nh ,D bit rng nh A c tung m.

Cu 9 (1,0 im).Gii phng trnh 2 1 2

2 22 log 1 4 log (3 ).x xx x x

Cu 10(1,0 im).Tm s thc mln nht sao cho tn ti cc s thc khng m , ,x y ztha mn

4x y z v 3 3 3 2 2 28 .x y z xy yz zx m

------------------ Ht ------------------

Ghi ch: 1. BTC s tr bi vo cc ngy 16, 17/4/2016. nhn c bi thi, th sinh phi np liphiu d thi cho BTC.

2. Thi th THPT Quc gia ln 3 s c t chc vo chiu ngy 07 v ngy 08/5/2016. ng kd thi ti Vn phng Trng THPT Chuyn t ngy 16/4/2016.

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P N THI TH THPT QUC GIA NM 2016 LN 2Mn: TON; Thi gian lm bi: 180 pht

Cu p n im1o. Tp xc nh: \ {2}. 2o. S bin thin:* Gii hn, tim cn: Ta c

2

limx

y

v2

lim .x

y

Do ng thng 2x l

tim cn ng ca th ( ).H

V lim lim 1x x

y y

nn ng thng 1y l tim cn ngang ca th ( ).H

* Chiu bin thin: Ta c 21

' 0,( 2)

yx

vi mi 2.x

Suy ra hm s ng bin trn mi khong ( ; 2), (2; ).

0,5

* Bng bin thin:

Cu 1(1,0

im)

3o. th: th ( )H ct Ox ti (1; 0), ct Oy

ti

10; ;

2 nhn giao im (2; 1)I

ca hai ng tim cn lm tm i

xng.

0,5

Hm s xc nh vi mi .x

Ta c3 2

1 2 3'( ) 12 12 24 ; '( ) 0 1, 0, 2.f x x x x f x x x x

0,5Cu 2(1,0

im)

2''( ) 12 3 2 2f x x x .Ta li c ''( 1) 0, ''(0) 0, ''(2) 0.f f f

Suy ra 1, 2x x l cc im cc tiu; 0x l im cc i ca hm s.

Ch . Hc sinh c th lp Bng bin thin a ra kt lun.

0,5

a) Hm s xc nh vi mi x v 2'( ) 2 , .x xf x e e x Khi

2 2'( ) 2 ( ) 3 2 2 2 3 1 0.x x x x x f x f x e e e e e x 0,5Cu 3

(1,0im)

b) T gi thit ta c

2

2 4 2 4 12 2 .

2(1 )

i iz i

i ii

Vy, phn thc ca zbng 2, phn o ca zbng 1.

0,5

x

'y

2

1 1

yO

1

y

1

2

I

x

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Ta c1 1

0 0

3 1sin d d

5

xI x x x

x

+)

11

00

1 2sin cos .xdx x

0,5

Cu 4(1,0

im)

+) Tnh

1

03 1 d .5

xx

x

t 3 1 .x t

Khi 0 1; 1 2x t x t v2 1 2

.3 3

t tx dx dt

Suy ra1 2 22

20 1 1

3 1 2 2d 2 2 1

5 4 416

x tx dt dt

x t tt

2

1

2 4 ln 4 4 ln 4 2 8 ln 3 4 ln5.t t t

T ta c2

2 8 ln 3 4 ln 5.I

0,5

Ta c , ( ) 3.R d I P Suy ra 2 2 2( ) : ( 1) ( 2) ( 3) 3.S x y y 0,5Cu 5(1,0

im) Gi Hl tip im ca ( )S v ( ).P Khi Hl hnh chiu ca Iln ( ).P

Ta c (1; 1; 1).IH P

u n

Suy ra1 2 3

: .1 1 1

x y zIH

Do ( 1; 2; 3).H t t t V ( )H P nn

( 1) ( 2) ( 3) 3 0 1.t t t t

Suy ra (0; 1; 2).H

0,5

a) Ta c

2sin 3 sin 2 cos 2 sin cos 2 2 cos 1 7.

sin 2 2 sin cos cos cos 3

a a a a a a P

a a a a a

0,5Cu 6(1,0

im)

b) Gi Xl bin c Nam thng cuc; ( 0, 1, 2)i

N i l bin c Nam thnh cng i

qu; ( 0, 1, 2)i

H i l bin c Hng thnh cng iqu.

Khi

1 0 2 0 2 1X N H N H N H .Theo gi thit ta c

1 0 1 0. 0, 9.0, 3 0, 1.0, 7 0, 3.0, 2 0, 0204.P N H P N P H

2 0 2 0. 0, 9.0, 7 0, 3.0, 2 0, 0378.P N H P N P H

2 1 2 1. 0, 9.0, 7 0, 7.0, 2 0, 3.0, 8 0, 2394.P N H P N P H

Suy ra (X) 0, 0204 0, 0378 0, 2394 0, 2976.P

0,5

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a

450

K N

MH

C

B

A

C'

B'

A'

Gi Hl trung im ca ' ' .A B Khi

( ' ' ').AH A B C Suy ra

0' ( ', ( ' ' ')) 45 .AA H AA A B C

Do ' .2

aAH A H Suy ra

30

. ' ' '

1 3. . . .sin60 .

2 2 8ABC A B Ca a

V a a

0,5

Cu 7(1,0

im)

Gi Nl trung im ca .BC Khi ( ' , ') ( , ').A M AB AN AB

Trong tam gic vung 'HAB ta c2 2

2 2 2' ' .2 2 2

a a aAB AH HB

Tam gic ABCu cnh ann

3

.2

a

AN

Gi Kl trung im ca .AB Khi ' / /B K AH nn ' .B K KN Suy ra

2 2

2 2 2' ' .2 2 2

a a aB N B K KN

p dng h qu ca nh l hm s csin trong tam gic 'AB Nta c

2 2 22 3 2

4 4 4 6cos( ' , ') cos ' .

42 32. .

2 2

a a a

A M AB NAB a a

0,5

11

I

F

E

D C

BA Gi .I EF CD Ta s chng minh tam

gic EAIvung cn ti .E

t , .AB a AD b

Khi a b

v

. 0.a b

Ta c 3 .AC AD DC b a

1 5 1 5 13 3 .4 6 4 6 12

FE AE AF AC AB b a a b a

Suy ra

2 21

. 3 3 0.12AC EF b a

Do .AC EF (1)

T (1) suy ra t gic ADIEni tip. Suy ra 01 1

45 .I D (2)

T (1) v (2) suy ra tam gic EAIvung cn ti .E

0,5

Cu 8(1,0

im)

Ta c (2; 6)AC

n EF

nn : 3 12 0 (3 12; ).AC x y A a a

Theo nh l Talet ta c

3EI EC CD

EF EA AB 3 ( 3; 15).EI FE I

0,5

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Khi

2 2 3

(3 9) ( 3) 360 .9

aEA EI a a

a

V A c tung m nn ( 15; 9).A

Ta c (20; 0)AD

n AF

nn : 15 : 15.AD x CD y Do ( 15; 15).D

iu kin: 0.x Phng trnh cho tng ng vi2 1 2 3

2 22 log 1 2 log (3 ).x x xx x x

(1)

Xt hai trng hp sau:

TH1.1

0 .3

x Khi 2 1 2 3

2 22 log 1 2 0 2 log (3 ).x x xx x x

Suy ra (1) khng tha mn.

0,5

Cu 9(1,0

im)

TH2.1

.3

x Ta c 2 1x x v 3xu thuc khong [1; + ).

Xt hm s2

( ) 2 logtf t t trn khong [1; + ).

Ta c2

1'( ) 2 ln 2. log 2 . 0

ln 2t tf t t

t vi mi tthuc khong [1; + ).

Suy ra ( )f t ng bin trn khong [1; + ).

Do (1) tng ng vi 2 1 3 .x x x T y gii ra c1

.3

x

Vy, phng trnh cho c nghim duy nht1

.3

x

0,5

Gi s tn ti cc s thc , ,x y ztha mn yu cu bi ton t ra.

Khng mt tnh tng qut ta gi s ynm gia xv .zKt hp vi gi thit ta c0 2y v ( )( ) 0.x y x y z

T y ta c 22 2 2 .xy yz zx y x z

Mt khc, do ,x zkhng m nn 33 3 .x z x z

Do

3 2 3 23 38 4 8 4m x z y y x z y y y y

3 28 52 80 64y y y . (1)

0,5

Cu

10(1,0im)

Xt hm s 3 2( ) 8 52 80 64, 0 2.f y y y y y Ta c

2 2( ) 24 104 80 8 3 13 10 .f y y y y y ( ) 0, 0 2 1.f y y y

Ta c (0) 64, (1) 100, (2) 80.f f f

Suy ra ( ) (1) 100, [0; 2].f y f y (2)

T (1) v (2) ta c 100.m Khi 0, 1, 3x y z ta c du ng thc.

Vy s mln nht cn tm l 100.

0,5

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S GD&T VNH PHC KSCL N THI THPT QUC GIA LN 1NM HC 2015-2016

MN THI: TONThi gian lm bi: 180 pht, khng k thi gian pht

Cu 1(1,0 im). Kho st s bin thin v v th ca hm s2 1

2y

Cu 2 (1,0 im). Tm cc im cc tr ca th hm s 3 23 6y x x

Cu 3 (1,0 im).

a) Gii bt phng trnh 22 2log log 44x

x

b) Gii phng trnh 5.9 2.6 3.4x x x

Cu 4(1,0 im). Tnh nguyn hm 2 sin 3I x xdx Cu 5 (1,0 im). Cho hnh chp .S ABC c 0, 90 , , 3, 2SA ABC ABC AB a BC a SA a .

Chng minh trung im I ca cnh SCl tm ca mt cu ngoi tip hnh chp .S ABCv tnh din

tch mt cu theo a.

Cu 6(1,0 im).

a) Gii phng trnh: 22cos sin 1 0x x .

b) i vn ngh ca nh trng gm 4 hc sinh lp 12A, 3 hc sinh lp 12Bv 2 hc sinh lp

12C. Chn ngu nhin 5 hc sinh t i vn ngh biu din trong l b ging nm hc. Tnh xc sut

sao cho lp no cng c hc sinh c chn v c t nht 2 hc sinh lp 12A.

