đề + đáp án tốt nghiệp môn toán thpt 2002-2014

b gio dc v o to ----------------------- chnh thc

k thi tt nghip trung hc ph thng nm hc 2002 2003

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

mn thi: ton Thi gian lm bi: 150 pht, khng k thi gian giao .

----------------- Bi 1 (3 im).

1. Kho st hm s 2

542

+

=x

xxy

2. Xc nh m th hm s 2

54)4( 22

+

+=

mxmmxmx

y c cc tim cn trng vi

cc tim cn tng ng ca th hm s kho st trn. Bi 2 (2 im).

1. Tm nguyn hm F(x) ca hm s

12133)( 2

23

++

++=

xxxxxxf

bit rng F(1) = 31 .

2. Tm din tch hnh phng gii hn bi th ca hm s

212102 2

+

=x

xxy

v ng thng y = 0. Bi 3 (1,5 im). Trong mt phng vi h to Oxy, cho mt elp (E) c khong cch gia cc

ng chun l 36 v cc bn knh qua tiu ca im M nm trn elp (E) l 9 v 15. 1. Vit phng trnh chnh tc ca elp (E). 2. Vit phng trnh tip tuyn ca elp (E) ti im M.

Bi 4 (2,5 im). Trong khng gian vi h to Oxyz, cho bn im A, B, C, D c to xc nh bi cc h thc:

A = (2; 4; - 1) ,

+

=

kjiOB 4 , C = (2; 4; 3) ,

+

=

kjiOD 22 . 1. Chng minh rng AB AC, AC AD, AD AB. Tnh th tch khi t din ABCD. 2. Vit phng trnh tham s ca ng vung gc chung ca hai ng thng AB v

CD. Tnh gc gia ng thng v mt phng (ABD). 3. Vit phng trnh mt cu (S) i qua bn im A, B, C, D. Vit phng trnh tip din () ca mt cu (S) song song vi mt phng (ABD).

Bi 5 (1 im). Gii h phng trnh cho bi h thc sau:

2:5:6:: 111 =+

+ CCCyx

yx

yx

-------- ht --------

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

Ch k ca gim th 1 v gim th 2: .........................................................................

B gio dc v o to -----------------

chnh thc

k thi tt nghip trung hc ph thngnm hc 2003 2004

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

mn thi: ton

Thi gian lm bi: 150 pht, khng k thi gian giao Bi 1 (4 im) Cho hm s 23

31 xxy = c th l (C).

1. Kho st hm s. 2. Vit phng trnh cc tip tuyn ca (C) i qua im . )A(3; 03. Tnh th tch ca vt th trn xoay do hnh phng gii hn bi (C) v cc ng y = 0, x = 0, x = 3 quay quanh trc Ox.

Bi 2 (1 im) Tm gi tr ln nht v gi tr nh nht ca hm s

xxy 3sin34sin2 =

trn on [ . ]0 ; Bi 3 (1,5 im) Trong mt phng vi h to Oxy cho elp

(E): 11625

22=+

yx

c hai tiu im , F . 1F 21. Cho im M(3; m) thuc (E), hy vit phng trnh tip tuyn ca (E) ti M khi m > 0. 2. Cho A v B l hai im thuc (E) sao cho A + B F = 8. Hy tnh A + B F .

1F 22F 1

Bi 4 (2,5 im) Trong khng gian vi h to Oxyz cho bn im A(1; -1; 2),

B(1; 3; 2), C(4; 3; 2), D(4; -1; 2). 1. Chng minh A, B, C, D l bn im ng phng. 2. Gi A l hnh chiu vung gc ca im A trn mt phng Oxy. Hy vit phng trnh mt cu (S) i qua bn im A, B, C, D. 3. Vit phng trnh tip din () ca mt cu (S) ti im A.

Bi 5 (1 im) Gii bt phng trnh (vi hai n l n, k N)

23

5 60!)(

++

+

kn

n Akn

P

------- ht -------

H v tn th sinh: S bo danh:

Ch k gim th 1: Ch k gim th 2:

B GIO DC V O TO

CHNH THC

K THI TT NGHIP TRUNG HC PH THNG NM HC 2004 - 2005

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

MN THI: TON

Thi gian lm bi: 150 pht, khng k thi gian giao .

Bi 1 (3,5 im).

Cho hm s 1x1x2y

++

= c th (C).

1. Kho st v v th hm s. 2. Tnh din tch hnh phng gii hn bi trc tung, trc honh v th (C). 3. Vit phng trnh tip tuyn ca th (C), bit tip tuyn i qua im A(-1; 3).

Bi 2 (1,5 im).

1. Tnh tch phn

+=2

0

2 xdxcos)xsinx(I .

2. Xc nh tham s m hm s y = x3 - 3mx2 + (m2 - 1)x + 2 t cc i ti im x = 2. Bi 3 (2 im). Trong mt phng vi h to Oxy, cho parabol (P): y2 = 8x.

1. Tm to tiu im v vit phng trnh ng chun ca (P). 2. Vit phng trnh tip tuyn ca (P) ti im M thuc (P) c tung bng 4. 3. Gi s ng thng (d) i qua tiu im ca (P) v ct (P) ti hai im phn bit A, B c honh tng ng l x1, x2. Chng minh: AB = x1 + x2 + 4.

Bi 4 (2 im). Trong khng gian vi h to Oxyz, cho mt cu (S): x2+ y2 + z2 - 2x + 2y + 4z - 3 = 0

v hai ng thng

==+

0z2x

02y2x:)( 1 , 1

z1y

11x:)( 2

==

.

1. Chng minh )( 1 v )( 2 cho nhau. 2. Vit phng trnh tip din ca mt cu (S), bit tip din song song vi hai ng thng )( 1 v ( 2 ).

Bi 5 (1im). Gii bt phng trnh, n n thuc tp s t nhin:

2n

n2n

1n2n A2

5CC >+ ++ .

.....HT.......

Th sinh khng c s dng ti liu. Gim th khng gii thch g thm.

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

Ch k ca gim th s 1: ....................................................... Ch k ca gim th s 2: ..................................................

B gio dc v o to

thi chnh thc

k thi tt nghip trung hc ph thng nm 2006 Mn thi: ton - Trung hc ph thng phn ban

Thi gian lm bi: 150 pht, khng k thi gian giao

I. Phn chung cho th sinh c 2 ban (8,0 im) Cu 1 (4,0 im) 1. Kho st v v th (C) ca hm s y = x3 + 3x2. 2. Da vo th (C), bin lun theo m s nghim ca phng trnh x3 + 3x2 m = 0. 3. Tnh din tch hnh phng gii hn bi th (C) v trc honh.

Cu 2 (2,0 im) 1. Gii phng trnh 2x 2 x2 9.2 2 0.+ + = 2. Gii phng trnh 2x2 5x + 4 = 0 trn tp s phc.

Cu 3 (2,0 im) Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, cnh bn SA vung gc vi y, cnh bn SB bng a 3 . 1. Tnh th tch ca khi chp S.ABCD. 2. Chng minh trung im ca cnh SC l tm mt cu ngoi tip hnh chp S.ABCD.

II. PHN dnh cho th sinh tng ban (2,0 im) A. Th sinh Ban KHTN chn cu 4a hoc cu 4b Cu 4a (2,0 im)

1. Tnh tch phn ln 5 x x

xln 2

(e 1)eI dx .

e 1

+=

2. Vit phng trnh cc tip tuyn ca th hm s 2x 5x 4

yx 2

+=

, bit cc tip

tuyn song song vi ng thng y = 3x + 2006. Cu 4b (2,0 im) Trong khng gian ta Oxyz cho ba im A(2; 0; 0), B(0; 3; 0), C(0; 0; 6). 1. Vit phng trnh mt phng i qua ba im A, B, C. Tnh din tch tam gic ABC. 2. Gi G l trng tm tam gic ABC. Vit phng trnh mt cu ng knh OG.

B. Th sinh Ban KHXH-NV chn cu 5a hoc cu 5b Cu 5a (2,0 im)

1. Tnh tch phn 1

x

0

J (2x 1)e dx.= +

2. Vit phng trnh tip tuyn ca th hm s 2x 3

yx 1

+=+

ti im thuc th c

honh x0 = 3. Cu 5b (2,0 im) Trong khng gian ta Oxyz cho ba im A( 1; 1; 2), B(0; 1; 1), C(1; 0; 4). 1. Chng minh tam gic ABC vung. Vit phng trnh tham s ca ng thng AB. 2. Gi M l im sao cho MB 2MC= . Vit phng trnh mt phng i qua M v vung gc vi ng thng BC.

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

Ch k ca gim th 1: ....................................................... Ch k ca gim th 2: ..................................................

B gio dc v o to

thi chnh thc



I. Phn chung cho th sinh c 2 ban (8,0 im) Cu 1 (3,5 im)

Cho hm s ,12 24 += xxy gi th ca hm s l (C). 1. Kho st s bin thin v v th ca hm s. 2. Vit phng trnh tip tuyn vi th (C) ti im cc i ca (C). Cu 2 (1,5 im) Gii phng trnh .5)4(loglog 24 =+ xx

Cu 3 (1,5 im)

Gii phng trnh 0742 =+ xx trn tp s phc. Cu 4 (1,5 im) Cho hnh chp tam gic S.ABC c y ABC l tam gic vung ti nh B, cnh bn SA vung gc vi y. Bit SA = AB = BC = a. Tnh th tch ca khi chp S.ABC.

II. PHN dnh cho th sinh tng ban (2,0 im) A. Th sinh Ban KHTN chn cu 5a hoc cu 5b Cu 5a (2,0 im)

1. Tnh tch phn +

=2

1 2 1

2

x

xdxJ .

2. Tm gi tr ln nht v gi tr nh nht ca hm s 9168)( 23 += xxxxf trn on [ ]3;1 .

Cu 5b (2,0 im) Trong khng gian vi h to Oxyz, cho im M ( )0;1;1 v mt phng (P) c phng trnh x + y 2z 4 = 0. 1. Vit phng trnh mt phng (Q) i qua im M v song song vi mt phng (P). 2. Vit phng trnh tham s ca ng thng (d) i qua im M v vung gc vi mt phng (P). Tm to giao im H ca ng thng (d) vi mt phng (P).

B. Th sinh Ban KHXH&NV chn cu 6a hoc cu 6b Cu 6a (2,0 im)

1. Tnh tch phn =3

1ln2 xdxxK .

2. Tm gi tr ln nht v gi tr nh nht ca hm s 13)( 3 += xxxf trn on [ ]2;0 . Cu 6b (2,0 im) Trong khng gian vi h to Oxyz, cho im E ( )3;2;1 v mt phng ( ) c phng trnh x + 2y 2z + 6 = 0. 1. Vit phng trnh mt cu (S) c tm l gc to O v tip xc vi mt phng ( ) . 2. Vit phng trnh tham s ca ng thng ( ) i qua im E v vung gc vi mt phng ( ) .

.........Ht......... Th sinh khng c s dng ti liu. Gim th khng gii thch g thm.

H v tn th sinh: ..................................................................... S bo danh:......................................................................................... Ch k ca gim th 1: ....................................................... Ch k ca gim th 2: ............................................................

B gio dc v o to

chnh thc

k thi tt nghip trung hc ph thng ln 2 nm 2007 Mn thi: ton - Trung hc ph thng phn ban


I. Phn chung cho th sinh c 2 ban (8,0 im) Cu 1 (3,5 im)

Cho hm s 21

+=xxy , gi th ca hm s l )(C .

1. Kho st s bin thin v v th ca hm s. 2. Vit phng trnh tip tuyn vi th )(C ti giao im ca )(C vi trc tung. Cu 2 (1,5 im) Gii phng trnh 097.27 1 =+ xx .

Cu 3 (1,5 im) Gii phng trnh 02562 =+ xx trn tp s phc.

Cu 4 (1,5 im) Cho hnh chp t gic ABCDS. c y ABCD l hnh vung cnh bng a , cnh bn SA vung gc vi y v ACSA = . Tnh th tch ca khi chp ABCDS. . II. PHN dnh cho th sinh tng ban (2,0 im)

A. Th sinh Ban KHTN chn cu 5a hoc cu 5b Cu 5a (2, 0 im)

1. Cho hnh phng )(H gii hn bi cc ng xy sin= , 0=y , 0=x , 2=x .

Tnh th tch ca khi trn xoay c to thnh khi quay hnh )(H quanh trc honh. 2. Xt s ng bin, nghch bin ca hm s 28 24 += xxy .

Cu 5b (2,0 im) Trong khng gian vi h to Oxyz , cho hai im ( )5;4;1 E v ( )7;2;3F . 1. Vit phng trnh mt cu i qua im F v c tm l E . 2. Vit phng trnh mt phng trung trc ca on thng EF .


