122 de on Thi Tot Nghiep Mon Toan 12

8/8/2019 122 de on Thi Tot Nghiep Mon Toan 12

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http://www.VNMATH.com122 N THI TT NGHIP MN TON THPT NM HC 2010 - 2011

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

Cu I.(3 im) Cho hm s y =2 1

1

+

x

x

c th (C).

1/ Kho st s bin thin v v th (C) ca hm s.2/ Vit phng trnh tip tuyn ca (C) ti giao im ca (C) vi trc tung.

Cu II. (3 im)1/ Gii phng trnh : log3(x + 1) + log3(x + 3) = 1.

2/ Tnh I =2

3

0

cos .

xdx .

3/ Xt s ng bin v nghch bin ca hm s y = -x3 + 3x -1Cu III. (1 im). Cho hnh chp S.ABC c ABC l tam gic vung cn ti B, aAC =

, SA ( ) ABC , gc gia cnh bn SB v y bng 600

. Tnh th tch ca khi chp.II. PHN RING (3 im).1.Theo chng trnh chun.Cu IVa. (2 im). Trong khng gian vi h ta Oxyz , cho im M(1; 1 ; 0) vmt phng (P): x + y 2z + 3 = 0.

1/ Vit phng trnh mt cu tm M v tip xc vi mp(P).2/ Vit phng trnh ng thng (d) i qua M v vung gc vi (P). Tm ta giao im.

Cu Va. (1 im). Tnh din tch hnh phng gii hn bi cc ng y = 3 v

y = x2

2x2. Theo chng trnh nng cao.Cu IVb (2 im) Trong khng gian vi h ta Oxyz, cho im M(-1 ; 2 ; 1) v

ng thng (d):1 2

2 1 1

+= =

x y z

.

1/ Vit phng trnh mt cu tm M v tip xc vi (d).2/ Vit phng trnh mt phng i qua M v vung gc vi (d). Tm ta giaoim.

Cu Vb. (1 im).Tnh din tch hnh phng gii hn bi cc ng y = 21

4x

v y =21 32 +x x

2

I.PHN CHUNG CHO TT C TH SINH.(7 im)Cu I.(3 im). Cho hm s y = x3 3x2 + 2 c th (C).

1/ Kho st s bin thin v v th (C) ca hm s.2/ Bin lun theo m s nghim ca phng trnh: x3 3x2 m = 0.

Cu II. (3 im).

1/ Gii phng trnh: 3x + 3x+1 + 3 x+2 = 351.

2/ Tnh I =1

0

( 1) .+ xx e dx

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3/ Tm gi tr ln nht v gi tr nh nht ca hm s y = x4 2x2 + 1 trn an[-1 ; 2].Cu III. (1 im). Tnh th tch khi t din u S.ABC c tt c cc cnh u bng a.II. PHN RING.(3 im)

1.Theo chng trnh chun.Cu IV a. (2 im). Trong khng gian vi h ta Oxyz, cho cc im A(-1 ; 2 ; 0),B(-3 ; 0 ; 2), C(1 ; 2 ; 3), D(0 ; 3 ; - 2).

1/ Vit phng trnh mt phng (ABC) v phng trnh ng thng AD.2/ Tnh din tch tam gic ABC v th tch t din ABCD.

Cu V a. (1 im). Tnh th tch khi trn xoay do hnh phng gii hn bi cc ng

y = tanx , y = 0, x = 0, x =4

quay quanh trc Ox.

2. Theo chng trnh nng cao.

Cu IV b.(2 im)Trong khng gian vi h ta Oxyz, cho cc im A(-2 ; 0 ; 1),B(0 ; 10 ; 2), C(2 ; 0 ; -1), D(5 ; 3 ; -1).1/ Vit phng trnh mt phng (P) i qua ba im A, B, C v vit phng trnh

ng thng i qua D song song vi AB.2/ Tnh th tch ca khi t din ABCD, suy ra di ng cao ca t din v t

nh D.Cu Vb. (1 im). Tnh th tch khi trn xoay do hnh phng gii hn bi cc ngy =

1

2 . xx e , y = 0, x = 0, x = 1 quay quanh trc Ox.

3I.PHN CHUNG CHO T C TH SINH. (7 im)Cu I. (3 im) Cho hm s y = - x3 + 3x -1 c th (C).

1/ Kho st s bin thin v v th (C) ca hm s.2/ Vit phng trnh tip tuyn ca (C) ti im cc tiu ca (C).

Cu II.(3 im)1/ Gii phng trnh: 26log 1 log 2= + xx

2/ Tnh I =2

2

0

cos 4 .

xdx

3/ Tm gi tr ln nht v gi tr nh nht ca hm s y =lnx

xtrn on [1 ;

e2 ]Cu III.(1 im). Cho hnh chp tam gic u S.ABC c cnh y bng a, cc cnh

bn u to vi y mt gc 600. Tnh th tch ca khi chp.II. PHN RING. (3 im)1.Theo chng trnh chun.Cu IV a.(2 im). Trong khng gian vi h ta Oxyz, cho mt phng(P): 2x + y z 6 = 0 v im M(1, -2 ; 3).

1/ Vit phng trnh mt phng (Q) i qua M v song song vi mp(P).Tnhkhang cch t M n mp(P).



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2/ Tm ta hinh chiu ca im M ln mp(P).Cu Va. (1 im). Gii phng trnh: x2 2x + 5 = 0 trong tp s phc C.2. Theo chng trnh nng cao.Cu IV b.(2 im). Trong khng gian vi h ta Oxyz , cho hai mt phng

(P): 3x 2y + 2z 5 = 0, (Q): 4x + 5y z + 1 = 0.1/ Tnh gc gia hai mt phng v vit phng tnh tham s ca giao tuyn ca

hai mt phng (P) v (Q).2/ Vit phng trnh mt phng (R) i qua gc ta O vung gc vi (P) v

(Q).Cu Vb.(1 im). Cho s phc z = x + yi (x, y )R . Tm phn thc v phn o ca s

phc z2 2z + 4i .

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I.PHN CHUNG CHO TT C TH SINH. (7 im)Cu I. (3 im). Cho hm s y =

2

1+x

xc th (C).

1/ Kho st s bin thin v v th (C) ca hm s.2/ Vit phng trnh tip tuyn ca(C) ti im c hanh x = -2.

Cu II. (3 im)1/ Gii phng trnh : 1 13 3 10+ + =x x .

2/ Tnh I =tan4

2

0cos

xe

dxx

3/ Tm gi tr ln nht v gi tr nh nht ca hm s y =2

1x .Cu III.(1 im).Cho hnh chp t gic u S.ABCD c cnh y bng a, cnh bnhp vi y mt gc 600 .

1/ Tnh th tch khi chp S.ABCD2/ Tm tm v tnh bn knh mt cu ngai tip hnh chp.

II. PHN RING. (3 im)1. Theo chng trnh chun.Cu IV a. (2 im). Trong khng gian vi h ta Oxyz, cho im D(-3 ; 1 ; 2) vmt phng (P) i qua ba im A(1 ; 0 ; 11), B(0 ; 1 ; 10), C(1 ; 1 ; 8).

1/ Vit phng trnh ng thng AB v phng trnh mt phng (P).2/Vit phng trnh mt cu tm D, bn knh R = 5. Chng minh rng mt cuny ct mt phng (P).Cu Va. (1 im). Tnh din tch hnh phng gii hn bi cc ng y = lnx ,y = 0, x

=1

e, x = e .

2.Theo chng trnh nng cao.Cu IV b.(2 im). Trong khng gian vi h ta Oxyz, cho mt phng (P): 2x +2y + z + 5 = 0 v mt cu (S): x2 + y2 + z2 2x 4y + 4z = 0.

1/ Tm tm v bn knh ca mt cu (S).

2/ Vit phng trnh mt phng (Q) song song vi (P) v tip xc vi (S). Tmta ca tip im.



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Cu Vb.(1 im). Tm m ng thng d: y = mx + 1 ct th (C): y =2 3

1

+

x

xti hai

im phn bit.

5I.PHN CHUNG CHO TT C TH SINH.(7 im)Cu I. (3 im). Cho hm s y = - x4 + 2x2 +3 c th (C).

1/ Kho st s bin thin v v th (C) ca hm s.2/ Da vo th (C), tm cc gi tr ca m phng trnh x 4 2x2 + m = 0 c

bn nghim thc phn bit.Cu II. (3 im)

1/ Gii bt phng trnh: 2 4log log ( 3) 2 =x x

2/ Tnh I =4

0

sin2

1 cos 2

+

xdx

x

.

3/ Cho hm s y = 25log ( 1)+x . Tnh y(1).Cu III. (1 im).Cho hnh chp S.ABC c y ABC l tam gic vung ti B, cnh

bn SA (ABC), bit AB = a, BC = 3a , SA = 3a.1/ Tnh th tch khi chp S.ABC theo a.2/ Gi I l trung im ca cnh SC, tnh di ca cnh BI theo a.

II. PHN RING. (3 im)1.Theo chng trnh chun.Cu IV a. (2 im) Trong khng gian vi h ta Oxyz , cho ba im A(1 ; 4 ; 0),

B(0 ; 2 ; 1), C(1 ; 0 ; -4).1/ Tm ta im D ABCD l hnh bnh hnh v tm ta tm ca hnhbnh hnh .

2/ Vit phng trnh ng thng (d) i qua trng tm ca tam gic ABC vvung gc vi mp(ABC).Cu V a. (1 im). Tnh th tch ca khi trn xoay to thnh khi quay quanh trctung hnh phng gii hn bi cc ng y = lnx, trc tung v hai ng thng y = 0,y = 1.2. Theo chng trnh nng cao.

Cu IV b. (2 im) Trong khng gian vi h ta Oxyz, cho hai ng thng d:1 2 3

2 1 1

= =

x y z

, d: 1 51 3

= = =

x t

y t

z t

1/ Chng minh d v d cho nhau.2/ Vit phng trnh mt phng (P) cha d v song song vi d.Tnh khang

cch gia d v d.Cu V b. (1 im). Tnh th tch khi trn xoay to thnh khi quay quanh trc hanhhnh phng gii hn bi cc ng y = lnx, y = 0, x = 2.

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I.PHN CHUNG CHO TT C TH SINH. (7im)Cu I.(3 im) Cho hm s y = x(x 3)2 c th (C).

1/ Kho st s bin thin v v th (C) ca hm s.2/ Vit phng trnh ng thng i qua hai im cc tr ca th hm s.

Cu II. (3 im)1/ Gii bt phng trnh: 2 22 2log 5 3log+ x x .

2/ Tnh I =2

2

0

sin 2 .

xdx .

