tuyen tap 150 de thi dai hoc 3403

V Vn Ninh - THPT L Thng Kit - Hi Phng

Trang:1

S 1 CU1: (2,5 im)

Cho hm s: y = -x3 + 3mx2 + 3(1 - m2)x + m3 - m2 1) Kho st s bin thin v v th ca hm s trn khi m = 1. 2) Tm k phng trnh: -x3 + 3x2 + k3 - 3k2 = 0 c 3 nghim phn bit. 3) Vit phng trnh ng thng i qua 2 im cc tr ca th hm s trn. CU2: (1,75 im)

Cho phng trnh: 01212323 =++ mxlogxlog (2)

1) Gii phng trnh (2) khi m = 2.

2) Tm m phng trnh (2) c t nht 1 nghim thuc on

331; . CU3: (2 im)

1) Tm nghim (0; 2) ca pt : 32221

335 +=

+

++ xcosxsin

xsinxcosxsin

2) Tnh din tch hnh phng gii hn bi cc ng: y = 342 + xx , y = x + 3 CU4: (2 im)

1) Cho hnh chp tam gic u S.ABC nh S c di cnh y bng a. Gi M v N ln lt l trung im ca cc cnh SB v SC. Tnh theo a din tch AMN bit rng mt phng (AMN) vung gc mt phng (SBC).

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

=++=+

0422042

zyx

zyx

v 2:

+=+=+=

tz

ty

tx

2121

a) Vit phng trnh mt phng (P) cha ng thng 1 v song song vi ng thng 2. b) Cho im M(2; 1; 4). Tm to im H thuc ng thng 2 sao cho on thng MH c di nh nht. CU5: (1,75 im)

1) Trong mt phng vi h to cc vung gc Oxy xt ABC vung ti A, phng trnh ng thng BC l: 033 = yx , cc nh A v B thuc trc honh v bn knh ng trn ni tip bng 2. Tm to trng tm G ca ABC 2 Khai trin nh thc:

nxnn

nxxnn

xnx

n

nx

n

nxx

CC...CC

+

++

+

=

+

3

1

321

13

1

21

121

0321

22222222

Bit rng trong khai trin 13 5 nn CC = v s hng th t bng 20n, tm n v x S 2

CU1: (2 im)

Cu Cho hm s: y = mx4 + (m2 - 9)x2 + 10 (1)

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


Trang:2

2) Tm m hm s (1) c ba im cc tr.

CU2: (3 im)

1) Gii phng trnh: sin23x - cos24x = sin25x - cos26x

2) Gii bt phng trnh: logx(log3(9x - 72)) 1

3) Gii h phng trnh:

++=+=

2

3

yxyx

yxyx

CU3: (1,25 im)

Tnh din tch hnh phng gii hn bi cc ng: y = x

y v x2

2444

2=

CU4: (2,5 im)

1) Trong mt phng vi h to cc vung gc Oxy cho hnh ch nht ABCD c tm

I

021

; , phng trnh ng thng AB l x - 2y + 2 = 0 v AB = 2AD. Tm to cc nh A, B,

C, D bit rng nh A c honh m

2) Cho hnh lp phng ABCD.A1B1C1D1 c cnh bng a

a) Tnh theo a khong cch gia hai ng thng A1B v B1D.

b) Gi M, N, P ln lt l cc trung im ca cc cnh BB1, CD1, A1D1. Tnh gc gia hai ng

thng MP v C1N.

CU5: (1,25 im)

Cho a gic u A1A2...A2n (n 2, n Z) ni tip ng trn (O). Bit rng s tam gic c cc nh l 3 im trong 2n im A1, A2, ... ,A2n nhiu gp 20 ln s hnh ch nht c cc nh l 4

im trong 2n im A1, A2, ... ,A2n . Tm n.

S 3

CU1: (3 im)

Cho hm s: y = ( )

112 2

xmxm

(1) (m l tham s)

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

2) Tnh din tch hnh phng gii hn bi ng cong (C) v hai trc to .

3) Tm m th ca hm s (1) tip xc vi ng thng y = x.

CU2: (2 im)

1) Gii bt phng trnh: (x2 - 3x) 0232 2 xx .


Trang:3

2) Gii h phng trnh:

=++

=+

y

yy

x

xx

x

2224

4521

23

CU3: (1 im)

Tm x [0;14] nghim ng phng trnh: cos3x - 4cos2x + 3cosx - 4 = 0 . CU4: (2 im)

1) Cho hnh t din ABCD c cnh AD vung gc vi mt phng (ABC); AC = AD = 4 cm

; AB = 3 cm; BC = 5 cm. Tnh khong cch t im A ti mt phng (BCD).

2) Trong khng gian vi h to cc vung gc Oxyz, cho mt phng

(P): 2x - y + 2 = 0 v ng thng dm: ( ) ( )

( )

=++++=+++

0241201112

mzmmx

mymxm

Xc nh m ng thng dm song song vi mt phng (P) .

CU5: (2 im)

1) Tm s nguyn dng n sao cho: 243242 210 =++++ nnnnnn C...CCC . 2) Trong mt phng vi h to cc vung gc Oxy cho Elp (E) c phng trnh:

1916

22=+ yx . Xt im M chuyn ng trn tia Ox v im N chuyn ng trn tia Oy sao cho

ng thng MN lun tip xc vi (E). Xc nh to ca M, N on MN c di nh nht.

Tnh gi tr nh nht .

S 4 CU1: (2 im)

Cho hm s: y = 132

+

xx

1) Kho st s bin thin v v th hm s. 2) Tm trn ng thng y = 4 cc im m t k c ng 2 tip tuyn n th hm s. CU2: (2 im)

1) Gii h phng trnh:

=++=++

0

123

yxyx

yxyx

2) Gii bt phng trnh: ( ) 012

1 2 >++ xxlnxln CU3: (2 im)

1) Gii phng trnh: cosx+ cos2x + cos3x + cos4x + cos5x = -21


Trang:4

2) Chng minh rng ABC tho mn iu kin

224

22

27 B

cosA

cosC

sinCcosBcosAcos ++=+ th ABC u CU4: (2 im)

1) Trn mt phng to cho A(1, 0); B(0, 2); O(0, 0) v ng trn (C) c phng trnh: (x

- 1)2 + 2

21

y = 1. Vit phng trnh ng thng i qua cc giao im ca ng thng (C) v ng trn ngoi tip OAB. 2) Cho hnh chp S.ABC c y ABC l tam gic vung cn vi AB = AC = a, SA = a, SA vung gc vi y. M l mt im trn cnh SB, N trn cnh SC sao cho MN song song

vi BC v AN vung gc vi CM. Tm t s MBMS

.

CU5: (2 im) 1) Tnh din tch phn mt phng gii hn bi cc ng cong: y = x3 - 2 v

(y + 2)2 = x. 2) Vi cc ch s 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s c 3 ch s khc nhau, bit rng cc s ny chia ht cho 3.

S 5

CU1: (2 im)

Cho hm s: y = x + 1 + 1

1x .

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

2) T mt im trn ng thng x = 1 vit phng trnh tip tuyn n th (C).

CU2: (2 im)

1) Gii phng trnh: 1635223132 2 +++=+++ xxxxx

2) Tm cc gi tr x, y nguyn tho mn: ( ) yyxxlog y 3732 2822 2 +++ + CU3: (2 im)

1) Gii phng trnh: (cos2x - 1)(sin2x + cosx + sinx) = sin22x

2) ABC c AD l phn gic trong ca gc A (D BC) v sinBsinC 2

2 Asin . Hy chng

minh AD2 BD.CD . CU4: (2 im)

1) Trn mt phng to vi h to cc vung gc Oxy, cho elip c phng trnh: 4x2

+ 3y2 - 12 = 0. Tm im trn elip sao cho tip tuyn ca elip ti im cng vi cc trc to

to thnh tam gic c din tch nh nht.


Trang:5

2) Trong khng gian vi h trc to cc vung gc Oxyz, cho hai mt phng (P): x - y

+ z + 5 = 0 v (Q): 2x + y + 2z + 1 = 0. Vit phng trnh mt cu c tm thuc mt phng (P) v

tip xc vi mt phng (Q) ti M(1; - 1; -1).

CU5: (2 im)

1) Tnh din tch hnh phng gii hn bi cc ng: y = 2 - 4

2x v x + 2y = 0

2) a thc P(x) = (1 + x + x2)10 c vit li di dng: P(x) = a0 + a1x + ... + a20x20. Tm h

s a4 ca x4.

S 6 CU1: (2 im)

Cho hm s: y = 1

2

++

xmxmx

(1) (m l tham s)

1) Kho st s bin thin v v th ca hm s (1) khi m = -1. 2) Tm m th hm s (1) ct trc honh ti hai im phn bit v hai im c honh dng. CU2: (2 im)

1) Gii phng trnh: cotgx - 1 = tgx

xcos+1

2 + sin2x -

21

sin2x

2) Gii h phng trnh:

+=

=

12

11

3xy

yy

xx

CU3: (3 im) 1) Cho hnh lp phng ABCD.A'B'C'D'. Tnh s o ca gc phng nh din

[B, A'C, D]. 2) Trong khng gian vi h to cc Oxyz cho hnh hp ch nht ABCD.A'B'C'D' c A trng vi gc ca h to , B(a; 0; 0), D(0; a; 0), A'(0; 0; b) (a > 0, b > 0). Gi M l trung im cnh CC'. a) Tnh th tch khi t din BDA'M theo a v b.

b) Xc nh t s ba

hai mt phng (A'BD) v (MBD) vung gc vi nhau.

CU4: (2 im) 1) Tm h s ca s hng cha x8 trong khai trin nh thc Niutn ca:


Trang:6

n

xx

+ 531

, bit rng: ( )37314 += +++ nCC nnnn (n N*, x > 0)

2) Tnh tch phn: I = +32

52 4xx

dx

CU5: (1 im) Cho x, y, z l ba s dng v x + y + z 1. Chng minh rng:

82111 22

22

22 +++++

zz

yy

xx

S 7 CU1: (2 im)

Cho hm s: y = x3 - 3x2 + m (1) 1) Tm m th hm s (1) c hai im phn bit i xng vi nhau qua gc to . 2) Kho st s bin thin v v th ca hm s (1) khi m = 2 . CU2: (2 im)

1) Gii phng trnh: cotgx - tgx + 4sin2x = xsin2

2

2) Gii h phng trnh:

+=

+=

2

2

2

2

23

23

y

xx

x

yy

CU3: (3 im) 1) Trong mt phng vi h ta cc vung gc Oxy cho ABC c: AB = AC, = 900. Bit M(1; -1) l trung im cnh BC v G

0

32

; l trng tm ABC. Tm to cc nh A, B, C .

2) Cho hnh lng tr ng ABCD.A'B'C'D' c y ABCD l mt hnh thoi cnh a, gc = 600 . gi M l trung im cnh AA' v N l trung im cnh CC'. Chng minh rng bn im B', M, D, N cng thuc mt mt phng. Hy tnh di cnh AA' theo a t gic B'MDN l hnh vung. 3) Trong khng gian vi h to cc Oxyz cho hai im A(2; 0; 0) B(0; 0; 8) v im C

sao cho ( )060 ;;AC = . Tnh khong cch t trung im I ca BC n ng thng OA. CU4: (2 im)

1) Tm gi tr ln nht v nh nht ca hm s: y = x + 24 x

2) Tnh tch phn: I =

+4

0

2

2121

dxxsinxsin

CU5: (1 im) Cho n l s nguyn dng. Tnh tng:


Trang:7

nnn

nnn Cn

...CCC1

123

122

12 123120+++++

+

( knC l s t hp chp k ca n phn t) S 8

CU1: (2 im)

1) Kho st s bin thin v v th ca hm s: y = 2

422

+

xxx

(1)

2) Tm m ng thng dm: y = mx + 2 - 2m ct th ca hm s (1) ti hai im phn bit. CU2: (2 im)

1) Gii phng trnh: 0242

222 =

xcosxtgxsin

2) Gii phng trnh: 32222 2 = + xxxx

CU3: (3 im) 1) Trong mt phng vi h ta trc cc vung gc Oxy cho ng trn:

(C): (x - 1)2 + (y - 2)2 = 4 v ng thng d: x - y - 1 = 0 Vit phng trnh ng trn (C') i xng vi ng trn (C) qua ng thng d. Tm ta cc giao im ca (C) v (C'). 2) Trong khng gian vi h to cc vung gc Oxyz cho ng thng:

dk:

=++=++01

023zykx

zkyx

Tm k ng thng dk vung gc vi mt phng (P): x - y - 2z + 5 = 0. 3) Cho hai mt phng (P) v (Q) vung gc vi nhau, c giao tuyn l ng thng . Trn ly hai im A, B vi AB = a. Trong mt phng (P) ly im C, trong mt phng (Q) ly im D sao cho AC, BD cng vung gc vi v AC = BD = AB. Tnh bn knh mt cu ngoi tip t din ABCD v tnh khong cch t A n mt phng (BCD) theo a. CU4: (2 im)

1) Tm gi tr ln nht v gi tr nh nht ca hm s: y = 1

12 ++

x

x

trn on [-1; 2]

2) Tnh tch phn: I = 2

0

2 dxxx

CU5: (1 im) Vi n l s nguyn dng, gi a3n - 3 l h s ca x3n - 3 trong khai trin thnh a thc ca (x2

+ 1)n(x + 2)n. Tm n a3n - 3 = 26n. S 9

CU1: (2 im)


Trang:8

Cho hm s: y = ( )12332

+

xxx

(1)

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

2) Tm m ng thng y = m ct th hm s (1) ti hai im A, B sao cho AB = 1.