1. Tnh din tch hnh phng gii hn bi cc ng xxy 62 += , 0=y . 2. Xt s ng bin, nghch bin ca hm s 133 += xxy .

Cu 6b (2,0 im) Trong khng gian vi h to Oxyz , cho hai im )2;0;1(M , )5;1;3(N v ng thng

)(d c phng trnh

=+=

+=

.6321

tztytx

1. Vit phng trnh mt phng )(P i qua im M v vung gc vi ng thng )(d . 2. Vit phng trnh tham s ca ng thng i qua hai im M v .N

.........Ht.........


H v tn th sinh: ..................................................................... S bo danh:......................................................................................... Ch k ca gim th 1: ....................................................... Ch k ca gim th 2: ............................................................

B gio dc v o to

thi chnh thc



I. Phn chung cho th sinh c 2 ban (8 im)

Cu 1 (3,5 im) Cho hm s 1x3x2y 23 += . 1) Kho st s bin thin v v th ca hm s. 2) Bin lun theo m s nghim thc ca phng trnh 3 22x 3x 1 m.+ =

Cu 2 (1,5 im) Gii phng trnh 2x 1 x3 9.3 6 0+ + = .

Cu 3 (1,0 im) Tnh gi tr ca biu thc 22 )i31()i31(P ++= .

Cu 4 (2,0 im) Cho hnh chp tam gic u S.ABC c cnh y bng a, cnh bn bng 2a. Gi I l trung im ca cnh BC. 1) Chng minh SA vung gc vi BC. 2) Tnh th tch khi chp S.ABI theo a. II. PHN dnh cho th sinh tng ban (2 im) A. Th sinh Ban KHTN chn cu 5a hoc cu 5b

Cu 5a (2,0 im)

1) Tnh tch phn dx)x1(xI 431

1

2 =

.

2) Tm gi tr ln nht v gi tr nh nht ca hm s f (x) x 2 cos x= + trn on

2;0 .

Cu 5b (2,0 im) Trong khng gian vi h to Oxyz, cho im )2;2;3(A v mt phng (P) c phng trnh

01zy2x2 =+ . 1) Vit phng trnh ca ng thng i qua im A v vung gc vi mt phng (P). 2) Tnh khong cch t im A n mt phng (P). Vit phng trnh ca mt phng (Q) sao cho (Q) song song vi (P) v khong cch gia (P) v (Q) bng khong cch t im A n (P). B. Th sinh Ban KHXH-NV chn cu 6a hoc cu 6b

Cu 6a (2,0 im)

1) Tnh tch phn 2

0

J (2x 1)cos xdx

= .

2) Tm gi tr ln nht v gi tr nh nht ca hm s 1x2x)x(f 24 += trn on [ ]2;0 . Cu 6b (2,0 im)

Trong khng gian vi h to Oxyz, cho tam gic ABC vi A(1;4; 1), )3;4;2(B v C(2;2; 1) . 1) Vit phng trnh mt phng i qua A v vung gc vi ng thng BC. 2) Tm to im D sao cho t gic ABCD l hnh bnh hnh.

.........Ht......... Th sinh khng c s dng ti liu. Gim th khng gii thch g thm. H v tn th sinh: ..................................................................... S bo danh:..............................................................................

Ch k ca gim th 1: ....................................................... Ch k ca gim th 2: ..................................................

B GIO DC V O TO

CHNH THC

K THI TT NGHIP TRUNG HC PH THNG NM 2008 LN 2 Mn thi: TON Trung hc ph thng phn ban


I. PHN CHUNG CHO TH SINH C 2 BAN (8,0 im)

Cu 1 (3,5 im)

Cho hm s 3x 2yx 1

=+

,

log x 2 log x 2 log 5+ + = x .

)

gi th ca hm s l ( )C .1. Kho st s bin thin v v th ca hm s cho. 2. Vit phng trnh tip tuyn ca th ti im c tung bng ( )C 2.

Cu 2 (1,5 im) Gii phng trnh ( ) ( ) ( )3 3 3

Cu 3 (1,0 im) Gii phng trnh trn tp s phc. 2x 2x 2 0 + =

Cu 4 (2,0 im) Cho hnh chp c y l tam gic vung ti B, ng thng SA vung gc vi mt phng ( Bit

S.ABC ABCABC . AB a,= BC a 3= v SA 3a.=

1. Tnh th tch khi chp S.ABC theo a. 2. Gi I l trung im ca cnh SC, tnh di on thng BI theo a.

II. PHN DNH CHO TH SINH TNG BAN (2,0 im) A. Th sinh Ban KHTN chn cu 5a hoc cu 5b Cu 5a (2,0 im)

1. Tnh tch phn ( )1

x

0

I 4x 1 e d= + x.

x 2x 4x 3= + +2. Tm gi tr ln nht v gi tr nh nht ca hm s f trn on ( ) 4 2 [ ]0; 2 . Cu 5b (2,0 im)

Trong khng gian vi h ta Oxyz, cho cc im v mt phng (P) c phng trnh

( )M 1; 2; 0 , (N 3; 4; 2 )

6x 4x 1 dx.= +x 2x 6x 1= +

2x 2y z 7 0.+ + =1. Vit phng trnh ng thng MN. 2. Tnh khong cch t trung im ca on thng MN n mt phng (P).


1. Tnh tch phn J ( )2

2

1

2. Tm gi tr ln nht v gi tr nh nht ca hm s f trn on ( ) 3 2 [ ]1; 1 . Cu 6b (2,0 im)

Trong khng gian vi h ta Oxyz, cho im v mt phng (P) c phng trnh .

(A 2; 1; 3 ) =x 2y 2z 10 0

1. Tnh khong cch t im A n mt phng (P). 2. Vit phng trnh ng thng i qua im A v vung gc vi mt phng (P).

...............Ht...............

Th sinh khng c s dng ti liu. Gim th khng gii thch g thm

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

Ch k ca gim th 1: .......................................... Ch k ca gim th 2: ..........................

B GIO DC V O TO

THI CHNH THC

K THI TT NGHIP TRUNG HC PH THNG NM 2009

Mn thi: TON Gio dc trung hc ph thng Thi gian lm bi: 150 pht, khng k thi gian giao

I. PHN CHUNG DNH CHO TT C CC TH SINH (7,0 im)

Cu 1 (3,0 im). Cho hm s 2 12

xyx

+=

.

1) Kho st s bin thin v v th (C) ca hm s cho. 2) Vit phng trnh tip tuyn ca th (C), bit h s gc ca tip tuyn bng 5. Cu 2 (3,0 im)

1) Gii phng trnh . 25 6.5 5 0x x + =

2) Tnh tch phn 0

(1 cos ) d .I x x

= + x

3) Tm gi tr nh nht v gi tr ln nht ca hm s 2( ) ln(1 2 )f x x x= trn on [ 2 ; 0].

Cu 3 (1,0 im). Cho hnh chp S.ABC c mt bn SBC l tam gic u cnh a, cnh bn SA vung gc vi mt phng y. Bit , tnh th tch ca khi chp S.ABC theo a. 0120BAC =

II. PHN RING (3,0 im) Th sinh hc chng trnh no th ch c chn phn dnh ring cho chng trnh (phn 1 hoc phn 2). 1. Theo chng trnh Chun:

Cu 4a (2,0 im). Trong khng gian Oxyz, cho mt cu (S) v mt phng (P) c phng trnh: (S): v (P): 2 2 2( 1) ( 2) ( 2) 3x y z + + = 6 02 2 18x y z+ + + = .

1) Xc nh to tm T v tnh bn knh ca mt cu (S). Tnh khong cch t T n mt phng (P). 2) Vit phng trnh tham s ca ng thng d i qua T v vung gc vi (P). Tm to giao im ca d v (P).

Cu 5a (1,0 im). Gii phng trnh 8 42 1 0z z + = trn tp s phc.

2. Theo chng trnh Nng cao: Cu 4b (2,0 im). Trong khng gian Oxyz, cho im A(1; 2; 3) v ng thng d c phng trnh

1 22 1

x y z 31

+ += =

.

1) Vit phng trnh tng qut ca mt phng i qua im A v vung gc vi ng thng d. 2) Tnh khong cch t im A n ng thng d. Vit phng trnh mt cu tm A, tip xc vi d.

Cu 5b (1,0 im). Gii phng trnh 22 1z iz 0 + = trn tp s phc. ......... Ht .........


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

Ch k ca gim th 1: ................................ Ch k ca gim th 2: ................................

B GIO DC V O TO

THI CHNH THC

K THI TT NGHIP TRUNG HC PH THNG NM 2010 Mn thi: TON Gio dc trung hc ph thng

Thi gian lm bi: 150 pht, khng k thi gian giao I. PHN CHUNG CHO TT C TH SINH (7,0 im)

Cu 1 (3,0 im). Cho hm s 3 21 3 5.4 2

y x x= +

1) Kho st s bin thin v v th ca hm s cho. 2) Tm cc gi tr ca tham s m phng trnh x3 6x2 + m = 0 c 3 nghim thc phn bit.

Cu 2 (3,0 im).

1) Gii phng trnh 22 42log 14log 3 0.x x + =

x2) Tnh tch phn 1

2 2

0

( 1)I x x d= .

3) Cho hm s 2( ) 2 12.f x x x= + Gii bt phng trnh '( ) 0.f x

Cu 3 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, cnh bn SA vung gc vi mt phng y, gc gia mt phng (SBD) v mt phng y bng 60o. Tnh th tch khi chp S.ABCD theo a. II. PHN RING - PHN T CHN (3,0 im) Th sinh ch c lm mt trong hai phn (phn 1 hoc phn 2). 1. Theo chng trnh Chun Cu 4.a (2,0 im). Trong khng gian vi h to Oxyz, cho 3 im A(1; 0; 0), B(0; 2; 0) v C(0; 0; 3).

1) Vit phng trnh mt phng i qua A v vung gc vi ng thng BC. 2) Tm to tm mt cu ngoi tip t din OABC.

Cu 5.a (1,0 im). Cho hai s phc v Xc nh phn thc v phn o ca s phc

1 1 2z i= + 2 2 3 .z = i

1 22 .z z

2. Theo chng trnh Nng cao Cu 4.b (2,0 im). Trong khng gian vi h to Oxyz, cho ng thng c phng trnh

1 1.2 2 1x y z+ = =

1) Tnh khong cch t im O n ng thng . 2) Vit phng trnh mt phng cha im O v ng thng .

Cu 5.b (1,0 im). Cho hai s phc v Xc nh phn thc v phn o ca s phc

1 2 5z i= + 2 3 4 .z = i

1 2. .z z--------------------------------------------- Ht ---------------------------------------------


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

Ch k ca gim th 1: Ch k ca gim th 2:

B GIO DC V O TO

THI CHNH THC



I. PHN CHUNG CHO TT C CC TH SINH (7,0 im)

Cu 1. (3,0 im) Cho hm s 2 12 1

xyx

+=

.

1) Kho st s bin thin v v th ca hm s cho. ( )C2) Xc nh ta giao im ca th ( vi ng thng . )C 2y x= +

Cu 2. (3,0 im)

1) Gii phng trnh 2 1

7 8.7 1x x+

+ = 0 .

2) Tnh tch phn 1

4 5lne xI dxx

+= .

3) Xc nh gi tr ca tham s hm s t cc tiu ti m3 2

2y x x mx= + +1 1x = .

Cu 3. (1,0 im) Cho hnh chp c y .S ABCD ABCD l hnh thang vung ti A v D vi AD CD a= = , 3AB = a

ABCD

0

. Cnh bn vung gc vi mt y v cnh bn to vi mt

y mt gc . Tnh th tch khi chp S theo a .

SA SC

45o

.

II. PHN RING - PHN T CHN (3,0 im) Th sinh ch c lm mt trong hai phn (phn 1 hoc phn 2). 1. Theo chng trnh Chun (3,0 im)

Cu 4.a. (2,0 im) Trong khng gian vi h ta , cho im Oxyz (3;1;0)A v mt phng c phng trnh . ( )P 2 2 1x y z+ + =

1) Tnh khong cch t im A n mt phng . Vit phng trnh mt phng i qua im

( )P ( )QA v song song vi mt phng . ( )P

2) Xc nh ta hnh chiu vung gc ca im A trn mt phng . ( )PCu 5.a. (1,0 im) Gii phng trnh (1 trn tp s phc. ) (2 ) 4 5i z i i + =

2. Theo chng trnh Nng cao (3,0 im)

Cu 4.b. (2,0 im) Trong khng gian vi h ta , cho ba im , v C .

Oxyz (0;0;3)A ( 1; 2;1)B ( 1;0;2)

1) Vit phng trnh mt phng . ( )ABC2) Tnh di ng cao ca tam gic ABC k t nh A .

Cu 5.b. (1,0 im) Gii phng trnh ( ) trn tp s phc. 2

4 0z i + =------------------------ Ht ------------------------


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

Ch k ca gim th 1: ................................................................... Ch k ca gim th 2: .................................................