Cu 7 (1,0 im). Cho hnh chp .S ABCD c y l hnh vung cnh a,3

2

aSD . Hnh chiu vung

gcHca nh Sln mt phng (ABCD) l trung im ca onAB . Gi Kl trung im ca on

AD . Tnh theo ath tch khi chp .S ABCD v khong cch gia hai ng thng HKv SD .

Cu 8 (1,0 im). Trong mt phng vi h to Oxy cho hnh thang ABCD vung ti A v D c

AB AD CD , im (1;2)B , ng thng BD c phng trnh l 2 0y . ng thng qua B

vung gc vi BCct cnh AD ti M. ng phn gic trong gcMBCct cnh DCti N. Bit

rng ng thng MNc phng trnh 7 25 0x y . Tm ta nh D .

Cu 9 (1,0 im). Gii h phng trnh:

2

2

2 1 11 ,

3 8 3 4 1 1

xx y x y

x x y

x x x y

Cu 10 (1,0 im). Cho ,x y tha mn2

2

2

2 3

y x

x x

. Tm gi tr nh nht ca biu thc:

4 4

2

2P x y

y

-------------HT------------

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S GD&T VNH PHC HNG DN CHM KSCL N THI THPT QUC GIA LN 1NM HC 2015-2016

MN THI: TON

I. LU CHUNG:- Hng dn chm ch trnh by mt cch gii vi nhng c bn phi c. Khi chm bi hc sinh lmtheo cch khc nu ng v th vn cho im ti a.- im ton bi tnh n 0,25 v khng lm trn.- Vi bi hnh hc khng gian nu th sinh khng v hnh hoc v hnh sai th khng cho im tngng vi phn .II. P N:

Cu Ni dung trnh by im1

Kho st s bin thin v v th ca hm s2 1

2y

1,0

2 1

2

xy

x

1. Tp xc nh: \ {2}D

2. S bin thin.

2

3' 0,

( 2) x D

x

Suy ra hm s nghch bin trong cc khong ( ;2) v (2; ) Hm s khng c cc tr

0,5

Cc gii hn2 2

lim 2; lim 2; lim ; limx x x x

y y y y

Suy ra 2x l tim cn ng, 2y l tim cn ngang ca th.0,25

Bng bin thin

0,25

3. th: Giao vi trc Oxti1

;02

, giao vi trc Oyti1

0;2

, th c tm i

xng l im (2;2)I

0,25
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2 Tm cc im cc tr ca th hm s 3 23 6y x x 1,0

* Tp xc nh: 0,25

2 0' 3 6 , ' 02

xy x x y

x

0,25

Bng xt du o hm

0 2 + 0 - 0 +

0,25

T bng xt u o hm ta cHm s t cc i ti 0x v gi tr cc i 6y ; t cc tiu ti 2x v gi trcc tiu 2y .

Vy im cc i ca th hm s l M 0;6 , im cc tiu ca th hm s l

N 2;2

0,25

3 aGii bt phng trnh 22 2log log 44

xx (1) 0,5

+) iu kin ca bt phng trnh (1) l: 0x (*)+) Vi iu kin (*),

2 2

2 2 2 2 2(1) log log log 4 4 log log 2 0x x x x

2 2(log 2)(log 1) 0x x

0,25

2

2

4log 2

1log 1 0

2

xx

x x

+) Kt hp vi iu kin (*), ta c tp nghim ca bt phng trnh (1) l

1

0; 4;2

S

0,25

b Gii phng trnh 5.9 2.6 3.4x x x (1) 0,5

Phng trnh cho xc nh vi mi x

Chia c hai v ca phng trnh (1) cho 4 0x ta c :2

3 35.9 2.6 3.4 5. 2. 3

2 2

x xx x x

23 3

5. 2. 3 02 2

x x

3 31 5. 3 0

2 2

x x

(2)

0,25

V3

5. 3 02

x

x

nn phng trnh (2) tng ng vi

31 0

2

x

x

.

Vy nghim ca phng trnh l: 0x

0,25

4 Tnh nguyn hm 2 sin 3I x xdx 1,0

t2

sin3

u x

dv xdx

0,25

ta c cos33

du dx

v

0,25

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Do : 2 cos 3 1

cos33 3

x xI xdx

0,25

2 cos 3 1sin3

3 9

x xC

0,25

5 Cho hnh chp .S ABC c 0, 90 , , 3, 2SA ABC ABC AB a BC a SA a .Chng minh trung im I ca cnh SC l tm ca mt cu ngoi tip hnh chp

.S ABCv tnh din tch mt cu theo a.

I

A C

B

S

1,0

V SA ABC SA BC

Mt khc theo gi thit AB BC , nn BC SAB v do BC SB 0,25Ta c tam gic SBC vung nhB; tam gic SABvung nhAnn

2

SCIA IB IS IC (*)

Vy imIcch u bn nh ca hnh chp, do Il tm mt cu ngoi tip cahnh chp .S ABC

0,25

T (*) ta c bn knh ca mt cu l2

SCR

Ta c 2 2 2AC AB BC a 2 2 2 2 2SC SA AC a R a

0,25

Din tch mt cu l 2 24 8R a 0,256 a Gii phng trnh 22cos sin 1 0x x . 0,5

Ta c: 2 22cos sin 1 0 2sin sin 3 0 (sin 1)(2sin +3)=0x x x x x x 0,25sin 1x (do 2sin 3 0x x )

sin 1 22

x x k k

Vy nghim ca phng trnh cho l: 22

x k k

0,25

b i vn ngh ca nh trng gm 4 hc sinh lp 12A, 3 hc sinh lp 12Bv 2 hcsinh lp 12C. Chn ngu nhin 5 hc sinh t i vn ngh biu din trong l b

ging nm hc. Tnh xc sut sao cho lp no cng c hc sinh c chn v c tnht 2 hc sinh lp 12A.

0,5

Gi khng gian mu ca php chn ngu nhin l S phn t ca khng gian mu l: 59 126C

Gi A l bin c Chn 5 hc sinh t i vn ngh sao cho c hc sinh c ba lp vc t nht 2 hc sinh lp 12A.Ch c 3 kh nng xy ra thun li cho bin c A l :+ 2 hc sinh lp 12A, 2 hc sinh lp 12B, 1 hc sinh lp 12C+ 2 hc sinh lp 12A, 1 hc sinh lp 12B, 2 hc sinh lp 12C+ 3 hc sinh lp 12A, 1 hc sinh lp 12B, 1 hc sinh lp 12C

0,25

S kt qu thun li cho bin c A l: 2 1 2 2 2 1 3 1 14 3 2 4 3 2 4 3 2. . . . . . 78C C C C C C C C C . 0,25

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Xc sut cn tm l78 13

126 21P .

7Cho hnh chp .S ABCD c y l hnh vung cnh a,

3

2

aSD . Hnh chiu vung

gcHca nh Sln mt phng (ABCD) l trung im ca onAB . Gi Kl trungim ca on AD . Tnh theo ath tch khi chp .S ABCD v khong cch giahai ng thng HKv SD .

1,0

T gi thit ta c SHl ng cao ca hnh chp S.ABCDv

2 2 2 2 2 2 2 23( ) ( ) ( )2 2

a aSH SD HD SD AH AD a a

0,25

Din tch ca hnh vungABCDl 2a ,3

2.

1 1. .

3 3 3S ABCD ABCDa

V SH S a a 0,25

T gi thit ta c / / / /( )HK BD HK SBD

Do vy: ( , ) ( ,( ))d HK SD d H SBD (1)

GiEl hnh chiu vung gc caHlnBD, F l hnh chiu vung gc caHln SETa c , ( )BD SH BD HE BD SHE BD HF m HF SE nn suy ra

( ) ( , ( ))HF SBD HF d H SBD (2)

0,25

+) 02

.sin .sin 452 4

a aHE HB HBE

+) Xt tam gic vung SHE c:

2 2

2.. 4. .

32( )

4

aaSH HE a

HF SE SH HE HFSE a

a

(3)

+) T (1), (2), (3) ta c ( , )3ad HK SD .

0,25

8 Trong mt phng vi h to Oxy cho hnh thang ABCD vung ti A v D cAB AD CD , im (1;2)B , ng thng ng thng BD c phng trnh l

2 0y .. ng thng qua B vung gc vi BC ct cnh AD ti M . ngphn gic trong gc MBC ct cnh DC ti N. Bit rng ng thng MN cphng trnh 7 25 0x y . Tm ta nh D .

1,0

0,25

E

O

K

H

B

A D

C

S

F

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T gic BMDCni tip 045BMC BDC DBA BMC vung cn tiB,BNl

phn gic trongMBC,M C i xng quaBN

4( , ) ( , )

2AD d B CN d B MN 0,25

Do 2 4AB AD BD AD 0,25

: 2 0 ( ; 2)BD y D a ,

5 5; 24

3 3;2 ( )

a DBD

a D loai cung phia B so voi MN

Vy c mt im tha mn l: (5;2)D

0,25

9 Gii h phng trnh:

2

2

2 1 11 ,

3 8 3 4 1 1

xx y x y

x x y

x x x y

1,0

iu kin:1

1

x

y

33 2 11 2 1 1 2 1

1 1 1

x x xx x xx y y y

x x x

3

31 11 1

x x yx x

.

0,25

Xt hm s 3f t t t trn c 23 1 0f t t t suy ra f(t)ng bin

trn . Nn 1 11 1

x xf f y y

x x

. T y suy ra 0x Thay

vo (2) ta c 23 8 3 4 1x x x .

0,25

22

2 1 2 1x x x

2

2

1

6 3 0 3 2 32 1 1

1 5 2 132 1 1 3

3 99 10 3 0

x

x x xx x

xx x x

x x

0,25

Ta c2

11

xy

Vi4 3 3

3 2 32

x y

. Vi5 2 13

( 0)9

x loai do x

.

0,25

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KL: H phng trnh c mt nghim 4 3 3

; 3 2 3;2

x y

.

10Cho ,x y tha

2

2

2

2 3

y x

x x

. Tm gi tr nh nht ca biu thc

4 4

2

2P x y

x y

1,0

T gi thit ta c 0y v

2

2 62 3 02 5x x x x v

22 2 2 2 2 22 3 2 2 6 5x y x x x x x x

Xt hm s 2 26

( ) 2 2 6 5 ; 0;5

f x x x x x ta c

60;

5

Max

f(x) = 2

2 2 2x y

0,25

22 22 22 2 2 2 2 2

2 2 2

2 22

2

x yP x y x y x y

yx y

t 2 2t x y 2

2 , 0 22tP tt

0,25

Xt hm s:

2 2

( ) , 0;22

tg t t

t

33

2 2

2 2'( ) ; '( ) 0 2

tg t t g t t

t t

0,25

Lp bng bin thin ta cMin3 63 4 16

2 2P khi x y

0,25

------------Ht------------

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TR NG THPT CHUY N V NH PH C

CHNH THC

THI THPT QUC GIA NM HC 2015-2016-LN 3Mn: TON

Thi gian lm bi: 180 pht, khng k thi gian pht .