CU2: (2 im)

1) Gii bt phng trnh: ( )

373

3162 2

>+

x

xxx

x

2) Gii h phng trnh: ( )

=+

=

25

11

22

441

yx

ylogxylog

CU3: (3 im)

1) Trong mt phng vi h ta cac Oxy cho im A(0; 2) v B ( )13 ; . Tm to trc tm v to tm ng trn ngoi tip OAB. 2) Trong khng gian vi h to cc Oxyz cho hnh chp S.ABCD c y ABCD l

hnh thoi, AC ct BD ti gc to O. Bit A(2; 0; 0) B(0; 1; 0)

S(0; 0; 2 2 ). Gi M l trung im ca cnh SC. a) Tnh gc v khong cch gia hai ng thng SA v BM.

b) Gi s mt phng (ABM) ct SD ti N. Tnh th tch hnh chp S.ABMN.

CU4: (2 im)


1 11dx

xx

2) Tm h s ca x8 trong khai trin thnh a thc ca: ( )[ ]82 11 xx + CU5: (1 im)

Cho ABC khng t tho mn iu kin: cos2A + 2 2 cosB + 2 2 cosC = 3 Tnh cc gc ca ABC.

S 10 CU1: (2 im)

Cho hm s: y = xxx 3231 23 + (1) c th (C)

1) Kho st s bin thin v v th ca hm s (1). 2) Vit phng trnh tip tuyn ca (C) ti im un v chng minh rng l tip tuyn ca (C) c h s gc nh nht. CU2: (2 im)


Trang:9

1) Gii phng trnh: 5sinx - 2 = 3(1 - sinx)tg2x

2) Tm gi tr ln nht v gi tr nh nht ca hm s: y = x

xln2 trn on [ ]31 e; .

CU3: (3 im) 1) Trong mt phng vi h ta cc Oxy cho im A(1; 1), B(4; -3). Tm im C thuc ng thng y = x - 2y - 1 = 0 sao cho khong cch t C n ng thng AB bng 6. 2) Cho hnh chp t gic u S.ABCD c cnh y bng a, gc gia cnh bn v mt y bng (00 < < 900). Tnh tang ca gc gia hai mt phng (SAB) v (ABCD) theo a v . 3) Trong khng gian vi h to cc Oxyz cho im A(-4; -2; 4) v ng thng d:

+==+=

tz

ty

tx

411

23 (t R). Vit phng trnh ng thng i qua im A, ct v vung gc vi

ng thng d. CU4: (2 im)

1) Tnh tch phn I = +e

xdxlnx

xln

1

31

2) Trong mt mn hc, thy gio c 30 Cu hi khc nhau gm 5 Cu hi kh, 10 Cu hi trung bnh, 15 Cu hi d. T 30 Cu hi c th lp c bao nhiu kim tra, mi gm 5 Cu hi khc nhau, sao cho trong mi nht thit phi c 3 loi Cu hi (kh, d, trung bnh) v s Cu hi d khng t hn 2? CU5: (1 im) Xc nh m phng trnh sau c nghim:

22422 1112211 xxxxxm ++=

++ S 11

CU1: (2 im)

Cho hm s y = x3 - 3mx2 + 9x + 1 (1) (m l tham s)


2) Tm m im un ca th hm s (1) thuc ng thng y = x + 1.

CU2: (2 im)

1) Gii phng trnh: ( )( ) xsinxsinxcosxsinxcos =+ 2212 2) Tm m h phng trnh sau:

=+=+

myyxx

yx

31

1 c nghim.

CU3: (3 im)

1) Trong mt phng vi h ta cc Oxy cho ABC c cc nh A(-1; 0); B(4; 0); C(0; m) vi m 0. Tm to trng tm G ca ABC theo m. Xc nh m GAB vung ti G.


Trang:10

2) Trong khng gian vi h to cc Oxyz cho hnh lng tr ng ABC.A1B1C1. Bit

A(a; 0; 0); B(-a; 0; 0); C(0; 1; 0); B1(-a; 0; b) a > 0, b > 0.

a) Tnh khong cch gia hai ng thng B1C v AC1 theo a, b.

b) Cho a, b thay i nhng lun tho mn a + b = 4. Tm a, b khong cch gia 2 ng

thng B1C v AC1 ln nht.

3) Trong khng gian vi h to cc Oxyz cho 3 im A(2; 0; 1) B(1; 0; 0) C(1; 1; 1) v

mt phng (P): x + y + x - 2 = 0. Vit phng trnh mt cu i qua 3 im A, B, C v c tm thuc

mt phng (P).

CU4: (2 im)

1) Tnh tch phn I = ( ) 32

2 dxxxln

2) Tm cc s hng khng cha x trong khai trin nh thc Newtn ca 7

43 1

+

xx vi x

> 0

CU5: (1 im)

Chng minh rng phng trnh sau c ng 1 nghim: x5 - x2 - 2x - 1 = 0

S 12 Cu1: (2 im)

Gi (Cm) l th ca hm s: y = mx + 1x

(*) (m l tham s)

1. Kho st s bin thin v v th ca hm s (*) khi m = 14

2. Tm m hm s (*) c cc tr v khong cch t im cc tiu ca (Cm) n tim cn

xin ca (Cm) bng 12

Cu2: (2 im)

1. Gii bt phng trnh: 5 1 1 2 4x x x > 2. Gii phng trnh: cos23xcos2x - cos2x = 0

Cu3: (3 im) 1. Trong mt phng vi h to Oxy cho hai ng thng

d1: x - y = 0 v d2: 2x + y - 1 = 0 Tm to cc nh ca hnh vung ABCD bit rng nh A thuc d1, nh C thuc d2 v cc nh B, D thuc trc honh.

2. Trong khng gian vi h to Oxyz cho ng thng d: 1 3 3

1 2 1x y z + = = v

mt phng (P): 2x + y - 2z + 9 = 0. a. Tm to im I thuc d sao cho khong cch t I n mt phng (P) bng 2


Trang:11

b. Tm to giao im A ca ng thng d v mt phng (P). Vit phng trnh tham s ca ng thng nm trong mt phng (P), bit i qua A v vung gc vi d.

Cu4: (2 im)

1. Tnh tch phn I = 2

0

sin 2 sin1 3cos

x x dxx

+

+ 2. Tm s nguyn dng n sao cho:

( )1 2 2 3 3 4 2 12 1 2 1 2 1 2 1 2 12.2 3.2 4.2 ... 2 1 2 2005n nn n n n nC C C C n C2 ++ + + + + + + + + = Cu5: (1 im)

Cho x, y, z l cc s dng tho mn: 1 1 1 4x y z+ + = . Chng minh rng:

1 1 1 1

2 2 2x y z x y z x y z+ + + + + + + +

S 13 Cu1: (2 im)

Gi (Cm) l th hm s y = ( )2 1 1

1x m x m

x+ + + +

+ (*) m l tham s 1. Kho st s bin thin v v th ca hm s (*) khi m = 1. 2. Chng minh rng vi m bt k, th (Cm) lun lun c im cc i, cc tiu v

khong cch gia hai im bng 20 Cu2: (2 im)

1. Gii h phng trnh: ( )2 39 31 2 1

3log 9 log 3

x y

x y

+ = =

2. Gii phng trnh: 1 + sinx + cosx + sin2x + cos2x = 0 Cu3: (3 im)

1. Trong mt phng vi h to Oxy cho A(2; 0) v B(6; 4). Vit phng trnh ng trn (C) tip xc vi trc honh ti hai im v khong cch t tm ca (C) n im B bng 5.

2. Trong khng gian vi h to Oxyz cho hnh lng tr ng ABC.A1B1C1 vi A(0; -3; 0) B(4; 0; 0) C(0; 3; 0) B1(4; 0; 4)

a. Tm to cc nh A1, C1. Vit phng trnh mt cu c tm l A v tip xc vi mt phng (BCC1B1).

b. Gi M l trung im ca A1B1. Vit phng trnh mt phng P) i qua hai im A, M v song song vi BC1. mt phng (P) ct ng thng A1C1 ti im N. Tnh di on MN

Cu4: (2 im)


Trang:12

1. Tnh tch phn: I = 2

0

sin 2 cos1 cos

x x dxx

+ 2. Mt i thanh nin tnh nguyn c 15 ngi, gm 12 nam v 3 n. Hi c bao nhiu

cch phn cng i thanh nin tnh nguyn v gip 3 tnh min ni, sao cho mi tnh c 4 nam v 1 n?

Cu5: (2 im) Chng minh rng vi mi x thuc R ta c:

12 15 20 3 4 55 4 3

x x xx x x + + + +

Khi no ng thc xy ra? S 14

Cu1: (2 im)

Gi (Cm) l th hm s: y = 3 21 13 2 3

mx x + (*) (m l tham s) 1. Kho st s bin thin v v th ca hm s (*) khi m = 2 2. Gi M l im thuc (Cm) c honh bng -1. Tm m tip tuyn ca (Cm) ti im

M song song vi ng thng 5x - y = 0 Cu2: (2 im) Gii cc phng trnh sau:

1. 2 2 2 1 1 4x x x+ + + + = 2. 4 4

3cos sin cos sin 3 04 4 2

x x x x + + = Cu3: (3 im)

1. Trong mt phng vi h to Oxy cho im C(2; 0) v Elip (E): 2 2

14 1x y+ = . Tm

to cc im A, B thuc (E), bit rng A, B i xng vi nhau qua trc honh va ABC l tam gic u.

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

d1: 1 2 1

3 1 2x y z + += = v d2:

2 03 12 0

x y zx y+ = + =

a. Chng minh rng: d1 v d2 song song vi nhau. Vit phng trnh mt phng (P) cha c hai ng thng d1 v d2

b. mt phng to Oxz ct hai ng thng d1, d2 ln lt ti cc im A, B. Tnh din tch OAB (O l gc to )

Cu4: (2 im)

1. Tnh tch phn: I = ( )2 sin0

cos cosxe x xdx

+


Trang:13

2. Tnh gi tr ca biu thc M = ( )4 3

1 31 !

n nA An+ ++ bit rng

2 2 2 21 2 3 42 2 149n n n nC C C C+ + + ++ + + =

Cu5: (1 im) Cho cc s nguyn dng x, y, z tho mn xyz = 1. Chng minh rng:

3 3 3 3 3 31 1 1 3 3x y y z z x

xy yz zx+ + + + + ++ +

Khi no ng thc xy ra? S 15

PHN CHUNG C TT C CC TH SINH Cu1: (2 im)

1. Kho st s bin thin v v th ca hm s: y = 2x3 - 9x2 + 12x - 4

2. Tm m phng trnh sau c 6 nghim phn bit: 3 22 9 12x x x m + =

Cu2: (2 im)

1. Gii phng trnh: ( )6 62 sin sin .cos

02 2sin

cos x x x x

x

+ =

2. Gii h phng trnh: 3

1 1 4

xy xy

x y

= + + + =

Cu3: (2 im) Trong khng gian vi h to Oxyz. Cho hnh lp phng ABCD.ABCD vi A(0; 0; 0) B(1; 0; 0) D(0; 1; 0) A(0; 0; 1). Gi M v N ln lt l trung im ca AB v CD.

1. Tnh khong cch gia hai ng thng AC v MN. 2. Vit phng trnh mt phng cha AC v to vi mt phng Oxy mt gc bit cos =

16

Cu4: (2 im)

1. Tnh tch phn: I = 2

2 20

sin 2cos 4sin

x dxx x

+ 2. Cho hai s thc x 0, y 0 thay i v iu kin: (x + y)xy = x2 + y2 - xy. Tm GTLN

ca biu thc A = 3 31 1x y

+ PHN T CHN: Th sinh chn Cu 5.a hc Cu 5.b

Cu5a: Theo chng trnh khng phn ban: (2 im) 1. Trong mt phng vi h to Oxy cho cc ng thng: d1: x + y + 3 = 0 d2: x - y - 4 = 0 d3: x - 2y = 0.

Tm to im M nm trn ng thng d3 sao cho khong cch t M n ng thng d1 bng hai ln khong cch t M n ng thng d2

2. Tm h s ca s hng cha x26 trong khai trin nh thc: 741 nxx

+ , bit rng: 1 2 02 1 2 1 2 1... 2 1

nn n nC C C

2+ + ++ + + =


Trang:14

Cu5b: Theo chng trnh phn ban: (2 im) 1. Gii phng trnh: 3.8x + 4.12x - 18x - 2.27x = 0 2. Cho hnh lng tr c cc y l hai hnh trn tm O v O, bn knh bng chiu cao v bng a. Trn ng trn y tm O ly im A, trn ng trn y tm O ly im B sao cho AB = 2a. Tnh th tch ca khi t din OOAB.