I. PHN CHUNG CHO TT C TH SINH (7,0 im)

Cu 1. (3,0 im) Cho hm s ( ) 4 214

2y f x x x .= =

1) Kho st s bin thin v v th ( )C ca hm s cho. 2) Vit phng trnh tip tuyn ca th ( )C ti im c honh 0x , bit ( )0 1f " x .=

Cu 2. (3,0 im)

1) Gii phng trnh ( )2 4 33 2 3 2.log x log .log x + =

2) Tnh tch phn ( )2 2

01

lnx x .I e e dx=

3) Tm cc gi tr ca tham s m gi tr nh nht ca hm s ( )2

1x x m mf

x +=

+ trn

on [ ]0;1 bng 2. Cu 3. (1,0 im) Cho hnh lng tr ng ABC.A B C c y ABC l tam gic vung ti B

v BA BC a.= = Gc gia ng thng A B vi mt phng ( )ABC bng 60 . Tnh th tch khi lng tr ABC.A B C theo a.

II. PHN RING - PHN T CHN (3,0 im) Th sinh ch c lm mt trong hai phn (phn 1 hoc phn 2). 1. Theo chng trnh Chun

Cu 4.a. (2,0 im) Trong khng gian vi h ta Oxyz, cho cc im ( )2;2;1A , ( )0;2;5B v mt phng ( )P c phng trnh 2 5 0.x y + =

1) Vit phng trnh tham s ca ng thng i qua A v B.

2) Chng minh rng ( )P tip xc vi mt cu c ng knh AB.

Cu 5.a. (1,0 im) Tm cc s phc 2z z+ v 25i ,z

bit 3 4 .z i=

2. Theo chng trnh Nng cao

Cu 4.b. (2,0 im) Trong khng gian vi h ta Oxyz, cho im ( )2;1;2A v ng thng

c phng trnh 1 32 2 1

x y z . = =

1) Vit phng trnh ca ng thng i qua O v A.

2) Vit phng trnh mt cu ( )S tm A v i qua O. Chng minh tip xc vi ( )S .

Cu 5.b. (1,0 im) Tm cc cn bc hai ca s phc 1 9 51

iz i.i

+=

-------------- Ht -------------- Th sinh khng c s dng ti liu. Gim th khng gii thch g thm.

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

Ch k ca gim th 1: ............................................. Ch k ca gim th 2: ..................................

B GIO DC V O TO

THI CHNH THC



B GIO DC V O TO

THI CHNH THC


Thi gian lm bi: 150 pht, khng k thi gian giao I. PHN CHUNG CHO TT C TH SINH (7,0 im)

Cu 1 (3,0 im). Cho hm s 3 3 1y x x= .

2 0.x x + =

1) Kho st s bin thin v v th ( ca hm s cho. )C

2) Vit phng trnh tip tuyn ca bit h s gc ca tip tuyn bng 9. ( ),C

Cu 2 (3,0 im)

1) Gii phng trnh 3 3 1

2) Tnh tch phn ( )2

01 cos d .I x x

= + x

3) Tm gi tr ln nht v gi tr nh nht ca hm s 2 3 lny x x= + x trn on [ ]1; 2 . Cu 3 (1,0 im). Cho hnh chp c y l hnh vung cnh cnh bn vung gc vi mt phng y. ng thng to vi mt phng mt gc Tnh th tch ca khi chp theo

.S ABCD ABCD ,a SASD (SAB)

yz.

o30 ..S ABCD .a

II. PHN RING - PHN T CHN (3,0 im) Th sinh ch c lm mt trong hai phn (phn 1 hoc phn 2). 1. Theo chng trnh Chun Cu 4.a (2,0 im). Trong khng gian vi h ta Ox cho im v mt phng c phng trnh

, ( 1; 2; 1)M ( )P2 2 3 0x y z+ + =

1) Vit phng trnh tham s ca ng thng d i qua M v vung gc vi ( ).P

2) Vit phng trnh mt cu c tm l gc ta v tip xc vi ( )S ( ).PCu 5.a (1,0 im). Cho s phc tha mn (1z ) 2 4 0.i z i+ = Tm s phc lin hp ca .z

2. Theo chng trnh Nng cao Cu 4.b (2,0 im). Trong khng gian vi h ta cho im v ng

thng c phng trnh

,Oxyz ( 1; 1; 0)A

d 1 1.1 2 1

x y z += =

1) Vit phng trnh mt phng i qua gc ta v vung gc vi ( )P .d

2) Tm ta im M thuc sao cho di on d AM bng 6 .

Cu 5.b (1,0 im). Gii phng trnh 2 (2 3 ) 5 3 0z i z i + + + = trn tp s phc. -------------------- Ht --------------------


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


B GIO DC V O TO

THI CHNH THC



Cu 1 (3,0 im). Cho hm s 2 3 .1

xyx

+=

1) Kho st s bin thin v v th (C) ca hm s cho.

2) Vit phng trnh tip tuyn ca (C) ti cc giao im ca (C) v ng thng 3.y x=

Cu 2 (2,5 im)

1) Gii phng trnh ( )22 2log 3log 2 1 0x x+ = trn tp hp s thc.

2) Tm gi tr ln nht v gi tr nh nht ca hm s ( ) 2 21 4 .4

f x x x x x=

Cu 3 (1,5 im). Tnh tch phn ( )1

0

1 .xI xe dx=

Cu 4 (1,0 im). Cho hnh chp S.ABC c y ABC l tam gic vung cn ti A v 2 5SC a= .

).

Hnh chiu vung gc ca S trn mt phng (ABC) l trung im M ca cnh AB. Gc gia ng thng SC v (ABC) bng Tnh th tch khi chp S.ABC theo a. 60 .

Cu 5 (2,0 im). Trong khng gian vi h to Oxyz, cho im v mt phng (P) c phng trnh

(1; 1;0A 2 2 1 0x y z + =

1) Vit phng trnh tham s ca ng thng i qua A v vung gc vi (P).

2) Tm to im M thuc (P) sao cho AM vung gc vi OA v di on AM bng ba ln khong cch t A n (P).

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


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


1

B gio dc v o to

thi chnh thc

k thi tt nghip trung hc ph thng nm 2006 Mn thi: Ton - Trung hc ph thng phn ban

hng dn chm THi

Bn hng dn chm gm: 05 trang

I. Hng dn chung 1. Nu th sinh lm bi khng theo cch nu trong p n m vn ng th cho

im tng phn nh hng dn quy nh. 2. Vic chi tit ho thang im (nu c) so vi thang im trong hng dn chm

phi m bo khng sai lch vi hng dn chm v c thng nht thc hin trong Hi ng chm thi.

3. Sau khi cng im ton bi mi lm trn im thi theo nguyn tc: im ton bi c lm trn n 0,5 im ( l 0,25 lm trn thnh 0,5; l 0,75 lm trn thnh 1,0 im)

II. p n v thang im

p n im

Cu 1 (4,0 im)

1. (2,5 im) a) Tp xc nh: R. b) S bin thin: Chiu bin thin: 2y ' 3x 6x= + . y' = 0 x = 0 hoc x = 2. Trn cc khong ( ); 0 v ( )2;+ , y ' 0< hm s nghch bin. Trn khong (0; 2), y ' 0> hm s ng bin. Ch : Nu ch xt du y' hoc ch nu cc khong ng bin, nghch bin th vn cho 0,25 im. Cc tr: Hm s t cc tiu ti x = 0; yCT = y(0) = 0. Hm s t cc i ti x = 2; yC = y(2) = 4. Gii hn v cc:

+= + =

x xlim y ; lim y .

Bng bin thin: x 0 2 + y' 0 + 0 + 4 y 0

0,25 0,25 0,25

0,25 0,25 0,25 0,50

2

c) th: Giao im vi cc trc ta : (0; 0) v (3; 0). 2. (0,75 im) 3 2 3 2x 3x m 0 x 3x m + = + = (1) S nghim ca phng trnh (1) l s giao im ca th (C) v ng thng y = m. Da vo s tng giao ca th (C) v ng thng y = m ta c: Nu m < 0 hoc m > 4 th phng trnh c 1 nghim. Nu m = 0 hoc m = 4 th phng trnh c 2 nghim. Nu 0 < m < 4 th phng trnh c 3 nghim. 3. (0,75 im) Gi S l din tch hnh phng cn tm.

T th ta c: S = 3

3 2

0

x 3x dx +

33 43 2 3

0 0

x( x 3x )dx x

4

= + = +

= 27

4 (vdt).

0,50

0,25 0,50 0,25 0,25

0,25

Cu 2 (2,0im)

1. (1,0 im) 2x + 2 x x 2 x2 9.2 + 2 = 0 4.(2 ) 9.2 2 0 + =

x

x

2 2

12

4

= =

x 1 = hoc x 2= . Phng trnh cho c hai nghim x = 1; x = 2. 2. (1,0 im)

7. = += = +

= =

1

2

5 i 7 5 7x i ;

4 4 4

5 i 7 5 7x i .

4 4 4

Phng trnh c hai nghim 1 25 7 5 7

x i ; x i .4 4 4 4

= + =

0,25 0,25

0,25

0,25

0,25 0,25 0,25

0,25

x

4 m

O 2 3

(C)y

3

Cu 3 (2,0 im)

Ch : Nu bi lm khng c hnh v ng th khng cho im. 1. (1,0 im) Gi di ng cao hnh chp l h, din tch y hnh chp l ABCDS .

Ta c: 2 2h SA SB AB a 2;= = = 2ABCDS a= .

Gi V l th tch ca khi chp. Ta c: 3ABCD1 1

V S .h a 23 3

= = (vtt).

2. (1,0 im) Gi I l trung im cnh SC. SA(ABCD) SAAC SAC vung ti A IA = IC = IS (1). CB AB, CB SA CB (SAB) CB SB SBC vung ti B IB = IC = IS (2). Chng minh tng t: SDC vung ti D ID = IC = IS (3). T (1), (2), (3) suy ra: trung im I ca cnh SC cch u cc nh ca hnh chp S.ABCD I l tm mt cu ngoi tip hnh chp S.ABCD.

0,25

0,25

0,50 0,25 0,25 0,25 0,25

Cu 4a (2,0 im)

1. (1,0 im)

t x x 2 xt e 1 e t 1, e dx 2tdt= = + = . x = ln2 t = 1; x = ln5 t = 2.

22

1

I 2 (t 2)dt= +

=

23

1

t2 2t

3

+

= 26

3.

0,25 0,25 0,25 0,25

CD

S

AB

. I

4

2. (1,0 im) Gi x l honh tip im, theo gi thit ta c: y '(x) 3= (1)

(1)( )2

2

x 4x 63

x 2

+ =

x = 1 hoc x = 3.

Ta cc tip im: A(1; 0), B(3; 2). Phng trnh tip tuyn ti A: y 3(x 1) y 3x 3.= = Phng trnh tip tuyn ti B: y 3(x 3) 2 y 3x 11.= = (Tha mn yu cu bi).

0,25 0,25 0,25 0,25

Cu 4b (2,0 im)

1. (1,0 im)

Mt phng i qua ba im A, B, C c phng trnh: x y z

12 3 6

+ + =

3x + 2y + z 6 = 0. AB ( 2; 3; 0), AC ( 2; 0; 6)= = .

AB AC (18; 12; 6) = ABC1

S AB AC 3 142

= = (vdt).

2. (1,0 im)

G l trng tm tam gic ABC: 2

G ; 1; 2 .3

=

Tm I ca mt cu l trung im OG: 1 1

I ; ; 1 .3 2

=

Bn knh mt cu: 7

R OI .6

= =

Phng trnh mt cu: ( )2 2

21 1 49x y z 1 .

3 2 36 + + =

0,50 0,25 0,25 0,25

0,25

0,25 0,25

Cu 5a (2,0 im)

1. (1,0 im)

t x x

u 2x 1 du 2dx

dv e dx v e .

= + = = =

11

x x

00

J (2x 1)e 2 e dx = +

= 1 1x x

00(2x 1)e (2e ) +

= e + 1. 2. (1,0 im)

Tnh c 2

1y '

(x 1)

=+

.

0

3 1y y( 3) ; y '( 3) .

2 4

= = =

Phng trnh tip tuyn: 1 3

y x .4 4

= +

0,25 0,25

0,25 0,25

0,25 0,50

0,25

5

Cu 5b (2,0 im)

1. (1,0 im) AB (1;0; 1), AC (2; 1;2)= = .

AB.AC 0. = Suy ra iu phi chng minh. Vect ch phng ca ng thng AB: AB (1;0; 1).=

Phng trnh tham s ca ng thng AB:

x 1 t

y 1

z 2 t .

= + = =

2. (1,0 im) Gi M(x; y; z). MB (0 x;1 y;1 z),MC (1 x;0 y;4 z).= =

0 x 2(1 x)

MB 2MC 1 y 2(0 y)

1 z 2(4 z)

= = = =

2x

31

y3

z 3

= =

=

2 1M ; ;3 .3 3

Gi (P) l mt phng qua M v vung gc vi ng thng BC. Vect php tuyn ca (P): BC (1; 1;3).=

Phng trnh mt phng (P): 28

x y 3z 03

+ = .

