ON TAP HK1 THANG LONG HN

<span class='text_page_counter'>(1)</span><div class='page_container' data-page=1>

i'\T

[\ci, Lrrn

THPT Thdng Long Hd ndi - Ndm hoc 2010-2011

pi

_crloNc

_ON

_{rAp roAN Hec}

_xi

_I

_Lop

_rz

Trtfrng

THPT Thang

Long

Hd

ngi

-

Ndm

hoc

2010-2011

Bi6n

soqn:

Nguydn

Thiii

phtryng

_^\r.

PNANI:GIAITICH

Bni

f:-Tim

m dd hdm sd sau

c6

ba di?dm cuc

tri

I y ₌mxo

*(*,

_{-9)x, +10( m}_{ld tham}_{sd )}

l

Biri 2: Cho hdm so

:

.r,-

'

-x. + ,,''\ (1) ( rn ld tham so )

l-x

l.Tim

m dd hhm sd

(l)

c6 cuc clai vir cr,rc tidu. Vdi girl tri

nio

cira m rhi khoang cdch giffa hai

didm cuc tri ctia dd thi hdm so

(t)

birng l0?

Biri3:

fim

m ddhhm so sau c6 cuc

tri

_{r .y--,r3}+3mx1+:(r-

n')*+mt

*nf(

m Idtham

sdXl).

Vidt phucng trinh ducrng thang di qua hai ctidrn cuc rri cua dd rhi h)m sd ( I _).

Bii

4: Cho hhm

so:

'

r,- -'vr+4,r-'l

111

x-2

\

/

l.Khrlo sdt su bien thi6n v:\ vd dd thi h)rn sd ( I ₎

2.Chung minh

rlng

tich

cdc khoang cdch tir mot didm bdt

k]'

tr€n dd rhi

h]m

sd ddn cdc

duong ti€m cAn cria n6

ll

m6r hang so .

Bdri 5: Cho

hlm

so: _.r,=__:!_

₍₍

)(1)

"l-l

l.Khrlo s6t su bien thi€n vir vd dd thi (C ₎cua hfun so ( I ₎

2.Yi6t phucng trinh

tiip

tuyen cl cira

(C

) sao cho

d

vl

hai tiem can cua (C

₎

cit

nhau tao

thlnh

m6t tam giiic can.

Biri 6: Cho

hlm

sii:

r'=

I1

1H;

r*l

Chtmg minh rang :

Tich

cdc khoang cdch tu mOt clidm Mu (x,,;!o)bdr

k''

rhu6c (H) ddn cdc

duong tiOm cAn cua n6 li,r mOt hang so .
Biri 7: Cho hirm so: _"1₌

t,tx+!

(,,tlurlt,,nr rr,)(1)

J

fim

m 0d hhm so

(l)

cd cuc

tri r,i

khoirng cdch tU ciidm cuc ridu cua dd thi

hlm

so

(l)

den

ti€m cAn xiCn cua n5 bang

_-Jz

{

BiriS: Chohdm so y ₌xr + nrx +

2

(l)

l.

Kh6o siit su bi€n thi€n va v0 dd thi cua ham 15 ( t ) khi m ₌-3.

2.

Tim m d6 ti6p tuy€n

voi

d6

thi

hanr rO ( t _{) tai}

A(0,2)

tao vdi hai truc toa dQ m6t tam

</div>
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THPT Thdng Long Hd ndi - Ndm hoc 2010-2011
Biri 9: Cho hdm

sd.

y:?x+3

_'

x-2

1.

KhAo s6t su bidn thiOn

vi

v€ dd rhi (C ) cria

him

sd d6 cho.

2.

Tim

tdt ch cdc gi6

tri

cira rham sd m dd duong

thing

y=2x+m c6t (C

₎

tai hai didm

phAn bi€t mh hai tidp truydn cta (C ₎tai hai didm d6 song song vdi nhau.

Biri

10: Cho

him

so y ₌_{xt -3xt}+4

l.Khrio si{t v}r vE dd thi (C)

2.Goi (d)

le

duong

thing

qua A(3;4) vt\ c6 h€ so g6c k.Tim

k

dd (d)

cit

(C) tai 3 didm

phan bict

A,M,N

sao cho hai

tiip

ruydn cua (C ₎rai M

vi

N vu6ng g6c v6i nhau.