Cu 1 (1,0 im). Kho st s bin thin v v th ca hm s : 3 23 2y x x

Cu 2 (1,0 im).Tm gi tr ln nht v gi tr nh nht ca hm s : 2 1

1f x

x

trn on 3; 5

Cu 3 (1,0 im).a) Cho ;

2

v

1sin

3 . Tnh gi tr biu thc sin 2 cos 2P

b) Gii phng trnh : 2sin 2 2sin sin cosx x x x

Cu 4 (1,0 im).Tnh tch phn sau : 4

2 2

0

2 2 ln 9I x x x dx

Cu 5 (1,0 im).a) Gii bt phng trnh : 2 2log 3 2 log 6 5 0x x .

b) Cho tp hp 1;2;3;4;5;6E vMl tp hp tt c cc s gm hai ch s phn bit lp t

E. Ly ngu nhin mt s thucM. Tnh xc sut tng hai ch s ca s ln hn 7 .

Cu 6 (1,0 im). Trong khng gian vi h ta Oxyz , cho cc im 1; 2;0 , 3;4; 2M N v

mt phng : 2 2 7 0P x y z . Vit phng trnh ng thng MNv tnh khong cch t

trung im ca on thng MNn mt phng P .

Cu 7 (1,0 im). Cho hnh chp .S ABCc y ABCl tam gic u cnh a GiIl trung imcnhAB .Hnh chiu vung gc ca nh S trn mt phng y l trung im H ca ,CI gc

gia ng thng SA v mt y bng 060 . Tnh theo a th tch khi chp .S ABC v khongcch t im Hn mt phng SBC .

Cu 8 (1,0 im)..

Trong mt phng vi h to Oxy , cho hai ng thng 1 2:3 4 8 0 , :4 3 19 0d x y d x y .Vit phng trnh ng trn C tip xc vi hai ng thng 1d v 2d , ng thi ct ng

thng :2 2 0x y ti hai im ,A B sao cho 2 5AB .Cu 9 (1,0 im).

Gii bt phng trnh : 2

2 2 1

26 2 4 2 2

x

x x x

Cu 10 (1,0 im).

Cho cc s thc dng ,x y tha mn iu kin 2016x y .Tm gi tr nh nht ca biu thc

2 2 2 2 2 2 2 25 3 3 5 2 2P x xy y x xy y x xy y x xy y

--------Ht-------

Th sinh khng c s dng ti liu. Cn b coi thi khng gii thch g thm.

H v tn th sinh:.......; S bo danh:

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21/117

TR NG THPT CHUY N V NH PH C HNG DN CHM THI THPT QUC GIA LN 3NM HC 2015-2016

Mn: TON ( Gm 5 trang)

Cu p n im

Cu 1. Kho st s bin thin v v th ca hm s : 3 23 2y x x 1,0

Tp xc nh: D .

Ta c 23 6y' x x. ;0

02

xy'

x

0,25

1 (1,0 ) - Hm s ng bin trn cc khong ( ;0) v (2; ) ; nghch bin trn khong (0;2) .

- Cc tr: Hm s t cc i ti 0, 2CDx y ; t cc tiu ti 2, 2CTx y

- Gii hn: lim , limx xy y

0,25

Bng bin thin:x 0 2 y' + 0 - 0 +y 2

-2

0.25

th:

f(x)=(x^3)-3*(x)^2+2

-8 -6 -4 -2 2 4 6 8

-5

5

x

y

0,25

Cu2.Tm gi tr ln nht v gi tr nh nht ca hm s : 2 1

1f x

trn on 3; 5 1,0

Hm s xc nh v lin tc trn 3; 5D 0,25

2 (1,0 ) Ta c

23

0, 3;51

f x xx

0,25

Do hm s ny nghch bin trn on 3; 5 0,25

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Suy ra

3;53;5

7 11max 3 ; min 5

2 4xxf x f f x f

0,25

Cu 3a. Cho ;2

v

1sin

3 . Tnh gi tr biu thc sin 2 cos 2P 0,5

V ;2

nn cos 0 , suy ra 2

2 2cos 1 sin

3 0,25

3.(1,0) Do 2sin 2 cos 2 2sin cos 1 2sinP

21 2 2 1 7 4 2

2 1 23 3 3 9P

0,25

Cu 3b) Gii phng trnh : 2sin 2 2sin sin cosx x x 0,5

Phng trnh cho 2sin sin cos sin cosx x x x x

sin cos 0 1

2sin 1 2

x x

x

0,25

1 tan 1 ,x x k k

1 5

2 sin 2 2 ,

2 6 6

x x k x k k

0,25

Vy phng trnh c ba h nghim5

, 2 , 24 6 6

x k x k x k

vi k

Cu 4. Tnh tch phn sau : 4

2 2

0

2 2 ln 9I x x x dx 1,0

4 4

3 21 2

0 0

4 2 ln 9I x dx x x dx I I 0,25

443 4

1 00

4 256I x dx x 0,25

4 .(1,0 )

42

2

0

2 ln 9I x x dx . t 2

2

2

2ln 9 92 9

xdu dxu x xdv xdx v x

4

4 4 42 2 2 2 22 00 0

0

9 ln 9 2 9 ln 9I x x xdx x x x

2 25ln 25 9ln 9 16 50ln 5 18ln 3 16I

0,25

Vy 1 2 240 50ln 5 18ln 3I I I 0,25

Cu 5 a) Gii bt phng trnh : 2 2log 3 2 log 6 5 0x x . 0,5

Bt phng trnh cho 2 2

3 2 0

log 3 2 log 6 5 6 5 0

3 2 6 5

x

x x x

x

0,25

2

36 6

15 51

x

x x

x

.Vy nghim ca bt phng trnh l :6

15

x 0,25

5 (1,0 ) Cu 5 b) Cho tp hp 1;2;3;4;5;6E v M l tp hp tt c cc s gm hai ch s

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phn bit thuc E. Ly ngu nhin mt s thuc M . Tnh xc sut tng hai ch sca s ln hn 7.

S phn t ca tp Ml 26 30A 0,25

Cc s c tng hai ch s ln hn 7 gm 26,62,35,53,36,63,45,54,46,64,56,65

C 12 s nh vy . Suy ra xc sut cn tm l12 2

30 5P

0,25

Cu 6. Trong khng gian vi h ta Oxyz , cho cc im 1; 2;0 , 3;4; 2M N

v mt phng : 2 2 7 0P x y z . Vit phng trnh ng thng MN v tnhkhong cch t trung im ca on thng MN n mt phng P .

1,0

ng thng MNc vect ch phng 4;6;2MN

hay 2;3;1u

0,25

6 .(1,0 ) Phng trnh ng thng MN:1 2

2 3 1

y z

(c th vit di dng pt tham s) 0,25

Trung im ca on thng MNl 1;1;1I 0,25

Khong cch t In mt phng P l :

2 2 1 7

, 24 4 1d I P

0,25

Cu 7. Cho hnh chp .S ABC c y ABC l tam gic u cnh a GiI l trungimcnhAB .Hnh chiu vung gc ca nh Strn mt phng y l trung im Hca

,CI gc gia ng thng SA v mt y bng 060 . Tnh theo a th tch khi chp

.S ABC v khong cch t im Hn mt phng SBC .

1,0

7. (1,0 )

K

H

KH

S

A

B

C

A

B C

I

I

A'I'

H'

E

H'

0,25

Ta c 2 23

2

aCI AC AI

Do 2 27

4

aAH AI IH , suy ra 0

21.tan60

4

aSH AH .

Vy3

.

1 7.

3 16S ABC ABC a

V SH S

0,25

Gi ', ', 'A H I ln lt l hnh chiu ca , ,A H ItrnBC;El hnh chiu caHtrn SH'

th ( ) ;( )HE SBC d H SBC HE . Ta c1 1 3

' ' '2 4 8

aHH II AA

0,25

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T2 2 2

1 1 1

'HE HS HH , suy ra

21

4 29

aHE . Vy

21;( )

4 29

ad H SBC . 0,25

Cu8.Trong mt phng vi h to Oxy, cho hai ng thng 1 :3 4 8 0d x y ,

2 :4 3 19 0d x y .Vit phng trnh ng trn C tip xc vi hai ng thng 1d

v 2d, ng thi ct ng thng :2 2 0x y ti hai im ,A B sao cho 2 5AB

1,0

Gi ;I a b l ta tm v R l bn knh ng trn C .

Do ng thng ct ng trn C ti hai im ,A B sao cho 2 5AB nn ta c

2 22 2, 5 2 5 *5

a bd I R R

0,25

8 .(1,0 )ng trn C tip xc vi 1d , 2d khi :

1

2

,

,

d I d R

d I d R

3 4 83 4 8

55

4 3 194 3 19 3 4 8

5

a ba bR

R

a ba b a bR

7 27

5 20

7 11

5 5

b a

R a

a b

R b

0,25

-Vi 7 275 20

b aR a

thay vo * ta c 2 95 5 5 20 5 3

2a a a a

Vy phng trnh ng trn l

2 2

: 3 6 25C x y hoc 2 2

9 9 25:

2 2 4C x y

0,25

-Vi7 11

5 5

a b

R b

thay vo * ta c

2 35 3 4 5 5 5 2

2b b b b

Vy phng trnh ng trn l

2 2

: 3 2 25C x y

hoc

2 21 3 25

: 2 2 4C x y

0,25

Cu 9.Gii bt phng trnh : 2

2 2 1

26 2 4 2 2

x

x x x

1,0

iu kin : 2x

Ta c

2

2

2

2 2 46 2 4 2 2 0, 2

6 2 4 2 2

x xx x x x

x x x

0,25

9 .(1,0 )

Do bt phng trnh 22 2 2 6 2 4 2 2x x x x

2

2 2 2 12 2 6 1x x x x

Nhn xt 2x khng l nghim ca bt phng trnhKhi 2x chia hai v bt phng trinh 1 cho 2 0x ta c

0,25

2

2 2 12 6 22 2

x x

x x

. t

2t

x

th bt phng trnh 2 c

2

22 2

12 2 02 2 12 6 2

4 8 4 12 6 2 2 0

ttt t t

t t t t

0,25

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Lu khi chm bi:

- p n ch trnh by mt cch gii bao gm cc bt buc phi c trong bi lm ca hc sinh. Khi chm

nu hc sinh b qua bc no th khng cho im bc .