3/ Tm gi tr ln nht v gi tr nh nht ca hm s y = x2e2x trn na khong(- ; 0 ]Cu III.(1 im). Cho hnh chp S.ABC c y ABC l tam gic vung ti A. BitAB = a, BC = 2a, SC = 3a v cnh bn SA vung gc vi y. Tnh th tch khi chpS.ABC theo a.II. PHN RING. (3 im)1.Theo chng trnh chun.Cu IV a. (2 im). Trong khng gian Oxyz, cho bn im A(1 ; -2 ; 2), B(1 ; 0 ; 0),C(0 ; 2 ; 0), D(0 ; 0 ; 3).

1/ Vit phng trnh mt phng (BCD). Suy ra ABCD l mt t din.2/ Tm im A sao cho mp(BCD) l mt phng trung trc ca an AA.

Cu V a. (1 im). Tnh th tch khi trn xoay to thnh khi quay quanh trc hanh

hnh phng gii hn bi cc ng y = sinx.cosx, y = 0, x = 0, x =2

.

2. Theo chng trnh nng cao.Cu IV b. (2 im). Trong khng gian vi h ta Oxyz, cho ng thng d:1 1

2 1 2

+= =

x y zv hai mt phng (P1): x + y 2z + 5 = 0, (P2): 2x y + z + 2 = 0.

1/ Tnh gc gia mp(P1) v mp(P2), gc gia ng thng d v mp(P1).2/ Vit phng trnh mt cu tm I thuc d v tip xc vi mp(P1) v mp(P2).

Cu Vb. (1 im). Tnh th tch khi trn xoay to thnh khi quay quanh trc tunghnh phng gii hn bi cc ng y = x2 v y = 6 - | x | .

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I.PHN CHUNG CHO TT C TH SINH.(7 im).Cu I. (3 im). Cho hm s y =

1x

xc th l (C).

1/ Kho st s bin thin v v th (C) ca hm s.2/ Tm m ng thng d: y = -x + m ct th (C) ti hai im phn bit.

Cu II.(3 im)1/ Gii phng trnh: 4x + 10x = 2.25x.

2/ Tnh I =9

24 ( 1)

dx

x x

3/ Tm gi tr ln nht v gi tr nh nht ca hm s y = .lnx x trnan [ 1; e ].Cu III.(1 im). Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, cnh

bn SA = a 3 v vung gc vi y.



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1/ Tnh th tch khi chp S.ABCD.2/ Chng minh trung im I ca cnh SC l tm ca mt cu ngai tip hnh

chp S.ABCD.II. PHN RING. (3 im)

1. Theo chng trnh chun.Cu IV a.(2 im). Trong khng gian vi h ta Oxyz,cho hai im A(2 ; 1 ; 1),B(2 ; -1 ; 5).

1/ Vit phng trnh mt cu (S) ng knh AB.2/ Tm im M trn ng thng AB sao cho tam gic MOA vung ti O.

Cu V a. (1 im). Gii phng trnh sau trn tp s phc : z4 1 = 0.2. Theo chng trnh nng cao.Cu IV b.(2 im). Trong khng gian vi h ta Oxyz, cho mt cu (S): x2 + y2 +z2 2x 4y 6z = 0 v hai im M(1 ; 1 ; 1), N(2 ; -1 ; 5).

1/ Tm tm I v bn knh R ca mt cu (S).Vit phng trnh mt phng (P)qua cc hnh chiu ca tm I trn cc trc ta .2/ Chng t ng thng MN ct mt cu (S) ti hai im. Tm ta cc giao

im .Cu V b.(1 im). Biu din s phc z = 1 i. 3 di dng lng gic.

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

Cu I. (3 im). Cho hm s y =4 21 5

32 2 +x x c th l (C).1/ Kho st s bin thin v v th (C) ca hm s.2/ Vit phng trnh tip tuyn ca (C) ti im M(1; 0).

Cu II. (3 im)

1/ Gii bt phng trnh:2

2 33 4

4 3

x x

. 2/ Tnh I =2

2

0

cos2

1 sin

+x

dxx

.

3/ Tm gi tr ln nht v gi tr nh nht ca hm s y = sin2x x trn an

;6 2

.

Cu III. (1 im). Cho hnh chp S.ABCD c y ABCD l hnh vung, cnh bn2aSA = v vung gc vi y, gc gia SC v y l 45 0 .Tnh th tch ca khi

chp.II. PHN RING. (3 im)1.Theo chng trnh chun.Cu IV a. (2 im).Trong khng gian vi h ta Oxyz, cho hai im A(3 ; 0 ; -2),B(1 ; -2 ; 4).

1/ Vit phng trnh ng thng AB v phng trnh mt phng trung trc caan AB.

2/ Vit phng trnh mt cu tm A v i qua im B. Tm im i xng ca Bqua A.Cu V a.(1 im). Tnh th tch ca khi trn xoay c to thnh khi quay quanhtrc tung hnh phng gii hn bi cc ng y = 2 x2 v y = | x | .

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2. Theo chng trnh nng cao.Cu IV b. (2 im) Trong khng gian vi h ta Oxyz, cho hai ng thng d:

1 1 2

2 3 4

+ = =

x y zv d:

2 2

1 3

4 4

= + = +

= +

x t

y t

z t

.

1/ Chng minh d song song vi d. Tnh khang cch gia d v d.2/ Vit phng trnh mt phng (P) cha d v d.

Cu V b.(1 im).Cho hm s y =2 3 6

2

+ ++

x x

x(1). Vit phng trnh ng thng

d i qua im A(2 ; 0) v c h s gc l k. Vi gi tr no ca k th ng thng dtip xc vi th ca hm s (1).


I.PHN CUNG CHO TT C TH SINH. (7 im).Cu I.(3 im). Cho hm s y = -x3 + 3x2 2 c th (C).

1/ Kho st s bin thin v v th (C) ca hm s.2/ Vit phng trnh tip tuyn vi (C) bit tip tuyn c h s gc k = -9.

Cu II.(3 im).1/ Gii phng trnh: 12 2log (2 1).log (2 2) 6++ + =x x

2/ Tnh I =2

0

sin2.

1 cos

+x

dxx

3/ Tm gi tr nh nht ca hm s y = x lnx + 3.Cu III. (1 im). Cho hnh chp S.ABC c SA, AB, BC vung gc vi nhau tngi mt. Bit SA = a, AB = BC = a 3 .Tnh th tch ca khi chp v tm tm ca mtcu ngai tip hnh chp.II. PHN RING. (3 im).1. Theo chng trnh chun.Cu IV a. (2 im). Trong khng gian vi h ta Oxyz, cho im A(2 ; -1 ; 3), mt

phng (P): 2x - y - 2z + 1 = 0 v ng thng d:1 2

2 1 3

= =

x y z

.

1/ Tm ta im A i xng ca A qua mp(P).

2/ Tm ta ca im M trn ng thng d sao cho khang cch t M nmp(P) bng 3.Cu V a.(1 im). Gii phng trnh sau trn tp s phc: z4 z2 6 = 02. Theo chng trnh nng cao.Cu IV b. (2 im). Trong khng gian vi h ta Oxyz, cho im A(1 ; 1 ; 1),

mp(P): x + y z 2 = 0 v ng thng d:2 1

1 1 1

= =

x y z

.

1/ Tm im A i xng ca A qua d.2/ Vit phng trnh ng thng i qua A, song song vi mp(P) v ct d.

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

2 42

2 4

5log log 8

5log log 19 = =

x y

x y

10



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http://www.VNMATH.comI.PHN CHUNG CHO TT C TH SINH. (7 im).Cu I.(3 im). Cho hm s y = (x 1)2(x +1)2 c th (C).

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

2/ Tm m ng thng d: y = m ct th (C) ti ba im phn bit.Cu II.(3 im)1/ Gii phng trnh: log(x 1) log(x2 4x + 3) = 1.

2/ Tnh I =3

1

(1 ln ).

+e x

dxx .

3/ Cho hm s y = x3 (m + 2)x + m ( m l tham s). Tm m hm s c cctr ti x = 1.Cu III.(1 im). Cho hnh lng tr ABC. ABC c y l tam gic u cnh a, cnh

bn bng a 3 v hnh chiu ca A ln mp(ABC) trng vi trung im ca BC.Tnh

th tch ca khi lng tr .II. PHN CHUNG. (3 im)1. Theo chng trnh chun.Cu IV a.(2 im). Trong khng gian vi h ta Oxyz, cho hai im A, B c ta xc nh bi cc h thc 2 , 4 4

= = uuur uuurOA i k OB j k v mt phng (P): 3x 2y + 6z + 2

= 0.1/ Tm giao im M ca ng thng AB vi mp(P).2/ Vit phng trnh hnh chiu vung gc ca AB trn mp (P).

Cu V a.(1 im). Tnh th tch khi trn xoay tao thnh khi quay quanh trc Ox hnh

phng gii hn bi cc ng y =1

2

+xx , y = 0, x = -1 v x = 2.

2/ Theo chng trnh nng cao.Cu IVb. (2 im). Trong khng gian vi h ta Oxyz, cho ng thng d:

1 2

2

= + = =

x t

y t

z tv mt phng (P): x + 2y 2z + 3 = 0.

1/ Vit phng trnh ng thng i qua gc ta O vung gc vi d v songsong vi (P).

2/ Vit phng trng mt cu c tm thuc d, tip xc (P) v c bn knh bng 4.

Cu Vb.(1 im). Tnh ( )83 + i

11I/_ Phn dnh cho tt c th sinh

Cu I ( 3 im) Cho hm s ( )1

11

+=

x

yx

c th l (C)

1) Kho st hm s (1)2) Vit phng trnh tip tuyn ca (C) bit tip tuyn i qua im P(3;1).

Cu II ( 3 im)1) Gii bt phng trnh: 2.9 4.3 2 1+ + >x x

2) Tnh tch phn:1

5 3

0

1= I x x dx

3) Tm gi tr ln nht v gi tr nh nht ca hm s2 1+ +

=x x

yx

vi 0>x



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Cu III (1 im). Xc nh tm v bn knh mt cu ngoi tip mt hnh lng tr tamgic u c 9 cnh u bng a.II/_Phn ring (3 im)

1) Theo chng trnh chun

Cu IV. a (2 im) Trong khng gian cho h ta Oxyz, im A (1; -1; 1) v haing thng (d1) v (d2) theo th t c phng trnh:

( ) ( )1 23 3 0

: 1 2 ; :2 1 0

3

= + = = + = =

x tx y z

d y t d x y

z t

Chng minh rng (d1), (d2) v A cng thuc mt mt phng.Cu V. a (1 im) Tm mun ca s phc ( ) 22 2= + z i i

2) Theo chng nng cao.Cu IV. b (2 im) Trong khng gian cho h ta Oxyz, cho mt phng ( ) ( ) v ln

lt c phng trnh l: ( ) ( ): 2 3 1 0; : 5 0 + + = + + =x y z x y z v im M (1; 0; 5).1. Tnh khong cch t M n ( )2. Vit phng trnh mt phng i qua giao tuyn (d) ca ( ) ( ) v ng thi

vung gc vi mt phng (P): 3 1 0 + =x yCu V. b (1 im) Vit dng lng gic ca s phc 1 3= +z i

12

I. Phn chung cho tt c th sinh (7,0 im)

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

3 3= + +y x mx x m ( )mC

1. Kho st s bin thin v v th ( C) ca hm s khi m =0.2.Tm im c nh ca th hm s ( )mC .

Cu II.(3,0 im)

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

4 28 16= +y x x

trnon [ -1;3].

