tai lieu toan

Li ni u I. Kinh nghim lm bi Bn cnh vic cng c kin thc th kinh nghim thc t cho cc bn th sinh khi bc vo phng thi l v cng quan trng. Hy nh rng khi vo phng thi, c ngha y khng phi l bui din tp na. Chnh v vy, nhm tc gi c kt li nhng kinh nghim qu bu, nhng li khuyn b ch trong phn cui cun sch ny. Hy ch : iu u tin, v cng quan trng: bn phi chc chn rng khi vo phng thi,

ng nn ti gim th hay nhng th sinh khc, hy tp trung vo bi lm ca chnh mnh. S tp trung s gip bn tit kim thi gian v n chnh l ci ni ca nhng tng. Tip theo, sau khi c thi mt lt, hy lm ln lt nhng cu t d n

kh. Nh vy bn s nm chc im ca nhng cu v to s t tin lm tip nhng cu kh hn. To ra s thoi mi ban u l mt yu t rt quan

trng gip th sinh hon thnh tt nht bi thi ca mnh. Thng thng th t t d n kh s l:

Kho st hm s ( u tin ca cu I). S phc (cu VII.b). Tch phn (cu III)

Phng trnh lng gic ( u tin ca cu II). Phng trnh h phng trnh bt phng trnh ( th hai ca cu II).

Hnh gii tch trong khng gian, hnh gii tch trong phng (cu VI.a). Cu hi ph hm s ( th hai ca cu I). Hnh khng gian (cu IV). Cui cng l bt ng thc (cu V).

Sau khi gii quyt trn vn nhng cu d, nu thy nhng cu cn li u ri vo nhng dng l, lc bn cng ng nn mt bnh tnh. Hy th su, c k nhiu ln, tp trung nh li nhng kin thc lin quan hc, hay nhng kinh nghim, nhng dn d tng c nghe, chc chn bn s c nhng

tng n c vi li gii. Nhng lc cng thng hy ly li s t tin bng cch th su v t nh

rng: Mnh c th lm c!. Vi nhng bn hc sinh trung bnh kh, chng ti khuyn cc bn nn b nhng cu nh hnh khng gian v c bit l bt ng thc. Bi nu khng

c kin thc tht vng th xc sut bn lm c hai mng gn nh bng 0. Thm ch, nu nh th tm vn may vi chng, chc chn bn s mt

mt khong thi gian v ch v iu s ch khin tm l ca bn b dao ng, nh hng khng tt ti vic gii quyt nhng phn cn li. Mt khc,

hin nhin l nu b qua hai phn th bn s c thm kh nhiu thi gian ngh ra nhng tng mi cng nh kim tra li cn thn nhng bi mnh lm. Mt lu na, l khng nn lm trc vo giy nhp. Vi nhng bi ton

m th sinh nh hng c cch gii th khng nn gii hon ton trn giy nhp ri mi vit vo giy thi. Lm nh vy va mt thi gian va d sai

st. Bi khi gii trc tip bi ton l "vit ra nhng g trong u", ta s rt ch ng. Cn khi chp li (k c chp nhng g mnh va vit) th sinh li tr thnh th ng v vy rt d vit nhm, b st. Do , ch nn s dng giy nhp nhng phn cn tnh ton. C gng luyn cho mnh k nng tnh ton

linh hot, bi tc tnh ton l mt yu t quan trng nh hng ti thi gian cng nh nim tin khi lm bi ca bn.

Ch , vi nhng bi ton kh, nu ch lm c mt phn m cha lm c trn vn th cng nn vit vo bi lm. V nhng phn bn lm c, nu ng theo ba-rem chm th vn c im.

Chng ti xin c ly mt v d n gin l phn cu hi ph ca bi kho

st hm s, chng hn bi hi v cc tr: Tm iu kin ca hm s c cc i v cc tiu tha mn. Hy nh rng c nhc n cc tr, ta phi ngh ngay n k nng tnh nhanh cc tr. V sau khi s dng k nng ny, vit c phng trnh ng thng i qua cc i, cc tiu th bn nm chc c mt na s im ca cu . Cn vic tha mn mt iu kin no, nu iu kin kh qu bn khng gii tip c, khng sao c, v chng ta c mt na s im ri. Hy coi y l mt thnh cng mc chp nhn c.

iu cui cng, mi bi ton cc bn th sinh nn c mt phn kt lun. C th l vit li p s hoc tr li cu hi ca bi ngi chm thi bit

c th sinh kt thc bi hay cha. Vi kinh nghim nhiu nm chm bi thi i hc th b phn kt lun l mt trong nhng li kh ph bin ca cc th sinh. n gin ch thm mt cu kt lun nhng chc chn cc bn s gy c cm tnh vi ngi chm bi tnh r rng, mch lc trong bi lm ca

mnh. II. Kinh nghim chun b tm l trc khi thi

Tm l chc chn s l vn m bt c th sinh no cng u phi tri qua. N c

tm nh hng khng nh ti kt qu thi m bn s t c. Cu hi t ra l lm th no duy tr c mt tinh thn thoi mi v mt sc khe tt trc

th thch kh khn ny ? Sau y l mt vi kinh nghim m chng ti mun chia s. Nhng kinh nghim ny trn thc t mang li s thnh cng cho nhng ngi bit p dng chng mt cch hp l. Trc tin, hy nh khng nn np qu nhiu kin thc trong giai on

nc rt. Do tm l thi c nng n nn nhiu bn gp rt n tp ba mn thi mt lc. Nhng chnh lng kin thc qu ln khin h ri vo trng thi

stress v lm gim hiu qu n tp. trnh hin tng ny, cc bn cn phi bit loi ra nhng kin thc khng cn thit. Ni cch khc, hy ti u ha chng trnh hc ca mnh.

Tip theo, ng quan tm n nhng thng tin gy nhiu. Nhiu thng tin gy nhiu vo giai on cui c kh nng s lm cho th sinh mt tp trung.

Vi nhng n o v chn trng, ch i t l chi chm trng d thi, hay ch i vo nhng tin vt kiu l trn mng s lm bn mt tp trung khi n tp. Do cn loi b tt c nhng thng tin ny vi nim tin mnh s trng tuyn.

Cui cng, hy iu ha cm xc ca mnh, gim sc p - trt. Sc

p ny s gy au u, cng thng, nh hng trc tip ti tm l v sc khe ca bn trc khi thi. Do , bn cn xc nh mc tiu hc ht mnh thi ht sc. Bn cnh , bn ng bao gi ngh n sc p ca gia nh v ngi thn. Bi nu khng may, bn c thi trt, h cng khng bao gi trch bn v

bn n lc ht mnh. Vi suy ngh nh vy, bn s c mt tm l thoi mi trc khi bc vo k thi chnh thc.

III. Nhng li khuyn chn thnh ca nhm tc gi

Mi con ngi u c mt cu chuyn nh ca ring mnh. Nhng hm nay, chng ti khng ni v chng ti. Bi ngy hm nay, c mt cu chuyn chung ca cc bn hc sinh, mt cu chuyn m chng ta khng ngng quan tm. l cu chuyn v c m, v mc tiu thc s bn mun c c trong cuc i ny. Hy c gng la chn tht chnh xc trng i hc hay bt c con ng no bn mun i. Bi nu chn sai, s l s lng ph rt ln cho

bn thn, cho gia nh v cho c x hi. Chng ti li nh n nhng phn tch ca chuyn gia tuyn sinh Trn nh L:

L ra, v nguyn tc vic chn ngnh, ngh ca th sinh phi t rt sm, ngha l n mang tnh nh hng, ch khng phi ngay trc k thi tuyn sinh H-C. Th so snh v mc kinh ph cho vic nh hng ngh nghip thi v hc cng ging nh 3 mc trong ngnh y: phng bnh, iu tr v cp cu. Nu lo 1 ng

chi ph d phng s gim c 10 ng iu tr trc khi trnh c 100 ng cp cu. Nu vic chn la xut pht t s thch, nguyn vng ca bn thn th s bn vng hn. Vic chn ngnh khng ph hp nh hng rt ln (theo hng tiu cc) i vi vic hc hnh v cng vic ca cc th sinh sau ny. Hy nh, i hc khng phi l con ng duy nht vo i. Nu xc nh nh vy, cc bn s lun thoi mi v t tin vo chnh bn thn mnh. Vi cc th sinh np nhiu h s, cc bn cn lng sc mnh chn trng cho ph hp.

V sau khi cc bn a ra c la chn ti u, hy sn sng chin u v n. Hy t nh rng y l th thch cn phi vt qua trn con ng chinh

phc c m ca mnh. Bn cm thy stress v p lc qu ln? Bn cm thy mt mi v n luyn qu nhiu? Hy tp qun ht iu i, v thnh cng ch n khi ta vt qua thi khc kh khn nht. iu c ngha l khi mi th tr nn ti t, thng l du hiu cho thy gi sp i chiu. Bn c bit rng giy pht mt tri sp xut hin l lc mn m tr nn en ti nht? Hy nh li nhng b phim tng xem v bn cng c th nhn thy hin tng ny. Ch ngay trc ci kt c hu, ngi anh hng thng phi tri qua mt giai on cc k kh khn v v vng. l khi mi th t nhin xoay chiu v b phim kt thc. cng l cuc sng. Hy bm vo nhng mc tiu bn d nh, ng bao gi qun n, ng bao gi lm tt i ngn la am m ca mnh, ng bt c ai thc hin c m ca bn thay bn. Khi vt qua c tt c nhng kh khn, nhng p lc th thnh cng s n tht ngt ngo. C th bn khng phi mt vng dng chi li nhng bn c th tr thnh mt v sao lp lnh. Mt ln na, xin chc nhng con ngi c c m s sm t c thnh cng trn con ng mnh chn.

==================================================

S GD & T NGH AN TRNG THPT CHUYN PHAN BI CHU CHNH THC

THI TH TUYN SINH I HC LN 1- NM 2012 Mn: TON; Khi: A

Thi gian lm bi: 180 pht, khng k thi gian pht PHN CHUNG CHO TT C TH SINH (7,0 im) Cu I (2,0 im) Cho hm s 3 23 3 ( 2) 1y x x m m x mC 1. Kho st s bin thin v v th (C) ca hm s cho khi m = 0. 2. Tm m mC c cc i , cc tiu ng thi gi tr cc i v gi tr cc tiu cng du. Cu II (2,0 im) 1. Gii phng trnh 4 2sin cos 2 3cox x x

2. Gii h phng trnh: 2 2 3 5 7

;3 5 2 3 1

x y x yx y

x y x y

Cu III (1,0 im) Tnh tch phn 24

0

2 cos1 sin 2

I dxx xx

Cu IV (1,0 im) Cho hnh chp SABC c tam gic ABC u cnh a . Hnh chiu ca S xung mt phng (ABC) trng vi trng tm G ca tam gic ABC. Mt phng i qua BC vung gc vi SA. Tnh th tch khi chp S.ABC

bit ct hnh chp theo mt thit din c din tch bng 2 38

a .

Cu V (1,0 im) Cho , ,a b c l nhng s thc tho mn iu kin 2 2 2 3

0

b c

a b ca

.

Tm gi tr ln nht ca biu thc: 5 4M a abc . PHN RING (3,0 im): Th sinh ch c lm mt trong hai phn (phn A hoc B)

A. Theo chng trnh Chun Cu VI.a (2,0 im)

1. Trong mt phng to Oxy, cho tam gic ABC c 90BAC bit 5;0 , 7;0B C , bn knh ng trn ni tip 2 13 6r . Xc nh to tm ng trn ni tip I ca tam gic ABC bit I c tung dng. 2. Trong khng gian vi h ta Oxyz, cho mt phng : 2 6 0P x y z v ng thng

3 3 3:1 2 4

x y zd . Vit phng trnh hnh chiu ca d ln mt phng (P).

Cu VII.a (1,0 im) Tm m un ca s phc z bit 22 31 2

i zi iz z

B. Theo chng trnh Nng cao Cu VI.b (2,0 im)

1. Trong mt phng ta Oxy, cho tam gic ABC c 1;2B , ng phn gic trong AK c phng trnh: 2 1 0x y v khong cch t C n ng thng AK bng 2 ln khong cch t B n ng thng AK. Tm to cc nh A v C bit C thuc trc tung.

2. Trong khng gian vi h ta Oxyz, cho tam gic ABC c 1;0;1 , 2; 3;1 , 1; 3; 1A B C . Vit phng trnh ng thng d bit d i qua tm ng trn ngoi tip tam gic ABC v vung gc vi mt phng ABC .

Cu VII.b (1,0 im) Gii h phng trnh

2

2 21 log .log (1 ) 1

8 8 2 1 00y

y xy x yx

----------- 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:................................

I. PHN CHUNG CHO TT C CC TH SINH (7,0 im) Cu I (2,0 im) Cho hm s y = x4 2mx2 + m (1) , m l tham s 1. Kho st s bin thin v v th hm s khi m = 1 .

2. Bit A l im thuc th hm s (1) c honh bng 1. Tm m khong cch t im 3

; 14

B

=

n tip tuyn ca th hm s (1) ti A ln nht . Cu II (2,0 im)

1. Gii phng trnh 5

2.cos5 sin( 2 ) sin 2 .cot 3 .2

x x x x

+ = +

2. Gii h phng trnh 2 2

4 2 2

2 3 15 0

2 4 5 0

x y x y

x y x y

+ + =

+ =

Cu III (1,0 im) Tnh 23 9 1

xI dx

x x=

+

Cu IV (1,0 im) Cho hnh chp S.ABCD c y ABCD l hnh thoi cnh a, hnh chiu vung gc ca

nh S trn (ABCD) l trung im H ca AB, ng trung tuyn AM ca ACD c di 3

2

a, gc gia

(SCD) v (ABCD) bng 300. Tnh th tch khi chp S.ABCD v tnh din tch mt cu ngoi tip hnh chp S.ABC. Cu V (1,0 im) Cho , ,x y z l cc s thc dng tho mn x y z v 3x y z+ + = .

Tm gi tr nh nht ca biu thc: 3x z

P yz y

= + +

II. PHN RING (3 im) Th sinh ch chn mt trong hai phn (phn A hoc phn B) A. Theo chng trnh chun Cu VI.a (2,0 im) 1. Trong mt phng ta Oxy cho hnh thoi ABCD bit phng trnh ca mt ng cho l: 3 7 0x y+ = , im B(0;-3). Tm ta cc nh cn li ca hnh thoi bit din tch hnh thoi bng 20.