Biri

1l:Cho

hlm

so

y=

'

i*2

( l )

2x+3

Khrio sdt su bien thi0n

vi

vE d<i thi ( I ₎

1.

2.

Viet

phuong trinh

tiip

tuyin

cua drj

thi

hlm

so

(l),

bidt

truc tung tai hai didm

A

,B sao cho tam gi6c

OAB

cAn tai

tiep tuydn d6 c6t

goc toa dQ O

truc hohnh,
Bni 12: Cho hdm

sd

y:

x4- (3nt+2).rr + 3m

(l)

_,m lb tham so

I

.

KhAo sdt su bidn thien

vI

vC dd thi ( _{I )}_{vdi m=0}

2.

Vdi

gi6

tri

nho ciia m , ducing thAng

y=-l cit

dd thi

hlm

so

(l

_{) tai 4 didm}phAn biet c6

hohnh dO nh6 hon 2.

Bii

13: Tim gi6

tri

ldn nhat

vI

gi6

tri

nho nhat cua

him

so

:

_/(x)=

?+

rrenf-z.ol.

Biri

14: 3.Tim gi6

tri

ldn nh6t vd gi6

tri

nho nhdt cua hdrn

s6

_!

₌x6 ++(r

-

t')'

trrr[-r;r]

Bii l5:

Tim gi6

tri

ldn nhAt vd gi6 tri nho nhAt cua hdm

s6

_,u=

("

+

l)Jl

_-

t'

.

Biri

16: Tim gi6

tri

ldn nh6t vd gia

tri

nhtj nhAt cua hdm

s6

y = x

Biri

17: Tim gi6

tri

lcrn nh6t va gia tri nho rrhAt cua hdm

sd

,u

-

=

_^U*"

-

Vx e

_[0;z].

2+cosx

Biri l8:Tim

gi6

tri

nho nhat cua h)nr

s0:.r'

)1

-*-(0.r.1)

I _-X X

nho nhAt cua hdrn

sd

.v = x + J4

-

_"'

nho nhAt cua hdm

s6

.1, = ..f +

si*

*

J

*

rrr"

.

Bdri

2l

:Cho a > 2;b > 3;c > 4 . Tim gia

tri

ldn nhdt cua bidu rhfc:

f=

ab,1;4

a

6r,{n

_ ':t

*

,r,Ji

1

ahc

Biri

19: Tim giri

trl

l6n

Biri 20: Tim gi6

tri

l6n

Biri 20: Tim gi6

tri

lon nhat

vI

girl _{/'('r) =.r+2+}

),,rrr(l;+-).

nhAt vd gid tri

,^

nnat va gra

til

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)---l

THPT Thdng Long Hd n6i - Ndm hoc 2010-2011

Bii

21: Tim gir{ tri l6n nhat vd gir{ tri nh6 nhat cira

him

so

:

/

= cos3 x-6cos2 x+9cosx+5

./ = sin3 x _-cos 2x + sin x + 2

Biri2l:Tim

gi6 tri ldn nhat

vi

gid,tri.nhd nhat cira

him

sd

:

_/(x)

₌

x-er*

tren[-t;ol.

PHUONG

TRiNH, HE

PHUONG

TRiNH

sAr

pHUoNG TRiNH

naU

vA

LocARiT

Biri 1: Tim ci{c gidi

han:-o5'

_r

,

,,.

\

(r0,, _

,). ,.

_{c,, _,j,,}

t'li$ .

=

'-:;:;

:,

z.lim--=]"0

ia

';'-i,^

5'r

Bii

2: Tim dao hdm c[ra cdc

hlm

so sau:

3.

_1i*

ln(5x+ lF- ln(3x+ t)

r+0 X

l. y:5t''(rt

+2x?

_-x+l)

r _-r

^

e

-e

5.

_'

v=

e'+e-'

Bii

3:

Giii

c6c phuong rrinh sau :

t.