2.Tnh tch phn7 3

3 20 1

=+

x

I dxx

3. Gii bt phng trnh0,5

2 12

5log +

+x

x

Cu III.(1,0 im)Cho t din S.ABC c SA vung gc vi mt phng (ABC), SA = a; AB = AC=

b, 60=BAC . Xc nh tm v bn hnh cu ngoi tip t din S.ABC.II.Phn ring(3,0 im)

Th sinh hc chng trnh no th ch c lm phn dnh ring cho chng trnh .1. Theo chng trnh Chun:Cu IV.a(2,0 im) Trong khng gian vi h to Oxyz:



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a)Lp phng trnh mt cu c tm I(-2;1;1) v tip xc vi mt phng2 2 5 0+ + =x y z

b) Tnh khong cch gia hai mt phng:0124801224 ==+ zyxvzyx

Cu V.a(1,0 im) Gii phng trnh :4 2

3 4 7 0+ =z z trn tp s phc.2.Theo chng trnh nng cao.Cu IV.b(2,0 im)

Trong khng gian vi h to Oxyz,cho ng thng d c phng trnh:1 1

2 1 2

+= =

x y zv hai mt phng 052:)( =++ zyx v 022:)( =++ zyx . Lp phng

trnh mt cu tm I thuc ng thng d v tip xc vi c hai mt phng ( ) ( ), .Cu V.b(1 im)Tnh din tch hnh phng gii hn bi h cc hm s

, 2 , 0= = =y x y x y

13

I. Phn chung cho tt c th sinh (7,0 im)Cu I.( 3,0 im)

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

3

+=

x

yx

2.Tm trn th im M sao cho khong cch t M n ng tim cn ngbng khong cch t M n tim cn ngang.Cu II.(3,0 im)

1. Gii phng trnh2 1

3 .5 7 245

=

x x x

. 2.Tnh tch phn a) 11 ln+

=

e x

I dxx Cu III.(1,0 im)Mt hnh tr c thit din qua trc l hnh vung, din tch xung quanh l 4 .

1.Tnh din tch ton phn ca hnh tr.2. Tnh th tch ca khi tr.

II.Phn ring(3,0 im)Th sinh hc chng trnh no th ch c lm phn dnh ring cho chng trnh .1. Theo chng trnh Chun:Cu IV.a(2,0 im) Trong khng gian vi h to Oxyz:cho A(1;0;0), B(1;1;1),

1 1 1; ;3 3 3

C

a)Vit phng trnh tng qut ca mt phng ( ) i qua O v vung gc vi OC.b) Vit phng trnh mt phng ( ) cha AB v vung gc vi ( ) CuV.a(1,0 im)

Tm nghim phc ca phng trnh 2 2 4+ = z z i

14

I. PHN CHUNG CHO TT C CC TH SINHCu 1 (4,0 im):1. Kho st v v th (C) ca hm s 3 23= y x x2. Da vo th (C) bin lun theo m s nghim ca phng trnh



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3 23 0 + =x x m

3. Tnh din tch hnh phng gii hn bi th (C) v trc honh.Cu 2 ( 2,0 im)

1. Gii phng trnh: 23 5.3 6 0 + =x x

2. Gii phng trnh:2

4 7 0 + =x x

Cu 3 (2,0 im)Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, cnh bn SB vung

gc vi y, cnh bn SC bng 3a .1. Tnh th tch ca khi chp S.ABCD.2. Chng minh trung im ca cnh SD l tm mt cu ngoi tip hnh chp

S.ABCD.II. PHN DNH CHO TNG TH SINHA. Dnh cho th sinh Ban c bn:

Cu 4 (2,0 im)1.Tnh tch phn:

1

0

( 1).= + xI x e dx

2. Trong khng gian vi h ta Oxyz cho ba im A(5;0;4), B(5;1;3),C(1;6;2), D(4;0;6)a. Vit phng trnh tham s ca ng thng AB

b. Vit phng trnh mt phng ( ) i qua im D v song song vi mt phng (ABC).B. Dnh cho th sinh Ban nng cao

Cu 5 (2,0 im)

1. Tnh tch phn:

2

32 3

1 1= +I x x dx2. Trong khng gian vi h ta Oxyz, cho im M(1;2;3) v mt phng (P)

c phng trnh: x - 2y + z + 3 = 0a. Vit phng trnh mt phng (Q) i qua im M v song song vi mtphng (P).

b. Vit phng trnh tham s ca ng thng (d) i qua im M v vung gcvi mt phng (P). Tm ta giao im H ca ng thng (d) vi mt phng(P)

15I . PHN CHUNG CHO TT C TH SINH ( 7 im )Cu 1 ( 3 im )

Cho hm s y =4

2x 5- 3x +2 2

(1)

1. Kho st v v th hm s (1).2. Vit phng trnh tip tuyn ti im c honh x = 1

Cu 2 ( 3 im )

1. Tnh tch phn ( )1+

13

2

0

I = 2x xdx

2. Tm gi tr ln nht v gi tr nh nht ca hm sy = 3 22 4 2 2 + + x x x trn [ 1; 3] .



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3. Gii phng trnh: 16 17.4 16 0 + =x x

Cu 3 ( 1 im )Cho khi chp S.ABC c ng cao SA= a, (a > 0 ) v y l tam gic u. Gc giamt bn (SBC) v mt dy bng 600 . Tnh th tch ca ca khi chp S.ABC theo a.

II. PHN RING (3 im)1. Theo chng trnh Chun:

Cu 4. a ( 2 im)Trong khng gian vi h to Oxyz cho A(2 ; 0; 0) , B( 0; 4; 0 ) v C(0; 0; 4).

1.Vit phng trnh mt cu qua 4 im O, A, B, C. Xc nh to tm I vtnh bn knh R ca mt cu.2.Vit phng trnh mt phng ( ABC) v ng thng d qua I vung gc vi(ABC).

Cu 4. b (1 im ) Tm s phc z tho mn 5=z v phn thc bng 2 ln phn o ca

n.Theo chng trnh nng cao:Cu 4. a ( 2 im) Trong khng gian vi h to Oxyz, cho 2 ng thng c

phng trnh

11

: 1

2

= + = =

x t

y t

z 2

3 1:

1 2 1

= =

x y z

1.Vit phng trnh mt phng qua ng thng 1 v song song vi ng thng 2

2.Xc nh im A trn 1 v im B trn 2 sao cho AB ngn nht .Cu 4. b (1 im ) Gii phng trnh trn tp s phc: 2z2 + z +3 = 0

16ht tp://VNMATH.com

I . PHN CHUNG CHO TT C TH SINH ( 7 im )Cu 1 ( 3 im )

Cho hm s y = 4 2x + 2(m+1)x + 1 (1)1. Kho st v v th hm s (1) khi m = 1.2. Tm m hm s c 3 cc tr.Cu 2 ( 3 im )

1. Tnh tch phn ( )1+1

32

0

I = 4x .xdx

3. Tm gi tr ln nht v gi tr nh nht ca hm sy = 3 22 4 2 1 + +x x x trn [ 2; 3] .3. Gii phng trnh: 2 33.2 2 2 60+ ++ + = x x x

Cu 3 ( 1 im )Cho khi chp S.ABC c y l tam gic u cnh a, (a >0). Tam gic SAC cn tiS gc SAC bng 600 ,(SAC) (ABC) . Tnh th tch ca ca khi chp S.ABC theoa.II. PHN RING (3 im)

2. Theo chng trnh Chun:

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Cu 4. a ( 2 im)Trong khng gian vi h to Oxyz cho A(2 ; 4; -1) , B( 1; 4; -1 ) , C(2; 4; 3) vD(2; 2; -1).1.CMR AB AC, AC AD, AD AB . Tnh th tch ca t din ABCD.2.Vit phng trnh mt cu qua 4 im A, B, C, D. Xc nh to tm I v tnh

bn knh R ca mt cu.Cu 4. b (1 im )

Tnh T =5 6

3 4

+

i

itrn tp s phc.

Theo chng trnh nng cao:Cu 4. a ( 2 im)Trong khng gian vi h to Oxyz cho A(4 ; 3; 2) , B( 3; 0; 0 ) , C(0; 3; 0) vD(0; 0; 3).

1. Vit phng trnh ng thng i qua A v G l trng tm ca tam gic BCD.2.Vit phng trnh mt cu tm Av tip xc (BCD).Cu 4. b (1 im )

Cho s phc 1 32 2

= +z i , tnh z2 + z +3

17http://VNMATH.com

I . PHN CHUNG CHO TT C TH SINH (7 im)Cu I.(3 im)Cho hm s 3 3 2= + y x x

1. Kho st s bin thin v v th hm s cho.2. Bin lun theo m s nghim ca phng trnh 3 3 2 + =x x m

Cu II.(3 im)1. Gii phng trnh:

12

3 63 3 80 0

=x x

2. Tnh nguyn hm: ln(3 1) x dx3. Tm gi tr ln nht v nh nht hm s 3 2( ) 3 9 3= + +f x x x x trn on [ ]2;2

Cu 3.(1 im)Cho t din S.ABC c ba cnh SA, SB, SC i mt vung gc v SA=a, SB=b,

SC=c. Hai im M, N ln lt thuc 2 cnh AB, BC sao cho1 1

,3 3

= =AM AB BN BC . Mt

phng (SMN) chia khi t din S.ABC thnh 2 khi a din (H) v (H) trong (H)l khi a din cha nh C. Hy tnh th tch ca (H) v (H)II . PHN RING (3 im) :

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



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1. Hy tm ta ca hnh chiu vung gc ca A trn mt phng (P).