2. Gii phng trnh: .25log)20.155.10log( +=+ xxx Cu VII.a (1,0 im) Cho khai trin (1 + 2x)10 (x2 + x + 1)2 = a0 + a1x + a2x

2 + + a14x14

. Hy tm gi tr ca a6. B. Theo chng trnh nng cao Cu VI.b (2,0 im) 1. Cho hnh thang vung ABCD vung ti A v D c y ln l CD, ng thng AD c phng trnh 3x y = 0, ng thng BD c phng trnh x-2y=0, gc to bi hai ng thng BC v AB bng 450. Vit phng trnh ng thng BC bit din tch hnh thang bng 24 v im B c honh dng.

2. Gii bt phng trnh 2 2 21 log log ( 2) log (6 )x x x+ + + >

Cu VII.b (1,0 im) Cho *Nn Chng minh rng 0 1 2 3 22 2 2 2 22 3 4 ... (2 1) 0n

n n n n nC C C C n 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...............................

S GD&T H TNH

TRNG THPT C TH

THI TH I HC, CAO NG LN I, NM 2012

Mn: TON; Khi A, B, D

Thi gian lm bi: 180 pht(Khng k thi gian pht )

TRUNG TM LUYN THI I HC THI TH TUYN SINH I HC NM 2011 THPT CHUYN L T TRNG CN TH Mn thi: TON; khi A Thi gian lm bi: 180 pht, khng k pht PHN CHUNG CHO TT C TH SINH (7 im) Cu I (2 im)

Cho hm s 133 xxy (1)

1. Kho st s bin thin v v th (C) ca hm s (1). 2. nh m phng trnh sau c 4 nghim thc phn bit:

mmxx 33 33

Cu II (2 im)

1. Gii phng trnh: 2 2

4

4

(2 sin 2 )(2cos cos )cot 1

2sin

x x xx

x

2. Gii h phng trnh: 2 2

2

5 0( , )

2 5 1 0

x y xy x yx y

xy y y

Cu III (1 im)

Tnh

2cos8

sin 2 cos 2 2

x

dxx x

Cu IV (1 im) Cho hnh chp S.ABC c mt phng (SAC) vung gc vi mt phng (ABC), , 2SA AB a AC a v

090 .ASC ABC Tnh th tch khi chp S.ABC v cosin ca gc gia hai mt phng (SAB), (SBC). Cu V (1 im) Cho ba s thc dng a, b, c tha mn: a.b.c = 1. Tm gi tr ln nht ca biu thc:

ab bc ca

Ta b ab b c bc c a ca

PHN T CHN (3 im) - Th sinh ch c lm mt trong hai phn (phn A hoc phn B) A. Theo chng trnh Chun Cu VI.a (2 im)

1. Trong mt phng ta Oxy , cho hai im (4; 1), ( 3; 2)A B v ng thng : 3 4 42 0x y .

Vit phng trnh ng trn ( )C i qua hai im ,A B v tip xc vi ng thng .

2. Trong khng gian ta Oxyz, cho bn im A(6; 6; 6), B(4; 4; 4), C( 2; 10; 2) v S(2; 2; 6). Chng minh O, A, B, C l bn nh ca mt hnh thoi v hnh chiu vung gc ca S trn mt phng (OABC) trng vi tm I ca OABC. Tnh khong cch gia hai ng thng SO v AC.

Cu VII.a (1 im)

Gii phng trnh: 23 3(2 1) log (4 9) log 14 0x x x x

B. Theo chng trnh Nng cao Cu VI.b (2 im)

1. Trong mt phng ta Oxy , cho hnh thoi ABCD c A(1; 0), B(3; 2) v 0120 .ABC Xc nh ta

hai nh C v .D 2. Trong khng gian ta Oxyz, cho ba im A, B, C ln lt di ng trn cc tia Ox, Oy v Oz sao cho

mt phng (ABC) khng i qua O v lun i qua im M(1; 2; 3). Xc nh ta cc im A, B, C th tch khi t din OABC t gi tr nh nht.

Cu VII.b (1 im)

Gii h phng trnh: 2 2 2

3 3

3 3 27 9( , )

log ( 1) log ( 1) 1

x y x y x y

x yx 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

Cu I. Cho hm s

2 3

2

xy

x

=

1) Kho st v v th ( )C hm s cho 2) Tm trn ( )C nhng im M tip tuyn ca ( )C ti M ct hai tim cn ti ,A B sao

cho AB ngn nht. Cu II.

1. Gii phng trnh: 3(cot cos )

2(1 sin )cot cos

x xx

x x

+ = +

2. Gii h phgn trnh: 2

2

2 1 2 2 2(2 )

4 1 17

x y x x x y

y x x

+ = +

+ =

Cu III. Tm nguyn hm ca hm s tan

4( )

cos 2

xf x

x

=

Cu IV. Cho lng tr ng ' ' 'ABCA B C c y ABC l tam gic cn B , 'AA AC a= = , gc gia ng thng 'BC v mt phng ( )ABC bng 060 . Gi ,P M ln lt l trung im ca

', 'BB CC , N l im thuc ' 'A C sao cho '4

aNC = . Tnh th tch khi t din ' 'AB C B theo a

v chng minh 'PN A M Cu V. Cho cc s thc dng , ,x y z tha mn 1xyz = . Chng minh rng

2 2 21

2 2 2

x y z

x y z+ +

+ + +

Cu VI. 1. Trong mt phng ta Oxy cho tam gic ABC cn A c (2;1)H l trung im ca

BC ,5

2AB BC= v : 2 2 0AC x y + = . Tm ta im A

2. Trong khng gian vi h ta Oxyz cho tam gic ABC cn ti C c din tch bng 6 . Bit (1; 1;2), (3;1;0)A B . Tm ta im C bit C thuc mt phng ( ) : 2 4 8 0P x y z + = Cu VII. Tm m phng trnh sau c nghim thc: 22 1

2

log ( ) log ( 3) 0x mx x+ + =

TRNG THPT CHUYN NGUYN HU

K THI TH I HC LN 1 NM 2012 MN TON

(Thi gian lm bi 180 pht)
AdministratorRectangle

THI TH I HC LN 2 NM HC 2010-2011 MN TON (Khi A - B - D) - Thi gian lm bi: 180 pht

PHN CHUNG (7 im). Dnh cho tt c cc th sinh.

Cu I (2 im). Cho hm s 1x

yx m

(1)

1. Kho st s bin thin v v th ca hm s (1) khi 1m . 2. Tm cc gi tr ca tham s m sao cho ng thng (d): 2y x ct th hm s (1) ti hai im A v B

sao cho 2 2AB . Cu II (2 im)

1. Gii phng trnh lng gic:2

4sin .sin .sin 4 3.cos .cos .cos 23 3 3 3

x x x x x x

.

2. Gii h phng trnh: 2

2

(1 ) ( 2 ) 5

(1 )( 2 2) 2

y x x y x

y x y x

Cu III (1 im). Tnh tch phn sau: 3

2

0

cos cos sin( )

1 cos

x x xI x dx

x

Cu IV (1 im).

Cho hnh thoi ABCD cnh a, gc 060BAD . Gi G l trng tm tam gic ABD, ( )SG ABCD v 6

3

aSG .

Gi M l trung im CD. 1. Tnh th tch khi chp S.ABMD theo a. 2. Tnh khong cch gia cc ng thng AB v SM theo a.

Cu V (1 im). Cho cc s thc dng , ,x y z tha mn 3x y z . Tm gi tr nh nht ca biu thc:

3 3 3

4 4 4

(2 1 8 4 2) (2 1 8 4 2) (2 1 8 4 2)

x y zA

y y x z z y x x z

.

PHN RING (3 im). Th sinh ch lm mt trong hai phn A hoc B. A. Theo chng trnh Chun Cu VI.a (2 im)

1. Trong mt phng ta Oxy, cho hnh ch nht ABCD c phng trnh cnh AB: 2 1 0x y , ng cho

BD: 7 14 0x y v ng cho AC i qua im (2;1)E . Tm ta cc nh ca hnh ch nht.

2. Trong khng gian Oxyz cho cc ng thng 1 21 1 1 4

: , :1 2 1 1 2 3

x y z x y zd d

.

a. Chng minh rng hai ng thng cho nhau v vung gc vi nhau. b. Vit phng trnh ng d ct c hai ng thng 1 2, d d ng thi song song vi ng thng

4 7 3:

1 4 2

x y z

.

Cu VII.a (1 im).

Tm s phc z tha mn ng thi cc iu kin 2 1z i z i v 1

2

z i

z i

l mt s thun o.

B. Theo chng trnh Nng cao. Cu VI.b (2 im)

1. Trong mt phng ta Oxy cho elip 2 2

( ) : 116 9

x yE v ng thng :3 4 12 0d x y . Chng minh rng

ng thng d ct elip (E) ti hai im A, B phn bit. Tm im ( )C E sao cho ABC c din tch bng 6.

2. Trong khng gian Oxyz cho cc ng thng 12 4

:1 1 2

x y zd

v 2

8 6 10:

2 1 1

x y zd

.

a. Chng minh rng 1 2,d d cho nhau. Tnh khong cch gia hai ng thng .

b. Gi AB l ng vung gc chung ca 1d v 2d ( 1 2, A d B d ). Vit phng trnh mt cu ng knh AB.

Cu VII.b (1 im). Gii h phng trnh:

3 3log ( ) log ( )

2 24 4 4

4 2 2

1log (4 4 ) log log ( 3 )

2

xy xy

x y x x y

S GD& T PH TH

Trng thpt hng vng THI TH I HC NM 2011

( c 01 trang) Mn:Ton- Khi A+ B (Thi gian lm bi 180 pht khng k thi gian giao )

I.phn chung cho tt c th sinh:

CuI:(2,0 im) Cho hm s 3 23 2y x x mx (1)

1.Kho st s bin thin v v th ca hm s (1) khi m = 0. 2.Tm m hm s c cc i , cc tiu v cc im cc i, cc tiu cch u ng thng (d) y = x - 1. Cu II:(2,0 im) 1.Gii phng trnh: Cos2x + 3sin2x +5Sinx 3Cosx =3

2.Gii h phng trnh:2 2 2 2 2 2

( )(1 ) 4

( )(1 ) 4

x y xy xy

x y x y x y

Cu III:(1,0im): Tnh tch phn : 5

2

ln( 1 1)

1 1

xI dx

x x

Cu IV:(1,0 im):Cho hnh chp S.ABCD c y ABCD l hnh thang vung ti A v B vi AB =BC =a AD = 2a.Cc mt (SAC) v (SBD) cng vung gc vi mt y(ABCD).Bit gc gia hai mt phng(SAB)

v (ABCD) bng 060 .Tnh th tch khi chp SABCD v khong cch gia hai ng thng CD v SB. Cu V:(1,0 im) Cho x,y, z l cc s thc dng tho mn xy + yz + xz = 3xyz. Hy chng minh rng:

2 2 2

2 2 2 2 2 21

2 2 2

y x z

xy x zx z yz y

.

II. phn ring (3im) Th sinh ch c chn mt trong hai phn ( phn 1 hoc 2) 1.Theo chng trnh chun: Cu VI.a(2 im)

1.Trong h to Oxy ng thng (d): x y +1 =0 v ng trn (C): 2 2 2 4 0x y x y .Tm im

M thuc ng thng (d) m qua M k c hai ng thng tip xc vi ng trn (C) ti A v B sao

cho 060 .AMB 2.Trong khng gian vi h to Oxyz cho A(2;0;1),B(3;1;2),C(2;0;-2) ,D(0;4;2).Lp phng trnh mt phng (P) i qua A , B v cch u C v D.

Cu VII.(1im): Tm h s 4a ca 4x trong khai trin Niutn a thc 2( ) ( 1)nf x x x vi n l s t

nhin tha mn:2 3 1 11

0 1 23 3 3 4 13 ...2 3 1 1

nn

n n n nC C C Cn n

.

2.Theo chng trnh nng cao: Cu VI.b(2,0 im) 1.Trong mt phng vi h to Oxy cho hnh ch nht ABCD bit phng trnh cnh BC:x + 2y - 4 = 0 phng trnh ng cho BD:3x + y 7 = 0,ng cho AC i qua M(-5;2).Hy tm ta cc nh ca hnh ch nht ABCD. 2.Trong khng gian vi h to Oxyz cho A(-3;5;-5), B(5;-3;7) v mt phng (P) c phng trnh: x +y + z - 6 = 0 . a)Lp phng trnh mt phng (P) i qua A,B v vung gc vi (P).

b)Tm im M nm trn mt phng (P) sao cho 2 2MA MB nh nht.

Cu VII.b(1,0 im :Gii bt phng trnh: 31 12 2

1log ( 1) log (1 2)2

x x

.Ht

TRNG THPT CHUYN HA LONG -------------------

THI TH I HC LN I NM HC 2011-2012

MN TON KHI A THI GIAN: 180 PHT

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

Cu I (2,0) im

Cho hm s 4 2 22y x mx m m= + + + c th l ( )mC vi m l tham s. 1. Kho st s bin thin v v th hm s vi 1m = . 2. Tm m ( )mC c 3 im cc tr v 3 im cc tr ny lp thnh mt tam gic c mt gc bng

0120 . Cu II (2,0 im)

1. Gii phng trnh lng gic 3 3

2

sin cos 1sin 4

1 (cos sin ) 16

x xx

x x

+=

+

2. Gii h phng trnh 2

4 2 2 2

3 0

3 5 0

x xy x y

x x y x y

+ + =

+ + =

Cu III (1,0 im) Tnh gii hn

2 3

20

cos ln(1 )lim

x

x

e x xL

x + +

=

Cu IV (1,0 im) Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a , tam gic SAB u v 090SAD = . J l trung im SD. Tnh theo a th tch t din ACDJ v khong cch t D n mt phng (ACJ).

Cu V (1,0 im) Cho cc s thc dng a, b, c tha mn 2 2 2 3ab bc ca+ + = . Chng minh rng 4 4 43 3 37 7 7 2( )a b c a b c+ + + + + + +

PHN RING (3,0 im) Th sinh ch c lm mt trong hai phn (phn A hoc B) A. Theo chng trnh Chun

Cu VI.a (2,0 im) 1. Trong mt phng Oxy cho im (1;1)A . Tm ta im B thuc ng thng 3y = v im C thuc trc honh sao cho tam gic ABC l tam gic u. 2. Trong mt phng Oxy cho (1;2)A v (3;1)B . Vit phng trnh ng trn qua A, B v c tm nm trn ng thng 7 3 1 0x y+ + = . Cu VII.a (1,0 im) Cho s t nhin 2n , chng minh h thc

1 2 2 2 3 2 22

1( ) 2( ) 3( ) ... ( )

2n n

n n n n nC C C n C nC+ + + + =


1. Trong mt phng Oxy cho (1;0)A , ( 2;4)B , ( 1;4)C , (3;5)D , tm ta im M trn ng thng 3 5 0x y = sao cho hai tam gic MAB v MCD c din tch bng nhau. 2. Trong mt phng Oxy, vit phng trnh ng trn c tm nm trn ng thng 4 3 2 0x y+ = v tip xc vi c hai ng thng 4 0x y+ + = v 7 4 0x y + = . Cu VII.b (1,0 im) Gii bt phng trnh 2 3 3 2log .log 2 log .log 3 0x x x x+ .