(t-o..6)'

+(t

+a..6)'= ra

3.(++Jrs)'*(+-fi)'=oz

t

_(Jt-')'*(Jt.

t)'

_-zJi

_=o

Bhi 4: GiAi cr{c phuong rrinh sau :

3.4'+9'

= 25'

5. 4'-l.i-! _rr.2,

vi,lit

* g

= 0

T.gfinr'+gl'""'=30

2. _.y

_{="" (r'}

+2x2 _-x

+l)

4. _)'= e5' (sin x + cos.r)

l.(z+f)"-"*'

*(2-S)"

" --1-

_{2.23\+1 -7.21,}_+7.2,

_-2:0

2.

5t*'+6.5.'-3.5'-r ₌52

4

_(.6;rG)

'+(S-rG)

' _=,0

o.(zJI*

_Jil)"-'

*(2..6

_-

J,

i)"-'

=

4Ji

4

_{(,1;J)'.(#;F)'=oi}

6

_{(..6_ffi)'.('6.rG)'}

_=,0;

8. 3.25'-r +(3x-10).5'-']

+3-x

= 0

14..2log,x + logrr,r+ log, x = 9

/r

o. (za+rs.,6) '

+z(t++.6)

'

-r(r-.6)'

='

10.

log(x+l)

_-

log

( l-x

):

log(2x+3)

I

l.

log.,('-r).*:=1+lns.

"t.r

12. tog,(+'+r5.2.+

27)+2tog,-J-=6

t

13. _{log,, x = log,}

_(J;

- o)

</div>
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THPT Thdng Long Hd ndi - _{Ndm hoc}_2010-201t

Bii

S:Cho phuong _{trinh: 9r*JF -}(o * 2)3r."[} + 2a +t= 0(1) a ld tham sd

l.

Giei phucng trinh v6i a=4

2.

fim

a dd phuong trinh (1) c6 nghi€m.

Bii

6:

Tim m dd phuong

trinh:

_/t"g!"irug,

;t

₌rn (logo

,'

4)

c6 nghiom x e

_[:z;r--)

17

.

log,(s. _-t)log,

(5'.'

_-2s)= 3

19. log, (+.3. _-6)- los,

(l'

_-o) = r

1.4x2 +x.3tr +3r+Ji.2x'3ji +2x+6

3.

log.-,(.r'-r)>

z

5. (x+l)log] x+(2x+5)log, x+6 > ₀
27

7.

logo(s' +

t)*

toe1,,,,) a

r

i

2'-r +4x-16

x-2

18. tog,

{togn

(:'.'

-

t)}

= t

29 4x1-3x+2 * 4x2 +6x+5 _ 42x2 +3x+7 *1

2. log.

(tr'

_-sx + 3) > 2

O.

s'-'-(:x-g)

,u

Bii

7:Giii

ci{c h0 phuong trinh:

,.{tn(l+')-ln(l+

y)=x-y

2.1

to*,(.r'+.v')=s

I

x' -l2xy +20y' =

g

_[Ztog _,r.+ log, y ₌4

^

fr"

=5y'-4y

I

(r'n

, _).3,,' = l

3'

j

_4'₊

2'"'

4.1

lji;=,

l*(,'*

,)-6

=o

.r.

_{(t

*.v)' = ("

-.u)'

6. J los, ,,lo = log, .v

llog,x-log,.v=l

[

:'+2'=3

^f 3'.2'=972

t

.i-J1r'l+3=o

t

t,orn

g-.v)=2

t

i/*,

-u/ing,1,=o

n.l'o*,(x'+2x'-3.r-5.u)=3,,,

Jff+Jtog.r,=3

[log,(y'+2y'-3y-5x)=;

I

t-3log.r'2=l

Bii

8:Girii cr{c bAt phucrng trinh:

g'-3

6.

log,

Itog, (+' - u)] =

g. _f

l

_)''*ll"'*l[

t'

":ror't-r]n

\3,

10. log._,(x+l)> log,,

9.