0,25

0,25

0,25 0,25 0,25 0,25

0,25

0,25

...Ht...

1

b gio dc v o to

thi chnh thc

k thi tt nghip trung hc ph thng nm 2007 Mn thi: ton Trung hc ph thng phn ban

Hng dn chm thi Bn hng dn chm gm 04 trang

I. Hng dn chung

1) Nu th sinh lm bi khng theo cch nu trong p n m vn ng th cho im tng phn nh hng dn quy nh.

2) Vic chi tit ho thang im (nu c) so vi thang im trong hng dn chm phi m bo khng sai lch vi hng dn chm v c thng nht thc hin trong Hi ng chm thi.

3) Sau khi cng im ton bi, lm trn n 0,5 im (l 0,25 lm trn thnh 0,5; l 0,75 lm trn thnh 1,0 im).

II. p n v thang im

Cu p n im1. (2,5 im) 1) Tp xc nh: R 0,25

2) S bin thin: Chiu bin thin: Ta c: )1(444' 23 == xxxxy ; 0'=y x = 0, x = 1. Trn cc khong ( )0;1 v ( )+;1 , y > 0 nn hm s ng bin. Trn cc khong ( )1; v ( )1;0 , y < 0 nn hm s nghch bin.

0,50

Cc tr: T cc kt qu trn suy ra: Hm s c hai cc tiu ti x = 1; yCT = y( 1) = 0. Hm s c mt cc i ti x = 0; yC = y(0) = 1. Gii hn v cc:

+=

yx

lim ; +=+

yx

lim .

0,75

Cu 1 (3,5 im)

Bng bin thin:

0,50

x 1 0 1 +

y - 0 + 0 - 0 +

+ 1 +

y

0 0

2

3) th: Hm s cho l chn, do th nhn trc Oy lm trc i xng. th ct trc tung ti im (0; 1). im khc ca th: ( )9;2 .

0,50

2. (1,0 im) - H s gc ca tip tuyn ti im cc i (0; 1) ca th cho l y(0) = 0. - Phng trnh tip tuyn ca (C) ti im cc i l y = 1.

1,00

iu kin xc nh ca phng trnh l x > 0. Phng trnh cho tng ng vi

5log4loglog21

222 =++ xx 0,75

Cu 2 (1,5 im)

3log23

2 =x

2log2 =x x = 4 (tho mn iu kin). Vy phng trnh cho c nghim x = 4.

0,75

Ta c: ' = .33 2i= 0,50 Cu 3

(1,5 im) Phng trnh c hai nghim phn bit l: ix 32 = v ix 32 += . 1,00 Cu 4

(1,5 im) Gi thit SA vung gc vi y suy ra ng cao ca hnh chp l

SA = a. y l tam gic vung (nh B), c din tch l 221 a .

Vy th tch khi chp S.ABC l:

3261.

21.

31 aaaV == (vtt).

1,50

A

B

a

aa

C

S

-2 -1 O 1 2 x

1

9

y

3

1. (1,0 im)

t tx =+12 2xdx = dt. Vi x = 1 th t = 2; vi x = 2 th t = 5.

0,50

Do J = 5

2

21

dtt = 25

.2 21

t = 2 )25( . 0,50

Cu 5a (2,0 im)

2. (1,0 im)

- Ta c 16163)(' 2 += xxxf .

- Xt trn on [ ]3;1 ta c 0)(' =xf 34=x .

- Ta c f(1) = 0,

34f =

2713

, f(3) = - 6.

Vy [ ] 27

1334)(max

3;1=

= fxf ,

[ ]6)3()(min

3;1== fxf .

1,00

1. (1,0im)

V mt phng (Q) song song vi mt phng (P) nn phng trnh mt phng (Q) c dng x + y 2z + m = 0 (m - 4).

0,50

Mt phng (Q) i qua im M(-1; -1; 0) 1 1 + m = 0 m = 2. Vy phng trnh mt phng (Q) l: x + y 2z + 2 = 0.

0,50

2. (1,0im) - V ng thng (d) vung gc vi mt phng (P) nn vct php

tuyn )2;1;1( =n ca mt phng (P) cng l vct ch phng ca ng thng (d).

- ng thng (d) i qua im M(-1; -1; 0) nhn )2;1;1( =n lm

vct ch phng nn c phng trnh tham s l:

=+=+=

.211

tztytx

0,50

Cu 5b

(2,0 im)

- To H(x; y; z) tho mn h:

=+=+=+=

0422

11

zyxtztytx

====

.200

1

zyxt

Vy H(0; 0; - 2).

0,50

Cu 6a

(2,0 im)

1. (1,0 im)

t u = lnx v dv = 2xdx; ta c du = x1

dx v v = 2x .

Do =3

1ln2 xdxxK =

3

1

213

)ln( xdxxx

= 13

213

)ln(2

2 xxx = 43ln9 .

1,00

4

2. (1,0 im)

- Ta c 33)(' 2 = xxf . - Xt trn on [ ]2;0 ta c f(x) = 0 x = 1. - Ta c f(0) = 1, f(1) = -1, f(2) = 3. Vy

[ ]3)2()(max

2;0== fxf ,

[ ]1)1()(min

2;0== fxf .

1,00

1. (1,0 im)

- Mt cu (S) c tm l gc to O v tip xc vi mt phng ( ) nn bn knh mt cu bng khong cch t O n ( ).

d(O; ( )) = 222 )2(21

6000

++

++ = 2.

0,50

Mt cu (S) c tm l gc to O v bn knh bng 2 c phng

trnh l: 4222 =++ zyx . 0,50

Cu 6b

(2,0 im)

2. (1,0 im)

V ng thng ( ) vung gc vi mt phng ( ) nn vct php tuyn )2;2;1( =n ca mt phng ( ) cng l vct ch phng ca ng thng ( ). ng thng ( ) i qua im E(1; 2; 3) nhn )2;2;1( =n lm vct

ch phng c phng trnh tham s l:

=+=+=

.2322

1

tzty

tx

1,00

.Ht.

1

b gio dc v o to

chnh thc

k thi tt nghip trung hc ph thng ln 2 nm 2007Mn thi: ton Trung hc ph thng phn ban

Hng dn chm thi Bn hng dn chm gm 04 trang

I. Hng dn chung

1) Nu th sinh lm bi khng theo cch nu trong p n m vn ng th cho

im tng phn nh hng dn quy nh. 2) Vic chi tit ho thang im (nu c) so vi thang im trong hng dn

chm phi m bo khng sai lch vi hng dn chm v c thng nht thc hin trong Hi ng chm thi.


II. p n v thang im

Cu p n im1. (2,5 im) 1) Tp xc nh: { }2\ =RD . 0,25

2) S bin thin: Chiu bin thin:

Ta c: 2)2(

3'+

=x

y ; 0'>y vi mi Dx .

Hm s ng bin trn cc khong ( )2; v ( )+ ;2 .

0,50

Cc tr: Hm s khng c cc tr. Tim cn:

1lim =y

x v 1lim =

+y

x tim cn ngang: 1=y .

+=y

x 2lim v =

+y

x 2lim tim cn ng: 2=x .

0,75

Cu 1 (3,5 im)

Bng bin thin:

0,50

y + +

y 1

1

x -2 +

+

-

2

3) th: - th ct Ox ti im )0;1( v ct Oy ti im )21;0( .

th nhn giao im )1;2(I ca hai ng tim cn lm tm i xng.

0,50

2. (1,0 im)

- Giao im ca th )(C vi trc tung l )21;0( M .

- H s gc ca tip tuyn ti im M l 43)0(' =y .

- Phng trnh tip tuyn ca )(C ti im M l 21

43 = xy .

1,00

Bin i phng trnh v dng 0147.972 =+ xx . t )0(7 >= ttx .

Phng trnh cho tr thnh: 01492 =+ tt

==

.72

tt

0,75

Cu 2 (1,5 im)

Vi 2log2 7== xt . Vi .17 == xt Phng trnh c hai nghim 2log7=x v .1=x

0,75

Ta c: ' = 016

3

Cu 4 (1,5 im)

- Din tch y ABCD bng 2a . - ABC vung cn ti nh 2aACB = . - ng cao hnh chp 2aSA= . Vy th tch khi chp ABCDS. l

322..

31 32 aaaV == (vtt).

1,50

1. (1,0 im)

Ta c ==2

0

2

0

2 )2cos1(2

sin

dxxxdxVx

0,50

= 42

2sin2

22

0

=

xx (vtt). 0,50

Cu 5a (2,0 im)

2. (1,0 im) Tp xc nh: R. 2,00';164' 3 ==== xxyxxy .

Trong cc khong )0;2( v );2( + , 0'>y nn hm s ng bin. Trong cc khong )2;( v )2;0( , 0'

4

1. (1,0 im) -Honh giao im ca ng cong xxy 62 += v ng thng

0=y l nghim ca phng trnh .6,0062 ===+ xxxx

-Din tch hnh phng cho l +=+6

0

6

0

22 )6(6 dxxxdxxx

3633

6

0

23

=

+= xx (vdt).

1,00

Cu 6a

(2,0 im)

2. (1,0 im) Tp xc nh: R . 10';33' 2 === xyxy .

Trn cc khong )1;( v );1( + , 0'>y nn hm s ng bin. Trn khong )1;1( , 0'

1

B gio dc v o to thi chnh thc

k thi tt nghip trung hc ph thng nm 2008 Mn thi: ton Trung hc ph thng phn ban

Hng dn chm thi

Bn hng dn chm gm 04 trang

I. Hng dn chung

1) Nu th sinh lm bi khng theo cch nu trong p n m vn ng th cho im tng phn nh hng dn quy nh.

2) Vic chi tit ho thang im (nu c) so vi thang im trong hng dn chm phi m bo khng sai lch vi hng dn chm v c thng nht thc hin trong Hi ng chm thi.


II. p n v thang im

cu p n im1. (2,5 im)

a) Tp xc nh: R 0,25

b) S bin thin: Chiu bin thin:

)1x(x6x6x6y 2 +=+= . Phng trnh 0y = c nghim: x = -1, x = 0. 0,50

( ) ( )+> ;01;x0y , ( )0;1x0y

2

c) th: Giao im vi Oy: (0; -1).

Giao im vi Ox: (-1; 0) v ( )0;21

0,50

2. (1,0 im) S nghim thc ca phng trnh 3 22x +3x -1= m bng s giao im ca

th (C) ca hm s 1x3x2y 23 += v ng thng (d): y = m. Da vo th ta c: Vi m < -1 hoc m > 0, (d) v (C) c mt im chung, do phng trnh c mt nghim. Vi m = -1 hoc m = 0, (d) v (C) c hai im chung, do phng trnh c hai nghim. Vi -1 < m < 0, (d) v (C) c ba im chung, do phng trnh c ba nghim.

1,0

t 0t3x >= ta c phng trnh 3t2 9t + 6 = 0 phng trnh trn c hai nghim t = 1 v t = 2 (u tho mn).

0,75

Cu 2

(1,5 im)

Nu t =1 th 3x = 1 x = 0. Nu t = 2 th 3x = 2 x = log32. Vy phng trnh cho c hai nghim: x = 0, x = log32.

0,75

Khai trin ng: + = + 2(1 3 i ) 1 2 3 i 3 v = 2(1 3 i ) 1 2 3 i 3 0,50 Cu 3 (1,0 im)

Rt gn c = P 4 0,50

Cu 4

(2,0 im)

1. (1,0 im) Tam gic SBC cn ti S, I l trung im BC suy ra SIBC . Tam gic ABC u suy ra AIBC .

0,50

Ox

y

-1-1

21

O

S

A C

B

I

3

V BC vung gc vi hai cnh AI v SI ca tam gic SAI nn SABC . 0,50

2. (1,0 im)

Gi O l tm ca y ABC, ta c 33a

23a

32AI

32AO === . V S.ABC l

hnh chp tam gic u nn ).ABC(SO

0,50

Xt tam gic SOA vung ti O:

333aSO

9a33)

33a()a2(AOSASO

222222 ====

Th tch khi chp S.ABI l: 3

S.ABI ABI

1 1 1 1 a 3 a a 33 a 11V S .SO AI.BI.SO

3 3 2 6 2 2 3 24= = = = (vtt).

0,50

1. (1,0 im) t u = 1 x3 du = -3x2dx. Vi x = -1 u = 2, x = 1 u = 0. 0,50

= = = = 0 2

4 4 5

2 0

21 1 1 32I ( u )du u du u

3 3 15 0 5. 0,50

2. (1,0 im)

Xt trn on

2;0 , hm s cho c: xsin21)x(f = ;

4x0)x(f == .

0,50

Cu 5a

(2,0 im)

2)2(f;1

4)4(f;2)0(f =+== .