2. Vit phng trnh ca mt cu tm A, tip xc vi (P).

Cu V.a(1 im)Tnh th tch khi trn xoay c to bi php quay quanh trc Ox

hnh phng gii hn bi cc ng2

2 1, 0, 2, 0= + = = =y x x y x x .2.Theo chng trnh nng cao :Cu IV.b(2 im)

Cho mt phng (P): 2x+y-z-3=0 v ng thng (d):2 3

1 2 2

+ += =

x y z

1. Tm ta giao im M ca ng thng (d) v mt phng (P).2. Vit phng trnh hnh chiu ca ng thng (d) trn mt phng (P).Cu Vb. (1 im)

Xc nh ta giao im ca tim cn xin ca th hm s2 3 1

2

+=

x x

yx

vi

parabol (P):2

3 2= +y x x 18

http://www.VNMATH.comCu I:(3 im):

1/Kho st s bin thin v v th (C ) ca hm s y=1

1

+

x

x

2/Vit phng trnh tip tuyn vi(C) ti giao im ca ( C) vi trc tungCu II:(3im)

1/Tnh I= ( )cos0

sin

+ xe x xdx

2/Gii bt phng trnh log 3 ( )2+x log 9 ( )2+x3/Tnh cc cnh ca hnh ch nht c chu vi nh nht trong tt c cc hinh ch nht

c din tch 48m 2

Cu III: (2im)Trong khng gian Oxyz cho 3 im A(2;2;3) ;B(1;2;-4) ;v C(1;-3;-1)1/Vit phng trnh mt phng ABC

2/Vit phng trnh mt cu ngoi tip t din OABC.Tm ca mt cu ctrng vi trng tm ca t din khng?

Cu IV:(1 im)Cho hnh chp t gic u S.ABCD c cnh y bng a;gc SAB bng 30 0

.Tnh din tch xung quanh ca hnh nn nh S, y l hnh trn ngoi tip t gicABCD

Cu V: (1 im)Tnh2 15

3 2

+

i

i


I . PHN CHUNG CHO TT C TH SINH ( 7 im )Cu I ( 3,0 im )

Cho hm s 3 23 1= + xy x c th (C)1. Kho st s bin thin v v th (C).

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2. Dng th (C), xc nh k phng trnh 3 23 0 + =xx k c ng 3 nghimphn bit.

Cu II ( 3,0 im )1. Gii phng trnh: 4.9 12 3.16 0. ( )+ = x x x x

2. Tnh tch phn:2 2

30 1

=+

xI dxx

.

3. Tm gi tr ln nht, gi tr nh nht ca hm s: 24 4 .= + y xCu III ( 1,0 im )

Cho hnh chp S.ABCc yABCl tam gic vung ti A, , 3,= =AB a AC a mt bnSBCl tam gic u v vung gc vi mt phng y. Tnh theo a th tch ca khichp S.ABC.

II . PHN RING ( 3 im )1. Theo chng trnh chun :

Cu IV.a ( 2,0 im ) :Trong khng gian vi h ta Oxyz , cho ng thng (d):

2 3

1 2 2

+ += =

x y z

v mt

phng(P): 2 2 6 0+ + = x y z .1. Vit phng trnh mt cu tm (1; 2; 3)I v tip xc vi mt phng (P).2. Vit phng trnh mt phng ( ) cha ng thng (d) v vung gc vi mt

phng (P).Cu V.a ( 1,0 im ) :

Tnh mun ca s phc3(1 2 )

3

+=

i

zi

.

2. Theo chng trnh nng cao :Cu IV.b ( 2,0 im ) : Trong khng gian vi h ta Oxyz , cho ng thng (d):

2 3

1 2 2

+ += =

x y z

v mt phng (P): 2 2 6 0+ + =x y z .

1. Vit phng trnh mt cu tm (1; 2; 3)I v tip xc vi mt phng (P).2. Vit phng trnh hnh chiu vung gc ca ng thng (d) trn mt phng

(P).Cu V.b ( 1,0 im ) : Tm cn bc hai ca s phc 4= z i

20

http://www.VNMATH.comCu 1 : Cho hm s 3 3 2= +y x x (C)a.Kho st v v th hm s (C)

b.Da vo (C) bin lun theo m s nghim phng trnh : 3 3 1 0 + = x x mc.Tnh din tch hnh phng gii hn bi (C ) v trc Ox .

Cu 2 :a)Tnh o hm ca hm s sau : 4 2 os(1-3x)+= xy e c ; y = 5cosx+sinx

b) Tm GTLN, GTNN ca hm s 4 21

( ) 24

= +f x x x trn on [-2 ;0]

c) Tnh gi tr biu thc A = 9 21 log 4 2 log 3(3 ) : (4 )+

d) Gii cc phng trnh, bt phng trnh sau : 2 4 16log log log 7+ + =x x x

e) tnh cc tch phn sau : I =2

2

1

1+x x dx ; J =2

3

3

2cos 3

3

x dx



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Cu 3 : Tnh din tch xung quanh v th tch khi chp t gic u c di cnh bngp i cnh y v bng a ?Cu 4/ Cho 2 im A (0; 1; 2) v B (-3; 3; 1)

a/ Vit phng trnh mt cu tm A v i qua B

b/ Vit phng trnh tham s ca ng thng (d ) qua B v song song vi OAc/ Vit phng trnh mt phng ( OAB)

Cu 5/ a/ Gii phng trnh sau trong tp tp s phc : x2 x + 1 = 0b/ Tm moun ca s phc Z = 3 2i

21

Cu 1 : a)Kho st v v th hm s: y =2

2 1

+

x

x th (C)

b)Vit phng trnh tip tuyn ca (C) ti im c honh bng -1c.) Tnh din tch hnh phng gii hn bi (C) ; tim cnh ngang ; x=0 ; x=1

Cu2 : a) Tm GTLN GTNN ca hm s y = (x 6)2

4+x trn on [0 ; 3].b)Tm m hm s: y =

3

3

x- (m + 1)x2 + 4x + 5 ng bin trn R

c)Tnh o hm cc hm s sau:

a/ ( ) 21= xy x e b/ y = (3x 2) ln2x c/ ( )2ln 1+

=x

yx

d) tnh cc tch phn : I = ( )2

2

1

ln+e

x x xdx ; J =1

2

02+

dx

x x

e) Gii phng trnh :a) 2 2log ( - 3) +log ( - 1) = 3x x b) 3.4 21.2 24 0 =x x

Cu 3 : Thit din ca hnh nn ct bi mt phng i qua trc ca n l mt tam gicu cnh a

Tnh din tch xung quanh; ton phn v th tch khi nn theo a ?Cu 4 : Trong khng gian Oxyz

a) Cho 4 3= +a i j , b = (-1; 1; 1). Tnh1

2=

r r rc a b

b) Cho 3 im A(1; 2; 2), B(0; 1; 0), C(0; 0; 1)+ Tnh AB . AC

+ Chng minh A, B, C khng thng hng. Vit phng trnh mt phng

( ABC ).+ Vit phng trnh mt cu tm I ( -2;3;-1) v tip xc (ABC)Cu 5 : a/ Gii phng trnh : (3-2i)x + (4+5i) = 7+3i

b/ Tm x;y bit : (3x-2) + (2y+1)i = (x+1) (y-5)i .

22Cu1: Cho hm s y = x3 - 3x2 + 2 (C)

a).Kho st s bin thin v v th hm s.b).Tm gi tr ca m phng trnh : -x3 + 3x2 + m = 0 c 3

nghim phn bit.c) .Tnh din tch hnh phng gii hn bi (C); Ox ; Oy ; x=2.Cu 2: a)Tm gi tr ln nht, gi tr nh nht ca hm s: y = x+

21x



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b) nh m hm s: y = x3 + 3mx2 + mx c hai cc tr .c) Cho hm s f(x) = ln 1+ xe . Tnh f(ln2)d) Gii phng trnh , Bt phng trnh: 9x - 4.3x +3 < 0

e)2

2

0( sin )cos

= +E x x xdxCu 3 : Cho hnh chp t gic S.ABCD c y l hnh vung cnh a ,cnh bn SA vung gc vi y, cnh bn SC to vi y mt gc30o .

a) Tnh din tch xung quanh v th tch khi chp.b) Tm tm v bn knh mt cu ngoi tip hnh chp.

Cu 4: Trong khng gian cho hai ng thng (d1) v (d2) c phng trnh: (d1)2 1

2( )

3 1

= + = + =

x t

y t t R

z t

2)2

1 2 ( )

1

= + = + = +

x m

y m m R

z m

a. Chng t d1 v d2 ct nhaub. Vit phng trnh mt phng (p) cha (d1)v (d2)

c. Vit phng trnh mt cu ng knh OH vi H l giaoim ca hai ng thng trnCu 5 : a. Tm nghch o ca z = 1+2i

b. Gii phng trnh : (3+2i)z = z -1

23A. P hn chung cho th sinh c hai banCu 1: Cho hm s: 3 23 4= + y x x . Vi m l tham s.

1. Kho st v v th ( C ) ca hm s.2. Bin lun theo m s nghim ca phng trnh: 3 23 2 1 0+ + + =x x m

Cu 2: Gii h phng trnh sau: 12 3 0

5 5 10 + =

+ =

x y

x y

Cu 3: Tm phn thc v phn o ca s phc sau:

2 2(1 ) (2 1)

1

+ = +

+i i

zi i

Cu 4: Tnh th tch ca khi lng tr ng c y l tam gic u cnh a, gc giang cho mt bn v y l 30 .B. P hn ring cho th sinh tng banTh sinh ban khoa hc t nhin lm cu 5a hoc 5bCu 5a:

1. Tnh tch phn:2

0

3cos 1sin

= +

I x xdx

2. Tm m hm s:2 2 4

2

+ =

+x mx m

yx

c 2 cc tr nm cng mt pha so vi trc

honh.



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Cu 5b:Trong h to Oxyz cho cc im A(0,1,2), B(2,3,1), C(2,2,-1). Lp phngtrnh mt phng i qua A,B,C.Chng minh rng im O cng nm trn mt phng v OABC l hnh ch nht. Tnh th tch khi chp SOABC bit rng S(0,0,5)Th sinh ban khoa hcx hi lm cu 6a hoc 6b

Cu 6a:1. Tnh tch phn: 2

1

( 1) ln= +e

I x xdx

2. Tm m hm s: 4 218 5 2008= y x mx c 3 cc tr .Cu 6b:Trong h to Oxyz cho cc im: A(0,1,1), B(1,2,4), C(-1,0,2). Hy lp

phng trnh mt phng (Q) i qua A,B,C.Lp phng trnh tham s ca ng thngi qua B v M vi M l giao im ca mt phng (Q)( vi trc Oz.