S 16 PHN CHUNG C TT C CC TH SINH Cu1: (2 im)

Cho hm s: y = 2 1

2x x

x+ +

1. Kho st s bin thin v v th (C) ca hm s. 2. Vit phng trnh tip tuyn ca th (C), bit tip tuyn vung gc vi tim cn

xin ca (C). Cu2: (2 im)

1. Gii phng trnh: cotx + sinx 1 tan .tan 42xx + =

2. Tm m phng trnh sau c hai nghim thc phn bit:

2 2 2 1x mx x+ + = Cu3: (2 im)

Trong khng gian vi h to Oxyz cho im A(0; 1; 2) v hai ng thng :

d1: 1 1

2 1 1x y z += = d2:

11 2

2

x ty tz t

= + = = +

1. Vit phng trnh mt phng (P) qua A, ng thi song song vi d1 v d2. 2. Tm to cc im M d1, N d2 sao cho ba im A, M, N thng hng

Cu4: (2 im)

1. Tnh tch phn: I = ln 5

ln 3 2 3x x

dxe e+

2. Cho x, y l cc s thc thay i. Tm GTNN ca biu thc:

A = ( ) ( )2 22 21 1 2x y x y y + + + + + PHN T CHN: Th sinh chn Cu 5.a hc Cu 5.b

Cu5a: Theo chng trnh khng phn ban: (2 im) 1. Trong mt phng vi h to Oxy cho ng trn (C): x2 + y2 -2x - 6y + 6 = 0 v im M(-3; 1). Gi T1 v T2 l cc tip im ca cc tip tuyn k t M n (C). Vit phng trnh ng thng T1T2 2. Cho tp hp A gm n phn t (n 4). Bit rng s tp con gm 4 phn t ca A bng 20 ln s tp con gm 2 phn t ca A. Tm k {1, 2,..., n} sao cho s tp con gm k phn t ca A l ln nht. Cu5b: Theo chng trnh phn ban: (2 im) 1. Gii bt phng trnh: ( ) ( )25 5 5log 4 144 4log 2 1 log 2 1x x+ < + + 2. Cho hnh chp S.ABCD c y ABCD l hnh ch nht vi AB = a, AD = a 2 , SA = a v SA vung gc vi mt phng (ABCD). Gi M v N ln lt l trung im ca AD v SC; I l giao im ca BM v AC. Chng minh rng: mt phng (SAC) vung gc vi mt phng (SMB). Tnh th tch ca khi t din ANIB


Trang:15

S 17 PHN CHUNG C TT C CC TH SINH Cu1: (2 im) Cho hm s y = x3 - 3x + 2

1. Kho st s bin thin v v th (C) ca hm s cho. 2. Gi d l ng thng i qua im A(3; 2) v c h s gc l m. Tm m ng thng d

ct th (C) ti ba im phn bit. Cu2: (2 im)

1. Gii phng trnh: cos3x + cos2x - cosx - 1 = 0

2. Gii phng trnh: 22 1 3 1 0x x x + + = (x R) Cu3: (2 im) Trong khng gian vi h to Oxyz, cho im A(1; 2; 3) v hai ng thng

d1: 2 2 3

2 1 1x y z + = = d2:

1 1 11 2 1

x y z += = 1. Tm to im A i xng vi im A qua ng thng d1 2. Vit phng trnh ng thng i qua A vung gc vi d1 v ct d2

Cu4: (2 im)

1. Tnh tch phn: I = ( )1 20

2 xx e dx 2. Chng minh rng: vi mi a > 0, h phng trnh sau c nghim duy nht:

( ) ( )ln 1 ln 1x ye e x yy x a

= + + =

PHN T CHN: Th sinh chn Cu 5.a hc Cu 5.b Cu5a: Theo chng trnh khng phn ban: (2 im) 1. Trong mt phng vi h to Oxy cho ng trn (C): x2 + y2 - 2x - 2y + 1 = 0 v ng thng d: x - y + 3 = 0. Tm to im M nm trn d sao cho ng trn tm M, c bn knh gp i bn knh ng trn (C) tip xc ngoi vi ng trn (C) 2. i thanh nin xung kch ca mt trng ph thng c 12 hc sinh, gm 5 hc sinh lp A, 4 hc sinh lp B v 3 hc sinh lp C. Cn chn 4 hc sinh i lm nhim v, sao cho 4 hc sinh ny thuc khng qu 2 trong 3 lp trn. Hi c bao nhiu cch chn nh vy? Cu5b: Theo chng trnh phn ban: (2 im)

1. Gii phng trnh: 2 2 22 4.2 2 4 0x x x x x+ + =

2. Cho hnh chp S.ABC c y ABC l tam gic u cnh a, SA = 2a v SA vung gc vi mt phng (ABC). Gi M v N ln lt l hnh chiu vung gc ca A trn cc ng thng SB v SC. Tnh th tch ca khi chp A.BCNM

S 18 PHN CHUNG C TT C CC TH SINH Cu1: (2 im)

Cho hm s: y = ( )2 22 1 4

2x m x m m

x+ + + +

+ (1) m l tham s 1. Kho st s bin thin v v th ca hm s (1) khi m = -1.


Trang:16

2. Tm m hm s (1) c cc i v cc tiu, ng thi cc im cc tr ca th cng vi gc to to thnh mt tam gic vung ti O

Cu2: (2 im) 1. Gii phng trnh: ( ) ( )2 21 sin cos 1 cos sin 1 sin 2x x x x x+ + + = + 2. Tm m phng trnh sau c nghim thc: 243 1 1 2 1x m x x + + =

Cu3: (2 im) Trong khng gian vi h to Oxyz cho hai ng thng

d1: 1 2

2 1 1x y z += = v d2:

1 213

x ty tz

= + = + =

1. Chng minh rng: d1 v d2 cho nhau. 2. Vit phng trnh ng thng d vung gc vi mt phng (P): 7x + y - 4z = 0 v ct hai

ng thng d1, d2 Cu4: (2 im)

1. Tnh din tch hnh phng gii hn bi cc ng: y = (e + 1)x, y = (1 + ex)x 2. Cho x, y, z l cc s thc dng thay i v tho mn iu kin: xyz = 1. Tm GTNN

ca biu thc: P = ( ) ( ) ( )2 2 2

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

y y z z z z x x x x y y+ + ++ ++ + +

PHN T CHN: Th sinh chn Cu 5.a hc Cu 5.b Cu5a: Theo chng trnh khng phn ban: (2 im) 1. Trong mt phng vi h to Oxy cho ABC c A(0; 2) B(-2 -2) v C(4; -2). Gi H l chn ng cao k t B; M v N ln lt l trung im ca cc cnh AB v BC. Vit phng trnh ng trn i qua cc im H, M, N

2. Chng minh rng: 2

1 3 5 2 12 2 2 2

1 1 1 1 2 1...2 4 6 2 2 1

nn

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

Cu5b: Theo chng trnh phn ban: (2 im) 1. Gii bt phng trnh: ( ) ( )3 1

3

2log 4 3 log 2 3 2x x + + 2. Cho hnh chp S.ABCD c y l hnh vung cnh a, mt bn SAD l tam gic u v nm trong mt phng vung gc vi y. Gi M, N, P ln lt l trung im ca cc cnh SB, BC, CD. Chng minh AM vung gc vi BP v tnh th tch ca khi t din CMNP.

S 19 PHN CHUNG C TT C CC TH SINH Cu1: (2 im) Cho hm s: y = -x3 + 3x2 + 3(m2 -1)x - 3m2 - 1 (1) m l tham s

1. Kho st s bin thin v v th ca hm s (1) khi m = 1 2. Tm m hm s (1) c cc i, cc tiu v cc im cc tr ca th hm s (1) cch

u gc to O. Cu2: (2 im)

1. Gii phng trnh: 2sin22x + sin7x - 1 = sinx 2. Chng minh rng vi mi gi tr dng ca tham s m, phng trnh sau c hai nghim

thc phn bit: x2 + 2x - 8 = ( )2m x Cu3: (2 im)


Trang:17

Trong khng gian vi h to Oxyz cho mt cu (S): x2 + y2 + z2 - 2x + 4y + 2z - 3 = 0 v mt phng (P): 2x - y + 2z - 14 = 0

1. Vit phng trnh mt phng (Q) cha trc Ox v ct (S) theo mt ng trn c bn knh bng 3.

2. Tm to im M thuc mt cu (S) sao cho khong cch t M n mt phng (P) ln nht

Cu4: (2 im) 1. Cho hnh phng H gii hn bi cc ng: y = xlnx, y = 0, x = e. Tnh th tch ca khi

trn xoay to thnh khi quay hnh H quanh trc Ox. 2. Cho x, y, z l ba s thc dng thay i. Tm gi tr nh nht ca biu thc:

P = 1 1 1

2 2 2x y zx y z

yz zx xy + + + + +

PHN T CHN: Th sinh chn Cu 5.a hc Cu 5.b Cu5a: Theo chng trnh khng phn ban: (2 im) 1. Tm h s ca s hng cha x10 trong khai trin nh thc ca (2 + x)n bit

( )0 1 1 2 2 3 33 3 3 3 ... 1 2048nn n n n nn n n n nC C C C C + + + = 2. Trong mt phng vi h to Oxy cho im A(2; 2) v cc ng thng:

d1: x + y - 2 = 0 d2: x + y - 8 = 0 Tm to cc im B v C ln lt thuc d1 v d2 sao cho ABC vung cn ti A. Cu5b: Theo chng trnh phn ban: (2 im)

1. Gii phng trnh: ( ) ( )2 1 2 1 2 2 0x x + = 2. Cho hnh chp t gic u S.ABCD c y l hnh vung cnh a. Gi E l im i xng ca D qua trung im ca SA, M l trung im ca AE, N l trung im ca BC. Chng minh MN vung gc vi BD v tnh theo a khong cch gia hai ng thng MN v AC.

S 20 PHN CHUNG C TT C CC TH SINH

Cu1: (2 im) Cho hm s: y = 2

1x

x + 1. Kho st s bin thin v v th (C) ca hm s cho. 2. Tm to im M thuc (C), bit tip tuyn ca (C) ti M ct hai trc Ox, Oy ti A, B

v tam gic OAB c din tch bng 14

Cu2: (2 im)

1. Gii phng trnh: 2

sin cos 3 cos 22 2x x x + + =

2. Tm gi tr ca tham s m h phng trnh sau c nghim thc:

3 33 3

1 1 5

1 1 15 10

x yx y

x y mx y

+ + + = + + + =

Cu3: (2 im) Trong khng gian vi h to Oxyz cho hai im A(1; 4; 2 B(-1 2; 4) v ng thng :

1 21 1 2

x y z += =


Trang:18

1. Vit phng trnh ng thng d i qua trng tm G ca tam gic OAB v vung gc vi mt phng (OAB).

2. Tm to im M thuc ng thng sao cho MA2 + MB2- nh nht Cu4: (2 im)

1. Tnh tch phn: I = 3 2

1

lne

x xdx 2. Cho a b > 0. Chng minh rng:

1 12 22 2

b aa b

a b + +

PHN T CHN: Th sinh chn Cu 5.a hc Cu 5.b Cu5a: Theo chng trnh khng phn ban: (2 im) 1. Tm h s ca x5 trong khai trin thnh a thc ca: x(1 - 2x)5 + x2(1 + 3x)10 2. Trong mt phng vi h to Oxy cho ng trn (C): (x - 1)2 + (y + 2)2 = 9 v ng thng d: 3x - 4y + m = 0. Tm m trn d c duy nht mt im P m t c th k c hai tip tuyn PA, PB ti (C) (A, B l cc tip im) sao cho PAB u Cu5b: Theo chng trnh phn ban: (2 im)

1. Gii phng trnh: ( )2 2 1log 4 15.2 27 2log 04.2 3x x x+ + + = 2. Cho hnh chp S.ABCD c y l hnh thang, ABC = BAD = 900 , BA = BC = a, AD

= 2a. cnh bn SA vung gc vi y v SA = a 2 . Gi H l hnh chiu vung gc ca A trn SB. Chng minh tam gic SCD vung v tnh theo a khong cch t H n mt phng (SCD)

S 21

CU1: (2 im) Cho hm s: y = x4 - mx2 + m - 1 (1) (m l tham s)

1) Kho st s bin thin v v th ca hm s (1) khi m = 8. 2) Xc nh m sao cho th ca hm s (1) ct trc honh ti bn im phn bit. CU2: (2 im)

1) Gii bt phng trnh: ( ) ( )xxx 2.32log44log 1221

21 + +

2) Xc nh m phng trnh: ( ) 02sin24coscossin4 44 =+++ mxxxx c t nht mt nghim thuc on

2;0

CU3: (2 im) 1) Cho hnh chp S.ABC c y ABC l tam gic u cnh a v cnh bn SA vung gc vi mt

phng y (ABC). Tnh khong cch t im A ti mt phng (SBC) theo a, bit rng SA = 2

6a


02

3

1xdxx

CU4: (2 im) Trong mt phng vi h to cc vung gc Oxy, cho hai ng trn: (C1): x2 + y2 - 10x = 0, (C2): x2 + y2 + 4x - 2y - 20 = 0


Trang:19

1) Vit phng trnh ng trn i qua cc giao im ca (C1), (C2) v c tm nm trn ng thng x + 6y - 6 = 0. 2) Vit phng trnh tip tuyn chung ca cc ng trn (C1) v (C2). CU5: (2 im)

1) Gii phng trnh: 16212244 2 +=++ xxxx 2) i tuyn hc sinh gii ca mt trng gm 18 em, trong c 7 hc sinh khi 12, 6 hc sinh khi 11 v 5 hc sinh khi 10. Hi c bao nhiu cch c 8 hc sinh trong i i d tri h sao cho mi khi c t nht mt em c chn. CU6: ( Tham kho) Gi x, y, z l khong cch t im M thuc min trong ca ABC c 3 gc nhn n cc cnh BC,

CA, AB. Chng minh rng: R

cbazyx2

222 ++++ ; a, b, c l ba cnh ca , R l bn knh ng trn ngoi tip. Du "=" xy ra khi no?

S 22

CU1: (2 im)

1) Tm s n nguyn dng tho mn bt phng trnh: nCA nnn 9223 + , trong knA v knC

ln lt l s chnh hp v s t hp chp k ca n phn t.

2) Gii phng trnh: ( ) ( ) ( )xxx 4log1log413log

21

28

42 =++ CU2: (2,5 im)

Cho hm s: y = 2

22

+

xmxx

(1) (m l tham s)

1) Xc nh m hm s (1) nghch bin trn on [-1; 0]. 2) Kho st s bin thin v v th ca hm s (1) khi m = 1. 3) Tm a phng trnh sau c nghim:

( ) 012329 22 1111 =+++ ++ aa tt CU3: (1,5 im)

1) Gii phng trnh: x

xgx

xx2sin8

12cot21

2sin5cossin 44 =+

2) Xt ABC c di cc cnh AB = c; BC = a; CA = b. Tnh din tch ABC, bit rng: bsinC(b.cosC + c.cosB) = 20 CU4: (3 im)

1) Cho t din OABC c ba cnh OA; OB v OC i mt vung gc. Gi ; ; ln lt l cc gc gia mt phng (ABC) vi cc mt phng (OBC); (OCA) v (OAB). Chng minh rng:

3coscoscos ++ .