I

,l

,>l

_,(x+l)

On thi hoc ki

I

>4

</div>
<span class='text_page_counter'>(5)</span><div class='page_container' data-page=5>

-1-THPT Thdng Long Hd nili - Ndm hoc 2010-2011

psAn

_u:

_HiNn

_Hec

Biri

l:Cho khdi ch6p ddu S.ABCD c6

AB-a,

g6c gifia mdt b€n

vi

mat ddy bang 600. Tinh thd

tfch khOi ch6p S.ABCD theo a

Dii

2:cho khdi ch6p S.ABC c6 ddy ABC

l)

tam gi6c vu6ng tai A,

AR=a,

AC = aJ1, m6r b€n

!SBg)_ld tam gi6c ddu

vi

vuong gdc

v6i

mat phang ddy. Ttuh theo a rhd

tich khdi

ch6p

S.ABC.

Bii

3:Cho

ling

tru tam giiic

ABC.AIB'C'

cd tdt cA c6c canh d6y ddu bang a, g6c tao b&i canh

b€n

vi

mat di{y la 600

v}

hinh chieu H cua dinh

A

l€n

mp(A'B'C') trirnf

vditrung didm ctra

c4nh B'C'. T(nh th6 tich kh6i lang tru

ABC.A'B'C'.

BAi

4:Cho

hinh

hOp

dung

ABCD.A'B'C'D'

cd

ddy

ABCD

lI

hinh thoi

canh

a.

G6c

ffiouG69

_sfuaou,rng.h6o

A'c

vir

mit

phing d?y bang 600.Tinh tnd tich rri"r, nOp.

ia.

dinh vd tfnh dd

dli

doan vuong g6c chung cua

A'C

vI

BB'.

Biri S:Cho hinh ch6p S.ABCD c6 cl6y ABCD ld nua luc gi6c ddu vdi AD=2a,AB=Bc-CD-a,

SA I(ABCD);S.q ₌

oJT.Vot

_rnlr

_phing_{(P) di}_qua

_A

_vl

_{vu6ng g6c}_vdi_SD

_cit

_{SB, SC,SD}_tai

B',C',D'.Tinh

thd

tich khdi

ch6p

AD'DBB',X6c

dinh tAm vh br{n

kinh

m6t cdu ngoai tidp

hinh ch6p.

Blri

6:Cho

ling

trg clfng

ABC.A'B'C'

c(r tat cA c/rc canh ddu

bing

a.Goi

M

le rrung didm

AA'.Tfnh thtj tfch khoi tri cli€n

BMB'C'

thco a vi\ chfng minh rang BM vu6ng g6c vdi b'C.

Bii 7:

Cho

hinh

ch6gr

S.ABCD

c6

tliy

SA

I(ABCD):Str

₌oJ1 .Xac clinh ram vi) bitn

ABCD

le

hinh

vu6ng canh

a,

canh

b€n

kinh

mit

cdu ngoai tiep hinh ch6p S.ABCD.

Bii

8: Cho hinh ch6p S.ABCD c6 diry ABCD

lI

hinh vuOng canh a, SA=SB=AB _,mar phing

(SAB) vu6ng g6c

vdi

rnat

phing

(ABCD).

Tfnh

brin

kfnh

mdt cdu ngoai tidp hinh ch6p
S.ABCD.

Bii

9: Cho hinh ch6p S.ABCD c6

diy

ABCD

ll

hinh chfr nhAt

vI

Sl r(ABCD).Goi

B',C',D'

ldn luot

l)

hinh chieu vuong g6c cua

A

tren SB,SC,SD.

l.

Chtnrg minh ciic diCm A,B',C'.D'clOng phang.

2.

Chrnrg minh

7

didm A.B,C,D,B',C',D'

cing

thu6c mOt m4t cdu.

Biri

l0:

Cho hinh tru c(r birn kfnh diry

bing

R, thidt cliOn qua truc cua hinh tru

ll

mor hinh

vu6ng.

l.

Tinh diOn tfch vir thd tich hinh c.,iu ngoai

tiip

hinh tru.

2.

MQt mat ph.ang (P) song song

vtii

truc cua hinh tru.

cit

driy hinh tru theo rn6t dAy cung

c6 dO

dli

bing

b:in kfnh driy hinh tru. "finh cliOn tich ciic thiet dien cria h)nh tru ua ninn

</div>
<span class='text_page_counter'>(6)</span><div class='page_container' data-page=6>

THPT Thdng Long Hd ndi - Ndm hoc 2010-2011

Bii

11: Cho hinh tru c6 bdn kinh dr{y bang R,chidu

cao

Rr6

1. Tinh diOn tich xung quanh

vi

di€n tich

toin

phdn cfia hinh tru.