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

S GIO DC V O TO H NI THI TH I HC NM 2011 TRNG THPT NG QUAN Mn:TON; Khi :A Thi gian lm bi: 180 pht, khng k thi gian pht

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

Cu I (2,0im) Cho hm s 2 3

2

xy

x

th (C)

1. Kho st s bin thin v v th ( C). 2. Vit phng trnh tip tuyn ti im M thuc (C) bit tip tuyn ct tim cn ng v tim cn ngang ln

lt ti A, B sao cho csin gc ABI bng 4

17,vi I l giao 2 tim cn ca(C).

Cu II (2,0 im)

1. Gii phng trnh 3. 6. 2

22 1

cosx sinx sin x

cos x

.

2. Gii h phng trnh 22 3 ( 2011)(5 )

( 2) 3 3

x y y y

y y x x

( , )x y R

Cu III (1,0 im) Tnh tch phn I=4 3

2

1

(5 ) . 5ln x x xdx

x

.

Cu IV (1,0 im) Lng tr tam gic ABC.ABC c y l tam gic u cnh a, hnh chiu vung gc ca A ln mt phng (ABC) trng vi tm tam gic ABC. Tnh th tch lng tr ABC.ABC v khong cch gia cnh

AA v cnh BC theo a, bit gc gia mt phng (ABC) v (ABC) bng 060 .

Cu V (1,0 im) Cho x v y l cc s thc tha mn: 21 ( )y x x y .

Tm gi tr ln nht v nh nht ca biu thc: 6 6

3 3

1x yP

x y xy

II. PHN RING (3,0 im) Th sinh ch c lm mt trong hai phn (phn A hoc phn B) A. Chng trnh chun Cu VI.a (2,0 im) 1. Trong mt phng ta Oxy cho hnh thoi ABCD bit phng trnh ca mt ng cho l: 3 7 0x y , im B(0;-3), din tch hnh thoi bng 20(vdt). Tm ta cc nh cn li ca hnh thoi.

2. Trong khng gian ta Oxyz, cho mt cu (S) c phng trnh: 2 2 2 2 4 4 16 0x y z x y z , mt

phng (Q) c phng trnh: 2 2 3 0x y z . Vit phng trnh mt phng (P) song song mp(Q) sao cho

mp(P) giao vi mt cu (S) to thnh ng trn c din tch 16 (vdt).

Cu VII.a (1,0 im) Gii bt phng trnh: 9 13

2 log 9 9 log 28 2.3x xx .

B.Chng trnh nng cao Cu VI.b ( 2,0 im)

1. Trong mt phng ta Oxy, cho ng trn (C) 2 2 4 96 0x y x . Tm im M thuc d: 2 4 0x y sao cho t M k c 2 tip tuyn ti (C), vi A,B l tip im m tam gic MAB u.

2. Trong khng gian ta Oxyz, (0;2;0)A (0;0; 1)B v C thuc Ox . Vit phng trnh mt phng (ABC)

bit khong cch t C ti mt phng (P): 2 2 0x y z bng khong cch t C ti ng

thng :1 2

1 2 2

x y z .

Cu VII.b (1,0im) Cho hm s 2 2 9

2

x xy

x

( H ) v ng thng ( ) 2y x m

Tm m sao cho (H) ct ( ) ti A,B phn bit tha mn 4

(2; )3

I l trng tm tam gic OAB, vi O l gc ta .

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

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

Cu I (2,0 im) Cho hm s 4 24 1 2 1y x m x m c th mC

1. Kho st s bin thin v v th C ca hm s khi 3

2m .

2. Xc nh tham s m hm s c 3 cc tr to thnh 3 nh ca mt tam gic u

Cu II (2,0 im)

1. Gii phng trnh 2

4

2

1 tan8 os ( ) sin 4 2.

4 1 tan

xc x x

x

2. Gii h phng trnh sau trn R: 3

2 4 3

1 1 2

9 (9 )

x y

x y y x y y

Cu III (1,0 im) Tnh tch phn

1 22

20

( )4

x xI x e dxx

Cu IV (1,0 im). Cho hnh chp .S ABCD c y l hnh thang vung ti A v B vi BC l y nh. Bit rng tam gic SAB l tam gic u c cnh vi di bng 2a v nm trong mt

phng vung gc vi mt y, 5SC a v khong cch t D ti mt phng SHC bng 2 2a ( y H l trung im AB ). Hy tnh th tch khi chp theo .a Cu V(1,0 im) Cho a, b, c l cc s thc dng thay i tha mn: 3a b c .

Chng minh rng: 2 2 22 2 2

4ab bc ca

a b ca b b c c a

PHN RING (3 im) Th sinh ch chn mt trong hai phn (phn A hoc phn B)

A. Theo chng trnh chun

Cu VI.a (2,0 im).

1. Cho ng trn (C) ni tip hnh vung ABCD c phng trnh 2 2( 2) ( 3) 10x y . Xc

nh to cc nh A, C ca hnh vung, bit cnh AB i qua M(-3; -2) v xA > 0.

2. Trong khng gian vi h ta Oxyz, hy xc nh to tm v bn knh ng trn ngoi tip tam gic ABC, bit A(-1; 0; 1), B(1; 2; -1), C(-1; 2; 3).

Cu VII.a (1,0 im) Tm phn thc v phn o ca s phc sau: z = 1 + (1 + i) + (1 + i)2 + (1 + i)3 + + (1 + i)20

B. Theo chng trnh nng cao Cu VI.b (2,0 im) 1. Trong mt phng vi h trc to Oxy cho hnh ch nht ABCD c din tch bng 12, tm I l giao im ca ng thng 03:1 yxd v 06:2 yxd . Trung im M ca cnh AD l giao

im ca d1 vi trc Ox. Tm to cc nh ca hnh ch nht 2. Trong khng gian vi h ta Oxyz, Cho ba im A(0;1;2), B(2;-2;1), C(-2;0;1). Vit phng trnh mt phng (ABC) v tm im M thuc mt phng 2x + 2y + z 3 = 0 sao cho MA = MB = MC Cu VII.b (1,0 im

Xc nh tp hp cc im trong mt phng phc biu din cc s phc z tha mn h thc

2 1 2z z z

S GD&T H TNH

TRNG THPT C TH

THI TH I HC LN II

MN: TON

Thi gian lm bi: 180 pht

THI TH I HC NM 2012 -TRNG THPT NG ANThi gian lm bi: 180 pht

Cu I (2,0 im) Cho hm s y = x4 2mx2 + 1(Cm).

1. Kho st s bin thin v v th (C) ca hm s vi m = 1

2. nh m hm s c 3 cc tr v ng trn i qua 3 im cc tr c bn knh bng 1

Cu II (3,0 im)

1. Gii phng trnh 23 sin2 x. cosx

3 sinx 2 sinx. cos2 x+ cosx 2 sin 2x+ 2 = 0

2. Gii h phng trnh

{2x3 + x2y + xy2 +

(x+ y2

)2= y3 3y

2

42 + x+

2y 1 = 5

3. Gii phng trnh4x + (x 12).2x + 11 x = 0

Cu III (1,0 im) Tnh tch phn I =

2

0

sinx(2x 1 + cos5 x)dx.

Cu IV (1,0 im) Cho lng tr tam gic ABC.ABC c y l tam gic vung ti B, BAC = 30, cnhAC bng 2a. Cnh bn AA to vi y mt gc 60 v chn ng vung gc h t A xung mt phng(ABC) trng vi tm ng trn ngoi tip y. Tnh theo a th tch khi lng tr ABC.ABC v khongcch gia hai ng thng BC v AA v SC theo a.

Cu V (1,0 im) Cho ba s dng x, y, z Tnh gi tr nh nht ca biu thc

P =x

2(y + z)+

2(y + z)

x+

y

2(z + x)+

2(z + x)

y+

z

2(x+ y)+

2(x+ y)

z

Cu VI.a (2,0 im)

1. Trong h ta Oxy cho im A(1; 1) v ng thng d c phng trnh 4x+3y 12 = 0. Gi B, C ln ltl giao im ca d vi cc trc Ox,Oy. Tm ta trc tm H ca tam gic ABC.

2. Trong khng gian vi h ta Oxyz cho mt cu (S) c phng trnh: x2 + y2 + z2 + 8x 2y+ 4z 4 = 0,mt phng (P) ct mt cu (S) theo mt ng trn c bn knh bng

11 v tm H thuc trc Ox. Vit

phng trnh mt phng (P).

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

UserA1K37

S GD &T BC GIANG THI TH I HC LN 2 NM HC 2011-2012 TRNG THPT PHNG SN MN TON- KHI A, B, D

Thi gian lm bi: 180 pht (khng k thi gian pht ) A.PHN CHUNG CHO TT C CC TH SINH (7 im)

Cu I. (2 im) ) Cho hm s 4 21 2 14

y x mx m

1. Kho st s bin thin v v th hm s 1 khi 1m . 2. Tm cc gi tr ca tham s m th hm s 1 c ba im cc tr ; ng thi ba im cc tr to thnh mt tam gic c din tch bng 32 2 . Cu II. (2 im) 1. Gii phng trnh: 2cos 3(2sin 1) t anx

s inx 1 cosxx

x

2. Gii h phng trnh: 2

2 1 1 2 2 1 8

2 1 2 13

x y x

y y x x

Cu III. (1 im) Tnh nguyn hm 28 os sin 2 3

s inx cosc x xI dx

x

Cu IV. (1 im) Cho hnh chp S.ABCD c y ABCD l hnh ch nht vi AB = 2a, BC = a. Cc cnh bn ca hnh chp bng nhau v bng 2a . a) Tnh th tch khi chp S.ABCD theo a. b) Gi M, N, E, F ln lt l trung im ca cc cnh AB, CD, SC, SD. Chng minh ng thng SN vung gc vi mt phng (MEF).

Cu V. (1 im) Cho , ,x y z l cc s thc dng tho mn: 2 1xy xz

Tm gi tr nh nht ca biu thc: 3 4 5yz zx xyPx y z

B.PHN RING (3 im) Th sinh ch c chn mt trong hai phn (phn a, hoc phn b). a. Theo chng trnh chun. Cu VIa. (2 im)

1. Trong mt phng vi h trc ta Oxy cho hnh thoi ABCD vi A(1;0) ng cho BD c phng trnh : x y +1 = 0 Tm to cc nh B, C, D Bit 4 2BD . 2. Cho lng tr ng tam gic ABC.ABC c y l tam gic u. Mt phng (ABC) to vi y gc 300 v din tch tam gic ABC bng 18. Tnh th tch khi lng tr ABC.ABC.

Cu VII. (1 im) . Gii phng trnh: 84 221 1log 3 log 1 log 4 .2 4

x x x

b. Theo chng trnh nng cao. Cu VIb. (2 im)

1. Trong mt phng vi h trc ta Oxy cho tam gic ABC vi 1; 2B ng cao : 3 0AH x y . Tm ta cc nh A, C ca tam gic ABC bit C thuc ng thng :2 1 0d x y v din tch tam gic ABC bng 1. 2. Cho lng tr tam gic ABC.ABC c AB=a; AC=2a; 0AA ' 2 5; 120a BAC ; I l trung

im ca CC. Chng minh rng 'IB IA v tnh khong cch t im A n mt phng (IAB).

Cu VIIb. (1 im) Gii h phng trnh: 2

2 3 1

log 3 7 6

2.8 2 17.2x y y xy x

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

UserA1K37

S GD&T THANH HO TRNG THPT TRN PH

THI TH I HC LN I NM 2011 Mn thi: TON, kh i A+B

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

I.PHN CHUNG CHO TT C TH SINH (7,0 im) Cu I (2,0 im) Cho hm s

1

12

+

+=

x

xy

1. Kho st s bin thin v v th )(H ca hm s cho. 2. Vit phng trnh tip tuyn ca th )(H bit tip tuyn cch u hai im )4;2(A v

)2;4( B .

Cu II (2,0 im) 1. Gii phng trnh: 12cos.2sin33)cos(sin4 66 =+ xxxx 2. Gii phng trnh: )(424202 Rxxxxx +=+++

Cu III (1,0 im) Tnh tch phn: dxxx

xxxI

+

+=

2

02

23

1

32

Cu IV (1,0 im) Cho hnh lp phng ABCD.ABCD cnh a, gi M,N ln lt l trung im cc cnh AB, BC . Tnh theo a th tch khi t din ADMN v khong cch t A n ng thng DN. Cu V (1,0 im) Cho 0,, >cba v 3=++ cba . Tm gi tr nh nht ca biu thc:

ba

cca

ac

bbc

cb

aabP

+

++

+

++

+

+=

333

II.PHN RING (3,0 im) Th sinh ch c lm mt trong hai phn (phn A hoc phn B) A. Theo chng trnh chun Cu VI.a (2,0 im)

1. Trong mt phng to Oxy cho ABC c phng trnh cc cnh ACAB, ln lt l 032 = yx , 0=+ yx v trng tm )1;2( G . Lp phng trnh cnh BC .

2. Trong khng gian vi h to Oxyz cho hai im )9;8;1(A v )3;4;3( B . Tm to im C trn mt phng Oxy sao cho tam gic CAB cn ti C v c din tch bng 4182 .

Cu VII.a (1,0 im) Gii phng trnh )(033

log)2(log23

23 Rx

xx

xx =

++

B. Theo chng trnh nng cao Cu VI.b (2,0 im)

1. Trong mt phng to Oxy, lp phng trnh ng thng i qua )3;2(M v ct ng trn 022222 =+ yxyx ti hai im BA, sao cho 32=AB .

2. Trong khng gian vi h to Oxyz cho hai im )3;4;2(A v )15;2;4(B . Tm to im M trn mt phng Oxz sao cho tam gic MAB c chu vi nh nht.

Cu VII.b (1,0 im) Gii h phng trnh

=++=++

4)1(log3)2(log2

0222

22

2

yyx

xyxyy

----------Ht ---------- Th sinh khng s dng ti liu. Cn b coi thi khng gii thch g thm. H v tn th sinh.; S bo danh www.laisac.page.tl

SGD&TVNHPHCTHPTCHUYNVNHPHUC

THITHIHC,CAONGNM2012Mnthi:Ton,khiAln3

Thigianlmbi:180pht(khngkthigiangiao)

A.PHNCHUNGCHOTTCTHSINH(8im)Cu I(2im)Chohms: 3 2y x 3x 1 = + cthl ( )C .1)Khostsbinthinvvthhms(C)2)Tmhaiim ,A B thucth ( )C saochotiptuynca(C)ti A v B songsongvinhauvdion 4 2AB =CuII(2im)1)Giiphngtrnh: ( ) ( )cos 2 5 2 2 cos sin cosx x x x + =

2)Giihphngtrnh: ( )

( )2 2

2

58 4 13

12 1

x y xyx y

xx y

+ + + = + + = +

( , )x yR .