Trang:20

2) Trong khng gian vi h to cc Oxyz cho mt phng (P): x- y + z + 3 = 0 v hai im A(-1; -3; -2), B(-5; 7; 12). a) Tm to im A' l im i xng vi im A qua mt phng (P). b) Gi s M l mt im chy trn mt phng (P), tm gi tr nh nht ca biu thc: MA + MB. CU5: (1,0 im)

Tnh tch phn: I = ( ) +3ln

03

1x

x

e

dxe

S 23

CU1: (3,0 im)

Cho hm s: y = 3122

31 23 + mxmxx (1) (m l tham s)

1) Cho m = 21

a) Kho st s bin thin v v th (C) ca hm s (1) b) Vit phng trnh tip tuyn ca th (C), bit rng tip tuyn song song vi ng thng d: y = 4x + 2.

2) Tm m thuc khong

65;0 sao cho hnh phng gii hn bi th ca hm s (1) v cc

ng x = 0, x = 2, y = 0 c din tch bng 4. CU2: (2 im)

1) Gii h phng trnh:

==+

0loglog

034

24 yx

yx

2) Gii phng trnh: ( )

xxxxtg 4

24

cos3sin2sin21 =+

CU3: (2 im) 1) Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, SA vung gc vi mt phng (ABCD) v SA = a. Gi E l trung im ca cnh CD. Tnh theo a khong cch t im S n ng thng BE. 2) Trong khng gian vi h to cc Oxyz cho -ng thng

:

=+++=+++02012

zyxzyx

v mt phng (P): 4x - 2y + z - 1 = 0

Vit phng trnh hnh chiu vung gc ca ng thng trn mt phng (P). Cu4: (2 im)

1) Tm gii hn: L = x

xxx

3

0

11lim ++ 2) Trong mt phng vi h ta cac Oxy cho hai ng trn: (C1): x2 + y2 - 4y - 5 = 0 v (C2): x2 + y2 - 6x + 8y + 16 = 0 Vit phng trnh cc tip tuyn chung hai ng trn (C1) v (C2)


Trang:21

Cu5: (1 im)

Gi s x, y l hai s dng thay i tho mn iu kin x + y = 45

. Tm gi tr nh nht ca biu

thc: S = yx 4

14 + s 24

Cu1: (2 im)

1) Gii bt phng trnh: 12312 +++ xxx 2) Gii phng trnh: tgx + cosx - cos2x = sinx(1 + tgxtg

2x

)

Cu2: (2 im) Cho hm s: y = (x - m)3 - 3x (m l tham s) 1) Xc nh m hm s cho t cc tiu ti im c honh x = 0. 2) Kho st s bin thin v v th ca hm s cho khi m = 1.

3) Tm k h bt phng trnh sau c nghim: ( )

+


Trang:22

s 25 Cu1: (2 im)

Cho hm s: y = xmxx

+

1

2 (1) (m l tham s)

1) Kho st s bin thin v v th ca hm s (1) khi m = 0. 2) Tm m hm s (1) c cc i v cc tiu. Vi gi tr no ca m th khong cch gia hai im cc tr ca th hm s (1) bng 10. Cu2: (2 im)

1) Gii phng trnh: 0log3log16 2327 3 = xx xx 2) Cho phng trnh: a

xxxx =+++

3cos2sin1cossin2

(2) (a l tham s)

a) Gii phng trnh (2) khi a = 31

.

b) Tm a phng trnh (2) c nghim. Cu3: (3 im) 1) Trong mt phng vi h ta cac Oxy cho ng thng d: x - y + 1 = 0 v ng trn (C): x2 + y2 + 2x - 4y = 0. Tm to im M thuc ng thng d m qua ta k c hai ng thng tip xc vi ng trn (C) ti A v B sao cho gc AMB bng 600. 2) Trong khng gian vi h to cc Oxyz cho ng thng

d:

=+=+

04220122

zyxzyx

v mt cu (S): x2 + y2 + z2 + 4x - 6y + m = 0.

Tm m ng thng d ct mt cu (S) ti hai im M, N sao cho khong cch gia hai im bng 9. 3) Tnh th tch khi t din ABCD, bit AB = a; AC = b; AD = c v cc gc BAC; CAD; DAB u bng 600 Cu4: (2 im)


56 3 cossincos1

xdxxx

2) Tm gii hn: x

xxx cos1

1213lim23 2

0 ++

CU5: (1 im) Gi s a, b, c, d l bn s nguyn thay i tho mn 1 a < b < c < d 50. Chng minh bt ng

thc: b

bbdc

ba

50502 +++ v tm gi tr nh nht ca biu thc:

S = dc

da +

S 26

CU1: (2 im)

1) Kho st s bin thin v v th ca hm s: y = xxx 3231 23 +


Trang:23

2) Tnh din tch hnh phng gii hn bi th hm s (1) v trc honh.

CU2: (2 im)

1) Gii phng trnh: xx

sincos8

12 =

2) Gii h phng trnh: ( )( )

=+=+

3532log

3532log23

23

xyyy

yxxx

y

x

CU3: (2 im)

1) Cho hnh t din u ABCD, cnh a = 6 2 cm. Hy xc nh v tnh di on vung gc chung ca hai ng thng AD v BC.

2) Trong mt phng vi h ta cac Oxy cho elip (E): 149

22=+ yx v ng thng dm: mx -

y - 1 = 0. a) Chng minh rng vi mi gi tr ca m, ng thng dm lun ct elp (E) ti hai im phn bit. b) Vit phng trnh tip tuyn ca (E), bit rng tip tuyn i qua im N(1; -3)

CU4: (1 im) Gi a1, a2, ..., a11 l h s trong khai trin sau:

( ) ( ) 11921011110 ...21 axaxaxxx ++++=++ Hy tnh h s a5

CU5: (2 im)

1) Tm gii hn: L = ( )26

1 156lim

+ x

xxx

2) Cho ABC c din tch bng 23

. Gi a, b, c ln lt l di ca cc cnh BC, CA, AB v ha,

hb, hc tng ng l di cc ng cao k t cc nh A, B, C ca tam gic. Chng minh rng:

3111111

++

++cba hhhcba

S 27

CU1: (2 im)

1) Kho st s bin thin v v th ca hm s y = ( )12342 2

xxx

2) Tm m phng trnh: 2x2 - 4x - 3 + 2m 1x = 0 c hai nghim phn bit. CU2: (2 im)

1) Gii phng trnh: ( ) 0623 =++ xcosxsintgxtgx


Trang:24

2) Gii h phng trnh:

=+=322 yx

xy ylogxylog

CU3: (3 im)

1) Trong mt phng vi h ta cc Oxy cho parabol (P) c phng trnh y2 = x v im I(0;

2). Tm to hai im M, N thuc (P) sao cho INIM 4= . 2) Trong khng gian vi h to cc Oxyz cho t din ABCD vi A(2; 3; 2), B(6; -1; -2), C(-

1; -4; 3), D(1; 6; -5). Tnh gc gia hai ng thng AB v CD. Tm to im M thuc ng

thng CD sao cho ABM c chu vi nh nht. 3) Cho lng tr ng ABC. A'B'C' c y ABC l tam gic cn vi AB = AC = a v gc BAC =

1200, cnh bn BB' = a. Gi I l trung im CC'. Chng minh rng AB'I vung A. Tnh cosin ca gc gia hai mt phng (ABC) v (AB'I).

CU4: (2 im)

1) C bao nhiu s t nhin chia ht cho 5 m mi s c 4 ch s khc nhau?


0 2cos1

dx

xx

CU5: (1 im)

Tm gi tr ln nht v gi tr nh nht ca hm s: y = sin5x + 3 cosx ]

S 28

CU1: (2 im)

Cho hm s: y = ( )

( )mxmmxmx

++++++

2412 22

(1) (m l tham s)

1) Tm m hm s (1) c cc tr v tnh khong cch gia hai im cc tr ca th hm s (1). 2) Kho st s bin thin v v th ca hm s (1) khi m = 0 CU2: (2 im) 1) Gii phng trnh: cos2x + cosx(2tg2x - 1) = 2

2) Gii bt phng trnh: 11 21212.15 ++ ++ xxx CU3: (3 im) 1) Cho t din ABCD vi AB = AC = a, BC = b. Hai mt phng (BCD) v (ABC) vung gc vi nhau v gc BDC = 900. Xc nh tm v tnh bn knh mt cu ngoi tip t din ABCD thao a v b.


Trang:25

2) Trong khng gian vi h to cc Oxyz cho hai ng thng:

d1: 12

11

zyx =+= v d2:

=+=+

012013

yxzx

a) Chng minh rng d1, d2 cho nhau v vung gc vi nhau. b) Vit phng trnh tng qut ca ng thng d ct c hai ng thng d1, d2 v song song vi

ng thng : 23

47

14

== zyx

CU4: (2 im) 1) T cc ch s 0, 1, 2, 3, 4, 5 c th lp c bao nhiu s t nhin m mi s c 6 ch s khc nhau v ch s 2 ng cnh ch s 3?


0

23 1 dxxx

CU5: (1 im)

Tnh cc gc ca ABC bit rng: ( )

=

8332

2sin

2sin

2sin

4

CBA

bcapp

trong BC = a, CA = b, AB = c, p = 2

cba ++

S 29

CU1: (2 im)

Cho hm s: y = (x - 1)(x2 + mx + m) (1) (m l tham s)

1) Tm m th hm s (1) ct trc honh ti ba im phn bit.


CU2: (2 im)

1) Gii phng trnh: 032943 26 =++ xcosxcosxcos 2) Tm m phng trnh: ( ) 04

21

22 =+ mxlogxlog c nghim thuc khong (0; 1).

CU3: (3 im)

1) Trong mt phng vi h ta cc Oxy cho ng thng d: x - 7y + 10 = 0. Vit phng

trnh ng trn c tm thuc ng thng : 2x + y = 0 v tip xc vi ng thng d ti im A(4; 2).

2) Cho hnh lp phng ABCD.A'B'C'D'. Tm im M thuc cnh AA' sao cho mt phng

(BD'M) ct hnh lp phng theo mt thit din c din tch nh nht.


Trang:26

3) Trong khng gian vi h to cc Oxyz cho t din OABC vi A(0; 0; 3a ), B(0;

0; 0), C(0; a 3 ; 0) (a > 0). Gi M l trung im ca BC. Tnh khong cch gia hai ng thng AB v OM.

CU4: (2 im)

1) Tm gi tr ln nht v gi tr nh nht ca hm s: y = x6 + ( )3214 x trn on [-1; 1]. 2) Tnh tch phn: I =

5

2

2

1

ln

lnx

x

e

dxe

CU5: (1 im)

T cc ch s 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s t nhin, mi s c 6 ch s v

tho mn iu kin: Su ch s ca mi s l khc nhau v trong mi s tng ca ba ch s u

nh hn tng ca ba ch s cui mt n v?

S 30

CU1: (2 im)

Cho hm s: y = 112

xx

(1)

1) Kho st s bin thin v v th ca hm s (C) ca hm s (1). 2) Gi I l giao im ca hai ng tim cn ca (C). Tm im M thuc (C) sao cho tip tuyn ca (C) ti M vung gc vi ng thng IM. CU2: (2 im)

1) Gii phng trnh: ( )

11cos2

42sin2cos32 2

=

x

xx

2) Gii bt phng trnh: ( ) 06log1log2log 241

21 ++ xx

CU3: (3 im)

1) Trong mt phng vi h ta cc Oxy cho elip (E): 114

22=+ yx , M(-2; 3), N(5; n). Vit

phng trnh cc ng thng d1, d2 qua M v tip xc vi (E). Tm n trong s cc tip tuyn ca (E) i qua N v c mt tip tuyn song song vi d1 hoc d2 2) Cho hnh chp u S.ABC, y ABC c cnh bng a, mt bn to vi y mt gc bng (00 < < 900). Tnh th tch khi chp S.ABC v khong cch t nh A n mt phng (SBC). 3) Trong khng gian vi h to cc Oxyz cho hai im I(0; 0; 1), K(3; 0; 0). Vit phng trnh mt phng i qua hai im I, K v to vi vi mt phng xOy mt gc bng 300

CU4: (2 im)


Trang:27

1) T mt t gm 7 hc sinh n v 5 hc sinh nam cn chn ra 6 em trong s hc sinh n phi nh hn 4. Hi c bao nhiu cch chn nh vy?

2) Cho hm s f(x) = ( )xbxe

xa ++ 31 . Tm a v b bit rng

f'(0) = -22 v ( ) 510

= dxxf CU5: (1 im)

Chng minh rng: 2

2cos2xxxex ++ x R S 31

CU1: (2 im)

Cho hm s: y = 3

65 22

++++

xmxx

(1) (m l tham s)

1) Kho st s bin thin v v th ca hm s (1) khi m = 1. 2) Tm m hm s (1) ng bin trn khong (1; + ). CU2: (2 im)

1) Gii phng trnh: ( ) ( )xsin

xcosxsinxcosxcos +=+ 121

2

2) Cho hm s: f(x) = 2xlogx (x > 0, x 1) Tnh f'(x) v gii bt phng trnh f'(x) 0

CU3: (3 im) 1) Trong mt phng vi h ta cc Oxy cho ABC c nh A(1; 0) v hai ng thng ln lt cha cc ng cao v t B v C c phng trnh tng ng l: x - 2y + 1 = 0 v 3x + y - 1 = 0 Tnh din tch ABC. 2) Trong khng gian vi h to cc Oxyz cho mt phng (P): 2x + 2y + z - m2 - 3m = 0 (m l tham s)

v mt cu (S): ( ) ( ) ( ) 9111 222 =+++ zyx Tm m mt phng (P) tip xc vi mt cu (S). Vi m tm c, hy xc nh to tip im ca mt phng (P) v mt cu (S). 3) Cho hnh chp S.ABC c y ABC l tam gic vung ti B, AB = a, BC = 2a, cnh SA vung gc vi y v SA = 2a. Gi M l trung im ca SC. Chng minh rng AMB cn ti M v tnh din tch AMB theo a. CU4: (2 im) 1) T 9 ch s 0, 1, 2, 3, 4, 5, 6, 7, 8 c th lp c bao nhiu s t nhin chn m mi s gm 7 ch s khc nhau?