2.Tinh thd tich ctra khdi tru girli han boi hinh tru.

3.Ctro hai didm

A

vh B ldn lucn

nim

tr€n hai duong trdn d6y sao cho g6c giffa AB

vi

truc cira

hinh tru

blng

300.Tfnh khoang ciich gifra AB vd rruc ctra hinh tru.

Biri

_?:

Cho tam gi6c ABC ddu canh. a vd (P) ld

mit

phEng qua BC vd vuong g6c mp(ABC).

Gqi (C) lh duong trdn dudng kinh BC

vl

nam trong mp(P).

l.Tinhbr{nkinhm[tcdudiquaducrngtrdn(C)vhdidmA.

2.M0t hinh n6n ngoai tiep

mlt

cdu n5i [r'On sao cho cdc tidp didm gifra hinh n6n vh

mit

cdu

lh duong trbn (C).Tinh thd tich cua khoi n(rn.

Bii

13: Cho hinh n6n (^tD _c6_bdn_{kinh driy}_R_ducrng_cao_SO._Gqi_(P)_ld_mat_phing_vuong_g6c

vdi

SO tai

O'

sao choSO'=

_lt,

M6t

rlat

phing qua truc hinh n6n c6t phdn khdi n6n (^rD

3

nam gifra (P)

vl

d6y hinh n6n theo thiet (lien

ll

hinh

tf

giric c6 hai duong ch6o vuong g6c .

Tinh thd tich phdn hinh n6n (^il) nam

gita

rnp(P)

vi mit

phing chrla d6y ciia hinh n6n (^il)

</div>
<span class='text_page_counter'>(7)</span><div class='page_container' data-page=7>

---THPT Thdng Long Hd nili - Ndm hoc 2010-2AtI

A\

nE

rHI

HoC

ri

I

LoP

12

nnON

ToAN

THOI GIAN : 90 PHUT

(Oay

li

dA thi

cia

nim

hoc 2009-ZO1O e6 HS tham

khio)

CAu 1(2.5d): Cho hhm so

:

_-v_{= x4}

_-

6x2 + -5

a) Khio

s6t su bidn thi€n vh vE cld thi _{(C )}cria

hlm

sd

b)

Tim m dd phuong trinh xa

_-6.rr

_-2m=0

c6 4 nghi€m phAn bi€t.

Chu2Q

d\z GiAi crlc phuong trinh:

I) (8-rJ7)'

+(s+rJ7)'

= ro

2)

logn (n'

-

t)logn (9.'*2 _-81)= 3

CAu 3(1d): Tim gi6

tri

ldn nhat vi\ giri tri nho nhat cua hirm sd _-y

-

x +

Jr=t

.

CAU 4(2d):Cho

hinh

ch(rp S.ABCD

c(r day ABCD

le

hinh

vu6ng canh 2a, SA=a

!

SB =

oJl

A

mp(S,ll)

L

mp(ABCD)

,goi

H,K ldn luor

ll

trung didm ciia c6c canh AB,BC.

Tfnh theo

a

thd tfch cua khoi ch6p S.BI-ll)K. X:ic clinh tArn

vI

tfnh bi{n kinh mf,t cdu ngoai

tidp hinh ch6p S.AHOJ. O l)r giao cliCnr cua AC

vl

BD, J

l)

_{trung didm AD.}

CAu S(2d):Cho hinh hop cltrng

ABCD.A'lJ'C'D'

c6

dAy

ll

hinh thoi canh a, g6c

fEd=60",

g6c gifra mat phing

(A'BD)

_vi

mit

phiing cliry blng 600.Tfnh theo a thd tfch hinh h6p. T(nh
theo a khoang crlch til dudng thing CD'

din

mAt phing (A'BD).

Cdu 6(0.5d):Giei hc phuong trinh:

I

,t

xl+l

]

"

=t"*'

</div>

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