- Nu hc sinh gii cch khc, gim kho cn c cc trong p n cho im.

- Trong bi lm, nu mt bc no b sai th cc phn sau c s dng kt qu sai khng c im.- Hc sinh c s dng kt qu phn trc lm phn sau.

- Trong li gii cu 7 nu hc sinh khng v hnh th khng cho im.

- im ton bi tnh n 0,25 v khng lm trn.

2

02 2 2 2 3

4 8 02

xxt x

x xx

. Bt phng trnh c nghim duy

nht 2 2 3x . (Ch bi ny c nhiu cch gii khc nh dng vc t, dng btng thc ,dng php bin i tng ng)

0,25

Cu 10.Cho , 0x y tha mn iu kin 2016x y .Tm gi tr nh nht ca biu thc

2 2 2 2 2 2 2 25 3 3 5 2 2P x xy y x xy y x xy y x xy y

1,0

P A B . Trong 2 2 2 25 3 3 5A x xy y x xy y v2 2 2 22 2B x xy y x xy y

0,25

10.(1,0) 2 2 2 26 180 36 108 108 36 180A x xy y x xy y

2 2 2 2

11 7 59 11 7 59y x y y x y x

11 7 11 7 18y y x x y

3 3 2016 6048 *A x y du ng thc xy ra khi v ch khi 1008x y

0,25

2 2 2 24 16 16 32 32 16 16B x xy y x xy y

2 2 2 2

3 5 7 3 5 7x y x y y x y x

3 5 3 5 8y y x x y

2 2 2016 4032 **B x y du ng thc xy ra khi v ch khi 1008x y

0,25

T * v ** ta c 6048 4032 10080P A B ,du ng thc xy ra khi v

ch khi 1008x y . Vy min 10080 1008P x y 0,25

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SGIO DC V O TOTNH QUNG NINH

CHNH THC

THI KHO ST LP 12 NM HC 2015-2016Mn: TON

Thi gian lm bi: 180 pht khng kthi gian pht

(thi gm 01 trang)

Cu 1 (1,0 im)Kho st sbin thin v vthhm s 3 26 9 1= + y x x x

Cu 2 (1,0 im)Tm gi trln nht, gi trnhnht ca hm s

9

1= +

y x

xtrn on [2;5]

Cu 3 (1,0 im)a) Gi

1 2,x x l hai nghim trn tp sphc ca phng trnh 2 2 5 0x x+ + = . Tnh 1 2x x+ .

b) Gii phng trnh 22 12

log ( 2 8) 1 log ( 2)x x x = + .

Cu 4 (1,0 im) Tnh tch phn

= +2

2

0

( sin ) cos .I x x xdx

Cu 5 (1,0 im)Trong khng gian vi h ta Oxyz, cho hai im A(2; 3; 5), B(6; 1; 3) v mt phng (P) c

phng trnh 2 2 13 0x y z+ + = . Vit phng trnh ng thng AB v phng trnh mt cu c tm ltrung im ca on thng AB ng thi tip xc vi mt phng (P).

Cu 6 (1,0 im)

Cho hnh hp ngABCD.ABCDc y l hnh thoi cnh a, gc 060ACB= , mt phng (ABD)

to vi y mt gc 060 . Tnh theo athtch khi hp v khong cch gia hai ng thng CD,BD.

Cu 7 (1,0 im)

a) Cho 2sin3

= , vi2

< < . Tnh 2cos3

A = +

b) Gii bng chuyn VTV Cup gm 12 i bng tham d, trong c 9 i nc ngoi v 3 i caVit Nam. Ban tchc cho bc thm ngu nhin chia thnh 3 bng A, B, C mi bng 4 i. Tnh xcsut 3 i bng ca Vit Nam ba bng khc nhau.

Cu 8 (1,0 im)Trong mt phng vi h ta Oxy, cho tam gic nhn ABC ni tip ng trn (C): 2 2 25x y+ = ,

ng thng AC i qua im K(2; 1). Gi M, N ln l!t l chn ng cao k"t#$nh B v C. Tm ta cc $nh ca ABC bit phng trnh ng thng MN l 4 3 10 0x y + = v im A c honh m.

Cu 9 (1,0im) Gii phng trnh

2 2

1 2 9 18 14 33+ + = + +

x x x x x trn tp sthc.Cu 10 (1,0 im)

Chox, y, z l cc sthc tha mn 2 2 2 25 2 2 8 4 5 4 2x xy y x xz z x y z+ + + + + = + + v [0;5]x

Tm gi trln nht, gi trnhnht ca biu thc 2 21 10P z xy x z xy= + + + .

===============Ht===============

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uy cp http://dethithu.netthng xuyn cp nht nhiu Thi ThTHPT Quc Gia, ti liu HPT Quc Gia cc mn Ton, L, Ha, Anh, Vnnh , S, a c DeThiThu.Net cp nht hng ngy phc vst!

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SGIO DC V O TOTNH QUNG NINH

HNG DN CHM THI KHO ST LP 12NM HC 2015-2016

Mn: TON

Cu P N V BIU IM imCu 1

1,0

Kho st sbin thin v vthhm s 3 26 9 1= + y x x x

* Tp xc nh: D = R* Sbin thin

Chiu bin thin: )34(39123' 22 +=+= xxxxy

Ta c

>

1

30'

x

xy , 310'


28/117

Phna)

0,5

Tnh !c hai nghim phc 1 1 2x i= ; 2 1 2x i= + 0,25

1 2 1 25 2 5x x x x= = + = 0,25

Phnb)

0,5

b) &K:x> 4

PT cho tng ng vi 22 2 2log ( 2 8) log 2 log ( 2)x x x = + +

2

2 2log ( 2 8) log 2( 2)x x x = +

0,25

2

2 0

2 8 2( 2)

x

x x x

+ >

= +

2

2 06

4 12 0

xx

x x

+ > =

=

0,25

Cu 4

1,0 Tnh tch phn

= +2

2

0

( sin ) cos .I x x xdx

= + = +

2 2 22 2

0 0 0

( sin ) cos cos sin cos .

M N

I x x xdx x xdx x xdx 0,25

TnhM

&tcos sin

u x du dxdv xdx v x

= =

= =

2

0

sin sin cos 1.2 22 2

0 0

M x x xdx x

= = + =

0,25

TnhN

&t sin cost x dt xdx= =

&i cn1

2

0 0

x t

x t

= =

= =

1 32

0

1 1.

03 3

tN t dt= = =

0,25

Vy2

.2 3

I M N

= + = 0,25

Cu 5

1,0

Trong khng gian vi hta Oxyz, cho hai im A(2;3;5), B(6;1;3) v mt phng (P) cphng trnh 2 2 13 0x y z+ + = . Vit phng trnh ng thng AB v phng trnh mt cu ctm l trung im ca on thng AB ng thi tip xc vi mt phng (P).

Ta c (8; 2;8) 2BA u= =

vi (4;1;4)u =

Suy ra u

l VTCP ca ng thng AB 0,25

Phng trnh ng thng AB l:

2 4

3

5 4

x t

y t

z t

= +

= + = +

0,25

Gi I l trung im ca on thng AB. Ta c I(-2;2;1)V mt cu cn tm tip xc vi (P) nn bn knh R = d(I,(P))= 3

0,25

Phng trnh mt cu (S) cn tm l: 2 2 2( 2) ( 2) ( 1) 9x y z+ + + = 0,25

Cu 6 Cho hnh hp ng ABCD.ABCD c y l hnh thoi cnh a, gc 060ACB= , mt phng(ABD) to vi y mt gc 060 . Tnh theo athtch khi hp v khong cch gia hai ngthng CD, BD.

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1,0

Tnh thtch:

T# 060ACB = suy ra ABC u suy ra AC = a2

0 3. .sin 602

ABCD

aS AC CB = =

Gi O = ACBD. T#githit suy ra gc gia (ABD) vi mt

y l 0' 60A OA = O

A' B'

D'C'

A

B

D C

H

0,25

0 3' . tan 60 2

aA A OA = = Suy ra

33

. ' 4ABCD

aV S A A= = 0,25

Khong cch gia hai ng thng CD, BD

Trong AAO hAH AO. Do ( ' )'

BD ACBD A AO BD AH

BD A A

T# suy ra ( ' )AH A BD . Ta c CD//AB '/ /( ' )CD A BD

( ', ) ( ,( ' )) ( ,( ' ))d CD BD d C A BD d A A BD AH = = =

0,25

Trong AHO vung ti H c 03

.sin604

aAH OA= = . Vy

3( ', )

4

ad CD BD = 0,25

Cu 7Phna)

0.5

Cho

2

sin 3 = vi 2

< < . Tnh

2

cos 3A

= +

Ta c2 2 5 5cos 1 sin cos

9 3 = = = (v

2

< < nn cos 0 < ) 0,25

2 2 5 2 3cos .cos sin .sin

3 3 6A

= = 0,25

Cu 7Phnb)

0.5

Gii bng chuyn VTV Cup gm 12 i bng tham d, trong c 9 i nc ngoi v 3 i caVit Nam. Ban tchc cho bc thm ngu nhin chia thnh 3 bng A, B, C mi bng 4 i.Tnh xc sut 3 i bng ca Vit Nam ba bng khc nhau.Tnh scch chn 3 bng, mi bng 4 i:

B1) 12 i chn 4:4

12C B2) 8 i cn li chn 4: 48C

B3) 4 i cn li chn 4: 1Scch chn l: ( )4 4 4 412 8 12 8C C n C C =

0,25

Gi A l bin c Chn 3 bng, mi bng 4 i trong c ng 1 i Vit Nam.Tnh n(A):

B1) Chn 1 trong 3 i Vit Nam: c 3 cch, ri chn 3 trong 9 i nc ngoi: c3

9C3

93.C cch

B2) cn li 8 i (6 i nc ngoi v 2 i VN): Chn 1 trong 2 i VN: 2 cch, ri

chn 3 trong 6 i nc ngoi: 3 36 62.C C cch

B3) cn li 4 i (3 nc ngoi v 1 VN): c 1 cch

Scch chn l: ( )3 3 3 39 6 9 63 2 3 2C C n A C C = ( )3 3

9 6

4 4

12 8

6 16

55

C CP A

C C = =

0,25

Cu 8

1,0

Trong mt phng vi hta Oxy, cho tam gic nhn ABC ni tip ng trn (C): 2 2 25x y+ = ,ng thng AC i qua im K(2;1). Gi M, N ln l!t l chn ng cao k"t#$nh B v C. Tmta cc $nh ca ABC bit phng trnh ng thng MN l 4 3 10 0x y + = v im A c

honh m.