Vy 2)x(fmin]2;0[

=

, 14

)x(fmax]2;0[

+=

. 0,50

1. (1,0 im)

ng thng cn tm vung gc vi (P), nhn )1;2;2(n = l mt vect ch phng.

Phng trnh tham s ca ng thng l:

+==

+=

t2zt22yt23x

1,0

2. (1,0 im) Khong cch t im A n mt phng (P) l:

37

1)2(2

1)2.(1)2.(23.2))P(,A(d

222=

++

+= .

0,25

Cu 5b

(2,0 im)

Phng trnh mt phng (Q) song song vi mt phng (P) c dng 2x 2y + z + D = 0.

4

Chn im M(0; 0; 1) thuc mt phng (P). Khong cch t im M n mt

phng (Q) l: 3D1

1)2(2

D1.10.20.2))Q(,M(d

222

+=

++

++= .

Khong cch gia hai mt phng (P) v (Q) bng khong cch t im M n mt phng (Q).

Do t gi thit ta c: 7D137

3D1

=+=+

=

=

8D6D

Vy c hai mt phng (Q) tho mn bi: (Q1): 2x 2y + z + 6 = 0; (Q2): 2x 2y + z - 8 = 0.

0,75

1. (1,0 im)

t

==xdxcosdv1x2u

==

xsinvdx2du

[ ] 20

J (2x 1)sin x 2 sin xdx20

= 0,50

J ( 1) 2 cos x 20

= + = ( -1)+2(0-1)= -3.

0,50

2. (1,0 im) Xt trn on [0; 2], hm s cho c: )1x(x4x4x4)x(f 23 == ;

==

=1x0x

0)x(f 0,50

Cu 6a (2,0 im)

f(0) = 1; f(1) = 0; f(2) = 9. Vy

[0;2]min f(x)=0,

[0;2]max f(x)=9. 0,50

1. (1,0 im)

Mt phng cn tm vung gc vi BC, nhn )4;2;0(BC = l mt vect php tuyn.

0,50

Phng trnh mt phng cn tm l: 0(x -1) 2(y - 4) 4(z + 1) = 0 y + 2z 2 = 0.

0,50

2. (1,0 im)

ABCD l hnh bnh hnh khi v ch khi ADBC = (1). Gi to ca D l (x; y; z). Ta c )1z;4y;1x(AD +=

v )4;2;0(BC = .

0,50

Cu 6b

(2,0 im)

iu kin (1)

=+=

=

41z24y01x

===

5z2y1x

D(1; 2; -5). 0,50

.Ht.

B GIO DC V O TO

CHNH THC

K THI TT NGHIP TRUNG HC PH THNG NM 2008 LN 2 Mn thi: TON Trung hc ph thng phn ban

HNG DN CHM THI Bn Hng dn chm c 04 trang

I. Hng dn chung

1. Nu th sinh lm bi khng theo cch nu trong p n m vn ng th cho im tng phn nh hng dn quy nh.

2. Vic chi tit ho thang im (nu c) so vi thang im trong Hng dn chm phi m bo khng sai lch vi Hng dn chm v c thng nht thc hin trong Hi ng chm thi.

3. Sau khi cng im ton bi, lm trn n 0,5 im (l 0,25 lm trn thnh 0,5; l 0,75 lm trn thnh 1,0 im).

II. p n v thang im

CU

P N IM

1. (2,5 im)

a) Tp xc nh: { }D \ 1= . 0,25

b) S bin thin:

Chiu bin thin: ( )2

5y ' ,x 1

=+

vi y ' 0> x D

1

.

Hm s ng bin trn cc khong ( ) v ( ) ; 1 1; . + Cc tr: Hm s khng c cc tr.

0,75

Gii hn, tim cn: Tim cn ng: .

( ) ( )x 1 x 1lim y , lim y .

+ = + = x 1=

= Tim cn ngang: x xlim y 3, lim y 3. +

= y 3.=

0,50

Cu 1 (3,5 im)

Bng bin thin:

0,50 y

+

x 1y ' + +

3

+

3

2

c) th:

th ct trc Ox ti im 2 ; 0 ,3

ct trc Oy ti im ( )0; 2 .

0,50

y

O

2. (1,0 im)

im thuc th c tung l im ( ) y = 2 0; 2 ; ( )y ' 0 5.= 0,50

Phng trnh tip tuyn cn tm: hay ( )y 5 x 0 2= y 5x 2.= 0,50

Phng trnh cho tng ng

( )23 3

x 2 0x 2 0

log x 4 log 5

+ > > =

0,50

2

x 2x 4

>

= 5 0,50

Cu 2 (1,5 im)

x 2x 3.x 3

x 3

> = =

=

Nghim ca phng trnh l x 3.=

0,50

( )224 4i 2i . = = = 0,50 Cu 3 (1,0 im)

Nghim ca phng trnh l: v x 1 i= + x 1 i.= 0,50

1. (1,0 im) Cu 4 (2,0 im) Tam gic ABC vung ti B, nn

din tch ca tam gic ABC l: 2

ABC1 aS BA.BC2 2

= = 3 .

0,50

x1

3

2

S

A

B

C

I

3

S.ABC ABC1 aV SA.S3 2

= = 3 . 0,50 Th tch khi chp S.ABC:

2. (1,0 im)

( )SA ABC v BC (nh l ba ng vung gc). Tam gic SBC vung ti B, nn

AB BC SB 0,50 SCBI .

2=

Tam gic SBC vung ti B v tam gic SAB vung ti A, nn:

Vy 0,50 a 13BI .2

= 2 2 2 2 2 2SC SB BC SA AB BC 13a .= + = + + = 2

3

1. (1,0 im) Cu 5a

t u 4 ta chn x 1 du 4dx;= + = xdv e dx,= xv e .=

( )11x x

00

I 4x 1 e 4 e dx= + 0,50

(2,0 im)

1x0

5e 1 4e e 3.= = + 0,50

2. (1,0 im)

Trn on [ ]0; 2 , ta c: ( ) ( )3 x 0f ' x 8x 8x; f ' x 0x 1.

== + = =

0,50

Tnh v hoc lp bng bin thin ca hm

s, ta c: v ( ) ( )f 0 3, f 1 5= = ( )f 2 13=

0,50 [ ]

( ) ( )0; 2

max f x f 1 5= =[ ]

( ) ( )0; 2

min f x f 2 13.= =

1. (1,0 im) Cu 5b

( )MN 4;6;2= Vect ch phng ng thng MN: hay (u 2;3;1= ). 0,50 (2,0 im)

x 1 y 2 z .2 3

+= =

Phng trnh ng thng MN: 1

0,50

2. (1,0 im)

Trung im ca on thng MN: ( )I 1; 1; 1 . 0,50

( ) 2 2 1 7d I, (P) 2.4 4 1

+ + = =

+ +Khong cch t I n (P): 0,50

1. (1,0 im) Cu 6a

( ) 23 21

J 2x 2x x= + 0,50 (2,0 im)

( ) ( )16 8 2 2 2 1 9.= + + = 0,50

2. (1,0 im)

Trn on [ ]1;1 , ta c: ( ) ( )2f ' x 6x 12x; f ' x 0 x 0.= = = 0,50

Tnh v hoc lp bng bin thin ca

hm s, ta c: v ( ) ( )f 0 1, f 1 7= = ( )f 1 3=

0,50 [ ]

( ) ( )1; 1

max f x f 0 1

= =[ ]

( ) ( )1; 1

min f x f 1 7.

= =

4

1. (1,0 im) Cu 6b

Khong cch t n (P): A ( )( ) 2 2 6 10d A, P1 4 4

+ =

+ + 0,50

(2,0 im)

12 4.3

= = 0,50

2. (1,0 im)

l mt vect ch phng ca ng thng cn tm. (n 1; 2; 2 ) 0,50

Phng trnh ng thng cn tm: x 2 ty 1 2z 3 2t

= + = =

t.

0,50

.Ht.

B GIO DC V O TO

THI CHNH THC


Mn thi: TON Gio dc trung hc ph thng

HNG DN CHM THI Bn hng dn gm 05 trang

I. Hng dn chung

1) Nu th sinh lm bi khng theo cch nu trong p n nhng ng th cho s im

tng phn nh hng dn quy nh. 2) Vic chi tit ho (nu c) thang im trong hng dn chm phi m bo khng lm sai

lch hng dn chm v phi c thng nht thc hin trong ton Hi ng chm thi. 3) Sau khi cng im ton bi, lm trn n 0,5 im (l 0,25 lm trn thnh 0,5; l 0,75

lm trn thnh 1,0 im).

II. p n v thang im

CU P N IM

1. (2,0 im)

a) Tp xc nh: { }\ 2D = 0,25 b) S bin thin:

Chiu bin thin: y' = 25

( 2)x

< 0 x D.

Suy ra, hm s nghch bin trn mi khong ( ); 2 v . ( )2;+ Cc tr: Hm s cho khng c cc tr.

0,50

Lu : b), cho php th sinh khng nu kt lun v cc tr ca hm s.

Gii hn v tim cn:

2limx

y+

= + , 2

limx

y

= ; lim lim 2x x

y y +

= = .

Suy ra, th hm s c mt tim cn ng l ng thng 2x = v mt tim cn ngang l ng thng 2y = .

0,50

Cu 1 (3,0 im)

Bng bin thin:

x 2 + y'

y 2 + 2

0,25

1

c) th (C):

(C) ct trc tung ti im 10;2

v ct trc honh ti im 1 ;02

.

0,50

Lu : - Cho php th sinh th hin to giao im ca (C) v cc trc to ch trn hnh v. - Nu th sinh ch v ng dng ca th (C) th cho 0,25 im.

2. (1,0 im)

K hiu d l tip tuyn ca (C) v (x0; y0) l to ca tip im. Ta c: H s gc ca d bng 5 y'(x0) = 5

0,25

20

5 5( 2)x

=

00

13

xx==

0 0 0 01 3; 3x y x y= = = = 7 . 0,50

T , ta c cc phng trnh tip tuyn theo yu cu ca bi l: 5y x= + 2 2 v 5 2y x= + . 0,25

1. (1,0 im)

t 5x = t, t > 0, t phng trnh cho ta c phng trnh t2 6t + 5 = 0 (*) 0,50

Gii (*), ta c t v t1= 5= . 0,25

Vi t , ta c: 51= 1x = 0x = Vi t , ta c: 55= 5x = 1x = Vy, phng trnh cho c tt c 2 nghim l 2 gi tr x va nu trn.

0,25

2. (1,0 im)

t u v , ta c dx= d (1 cos )dv x= + x xdu = v v x sin x= + . 0,50

Do : 00

( sin ) ( sin )dI x x x x x= + + x 0,25

Cu 2 (3,0 im)

= 2 2

2

0

4cos2 2x x

=

. 0,25

y

2

x 2 O 1

2

12

2

Lu : Th sinh c php trnh by li gii va nu trn nh sau:

2 22

00 0 0

4d( sin ) ( sin ) ( sin )d cos2 2xI x x x x x x x x x x

= + = + + = =

Ngoi cch 1 nu trn, cn c th tnh I theo cch sau: Cch 2:

0 0

2 2

00 00

2 2

0

d cos d (*)

d(sin ) sin sin d (**)2 2

4cos .2 2

I x x x x x

x x x x x x x

x

= +

= + = +

= + =

Trong trng hp th sinh tnh I theo cch 2, vic cho im c thc hin nh sau: - Bin i v (*): 0,25 im; - Bin i t (*) v (**): 0,50 im; - Bin i tip t (**) n kt qu: 0,25 im.

3. (1,0 im)

Ta c: 2 2(2 1)( 1)2 1

x xx x

'( ) 21 2

f x x +

= + =

x ( 2; 0).

Suy ra, trn khong ( 2; 0): 1'( ) 02

f x x= = . 0,50

Ta c: , , (0) 0f = ( 2) 4 ln 5f = 1 1 ln 22 4

f =

. 0,25

V 4

44 ln 5 ln 0 (do 5)5e e = > > v

441 ln 2 ln 0 (do 2 )

4 2e e = < <

Nn [ ]2;0

1min ( ) ln 24x

f x

= v [ ]2;0

max ( ) 4 ln 5x

f x

= . 0,25

Lu : Gi tr nh nht v gi tr ln nht ca hm s f(x) trn on [ 2; 0] cn c k hiu tng ng bi

[ 2;0]min ( )f x

v ma[ 2;0]

x ( )f x

.

Cu 3 (1,0 im)

V SA mp(ABC) nn

SA AB v SA AC. Xt hai tam gic vung SAB v SAC, ta c

}chungSA SAB SACSB SC = = AB AC =

0,25

S

C

B

a

A

3

p dng nh l csin cho tam gic cn BAC, ta c 2 2 2 2 2 02 . .cos 2 (1 cos120 ) 3a BC AB AC AB AC BAC AB AB= = + = = 2

Suy ra 33

aAB = .