24

I. Phn chung:Cu I: (3) Cho hm s y = x3 3x1) Kho st s bin thin v v th (C) ca hm s2) Da vo th (C), bin lun theo m s nghim ca phng trnh : x3 3x +

m = 0Cu II : (3)

1) Gii phng trnh : lg2x lg3x + 2 = 0

2) Tnh tch phn : I =/ 2

0

osxdx

xe c

3) Cho hm s f(x) = x3 + 3x2 + 1 c th (C). Vit phng trnh tip tuynca (C) i qua gc ta .

Cu III : (1) Cho hnh chp t gic u, tt c cc cnh u bng a. Tnh th tch hnhchp S.ABCDII. Phn ring : (3)Chng trnh chun :Cu IVa: Trong khng gian Oxyz cho 4 im A(3 ;-2 ; -2), B(3 ;2 ;0),C(0 ;2 ;1), D(-1;1;2)

1) Vit phng trnh mt phng (BCD). Suy ra ABCD l 1 t din2) Vit phng trnh mt cu tm A tip xc vi mt phng (BCD)

Cu Va : Gii phng trnh : x2 + x + 1 = 0 trn tp s phcChng trnh nng cao :

Cu VIb: Cho 2 ng thng d1 :4

3

4

= + = =

x t

y t

z, d2 :

2

1 2 '

'

= = + =

x

y t

z t

1) Tnh on vung gc chung ca 2 ng thng d1 v d2

2) Vit phng trnh mt cu c ng knh l on vung gc chung ca d1 vd2

Cu Vb: Gii phng trnh: x2 + (1 + i)x ( 1 i) = 0 trn tp s phc

25



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I/ PHN CHUNG : (7im)C u I : (3 im)

Cho hm s Cho hm s y = (x 1)2 (4 x)1/ Kho st v v th (C) ca hm s. Vit phng trnh tip tuyn ca th

(C) ti A(2;2).2/ Tm m phng trnh: x3 6x2 + 9x 4 m = 0, c ba nghim phn bit.

C u II : ( 3 im)

1/ Tnh tch phn: I =3

0

(cos 4 .sin 6 )

x x x dx 2/ Gii phng trnh: 4x 6.2x+1 + 32

= 03/ Tm tp xc nh ca hm s: y = 31 log ( 2) x

C u III : (1 im)

Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, mt bn SAB ltam gic u v vung gc vi y. Gi H l trung im AB. Chng minh rng: SHvung gc mt phng (ABCD). Tnh th tch khi chp S.ABCD theo a.II/ PHN RING: (3im)1. Theo chng trnh chun:C u IV.a : (2 im)

Trong khng gian Oxyz cho mt cu (S): x2 + y2 + z2 2x 4y 6z = 0.1/ Xc nh tm v bn knh ca mt cu (S).2/ Gi A ; B ; C ln lt l giao im (khc gc to O) ca mt cu (S) vi

cc trc Ox ; Oy ; Oz. Tm to A ; B ; C. Vit phng trnh mt phng (ABC).C u V.a : (1im)

Gii phng trnh sau trn tp s phc: z2 + 4z + 10 = 02. Theo chng trnh nng cao:C u IV.b : (2 im)

Trong khng gian Oxyz cho ng thng (D):2 1 1

2 3 5

+ = =

x y zv mt phng (P):

2x + y + z 8 = 0.1/ Chng t ng thng (D) khng vung gc mp (P). Tm giao im ca

ng thng (D) v mt phng (P).

2/ Vit phng trnh ng thng (D) l hnh chiu vung gc ca ngthng (D) ln mt phng (P).C u V.b : (1im)

Gii phng trnh sau trn tp s phc: (z + 2i)2 + 2(z + 2i) 3 = 0.


PHN CHUNG CHO TT C CC TH SINH (7):Cu I (3):

1. Kho st v v th (C) ca hm s3

1

+

= +

x

y x

2. CMR vi mi gi tr ca m, ng thng (d) y = 2x + m lun ct (C) ti 2 imphn bit.



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3. Gi A l giao im ca (C) vi trc Ox. Vit phng trnh tip tuyn ca (C) tiA.Cu II (3): 1. Gii phng trnh: 32 log3 81 =x x

1) Tm gi tr ln nht v gi r nh nht ca hm s: y = 2sin2x + 2sinx 1Cu III (1):

Cho t din SABC c cnh SA vung gc vi mt phng (ABC) v c SA = a,AB = b, AC = c v 090=BAC . Tnh din tch mt cu v th tch khi cu ngoi tip tdin SABC.PHN RING (3):1.Theo chng trnh chun:Cu IV.a (2):

Trong khng gian Oxyz. Cho im M(-3;1;2) v mt phng (P) c phngtrnh: 2x + 3y + z 13 = 0

1) Hy vit phng trnh ng thng (d) i qua M v vung gc vi mt phmg(P). Tm ta giao im H ca ng thng (d) v mt phng (P).

2) Hy vit phng trnh mt cu tm M c bn knh R = 4. Chng t mt cu nyct mt phng (P) theo giao tuyn l 1 ng trn.

Cu V.a (1):Tnh din tch hnh phng gii hn bi cc ng (P): y = 4 x2, (d): y = -x + 2

2.Theo chng trnh Nng cao:Cu IV.b (2):

Trong khng gian Oxyz cho 4 im A(-2;1;2), B(0;4;1), C(5;1;-5), D(-2;8;-5) vng thng d:

5 11 9

3 5 4

+ + = =

x y z

.

1) Vit phng trnh mt cu (S) ngoi tip t din ABCD.

2) Tm ta giao im M, N ca (d) vi mt cu (S).

3) Vit phng trnh cc mt phng tip xc vi mt cu (S) ti M,N

Cu V.b (1): Tnh din tch hnh phng gii han bi cc ng (P): y = x2 + 1, tip

tuyn ca (P) ti M(2;5) v trc Oy


CuI: ( 3 im)1/Kho st s bin thin v v th(C ) ca hm s y= -x 3 +3x 2 -3x+2.2/Tnh din tch hnh phng gii hn bi (C ) v 2 trc ta .

Cu II: (3 im)1/Cho hm s y= xsinx .Chng minh rng : xy-2 ( )' siny x +xy=0

2/Gii phng trnh: log3 ( )3 1

x

.log3 ( )

13 3+ x

= 6.3/Tnh I=

3

3 2

0

1+x x dx

Cu III( 2 im)



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Trong khng gian Oxyz cho 2 mt phng( ) v ( ' ) c phng trnh: ( ) :2x-y+2z-1=0 v ( ):x+6y+2z+5=0

1/Chng t 2 mt phng cho vung gc vi nhau.2/Vit phng trnh mt phng( ) i qua gc ta v giao tuyn ca 2 mt

phng( ) , ( ' )Cu IV: (1 im):Cho khi hp ABCD.ABCD c th tch 2009 cm 3 .Tnh th tch khi t dinCABC

Cu V:( 1 im) Tnh mun ca s phc z bit Z = ( )2 3 i1

32

+

i


I.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 3 22 3 2= + y x x c th (C)

1. Kho st s bin thin v v th (C).2. Vit phng trnh tip tuyn ca (C) ti im c honh 2= ox .Cu 2 ( 3,0 im )

1. Gii phng trnh 13 18.3 29+ + =x x . 2. Tnh tch phn2

0

cos

= I x xdx

3. Tm GTLN, GTNN ca hm s 29 7= y x trn on [-1;1].

Cu 3 ( 1,0 im ) Cho t din u ABCD c cnh bng2

a

1. Tnh chiu cao ca t din ABCD.2. Tnh th tch ca t din ABCD.

II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )Cu 4a ( 2,0 im ) Cho bn im A(1;0;0), B(0;1;0), C(0;0;1), D(-2;1;-1)

1. Chng minh A, B, C, D l bn nh ca mt t din.2. Tnh th tch ca t din .3. Lp phng trnh mt cu ngoi tip t din ABCD.

Cu 5a ( 1,0 im ) Gii phng trnh 2 7 0+ + =x x trn tp s phc.

29http://VNMATH.com

I. PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 3 23 4= + y x x c th (C)1.Kho st s bin thin v v th (C).

2.Vit phng trnh tip tuyn ca (C) ti tm i xng.Cu 2 ( 3,0 im )1.Gii phng trnh 6 33. 2 0 + =x xe e .

http://www.vnmath.com/http://www.vnmath.com/http://www.vnmath.com/http://vnmath.com/http://www.vnmath.com/http://www.vnmath.com/http://vnmath.com/http://www.vnmath.com/http://www.vnmath.com/


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2.Tnh tch phn2

2

0

sin2 .sin

= I x xdx

3.Tm GTLN, GTNN ca hm s 3 22 3 12 10= +y x x x trn on [-3;3].

Cu 3 ( 1,0 im )Cho hnh chp tam gic u S.ABC c cnh y bng2

a, cnh bn bng a

1.Tnh chiu cao ca hnh chp S. ABC.2.Tnh th tch ca hnh chp S.ABC.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho mt cu (S) c ng knh AB, bit A(6;2;-5), B(-4;0;7).1. Lp phng trnh mt cu (S).2. Lp phng trnh mt phng (P) tip xc mt cu (S) ti im A.



I. PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 3 23 4= + y x x c th (C)

1.Kho st s bin thin v v th (C).2.Dng th (C), bin lun theo m s nghim ca phng trnh

3 23 4 + = +x x m .Cu 2 ( 3,0 im )

1.Gii phng trnh94log log 3 3+ =xx

.2.Tnh tch phn

1

0

ln(1 )= +I x dx

3.Tm GTLN, GTNN ca hm s 5 4= y x trn on [-1;1].Cu 3 ( 1,0 im )

Cho hnh chp S.ABCD c y ABCD l hnh ch nht, cnh bn SA vunggc vi mt phng y. SA = 3a, SB = 5a, AD = a

1.Tnh di AB.2.Tnh th tch ca hnh chp S.ABCD.

II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )Cu 4a ( 2,0 im ) Cho bn im A(-2;6;3), B(1;0;6), C(0;2;-1), D(1;4;0)

1. Vit phng trnh mt phng (BCD). Suy ra ABCD l mt t din.2. Tnh chiu cao AH ca t din ABCD.3. Vit phng trnh mt phng (Q) cha AB v song song vi CD.


31I.PHN CHUNG (7,0 im )

Cu 1 ( 3,0 im ) Cho hm s 3 23 1= + +y x x c th (C)1.Kho st s bin thin v v th (C).2.Vit phng trnh tip tuyn ca (C) ti im c honh 2= ox .



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Cu 2 ( 3,0 im )

1.Gii bt phng trnh2 4 6

1 1

3 27

+

x x

.