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T#githit suy ra tgic MNBC ni tip ng trn. Suy ra

ABC AMN= (1) (cng b vi NMC).Gi D l giao im thhai ca AO vi ng trn (C). Khi ABC ADC= (2)

T#(1) v (2) suy ra ADC AMN= .Mt khc

0 090 90ADC DAC DAC AMN OA MN+ = + = D

N

MO

A

B C

0,25

Khi phng trnh OA l 3 4 0x y+ =

Ta A l nghim ca hPT2 2

3 4 0

25

x y

x y

+ =

+ =A(-4;3) hoc A(4;-3) (loi)

0,25

Khi AC i qua A(-4;3) v K(2;1) nn c PT: 3 5 0x y+ =

Ta C l nghim ca hPT2 2

3 5 0 ( 4;3) A

C(5;0)25

x y C

x y

+ =

+ =

Ta M l nghim ca hPT3 5 0

( 1; 2)4 3 10 0

x yM

x y

+ =

+ =

0,25

Phng trnh BM: 3 5 0x y + =

Ta B l nghim ca hPT2 2

3 5 0 ( 3; 4)

(0;5)25

x y B

Bx y

+ =

+ =

Th)li ta thy A(4;3), B(0;5), C(5;0) loi v gc B tVy A(4;3), B(3;4), C(5;0)

0,25

Cu 9

1,0

Gii phng trnh 2 21 2 9 18 14 33+ + = + +x x x x x (1)

&K:3

11

x

x

PT ( ) ( )2 21 2 9 18 14 33 1x x x x x x + = + +

(2)

& r*ng hai phng trnh 2 9 18 0x x x + + = v ( )2 14 33 1 0x x x + + + = v

nghim nn nhn lin h!p hai vca (2) ta c:

( ) ( )

( )2 218 2 16 2

9 18 14 33 1

x x

x x x x x x

=

+ + + + +

0,25

( )2 2

2

9 8(3)

9 18 14 33 1

x

x x x x x x

=

= + + + + +

Pt (3)

2 2

8 9 18 9 14 33 9x x x x x + + = + (4)

0,25

Kt h!p (1) v (4) ta c h2 2

2

2 2

8 9 18 9 14 33 95 14 33 3 13

2 9 18 14 33 1

x x x x xx x x

x x x x x

+ + = + + =

+ + =

0,25

2

1317 5 5

32

17 41 0

xx

x x

+

= + =

Th)li thy tha mn

Vy PT cho c 2 nghimx= 2 v17 5 5

2x

+=

0,25

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Cu10

1,0

Chox, y, z l cc sthc tha mn 2 2 2 25 2 2 8 4 5 4 2x xy y x xz z x y z+ + + + + = + + (*) v

[0;5]x . Tm gi trln nht, gi trnhnht ca biu thc 2 21 10P z xy x z xy= + + + .

Vi mix, y, zta c2 2 2 2 25 2 2 (2 ) ( ) (2 ) 2 2x xy y x y x y x y x y x y+ + = + + + = + +

2 2 2 2 28 4 5 4(x z) ( 2 ) 4(x z) 2 x z 2(x z)x xz z z x+ + = + + + = + +

Suy ra 4 2VT x y z + +

Gt Du = xy ra 2

0

x y

z x

x

=

=

0,25

Thay vo biu thc ta c 2 24 21 3 10 ( )P x x x x f x= + + + + = lin t(c trn [0;5]

C2 2

2 3 2'( )

4 21 2 3 10

x xf x

x x x x

=

+ + + +

0,25

1'( ) 0

3f x x= = 0,25

Ta c (0) 21 10f = ; 1 23

f = ; (5) 4f =

Vy max 4P = khi 5; 10x y z= = = ; min 2P = khi1 2

;3 3

x y z= = =

0,25

Truy cp http://dethithu.netthng xuyn cp nht nhiu ThiThTHPT Quc Gia, ti liu n thi THPT Quc Gia cc mn Ton, LHa, Anh, Vn

Sinh , S, a c DeThiThu.Net cp nht hng ngy phc vst!

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Cu 1 (2,0 im) .

a ) Kho sthm s v v th ( C ) ca hm s3 23 1

2 2y x x

b) Tm ta ca im M trn ( C ) sao cho tip tuyn ca ( C ) ti M song song ving thng ( d ) : x6 4 0y Cu 2 (1,0 im).

a). Cho hm s 2( 1)xy e x x .Tnh1

'(ln )2

y

b) Gii bt phng trnh sau3 1

3

2 log (4x 3) log (2x 3) 2 .

Cu 3 (1,0 im).Tnh tch phn0

(2 1) sinI x xdx

.

Cu 4(1,0 im). Trong khng gian Oxyz,cho mp(P) v mt cu (S) ln lt c phngtrnh ( ) : 2 2 1 0P x y z v 2 2 2( ) : 4 6 6 17 0S x y z x y z .Chng

minh mt cu(S) ct mt phng(P) .Tm ta tm v bn knh ca ng trn giaotuyn ca mt cu v mt phng.

Cu 5(1,0 im).

a)Cho tan 3 . Tnh3 3

3sin 2 cos

5sin 4 cosA

.

b)Cho a gic u 20 nh ni tip ng trn tm O. Chn ngu nhin 4 nh ca agic .Tnh xc sut sao cho 4 nh c chn l 4 nh ca mt hnh ch nht.

Cu 6(1,0 im). Cho hnh chp S.ABCc mt phng (SAC) vung gc vi mt phng(ABC), , 2SA AB a AC a v 090 .ASC ABC Tnh th tch khichp S.ABCvcosin ca gc gia hai mt phng (SAB), (SBC).Cu 7 (1,0 im). Trong mt phng vi h trc ta Oxy, cho hnh bnh hnh ABCD

c 0135B AD , trc tm tam gic ABD l H(-1;0).ng thng i qua D v H c

phng trnh 3 1 0x y . Tm ta cc nh ca hnh bnh hnh bit im G(5

;23

)

l trng tm tam gic ADC.Cu 8 (1,0 im).Gii h phng trnh sau

3 3 2

33 3 6 4 0

( 2 3 7 13 ) 3( 1)x y y x yy x y x

Cu 9 (1,0 im).Cho x, , 0y z v 2 2 25( ) 9( 2 x)x y z xy yz z .

Tm gi tr ln nht ca biu thc2 2 3

1

( )

xP

y z x y z

.

TRNG THPT CHUYNNGUYN HU

KHO ST CHT LNG LN 2MN TON NM HC 2015-2016

Thi gian lm bi: 180 pht,khng k thi gian pht http://dethithu.net

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33/117

TRNG THPT CHUYNNGUYN HU

KHO ST CHT LNGLN TH 2

NM HC 2015 2016P N V BIU IM MN: TON

Cu 1 Ni dung 1,01a.

(1,0

im)

Hm s 3 23 1

2 2y x x

TX: D = R S bin thin:

- Chiu bin thin: 2' 3 3y x x ,0

' 01

xy

x

0,25

Hm s ng bin trn cc khong ( ;0) v (1;+ ) , nghch bin trn khong (0;1)

- Cc tr: Hm s t cc i ti 1

0; 2Cx y , t cc tiu ti x 1, 0

CTy .

- Gii hn:

lim ; limx x

y y

0,25

- Bng bi n thin:

x 0 1

y + 0 0 +

y

1

2

0

0,25

th:

y

x

1

2

O 1

0,25

1b.

(1 )+ ng thng 6xy4=0 c h s gc bng 6+Gi M0( x0; y0) l im m ti tip tuyn song song ng thng6x - y- 4=0 0'( ) 6f x

0,25

0,25http://dethithu.net

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34/117

2

0 0

0

0

3 3 6

1

2

x x

x

x

+Vi x0=2 y0= 5/2M0( 2; 5/2)

x0= -1y0= -2 M0( -1 ; -2 )+ Kim tra liM0( 2,5/2) tip tuyn ti M0c pt l y= 6(x2)+5/2( nhn)M0(-1;-2)tip tuyn ti M0c pt l 6( 1) 2y x =6x+4(nhn)

0,25

0,25

Cu

2(1,0

im)2a(0,5

im)

TX: D=R2 2 2

2

( ) ( 1) ( 1) ( 1) (2 1)

( 3 )

x x x x

x

y e x x e x x e x x e x

e x x

21'(ln ) 2( ln 2 3 ln 2)2

y

0,25

0,25

2b(0,

5

im)

iu kin3

4x

Bt phng trnh tng ng2

3

(4x 3)log 2

2x 3

216x 42x 18 0

33

8x

Kt hp iu kin ta c tp nghim ca bt phng trnh l S= 3

;34

0,25

0,25

Cu

3(1)

t2 1 2.

sin cos

u x du dx

dv xdx v x

0,25

0 0(2 1)cos ( 2 cos )I x x x dx

0,25

= 0(2 1) 1 2 sin x

0,25

=2 2 0,25

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Cu

4(1,0

)

Mt cu (S) c tmI(2;3;3), bn knh 2 2 22 ( 3) ( 3) 17 5R 0,25

Khong cch t tm I n mp(P):

2 2 2

2 2( 3) 2( 3) 1

( ,( )) 11 ( 2) 2d d I P R

V ( ,( ))d I P R nn (P) ct mt cu (S) theo giao tuyn l ng trn (C)

0,25

Gi dl ng thng qua tmIca mtcu v vung gc mp(P) th dc vtcp

(1; 2;2)u

nn c PTTS

2

: 3 2

3 2

x t

d y t

z t

(*). Thay (*) vo pt mt phng (P)

ta c1

(2 ) 2( 3 2 ) 2( 3 2 ) 1 0 9 3 0 3t t t t t

Vy, ng trn (C) c tm5 7 11

; ;3 3 3

H

0,25

Bn knh 2 2 5 1 2r R d 0,25

Cu 5

a(0,5

im) 3 3 2 33sin 2 cos 3 tan 2

5sin 4 cos cos 5 tan 4A

0,25

2

3

3 tan 2 701 tan5 tan 4 139

0,25

Cu

5b(0,

5)

-C 10 ng knh ca ng trn c ni bi 2 nh ca a gic u. - Mt hnh ch nht c 4 nh l nh ca a gic c to bi 2 ng knh ni trn.