CuIII(1im)Tnhtng:8 8 8 8 88 9 10 2011 2012

7.8 8.9 9.10 2010.2011 2011.2012C C C C CS = + + + + + L ,trong knC ls

thpchp k ca n phnt.CuIV.(2,0im)Chohnhhpng 1 1 1 1.ABCD A BC D ccccnh 12, 3AB AD AA = = = vgc 060BAD = .Gi ,M Nlnltltrungimcacccnh 1 1AD v 1 1A B .

1. Chngminhrng 1AC vunggcvimtphng ( )BDMN2. Tnhthtchkhichp .A BDMN

CuV.(1im) Cho , ,a b c lccsthckhngm thomn 3a b c + + = .Tmgitrlnnhtcabiuthc: 2 2 2S a b b c c a = + + .

B. PHNRING (2im).Thsinhchclmmttronghaiphn(phn1 hoc2)1.TheochngtrnhChunCuVIa.(1im)Trongmt phng vi h ta Oxy cho hnh vung ABCD c hai cnh l 1: 4 3 3 0x y + = v

2 : 4 3 17 0x y = ,nh ( )2 3A .Lpphngtrnhhaicnhcnlicahnhvung ABCD .CuVIIa.(1 im)Giiphngtrnh: 15 25 1x x x + = 2.TheochngtrnhNngcaoCuVIb.(1im)TrongmtphngvihtaOxy,lpphngtrnhchnhtccaelip ( )E bitrngcmtnhvhai tiu im ca ( )E to thnh mt tam gic u v chu vi hnh ch nht c s ca ( )E l

( )12 2 3 +CuVIIb.(1im)Giiphngtrnh: ( ) ( )22 4 1

2

log 2 log 5 log 8 0x x + + + =

HT

Ghich: Thsinhkhngcsdngbt ctiliug!Cnbcoithikhnggiithchgthm!

Cm n [email protected]
mailto:[email protected]://www.laisac.page.tl/

S GD&T NGH AN

TRNG THPT NG THC HA

THI TH I HC LN 1 - NM 2011 Mn thi: TON; Khi: A v B

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

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

Cu I (2,0 im) Cho hm s 3 2 2 33 3( 1) 4 1y x mx m x m m= + + (1) , m l tham s thc. 1. Kho st s bin thin v v th ca hm s (1) khi 1m = . 2. Tm cc gi tr ca m th hm s (1) c hai im cc tr ,A B sao cho tam gic OAB vung ti ,O trong

O l gc ca h trc to . Cu II (2,0 im)

1. Gii phng trnh 4 45

cos sin sin 2 .cos 2 tan tan4 4 4 4

x x x x x x

+ + = + .

2. Gii h phng trnh 2

4

16 2 3

x y x y x y

x y x

+ = = +

( , )x y .

Cu III (1,0 im) Tnh tch phn ( )4

1

ln 9 xI dx

x

= .

Cu IV (1,0 im) Cho hnh lng tr . ' ' 'ABC A B C c y ABC l tam gic cn ti A, 02 , 120 .AB a BAC= = Hnh chiu ca nh 'A ln y trng vi tm ng trn ngoi tip tam gic .ABC Bit tam gic 'A BC vung ti 'A . Tnh th tch khi lng tr . ' ' 'ABC A B C theo a .

Cu V (1,0 im) Cho cc s thc khng m , ,a b c tho mn iu kin 1.3

ab bc ca+ + = Chng minh rng

2 2 2

1 1 13

1 1 1a bc b ca c ab+ +

+ + +.

PHN RING (3,0 im): Th sinh ch c lm mt trong hai phn (phn A hoc B) A. Theo chng trnh Chun

Cu VI.a (2,0 im) 1. Trong mt phng vi h to ,Oxy cho tam gic ABC c : 2 3 0d x y = l ng phn gic trong gc

.A Bit 1 1( 6;0), ( 4;4)B C ln lt l hnh chiu vung gc ca nh ,B C trn cc ng thng ,AC AB .

Xc nh to cc nh , ,A B C ca tam gic .ABC

2. Trong khng gian vi h to ,Oxyz cho hai ng thng 1

1 2: ,1 1 1

x y z = =

2

3:1 1 3

x y z + = = v

mt phng ( ) : 1 0P x y z + = . Vit phng trnh ng thng song song vi mt phng ( ),P ct

ng thng 1, ng thi ng thng ct v vung gc vi ng thng

2.

Cu VII.a (1,0 im) Tm s phc z tho mn | | 2

| 2 . | 2

z

z i z

= + =

.


1. Trong mt phng vi h to ,Oxy cho hnh ch nht ABCD c cc nh ,AC ln lt thuc cc ng

thng 1: 0d x y+ = v

2: 2 7 0.d x y + = Bit giao im ca hai ng cho l (1; 4)I v ng thng

AB i qua im (0; 4)M , xc nh to cc nh , , ,A B C D ca hnh ch nht .ABCD

2. Trong khng gian vi h to ,Oxyz cho hai ng thng 1

1: ,1 1 1

x y z = =

2

2:1 1 2

x y z = =

v

im ( 1;0;1)A . Xc nh to im M trn ng thng 1 v im N trn ng thng

2 sao cho

6MN = v . 3.AM AN =

Cu VII.b (1,0 im) Gii phng trnh ( ) ( )2 2 23( 9 )log 9 log 9 2 ( ).x x x x x x x+ + + + + + = ---------------Ht---------------

Ch : p n c cp nht ti a ch http://dangthuchua.com

S GD V T NGH AN THI TH I HC, CAO NG LN I NM 2012 TRNG THPT LNG 4 Mn: Ton khi A, B Thi gian:180 pht khng k thi gian giao .

I. PHN CHUNG CHO TT C CC TH SINH

Cu I. (2im) Cho hm s ( )3 23 1 12 3 4y x m x mx m= + + + (C) 1. Kho st v v th hm s khi m = 0.

2. Tm m hm s c hai cc tr l A v B sao cho hai im ny cng vi im 9

1;2

C

lp thnh

tam gic nhn gc ta O lm trng tm. Cu II . (2im)

1. Gii phng trnh: ( ) 3cos 1 2 3 sin 2 cos3 4cos 22

x x x x + =

.

2. Gii h phng trnh. ( ) ( )

2 23 8 5

8 3 13

x y y x

x x y y

+ + + =

+ + + =

Cu III. (1im) Tnh tch phn: 24

2

3

sin 1 cos

cos

x xI dx

x

=

Cu IV. (1im) Cho hnh chp S.ABCD c y l hnh thang vung ti A v D. Bit AB = 2a, AD =a, DC = a (a > 0) v SA mt phng y (ABCD). Gc to bi gia mt phng (SBC) vi y bng 450. Tnh th tch khi chp S.ABCD v khong cch t B ti mt phng (SCD) theo a.

Cu V. (1im) Cho cc s dng a, b, c tho mn iu kin 2 2 2 4a b c abc+ + + = . Chng minh rng 3a b c+ + .

II. PHN RING. (3im) Th sinh ch c lm mt trong hai phn.

1. Theo chng trnh chun.

Cu VIa. (2im).

1. Trong mt phng Oxy cho hnh vung ABCD c tm 3 1;2 2

I

. Cc ng thng AB, CD ln lt i qua

cc im ( )4; 1M , ( )2; 4N . Tm to cc nh ca hnh vung bit B c honh m.

2. Tm m phng trnh sau c nghim thc: ( )29 2 4 2 2x m x x+ = + +

Cu VIIa . (1im). Trong mt phng to Oxy. gc phn t th nht ta ly 2 im phn bit, c th cc gc phn t th hai, th ba, th t ta ln lt ly 3, 4, 5 im phn bit (cc im khng nm trn cc trc to ). Trong 14 im ta ly 2 im bt k. Tnh xc sut on thng ni hai im ct c hai trc to .

2. Theo chng trnh nng cao.

Cu VIb. (2im).

1. Trong mt phng ta Oxy, cho hnh ch nht ABCD c din tch bng 12, tm I thuc ng thng

( ) : 3 0d x y = v c honh 92I

x = , trung im ca mt cnh l giao im ca (d) v trc Ox. Tm ta

cc nh ca hnh ch nht.

2. Trong khng gian Oxyz cho ng thng 1 1:1 2 1x y z

d += =

v hai im ( ) ( )1;1; 2 , 1;0;2 .A B

a. Vit phng trnh mt phng (P) cha A v B ng thi song song vi ng thng d. b. Qua A vit phng trnh ng thng ( ) vung gc vi d sao cho khong cch t B ti ( ) l nh nht.

Cu VIIb . (1im). Cho hai s phc lin hp nhau 1 2,z z tho mn iu kin 122

z

z l mt s thc v

1 2 2 3z z = . Tm s phc z1. ............................. Ht ............................

c DAI Hec vINH Dt o sAr cnAr tUqrrlc t 6p tz LAN r, NAwI zorrT c THPT CrrurEX UOX: TOAX; Thdi gian I m bii: 180 phrtt

r. prrn c c cHo rAr cA rff suvn e,o a$m')7CiuI.(2,04i6m1 Chohdms6 v= !*'-(2**l)x2+(m+2)t..,'+ cOd6th! (C^), m ldthams6.'3

I . Khio s6t sy bi6n thin vi vE dd thi cua ham s5 d[ cho l

TRNGTHPTCHUYNVNHPHC KTHITHIHCLN4NMHC20112012

Mn:Ton12.KhiA.

Thigianlmbi:180pht(Khngkthigiangiao)

A.PHNCHUNGCHOTTCTHSINH(7,0im)Cu I(2,0im)Chohms: 3 2y x 3x 1 = + cthl ( )C .1)Khostsbinthinvvthhms(C)2)Vigitrnoca m thngthngiquahaiimcctrcathhms(C)tipxcvingtrn ( ) ( ) ( )2 2: 1 5x m y m + =CuII(2,0im)1)Giiphngtrnh: ( )( )1 tan 1 sin 2 1 tanx x x + = +

2)Giihphngtrnh: ( )

3

4

2 1 27

2 1

x y x

x y

=

+ = ( , )x yR .

CuIII(1,0im)Tnh tchphn: ( )1

4 2

13

ln 3 2lnI x x x dx = +

CuIV. (1,0im)Cho lngtrtamgicu 1 1 1.ABC ABC cchncnhubng 5 .Tnhgcvkhongcchgiahaingthng 1AB v 1BC .CuV.(1,0im) Cho , ,a b c lccsthcdngthomn 7ab bc ca abc + + = .

Tmgitr nhnhtcabiuthc:4 5 6

2 2 2

8 1 108 1 16 1a b cSa b c

+ + + = + + .

B.PHNRING (3,0im).Thsinhchclmmttronghaiphn(phn1hoc2)1.Theochngtrnh ChunCuVIa.(2,0im)1)TrongmtphngvihtaOxychongtrn ( ) ( )2 2: 4 4C x y + = vim ( )41E .Tm to im M trn trc tung sao cho t im M k c hai tip tuyn

,MA MB nngtrn ( )C vi ,A B lcctipimsaochongthng AB iqua .E2)Trongkhnggianvihto Oxyz chomtphng ( ) : 2 2 1 0P x y z + = vccngthng

11 3:

2 3 2x y zd = =

v 2

5 5:6 4 5x y zd + = =

.Tmccim 1 2,M d N d saocho MN songsong

vi ( )P vcch ( )P mtkhongbng 2.CuVIIa.(1,0im)Giiphngtrnh: ( ) ( ) 33 5 12 3 5 2x x x+ + + =2.TheochngtrnhNngcaoCuVIb.(2,0im)1)TrongmtphngvihtaOxy,chongthng ( ) : 3 4 0d x y = vngtrn ( ) 2 2: 4 0.C x y y + = Tmim ( )M d vim ( )N C saochochngixngnhauquaim ( )31A .

2) Trong khng gian vi h to Oxyz , cho ng thng 2 4:3 2 2x y z

= =

v hai im

( )12 1 ,A ( )7 23B .Tm trn nhngim Msao cho khong cch t M nng thngcha AB lnhnht.

CuVIIb.(1,0im)Giiphngtrnh: ( ) ( ) ( )2 22 1log 1 log 1 log 22

x x x = + +

chnhthc(thigm01trang)

TTrrnngg

tthhpptt ccuu xxee nnmm 22001111

chnh thc

tthhii tthh ii hhcc MMnn tthhii:: TTOONN;; KKhhii AA

PPhhnn cchhuunngg cchhoo tttt cc cccc tthh ssiinnhh ((77,,00 iimm)) CCuu II ((22,,00 iimm)) CChhoo hhmm ss 3 3 2y x x= + 11.. KKhhoo sstt ss bbiinn tthhiinn vv vv tthh ((CC)) ccaa hhmm ss.. 22.. VViitt pphhnngg ttrrnnhh nngg tthhnngg cctt tthh ((CC)) ttii 33 iimm pphhnn bbiitt , ,A B C ssaaoo cchhoo iimm A

bbnngg 2 vv 2 2BC = .. CCuu IIII ((22,,00 iimm))

11.. GGiiii pphhnngg ttrrnnhh

22.. GGiiii hh pphhnngg ttrrnnhh

CCuu IIIIII ((11,,00 iimm))

CCuu IIVV ((11,,00 iimm)) CChhoo hhnnhh cchhpp .S ABCD cc yy ABCD ll hhnnhh tthhaanngg vvuunngg ttii A vv B vvii BC yy nnhh,, H ll ttrruunngg iimm ccaa AB .. BBiitt rrnngg ttaamm ggiicc SAB ll ttaamm ggiicc uu cc ccnnhh vvii ddii bbnngg vv nnmm ttrroonngg mmtt pphhnngg vvuunngg ggcc vvii yy,, 5SC a= vv kkhhoonngg cccchh tt D ttii mmtt pphhnngg ( )SHC bbnngg a .. TTnnhh tthh ttcchh ccaa kkhhii cchhpp .S ABCD

TThh ssiinnhh cchh cc llmm mmtt ttrroonngg hhaaii pphhnn ((pphhnn AA hhoocc BB)) AA.. TThheeoo cchhnngg ttrrnnhh CChhuunn CCuu VVII..aa

22.. TTrroonngg kkhhnngg ggiiaann ttoo Oxyz ,, cchhoo mmtt pphhnngg ( ) cc pphhnngg ttrrnnhh:: 2 3 0x y z + = vv hhaaii iimm (0; 2;1)A ,, (1;0;3)B .. GGii 'A ll iimm ii xxnngg vvii A qquuaa mmtt pphhnngg ( ) ,,

CCuu VVIIII..aa ((11,,00 iimm)) TTmm ss pphhcc lliinn hhpp ccaa ss pphhcc z bbiitt ( 1) 1z i + = vv 2z i ll mmtt ss tthhcc.. BB.. TThheeoo cchhnngg ttrrnnhh NNnngg ccaaoo

TThhii ggiiaann llmm bbii:: 118800 pphhtt,, kkhhnngg kk tthhii ggiiaann pphhtt

CCuu VVII..bb ((2,,00 iimm)) 11.. TTrroonngg mmtt pphhnngg ttoo Oxy ,, cchhoo EElliipp ( )E cc pphhnngg ttrrnnhh::

2 2

19 4x y+ = vv hhaaii iimm (3; 2),A ( 3;2)B ..