Do 2 2 63

aSA SB AB= = v SABC = 2

21 3.sin2 1

aAB BAC =2

.

0,50

V vy VS.ABC = 13

SABC.SA = 3 236

a . 0,25

Lu : cu ny, khng cho im hnh v.

1. (0,75 im)

Tm T v bn knh R ca (S): (1;2;2)T = v 6R = . 0,25

Khong cch h t T n (P): 2 2 2

|1.1 2.2 2.2 18 | 91 2 2

h + + += =+ +

0,50

2. (1,25 im)

Phng trnh tham s ca d: V d (P) nn vect php tuyn n ca (P) l vect ch phng ca d. T phng trnh ca (P), ta c ( )1;2;2n = .

0,25

Do , phng trnh tham s ca d l: 12 22 2

x ty tz t

= + = += +

0,25

To giao im H ca d v (P): Do H d nn to ca H c dng (1 + t ; 2 + 2t ; 2 + 2t).

0,25

V H (P) nn 1 + t + 2(2 + 2t) + 2(2 + 2t) + 18 = 0, hay . 3t = 0,25

Cu 4a (2,0 im)

Do ( 2; 4; 4)H = . 0,25

Ta c: . 216 32 16 (4 )i = = = 0,50

Do , phng trnh cho c 2 nghim l:

14 4 1 1

16 4 4iz i+= = + v 2

4 4 1 116 4 4

iz i= = . 0,50

Cu 5a (1,0 im)

Lu : Cho php th sinh vit nghim dng 1, 21

4iz = hoc 1, 2

4 416

iz = .

1. (0,75 im)

Gi (P) l mt phng i qua A v vung gc vi d. V d (P) nn vect ch phng u ca d l vect php tuyn ca (P). T phng trnh ca d, ta c ( )2;1; 1u = .

0,25

Cu 4b (2,0 im)

Do , phng trnh tng qut ca mp(P) l: 2.( 1) 1.( 2) ( 1)( 3) 0x y z + + + = hay 2 3 0x y z+ + = . 0,50

4

2. (1,25 im)

Khong cch h t A n d: T phng trnh ca d suy ra im B(1; 2; 3) thuc d.

Do ,

| |

BA uh

u

= .

0,50

Ta c . Do : (2; 4;6)BA =

( )1 1 1 2 2 1, ; ; (2; 14; 10)4 6 6 2 2 4BA u = = 0,25

V vy 2 2 2

2 2 2

2 ( 14) ( 10) 5 22 1 ( 1)

h + + = =+ +

. 0,25

Phng trnh mt cu (S) tm A(1; 2; 3), tip xc vi d: V (S) tip xc vi d nn c bn knh bng h. Do , phng trnh ca (S) l:

2 2 2( 1) ( 2) ( 3) 5x y z + + + = 0 0,25

Lu : C th s dng kt qu phn 1) tnh khong cch h t A n d. Di y l li gii tm tt theo hng ny v thang im cho li gii :

Gi H l giao im ca d v mt phng (P), ta c H l hnh chiu vung gc ca A trn (P). Do h AH= . 0,25

To ca H l nghim ca h phng trnh 1 2

2 12 3

x y z

x y z

31

0

+ + = = + + =

T kt qu gii h trn ta c ( )3 ; 1 ; 2H = .

0,50

V vy ( ) ( ) ( )2 2 21 3 2 1 3 2 5 2h AH= = + + + + = . 0,25

Ta c: . ( )22 8 9 3i i = = = 0,50 Cu 5b (1,0 im)

Do , phng trnh cho c 2 nghim l:

13

4i iz i+= = v 2

3 14 2

i iz i= = . 0,50

- Ht -

5

1

B GIO DC V O TO

THI CHNH THC



HNG DN CHM THI

(Vn bn gm 04 trang)

I. Hng dn chung

1) Nu th sinh lm bi khng theo cch nu trong p n nhng ng th cho s im tng phn nh hng dn quy nh.

2) Vic chi tit ho (nu c) thang im trong hng dn chm phi m bo khng lm sai lch hng dn chm v phi c thng nht thc hin trong ton Hi ng chm thi.


II. p n v thang im

CU P N IM

1. (2,0 im)

a) Tp xc nh: D = . 0,25

b) S bin thin:

Chiu bin thin: 'y = 234

x 3x. Ta c:

'y = 0 04xx==

; 'y > 0 04xx

v 'y < 0 0 < x < 4.

Do : + Hm s ng bin trn mi khong ( ;0) v (4; );+

+ Hm s nghch bin trn khong (0; 4).

0,50

Cc tr:

+ Hm s t cc i ti x = 0 v yC = y(0) = 5;

+ Hm s t cc tiu ti x = 4 v yCT = y(4) = 3. 0,25

Gii hn: lim ; limx x

y y +

= = + . 0,25

Cu 1 (3,0 im)

Bng bin thin:

0,25

x 0 4 + y + 0 0 +

y 5

3

+

2

c) th (C):

0,50

2. (1,0 im)

Xt phng trnh: 3 26 0x x m + = (). Ta c:

() 3 21 3 5 5 .4 2 4

mx x + = 0,25

Do :

() c 3 nghim thc phn bit ng thng 54my = ct th (C) ti 3 im phn bit 0,25

3 < 5 4m < 5 0 < m < 32. 0,50

1. (1,0 im)

iu kin xc nh: x > 0. Vi iu kin , phng trnh cho tng ng vi phng trnh

22 22 log 7 log 3 0x x + =

0,50

2

2

log 31log2

x

x

=

= 0,25

8 2.xx=

= 0,25

Lu : Nu th sinh ch tm c iu kin xc nh ca phng trnh th cho 0,25 im.

2. (1,0 im)

( )1

4 3 2

0

2 dI x x x x= + 0,25

= 1

5 4 3

0

1 1 15 2 3

x x x +

0,50

= 1 .30

0,25

3. (1,0 im)

Cu 2 (3,0 im)

Trn tp xc nh D = R ca hm s f(x), ta c: '( )f x = 2

2112

x

x

+. 0,25

5

3

O x

y

64 2

3

Do : '( )f x 0 2 12 2x x+ 0,25

204

xx

0,25

x 2. 0,25

Gi O l giao im ca AC v BD. V ABCD l hnh vung nn AO BD. (1)

V SA mp(ABCD) nn: + SA l ng cao ca khi chp S.ABCD; + SA BD. (2) T (1) v (2) suy ra BD mp(SOA). Do SO BD. (3) T (1) v (3) suy ra SOA l gc gia mp(SBD) v mp(ABCD). Do SOA = 60o.

0,50

Xt tam gic vung SAO, ta c:

SA = OA. tan SOA = 2

AC .tan60o = 2 .2

a 3 = 6 .2

a 0,25

Cu 3 (1,0 im)

V vy VS.ABCD = 13

SA. ABCDS = 13

. 6 .2

a 2a = 3 66

a . 0,25

1. (1,0 im)

Gi (P) l mt phng i qua A(1; 0; 0) v vung gc vi BC. V BC (P) nn BC l mt vect php tuyn ca (P).

0,25

Ta c: BC = (0; 2; 3). 0,25

Do , phng trnh ca (P) l: 2y + 3z = 0. 0,50

2. (1,0 im)

Gi (S) l mt cu ngoi tip t din OABC. V O(0; 0; 0) (S) nn phng trnh ca (S) c dng:

x2 + y2 + z2 + 2ax + 2by + 2cz = 0. () 0,25

V A(1; 0; 0), B(0; 2; 0), C(0; 0; 3) (S) nn t () ta c: 1 2 04 4 09 6 0.

abc

+ = + = + =

Suy ra: a = 12

; b = 1; c = 3 .2

0,50

V vy, mt cu (S) c tm 1 3; 1;2 2

I =

. 0,25

Cu 4.a (2,0 im)

Lu : Th sinh c th tm to ca tm mt cu (S) bng cch da vo cc nhn xt v tnh cht hnh hc ca t din OABC. Di y l li gii theo hng ny v thang im cho li gii :

B

A

C

DO

S

4

Tm I ca mt cu (S) l giao im ca ng trc ca ng trn ngoi tip tam gic OAB v mt phng trung trc ca on thng OC. 0,25

T , v tam gic OAB vung ti O, cc im A, B thuc mp(Oxy) v im C thuc trc Oz nn honh , tung ca I tng ng bng honh , tung ca trung

im M ca on thng AB v cao ca I bng 12

cao ca C. 0,50

Ta c M = 1 ; 1; 02

v C = (0; 0; 3) (gi thit). V vy 1 3; 1;2 2

I =

. 0,25

Ta c 1 22 3 8 .z z i = + 0,50 Cu 5.a (1,0 im) Do , s phc 1 22z z c phn thc bng 3 v phn o bng 8. 0,50

1. (1,0 im)

T phng trnh ca suy ra i qua im M(0; 1; 1) v c vect ch phng u = (2; 2; 1).

Do d(O, ) = ,MO u

u

. 0,50

Ta c MO = (0; 1; 1). Do ( ), 1; 2; 2MO u = . 0,25

V vy d(O, ) = 2 2 2

2 2 2

( 1) ( 2) ( 2)

2 ( 2) 1

+ +

+ + = 1. 0,25

2. (1,0 im)

Gi (P) l mt phng cha im O v ng thng .

Do vect ,n MO u = c phng vung gc vi (P) nn n l mt vect php

tuyn ca (P).

0,50

Cu 4.b (2,0 im)

Suy ra phng trnh ca (P) l: x 2y 2z = 0, hay x + 2y + 2z = 0. 0,50

Ta c: 1 2.z z = 26 + 7i. 0,50 Cu 5.b (1,0 im) Do , s phc 1 2.z z c phn thc bng 26 v phn o bng 7. 0,50

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

B GIO DC V O TO

THI CHNH THC


HNG DN CHM THI

(Vn bn gm 04 trang)

I. Hng dn chung

1) Nu th sinh lm bi khng theo cch nu trong p n nhng ng th cho s im tng phn nh hng dn quy nh.



II. p n v thang im

CU P N IM

1. (2,0 im)

a) Tp xc nh : 1\2

D =

. 0,25

b) S bin thin :

Chiu bin thin : ( )2

4' 0,2 1

y xx

= <

D .

Hm s nghch bin trn mi khong 1;2

v 1 ;2

. +

0,50

Tim cn :

12

limx

y

= ;12

limx

y+

= + 12

x = l tim cn ng.

lim 1x

y

= ; lim 1x

y+

= 1y = l tim cn ngang.

0,50

Cu 1 (3,0 im)

Bng bin thin :

1

x 12

+

'y

y

+

1

1

0,25

c) th (C):

0,50

2. (1,0 im)

Honh giao im ca vi ng thng l nghim ca phng

trnh

( )C 2y x= +2 1 22 1

x xx

+ = +

(1)

(1) + (2 (v 2 1 (2 1)( 2x x x

2

)= + ) 12

x = khng l nghim ca (2))

2

2 3 0= 1x x + x = hoc 32

x = .

0,50

Vi 32

x = th 12

y = .

Vi 1x = th . 3y =

Vy ta giao im cn tm l 3 1;2 2

v ( . 1;3)

0,50

1. (1,0 im)

t 7x

t ( t ). = > 0 0,25

Phng trnh cho tr thnh 7 8 hoc 2

1 0 1t t t + = = 17

t = . 0,25

Vi t , ta c 7 . 1= 1 0x

x= =

Vi 17

t , ta c = 17 1. 7

xx= =

Vy nghim ca phng trnh l hoc . 0x = 1x =

0,50

2. (1,0 im)

t 2 54 5ln 4 5ln 2t x t x tdt dx

x= + = + = . 0,25

Cu 2 (3,0 im)

i cn : 1 2x t= = v 3x e t= = . 0,25

Do 3 32 3 3 3

22

2 2 2 3 25 15 15

I t dt t = = = =

3815

. 0,50

3. (1,0 im)

Ta c 2

' 3 4y x x= + m . 0,25

Nu hm s t cc tiu ti 1x = th , suy ra . '(1) 0y = 1m = 0,25

Vi th 1m =3 2

2 1y x x x= + + , 2

' 3 4 1y x x= + v " 6 4y x= . M v nn hm s t cc tiu ti '(1) 0y = ( )" 1 2 0y = > 1x = . 0,25

Vy l gi tr cn tm. 1m = 0,25

Ta c nn l hnh chiu ca trn ( )SA ABCD AC SC ( )ABCD .

Do . 45o

SCA =Tam gic vung cn ti nn ACD D 2AC a= . Tam gic vung cn ti SAC A nn 2SA a= .

0,50

Cu 3 (1,0 im)

Din tch ca hnh thang vung ABCD l 2( 3 ) 2

2a a a a+ = .

Vy 3

.2 2

3S ABCDaV = .