Trang:28


0

3 2 dxex x

CU5: (1 im)

Tm cc gc A, B, C ca ABC biu thc: Q = CsinBsinAsin 222 + t gi tr nh nht.

S 32

CU1: (2 im) 1) Kho st s bin thin v v th ca hm s (C) ca hm s: y = 2x3 - 3x2 - 1 2) Gi dk l ng thng i qua im M(0 ; -1) v c h s gc bng k. Tm k ng thng dk ct (C) ti ba im phn bit. CU2: (2 im)

1) Gii phng trnh: xsinxcos

tgxgxcot242+=

2) Gii phng trnh: ( ) xlog x = 1455 CU3: (3 im) 1) Trong khng gian vi h to cc Oxyz cho hai im A(2; 1; 1), B(0; -1; 3) v ng thng

d:

=+=083

01123zy

yx

a) Vit phng trnh mt phng (P) i qua trung im I ca AB v vung gc vi AB. Gi K l giao im ca ng thng d v mt phng (P), chng minh rng d vung gc vi IK. b) Vit phng trnh tng qut ca hnh chiu vung gc ca d trn mt phng c phng trnh: x + y - z + 1 = 0. 2) Cho t din ABCD c AD vung gc vi mt phng (ABC) v ABC vung ti A, AD = a, AC = b, AB = c. Tnh din tch ca BCD theo a, b, c v chng minh rng: 2S ( )cbaabc ++ CU4: (2 im)

1) Tm s t nhin n tho mn: 1002 333222 =++ nnnnnnnn CCCCCC trong knC l s t hp chp k ca n phn t.

2) Tnh tch phn: I = +e

xdxlnx

x

1

2 1

CU5: (1 im)

Xc nh dng ca ABC, bit rng: ( ) ( ) BsinAsincBsinbpAsinap =+ 22 trong BC = a, CA = b, AB = c, p =

2cba ++


Trang:29

S 33

CU1: (2,5 im)

1) Cho hm s: y = 1

12

+

xmxx

(*)

a) Kho st s bin thin v v th (C) ca hm s khi m = 1. b) Tm nhng im trn (C) c to l nhng s nguyn. c) Xc nh m ng thng y = m ct th ca hm s (*) ti hai im phn bit A, B sao cho OA vung gc vi OB. CU2: (1 im) Cho ng trn (C): x2 + y2 = 9 v im A(1; 2). Hy lp phng trnh ca ng thng cha dy cung ca (C) i qua A sao cho di dy cung ngn nht. CU3: (3,5 im)

1) Cho h phng trnh:

+=+=+

123

mymx

myx

a) Gii v bin lun h phng trnh cho. b) Trong trng hp h c nghim duy nht, hy tm nhng gi tr ca m sao cho nghim

(x0; y0) tho mn iu kin

>>

00

0

0

y

x

2) Gii cc phng trnh v bt phng trnh sau: a) sin(cosx) = 1 b) 11252 5


Trang:30

2) Tm cc im trn th hm s c to l cc s nguyn.

CU2: (2 im)

1) Gii phng trnh: xsinxcostgxxtg 3312 =

2) Gii bt phng trnh: ( ) ( ) ( ) 04221 331

31


Trang:31

Cho phng trnh: 04 22 =+ mxx (2) 1) Gii phng trnh (2) khi m = 2.

2) Xc nh m phng trnh (2) c nghim.

CU4: (1 im)

Cho cc ch s: 0, 1, 2, 3, 4. C bao nhiu s t nhin chn gm 5 ch s khc nhau lp t cc ch

s trn?

CU5: ( 2,5 im)

Cho elip (E) c hai tiu im l F1( 03; ); ( )032 ;F v mt ng chun c phng trnh: x =

34

.

1) Vit phng trnh chnh tc ca (E).

2) M l im thuc (E). Tnh gi tr ca biu thc:

P = MF.MFOMMFMF 2122

22

1 3 + 3) Vit phng trnh ng thng (d) song song vi trc honh v ct (E) ti hai im A, B

sao cho OA OB.

S 36

CU1: (2,5 im)

Cho hm s: y = x

xx 232 +

1) Kho st s bin thin v v th (C) ca hm s. 2) Tm trn ng thng x = 1 nhng im M sao cho t M k c hai tip tuyn ti (C) v hai tip tuyn vung gc vi nhau. CU2: (1,5 im) Gii cc phng trnh: 1) ( ) ( ) 24224 =+ xloglogxloglog

2) 55

33 xsinxsin =

CU3: (2 im) Gii cc bt phng trnh:

1) ( ) ( ) 06140252 1 + xxx


Trang:32

CU4: (2 im) Cho In = ( ) 10

22 1 dxxxn

v J n = ( ) 10

21 dxxxn

vi n nguyn dng.

1) Tnh Jn v chng minh bt ng thc: ( )121+ nIn

2) Tnh In + 1 theo In v tm n

n

x II

lim 1+

CU5: (2 im) 1) Trong mt phng (P) cho ng thng (D) c nh, A l mt im c nh nm trn (P) v khng thuc ng thng (D); mt gc vung xAy quay quanh A, hai tia Ax v Ay ln lt ct (D) ti B v C. Trn ng thng (L) qua A v vung gc vi (P) ly im S c nh khc A. t SA = h v d l khong cch t im A n (D). Tm gi tr nh nht ca th tch t din SABC khi xAy quay quanh A. 2) Trong mt phng vi h ta cc Oxy cho ABC. im M(-1; 1) l trung im ca cnh BC; hai cnh AB v AC theo th t nm trn hai ng thng c phng trnh l: x + y - 2 = 0; 2x + 6y + 3 = 0. Xc nh to ba nh A, B, C.

S 37

CU1: (3 im) Cho hm s: y = x3 - 3mx + 2 c th l (Cm) (m l tham s) 1) Kho st s bin thin v v th (C1) ca hm s khi m = 1. 2) Tnh din tch hnh phng gii hn bi (C1) v trc honh. 3) Xc nh m (Cm) tng ng ch c mt im chung vi trc honh.

CU2: (1 im) 1) Chng minh rng vi mi s nguyn dng n ta u c:

nnnnnnnnnn C...CCCC...CCC

22

42

22

02

122

52

32

12 ++++=++++

2) T cc ch s 1, 2, 3, 4, 5 c th lp c bao nhiu s gm 3 ch s khc nhau nh hn 245.

CU3: (1,5 im)

1) Gii h phng trnh: ( )( )( )( )

=++=

15

322

22

yxyx

yxyx

2) Gii phng trnh: xx +=+ 173 CU4: (1,5 im)

Cho phng trnh: ( ) 01122 =++ mxcosmxcos (m l tham s) 1) Gii phng trnh vi m = 1.


Trang:33

2) Xc nh m phng trnh c nghim trong khong

;2

.

CU5: (3 im) 1) Cho khi chp t gic u S.ABCD c cc cnh bn v cnh y u bng a. Gi M, N v P ln lt l trung im ca cc cnh AD, BC v SC. Mt phng (MNP) ct SD ti Q. Chng minh rng MNPQ l hnh thang cn v tnh din tch ca n. 2) Trong khng gian vi h to cc Oxyz cho hai ng thng:

(D1):

==

=

tz

ty

tx 1 v (D2):

==

=

'tz

'ty

'tx

12

(t, t' R)

a) Chng minh (D1), (D2) cho nhau v tnh khong cch gia hai ng thng y. b) Tm hai im A, B ln lt trn (D1), (D2) sao cho AB l on vung gc chung ca (D1) v (D2).

S 38

CU1: (3 im)

Cho hm s: y = 1

12

+

xmxx

1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Xc nh m hm s ng bin trn cc khong (- ; 1) v (1; + ) 3) Vi gi tr no ca m th tim cn xin ca th hm s to vi cc trc to mt tam gic c din tch bng 4 (n v din tch).

CU2: (2 im)

Cho phng trnh: ( ) ( ) mtgxtgx =++ 223223 1) Gii phng trnh khi m = 6.

2) Xc nh m phng trnh c ng hai nghim phn bit nm trong khong

22

; .

CU3: (2 im)

1) Gii bt phng trnh: ( )43

161313

414

xx loglog


032 xdxsinxsinxsin

CU4: (2 im)


Trang:34

Trong mt phng vi h ta cc Oxy cho ABC v im M(-1; 1) l trung im ca AB. Hai cnh AC v BC theo th t nm trn hai ng: 2x + y - 2 = 0 v x + 3y - 3 = 0 1) Xc nh ta ba nh A, B, C ca tam gic v vit phng trnh ng cao CH.

2) Tnh din tch ABC. CU5: (1 im)

Gi s x, y l cc nghim ca h phng trnh:

+=+=+

32

12222 aayx

ayx

Xc nh a tch P = x.y t gi tr nh nht.

S 39

CU1: (2 im)

Cho hm s: y = 2

52

+

xxx

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

2) Bin lun theo m s nghim ca phng trnh: 2

52

+

x

xx = m

CU2: (2 im)

1) Gii phng trnh: 01 =++ xcosxsin 2) Gii bt phng trnh: ( ) xlogxlog x 2222 + 4 CU3: (1 im)

Gii h phng trnh: ( )

++=+=

2

722

33

yxyx

yxyx

CU4: (1,5 im)

Tnh cc tch phn sau: I1 = ( )

+20

442 dxxcosxsinxcos I2 = 2

0

5 xdxcos

CU5: (3,5 im) 1) Trong mt phng vi h ta cc Oxy cho ng trn (S) c phng trnh: x2 + y2 - 2x - 6y + 6 = 0 v im M(2 ; 4) a) Chng minh rng im M nm trong ng trn. b) Vit phng trnh ng thng i qua im M, ct ng trn ti hai im A v B sao cho M l trung im ca AB. c) Vit phng trnh ng trn i xng vi ng trn cho qua ng thng AB.


Trang:35

2) Cho hnh chp t gic S.ABCD c di tt c cc cnh u bng a. Chng minh rng: a) y ABCD l hnh vung. b) Chng minh rng nm im S, A, B, C, D cng nm trn mt mt cu. Tm tm v bn knh ca mt cu .

S 40

CU1: (2 im)

Cho hm s: y = ( )

( )11322

++

mxmxmx

1) Kho st s bin thin v v th ca hm s khi m = 2.

2) Tm tt c cc gi tr ca m hm s cho ng bin trong khong (0; + ). CU2: (2 im)

1) Tnh tch phn: I = ( )

20

33 dxxsinxcos

2) T 5 ch s 0, 1, 2, 5, 9 c th lp c bao nhiu s l, mi s gm 4 ch s khc nhau.

CU3: (3 im)

1) Gii phng trnh: ( ) 442 =+ xsinxcosxsin 2) Gii h phng trnh:

+=+=

432

43222

22

yxy

xyx

3) Cho bt phng trnh: ( ) ( ) 114 2525


Trang:36

Cho hm s: y = x3 - mx2 + 1 (Cm)

1) Khi m = 3

a) Kho st s bin thin v v th ca hm s.

b) Tm trn th hm s tt c cc cp im i xng nhau qua gc to .

2) Xc nh m ng cong (Cm) tip xc vi ng thng (D) c phng trnh

y = 5. Khi tm giao im cn li ca ng thng (D) vi ng cong (Cm).

CU2: (1,5 im)

1) Gii bt phng trnh: ( ) ( ) 1331 310310 ++ + xxxx 0 2) Gii phng trnh: ( ) 01641 323 =++ xlogxxlogx CU3: (2 im)

1) Gii phng trnh: ( )( ) 45252 =++++ xxxx 2) Gii phng trnh:

xcosxcosxcos

17822 =+ CU4: (2 im)

1) Trong khng gian vi h to cc Oxyz cho im A(-1; 2; 5), B(11; -16; 10). Tm trn mt

phng Oxy im M sao cho tng cc khong cch t M n A v B l b nht.


248

7

21dx

xx

x

CU5: (2 im)

Trn tia Ox, Oy, Oz i mt vung gc ln lt ly cc im khc O l M, N v S vi OM = m,

ON = n v OS = a.

Cho a khng i, m v n thay i sao cho m + n = a.

1) a) Tnh th tch hnh chp S.OMN

b) Xc nh v tr ca cc im M v N sao cho th tch trn t gi tr ln nht.

2) Chng minh:

S 42

CU1: (2 im)

1) Kho st s bin thin v v th (C) ca hm s: y = 21

+

xx

2) Tm cc im trn th (C) ca hm s c to l nhng s nguyn.


Trang:37

3) Tm cc im trn th (C) sao cho tng khong cch t im n hai tim cn l nh nht.

CU2: (2 im)

1) Gii phng trnh: 012315 = xxx 2) Gii h phng trnh:

( )( )

=+=+

223223

xylog

yxlog

y

x

CU3: (1 im)

Gii phng trnh lng gic: 022 3 =+ xcosxcosxsin CU4: (2 im)

Cho D l min gii hn bi cc ng y = tg2x; y = 0; x = 0 v x = 4

.

1) Tnh din tch min D.

2) Cho D quay quanh Ox, tnh th tch vt th trn xoay c to thnh.

CU5: (1,5 im)

Trong khng gian vi h to cc Oxyz cho ba im A(1; 4; 0), B(0; 2; 1),

C(1; 0; -4).