2.Tnh tch phn

2

1

ln=

e

I x xdx

3.Tm GTLN, GTNN ca hm s1

=x

yx

trn on [-2;-1].

Cu 3 ( 1,0 im )Cho hnh chp S.ABCD c y ABCD l hnh bnh hnh.

( )SA ABCD .SA =2

a, AB = 2a, AD = 5a, gc BAD c s o 30o

Tnh th tch ca hnh chp S.ABCD.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho mt phng ( ) : 3 5 2 0 + =x y z v ng thng12 4

( ) : 9 3

1

= + = + = +

x td y t

z t.

1. Tm giao im M ca ng thng (d) v mt phng ( ) .2. Vit phng trnh mt phng ( ) cha im M v vung gc vi ng thng

(d).Cu 5a ( 1,0 im ) Gii phng trnh 2 2 7 0+ + =x x trn tp s phc.

32

I.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 3 23 1= + +y x x c th (C)1.Kho st s bin thin v v th (C).2.Vit phng trnh tip tuyn ca (C) ti im c honh 1= ox .Cu 2 ( 3,0 im )1.Gii phng trnh log( 1) log(2 11) log 2 =x x .

2.Tnh tch phnln 3

30 ( 1)

=+

x

x

eI dx

e

3.Tm GTLN, GTNN ca hm s 3 21

2 3 43

= + + y x x x trn on [-4;0].

Cu 3 ( 1,0 im )

Cho hnh chp t gic u S.ABCD c cnh y bng2

a, cnh bn bng 3a

1.Tnh chiu cao ca hnh chp S.ABCD.2.Tnh th tch ca hnh chp S.ABCD.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho hai ng thng 11

( ) : 2 2

3

= = + =

x t

d y t

z tv

/

/

2

1

( ) : 3 2

1

= + = =

x t

d y t

z.

Chng minh rng (d1) v (d2) cho nhau.

Cu 5a ( 1,0 im ) Gii phng trnh 22 3 7 0+ + =x x trn tp s phc.



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33I. PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 3 23 4= + y x x c th (C)

1.Kho st s bin thin v v th (C).2.Vit phng trnh tip tuyn ca (C) ti im c ta ( 1; 2) .Cu 2 ( 3,0 im )1.Gii phng trnh 16 17.4 16 0 + =x x .

2.Tnh tch phn2

3

2

2

( 1) = x xI x e dx


= +y xx

trn khong ( 0 ; + ).

Cu 3 ( 1,0 im )

Cho hnh chp S.ABCD c y ABCD l hnh ch nht. Cnh bn SA vunggc vi mt phng y. SB = 5a, AB = 3a , AC= 4a.1.Tnh chiu cao ca S.ABCD.2.Tnh th tch ca S.ABCD.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho mt cu 2 2 2( ) : 10 2 26 170 0+ + + + + =S x y z x y z .

1. Tm to tm I v di bn knh r ca mt cu (S).2. Lp phng trnh ng thng (d) qua im I vung gc vi mt phng

( ) : 2 5 14 0 + =x y z .Cu 5a ( 1,0 im ) Gii phng trnh

22 4 7 0 + =x x trn tp s phc.

34I.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 3 26 9= +y x x x c th (C)1.Kho st s bin thin v v th (C).2.Vit phng trnh tip tuyn ca (C) ti im cc i ca n.Cu 2 ( 3,0 im )1.Gii phng trnh 1 39 4.3 3 0+ + =x x .

2.Tnh tch phnln 5 2

ln 2 1=

x

x

eI dxe

3.Tm GTLN, GTNN ca hm s 3 28 16 9= + y x x x trn on [1;3].Cu 3 ( 1,0 im )

Cho t din u ABCD c cnh bng3

2

a

1.Tnh chiu cao ca t din ABCD.2.Tnh th tch ca t din ABCD.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho ba im A(1;0;-1), B(1;2;1), C(0;2;0). Gi G l trng tmtam gic ABC.1. Vit phng trnh ng thng OG.2. Vit phng trnh mt cu (S) i qua bn im O, A, B, C.



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3. Vit phng trnh cc mt phng vung gc vi ng thng OG v tip xcvi mt cu (S).

Cu 5a ( 1,0 im ) Gii phng trnh 2 3 9 0 + =x x trn tp s phc. 35

I.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 3 3= y x x c th (C)1.Kho st s bin thin v v th (C).2.Dng (C), tm cc gi tr ca m phng trnh sau c ba nghim thc

3 3 2 0 + =x x m .Cu 2 ( 3,0 im )1.Gii phng trnh 2 2 3+ =x x .

2.Tnh tch phn1

2

0

ln(1 )= +I x x dx


2 32 2= +xy x trn on [-1/2;2/3].Cu 3 ( 1,0 im )

Cho t din u ABCD c cnh bng2

3

b

1.Tnh chiu cao ca t din ABCD.2.Tnh th tch ca t din ABCD.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho ng thng2 1 1

( ) :1 2 3

+ = =

x y zd v mt phng

( ) : 3 2 0 + + =x y z .1. Tm to giao im M ca ng thng (d) v mt phng ( ) .2. Vit phng trnh mt phng cha (d) v vung gc vi mt phng ( ) .


36http:// www.VNMATH.com

I. PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 3 23 4 2= + +y x x x c th (C)1.Kho st s bin thin v v th (C).2.Vit phng trnh tip tuyn ca (C) ti im c honh 1= ox .Cu 2 ( 3,0 im )1.Gii phng trnh 1 15 5 24+ =x x .

2.Tnh tch phn2

5

1

(1 )= I x x dx

3.Tm GTLN, GTNN ca hm s2 3 6

1

+=

x x

yx

trn khong (1 ; + ).

Cu 3 ( 1,0 im )

Cho hnh chp t gic u S.ABCD c cnh y bng 2b , cnh bn bng 2b1.Tnh chiu cao ca S.ABCD.2.Tnh th tch ca S.ABCD.



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II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )Cu 4a ( 2,0 im ) Cho mt phng ( ) : 2 4 0 + =x y z v im

M(-1;-1;0).1. Vit phng trnh mt phng ( ) qua M v song song vi ( ) .

2. Vit phng trnh ng thng (d) qua M v vung gc vi ( ) .3. Tm to giao im H ca (d) v ( ) .

Cu 5a ( 1,0 im ) Gii phng trnh 2 2 0+ + =x x trn tp s phc. 37

http://www.VNMATH.comI.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 3 22 3 1= + y x x c th (C)1.Kho st s bin thin v v th (C).2.Vit phng trnh tip tuyn ca (C) ti im cc i ca n.

Cu 2 ( 3,0 im )1.Gii phng trnh

2

1 2

2

log log 2+ =x x .

2.Tnh tch phn3

1

2 ln= I x xdx

3.Tm GTLN, GTNN ca hm s 3 3 1= +y x x trn on [0;2].Cu 3 ( 1,0 im )

Cho hnh chp u S. ABC c cnh SA = AB = 32

1.Tnh chiu cao ca S.ABC.

2.Tnh th tch ca S.ABC.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )Cu 4a ( 2,0 im ) Cho bn im A(1;-1;2), B(1;3;2), C(4;3;2), D(4;0;0)1. Lp phng trnh mt phng (BCD). T suy ra ABCD l mt t din.2. Tnh th tch t din.3. Lp phng trnh mt phng ( ) qua gc to v song song mt phng (BCD).

Cu 5a ( 1,0 im ) Gii phng trnh 22 2 0+ + =x x trn tp s phc. 38

http://www.VNMATH.comI.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s

3 23 4= + y x x c th (C)1.Kho st s bin thin v v th (C).2.Tnh din tch hnh phng gii hn bi th (C) , trc honh v hai ng thng x= 0 v x =1.Cu 2 ( 3,0 im )

1.Gii bt phng trnh2 3

14

2

x x

.

2.Tnh tch phn1

2

0

= xI x e dx

3.Tm GTLN, GTNN ca hm s3 23 9 35= +y x x x trn on [-4;4].

Cu 3 ( 1,0 im )



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Cho hnh chp S. ABC c y ABC l tam gic vung ti A. Cnh bn SAvung gc vi mt phng y. SA = AB = 2a, BC = 3aTnh th tch ca S.ABC.

II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )Cu 4a ( 2,0 im ) Cho bn im A(0;-1;1), B(1;-3;2), C(-1;3;2), D(0;1;0)1. Lp phng trnh mt phng (ABC). T suy ra ABCD l mt t din2. Lp phng trnh ng thng (d) qua trng tm G ca tam gic ABC v i qua

gc ta .Cu 5a ( 1,0 im ) Gii phng trnh 2 9 0+ + =x x trn tp s phc.


I.PHN CHUNG (7,0 im )

Cu 1 ( 3,0 im ) Cho hm s3 2

3 2= + y x x c th (C)1.Kho st s bin thin v v th (C).2.Tnh din tch hnh phng gii hn bi th (C) , trc honh v hai ng thngx = -2 v x =-1.Cu 2 ( 3,0 im )


2 91

3 25

x x

2.Tnh tch phn2

sin

0

.cos

= xI e xdx

3.Tm GTLN, GTNN ca hm s3 22 3 1= + y x x trn on 12; 2

Cu 3 ( 1,0 im )Cho hnh chp S. ABC c y ABC l tam gic vung ti B. Cnh bn SA

vung gc vi mt phng y. SA = AB = 2a, BC = 3aTnh th tch ca S.ABC.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho im A(0;-1;1) v mt phng ( ) : 2 3 7 0 + =x y z1. Lp phng trnh ng thng (d) cha A v vung gc vi mt phng ( ) .

2. Tnh khong cch t A n mt phng( )

.Cu 5a ( 1,0 im ) Gii phng trnh 2 8 0+ + =x x trn tp s phc. 40

http://VNMATH.comI.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 3 3 4= + y x x c th (C)1.Kho st s bin thin v v th (C).2.Vit phng trnh tip tuyn ca (C) tai dim c honh x o l nghim ca phngtrnh // ( ) 6=oy xCu 2 ( 3,0 im )

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

.2.Tnh tch phn

1

ln= e

I x xdx



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3.Gii bt phng trnh 20,2 0,2log 5log 6 x xCu 3 ( 1,0 im )

Cho hnh chp S. ABC c y ABC l tam gic vung ti C. Cnh bn SAvung gc vi mt phng y. SA = AB = 5a, BC = 3a

Tnh th tch ca S.ABC.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho ba im A(1;0;4), B(-1;1;2), C(0;1;1)1. Chng minh tam gic ABC vung.2. Lp phng trnh ng thng (d) qua trng tm G ca tam gic ABC v i qua

gc ta .