-S cch chn 4 nh ca a gic l: 420

4845C

-S cch chn 4 nh ca a gic to thnh hnh ch nht l 210

45C

-Xc sut cn tm l : P= 45 3

4845 323

0,25

0,25http://dethithu.net

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A

S

C

B

M

H

Cu

6(1,0

)

+ K SH vung gc AC (HAC) SH (ABC)

3

3, ,2

aSC BC a SH

2 3

2ABC

aS

3

.

1.

3 4S ABC ABC

aV S SH

0,25

0,25

Gi M l trung im SB v l gc gia hai mt phng (SAB) v (SBC).

Ta c: SA = AB = a, SC BC a 3

AM SB v CM SB

cos cosAMC

0,25

+ SAC = BAC3 6

2 2

a aSH BH SB

AM l trung tuyn SAB nn:2 2 2 2

2 2 2 10

4 16

AS AB SB aAM

10

4

aAM

Tng t:42

4

aCM

2 2 2AM CM AC 105cosAMC

2.AM.CM 35

Vy:105

cos35

0,25

Cu

7(1,0

)

Ta c 0 0180 45B AD BH D BH D .

Gi ( ; )n a b

2 2( 0)a b l VTPT ca ngthng HB

Do ngthng HB to vi ng thng HD gc 045 nn

cos 0 2 2

2 2

2345 2a 3a 2 0

2a. 10

a ba bb b

ba b

0,25

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Nu a=-2b. Chn a=2,b=-1. Phng trnh ng thng HB: 2x-y+2=0B(b;2b+2), D(3d-1;d)

Do G l trng tm tam gic ADC nn BG=2GD 1

2 D1

bGB G

d

B(1;4),

D(2;1)

Phng trnh ng thng AB: 3x+y-7=0; phng trnh ng thng AD:x+2y-4=0Suy ra A(2;1)(loi)Nu b=2a. Phng trnh HB: x+2y+1=0

B(-2b-1;b), D(3d-1;d)2

2 D2

bGB G

d

B(-5;2), D(5;2)

Phng trnh AB: 3x+y+13=0; Phng trnhAD:2x-y-8=0. Suy ra A(-1;-10)

Do ABCD l hnh bnh hnh suy ra DA BC

suy ra C(1;14)

Th li: cos DAB =cos ( ; )AB AD

= 01

D 452

BA (LOI)

0,25

0,25

0,25

Cu

8(1,0

)

iu kin3

x2

T phng trnh (1) ta c 3 3x 3 ( 1) 3( 1)x y y

Xt hm s3

2

( ) 3

'( ) 3 3

f t t t

f t t

'( ) 0f t vi mi t suy ra hm s f(t) ng bin trn R.( ) ( 1) 1f x f y x y

0,25

0,25

Th x=y+1 vo phng trnh (2) ta c:3

( 1)( 2x 3 7x 6) 3( 1) (3)x x Ta c x=1 khng l nghim phng trnh.T

3 3( 1)( 2x 3 7x 6 )1

x

x

Xt hm s 3 3( 1)

( ) ( 2x 3 7x 6)1

xg x

x

TX: 3

; \ 12

D

223

1 7 6'( )

( 1)2x 3 3 (7x 6)g x

x

3'( ) 0 ; 1

2g x x x ,

3'( )

2g khng xc nh.

Hm s ng bin trn tng khong3

( ;1)2

v (1; ) .

Ta c g(-1)=0; g(3)=0. T phng trnh g(x)=0 c ng hai nghim x= -1 v x=3.Vy h phng trnh c hai nghim (-1;-2) v (3;2)

0,25

0,25

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Cu 9t y+z=t (t>0);

2 22 2 ;

2 4

t ty z yz .

2 2 2

2 2

2 2 2

5( ) 9( 2 z)

5x 5( ) 9x( ) 285x 5 9x 7

(5x )(x 2 ) 0

2

x y z xy yz x

y z y z yzt t t

t t

x t

0,25

0,25

2 3 3

2x 1 4 1

27 27P

t t t t

Xt hm s3

4 1( )

27f t

t t vi t>0

2 4

4 1'( )

9

'( ) 0 1

0 6

f tt t

f tt

t

Lp bng bin thin t suy ra GTLN ca P bng 16 t c ti1 1

x ;3 12

y z

0,25

0,25

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Truy cp http://dethithu.netthng xuyn cp nht nhi u Thi Th THPT QuGia, ti liu n thi THPT Quc Gia cc mn Ton, L, Ha, Anh, VnSinh , S, a c DeThiThu.Net cp nht hng ngy phc v s t!

Like Fanpage Thi Th THPT Quc Gia - Ti Liu n Thi:http://facebook.com/dethithu.net cp nht nhiu thi th v ti liu n thi hn

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TRNG THPT CHUYN BIN HA THI TH THPT QUC GIA LN 1Nm hc 2015 - 2016

Mn: TONThi gian lm bi : 180 pht, khng k thi gian pht

Cu 1 ( 1,0 im). Kho st s bin thin v v th ca hm s 3 3 1y x x .

Cu 2(1,0 im). Tm cc im cc tr ca th hm s 2

1f x x x .Cu 3(1,0 im).

a) Cho s phc z tha mn

31 4 1

1

i iz

i

. Tm modun ca s z.

b) Gii bt phng trnh 2 1 2 13 2 5.6 0x x x

Cu 4(1,0 im). Tnh tch phn 21

2lne xI dx

x

.

Cu 5(1,0 im).Trong khng gian vi h to Oxyz , cho im 4;1;3A v ng thng

1 1 3: 2 1 3x y zd . Vit phng trnh mt phng ( )P i qua A v vung gc vi ng

thng d. Tm ta im B thuc dsao cho 5AB . Cu 6(1,0 im).

a)

Gii phng trnh sin 2 1 6sin cos 2x x x b) cho mng ngy 26/03, trng t chc cm tri. Lp 10 A 19 hc sinh nam, 16 hc

sinh n. Gio vin cn chn 5 hc sinh trang tr tri. Tnh xc sut trong 5 hc sinhc chn c t nht 1 hc sinh n bit rng hc sinh no trong lp cng c kh nngtrang tr tri.

Cu 7(1,0 im). Cho hnh chp .S ABCD c y l hnh thang vung ti A v B. Cc mtbn (SAB) v (SAD) cng vung gc vi mt phng y. Cho AB = 2a, AD > a. SA = BC = a,

2 5CD a . Gi H l im nm trn on AD sao cho AH = a . Tnh th tch ca khi chpSABCD v khong cch gia hai ng thng BH v SC theo a.Cu 8(1,0 im). Trong mt phng vi h to Oxy cho tam gic ABC c AC=2AB, im

91;

2M

l trung im ca BC, D l im thuc cnh BC sao cho BAD CAM . Gi E l trung

im ca AC, ng thng DE c phng trnh: 2 11 44 0x y , im B thuc ng thng dc phng trnh: x + y 6 = 0. Tm ta 3 im A, B, C bit honh ca im A l mt snguyn.

Cu 9(1,0 im). Gii h phng trnh 2 2 2 2

2 2

2 5 2 4

3 2 2 9 0

xy y y xy y y xy

y x x x x y

Cu 10(1,0 im).Cho cc s , ,a b c khng m sao cho tng hai s bt k u dng. Chng

minh rng:9

6a b c ab bc ca

b c a c a b a b c

.

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P N THI TH THPT QUC GIA NM 2015-2016 LN 1

MN : TON

Ni dung imCu 1(1,0 im)

3 3 1x x TX: D R

2' 3 3y x , ' 0 1y x

0.25

limx

y

, limx

y

* Bng bin thinX -1 1 + y - 0 + 0 -

y+ 3

-1 -

0.25

Hm s nghch bin trn cc khong ; 1 v 1; , ng bin trn khong 1;1

Hm s t cc i ti 1 , 3CDy , t cc tiu ti 1x , 1CTy

0.25

th: th hm s i qua cc im (0;1), (-2;3), (2; -1). 0.25

Cu 2 (1,0 im)

:TXD

2

2 1'

2 1

xf x

x x

0.25

2

2

4

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1

' 02

f x x

0,25

Bng bin thinX 1/2 + y 0 +

y

3

2

0.25

th hm s c im cc tiu l1 3

;2 2

. th hm s khng c im cc i.0.25

Cu 3( 1,0 im)

Cu 3a) (0,5 im)

3 2 3

2

1 4 1 1 4 1 3 3 ( 1 2 )(1 )

1 1 (1 )(1 )

1 2 2 1 3

2 2

i i i i i i i iz i i i i

i i i i

0,25

2 21 3 10

2 2 2z

0,25

Cu 3b (0,5 im)

b)2

3 3 1 33 2 5 0 2

2 2 3 2

x x x

bpt

0,25

3 1

2 3

x

(lun ng)

3

2

32 log 2

2

x

x

.