0,50

1. (1,0 im)

Ta c ( )2 2 2

2.3 2.1 1.0 1, ( ) 3

2 2 ( 1)d A P

+ += =

+ + . 0,50

Ta c l vect php tuyn ca . (2;2; 1)n = ( )P

( )Q song song vi ( nn ( nhn )P )Q (2;2; 1)n = lm vect php tuyn. 0,25

Cu 4.a (2,0 im)

Mt khc ( qua nn c phng trnh )Q (3;1;0)A ( )Q

2( 3) 2( 1) 1( 0) 0 2 2 8 0x y z x y z + = + = . 0,25

3

2. (1,0 im)

Gi l ng thng qua v vung gc vi th d A ( )P (2;2; 1)n = l vect ch phng ca . d

Do phng trnh tham s ca l d3 21 2

x ty tz t

= + = + =

. 0,50

Gi l hnh chiu ca H A trn ( th l giao im ca v ( . )P H d )P

Do nn . H d (3 2 ;1 2 ; )H t t+ + t

Mt khc nn ta c . ( )H P 2(3 2 ) 2(1 2 ) ( ) 1 0t t t+ + + + = 1t =

Vy . (1; 1;1)H

0,50

Phng trnh cho tng ng vi phng trnh ( 1 ) 2 4i z i = 0,25

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

i iz zi i

= = )

ii

++

0,25

Cu 5.a (1,0 im)

(2 4 )(1 )2i iz + = 6 2 3

2iz z = = i .

Vy nghim ca phng trnh l . 3z i= 0,50

1. (1,0 im)

Ta c ( 1; 2; 2); ( 1;0; 1)AB AC= = , (2;1; 2)AB AC = . 0,50

Mt phng qua , nhn ( )ABC A ,AB AC lm vect php tuyn nn c

phng trnh . 2(x 0) 1(y 0) 2(z 3) 0 + = 2x y 2z 6 0 + + =0,50

2. (1,0 im)

Ta c: ABCS 1 ,2

AB AC = 2 2 21 32 1 ( 2)

2 2= + + = . 0,50

Cu 4.b (2,0 im)

2 2 2( 1 1) (0 2) (2 1) 5BC = + + + + = .

Gi AH l ng cao ca tam gic th ABC2 3

5ABCSAH

BC= = .

0,50

Phng trnh cho tng ng vi phng trnh 2

2 3 0z iz + = .

Ta c . ( )2 24 12 16 4i i = = =0,50

Cu 5.b (1,0 im)

Vy phng trnh c hai nghim l 12 4 3

2i iz i+= = v 2

2 42

i iz i= = . 0,50

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

4

1

B GIO DC V O TO

THI CHNH THC



HNG DN CHM THI

(Bn hng dn ny gm 04 trang)

I. Hng dn chung

1) Nu th sinh lm bi khng theo cch nu trong p n nhng ng th vn cho s im tng phn nh hng dn quy nh.



II. p n v thang im

CU P N IM

1. (2,0 im)

Tp xc nh: D .= 0,25

S bin thin:

Chiu bin thin: 30

4 ; 02

xy x x y'

x .= = = =

+ Trn cc khong ( )2 ; 0 v ( )2 ; 0, y+ > nn hm s ng bin. + Trn cc khong ( ); 2 v ( )0 ; 2 0, y < nn hm s nghch bin.

0,50

Cc tr: + Hm s t cc i ti x = 0 v yC 0.= + Hm s t cc tiu ti 2x = v yCT 4.=

0,25

Gii hn: ;x x

lim y lim y . +

= + = + 0,25

Cu 1 (3,0 im)

Bng bin thin:

0,25 +

4

x 2 0 2 +

y 0 + 0 0 +

y

4

+ 0

2

th:

Lu : Th sinh ch trnh by: th ct Ox ti O v ( )2 2 ;0 hoc th hin ( )2 2 ;0 trn hnh v th vn cho 0,50 im.

0,50

2. (1,0 im)

Ta c ( ) ( )3 24 ; 3 4f x x x f x x . = = 0,25

( ) 20 0 01 3 4 1 1f x x x . = = = 0,25

( )0 071 ; 1 3,4

x y f '= = = ta c phng trnh tip tuyn l 534

y x .= + 0,25

( )0 071 ; 1 3,4

x y f '= = = ta c phng trnh tip tuyn l 534

y x .= + 0,25

1. (1,0 im)

iu kin: 3x .> 0,25

Vi iu kin trn, phng trnh cho tng ng vi ( ) ( )2 4 2 23 2 32 2log x log x log x log x + += =

0,25

( ) 22 3 2 3 4 0log x x x x = = 0,25

14

xx=

= . Vy nghim ca phng trnh l 4x .= 0,25

2. (1,0 im)

t 1x xt e dt e dx.= = 0,25

i cn: 0 0x t= = ; 2 1x ln t .= = 0,25

Suy ra 11 3

2

0 03tI t dt .= = 0,25

Cu 2 (3,0 im)

Vy 13

I .= 0,25

(loi)

x

y

O 2

4

2 2 2 2 2

3

3. (1,0 im)

Trn on [ ]0 ; 1 , ta c ( )( )

2

21

1m mf x .

x + =+

0,25

M ( )2 1 0, 0m m m f x . + > > Nn hm s ng bin trn [ ]0 ; 1 . 0,25

Suy ra gi tr nh nht ca hm s trn [ ]0 ; 1 l ( ) 20f m m.= + 0,25

[ ]( ) 2

0;12 2min f x m m .= + = Vy 1m = v 2m = . 0,25

Ta c ( ) o60A A ABC A BA . =

0,25

Din tch y: 2

2ABCaS . = 0,25

Chiu cao lng tr: 60 3AA' atan a .= = 0,25

Cu 3 (1,0 im)

Vy th tch khi lng tr ABC.A B C l 3 32ABC.A B C ABC

aV S .A A' . = = 0,25

1. (1,0 im)

Ta c ( )2 ; 0 ; 4 ,AB = suy ra AB c vect ch phng l ( )1 ; 0 ; 2u .= 0,50

Vy phng trnh tham s ca ng thng AB l 221 2

x tyz t.

= = = +

0,50

2. (1,0 im)

Gi ( )S l mt cu c ng knh AB v I l trung im AB. Suy ra ( )1 ; 2 ; 3I l tm ca ( )S .

0,25

Bn knh ca ( )S l ( ) ( ) ( )2 2 22 1 2 2 1 3 5R IA .= = + + = 0,25

M ( )( ) ( )( )22 2

2 1 1 2 5, 5

2 1 0

. .d I P .

+ += =

+ + 0,25

Cu 4.a (2,0 im)

Nn ( )( ),d I P R= . Vy ( )P tip xc vi ( )S . 0,25

A

A' C'

C

B

B'

60

4

Ta c 2 6 8z i= v 3 4z i.= + 0,25

Suy ra 2 9 4z z i.+ = 0,25

Cu 5.a (1,0 im)

( )( )( )

( )25 3 4 25 4 325 4 33 4 3 4 9 16

i i ii i.z i i

+ += = = +

+ + 0,50

1. (1,0 im)

ng thng OA c vect ch phng l ( )2 ; 1 ; 2OA .= 0,50

Vy phng trnh ca ng thng OA l 2

2

x ty tz t

= = =

hoc 2 1 2x y z .= = 0,50

2. (1,0 im)

Bn knh mt cu ( )S l 2 2 22 1 2 3R OA .= = + + = 0,25

Suy ra ( )S : ( ) ( ) ( )2 2 22 1 2 9x y z . + + = 0,25

ng thng qua ( )1 ; 3 ; 0B v c vect ch phng ( )2 ; 2 ; 1u .=

Mt khc, ( )1 ; 2 ; 2BA = ( ), 6 ; 3 ; 6BA u . =

Nn ( ) ( )2 2 2

2 2 2

, 6 3 6, 3

2 2 1

BA ud A .

u

+ + = = =+ +

0,25

Cu 4.b (2,0 im)

Suy ra ( ),d A R = . Vy tip xc ( )S . 0,25

Ta c ( )( )( )( )1 9 11 9 8 10

1 1 1 2i ii i .

i i i+ ++ +

= = +

0,25

Suy ra 4 5 5 4z i i .= + = 0,25

Cu 5.b (1,0 im)

Mt khc, ( )24 2z i .= = V vy cc cn bc hai ca z l 2i v 2i. 0,50

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

B GIO DC V O TO K THI TT NGHIP TRUNG HC PH THNG NM 2013 Mn thi: TON Gio dc trung hc ph thng THI CHNH THC

HNG DN CHM THI

(Bn Hng dn chm thi gm 04 trang)

I. Hng dn chung

1) Nu th sinh lm bi khng theo cch nu trong p n nhng ng th vn cho s im tng phn nh Hng dn chm thi quy nh.

2) Vic chi tit ha im s ca tng cu (nu c) trong Hng dn chm thi phi m bo khng lm sai lch Hng dn chm thi v phi c thng nht thc hin trong Hi ng chm thi.


II. p n v thang im

CU P N

IM

1. (2,0 im)

a) Tp xc nh: . D = 0,25

Cu 1 (3,0 im)

b) S bin thin:

Chiu bin thin: 21

' 3 3; ' 01.

== = =

xy x y

x

Trn cc khong v ( ; 1 ) (1; ),+ ' 0y > nn hm s ng bin. Trn khong nn hm s nghch bin. ( 1; 1), ' 0y x t t ta .2 2 3 0 =t tGii phng trnh (*) vi iu kin ta c 0,t > 3.t =

0,50

Vi t ta c Phng trnh c nghim duy nht 3,= 1.x = 1.x = 0,25

2. (1,0 im)

t u x ta c = + =1v d cos d ,v x x = =d d v sin .u x v x 0,25

Do ( )

= + 22

0 0

1 sin sin dI x x x x 0,25

= + + =

2

0

1 cos .2 2

x 0,50

3. (1,0 im)

Trn on [ ]1; 2 , ta c ( )= ++2

' 13

xln .y x

x 0,25

Vi mi x thuc on [ ]1; 2 , ta c: 2

13

x

x 0,25

Vi iu kin trn, phng trnh cho tng ng vi 22 2log 3log 2 0x x+ + = 0,25

2

2

log 1log 2.

xx=

= 0,50

21log 12

x x= = (tho mn iu kin). 0,25

Cu 2 (2,5 im)

21log 24

x x= = (tho mn iu kin).

Vy nghim ca phng trnh l 1 1, .2 4

x x= = 0,25

y

132

O x

2

3

2

2) (1,0 im)

Tp xc nh: [ ]0;4 .D = 0,25

Trn ta c ( )0;4 , ( )2

2' 1 .2 4

x xf xx x

= +

0,25

( ) ( )2

1 1' 0 2 0 2.x2 4f x x

x x

= + =

= 0,25

Ta c: ( ) ( ) ( )0 0, 2 3, 4 0f f f= = .=

T , gi tr ln nht ca ( )f x bng v gi tr nh nht ca 0 ( )f x bng 3.0,25

Ta c 1 1

0 0

xI dx xe dx= . 0,25

Ta c: I1 =1

10

0

1.dx x= = 0,25

Tnh I2 = 1

0

.xxe dx t u x= v ta c ,xdv e dx= du dx= v .xv e= Do : 0,25

I2 =1 11 1

0 00 0

1.x x x xxe dx xe e dx e e= = = 0,50

Cu 3 (1,5 im)

Vy 1 2 0.I I I= = 0,25

( )( ; ( )) 60

SM ABC

SCM SC ABC

= = . 0,25

0

0

.sin 60 15;

.cos 60 5.

SM SC a

MC SC a

= =

= = 0,25

Xt tam gic vung MAC, ta c: 2 2 2AC AM MC+ =

22 5

2AC 2AC a + =

2 .AC a =

0,25

Cu 4 (1,0 im)

Suy ra 2 21 2 .2ABC

S AC = = a

Vy 3

.1 2. .3 3S ABC ABC

a 15M S= =V S

0,25

B60

C

M A

S

3

1) (1,0 im)

Gi d l ng thng i qua A v vung gc vi (P).

Vect php tuyn ca (P) l vect ch phng ca d. (2; 2;1n = )0,50

Do phng trnh tham s ca d l 1 2

1 2.

x ty tz t

= + = =

0,50

2) (1,0 im)

Ta c:

( ) ( ); ; 2 2 1 0 2 2 1M a b c P a b c c b a + = = + (1)

2AM OA a b = (2)

0,25

Th (2) vo (1), ta c 3.c = 0,25

V ( ) ( ) ( ) ( )2 2 2 221 1 1 1 9AM a b c a b= + + + = + + + ( )( ), 1d A P = v 0,25

Cu 5 (2,0 im)

nn: ( )( ) ( ) ( )2 23 , 1 1 0 1,AM d A P a b a b 1= + + = = = (tha mn (2)). Vy c duy nht im M cn tm l ( )1; 1; 3 .M

0,25

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

4

b gio dc v o to --------------------


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

hng dn chm chnh thc

mn ton * Bn hng dn chm thi ny c 4 trang *

I. Cc ch khi chm thi

1) Hng dn chm thi (HDCT) ny nu biu im chm thi tng ng vi p n nu di y. 2) Nu th sinh c cch gii ng, cch gii khc vi p n, th ngi chm cho im theo s

im qui nh dnh cho cu ( hay phn ) . 3) Vic vn dng HDCT chi tit ti 0,25 im phi thng nht trong tt c cc t chm thi mn

Ton ca Hi ng. 4) Sau khi cng im ton bi mi lm trn im mn thi theo qui nh chung.