Cu 5a ( 1,0 im ) Tnh gi tr biu thc:2

2

( 3 )

( 3 )

+=

i

Pi

41

I.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 4 22 2= + y x x c th (C)1.Kho st s bin thin v v th (C).2.Dng th (C), bin lun theo m s nghim ca phng trnh

4 22 2 + =x x m

Cu 2 ( 3,0 im )

1.Gii phng trnh 22 2

6 43

log 2 log+ =

x x .

2.Tnh tch phn3

20

4

1=

+

x

I dxx

3.Tnh gi tr biu thc 2009 2009log(2 3) log(2 3)= + + A Cu 3 ( 1,0 im )

Cho hnh chp S. ABC c y ABC l tam gic vung ti A. Cnh bn SBvung gc vi mt phng y. SA = 5a, AB = 2a, BC = 3aTnh th tch ca S.ABC.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho hai im A(1;2;-1), B(7;-2;3) v ng thng1 3

( ) : 2 2

2 2

= + = = +

x t

d y t

z t

1. Lp phng trnh ng thng AB.2. Chng minh ng thng AB v ng thng (d) cng nm trong mt mt

phng.Cu 5a ( 1,0 im ) Gii phng trnh 22 9 0+ + =x x trn tp s phc.


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

23

= + y x x c th (C)

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

2.Vit phng trnh tip tuyn ca (C) ti tm i xng ca n.Cu 2 ( 3,0 im )1.Gii phng trnh 2 4log log ( 3) 2 =x x .



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2.Tnh tch phn2

2

1

3= +I x x dx

3.Tm GTLN, GTNN ca hm s 3 23 7 1= +y x x x trn on [0;3].Cu 3 ( 1,0 im )

Cho hnh chp S. ABC c y ABC l tam gic vung ti C. Cnh bn SAvung gc vi mt phng y. SA = BC, bit CA = 3a, BA = 5aTnh th tch ca S.ABC.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho ba im A(0;2;1), B(3;0;1), C(1;0;0)1. Lp phng trnh mt phng (ABC).2. Lp phng trnh ng thng (d) qua M(1;-2;1/2) v vung gc mt phng

(ABC).3. Tnh khong cch t im M n mt phng (ABC).

Cu 5a ( 1,0 im ) Tnh gi tr ca biu thc2

5 3 31 2 3

+= iPi


Cu 1 ( 3,0 im ) Cho hm s 4 21

4= +y x x c th (C)

1.Kho st s bin thin v v th (C).2.Dng th (C), tm cc gi tr ca m phng trnh sau c bn nghim thc

42 2 0

4 + =

xx m .

Cu 2 ( 3,0 im )1.Gii phng trnh 1 2

2

log (2 3) log (3 1) 1+ + + =x x .

2.Tnh tch phn2

1

ln=

e xI dx

x

3.Gii bt phng trnh 2 13 3 28+ + x x .Cu 3 ( 1,0 im )

Cho hnh chp S. ABC c y ABC l tam gic vung cn ti A. Cnh bn SAvung gc vi mt phng y. SA = AB = 2a.Tnh th tch ca S.ABC.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho hai im A(1;0;-2), B(0;1;1)1. Lp phng trnh ng thng i hai A v B.2. Lp phng trnh mt cu (S) c ng knh l AB.


1 +

i

i

44I.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 4 22 3= + +y x x c th (C)

1.Kho st s bin thin v v th (C).2.Dng th (C) , bin lun theo m s nghim ca phng trnh

4 22 0 =x x m

Cu 2 ( 3,0 im )



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1.Gii phng trnh 1 14 6.2 8 0+ + + =x x .

2.Tnh tch phn2

2 3

0

2.= + I x x dx

3.Tm GTLN, GTNN ca hm s 3 23 9= + y x x x trn on [-2;2].

Cu 3 ( 1,0 im )Cho hnh chp S. ABC c y ABC l tam gic vung ti B. Cnh bn SC

vung gc vi mt phng y. SC = AB = a/2, BC = 3aTnh th tch ca S.ABC.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho hai im M(3;-4;5), N(1;0;-2)1. Lp phng trnh cu i qua M v c tm l N.2. Lp phng trnh mt phng qua M tip xc vi mt cu.

Cu 5a ( 1,0 im ) Gii phng trnh 22 3 11 0+ + =x x trn tp s phc.


Cu 1 ( 3,0 im ) Cho hm s 4 21

12

= + y x x c th (C)

1.Kho st s bin thin v v th (C).2.Lp phng trnh tip tuyn ca (C) ti im c honh bng 2 .Cu 2 ( 3,0 im )


2 5

5 2

x x

.

2.Tnh tch phn2

0

1 3cos .sin

= +I x xdx

3.Gii phng trnh 3 3log log ( 2) 1+ + =x x Cu 3 ( 1,0 im )

Cho hnh chp S. ABCD c y ABCD l hnh vung cnh a. Cnh bn SAvung gc vi mt phng y, SA = 2a.Tnh th tch ca S.ABCD.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho im H(1;0;-2) v mt phng ( ) : 3 2 7 0 + + =x y z

1. Tnh khong cch t H n mt phng (

)2. Lp phng trnh mt cu c tm H v tip xc vi mt phng ( )Cu 5a ( 1,0 im ) Tnh gi tr ca 2010(1 )+ i



Cu 1 ( 3,0 im ) Cho hm s 4 21 3

4 2= +y x x c th (C)

1.Kho st s bin thin v v th (C).2.Dng th (C), bin lun theo m s nghim phng trnh

4 2

2 3 + =x x mCu 2 ( 3,0 im )1.Gii phng trnh 24 2.5 10 = x x x.2.Tm nguyn hm ca hm s 3cos .sin=y x x



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3.Tm GTLN, GTNN ca hm s22 5 4

2

+ +=

+x x

yx

trn on [0;1].

Cu 3 ( 1,0 im )Cho hnh chp S. ABCD c y ABCD l hnh ch nht. Cnh bn SA vung

gc vi mt phng y. SA = AC , AB = a, BC = 2AB.Tnh th tch ca S.ABCD.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho im M(1;4;2) v mt phng ( ) : 1 0 + + =x y z1. Lp phng trnh ng thng (d) qua M v vung gc vi mt phng ( )2. Tm to giao im H ca (d) v mt phng ( )

Cu 5a ( 1,0 im ) Tnh gi tr ca biu thc ( ) ( )2 2

3 3= + + P i i


I.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s

1

1

+=

x

yx

c th (C)

1.Kho st s bin thin v v th (C).2.Lp phng trnh tip tuyn ca (C) ti im c honh 2= ox .Cu 2 ( 3,0 im )1. Gii phng trnh 2.4 17.2 16 0 + =x x .

2.Tnh tch phn1

1 ln+=

e xI dx

x


1 5= + + y x x (x > 5 )Cu 3 ( 1,0 im )

Tnh th tch ca khi t din u c cnh bng aII. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho mt phng ( ) : 3 5 2 0 + =x y z v ng thng12 9 1

( ) :4 3 1

= =

x y zd

1. Tm to giao im H ca (d) v mt phng ( ) .2. Lp phng trnh mt cu (S) qua H v c tm l gc ta .

Cu 5a ( 1,0 im ) Gii phng trnh 22 11 0 + =x x trn tp s phc. 48

http://VNMATH.comI.PHN CHUNG (7,0 im )

Cu 1 ( 3,0 im ) Cho hm s2

2 1

+=

+x

yx

c th (C)

1. Kho st s bin thin v v th (C).2. Tnh din tch hnh phng gii hn bi th (C), trc honh v cc ng thngx = 0 v x = 2.

Cu 2 ( 3,0 im )1. Gii phng trnh 2 1 2

2

log (1 3 ) log ( 3) log 3 + =x x .



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2.Tnh tch phn5

2

2 ln( 1)= I x x dx

3.Tnh th tch vt th trn xoay, sinh bi mi hnh phng gii hn bi cc ng sauy khi n quay quanh trc Ox: 20; 2= = y y x x .

Cu 3 ( 1,0 im )Cho hnh chp t gic u S. ABCD c cnh y bng 3cm, cnh bn bng

5cm. Tnh th tch ca S.ABCD.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho ba im A(2;-1;-1), B(-1;3;-1), M(-2;0;1).1. Lp phng trnh ng thng (d) i qua A v B.2. Lp phng trnh mt phng ( ) cha M v vung gc vi ng thng AB.3. Tm to giao im ca (d) v mt phng ( )

Cu 5a ( 1,0 im ) Gii phng trnh 21

3 02

+ + =x x trn tp s phc.



Cu 1 ( 3,0 im ) Cho hm s3 2

2

+=

+x

yx

c th (C)

1. Kho st s bin thin v v th (C).2.Tm trn th (C) nhng im c to l cc s nguyn.Cu 2 ( 3,0 im )1. Gii phng trnh 2 24. 3 =x xe e .

2.Tnh tch phn2

2

1

ln= I x xdx


3 1=

x

yx

trn on [-1;-1/2].

Cu 3 ( 1,0 im )Cho hnh hp ch nht ABCD.A/B/C/D/ c chiu di 6cm, chiu rng 5cm,

chiu cao 3cm.1. Tnh th tch ca khi hp ch nht.2. Tnh th tch ca khi chp A/.ABD.

II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )Cu 4a ( 2,0 im ) Cho mt cu

2 2 2( ) : 4 8 2 4 0+ + + + =S x y z x y z v mt phng( ) : 3 5 1 0 + + =x y z

1. Xc nh ta tm I v di bn knh r ca mt cu (S).2. Lp phng trnh ng thng (d) qua im I v vung gc vi mt phng ( ) .

Cu 5a ( 1,0 im ) Tnh gi tr ca biu thc( )

( )

2

2

3

3

+=

iP

i


I.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s

1

2

=

+x

yx

c th (C)

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



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2. Lp phng trnh tip tuyn ca (C) ti giao im vi trc honh.Cu 2 ( 3,0 im )1. Gii phng trnh 1 15 5 26+ + =x x .

2. Tnh tch phn

2

2

1

ln(1 )= +

I x x dx

3. Tm GTLN, GTNN ca hm s2 1

1 3

+=

x

yx

trn on [-1;0].

Cu 3 ( 1,0 im )Cho hnh lng tr ng ABC.A/ C/ B/ c y ABC l tam gic vung ti A. AB

= 4cm, BC = 5cm, AA/ = 6cm.1. Tnh th tch ca khi lng tr .2. Tnh th tch ca khi chp A/ .ABC.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho ba im A(3;0;4), B(1;2;3), C(9;6;4)1. Tm to im D sao cho ABCD l hnh bnh hnh.2. Lp phng trnh mt phng (BCD).