1) Vit phng trnh tng qut ca mt phng () i qua im C v vung gc vi ng thng AB.

2) Tm to im C' i xng vi im C qua ng thng AB.

CU6: (1,5 im)

1) Gii phng trnh: xxCCC xxx 149662321 =++ (x 3, x N)

2) Chng minh rng: 1919201720

520

320

120 2=+++++ CC...CCC

S 43

CU1: (2,5 im)

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

2

xx

.

2) Bin lun theo tham s m s nghim ca phng trnh: mxx =1

2

CU2: (2,5 im)

1) Chng minh rng nu x, y l hai s thc tho mn h thc:

x + y = 1 th x4 + y4 81


Trang:38

2) Gii phng trnh: 12822324222 212 ++>++ + x.x..xx xxx

CU3: (2,5 im)

1) Gii phng trnh: 023962422

=+xcos

xcosxsinxsin

2) Cc gc ca ABC tho mn iu kin: ( )CcosBcosAcosCsinBsinAsin 222222 3 ++=++ Chng minh rng ABC l tam gic u. CU4: (2,5 im)

1) Tnh tch phn: e

xdxlnx1

22

2) Cho hnh lp phng ABCD.A'B'C'D' vi cc cnh bng a. Gi s M, N ln lt l trung

im ca BC, DD'. Tnh khong cch gia hai ng thng BD v MN theo a.

S 44

CU1: (3 im)

Cho hm s: y = x3 - 3mx2 + 3(2m - 1)x + 1 (1)


2) Xc nh m sao cho hm s (1) ng bin trn tp xc nh.

3) Xc nh m sao cho hm s (1) c mt cc i v mt cc tiu. Tnh to ca im cc tiu.

CU2: (2 im)

1) Gii phng trnh: 23sin2sinsin 222 =++ xxx 2) Tm m phng trnh: ( )33 242

21

22 =+ xlogmxlogxlog

c nghim thuc khong [32; + ). CU3: (2 im)

1) Gii h phng trnh:

=+=+

015132

93222

22

yxyx

yxyx

2) Tnh tch phn: e

dxx

xln

13


Trang:39

CU4: (1,5 im)

Cho hnh chp S.ABC c y ABC l tam gic u cnh a v SA vung gc vi mt phng (ABC).

t SA = h.

1) Tnh khong cch t A n mt phng (SBC) theo a v h.

2) Gi O l tm ng trn ngoi tip tam gic ABC v H l trc tm tam gic SBC. Chng minh:

OH (SBC). CU5: (1,5 im)

Trong khng gian vi h to cc Oxyz cho ng thng d v mt phng (P):

d:

==+03203

zy

zx (P): x + y + z - 3 = 0

1) Vit phng trnh mt phng (Q) cha ng thng d v qua im M(1; 0; -2).

2) Vit phng trnh hnh chiu vung gc ca ng thng d trn mt phng (P).

S 45

CU1: (3 im)

Cho hm s: y = 1

12

xxx

(C)

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

2) Lp phng trnh tip tuyn vi (C) ti im c honh x = 0.

3) Tm h s gc ca ng thng ni im cc i, cc tiu ca th (C).

CU2: (2,5 im)

1) Gii phng trnh: xxx .4269 =+ .

2) Tnh: ++2

02

3

123

xx

dxx

CU3: (2,5 im)

1) Gii h phng trnh:

=+=+

26

233 yx

yx

2) Tnh gc C ca ABC nu: ( )( ) 211 =++ gBcotgAcot CU4: (2 im)

Trong khng gian vi h to cc Oxyz :

1) Cho 2 ng thng:

(1):

==

00

y

x (2):

=

=+0

01z

yx

Chng minh (1) v (2) cho nhau.


Trang:40

2) Cho 2 im A(1 ; 1 ; -1), B(3 ; 1 ; 1) v mt phng (P) c phng trnh:

x + y + z - 2 = 0

Tm trn mt phng (P) cc im M sao cho MAB l tam gic u.

S 46 CU1: (2,5 im) Cho hm s: y = x3 - (2m + 1)x2 - 9x (1) 1) Vi m = 1; a) Kho st s bin thin v v th (C) ca hm s (1). b) Cho im A(-2; -2), tm to im B i xng vi im A qua tm i xng ca th (C). 2) Tm m th ca hm s (1) ct trc honh ti ba im phn bit c cc honh lp thnh mt cp s cng. CU2: (2 im) 1) Gii phng trnh: 03sin2cos4cossin =+ xxxx 2) Cho ABC cnh a, b, c tho mn h thc: 2b = a + c.

Chng minh rng: 32

cot2

cot =CgAg . CU3: (2 im)

1) Gii bt phng trnh: ( ) ( )12lg213lg 22 +> xxx

2) Tm a h phng trnh sau c nghim duy nht: ( )( )

=+=+

1

12

2

xayxy

yaxxy

CU4: (1,5 im)

1) Tnh tch phn: I = +++2

0 5cos3sin41sin3cos4

dx

xxxx

2) Tnh tng: P = 51054

1043

1032

1021

10110 33333 CCCCCC ++

1010109

1098

1087

1076

106 33333 CCCCC +++

CU5: (2 im) 1) Trong khng gian vi h to cc Oxyz cho mt phng (P) v mt cu (S) ln lt c phng trnh: (P): y - 2z + 1 = 0 (S): x2 + y2 + z2 - 2z = 0. Chng minh rng mt phng (P) v mt cu (S) ct nhau. Xc nh tm v bn knh ca ng trn giao tuyn. 2) Cho hnh chp u S.ABC nh S, chiu cao l h, y l tam gic u cnh a. Qua cnh AB dng mt phng vung gc vi SC. Tnh din tch thit din to thnh theo a v h.


Trang:41

S 47

CU1: (2,5 im)

Cho hm s: y = 1

2 222

+++

xmxmx

(m l tham s)


2) Tm m trn th c hai im i xng nhau qua gc to .

CU2: (2 im)

1) Gii phng trnh: 09328322 122 =+ +++ xxxx .

2) Cho ABC. Chng minh rng nu Csin

BsintgCtgB

2

2= th tam gic l tam gic vung hoc cn.

CU3: (2 im)

1) Tnh tch phn: 9

1

3 1 dxx x

2) Gii h phng trnh: ( )

+=++=+

yxyx

yyxx

322

22

CU4: (2,5 im)

1) Cho hnh chp tam gic u S.ABC c gc gia mt bn v mt y l v SA = a. Tnh th tch hnh chp cho.

2) Trong khng gian vi h to cc Oxyz vi h to vung gc Oxyz, cho hai ng

thng: 1: 3

32

21

1 == zyx 2:

=+=+

053202

zyx

zyx

Tnh khong cch gia hai ng thng cho.

CU5: ( 1 im)

Chng minh rng: P1 + 2P2 + 3P3 + ... + nPn = Pn + 1 - 1

Trong n l s t nhin nguyn dng v Pn l s hon v ca n phn t.

S 48

CU1: (3 im)

Cho hm s: y = x3 + 3x2 + 1 (1)



Trang:42

2) ng thng (d) i qua im A(-3 ; 1) c h gc l k. Xc nh k (d) ct th hm s (1) ti

ba im phn bit.

CU2: (2,5 im)

1) Gii phng trnh: 0221 =++++ xcosxsinxcosxsin


=++=++095

18322

2

yxx

yxxx

CU3: (2 im)

1) Gii bt phng trnh: ( )3824 1+ xlogxlog 1 2) Tm gii hn:

xcosxxlim

x ++

11213 23 2

0

CU4: (1,5 im)

Trong mt phng vi h ta cc Oxy cho hai im A(1; 2), B(3; 4). Tm trn tia Ox mt

im P sao cho AP + PB l nh nht.

CU5: (1 im)

Tnh tch phn: I = ++2

03 23

1 dxx

x

S 49

CU1: (2,5 im)

Cho hm s: y = ( ) ( ) 43131 23 +++ xmxmx (1) (m l tham s)


2) Xc nh m hm s (1) ng bin trong khong: 0 < x < 3

CU2: (2 im)

1) Gii phng trnh: 0322212 333 =+++++ xxx (1) 2) Cho phng trnh: ( ) 061232 2 =++ mxcosxsinmxsin a) Gii phng trnh vi m = 1.

b) Vi gi tr no ca m th phng trnh (1) c nghim.


Trang:43

CU3: (1 im)

Gii h bt phng trnh:

>+


Trang:44

CU4: (2,5 im) 1) Trong mt phng vi h ta cac Oxy cho im A(8; 6). Lp phng trnh ng thng qua A v to vi hai trc to mt tam gic c din tch bng 12. 2) Trong khng gian vi h to cc Oxyz Cho A(1; 2; 2), B(-1; 2; -1), C(1; 6; -1), D(-1; 6; 2) a) Chng minh rng ABCD l hnh t din v tnh khong cch gia hai ng thng AB v CD. b) Vit phng trnh mt cu ngoi tip t din ABCD. CU5: (1,5 im) Cho hai hm s f(x), g(x) xc nh, lin tc v cng nhn gi tr trn on [0; 1]. Chng minh

rng: ( ) ( ) ( ) ( )

1

0

1

0

21

0dxxgdxxfdxxgxf

S 51

CU1: (2 im)

Cho hm s: y = ( )( )

mmxmxxm

+++ 421 2

(Cm) (m l tham s, m 0, -41

)

1) Kho st s bin thin v v th ca hm s (C2) vi m = 2.

2) Tm m hm s (Cm) c cc i, cc tiu v gi tr cc i, cc tiu cng du.

CU2: (2 im)

1) Gii h phng trnh:

++=++=

22

223

3

yxy

xyx

2) Gii phng trnh: tg2x + cotgx = 8cos2x

CU3: (2,5 im)

1) Tnh th tch ca hnh chp S.ABC bit y ABC l mt tam gic u cnh a, mt bn

(SAB) vung gc vi y, hai mt bn cn li cng to vi y gc . 2) Trong khng gian vi h to cc Oxyz cho hai ng thng:

(D1):

=+=+

01040238

zy

zx (D2):

=++=

022032

zy

zx

a) Vit phng trnh cc mt phng (P) v (Q) song song vi nhau v ln lt i qua (D1) v (D2).

b) Vit phng trnh ng thng (D) song song vi trc Oz v ct c hai ng thng (D1), (D2)

CU4: (2 im)

1) Tnh tng: S = ( ) nnnnnnn nC....CCCC 1432 4321 +++ Vi n l s t nhin bt k ln hn 2, knC l s t hp chp k ca n phn t.


Trang:45


1 12xxdx

CU5: (1,5 im)

Cho ba s bt k x, y, z. Chng minh rng:

222222 zyzyzxzxyxyx +++++++ S 52

CU1: (2 im)

Cho hm s: y = 11

+

xx

(1) c th (C)


2) Chng minh rng ng thng d: y = 2x + m lun ct (C) ti hai im A, B thuc hai

nhnh khc nhau. Xc nh m on AB c di ngn nht.

CU2: (2,5 im)

Cho phng trnh: 03232322 224 =+ m. xx (1)

1) Gii phng trnh (1) khi m = 0.

2) Xc nh m phng trnh (1) c nghim.

CU3: (2,5 im)

Gii cc phng trnh v bt phng trnh sau:

1) xtgxsinxcos

xcosxsin 28

1322

66=

+

2) ( ) ( )2431243 2329 ++>+++ xxlogxxlog CU4: (1,5 im)

Trong khng gian vi h to cc Oxyz Cho A(1; 1; 1), B(1; 2; 0) v mt cu (S): x2 +

y2 + z2 - 6x - 4y - 4z + 13 = 0. Vit phng trnh mt phng cha ng thng AB v tip xc vi

(S).

CU5: (1,5 im)

Tnh tng: S = nnnnn Cn

...CCC1

131

21 211

+++++

Bit rng n l s nguyn dng tho mn iu kin: 7921 =++ nnnnnn CCC knC l s t hp chp k ca n phn t.


Trang:46

S 53

CU1: (2 im)

Cho hm s: y = -x3 + 3x2 - 2

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

2) Tm t phng trnh: 023 223 =+ tlogxx c 6 nghim phn bit.

CU2: (3 im)

1) Trong mt phng vi h ta cc Oxy cho ng trn

(C): ( ) ( ) 413 22 =+ yx . Vit phng trnh tip tuyn ca (C) bit rng tip tuyn ny i qua im M0(6; 3)

2) Trong khng gian vi h to cc Oxyz cho hnh hp ABCD.A'B'C'D' Vi A(2; 0;

2), B(4; 2; 4), D(2; -2; 2) v C'(8; 10; -10).

a) Tm to cc nh cn li ca hnh hp ABCD.A'B'C'D'.

b) Tnh th tch ca hnh hp ni trn.