TTmm

ttoo

iimm C cc hhoonnhh vv ttuunngg ddnngg tthhuucc EElliipp ( )E ssaaoo cchhoo ttaamm ggiicc ABC cc ddiinn ttcchh

llnn nnhhtt..

22.. TTrroonngg kkhhnngg ggiiaann ttoo Oxyz ,, cchhoo iimm (10;2; 1)A vv nngg tthhnngg cc pphhnngg ttrrnnhh ::

CCuu VVIIII..bb ((11,,00 iimm)) GGiiii pphhnngg ttrrnnhh 2 2log (2 4) 3 log (2 12)x xx+ = + + (( x )) ------------------------------------------------HHtt--------------------------------------------------

TThh ssiinnhh kkhhnngg cc ss ddnngg ttii lliiuu.. CCnn bb ccooii tthhii kkhhnngg ggiiii tthhcchh gg tthhmm.. HH vv ttnn tthh ssiinnhh::..............................................................................................................................;; ss bboo ddaannhh........................................................................

3

112

1 ==

zyx.. LLpp pphhnngg ttrrnnhh mmtt pphhnngg ((PP)) ii qquuaa AA,, ssoonngg ssoonngg vvii dd vv kkhhoonngg cccchh tt dd ttii

mmtt pphhnngg ((PP)) ll llnn nnhhtt..

cc

hhoonnhh

4 4sin cos1 tan tan sin2 cosx x xx x

x++ = +

2a l

CCuu VV ((11,,00 iimm))

TTmm ggii ttrr nnhh nnhhtt ccaa bbiiuu tthhcc::

11..

Tnh tch phn I = 2

1

ln +

x

x e exe x dxx

2 3

22

1 + 2 1 = 1

2. 2 2

+

+

=

y x

yy x xx

( ,x y )

2 2

((2,,00 iimm))

d

theo .a Cho x, y , z, l cc s thc dng tho mn iu kin : x y z 3+ + =

2 2 22 2 2

xy yz zxP x y z

x y y z z x

+ += + + +

+ +

hy tnh di on thng AC .

Bit rng im C thuc ng thng 'A B v ng thng AC song song vi mt phng ( ) .

Phn t chn (3,0 im)

TTrroonngg mmtt pphhnngg ttoo Oxy ,, cchhoo ttaamm ggiicc ABC cc ddiinn ttcchh bbnngg 22 vv nngg tthhnngg AB cc pphhnngg ttrrnnhh x y 0 = .. BBiitt rrnngg iimm I(2;1) ll ttrruunngg iimm ccaa oonn tthhnngg BC ,, hhyy ttmm ttoo ttrruunngg iimm K

ccaa oonn tthhnngg AC.

S GD &T BC GIANG THI TH I HC LN 2 NM HC 2011-2012 TRNG THPT PHNG SN MN TON- KHI A, B, D

Thi gian lm bi: 180 pht (khng k thi gian pht ) A.PHN CHUNG CHO TT C CC TH SINH (7 im)

Cu I. (2 im) ) Cho hm s ( )4 21 2 14

y x mx m= +

1. Kho st s bin thin v v th hm s ( )1 khi 1m = . 2. Tm cc gi tr ca tham s m th hm s ( )1 c ba im cc tr ; ng thi ba im cc tr to thnh mt tam gic c din tch bng 32 2 . Cu II. (2 im) 1. Gii phng trnh: 2cos 3(2sin 1) t anx

s inx 1 cosxx

x = +

2. Gii h phng trnh: ( )2

2 1 1 2 2 1 8

2 1 2 13

x y x

y y x x

+ = + + =

Cu III. (1 im) Tnh nguyn hm 28 os sin 2 3

s inx cosc x xI dx

x

=

Cu IV. (1 im) Cho hnh chp S.ABCD c y ABCD l hnh ch nht vi AB = 2a, BC = a. Cc cnh bn ca hnh chp bng nhau v bng 2a . a) Tnh th tch khi chp S.ABCD theo a. b) Gi M, N, E, F ln lt l trung im ca cc cnh AB, CD, SC, SD. Chng minh ng thng SN vung gc vi mt phng (MEF).

Cu V. (1 im) Cho , ,x y z l cc s thc dng tho mn: 2 1xy xz+ =

Tm gi tr nh nht ca biu thc: 3 4 5yz zx xyPx y z

= + +

B.PHN RING (3 im) Th sinh ch c chn mt trong hai phn (phn a, hoc phn b). a. Theo chng trnh chun. Cu VIa. (2 im)

1. Trong mt phng vi h trc ta Oxy cho hnh thoi ABCD vi A(1;0) ng cho BD c phng trnh : x y +1 = 0 Tm to cc nh B, C, D Bit 4 2BD = . 2. Cho lng tr ng tam gic ABC.ABC c y l tam gic u. Mt phng (ABC) to vi y gc 300 v din tch tam gic ABC bng 18. Tnh th tch khi lng tr ABC.ABC.

Cu VII. (1 im) . Gii phng trnh: ( ) ( ) ( )84 221 1log 3 log 1 log 4 .2 4

x x x+ + =

b. Theo chng trnh nng cao. Cu VIb. (2 im)

1. Trong mt phng vi h trc ta Oxy cho tam gic ABC vi ( )1; 2B ng cao : 3 0AH x y + = . Tm ta cc nh A, C ca tam gic ABC bit C thuc ng thng :2 1 0d x y+ = v din tch tam gic ABC bng 1. 2. Cho lng tr tam gic ABC.ABC c AB=a; AC=2a; 0AA ' 2 5; 120a BAC= = ; I l trung

im ca CC. Chng minh rng 'IB IA v tnh khong cch t im A n mt phng (IAB).

Cu VIIb. (1 im) Gii h phng trnh: ( )2

2 3 1

log 3 7 6

2.8 2 17.2x y y xy x

+ +

+ + =

+ =

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

SGD&TNGTHP THITHTUYNSINHIHCNM2012 LN1THPTChuynNguynQuangDiu Mn:TONKhi:A+B

Thigianlmbi:180pht,khngkthigianphtPHNCHUNGCHOTTCTHSINH(7,0 im)CuI(2,0 im)Chohms 3 23 2y x x = + .1.Khostsbinthinvvth(C)cahmscho.2.Tmhai imA,Bthucth(C)saochotiptuynca(C)tiAvBsongsongvinhauvdionthngABbng 4 2.

CuII(2,0 im)

1.Giiphngtrnh ( )2 2

2

sin cos 2sin 2 sin sin 31 cot 2 4 4

x x xx x

x + = +

.

2.Giihphngtrnh ( )

322

72 2 24

x y

y x x

+ = + + =

( ),x y .

CuIII(1,0im)Tnhtchphn ( )3 2

1

1 ln 2 12 ln

e x x xI dx

x x + + +

= + .

CuIV(1,0 im) ChohnhlngtrngABC.ABCc 0, 2 , 120AC a BC a ACB = = = vngthng'A C tovimtphng ( )' 'ABB A gc 030 .Tnhthtchkhilngtrchovkhongcchgiahai

ngthng ' , 'A B CC theoa.

CuV(1,0 im)Chophngtrnh ( )24 6 3 2 2 3x x x m x x + = + + Tmm phngtrnhcnghimthc.

PHNRING(3,0 im): Thsinhchclmmttronghaiphn(phnAhocB)A.TheochngtrnhChunCuVI.a(2.0im)1.Trongmtphng Oxy,chohaingtrn ( ) 2 2: 18 6 65 0C x y x y + + = v ( ) 2 2' : 9C x y + =TimM thucngtrn(C)khaitiptuynvingtrn(C),giA,Blcctipim.TmtaimM,bit dionABbng 4,8.

2.TrongkhnggianOxyz,chongthng ( ) : 1 21

x td y t

z

= = + =

vim ( )123A .Vitphngtrnh

mtphng(P)changthng(d)saochokhongcchtimA n mtphng(P)bng 3 .

CuVII.a(1.0 im)Giibt phngtrnh ( ) ( )2 22 21log 2 1 log 2 02 x x x .B.TheochngtrnhNngcaoCuVI.b(2.0 im)

1.TrongmtphngOxy,chohnhthoiABCDctm ( )33I v 2AC BD = .im 423

M

thucng

thng AB ,im 1333

N

thucngthngCD .Vitphngtrnh ngcho BD bitnh B c

honhnhhn3.

2.TrongkhnggianOxyz,chohaingthng ( ) ( )1 2x 1 y 2 z x 2 y 1 z 1d : d :1 2 1 2 1 1 + +

= = = = vmt

phng ( )P : x y 2z 5 0 + + = .Lpphngtrnh ngthng (d) songsongvimtphng(P)vct ( ) ( )1 2d , d lnlttiA,BsaochodionABnhnht.

CuVII.b(1.0im)Giiphngtrnh ( ) ( ) ( )233 9 31log 1 log 2 1 log 12

x x x + = + + .

HtThsinhkhngcsdngtiliu.Cnbcoithikhnggiithchgthm.
http://boxmath.vn/http://www.laisac.page.tl/

1

S GIO DC V O TO NGH AN TRNG THPT CHUYN PHAN BI CHU

THI TH I HC LN I NM 2011 MN TON; KHI A, B

Thi gian lm bi : 180 pht; khng k thi gian giao

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

Cu 1 (2,0 im) Cho hm s 3 2y x 3x mx 2 (Cm) 1. Kho st s bin thin v v th hm s (Cm) khi m = 0 2. Tm m hm s (Cm) c cc i v cc tiu, ng thi cc im cc tr ca thi hm s cch u ng thng d: x y 1 = 0 Cu II (2,0 im)

1. Gii phng trnh: sin 3x sin 2x.sin x4 4

2. Gii phng trnh: 24x 8x 2x 3 1 (x )

Cu III (1,0 im) Tnh tch phn 1

ln x. 1 ln xI dxx 1 ln x

e

Cu IV (1,0 im) Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, tam gic SAB u v tam gic SCD vung cn ti S. Gi I, J ln lt l trung im ca AB, CD. Tnh th tch khi chp S.AICJ. Cu V (1,0 im) Cho a, b, c l cc s thc khng m tha mn a + b + c = 1. Tm gi tr ln nht ca biu thc:

2 2 2

2 2 2

1 a 1 b 1 cM1 b 1 c 1 a

II. PHN RING (3,0 im) Th sinh ch c lm mt trong hai phn (phn A hoc phn B) A. Theo chng trnh chun Cu VI.a (2,0 im) 1. Trong mt phng ta Oxy, cho tam gic ABC c trc tm l H(-1;4), tm ng trn ngoi tip l I(3;0) v trung im ca cnh BC l M(0;3). Vit phng trnh ng thng AB, bit B c honh dng. 2. Trong khng gian Oxyz, cho hai im A(1; 2; 2) v B(5; 4; 4) v mt phng (P): 2x + y z + 6 = 0. Tm im M nm trn (P) sao cho MA2 + MB2 nh nht. Cu VII.a (1,0 im) Tm mun ca s phc x bit 4z 1 3i z 25 21i B. Theo chng trnh nng cao Cu VI.b (2,0 im) 1. Trong mt phng ta Oxy, cho ba im A(1;1), B(3;2) v C(7;10). Vit phng trnh ng thng d i qua A sao cho tng khong cch t B n ng thng d v C n ng thng d l ln nht. 2. Trong khng gian ta Oxyz, cho mt phng (P): 2x + y z + 6 = 0 v ng thng d: x 2 y 1 z 1

5 4 2

. Vit phng trnh hnh chiu vung gc ca d ln mt phng (P).

Cu VII.b (1 im) Gii h phng trnh

2

2 2

y 4xy 4x 2y 1x, y

log x.log 1 y 1

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

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

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

SGIODC&OTOHIDNGTRNGTHPTTK

*****

THITHIHCLN2 NGY10 4 NM2011MnTON KhiA,B,D

(Thigianlmbi180pht,khngkpht)

PHNCHUNGCHOTTCTHSINH (7im)Cu1(2im):Chohms 3 2 1y x mx m = + + (1)vi m lthamsthc

1) Khostsbinthinvvthhms(1)khi 3m = .2) Tmccgitrcathams m tiptuyncathhms(1)tiimchonh 1x = tipxcving

trn ( ) ( ) ( )2 2: 1 2 10K x y + + + =

Cu2(2im): 1)Giiphngtrnhlnggicsau: sin coscos2 sin 2 cos 01 cotx xx x x

x

+ =

( )x

2)Giihphngtrnh: ( )

2 2

1 3 3 1

3 4 4 2

x x x y x x y y

xy x y x

+ + = + + = + +

( ),x y

Cu3(1im):Tnhdintchhnhphnggiihnbiccthsau: sin 2 , cos , 0,2

y x y x x x = = = =

Cu4(1im): Chohnhchp .S ABCD cy ABCD lhnhthoicnha ,hnhchiuvunggccanh S trn ( )ABCD

ltrungim HcaAB,ngtrungtuyn AMca ACD cdi 32a ,gcgia ( )vSCD ( )ABCD bng 030 . Tnhth

tchkhichp .S ABCD vxcnhtm,bnknhmtcungoitiphnhchp .S ABC.

Cu5(1im):Chobas thcdng , ,a b c.Chngminhrng: ( )3 3 3 3 3 31 1 1 32a b b c c aa b c

a b c c a b + + + + + + + + +

PHNTCHN(3im)(ThsinhchcchnmttronghaiphnAhocB)PhnA:Cu6a (2im):

1) Trong mt phng vi h ta Oxy cho hnh vung ABCD c ( )26A , nh B thuc ng thng

: 2 6 0d x y + = .GiM,Nlnltlhaiimtrn2cnhBC,CDsaocho BM CN = .Bit AMctBNti 2 145 5

I

.Xc

nhtanh C.