Vy tp nghim ca bt phng trnh l 32

, log 2

0,25

Cu 4( 1,0 im)

2 2 211 1 1 1

1 ln ln ln2 ln 2 1 2e e e e

e x xI dx dx x dx dxx x x x

0.25

Tnh2

1

lneJ dx

x

t2

1ln ,u x dv dx . Khi

1 1,du dx v

x

0.25

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Do 2

1 1

1 1ln

e e

J x dxx x

1

1 1 21

e

Je x e

0.25

Vy4

1I

e

0.25

Cu 5( 1,0 im)

ng thng d c VTCP l 2;1;3du

V P d nn P nhn 2;1;3du

lm VTPT

0,25

Vy PT mt phng P l : 2 4 1 1 3 3 0x y z

2 3 18 0x y z

0,25

V B d nn 1 2 ;1 ; 3 3B t t t

5AB 2 22 25 3 2 6 3 5AB t t t 27 24 20 0t t

0,25

210

7

tt

Vy 5;3;3B hoc 27 17 9; ;7 7 7

B

0,25

Cu 6 ( 1,0 im)

Cu 6a(0,5 im)sin 2 1 6sin cos 2x x x

(sin 2 6sin ) (1 cos2 ) 0x x x 22sin cos 3 2sin 0x x x

2sin cos 3 sin 0x x x

0.25

sin 0

sin cos 3( )

x

x x Vn

x k . Vy nghim ca PT l ,x k k Z

0. 25

Cu 6b(0,5 im)

Gi A l bin c :trong 5 hc sinh c chn c t nht mt hc sinh n

535n C

0.25

S cch chn 5 hc sinh trong c t nht 1 hc sinh n l n(A)=5 535 19C C

Do xc sut 5 hc sinh c chn trong c t nht 1 hc sinh n l5 535 19

535

0,96C C

PC

0,25

Cu 7(1,0 im)

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Do (SAB) v (SAD) cng vung gc vi ynn ( )SA ABCD . AHCB l hnh bnh hnh,suy ra CH=AB=2a,

2 2 4 5 .HD CD CH a AD a

0.25

2

3

15 2 6

21

. 2 .3

ABCD

SABCD ABCD

S a a a a

V SA S a

0.25

Trong mt phng (ABCD), k CE//BH (E thuc AD), ta c

, , , ,

1

2BH SC BH SCE H SCE A SCE d d d d

K

,

,

A SCE

AF CE AJ SF AJ SCE

d AJ

0.25

Gi K l giao im ca BH v A F.

2 2 2

2 2 2

, ,

1 1 1 2 4

5 5

1 1 1 421

1 2

2 21BH SC A SCE

a aAK AF

AK AH AB

aAJAJ AS AF

ad d

0,25

Cu 8(1,0 im)Gi I l giao im ca BE v AD, G l giaoim ca AM v BE. ( . . )ABI AEG g c g suy ra BI=GE m BG=2GE( do G l trng tmca tam gic ABC) suy ra BI=IG=GE.

K EH //BC ( H thuc on AD). Chng minhc CD=2HE, HE=2BD, suy ra CB=5BD

0,25

K

B C

D

S

H E

F

J

G

E

B CD

H

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92 5 , ( ;6 ), (22 11 ;2 ), 1;

2

9 11 18;55 3 108

5 5 510 3 27

3 3;3

BM BD B b b D d d M

d Dd b

d bb B

0,25

91;2

M

l trung im ca BC suy ra C(-1;6). 0,25

Gi E(22-11e;2e), E l trung im ca AC suy ra A(45-22e; 4e-6)

2

2( ) (1;2)2 75 278 256 0 128

( )75

e tm AAC AB e e

e l

Vy A(1;2), B(3;3), C(-1;6).

0,25

Cu 9(1,0 im).

: 4 2 0DK y x y

Vi y=0 th x=0.

2 2 2 22

0, (1) 2 5 2 4 0

2 5 1 2 4 0

y x xy y y xy y y xy

x x x x

y y y y

0.25

t 2;4x

t ty

22 5 1 2 4 0 2 3 2 2 1 1 4 0

3 2 32 3 0 3 3

2 1 1 4

t t t t t t t t t

t t tt t t x y

t t

0,25

Thay 3x y , thay vo (2) ta c :

2 2

2

2 2 0

1 2 1 2

x x x x x x

x x x x

Xt hm s 2

2 2

21 2 , ' 1 2 0

2

tf t t t f t t t

t

0.25

0 0

11

3

x yf x f x x x

x y

0.25

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Vy h phng trnh c nghim 1

0;0 , 1;3

Cu 10 (1,0 im) .

t9a b c ab bc ca

Pb c a c a b a b c

Gi s a b c , khi . .ab ac b b c c b ca c a b b c c b

0,25

Suy rab c b c

a c a b a

.

0,25

t t b c th9a t at

Pt a a t

.0,25

Ta c9 9

6a t at a t at

t a a t a t at

(AM-GM). Do 6P (pcm).

ng thc xy ra khi 3a t at v chng hn mt b ( , , )a b c tha mn l

7 3 5( ; ; ) ;1;0

2a b c

0,25

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ruy cp http://dethithu.netthng xuyn cp nht nhiu Thi ThTHPT Quc Gu n thi THPT Quc Gia cc mn Ton, L, Ha, Anh, Vninh , S, a c DeThiThu.Net cp nht hng ngy phc vst!

ike Fanpage Thi ThTHPT Quc Gia - Ti Liu n Thi:ttp://facebook.com/dethithu.netcp nht nhiu thi thv ti liu n thi hn

ham gia Group: n Thi H Ton - Anh cng nhau hc tp, n thi:http://facebook.com/groups/onthidhtoananhvan


46/117

S GD&T BC GIANG THI TH THPT QUC GIA LN 3TRNG THPT NG S LIN Mn: Ton 12

- Thi gian: 120 pht, khng k thi gian pht CHNH THC ( gm01trang)

Cu 1 (2,0 im).Cho hm s y=2x + m

x 1 (C)(vi ml tham s thc).

1)Kho st s bin thin v v th hm s (C)vi m= 1.2)Tm m ng thng d : y = x+ 2ct th hm s(C)ti hai imA, Bphn bit.

Cu 2 (1,5 im). 1)Gii phng trnh: sinx + 2sin3x= sin5x

2)Gii phng trnh: log3(x 1)2 + log3(2x 1) = 2

Cu 3 (1,0 im).Tm nguyn hm sau: I=

(x+cosx)xdx

Cu 4 (1,0 im). Trong khng gian Oxyz, cho 4 im A(1;1;3), B(2;0;4), C(2;3;5),D(0;4;5Chng minh rng 4 im cho khng ng phng v tnh th tch ca t din ABCD.

Cu 5 (1,0 im).Cho hnh chpS.ABCDc y ABCDl hnh ch nht AB =a, AD=a

2. Hnh chiu vung gc ca nh Sln (ABCD)trng vi trng tm ca tam gic ABC, ngthng SD to vi mt y ABCDmt gc 450. Tnh th tch ca hnh chp S.ABCDv tnhkhong cch gia hai ng thng SCvBD theo a.

Cu 6 (1,0 im). chun b cho L k nim 70nm thnh lp ca mt trng THPT, nhtrng cn lp mt i tnh nguyn vin gm 40 em hc sinh thng qua n ng k. Qua ngk c 150 em hc sinh mun tham gia i tnh nguyn vin, bit rng trong 150 em c 60 emc hc lc gii. m bo cng bng nh trng quyt nh chn ngu nhin 40 hc sinh t 150hc sinh ni trn. Tnh xc sut trong s 40 em hc sinh c chn c ng 80%hc sinh chc lc gii.

Cu 7 (1,0 im).Trong mt phng Oxy, cho hnh ch nht ABCDc AB = 2BC. Gi Hl hnh chiu ca Aln ng thng BD. Gi E, F ln lt l trung im ca cc on thngCDvBH. Bit im A(1;1), phng trnh ng thngEF : 3x y 10 = 0v im Ec tung m. Tm ta cc nh B,C,Dca hnh ch nht.

Cu 8 (1,0 im).Gii h phng trnh sau:(x +

x2 + 1)(y+

y2 + 1) = 1

3x2 + y+ 3 =

3x + 1 +

4 5y, (x, y R)

,Cu 9 (0,5 im).Cho cc s a,b, cl cc s thc dng v tha mn: a + b+ c= 3. Tm gi

tr ln nht ca biu thc:

P = 2

3 + ab + bc + ca+ 3

abc

(1 + a)(1 + b)(1 + c)

Ht Th sinh khng c s dng ti liu. Cn b coi thi khng gii thch g thm.H v tn th sinh: . . . . . . . . . . . . . . . . . . . . . . . ;S bo danh: . . . . . . . . . . . . . . . . . . . .

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S GD&T BC GIANG P N - THANG IMTRNG THPT NG S LIN THI TH THPT QUC GIA LN 3

NM HC 2015 - 2016 CHNH THC Mn: Ton 12

(p n gm 6 trang)

CU NI DUNG IM

Cho hm s y=2x + m

x 1 ... (1) 1,00

Tp xc nh: D= R\{1}

Gii hn: limx1

y= , limx1+

y= +, limx

y= 2, limx+

y= 2.

th hm s c TC:x = 1, TCN: y= 2

0,25

Cu 1

S bin thin: y = 3(x 1)2 ; y

< 0,x D.

Hm s nghch bin trn cc khong (; 1)v(1;+).

0,25

- Bng bin thin: 0,25

th: th ct trc Oyti im (0;1), ct trc Oxti im(12

; 0) 0,25

2)Tm m ... 1,00

Phng trnh honh giao im: 2x + m

x 1 =x + 2(iu kin: x = 1) (2)(2) x2 x 2 m= 0 (3)

0,5

1

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ng thng dct th(C)ti hai im A, Bphn bit khi v ch khi phng trnh(3) c hai nghim phn bit khc 1.

> 0

1 1 2 m = 0

m > 94

m = 2

0,5

1) Gii phng trnh: sinx+2sin3x= sin5x (1) 0,5(1) 2sin3xcos2x + 2sin3x= 0 2sin3x(cos2x + 1)= 0 0,25Gii phng trnh tch ta tm c nghimx= k

3 vx=

2+ l, (k, l Z)

0,25

Cu 2 2) Gii phng trnh: log3(x 2)2 + log3(2x 1) = 2 1,0

iu kin:

x 1 = 02x 1> 0

x = 1x >

1

2

0,25

PT log3|x 1| + log

3(2x 1) = 1

log3[|x 1| (2x 1)] = 1 |x 1| (2x 1) = 3 ()

0,25

Vi 12

< x 1. PT (*) (x 1)(2x 1) = 3 2x2 3x 2 = 0 x= 1

2, x= 2

i chiu iu kin ta c nghim x = 2

0,25

Tm nguyn hm: I=

(x+cosx)xdx 1,00

Ta c: I=

x2dx +

xcosxdx= I1+ I2 0,25

Cu 3

Tnh I1=

x2dx= x3

3 + C1 0,25

Tnh I2=

xcosxdx. Bng cch s dng TPTP ta tnh c

I2= xsinx+cosx + C2

0,25

Kt lun: I= x3

3 + xsinx+cosx + C 0,25

Trong khng gian Oxyz, cho 4 im A(1;1;3), B(2;0;4), C(2;3;5), D(0;4;5).Chng minh rng 4 im cho khng ng phng v tnh th tch ca t dinABCD.