II. p n v cch cho im Bi 1 (3 im). 1. (2, 5 im) - Tp xc nh R \ { 2}. (0, 25 im)- S bin thin: a) Chiu bin thin:

2

12

+=x

xy , y ' = 2

2

)2(

34

+

x

xx,

=

==

31

0'xx

y

y< 0 vi ( ) ( ) ;31;x : hm s nghch bin trn cc khong ( ) ( )+ ;3,1; . y > 0 vi ( )2;1x (2; 3): hm s ng bin trn cc khong (1; 2), (2; 3).

(0, 75 im)b) Cc tr: Hm s c hai cc tr: cc tiu yCT = y(1) = 2 , cc i yC = y(3) = - 2.

(0, 25 im)

c) Gii hn:

.2

542

2lim

2lim,

2

542

2lim

2lim =

+

+=

++=

+

=

x

xx

xy

xx

xx

xy

x th c

tim cn ng x = - 2.

0)2

1(lim)]2([lim =

=+

xxxy

x. th c tim cn xin y = - x + 2.

(0, 25 im) (0, 25 im)

d) Bng bin thin:

(0, 25 im)

- th:

x + 321

y - 0 + + 0 -

y + + - 2 C CT 2 - - -

Hng dn chm thi TNTHPT nm 2003: chnh thc

2

(0, 50 im)

2. ( 0, 5 im)

2162

2+

++=mxmm

xy , th c tim cn ng l x = 2 khi v ch khi =

yx 2lim

=+

2162

2lim

mxmm

x. Qua gii hn c 2 + m 2 = 0 hay m = 0.

Vi m = 0 ta c 2

12

2542

+=

+

=x

xx

xxy ; nn th hm s c tim cn

xin l y = - x +2. Vy gi tr cn tm ca m l m = 0.

(0, 25 im)

(0, 25 im)Bi 2 (2 im ) 1. (1 im)

22

23

)1(

21

)1(

133)(

++=

+

++=

xx

x

xxxxf

;12

2

2

2)1(

13233 Cx

xxdxx

xxx+

+++= +

++

V 31

)1( =F nn613

=C . Do 613

12

2)(

2

+++=x

xxxF .

(0, 75 im)

(0, 25 im)

2. ( 1 im)

Din tch hnh phng S cn tm

+

+

++

+ ===

6

1

6

1

26

1

2)2

16214(

212102

0212102

dxx

xdxx

xxdx

xxxS

(0, 25 im)

V ng dng th : + Giao vi Oy: ti im (0; 2,5) + th c tm i xng ti im ( 2 ; 0). + th c hai tim cn: x = 2 v y = - x + 2.

Gii phng trnh:

212102 2

+

xxx = 0

ta tm c cc cn ly tch phn l: - 1 v 6.


3

.8ln1663)2ln1614( 61

2 =+=

xxx

(0, 75 im)

Bi 3 (1, 5 im)

1. (1 im). Gi s im M gc phn t th nht v M = (x; y). Khi theo u bi ta c cc h thc: cc bn knh qua tiu

1MF = a + ex = 15,

2MF = a - ex = 9, khong

cch gia cc ng chun: 2 . ea

= 36. Vy a = 12, e = 32

, x = 29

.

V c = a.e = 8 v c b2= a2- c2= 80 nn phng trnh chnh tc ca elp (E) l

180

2

144

2=+

yx

(0, 75 im) (0, 25 im)

2. (0, 5 im).

Tip tuyn vi elp (E) ti im M(29

; 2115

) l 3211 =+ yx .

Trn elp (E) cn 3 im c to l (- 29

; 2115

), (29

; - 2115

), (- 29

; - 2115

)

cng c cc bn knh qua tiu l 9 v 15. Do ta cn c 3 phng trnh tip tuyn vi elp (E) ti cc im (tng ng) l : - 3211 =+ yx , 3211 = yx ,

3211 =+ yx

(0, 25 im)

(0, 25 im) Bi 4 (2, 5 im) 1. (1 im) Theo u bi ta c A= (2; 4; -1), B = (1; 4; -1), C = (2; 4; 3), D = (2; 2; -1). Do :

ADABADAB

ADACADAC

ACABACAB

=++=

=++=

=++=

00.0)2.(00).1(.

00.4)2.(00.0.

04.00.00).1(.

Th tch khi t din ABCD tnh theo cng thc

VABCD =ADACAB ].,[

6

1 = 3

4 (do )0;4;0(],[ =ACAB )

(0, 75 im)

(0,2 5 im)2. (0, 75 im) ng thng CD nm trn mt phng (ACD) m mt phng (ACD) AB nn ng vung gc chung ca AB v CD l ng thng qua A v vung gc vi CD.

Vy ng thng c vect ch phng )1;2;0(],[2

1=

=

CDABu v phng trnh

tham s l:

+=

=

=

tzty

x

124

2

(0, 50 im)

Mt phng (ABD) c vect php tuyn [=

nAB ,

AD ] = (0; 0; 2). Vy gc nhn

gia v mt phng (ABD) xc nh bi biu thc:


4

sin =

un

un

.

.

55

522

1)2(.2

1.2)2.(00.0

222==

+

++=

(0, 25 im)

3. (0, 75 im) Phng trnh mt cu (S) c dng:

0222222 =++++++ dczbyaxzyx Bn im A, B, C, D nm trn mt cu nn c to tho mn phng trnh trn. Do cc h s a, b, c, d l nghim ca h phng trnh sau:

=+++=++++=+++=+++

)(02449)(068429)(028218)(028421

SDdcbaSCdcbaSBdcbaSAdcba

Gii h ny c a = 23

, b = -3, c = - 1, d = 7. Do phng trnh mt cu (S) l:

07263222 =+++ zyxzyx .

(0, 50 im)

Mt cu (S) c tm K = (23

; 3; 1) v bn knh R = 2

21; phng trnh ca mt

phng (ABD) l: z + 1 = 0. Phng trnh mt phng song song vi mt phng (ABD) c dng z + d = 0. Mt phng l tip din ca mt cu (S) khi v ch khi khong cch t tm K n mt phng bng R:

2

2212,2

22112

21212020

1.1 +=

==

++

+dd

d.

Vy c hai tip din ca mt cu (S) cn tm l:

(1): z + 2221

= 0

(2): z 2221+

= 0

(0, 25 im)

Bi 5 (1 im).

H thc 2:5:61:1:1 =+

+ CyxC

yxC

yx vi x v y l cc s nguyn dng m

2 y+1 x cho h phng trnh sau:

=+

+=+

2

1yxC

6

y1xC

5

1yxC

6

y1xC

Gii h:

=

=

=+

+=

+

+

+=

+

+

+=

+

+

3

8

26

1

)1(5

1

)1)((6

1

)!1()!1(2

!

)!1(!6

)!1(

)!1()!1(5

!

)!1(!6

)!1(

1 yx

y

x

yyxyx

x

yxy

x

yxy

x

yxy

x

yxy

x

(0, 50 im) (0, 50 im)

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

b gio dc v o to .......................

hng dn chm

chnh thc


..................... Mn thi: Ton

Bn hng dn chm c 4 trang

I. Cc ch khi chm thi

1) Hng dn chm thi (HDCT) ny nu biu im chm thi tng ng vi p n di y.

2) Nu th sinh c cch gii ng khc vi p n, th ngi chm cho im theo s im qui nh dnh cho cu ( hay phn ) .

3) Vic vn dng HDCT chi tit ti 0,25 im phi thng nht trong tt c cc t chm thi mn Ton ca Hi ng.

4) Sau khi cng im ton bi mi lm trn im mn thi theo qui nh chung.

II. p n v cch cho im

Bi 1 (4 im)

1. (2, 5 im)

- Tp xc nh R . 0, 25 - S bin thin:

a) Chiu bin thin:

2331 xxy = , y ' = , ; xx 22

=

==

20

0'xx

y

y< 0 vi : hm s nghch bin trn khong( 2;0x ) ( )2;0 , y > 0 vi (2; +): hm s ng bin trn cc khong (- ; 0), (2; +).

( 0;x )

0, 75

b) Cc tr:

Hm s c hai cc tr: cc i yC = y(0) = 0, cc tiu yCT = y(2) = 3

4 .

0, 25

c) Gii hn:

+=+= yxyx lim,lim , th khng c tim cn.

0, 25

d) Bng bin thin:

0, 25

x - 0 2 + y + 0 - 0 +

y

0 + C CT

- 3

4

1

e) Tnh li, lm v im un ca th:

y= 2x 2, y = 0 x = 1. Ta c y(1) = 3

2 ,

x - 1 + y - 0 +

th li . un lm

0, 25

- th:

0, 50

2. (1,0 im) Nu c iu kin cn v ng thng d vi h s gc k i qua

im (3; 0) c phng trnh y = k(x-3) tip xc vi (C) l h phng trnh sau c nghim

=

=

kxx

xkxx

2

)3(

2

2331

Tm c hai nghim (x; k) l: (0 ; 0) , (3 ; 3) . Vit c hai phng trnh tip tuyn: y = 0 , y = 3x 9 .

0, 25 0, 50 0, 25

3. (0,50 im)

+ ==3

0

4563

0

223 )32

91()

31( dxxxxdxxx V

35

81)5963

(0

3567 =+= xxx (vtt).

0, 25

0, 25

Bi 2 (1 im)

Tnh ng o hm ca hm s y = 2sinx :xsin3

4 3

cosx.x4sincosx2y' 2=

Tm c cc im ti hn trn on [0; ] : y = 0 x {4

3,4

,2

}.

0, 25

0, 25

)2

;1(U 3

3

4

3

2

y

-1 O 1 2 3 x V ng dng th : + Giao vi Oy: (0; 0) + Giao vi Ox: (0; 0) , (3; 0) + Tm i xng ca th:

U(1; ) 3

2

2

Tnh cc gi tr y(0), y(), y( )4

3(,)4

(,)2

yy

3

22,0 == yy][0;][0;

maxmin

.

0, 50

Bi 3 (1,5 im)

1. (0,75 im).

Tm ta im M(3; m) thuc (E), m>0: M = (3; 5

16 ).

Vit c phng trnh tip tuyn ca (E) ti M: 116.5.16

25.3

=+yx

Hay 1525

3=+

yx .

0, 50

0, 25

2. (0, 75 im). Tm c A + A F = B + B = 10 . 1F 2 1F 2F

Tnh c A + B = 20 (A + B ) = 12. 2F 1F 1F 2F

0, 50

0, 25

Bi 4 (2,5 im)

1. (1 im)

Nu c ba vect ng phng = 0, ADACAB ,,

ADACAB ].,[

Tnh c: ; ,)0;4;0(=AB )0;0;3(,)0;4;3( =

=

ADAC

; = 3.0 + 0.0 + 0.(-12) = 0. )12;0;0(],[ =ACAB

ADACAB ].,[

( Ghi ch: Nu th sinh lp lun bn im cho cng nm trn mt phng z = 2 th chm t im ti a)

0,2 5

0, 75

2. (1,0 im)

Nu c A = (1; -1; 0), phng trnh mt cu (S) cn tm c dng: 0222222 =++++++ dczbyaxzyx (*)

Nu c bn im A, B , C , D nm trn mt cu (S) nn c to tho mn phng trnh (*) v cc h s a, b, c, d l nghim ca h phng trnh :

=+++

=++++

=++++

=++

(S)D0d4c2b8a21

(S)C0d4c6b8a29

(S)B0d4c6b2a14

(S)A'0d2b2a2

Gii h tm c: a = 25

, b = -1, c = - 1, d = 1; phng trnh mt cu

(S) : . 01225222 =+++ zyxzyx

0, 50

0, 50

3

3. (0,50 im)

Tm c tm I = (25

; 1; 1) ca mt cu (S) v vect php tuyn

1)2;;23

(IA' =

ca tip din ().

Vit c phng trnh tip din () ca mt cu (S) ti im Al: 3x + 4y + 2z +1= 0.

0, 25

0, 25

Bi 5 (1 im)

Vit c:

+++

++

+60)1)(4)(5(

60!)(

23

5knnn

nkA

kn

P kn

n

Xt vi n > 4 : khng nh bt phng trnh v nghim. Xt vi n {0, 1, 2 , 3} tm c cc nghim (n; k) ca bt phng trnh

l: (0; 0) ,

đề + đáp án tốt nghiệp môn toán thpt 2002-2014

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