2) Trong khng gian vi h ta Oxyz cho ng thng 3 2 1:2 1 1x y zd + + = =

v mt phng

( ) : 2 0P x y z + + + = .Lpphngtrnh ngthng nmtrongmtphng ( )P ,ctdvto vi dgclnnht.Cu7a (1im):TrnmtphngtaOxy tmtphpccimbiudinchosphc ( )' 1 3 2z i z = + + trong z lsphcthamn 1 2z = .PhnB:Cu6b (2im):

1) Trong mt phng vi h ta Oxy cho tam gic ABC c phng trnh ng cao xut pht t A l

1: 2 3 0d x y + = ,phngtrnh ngphngictronggcCl 2 : 1 0d x y + + = .Bit ( )23H lchnngcaoxutphttnh B.Tmta A,B, C.

2) Trong khng gian vi h ta Oxyz cho mt cu ( ) ( ) ( ) ( )2 2 2: 1 2 3 16S x y z + + = v ng thng3 2:

2 1 2x y zd = =

.Lpphngtrnhmtphng ( )P changthngdvctmtcutheongtrncbnknhnh

nht.

Cu7b(1im):Tmhscashngcha 4x trongkhaitrinnhthcNiutncabiuthc ( )21 nx x + + bit n lstnhinthamn ( ) ( )

233 log 5log 5 152 25 15 4 5 15

n nn n n n

+ = .

Ht

Hvtnthsinh:

Trng thpt trn ph nga sn kho st cht lng cc mn thi i hc ln 2 chnh thc nm hc 2010 -2011 ( gm 1 trang) Mn : Ton ; khi A+B

Thi gian lm bi 180 pht, khng k thi gian giao I, Phn chung cho tt c cc th sinh (7,0 im)

Cu I ( 2.0 im)

Cho hm s 3 23 2y x x = + (C) 1. Kho st v v th (C) hm s cho.

2. Tm m phng trnh 3 2 23 2 logx x m + = c 8 nghim phn bit. Cu II (2,0 im).

1.Gii phng trnh :2( cos )1

cot 2 1x sinx

tanx x cotx

= +

2.Gii h phng trnh :

3 2 2 3

2 2

(1 ) (2 ) 30 0(1 ) 11 0

x y y x y y xyx y x y y y

+ + + + =

+ + + + = ( )x y R

Cu III (1,0 im). Tnh tch phn3

2

4

( )1tanxI dx

cosx cos x

= +

Cu IV (1,0 im). Cho hnh chp t gic u S.ABCD c cnh y AB = a ; chiu cao SO =62a

.Mt phng (P) qua A

vung gc vi SC ct SB, SC, SD ln lt ti B, C, D.Chng minh rng AC vung gc vi BD v tnh th tch khi chp S.ABCD.

Cu V (1,0 im).Cho a,b,c l cc s dng tho mn abc =1 . Tm gi tr nh nht ca biu thc :

2 2 2 2 2 2

bc ca abMa b a c b c b a c a c b

= + + + + +

II.Phn ring(3.0 im) Th sinh ch c lm mt trong hai phn (phn A hoc phn B)

A. Theo chng trnh Chun Cu VI.a (2,0 im). 1. Trong mt phng to Oxy , cho ng thng (d 1 ) :3 4 5 0x y + + = v (d 2 ) :4 3 5 0x y = . Vit phng trnh ng trn c tm nm trn ng thng ( ): 6 10 0x y = v tip xc vi hai ng thng (d 1 ) v (d 2 ).

2. Trong khng gian to Oxyz cho hai ng thng:

(d 1 ) :2 4

1 1 2x y z +

= =

v (d 2 ):8 6 10

2 1 1x y z +

= =

. Lp phng trnh ng thng (d) ct (d 1 ) , (d 2 ) v (d) song

song vi trc Ox

Cu VIIa(1,0 im). Cho hai s phc 1z v 2z tho mn 1 2 1z z = = ; 1 2 3z z + = . Tnh 1 2z z . B. Theo chng trnh nng cao Cu VI.b (2,0 im). 1. Trong mt phng to Oxy , cho tam gic ABC c nh A(-1;3), ng cao BH nm trn ng thng y x = , phn gic

trong gc C nm trn ng thng : 3 2 0x y + + = . Vit phng trnh cnh BC.

2. Trong khng gian to Oxyz cho im M(1;2;-1) v N(7;-2;3) ng thng (d) c phng trnh :1 2 2

3 2 2x y z +

= =

. Tm im I thuc (d) sao cho IM + IN nh nht.

Cu VIIb (1,0 im). Gii phng trnh : 5 4log (3 3 1) log (3 1)x x + + = +

tl ..
http://www.laisac.page.tl/

SGD&TVNHPHCo0o

TRNGTHPTPCB

THITHIHC,CAONGNM2011Mnthi:TON,khiA(Ln2)

Thigianlmbi180pht,khngkthigiangiao

CuI(2im):Chohms 2 1

1xyx

=

+(1).

1)Khostvvth(C)cahms(1).2)GiMlimthucth(C),Ilgiaoimhaingtimcnca(C).TmtoimM

saochotiptuynca(C)tiMvingthngIMctchhsgcbng 9.CuII(2im):

1)Giiphngtrnh: 22 os3x.cosx+ 3(1 s in2x)=2 3 os (2 )4

c c x + +

2)Tmgitrcam phngtrnhsaucnghimduynht:0)23(log)6(log 225,0 = + + xxxm

CuIII(1im):

Tnhtchphn:ln3

ln 2x x

dxIe e

=

CuIV(1im):Chohnhlngtr ABC.ABCcyltamgicucnh a,hnhchiuvunggccaA

lnmtphng (ABC)trngvitm Ocatamgic ABC.Tnhthtchkhilngtr ABC.ABC

bitkhongcchgiaAA vBCl a 34

.

CuV(1im):Cho , ,a b c lbasthcdng.

Chngminhrng: 2 3 6 6 5 3( )( ) ( )( ) ( )( )

a b ca b a c b a b c c a c b

+ + + + + + + +

CuVI(2im):1)Trongmp(Oxy)cho4imA(10),B(24),C(14),D(35).TmtoimMthucngthng ( ) : 3 5 0x y = saochohaitamgicMAB,MCDcdintchbngnhau.

2)TrongkhnggianvihtaOxyz,choimM(111)vhaingthng1( ) :

1 2 3x y z

d +

= =

v 1 4( ') :1 2 5

= =

x y zd

Chngminh:imM,(d),(d) cngnmtrnmtmtphng.Vitphngtrnhmtphng.CuVII(1im):

Tmsphczthomn: z 2 i 2 + = .Bitphnonhhnphnthc3nv.

HTCnbcoithikhnggithchg thm.

Htnthsinh:......................................................Sbodanh:.........................

S GIO DC V O TOTHNH PH NNG

TRNG THPT PHAN CHU TRINH

THI TH I HC CAO NG NM 2011-LN 1Mn thi: TON Khi A

Thi gian: 180 pht, khng k thi gian giao

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

Cu I (2,0 im) Cho hm s 4 2 1y x mx m (1) vi m l tham s, c th mC .1. Kho st s bin thin v v th ca hm s (1) khi 1m .2. Tm m cc tip tuyn ca th mC ti cc im c nh ca mC vung gc vi nhau.

Cu II (2,0 im)

1. Gii phng trnh 4cos 3 sin 2 2 1 sin1 sin

x xx

x

.

2. Gii h phng trnh2

2

5 2 4

3 2 2

x x y

y y x

.

Cu III (2,0 im)1. Gii phng trnh 3 1 5 4x x x .

2. Cho cc s thc dng a, b, c tho iu kin 1a b c . Tm gi tr nh nht ca biu thc

2 2 2

2 3P

ab bc caa b c

.

Cu IV (1,0 im) Cho lng tr ng . ' ' 'ABC A B C , c cc cnh ' 3AA AB a , 4BC a , 5CA a v Ml trung im cnh bn BB'. Tnh theo a th tch khi lng tr . ' ' 'ABC A B C v din tch thit din ca hnhlng tr . ' ' 'ABC A B C khi ct bi mt phng (P) qua A' v vung gc vi AM.

II. PHN RING (3,0 im) Tt c th sinh ch c lm mt trong hai phn: A hoc B

A. Theo chng trnh Chun

Cu Va (1,0 im) Trong mt phng ta Oxy, cho tam gic ABC c 1;2 , 1;0A B v 0;3C . Tnh bnknh ng trn ni tip tam gic ABC.

Cu VI.a (2,0 im)1. Cho hm s xf x xe . Gii bt phng trnh ' 0f x .

2. Vit phng trnh ng thng qua gc ta O v ct th hm s 12

xy

x

ti hai im phn

bit nhn O lm trung im ca n.

B. Theo chng trnh Nng caoCu Vb (1,0 im) Trong mt phng ta Oxy, cho tam gic ABC c nh B trn trc honh, nh A trn ng thng : 3 1 0x y v 2;1G l trng tm ca n. ng thng 3 0y l trung trccnh BC. Tm ta cc nh ca tam gic ABC.

Cu VI.b (2,0 im)1. Gii phng trnh 12 0,25 4log 7 log 3 4 log 2 3x x .2. Ty thuc vo tham s m, hy tm cc ng tim cn ca thi hm s

2 1mx xy

x

.

-----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:......................................

S GIO DC V O TOTHNH PH NNG

Trng THPT Phan Chu Trinh

THI TH I HC CAO NG NM 2011-LN 2Mn thi: TON Khi A&B

Thi gian: 180 pht, khng k thi gian giao I. PHN CHUNG CHO TT C TH SINH (7,0 im)

Cu I (2,0 im) Cho hm s 2 31

xy

x

.

1. Kho st s bin thin v v th (C) ca hm s cho.2. Vit phng trnh tip tuyn ca th (C) sao cho khong cch t tm i xng ca th (C)n t gi tr ln nht.

Cu II (2,0 im)

1. Gii phng trnh 2 2cos 24sin sin 2 1 sin2 1 cot

x xx x

x

.

2. Gii h phng trnh

3 2

2 1 1

7

x y x y

x y

.

Cu III (2,0 im)

1. Cho hnh phng gii hn bi th hm s3

1, 0

sin cosy y

x x v hai ng thng

4x

,

3x

.

Tnh din tch ca hnh phng .

2. Tm gi tr nh nht ca biu thc 2 2 2 2 2 2x xy y y yz z z zx x , vi mi s thcx, y, z tha mn iu kin 1x y z .

Cu IV (1,0 im) Cho hnh lng tr ABC.ABC, c 4AB AC a , 0120BAC v hnh chiu vung gcca A ln mt phng (ABC) trng vi tm ca ng trn ngoi tip tam gic ABC. Gc gia cnh bnvi y l 030 . Tnh theo a th tch khi lng tr ABC.ABCv khong cch gia AA vi BC.

II. PHN RING (3,0 im) Th sinh ch c lm mt trong hai phn: A hoc BA. Theo chng trnh ChunCu Va (1,0 im) Tnh mun ca s phc z i , bit 2z i z i iz ( i l n v o ).Cu VI.a (2,0 im)

1. Trong mt phng Oxy, cho im M trn elip 2 2: 5 9 45 0E x y v tch cc khong cch t M

n hai tiu im ca (E) bng 659

. Hy tm ta im M, bit im M gc phn t th hai.

2. Trong khng gian Oxyz, cho im 5;3; 1A . Vit phng trnh mt phng (P) cha trc Ox sao cho khong cch t A n (P) bng 1.B. Theo chng trnh Nng cao

Cu Vb (1,0 im) Tm s phc z, bit 22 31

2i zi

iz z

( i l n v o ).

Cu VI.b (2,0 im)1. Trong mt phng Oxy, cho elip 2 2: 4 1E x y . Tnh tm sai ca (E) v vit phng trnh chnhtc ca hypebol (H) nhn cc tiu im ca (E) lm nh v c hai tiu im l hai nh ca (E).2. Trong khng gian Oxyz, cho im 5;3; 1A . Vit phng trnh mt phng (P) i qua A v song

song vi trc Ox, bit khong cch gia Ox v (P) bng 1.-----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: ..........................................

CuI:(2,0im) Chohms 1)34()1(31 23 + + + = xmxmmxy cthl(Cm)

1.Khostsbinthinvvth(C1)cahmskhim=12.Tm tt c ccgi trmsao cho trn th (Cm) tn ti duy nhtmtimAc

honhmmtiptuynvi(Cm)tiAvunggcvingthng: x 2y 3 0. + =CuII:(2,0im)

1. Giiphngtrnh: 2 22sin 2sin t anx4

x x =

2. Giihphngtrnh:2 2

2

2 1xyx yx y

x y x y

+ + = + + =

(x,yR)

CuIII:(1,0im) Tnhtchphn:4

0

tan .ln(cos )cosx x

dxx

CuIV:(1,0im) ChohnhhpABCD.ABCD cyABCDlhnhthoicnhagc 060DAB = cnhbnBB= a 2.HnhchiuvunggccaimDtrnBBlimK

nmtrncnhBBv 1BK= BB'4

hnhchiuvunggccaimBtrnmtphng(ABCD)

limHnmtrnonthngBD.TnhtheoathtchkhihpABCD.ABCDvkhongcchgiahaingthngBCvDC.CuV:(1,0im)Xtccsthc a,b,c,dthamniukin 2 2a b 1 c d 3. + = =TmgitrnhnhtcaM ac bd cd = + .CuVI(2,0im)1.TrongmtphngvihtaOxy chongtrn:(C): 2 2x y 16 + = .Vitphngtrnh

chnhtccaelipctmsai 12

e = bitelipctngtrn(C)tibnimA,B,C,Dsaocho

ABsongsongvitrchonhvAB=2.CD.2. TrongkhnggianvihtaOxyzhaingthng:

11 1:

2 1 1x y zd + = = 2

1 2:1 2 1x y zd = = vmtphng(P): 2 3 0x y z + + = .

Vit phng trnh ng thng song song vi (P) v ct 1 2,d d ln lt tiA, B sao cho29AB =

CuVII(1,0im) Chohaisphcz,zthamn ' 1z z = = v ' 3z z + = .Tnh 'z z

HtThsinhkhngcsdngtiliu.Cnbcoithikhnggiithchgthem

Hvtn:..SBD:

TRNGTHPTCHUYN

NGUYNHU

KTHITHIHCLNTHBANMHC2010 2011THIMN:TON

KHIA,BThigianlmbi:180pht,khngkthigiangiao

SGD&TPHTH Trng thpt hng vng THI TH I HC NM 2011

( c 01 trang) Mn:Ton- Khi A+ B (Thi gian lm bi 180 pht khng k thi gian giao )

I.phn chung cho tt c th sinh: CuI:(2,0 im) Cho hm s 3 23 2y x x mx = + (1) 1.Kho st s bin thin v v th ca hm s (1) khi m = 0. 2.Tm m hm s c cc i , cc tiu v cc im cc i, cc tiu cch u ng thng (d) y = x - 1. Cu II:(2,0 im) 1.Gii phng trnh:

Cos2x + 3sin2x +5Sinx 3Cosx =3

2.Gii h phng trnh:2 2 2 2 2 2

( )(1 ) 4

( )(1 ) 4x y xy xyx y x y x y

+ + =

+ + =

Cu III:(1,0im): Tnh tch phn :5

2

ln( 1 1)1 1xI dx

x x +

= +

Cu IV:(1,0 im):Cho hnh chp S.ABCD c y ABCD l hnh thang vung ti A v B vi AB =BC =a AD = 2a.Cc mt (SAC) v (SBD) cng vung gc vi mt y(ABCD).Bit gc gia hai mt phng(SAB) v (ABCD) bng 060 .Tnh th tch khi chp SABCD v khong cch gia hai ng thng CD v SB.