1,00

Ta c:AB= (3;1;1),AC= (1;4;2),AD = (1;3;8) 0,25

2

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Cu 4 Tnh:

AB,AC

= (2; 7;13) =

AB,

AC

AD= 85 = 0.

Suy ra 4 im A, B,C,D khng ng phng.0,5

Suy ra: VABCD =85

6 (vtt) 0,25

Cho hnh chp S.ABCD c yABCD l hnh ch nhtAB = a,AD = a

2. Hnh

chiu vung gc ca nhSln(ABCD)trng vi trng tm ca tam gicABC, ngthngS Dto vi mt yABCDmt gc450. Tnh th tch ca hnh chp S.ABCD

v tnh khong cch gia hai ng thngSCvBD theo a.

1,00

S

A D

CB

E

F I

HO

K

Cu 5 - GiHl trng tm ca ABC, nn theo gi thit:S H (ABCD)v SDH= 450.Suy ra: SH=H D=

2

3

BD =2a

3

3-ABCDl hnh ch nht nn: SABCD = a22

0,25

- Th tch ca khi chp: V =1

3SH.SABCD=

2a3

6

9 (vtt). 0,25

- Gi El im i xng vi A qua B, ta c:

BD EC

BD (SC E)SC (SC E)

d(BD,SC) =d(BD, (SC E)) =d(H, (SC E)) (1)- Gi I , Kln lt l hnh chiu vung gc ca Hln EC,SIta c:

EC SHEC HI

HK ECKH SI

HK (SEC) d(H, (SE C)) =H K (2)

0,25

- Gi Fl hnh chiu ca B ln E C, ta c:B F =H Iv 1

BF2 =

1

BE2+

1

BC2, trong

HSC: 1HK2

= 1

HS2+

1

HI2 =

1

HS2+

1

BE2+

1

BC2

HK=2a3

(3)

T (1), (2), (3) ta suy ra: d(BD,SC) =2a

3

0,25

3

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Cu 6 chun b cho L k nim70nm thnh lp ca mt trng THPT, nh trng cnlp mt i tnh nguyn vin gm 40 em hc sinh thng qua n ng k. Qua ng k

c 150 em hc sinh mun tham gia i tnh nguyn vin, bit rng trong 150 em c

60 em c hc lc gii. m bo cng bng nh trng quyt nh chn ngu nhin

40 hc sinh t 150 hc sinh ni trn. Tnh xc sut trong s 40 em hc sinh c

chn c ng80% hc sinh c hc lc gii.

1,00

GiAl bin c ngu nhin: "Chn c80% hc sinh c hc lc gii".Chn ngu nhin 40 em hc sinh t 150 em hc sinh c n() =C40150

0,25

Chn c80%hc sinh c hc lc gii, tc l chn c 32 em. Chn 32 em trong 60em c hc lc gii c C3260 cch.Chn 8 em cn li trong 90 em c C890cch.Khi : n(A) =C890C3260

0,5

Theo cng thc tnh xc sut, ta tnh c: P(A) = C890C

3260

C40150 1, 823.109 0,25

Cu 7 Trong mt phngOxy, cho hnh ch nhtABCDc AB = 2BC. GiH l hnh chiuca A ln ng thngBD. GiE, F ln lt l trung im ca cc on thngCD

vB H. Bit imA(1;1), phng trnh ng thngE F : 3x y 10 = 0v imEc tung m. Tm ta cc nhB , C, D ca hnh ch nht.

1,00

A G B

CE

F

H

D

GiE , F , Gln lt l trung im ca cc onCD, BH, AB. Ta chng minh: AFEF.

Ta thy cc t gic ADEG, ADF Gni tip, nn t gic ADEFcng ni tip, do :AF EF. T ta suy ra: AF :x + 3y 4 = 0

0,25

- Tm ta ca F =AF EF, ta c F(175

;1

5), suy ra: AF =

32

5.

- Ta cAF E DC B EF =12

AF = 2

2

5.

- Gi E(a; 3a 10) EF EF2

= (a 17

5)2

+ (3a 51

5)2

=

8

5

5a2 34a + 57 = 0 a= 3 t= 195

. Hay E(3;1)hoc E(195

;7

5)

0,25

4

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Theo gi thit ta c E(3;1), phng trnh: AE : x+ y 2 = 0. Gi D(x; y),ADEvung, cn ti D nn:

AD= DE

AD DE

(x 1)2 + (y 1)2 = (x 3)2 + (y+ 1)2

(x 1)(x 3) = (y 1)(y+ 1)

y = x 2(x 1)(x 3) = 0

x= 1y= 1

x= 3y= 1 .

T D(1;1)hoc D(3;1)

0,25

V D, Fnm v hai pha so vi ng thng AEnn D(1;1).Khi : C(5;1), B(1;5). Vy B (1;5), C(5;1), D(1;1)

0,25

Cu 8 Gii h:

(x +

x2 + 1)(y+

y2 + 1) = 1 (1)

3x2 + y+ 3 =

3x + 1 +

4 5y (2), (x, y R)

1,00

- iu kin:

x 13

y45

Vix, y R, ta c:

(x +

x2 + 1)(x +x2 + 1) = 1(y+

y2 + 1)(y+ y2 + 1) = 1

.

Kt hp vi phng trnh (1) ta c:

y+

y2 + 1 = x +x2 + 1x +

x2 + 1 = y+ y2 + 1

. Cng hai v ca h ny ta c x = y

0,25

Thx=

y vo phng trnh (2) ta c:

3x2 x + 3 = 3x + 1 + 5x + 4 ()Ta c:() 3x2 x + 3 3x + 1 5x + 4 = 0 3x2 3x + (x + 1 3x + 1) + (x + 2 5x + 4) = 0 3(x2 x) + x

2 xx + 1 +

3x + 1

+ x2 x

x + 2 +

5x + 4= 0

(x2 x)(3 + 1x + 1 +

3x + 1

+ 1

x + 2 +

5x + 4) = 0

0,50

Vx 13

nn3 + 1

x + 1 +

3x + 1+

1

x + 2 +

5x + 4 >0.

Do vy:x2 x= 0 x= 0 x= 1.Kt lun h c hai nghim: (x; y) = (0;0), (1;1)

0,25

Cu 9 Cho cc sa,b, cl cc s thc dng v tha mn: a + b + c= 3. Tm gi tr ln nht

ca biu thc:

P = 2

3 + ab + bc + ca+ 3

abc

(1 + a)(1 + b)(1 + c)

0,5

5

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Ta c: (ab + bc + ca)2 3abc(a + b + c) = 9abc >0 = ab + bc + ca 3abc.Chng minh c: (1 + a)(1 + b)(1 + c)(1 + 3abc)3, a,b,c >0. Du "=" xy rakhia= b = c. Khi :

P 23(1 +

abc)

+3

abc

1 + 3

abc=Q.

t 6abc= t. V a,b,c >0nn 0 < abc ( a + b + c3 )3 = 1

0,25

Xt hm s: Q = 2

3(1 + t3)+

t2

1 + t2, t (0;1] Q(t) = 2t(t 1)(t

5 1)(1 + t3)2(1 + t2)2

0, t (0; 1].

Do hm s ng bin trn (0; 1]nn P Q(1) =56

. V t ta tm c MaxP =5

6,

khia= b = c = 1

0,25

Ch : Hc sinh lm theo cch khc nu ng vn cho im ti a theo tng bi.

6

ruy cp http://dethithu.netth

ng xuyn

cp nht nhiu

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53/117

TRNG THPTCHUYN H LONG

THI TH THPT QUC GIA LN 2 NM 2015 - 2016Mn: TON


Cu 1 (1,0 im).Kho st s bin thin v v th hm s3

.2 1

xy

x

Cu 2 (1,0 im). Cho hm s 3 3 2y x x c th l ( ).C Vit phng trnh tip tuyn ca

th ( )C ti cc giao im ca n vi ng thng c phng trnh 2.y x Cu 3 (1,0 im).

a) Cho s phc z tha mn (2 3 ) 1 9 .z i z i Tm mun ca s phc 2 1.w z z

b) Gii phng trnh 2 23 3 82.x x

Cu 4 (1,0 im). Tnh tch phn1

0

2( )d .

1xI x e x

x

Cu 5 (1,0 im). Trong khng gian vi h to ,Oxyz cho ba im (1;1;1), (3;5;2), (3;1; 3).A B C Lp phng trnh ng thng di qua gc ta ,O vung gc vi mt phng ( )ABC v lp

phng trnh mt cu ( )S ngoi tip t din .OABC

Cu 6 (1,0 im).

a) Tnh gi tr biu thc 2 2sin ( ) cos ( ),4 3

A

bit3

cos =5

v3

.2

b) Chng trnh To Qun nm 2016 (Gp nhau cui nm) c mt tr chi tn l Vng quay k diudnh cho cc To tng t nh tr chi truyn hnh Chic nn k diutrn knh VTV3. Chic nnc hnh trn c chia u thnh cc hnh qut, trong c 10 c tn Tham nhng, 4 c tnTrong schv 2 c tn Phn thng. C 4 To (Kinh t, X hi, Gio dc v Tinh thn) cngtham gia tr chi ny, mi To ch c quay ngu nhin mt ln. Tnh xc sut c 4 To uquay vo Trong sch.

Cu 7 (1,0 im). Cho hnh chp .S ABCc y ABCl tam gic u cnh ,a mt bn SACl tamgic cn ti Sv nm trong mt phng vung gc vi mt phng ( ),ABC ng thng SB to vi

mt phng ( )ABC mt gc 060 , Ml trung im cnh .BCTnh theo a th tch khi chp .S ABCv khong cch gia hai ng thng , .SM AC

Cu 8 (1,0 im). Trong mt phng vi h to ,Oxy cho hnh vung ABCD c (4;6).A Gi

,M Nln lt l cc im nm trn cc cnh BCv CD sao cho 045 , ( 4;0)MAN M v ngthng MNc phng trnh 11 2 44 0.x y Tm ta cc im , , .B C D

Cu 9 (1,0 im). Gii h phng trnh2 2 2 21 97 1 97 97( )

( , ).27 8 97

x y y x x yx y

x y

Cu 10 (1,0 im). Cho , ,a b c l cc s thc dng tha mn

2

4 .2016

a b cabc

Tm gi tr ln

nht

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