CU3: (2 im)

1) Gii phng trnh: 21 +=++ xxx

2) Gii h phng trnh:

==+

22

122 yy

xx

ysinxsin

CU4: (2 im)

1) Chng minh rng: knkn

kn

kn CCCCCCC =++ 22221212202

n k + 2 ; n v k l cc s nguyn dng, knC l s t hp chp k ca n phn t. 2) Tnh din tch hnh phng gii hn bi parabol: y = -x2 - 4x; ng thng x = -1; ng thng

x = -3 v trc Ox

CU5: (1 im)

Cho 2 s nguyn dng m, n l s l

Tnh theo m, n tch phn: I = 2

0xdxcosxsin mn

S 54

CU1: (2 im)

1) Kho st s bin thin v v th ca hm s: y = xxx 323

23

+


Trang:47

2) Da v th (C) Cu trn, hy bin lun theo tham s m s nghim ca phng trnh:

meee xxx

=+ 323

23

CU2: (3 im) 1) Trong mt phng vi h ta cc Oxy cho elp (E) c phng trnh:

122

2

2=+

b

y

a

x (a > 0, b > 0)

a) Tm a, b bit Elip (E) c mt tiu im l F1(2; 0) v hnh ch nht c s ca (E) c din tch

l 12 5 (vdt). b) Tm phng trnh ng trn (C) c tm l gc to . Bit rng (C) ct (E) va tm c Cu trn ti 4 im lp thnh hnh vung. 2) Trong khng gian vi h to cc Oxyz tm theo a, b, c (a, b, c 0) to cc nh ca hnh hp ABCD.A'B'C'D'. Bit A(a; 0; 0); B(0; b; 0) C(0; 0; c) v D'(a; b; c). CU3: (2 im) 1) Gii v bin lun phng trnh sau theo tham s m: ( ) 012 333 = mlogxlogxlog 2) Gii phng trnh: ( ) 032332 =++++ xcosxcosxcosxsinxsinxsin CU4: (2 im) 1) Cho f(x) l hm lin tc trn on [0; 1]. Chng minh rng:

( ) ( )

= 20

2

0dxxcosfdxxsinf

2) Tnh cc tch phn:

I =

+2

020032003

2003

xcosxsin

xdxsin J =

+2

020032003

2003

xcosxsin

xdxcos

CU5: (1 im)

Gii bt phng trnh: ( ) nnnnnn C.C.C.!n 323 720 knC l t hp chp k ca n phn t.

S 55 CU1: (2 im)

1) Kho st s bin thin v v th ca hm s: y = x4 - 10x2 + 9 2) Tm tt c cc gi tr ca tham s m phng trnh: x - 3mx + 2 = 0 c nghim duy nht. CU2: (2 im)

1) Tm tt c cc ng tim cn xin ca th hm s: y = 2x + 21 x+


Trang:48

2) Tnh th tch ca vt th trn xoay c to ra khi cho hnh phng gii hn bi cc

ng: y = ex ; y = e1

; y = e v trc tung quay xung quanh Oy.

CU3: (2 im)

1) Cho a thc: P(x) = ( )20051516 x , khai trin a thc di dng: P(x) = 20052005

2210 xa...xaxaa ++++

Tnh tng: S = 2005210 a...aaa ++++ 2) Gii h phng trnh: ( )

=+

=5

115223

22 logyxlog

yx

CU4: (2 im) 1) Cho ABC c di cc cnh BC, CA, AB theo th t lp thnh cp s cng. Tnh gi tr

ca biu thc: P = 22CgcotAgcot

2) Trong mt phng vi h to cc vung gc Oxy cho hypebol (H): 1916

22= yx .

Lp phng trnh ca elp (E), bit rng (E) c cc tiu im l cc tiu im ca (H) v (E) ngoi tip hnh ch nht c s ca (H) CU5: (2 im)

1) Trong khng gian vi h to cc Oxyz cho ABC c im B(2; 3; -4), ng cao CH c phng trnh:

522

51

== zyx v ng phn gic trong gc A l AI c phng trnh:

21

13

75 +== zyx . Lp phng trnh chnh tc ca cnh AC.

2) CMR: trong mi hnh nn ta lun c: 26

V

3

32

S

(V l th tch hnh nn, S l din tch xung quanh ca hnh nn) S 56

CU1: (2 im)

Cho hm s: y = ( )

1112

+++

xmxmx

(1) (m l tham s)


2) Chng minh rng hm s (1) lun c gi tr cc i (yC) v gi tr cc tiu (yCT) vi m. Tm cc gi tr ca m (yC)2 = 2yCT

CU2: (2 im)

1) Gii phng trnh: 3cosx ( ) 1221 2 = xsinxsinxcosxsin


Trang:49

2) Gii h bt phng trnh:

+

045

0224

2

xx

xx

CU3: (2 im)


0

23 1 dxxx

2) Tm s nguyn dng n tho mn ng thc: nCA nn 16223 =+

CU4: (3 im)

1) Cho t din ABCD c di cnh AB = x (x > 0), tt c cc cnh cn li c di bng

1. Tnh d di on vung gc chung ca hai cnh AB v CD. Tm iu kin i vi x Cu ton

c ngha.

2) Trong khng gian vi h to cc Oxyz cho t din OABC c O l gc ta , A Ox, B Oy, C Oz v mt phng (ABC) c phng trnh: 6x + 3y + 2z - 6 = 0.

a) Tnh th tch khi t din OABC.

b) Xc nh to tm v tnh bn knh ca mt cu ngoi tip khi t din OABC.

CU5: (1 im)

Cho x, y l hai s thc dng khc 1.

Chng minh rng nu: ( ) ( )yloglogxloglog xyyx = th x = y.

S 57 CU1: (2 im)

Cho hm s: y = 252

xx

1) Kho st s bin thin v v th ca hm s. 2) Vit phng trnh tip tuyn ca th hm s, bit tip tuyn i qua im A(-2; 0). CU2: (3 im)

1) Gii phng trnh: xsinxsin 24

3 =

+ 2) Gii bt phng trnh: ( ) ( )11 11 2 +>+ xlogxlog xx

3) Gii h phng trnh:

==+

72

343222

22

yx

xyyx

CU3: (2 im)


Trang:50

1) Tnh tch phn: ++2

02

3

12dx

xx

x

2) Tm h s ln nht ca a thc trong khai trin nh thc Niutn ca: 15

32

31

+ x

CU4: (3 im) 1) Cho hnh lp phng ABCD.A'B'C'D'. Chng minh rng cc im gia ca 6 cnh khng

xut pht t hai u ng cho AC' l nhng nh ca mt lc gic phng u. 2) Trong mt phng vi h ta cc Oxy cho hai ng thng: x + y - 1 = 0 v 3x - y + 5 = 0 Hy tm din tch hnh bnh hnh c hai cnh nm trn hai ng thng cho, mt nh l giao im ca hai ng v giao im ca hai ng cho l I(3; 3). 3) Trong khng gian vi h to cc Oxyz cho hai ng thng:

d1:

=+=+053

0523zy

yx v d2:

252

12

=+= zyx

Chng minh rng hai ng thng cho nhau v tm phng trnh ng vung gc chung ca chng.

S 58

CU1: (4 im)

Cho hm s: y = mx

mx

+ 13 (1)

1) Xc nh m hm s (1) nghch bin trong khong (1; + ) 2) Kho st s bin thin v v th ca hm s (1) khi m = 1, gi th ca hm s ny l

(C).

3) Tm hai im A, B thuc (C) sao cho A v B i xng vi nhau qua ng thng (d): x +

3y - 4 = 0.

CU2: (2 im)

Cho phng trnh: x2 - 2ax + 2 - a = 0 (1)

1) Xc nh a phng trnh (1) c hai nghim x1, x2 sao cho: -2 < x1 < 3 < x2

2) Xc nh a phng trnh (1) c hai nghim x1, x1 sao cho: 22

21 xx + t gi tr nh

nht.

CU3: (1 im)

Cho ABC c 3 gc tho mn iu kin sau: sinA + cosA + sinB - cosB + sinC - cosC = 1. Chng minh rng: ABC l tam gic vung. CU4: (3 im)


Trang:51

Cho ABC c A(-1; 5) v phng trnh ng thng BC: x - 2y - 5 = 0 (xB < xC) bit I(0 ; 1) l tm ng trn ngoi tip ABC. 1) Vit phng trnh cc cnh AB v AC.

2) Gi A1, B1, C1 ln lt l chn ng cao v t cc nh A, B, C ca tam gic. Tm to

cc im A1, B1, C1

3) Gi E l tm ng trn ni tip A1B1C1. Tm to im E.

S 59

CU1: (2,5 im)

Cho hm s: y = 1

2

+

xmxx (1) (m l tham s)


2) Tm m th hm s (1) ct trc honh ti hai im A, B phn bit v cc tip tuyn

ca th hm s (1) ti A, B vung gc vi nhau.

CU2: (2 im)

1) Gii phng trnh: ( )1

22

1=+ gxcot

xsinxcosxgcottgx

2) Gii bt phng trnh:

( ) ( )23233323 43282 xlogxxxlogxlogxlogx ++ CU3: (2 im)

1) Tnh din tch hnh phng gii hn bi cc ng y = 4 - x2 v y = xx 22 .

2) Tnh tch phn: I = ( ) ++1

021

1x

dxxln

CU4: (1,5 im)

Trong mt phng vi h ta cc Oxy cho ABC c nh A(2; -3) , B(3; -2) v din tch

ABC bng 23

. Bit trng tm G ca ABC thuc ng thng d: 3x - y - 8 = 0. Tm to im

C.

CU5: (2 im)


Trang:52

Trong khng gian vi h to cc Oxyz cho im A(1; 2; -1) , B(7; -2; 3)

v ng thng d:

=+=+

040432

zy

yx

1) Chng minh rng hai ng thng d v AB dng phng.

2) Tm to giao im ca ng thng d vi mt phng trung trc ca on thng AB.

3) Trn d, tm im I sao cho di ng gp khc IAB ngn nht.

S 60 CU1: (2,5 im)

Cho hm s: y = mx

mmxx+

+ 22 (1) 1) Kho st s bin thin v v th ca hm s (1) vi m = 1. 2) Chng minh rng nu th (Cm) ca hm s (1) ct Ox ti im x0 th cc

tip tuyn ct (Cm) ti im c h s gc l k = mx

mx+

0

0 22

p dng: Tm m th (Cm) ct Ox ti hai im phn bit v tip tuyn ti hai im ca (Cm) vung gc vi nhau. CU2: (1,5 im)

Gii phng trnh: 1) sinx.cosx + cosx = -2sin2x - sinx + 1 2) ( ) 161 12 +=+ xlogxlog CU3: (2 im)

1) Bng cch t x = t2

, hy tnh tch phn: I =

+2

0dx

xcosxsinxsin

2) Tm m bt phng trnh: mx - 3x m + 1 c nghim. CU4: (3 im)

1) Cho hnh lp phng ABCD.A'B'C'D'. Gi I, J ln lt l trung im ca A'D' v B'B. Chng minh rng IJ AC' 2) Trong khng gian vi h to cc Oxyz cho cc ng thng:

(d1):

+=+=

=

tz

ty

x

324

1 v (d2):

=+=

=

223

3

z

'ty

'tx

(t, t' R)

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


Trang:53

b) Vit phng trnh mt cu (S) c ng knh l on vung gc chung ca (d1) v (d2). CU5: (1 im)

Chng minh rng: 02

332 >++ xgxcotxcos vi x

2

0;

S 61 CU1: (2 im)

Cho hm s: y = 1

22

++

xxx

1) Kho st s bin thin v v th (C) ca hm s. 2) Chng minh rng trn th (C) tn ti v s cp im ti cc tip tuyn ca th song song vi nhau. CU2: (2 im)

1) Gii phng trnh:

=33

4 2 xcosxcos


=+=+

3141131411

xylog

yxlog

y

x

CU3: (3 im) 1) Trong mt phng vi h ta cc Oxy cho im F(3; 0) v ng thng

(d) c phng trnh: 3x - 4y + 16 = 0 a) Vit phng trnh ng trn tm F v tip xc vi (d). b) Chng minh rng parabol (P) c tiu im F v nh l gc to tip xc vi (d). 2) Cho t din ABCD c AB, AC, AD vung gc vi nhau tng i mt. Gi H l hnh chiu ca A ln mt phng (BCD) v S, S1, S2, S3 ln lt l din tch ca cc mt (BCD), (ABC), (ACD), (ABD). Chng minh rng:

a) 22221111

ADACABAH++=

b) 2322

21

2 SSSS ++= CU4: (2 im)

1) Tnh tch phn: I = ( )e

dxxlncos1

2) Tm gi tr ln nht v gi tr nh nht ca hm s F(t) xc nh bi:

F(t) = t

dxxcosx0

2


Trang:54

CU5: (1 im) T cc ch s 0, 1, 2, 3, 4, 5, 6, 7 c th lp c bao nhiu s t nhin chia

ht cho 5, mi s c 5 ch s phn bit.

2) Gii phng trnh: sin4x + cos4x - cos2x + 41

sin22x = 0

S 62

CU1: (3,5 im)

Cho hm s: y = x3 - 3x2 1) Kho st s bin thin v v th (C) ca hm s cho.

2) Tnh din tch ca hnh phng gii hn bi ng cong (C) v trc honh.

3) Xt ng thng (D): y = mx, thay i theo tham s m. Tm m ng thng (D) ct

ng cong (C) ti 3 im phn bit, trong c hai im c honh dng.

CU2: (2 im)

Tnh cc tch phn sau y:

1) I =

0xdxsinx 2) J =

2

0

32 xdxcosxsin

CU3: (2,5 im)

1) Trong mt phng vi h ta cc Oxy cho hypebol (H): 1916

22= yx .

Gi F l mt tiu im ca hypebol (H) (xF < 0) v I l trung im ca on OF. Vit

phng trnh cc ng thng tip xc vi hypebol (H) v i qua I. 2) Trong khng gian vi h to cc Oxyz cho im A(3; -3; 4) v mt phng (P): 2x -

2y + z - 7 = 0. Tm im i xng ca im A qua mt phng (P).