Cu V:(1,0 im) Cho x,y, z l cc s thc dng tho mn xy + yz + xz = 3xyz. Hy chng minh rng:

2 2 2

2 2 2 2 2 2 12 2 2y x z

xy x zx z yz y + +

+ + + .

II. phn ring (3im) Th sinh ch c chn mt trong hai phn ( phn 1 hoc 2) 1.Theo chng trnh chun: Cu VI.a(2 im) 1.Trong h to Oxy ng thng (d): x y +1 =0 v ng trn (C): 2 2 2 4 0x y x y + + = .Tm im M thuc ng thng (d) m qua M k c hai ng thng tip xc vi ng trn (C) ti A v B sao

cho 060 .AMB = 2.Trong khng gian vi h to Oxyz cho A(2;0;1),B(3;1;2),C(2;0;-2) ,D(0;4;2).Lp phng trnh mt phng (P) i qua A , B v cch u C v D. Cu VII.(1im): Tm h s 4a ca

4x trong khai trin Niutn a thc 2( ) ( 1)nf x x x = + + vi n l s t

nhin tha mn:2 3 1 11

0 1 23 3 3 4 13 ...2 3 1 1

nn

n n n nC C C Cn n

+ + + + + =

+ + .

2.Theo chng trnh nng cao: Cu VI.b(2,0 im) 1.Trong mt phng vi h to Oxy cho hnh ch nht ABCD bit phng trnh cnh BC:x + 2y - 4 = 0 phng trnh ng cho BD:3x + y 7 = 0,ng cho AC i qua M(-5;2).Hy tm ta cc nh ca hnh ch nht ABCD. 2.Trong khng gian vi h to Oxyz cho A(-3;5;-5), B(5;-3;7) v mt phng (P) c phng trnh: x +y + z - 6 = 0 . a)Lp phng trnh mt phng (P) i qua A,B v vung gc vi (P). b)Tm im M nm trn mt phng (P) sao cho 2 2MA MB + nh nht.

Cu VII.b(1,0 im :Gii bt phng trnh: 31 12 2

1log ( 1) log (1 2)2

x x > +

SGD&TTHANHHATRNGTHPTBMSN

KTHITHIHCLN2NM2011MN:TONKHI:A

(Thigianlmbi180khngkthigianpht)

I.PHNCHUNGCHOTTCCCTHSINH(7im)CuI(2im)Chohms ( )3 3 2 my x mx C = +

1. Khostsbinthinvvthcahms ( )1C2. Tmm ngthngiquaimcci,cctiuca ( )mC ctngtrntm

( )11 ,I bnknhbng1tihaiimphnbitA,BsaochodintchtamgicIABtgitrlnnhtCuII(2im)

1. Giiphngtrnh ( ) 22cos3 cos 3 1 sin 2 2 3 os 24

x x x c x + + = +

2. Giiphngtrnh ( )22 21 5 2 4x x x + = +CuIII(1im)Tnhtchphn

+

+ =

e

dx x x x x

x I

1

2 ln 3 ln 1

ln

CuIV(1im)ChohnhchpS.ABCcyABCltamgicvungcnnhA, 2AB a = .GiIltrungimcacnhBC.HnhchiuvunggcHcaSlnmtphng(ABC)thamn 2IA IH =

uur uuur.GcgiaSCv

mty(ABC)bng 060 .HytnhthtchkhichpS.ABCvkhongcchttrungimKcaSBnmtphng(SAH).CuV(1im)Cho3sthcdnga,b,cthamn 2 2 2 1a b c + + = .

Chngminhrng5 3 5 3 5 3

2 2 2 2 2 2

2 2 2 2 33

a a a b b b c c cb c c a a b + + +

+ + + + +

II.PHNRING(3,0im)ThsinhchclmmttronghaiphnAhocBA.TheochngtrnhchunCuVI.a(2,0im)

1.TrongmtphngvihtrctaOxychohnhchnhtABCDcdintchbng12,tmIlgiaoimcangthng : 3 0d x y = v ' : 6 0d x y + = .TrungimmtcnhlgiaoimcadvitrcOx.Tmtaccnhcahnhchnht.

2. Trongkhng gianvi h trc taOxyzcho haiim (0 12)M v ( 113)N .Vit phngtrnhmtphng(P)iquaM,Nsaocho khongccht ( )00 2K n(P)tgitrlnnht

CuVII.a(1,0im)Chokhaitrin ( )0

nn k n k k

nk

a b C a b =

+ = .Quycshngthicakhaitrin lshngngvik=i1.

Hytmccgitrcaxbitrngshngth6trongkhaitrin

81 13 1 log 3 1log 9 7 2522 2xx

+ +

+

l224.

B.TheochngtrnhnngcaoCuVI.b(2,0im)

1.TrongmtphngvihtrctaOxychohnhchnhtABCDcphngtrnhcnhABvngchoBDlnltl 2 1 0x y + = v 7 14 0x y + = ,ngthngACiquaim ( )21M .Tmtaccnhcahnhchnht.

2.TrongkhnggianvihtrctaOxyzchobaim ( ) ( ) ( )231 , 1 20 , 11 2A B C .TmtatrctmHvtmngtrnngoitiptamgicABCCuVII.a(1,0im)Giibtphngtrnh ( )2 23log 2 9log 2x x x >

1

SGD_DTNGAN THITHIHCNM2011(LNII)TRNGTHPTYNTHNHII MnTON: KhiA

Thigianlmbi180pht,khngkthigianpht

I. PHNCHUNGCHOMITHSINH( 7im)CuI( 2,0im)Chohmsy=x3+mx+2(1),mlthamsthc.

1. khostsbinthinvvthcahmskhim= 32. Tmm thcahms(1)cttrchonhtimtimduynht

CuII(2,0im)1. Giiphngtrnh 32cos cos 2 4sin 3 0x x x + + =2. Tmm phngtrnh 3( 7 3 5) (7 3 5) 2x x xm + + + = Cnghimduynht

CuIII( 1,0im)Tnhtchphn I=ln5

ln 4 3 4x x

dxe e +

CuIV ( 1,0im)ChohnhchpS.ABCDcylhnhvungcnha.Ccmtbntovimtphngymtgcbngnhauvbng60o.XcnhimMtrnSAvimNtrnBCsaochodionthngMNngnnhtvtnhdionthngMNtheoa

CuV (1,0im)Giiphngtrnh2

22

4 124 4xx

x x + =

+ +II. PHNRING(3,0im)

Thsinhchclmmttronghaiphn(phnAhocB)A.TheochngtrnhchunCuVI.a (2,0im)

1.Trong mtphngto0xychoimA(12)vngthngd:3x y 6=0.Tmhai imB,CtrndsaochotamgicABCvungcntiA2.Trongkhnggianto0xyzlpphngtrnhmtphng(Q)songsongvingthng

d: 2 11 1 4x y z

= = vvunggcvimtphng(P):x+3y 8z+2=0ngthitipxcvi

mtcu(S):(x 1)2+(y 3)2+(z 1)2=9CuVII.a(1,0im)GiZ1vZ2lhainghimphccaphngtrnhZ2+4Z+13=0

Tnhgitrca biuthcA= 2 21 2Z Z +

A.TheochngtrnhnngcaoCuVI.b( 2,0im)

1.Trongmtphngto0xy.Tmtrn ngthng(d):3x+4y+8=0nhngimmtcthktingtrn:(x1)2+(y 1)2=1nhngtiptuynmkhongcchttitipimcdinhnht2.Trongkhnggianto0xyzLpphngtrnhmtphng(P)ctcctia0x,0y,0zlnltticcimABCsaochotamgicABCnhnH(123)lmtrctm.

CuVII.b (1,0im)Cbaonhiustnhinchncnmchstngimtkhcnhaumnhthua50000

HT

SGIODCVOTONGHANTRNGTHPTCHUYNPHANBICHU

KTHI THIHCLN2NM2011Mnthi:TONKhiA,B

Thigianlmbi:180pht,khngkthigiangiao

I.PHNCHUNGCHOTTCTHSINH (7,0im)

CuI(2,0 im)Chohms2,1

xyx

=

+cthl ( ).C

1. Khostsbinthin cahmsvvth ( ).C2. Vitphngtrnhtiptuyncath ( ),C bittiptuyntovihaingtimcnca ( )C mttam

giccbnknhngtrn nitipln nht.CuII(2,0 im)

1. Giiphngtrnh 4 41(tan .cot 2 1)sin(4 ) (sin cos ).

2 2x x x x x + = +

2. Giihphngtrnh2 2

2 2

2 ( 1) 3

3 2 .x x y y yx xy y x y

+ =

+ =

CuIII(1,0im) Tnhtchphn2

21

1 .1

xI dxx x

+ =

+

CuIV(1,0im)Chohnhlngtr . ' ' 'ABC A B C c '.A ABC lhnhchptamgicu, .AB a = Gi lgcgiamtphng ( ' )A BC vmtphng ( ' ' ).C B BC Tnhtheoathtchkhichp '. ' ',A BCC B

bit1 .3

c = os

CuV(1,0 im)Chobasdng , , .a b c Chngminhrng2 2 2 2 2 2

3 .2

a b ca b b c c a

+ + + + +

II.PHNRING(3,0im): Thsinh chclmmttronghaiphnAhocB.A.TheochngtrnhcbnCuVIa(2,0im)

1. TrongmtphngtaOxy,choelip2 2

( ) : 1.8 2x yE + = Vitphngtrnhngthngdct ( )E ti

haiimphnbitctolccsnguyn.2. Trongkhnggian taOxyz,chohnh thoi ABCD cdintchbng12 2,nh AthuctrcOz, nh

Cthucmtphng ,Oxy hainh BvDthuc ngthng 1:1 1 2x y zd + = = vBchonhdng.

Tmto , , , .A B C D

CuVIIa(1,0im)Chosphczthomn 71 .2

zzz

+ =

Tnh

2 .z iz i +

B.TheochngtrnhnngcaoCuVIb(2,0im)1. Trong mt phng ta Oxy, cho hai ng trn 2 21( ) : ( 1) ( 2) 5C x y + + = v

2 22( ) : ( 1) ( 3) 9.C x y + + + = Vitphngtrnhng thng tipxcvi 1( )C vct 2( )C tihai

imA,B thomn 4.AB =

2. Trong khng gian ta Oxyz, cho ng thng1 2:

2 1 1x y zd + = = v mt phng

( ) : 2 3 0.P x y z + = Vitphngtrnh ngthng thuc(P),vunggcvidvckhongcchgiadv bng 2.

CuVIIb(1,0im)Tm m hms2

2x mx my

x + +

= +

cgitrccivgitrcctiutridu.

..................Ht.................Thsinhkhngcsdngtiliu.Gimthkhng giithchgthm.

TrngTHPTNgGiaT THITHIHCLNIVMnThi:Ton KhiAThigian:180pht,khngkthigiangiao

I.PHNCHUNGCHOTTCTHSINH(7,0im)CuI(2im)Chohms 3 23 2 y x x = + (C)

1)Khostsbinthinvvth(C).2)Tmtrnngthng(d):y=2ccimmtcthkcbatiptuynnth(C).

CuII(2im)1)Giiphngtrnh: 2 22 11 15 2 3 6 + + + + + x x x x x .

2)Giiphngtrnh: x x x x 3 2 2 cos2 sin 2 cos 4sin 0 4 4

+ + + =

.

CuIII(1im) Tnhtchphn: I x x x x dx 2

4 4 6 6

0 (sin cos )(sin cos )

= + + .

Cu IV (2 im) Chohnh chp S.ABC, y ABC l tam gic vung ti B cAB = a,BC=a 3 ,SAvunggcvimtphng(ABC),SA=2a.GiM,Nlnlt lhnhchiu vung gc ca imA trn cc cnhSB v SC. Tnh th tch ca khi chpA.BCNM.

CuV(1im)Choa,b,c,dlccsdng.Chngminhrng:

abcd a b c abcd b c d abcd c d a abcd d a b abcd 4 4 4 4 4 4 4 4 4 4 4 4 1 1 1 1 1

+ + + + + + + + + + + + + + +

II.PHNRING (3,0im)A.Theochngtrnhchun.

CuVI.a(2im)1) Trongmt phng vi h to Oxy, gi A, B l cc giao im ca ng thng

(d): 2x y 5= 0 v ng trn (C): 2 2 20 50 0 x y x + + = .Hy vit phng trnhngtrn(C) iquabaimA,B,CviC(11).

2)TrongkhnggianvihtrctaOxyz,choimA(456).Vitphngtrnhmtphng(P)quaA,ctcctrctalnlttiI,J,KmAltrctmcatamgicIJK.

CuVII.a(1im)Chngminhrngnu n a bi (c di) + = + th 2 2 2 2 n a b c d ( ) + = + .B.Theochngtrnhnngcao

CuVI.b(2im)

1)TrongmtphngvihtoOxy,chotamgicABCcdintchbng 3 2,A(23),

B(32),trngtm ca ABCnmtrnngthng(d):3xy8=0.Vitphngtrnhngtrniqua3imA,B,C.

2)TrongkhnggianvihtrctaOxyz,chobnimA(456)B(001)C(020)D(300).ChngminhccngthngABvCDchonhau.Vitphngtrnhngthng(D)vunggcvimtphngOxyvctccngthngAB,CD.