Cu 5a ( 1,0 im ) Tnh gi tr ca biu thc( )

( )

2

3

1 3

+=

iP

i




1

2

= +

+

y

x

c th (C)

1. Kho st s bin thin v v th (C).2. Tm trn th (C) nhng im c to l cc s nguyn.Cu 2 ( 3,0 im )

1. Gii phng trnh 47

log 2 log 06

+ + =x x

2. Tnh tch phn2

2

0

( sin )cos

= +I x x xdx

3.Tm GTLN, GTNN ca hm s 3 23 4= y x x trn on [-1;1/2].

Cu 3 ( 1,0 im )Cho hnh chp S. ABCD c y ABCD l hnh ch nht. Cnh bn SA vunggc vi mt phng y. SA = 2a , AB = 3a, BD = 5a.Tnh th tch ca S.ABCD.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho im I(-2;1;1) v mt phng ( ) : 2 2 5 0 + + =x y z1. Tnh khong cch t im I n mt phng ( )2. Lp phng trnh mt cu (S) c tm l I v tip xc vi mt phng ( )


4

1 3

=

+

iP

i

52



34/88


http://VNMATH.comI.PHN CHUNG (7,0 im )


2

=

+y

xc th (C)

1. Kho st s bin thin v v th (C).2. Lp phng trnh tip tuyn ca (C) ti giao im vi trc tung.Cu 2 ( 3,0 im )1. Gii phng trnh 2 2log log 3 =x x .

2. Tnh tch phn4

2

0

sin ( )4

= I x dx

3.Tm GTLN, GTNN ca hm s 24= y x Cu 3 ( 1,0 im )

Cho hnh chp S. ABC c y ABC l tam gic vung ti B. Cnh bn SC vung gc

vi mt phng y. SC = AB = a/3, BC = 3aTnh th tch ca S.ABC.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho im M(-2;3;1) v ng thng2 1 2

( ) :2 2 3

+ += =

x y z

d

1. Lp phng trnh tham s ca ng thng (d/) qua M v song song vi ngthng (d).

2. Tm to im M/ l hnh chiu vung gc ca M trn (d).


1 = +

iP

i



1

=

x

yx

c th (C)

1.Kho st s bin thin v v th (C).2.Tnh din tch hnh phng gii hn bi th (C), trc honh v cc ng thngx = -3 v x = -2.Cu 2 ( 3,0 im )1. Gii phng trnh 0,5 0,5 2 14 3 3 2 + = x x x x .

2. Tnh tch phn2

1

0

.= xI e xdx


1= +

y x

xtrn khong (1; )+ .

Cu 3 ( 1,0 im ) Cho hnh chp S. ABCD c y ABCD l hnh ch nht. Cnhbn SA vung gc vi mt phng y. SA = 2a , AB = a, AC = 3a.

1). Tnh th tch ca S.ABCD.2). Chng minh ( )BC SAB

II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )Cu 4a ( 2,0 im ) Cho mt phng ( ) : 1 0 + + =x y z v ng thng

2

( ) : 1

3

= = = +

x t

d y t

z t



35/88


1. Tm to giao im H ca (d) v mt phng ( ) .2. Lp phng trnh mt phng trung trc ca on OH.

Cu 5a ( 1,0 im ) Gii phng trnh 3 8 0+ =x trn tp s phc. 54



22

1

+=

xy

xc th (C)

1. Kho st s bin thin v v th (C).2. Lp phng trnh tip tuyn ca (C) ti giao im vi trc honh.Cu 2 ( 3,0 im )1. Gii bt phng trnh 20,5 0,5log log 2 0+ x x .

2. Tnh tch phn2

1

ln=

e xI dx

x

3.Tm GTLN, GTNN ca hm s3 3 3= +y x x trn on [-3;3/2].

Cu 3 ( 1,0 im )Cho hnh chp S. ABCD c y ABCD l hnh ch nht. Cnh bn SA vung

gc vi mt phng y. SA = AC , AB = 5cm, BC = 2AB.Tnh th tch ca S.ABCD.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho bn im A(6;-2;3), B(0;1;6), C(2;0;-1), D(4;1;0)1. Lp phng trnh mt phng (BCD). T suy ra ABCD l mt t din.2. Tnh th tch ca t din.

Cu 5a ( 1,0 im ) Tnh din tch hnh phng gii hn bi th ca hm s2 24; 2= = y x y x x


Cu 1 ( 3,0 im ) Cho hm s4 1

2 3

+=

+x

yx

c th (C)

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

2. Tm GTLN, GTNN ca hm s trn on5

; 22

Cu 2 ( 3,0 im )1. Gii bt phng trnh 20,5log ( 5 6) 1 + x x .

2. Tnh tch phn2

2

sin 2 .sin7

= I x xdx

3. Tnh din tch hnh phng gii hn bi th ca hm s2 1; 3= + + =y x x y

Cu 3 ( 1,0 im )Cho hnh chp S. ABC c y ABC l tam gic vung cn ti A. Cnh bn SA

vung gc vi mt phng y. SA = AB = 5a/2.

Tnh th tch ca S.ABC.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho im H(2;3;-4) v im K(4;-1;0)



36/88


1. Lp phng trnh mt phng trung trc ca on HK.2. Lp phng trnh mt cu (S) c ng knh l HK.

Cu 5a ( 1,0 im ) Tnh gi tr ca biu thc ( ) ( )2 2

3 3= + P i i


Cu 1 ( 3,0 im ) Cho hm s1 2

2 4

=

x

yx

c th (C)

1. Kho st s bin thin v v th (C).2. Tm trn th (C) nhng im c to l cc s nguyn.Cu 2 ( 3,0 im )1. Gii phng trnh 1 2 32 2 2 448 + + =x x x .

2.Tm nguyn hm ca hm s 21

cos (3 2)=

+y

x

3.Tm cc tr ca hm s2 1= + y x x

Cu 3 ( 1,0 im ) Cho hnh chp t gic u S.ABCD c cnh y bng3

a, cnh

bn bng 3a1.Tnh chiu cao ca S.ABCD.2.Tnh th tch ca S.ABCD.II. PHN DNH CHO TH SINH TNG BAN ( 3,0 im )

Cu 4a ( 2,0 im ) Cho im I(-2;1;0) v mt phng ( ) : 2 2 1 0 + + =x y z1. Lp phng trnh ng thng (d) qua I v vung gc vi mt phng ( )

2. Tm to hnh chiu vung gc ca I trn mt phng( )

Cu 5a ( 1,0 im ) Tnh din tch hnh phng gii hn bi th ca cc hm s; 2; 1= = =xy e y x


I. PHN CHUNG CHO TT C TH SINH (7 im)Cu I (3 im)

Cho hm s2

2

+=

+x

yx

.

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

2. Vit phng trnh tip tuyn ca (C), bit n vung gc vi ng thng1

422

= y x

Cu II (3 im).1. Gii phng trnh : 6.4 13.6 6.9 0 + = x x x

2. Tnh tch phn :2

3 3 2

1

3 4.= +I x x dx

3. Tm gi tr ln nht v gi tr nh nht ca hm s : 2( ) cos cos 3= + +f x x x .Cu III (1 im) Cho khi chp tam gic u S.ABCc y ABC l tam gicu cnh a v cc cnh bn to vi y mt gc 600. Hy tnh th tch ca khichp .

II. PHN RING (3 im)



37/88


Th sinh hc chng trnh no th ch c lm phn dnh ring cho chngtrnh (phn 1 hoc 2)

1. Theo chng trnh Chun :Cu IVa (2 im)

Trong khng gian vi h trc ta Oxyz, cho cc im A(1 ; 0 ; 2),B(-1 ; 1 ; 5), C(0 ; -1 ; 2) v D(2 ; 1 ; 1)

1.Lp phng trnh mt phng (P) cha AB v song song vi CD.2.Vit phng trnh mt cu (S) i qua 4 im A, B, C, D.

Cu Va (1 im)

Tm mun ca s phc8 3

1

=

i

zi

2. Theo chng trnh Nng cao :Cu IVb (2 im) Trong khng gian vi h trc ta Oxyz, cho ng

thng (d) v mt phng () ln lt c phng trnh :5 3 1

( ) : 1 2 3

+ = =

x y zd ,

( ) : 2 2 0 + =x y z

1. Vit phng trnh mt phng () i qua giao im Ica (d) v () vvung gc (d).

2. Cho A(0 ; 1 ; 1). Hy tm to im B sao cho () l mt trung trc caon AB.Cu Vb (1 im)

Tm s phcz sao cho3

1+

=+

z i

z i vz + 1 c acgumen bng 6

.

58http://www.VNMATH.comI.PHN CHUNG (7 im)Cu I (3 )Cho hm s y = x3 +(m -1) x2 (m +2)x -1 (1)a) Kho st v th (C) ca hm s khi m = 1

b) Vit phng trnh ng thng (d) vung gc vi ng thng y =3

xv tip xc

vi th (C) ca hm sa) Tm ta hnh chiu vung gc v t im A(2; 0; -1) ln ng thng (d).

b) Tm ta giao im B i xng ca A qua ng thng (d).Cu II (3 )1) Gii phng trnh 16x -17.4x +16 = 0;

2) Tnh tch phn ( )2

0

2 1 sin

+ x xdx

3) Tm gi tr ln nht ca biu thc ( )2

sin0,5

x

Cu III (1) Cho hnh chp tam gic S.ABC c SA, SB, SC i mt vung gc nhauv SA = a, SB = b, SC = c. Tnh di ng cao v t S ca hnh chp S.ABC.II.PHN RING (3 im)

1 THEO CHNG TRNH CHUN

Cu IV.a (2) Trong khng gian Oxyz cho ng thng (d):1 2

1

3

= + = + =

x t

y t

z t



38/88


a) Vit phng trnh mt phng (P) i qua A(2; 0; 0) v vung gc vi ng thng(d)

b) Tm ta giao im ca (d) vi mt phng (P).Cu IV.b (1) Gii phng trnh sau trn tp s phc ( )2 3 2 3 2 2 + = +i x i i

2. THEO CHNG TRNH NNG CAO

Cu IV.a (2) Trong khng gian Oxyz cho ng thng (d):1 2

1

3

= + = + =

x t

y t

z t

Cu IV.b (1) Tm gi tr ln nht ca biu thc 3 += x xy

59I- PHN CHUNG CHO TT C TH SINH (7.0 im)

Cu I (3.0 im):Cho hm s 4 22( 1) 2 1= + + y x m x m , c th (Cm)1) Kho st v v th (C) khi 0=m2) Vit pttt vi (C) ti im c honh 2=x

Cu II (3.0 im):

1) Gii bt phng trnh: 23

2 3log 0

1

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