CU4: (2 im)

1) Gii h phng trnh:

==+

93411

xy

yx


Trang:55

S 63

CU1: (2 im)

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

12

+

xxx

2) Tm m ng thng d: y = -x + m ct th (C) ti hai im phn bit. Khi chng minh rng c hai giao im cng thuc mt nhnh ca (C). CU2: (2,5 im)

1) Gii phng trnh: ( ) ( ) 43232 =++ xx 2) Cho ABC c ba gc nhn. Chng minh rng: tgA + tgB + tgC = tgAtgBtgC T tm gi tr nh nht ca biu thc E = tgA + tgB + tgC CU3: (1,5 im)

Chng minh rng nu: y = ln

++ 42xx th o hm y' = 4

12 +x

S dng kt qu ny tnh tch phn: I = +2

0

2 4dxx

CU4: (3 im)

1) Trong mt phng vi h ta cc Oxy cho parabol (P): y2 = 4x. T im M bt k trn ng chun ca (P) v hai tip tuyn n (P), gi T1, T2 l cc tip im. Chng minh rng T1, T2 v tiu im F ca (P) thng hng. 2) Trong khng gian vi h to cc Oxyz cho mt phng

(): x + y + z + 10 = 0 v ng thng :

+==

=

tz

ty

tx

312

(t R)

Vit phng trnh tng qut ca ng thng ' l hnh chiu vung gc ca ln mt phng (). 3) Cho t din OABC c OA, OB, OC vung gc vi nhau tng i mt, sao cho OA = a; OB = b; OC = 6 (a, b > 0). Tnh th tch t din OABC theo a v b. Vi gi tr no ca a v b th th tch y t gi tr ln nht, tnh gi tr ln nht khi a + b = 1. CU5: (1 im)

Hy khai trin nh thc Niutn (1 - x)2n, vi n l s nguyn dng. T chng minh rng:

1. ( ) nnnnnnnn nC...C.C.Cn...CC 2242221223212 242123 +++=+++

S 64

CU1: (2 im)


Trang:56

1) Kho st s bin thin v v th ca hm s: y = 1

2

xx . Gi th l (C)

2) Tm trn ng thng y = 4 tt c cc im m t c th ti th (C) hai tip tuyn

lp vi nhau mt gc 450.

CU2: (3 im)

Gii cc phng trnh sau y:

1) 11414 2 =+ xx 2) sin3x = cosx.cos2x.(tg2x + tg2x)

3) ( )xxxx PAAP 2672 22 +=+ trong Px l s hon v ca x phn t, 2xA l s chnh hp chp 2 ca x phn t (x l s nguyn dng).

CU3: (2 im)

1) Tu theo gi tr ca tham s m, hy tm GTNN ca biu thc:

P = (x + my - 2)2 + ( )[ ]21224 + ymx . 2) Tm h nguyn hm: I =

+

+ dxxgcotxtg63

CU4: (2 im)

Cho hnh chp SABC nh S, y l tam gic cn AB = AC = 3a, BC = 2a. Bit

rng cc mt bn (SAB), (SBC), (SCA) u hp vi mt phng y (ABC) mt gc

600. K ng cao SH ca hnh chp.

1) Chng t rng H l tm ng trn ni tip ABC v SA BC. 2) Tnh th tch hnh chp.

CU5: (1 im)

Chng minh rng vi x 0 v vi > 1 ta lun c: xx + 1 . T chng

minh rng vi ba s dng a, b, c bt k th: ac

cb

ba

a

c

c

b

b

a ++++ 33

3

3

3

3.

S 65

CU1: (2,5 im)

1) Kho st s bin thin v v th ca hm s: y = (x + 1)2(x - 2).

2) Cho ng thng i qua im M(2; 0) v c h s gc l k. Hy xc nh tt c gi tr ca k ng thng ct th ca hm s sau ti bn im phn bit:


Trang:57

y = 233 xx . CU2: (2 im)

Gii cc phng trnh:

1) 2

5122122 +=++++++ xxxxx

2) ( ) ( ) 1

12232 =

+++xsin

xsinxsinxsinxcosxcos

CU3: (2,5 im)

1) Gii v bin lun phng trnh sau theo tham s a: aaa xx =++ 22 2) Gii phng trnh:

( ) 222

22 2222

22 =

+++ xlogx

logx

logxlogxlogxlog xx

CU4: (2 im)

Cho t din SPQR vi SP SQ, SQ SR, SR SP. Gi A, B, C theo th t l trung im ca cc on PQ, QR, RP.

1) Chng minh rng cc mt ca khi t din SABC l cc tam gic bng nhau.

2) Tnh th tch ca khi t din SABC khi cho SP = a, SQ = b, SR = c.

CU5: (1 im)

Tnh tch phn: I =

+4

0 222

dxxcosxsin

xcos

S 66 CU1: (2,5 im)

Cho hm s: y = 2

2

+

xxx

(C)

1) Kho st s bin thin v v th ca hm s (C) 2) ng thng () i qua im B(0; b) v song song vi tip tuyn ca (C) ti im O(0; 0). Xc nh b ng thng () ct (C) ti hai im phn bit M, N. Chng minh trung im I ca MN nm trn mt ng thng c nh khi b thay i. CU2: (2 im)

1) Gii bt phng trnh: 113234 22 ++ xxxxx


Trang:58

2) Tnh tch phn: I =

32

0

3 dxxsin

CU3: (2 im)

1) Gii v bin lun phng trnh: 2m(cosx + sinx) = 2m2 + cosx - sinx + 23

2) Tam gic ABC l tam gic g nu:

=+=+

BsinAsinBsinAsin

BsinAcosabAsinbBsina

422422 22

CU4: (2 im) 1) Trong khng gian vi h to cc Oxyz cho cc im A(2; 0; 0),

B(0; 3; 0), C(0; 0; 3). Cc im M, N ln lt l trung im ca OA v BC; P, Q l hai im trn

OC v AB sao cho OCOP

= 32

v hai ng thng MN, PQ ct nhau. Vit phng trnh mt phng

(MNPQ) v tm t s ABAQ

?

2) Trong mt phng Oxy cho parabol (P) c nh ti gc to v i qua im A ( )222; . ng thng (d) i qua im I

1

25 ; ct (P) ti hai im M, N sao cho

MI = IN. Tnh di MN. CU5: (1,5 im)

Bit cc s a, b, c tho mn:

=++=++12222

cabcab

cba . Chng minh:

34

34 a ;

34

34 b ;

34

34 c

S 67 CU1: (2 im)

Cho hm s: y = x4 - 4x2 + m (C) 1) Kho st s bin thin v v th ca hm s vi m = 3. 2) Gi s (C) ct trc honh ti 4 im phn bit. Hy xc nh m sao cho hnh phng gii hn bi th (C) v trc honh c din tch phn pha trn v phn pha di trc honh bng nhau. CU2: (2 im)

1) Gii h phng trnh:

=+

=+

2

2

32

32

yxy

xyx

2) Gii phng trnh: ( )21 122 2 = xxxx


Trang:59

CU3: (2 im)

1) Gii phng trnh lng gic:

+=

23

1021

2103 x

sinx

sin

2) Cho ABC c di cc cnh l a, b, c v din tch S tho mn: S = (c + a - b)(c + b - a). Chng minh rng: tgC =

158

.

CU4: (2 im)

1) Tnh: 2

3

0

3121x

xxlimx

++

2) Tnh: I = ( )

+4

01 dxtgxln

CU5: (2 im) Trong khng gian vi h to trc trun Oxyz:

1) Lp phng trnh tng qut ca mt phng i qua cc im M(0; 0; 1) N(3; 0; 0) v to

vi mt phng (Oxy) mt gc 3

.

2) Cho 3 im A(a; 0; 0), B(0; b; 0), C(0; 0; c) vi a, b, c l ba s dng, thay i v lun tho mn a2 + b2 + c2 = 3. Xc nh a, b, c sao cho khong cch t im O(0; 0; 0) n mt phng(ABC) t gi tr ln nht.

S 68

CU1: (2,5 im)

Cho hm s: y = 1

12

++

xmmxx

(Cm)

1) Kho st s bin thin v v th ca hm s khi m = -1.

2) Chng minh rng h (Cm) lun i qua mt im c nh.

3) Tm m hm s (Cm) c cc tr. Xc nh tp hp cc im cc tr.

CU2: (3 im)

1) Gii phng trnh: 120002000 =+ xcosxsin 2) Gii bt phng trnh: 220001


Trang:60

1) Chng minh rng bn im A, B, C, D nm trn cng mt mt phng.

2) Tnh khong cch t im C n ng thng AB.

3) Tm trn ng thng AB im M sao cho tng MC + MD l nh nht.

CU4: (1 im)

Tnh tch phn: I =

+4

4

dxxcosxsinxcosxsin

B I5: (1,5 im)

Mt t hc sinh c 5 nam v 5 n xp thnh mt hng dc.

1) C bao nhiu cch xp khc nhau?

2) C bao nhiu cch xp sao cho khng c hc sinh cng gii tnh ng k nhau?

S 69

CU1: (2 im)

1) Gii bt phng trnh: 18184152158 222 ++++ xxxxxx 2) Xc nh gi tr ca a h bt phng trnh:

( )( )

+++

axyyx

ayxyx

3

32

2 c nghim duy

nht. CU2: (1 im)

Gii phng trnh: cos2x + cos4x + cos6x = cosxcos2xcos3x + 2 CU3: (3 im)

1) Cho hm s: y = 2x3 - 3(2m + 1)x2 + 6m(m + 1)x + 1 a) Vi cc gi tr no ca m th th (Cm) ca hm s c hai im cc tr i xng nhau qua ng thng y = x + 2. b) (C0) l th hm s ng vi m = 0. Tm iu kin ca a v b ng thng y = ax + b ct (C0) ti ba im phn bit A, B, C sao cho AB = BC. Khi chng minh rng ng thng y = ax + b lun i qua mt im c nh.

2) Tnh tch phn:

++2

011 dx

xcosxsin

CU4: (2 im) Cho cc ng trn: (C): x2 + y2 = 1 (Cm): x2 + y2 - 2(m + 1)x + 4my = 5

1) Chng minh rng c hai ng trn ( )1mC , ( )2mC tip xc vi ng trn (C) ng vi hai gi tr m1, m2 ca m. 2) Xc nh phng trnh cc ng thng tip xc vi c hai ng trn ( )1mC , ( )2mC trn.


Trang:61

CU5: (2 im) Cho hai ng thng cho nhau (d), (d') nhn on AA' = a lm on vung gc chung (A

(d), A' (d')). (P) l mt phng qua A' v vung gc vi (d'). (Q) l mt phng di ng nhng lun song song vi (P) v ct (d), (d') ln lt ti M, M'. N l hnh chiu vung gc ca M trn (P), x l khong cch gia (P) v (Q), l gc gia (d) v (P). 1) Tnh th tch hnh chp A.A'M'MN theo a, x, . 2) Xc nh tm O ca hnh cu ngoi tip hnh chp trn. Chng minh rng khi (Q) di ng th O lun thuc mt ng thng c nh v hnh cu ngoi tip hnh chp A.A'M'MN cng lun cha mt ng trn c nh.

S 70

CU1: (2,5 im)

Cho hm s: y = ( )1233

2

2

++=

xx

xxxf

1) Tm tp xc nh v xt s bin thin ca f(x);

2) Tm cc tim cn, im un v xt tnh li lm ca th f(x)

3) CMR o hm cp n ca f(x) bng: ( ) ( ) ( )

+ ++

11

1

12

1221

nn

nn

xx!n

CU2: (2 im)

1) Gii bt phng trnh: 0132

55lg


Trang:62

2) Cho 4 im A(0; 0; 0), B(3; 0; 0), C(1; 2; 1), D(2; -1; 2) trong h to cc trc trun Oxyz. Vit phng trnh mt phng i qua 3 im: C, D v tm mt cu ni tip hnh chp A.BCD. 3) Tm tp hp cc im M(x, y) trong h to cc trc trun Oxy, sao cho

khong cch t M n im F(0; 4) bng hai ln khong cch t M n ng thng

y = 1. Tp hp ng l g?

S 71

CU1: (2 im)

Cho hm s: y = f(x) = x3 + ax + 2, (a l tham s) 1) Kho st s bin thin v v th ca hm s khi a = -3. 2) Tm tt c gi tr ca a th hm s y = f(x) ct trc honh ti mt v ch mt im. CU2: (2 im)

1) Gii bt phng trnh: 431 +>+ xx 2) Gii phng trnh: ( ) ( )210010 3264 xlgxlgxlg .= CU3: (1 im)

Vi n l s t nhin bt k ln hn 2, tm x

2

0; tho mn phng trnh:

22

2n

nn xcosxsin

=+ CU4: (2 im)

Trong khng gian vi h to cc trc trun Oxyz cho ng thng

(d): 23

21

11

==+ zyx v mt phng (P): 2x - 2y + z - 3 = 0

1) Tm to giao im A ca ng thng (d) vi mt phng (P) . Tnh gc gia ng thng (d) v mt phng (P). 2) Vit phng trnh hnh chiu vung gc (d') ca ng thng (d) trn mt phng (P). CU5: (3 im)


Trang:63

1) Tm 2 s A, B hm s: h(x) = ( )222

xsin

xsin

+ c th biu din c di

dng: h(x) = ( ) xsinxcos.B

xsin

xcos.A+++ 22 2 , t tnh tch phn J = ( )

0

2

dxxh

2) Tm h nguyn hm ca hm s g(x) = sinx.sin2x.cos5x

3) Tnh tng: S = ( ) nnnnnnn C.n....CCCC 14321 1432 +++ (n l s t nhin bt k ln hn 2, knC l s t hp chp k ca n phn t)

S 72 CU1: (2 im)

1) Kho st s bin thin v v th ca hm s y = 32

+

xx

2) Tm trn th ca hm s im M sao cho khong cch t im M n

ng tim cn ng bng khong cch t M n ng tim cn ngang.

CU2: (3 im)

1) Vi nhng gi tr no ca m th h bt phng trnh:

+++

012

09102

2

mxx

xx

c nghim

2) Gii phng trnh: 1444 7325623222 +=+ +++++ xxxxxx

3) Cho cc s x, y tho mn: x 0, y 0 v x + y = 1. Hy tm gi tr ln nht v gi tr nh

nht ca biu thc: P = 11 +++ x

yy

x

CU3: (2 im)

1) Gii phng tr

tuyen tap 150 de thi dai hoc 3403

Documents