CuVII.b(1im) Giihphngtrnh: x y x x y

x xy y y x y

2 2 4 4 4

2 4 4 4

log ( ) log (2 ) 1 log ( 3 )

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

+ + = +

+ + + =

Ht

1

S gio dc v o to Thanh Ha THI KI M TRA CH T LNG NM HC 2010-2011 Trng THPT chuyn Lam Sn Mn thi :Ton kh i A ( thi gian 180 pht ) Ngy thi : 7 /5/2011 PHN CHUNG CHO TT C CC TH SINH ( 7,0 im ) Cu I (2,0 im) Cho hm s 3 22 3( 1) (1)y x m x m= + (m l tham s thc) 1. Kho st s bin thin v v th (C) c a hm s (1) khi 2.m= 2. Tm m th hm s c im cc tr, k hiu l A, B sao cho ba im A, B, (3;1)I thng hng. Cu II (2,0 im )

1. Gii phng trnh 2

2sin (7cos 3)cot .tan tan

4 4

xx x

x x

= +

2. Gii bt phng trnh 22 2 3 2 ( ).x x x x x+ +

Cu III ( 1,0 im ) Tnh din tch hnh phng gii hn bi cc ng: 22 2, 4 .y x y x x= + + = +

Cu IV (1,0 im) Cho hnh hp ng . ' ' ' 'ABCD A B C Dc , 2 , ' 3 ( 0)AB a AD a AA a a= = = > v 060 .BAD= Chng minh rng AB vung gc vi BD v tnh khong cch t im 'A n mt phng ( ').ABD

Cu V (1,0 im ) Cho cc s thc , ,x y z tha mn 2 2

0

0

2 1.

x

y

x y

+ =

Chng minh rng 1 1 2 1 2 1 2 4 2 6.x y+ + + + + +

PHN RING (3,0 im ): Th sinh ch c lm mt trong hai phn ( phn A hoc phn B ). A. Theo chng trnh chun

Cu VI.a (2,0 im ) 1.Trong mt phng ta Oxy cho hnh thoi ABCD c hai cnh ,AB CD ln lt nm trn hai

ng thng 1 2: 2 5 0, : 2 1 0.d x y d x y + = + = Vit phng trnh cc ng thng AD v ,BC

bit ( 3;3)M thuc ng thng AD v ( 1;4)N thuc ng thng BC . 2. Trong khng gian ta Oxyz, vi t phng trnh ng thng song song vi cc mt phng

( ) : 3 12 3 5 0, ( ) : 3 4 9 7 0P x y z Q x y z+ = + + = v ct hai ng thng

1 2

5 3 1 3 1 2: , : .

2 4 3 2 3 4

x y z x y zd d

+ + + = = = =

Cu VII.a (1,0 im ). T cc ch s 0;1;2;3;4;5 c th lp c bao nhiu s t nhin l, mi s gm 6 ch s khc nhau v tng ba ch s u ln hn tng ba ch s cui mt n v.

B. Theo chng trnh nng cao Cu VI.b (2,0 im )

1. Trong mt phng ta Oxy cho elp 2 2

( ) : 19 4

x yE + = v cc im ( 3;0), ( 1;0).A I Tm ta

cc im ,B C thuc ( )E sao cho I l tm ng trn ngoi tip tam gic .ABC 2. Trong khng gian ta Oxyz cho cc im (2;0; 5), ( 3; 13;7).A B Vit phng trnh mt phng ( )P i qua ,A B v to vi mt phng Oxz mt gc nh nht.

Cu VII.b (1,0 im ) Cho s phc 26(1 ) 4( 3 4 )

.1

i iz

i

+ + =

Tm dng lng gic ca s phc 3.z

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

TRNGTHPTCHUYNVNHPHC

CHNHTHC(thic01trang)

THITHIHC,CAONGNM2011Mnthi:Ton,khiA,B

Thigianlmbi:180pht(khngkthigiangiao)

A.PHN CHUNGCHOTTCTHSINH( 7,0im)

CuI:(2,0im). Chohms: 2x 1yx 1

=

+cthl ( )C .

1)Khostsbinthinvvthhms(C)2)Gi I lgiaoimcahaingtimcnca ( )C .Tmtrn th ( )C imMchonhdngsaochotiptuyntiMvith ( )C cthaingtimcnti A v B thomn :

2 2 40IA IB + = .CuII:(2,0im)1)Giiphngtrnh: 4 2 43sin 2cos 3 3 3 cos 1x x cos x cos x x + + = +

2)Giiphngtrnh: ( )24 1

5 2 4 227x

x x +

+ + =

CuIII:(1,0im).Tnhtchphn: ( )2

0

24xI x dxx

=

CuIV:(1,0im).Chohnhchp .S ABC c 04, 2, 4 3, 30AB AC BC SA SAB SAC = = = = = = .Tnhthtchkhichp .S ABC .

CuV:(1,0im).Cho , ,a b c lbasthckhngmthomn: 3a b c + + = .Tmgitrln nhtcabiuthc:P a b b c c a abc = + + .

B.PHNTCHN:(3,0im).(Thsinhchclm1trong2phn,phnAhocphnB)A.Theochngtrnhchun:CuVIA:(2,0im).1) TrongmtphngvihtoOxy chotamgic ABC cnti A ,bitphngtrnhcc ng thng ,AB BC lnltl 3 5 0x y + + = v 1 0x y + = ,ngthng ACiquaim ( )30M .Tmtoccnh , ,A B C .2) Trongkhnggianvihta0xyzchohaingthng:

11 1 1:

1 2 2x y zd = = v 2

1 3:1 2 2x y zd + = =

.

Tmtoim I lgiaoimca 1d v 2d ,lpphngtrnh ngthng 3d iquaim ( )0 12P ,ngthi 3d ct 1d v 2d lnltti ,A B khc I thomn AI AB = .

CuVIIA.(1,0im):Tnhtng 1 3 5 7 2009 20112011 2011 2011 2011 2011 2011S C C C C C C = + + + LB.Theochngtrnhnngcao

CuVIB:(2,0im).1)TrongmtphnghtoOxychoelp ( )2 2

: 125 9x yE + = vihaitiu

im 1 2,F F .im P thucelpsaochogc0

1 2 120PFF = .Tnhdintchtamgic 1 2PFF .

2) Trongkhnggian vihta0xyz,chohaingthng: 11 3:

2 3 2x y z

= =

v

25 5:

6 4 5x y z +

= =

,mtphng ( ) : 2 2 1 0P x y z + = .Tmccim 1 2,M N saochoMN

songsongvimtphng ( )P vcchmtphng ( )P mtkhongbng2.

CuVIIB:(1,0im):Tmphnthc,phnocasphc ( )

( )2012

2011

1

3

iz

i

+ =

+

S gio d c v o t o H n i Tr ng THPT Lin H THI TH AI HOC NM 2011 **************** Mn : TOAN; khi: A,B( Th i gian lam bai: 180 phut, khng k th i gian phat ) PHN CHUNG CHO TT CA CAC THI SINH (7,0 im)Cu I (2 im)

1. Kh o st s bi n thin v v th (C) c a ham s 2 11

xyx

=

2. Vi t ph ng trnh ti p tuy n c a (C), bi t kho ng cch t i m I(1;2) n ti p tuy n b ng 2 .Cu II (2 im)

1) Giai ph ng trnh 217sin(2 ) 16 2 3.s in cos 20sin ( )2 2 12

xx x x + + = + +

2) Giai h ph ng trnh : 4 3 2 2

3 2

1

1

x x y x y

x y x xy

+ =

+ =

Cu III (1 im) : Tinh tch phn: I = 4

0

tan .ln(cos )cos

x x dxx

Cu IV (1 im) : Cho hnh chp S.ABC c y ABC l tam gic vung t i A v i AB = a, cc m t bn l cc tam gic cn t i nh S. Hai m t ph ng (SAB) v (SAC) cng t o v i m t ph ng y gc 60 0. Tnh csin c a gc gi a hai m t ph ng (SAB) v (SBC) . Cu V: (1 im) Cho a,b,c la cac s d ng thoa man a + b + c = 1. Ch ng minh r ng:

3a b b c c aab c bc a ca b

+ + ++ + + + +

PHN RING (3 im) Thi sinh chi c lam mt trong hai phn (phn A hoc B) A. Theo ch ng trinh Chun

Cu VI.a (1 im) Trong mt phng toa Oxy cho im A(1;1) v ng th ng : 2x + 3y + 4 = 0. Tim t a i m B thu c ng th ng sao cho ng th ng AB v h p v i nhau gc 45 0.Cu VII.a (1 im ): Trong khng gian v i h toa Oxyz, cho im M(1;-1;1) va hai ng th ng 1( ) :

1 2 3x y zd += =

v

1 4( ') :1 2 5x y zd = =

Ch ng minh: im M, (d), (d) cung nm trn mt mt phng. Vit ph ng trinh mt phng o. Cu VIII.a (1 im)

Gi i ph ng trinh: 2 2 2 (24 1)(24 1) (24 1)log log xx x x xLog x x x++ ++ = Theo ch ng trinh Nng cao

Cu VI.b (1 im)Trong mt phng toa Oxy cho ng tron 2 2( ) : 1C x y+ = , ng thng ( ) : 0d x y m+ + = . Tim m ( )C ct ( )d tai A va B sao cho din tich tam giac ABO l n nht.

Cu VII.b (1 im)Trong khng gian v i h toa Oxyz, cho ba m t ph ng: (P): 2x y + z + 1 = 0, (Q): x y + 2z + 3 = 0, (R): x + 2y 3z + 1 = 0

v ng th ng 1 : 22

x

= 1

1+y =

3z

. G i 2 l giao tuy n c a (P) v (Q).

Vi t ph ng trnh ng th ng (d) vung gc v i (R) v c t c hai ng th ng 1 , 2 .Cu VIII.b (1 im) Gi i b t ph ng trnh: log x( log3( 9x 72 )) 1

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

TRIJONG EAI HQC VINHTRIIONG THPT CHUYTN

of xrrAo sAr cnArr,ugr\c t 6p tzr,An 3, NAna zorrm0n: TOAN; Thli gian lim bhi: IBA phrtt

r. rHAN cHUNc cHo rAr cA rHi slr.{H 1z,o a$q1

Ciu I. 1Z,O ei6m; Cho him s6 y = 1*o -(3m+l)xz +2(m+l), m ldtham s5."4l. Kh6o s6t sg bitin thi6n vd vC dO thi hdm sb c16 cho khi z = 0.2. Tinr llr A6 A6 fti ham sii da cho co 3 tli6m cgc ti l$p thanh mQt tam gf6c e6 trgng t6m ld g6c toa d0.

CAu II. (2,0 tti6m)I . Giai phuorg trinh 2Iogo(l a ,l2y a1= logz (5 - r) + log , (3 - x).

2

2. Gieiohuongtrinh lsinZx-cos2x)tanr* sin3x

=sinr+cosx.

Ciu III. (1,0 di6m) Tinh thc tich khdi trdn xoay tu":iah khi quay hinh phang gidi han boi d6 thi hem

1(:.1')L-

s6 y = E,trqc hoanh vi dudmg thdng x = 1 xung quanh truc hoanh.A* e- * -l-t-4r ttl

Ciu IV. (1,0 di6m) Cho hinh Hng tru dtmg ABC.A' B'C' c6 AC = a, BC =2a, ZACB = 1200 vd

Cu I: (2,0 im) Cho hm s 1)34()1(31 23 xmxmmxy c th l (Cm)

1. Kho st s bin thin v v th (C1) ca hm s khi m = 1 2. Tm tt c cc gi tr m sao cho trn th (Cm) tn ti duy nht mt im A c honh m m tip tuyn vi (Cm) ti A vung gc vi ng thng : x 2y 3 0. Cu II: (2,0 im)

1. Gii phng trnh: 2 22sin 2sin t anx4

x x

2. Gii h phng trnh: 2 2

2

2 1xyx yx y

x y x y

(x, y R)

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

0

tan .ln(cos )cos

x x dxx

Cu IV: (1,0 im) Cho hnh hp ABCD.ABCD c y ABCD l hnh thoi cnh a; gc 060DAB ; cnh bn BB= a 2 . Hnh chiu vung gc ca im D trn BB l im K

nm trn cnh BB v 1BK= BB'4

; hnh chiu vung gc ca im B trn mt phng (ABCD)

l im H nm trn on thng BD. Tnh theo a th tch khi hp ABCD.ABCD v khong cch gia hai ng thng BC v DC. Cu V: (1,0 im) Xt cc s thc a, b, c, d tha mn iu kin 2 2a b 1; c d 3. Tm gi tr nh nht ca M ac bd cd . Cu VI (2,0 im) 1. Trong mt phng vi h ta Oxy cho ng trn :(C): 2 2x y 16 . Vit phng trnh

chnh tc ca elip c tm sai 12

e bit elip ct ng trn (C) ti bn im A, B, C, D sao cho

AB song song vi trc honh v AB = 2.CD. 2. Trong khng gian vi h ta Oxyz hai ng thng:

11 1:

2 1 1x y zd ; 2

1 2:1 2 1

x y zd v mt phng (P) : 2 3 0x y z .

Vit phng trnh ng thng song song vi (P) v ct 1 2,d d ln lt ti A, B sao cho 29AB

Cu VII (1,0 im) Cho hai s phc z, z tha mn ' 1z z v ' 3z z . Tnh 'z z

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

H v tn:..SBD:

TRNG THPT CHUYN

NGUYN HU

K THI TH I HC LN TH BA NM HC 2010 2011 THI MN: TON

KHI A,B Thi gian lm bi: 180 pht, khng k thi gian giao

SGD TPhThTrngTHPTHHa

THITHIHCLN2NM2011MnTon KhiA,B,D.

Thigian150pht

PHNCHUNGCHOTTCTHSINH (7)CuI. (2im)Chohms 3 2y x 3mx m = + (1)

1. Khostvthhm s(1)khim=1.2. Tmccgitrmhms(1)c2cctr,ngthiccimcctrcngvigcta

Otothnhmttam giccdintchbng4.CuII.(3im)

1.Giiphngtrnh 2 22 1 4 14 2

log log ( 2 1) log ( 4 4) log ( 1) 0x x x x x x + + + = .

2.Tnhtchphn ( )

2

21

lnI dx1

xx

= +

3.Giiphngtrnh sautrntpsphc: ( 1)( 2)( 3) 10z z z z + + =CuIII.(1im)ChohnhchpS.ABCcyltamgicuABCcnha, ( )SA ABC vSA=3a.

GiM,NlnltlhnhchiuvunggccaAlncnhSB,SC.TnhthtchkhichpA.BCNMtheoa.

CuIV.(1im)Choccsthcdngx,y thamn: 2 2x xy y 1 + = . Tmgitrlnnhtvnh

nhtcabiuthc:4 4

2 2

11

x yPx y

+ + =

+ +PHNRING (3)PhndnhchothsinhkhiA,B:CuVa.

1. TrongmtphngvihtaOxy,chohnhbnhhnhABCDcdintchbng6vhainhA(1 2),B(2 3).Tmta2nhcnli,bitgiaoim2ngchocahnhbnhhnhnmtrntrcOxvchonhdng.

2. Trong khng gianvihtoOxyz, cho mt phng (P), mt cu (S) c phng trnh tng ng (P): 2x - 3y + 4z 5 = 0, (S): x 2 + y 2 + z 2 + 3x + 4y - 5z + 6 = 0.

a.CMR:Mtphng(P)ctmtcu(S)theogiaotuynlngtrn(C).b.Tmtmvbnknh cang

tai lieu toan

Documents