Đề đáp án thi thử ĐH lần 4 2011 THPT Lê Xoay

TRrtoNG THPT r,E

on rru

xoey

Niim hgc 2010-2011

l:

Cdu

rm rutlD4r

HQC- IAN

rv

uON: roAN- BAN KHoA Ho. c ryNnrcN
Thdigian: I80phfit

+m+2

(1)
Cho hdm s5 y = xo -4x2
Kh6o srlt sg biiln thi€n vd vE AO tni (C) cua hdm sO (l) vdi m 1.
2. Giast (C) cft Ox tai 4 di€m phen bi$t. Xdc ttinh m Oe ninn phing gidi han
bdi (C) vi tryc hoanh c6 diEn tich phdn phfa tr€n vd phia dudi tryc hodnh
bing nhau.

l.

:

Cf,u 2:

.-+=ry.

l.

-zGiei phuong trinh: Q+c?tzx,
tan x ' sint x cos" .r
2. Giaib6t phuong trinh: l' +.F+o > Jt'*'--*"

9

u.t'

*12

.

3:

Cho hinh hQp dung ABCD.A'B'C'D' c6 AEc =60'. G6c gita m{t phang (A'BD) vi mit phing elSy bing 60". Tinh theo a
khoang cdch gifa tlulng thlog CD' vd mflt phAng (A'BD).

Ciu

C0u 4:

Jz

x.dx
1. Tinh tich phdn I = I
!t7 -;;4
2. Cho x,y e R,x> 0. Tim gi6 tri lun nh6t cta biiiu thtic:

ryz

l{=
(xz

+3y2)(x*JV

*Wl

Ciu

1.
2.

5:
Trong

m{t phAng v6i h€ trgc to4 dQ Oxy, cho tam gi6c ABC vu6ng t4i A.
Hai titip tam gi6c ABC b6ng

I?

Trong kh0ng gian to4 d$ Oxyz. Cho m{t cAu (S)-c6 phuong.trinh:
(x+l)2 +{y-t)2 +(z+l)2 =25. Lap phumg trinh ti+5p tuyiSn vdi m{t cdu bi€t
titip tuyiin qua A(4; 0; 0) vi vuOng goc v6i Ox.

Cf,u 6:
Tim sd phrrc

5'r)
grta
zbi€xl{=l
v
-3-Mr(
rr vi m$t acgum€n

5

-1+i l-?J'

Hg

/>

r
,f

st

DAP AI{

mrr rrtrlDAtrrgc - r.*N'rv
nmtx
MON: rorix- BAI\I KHoA Hgc

Vdim =
TXE: R

1. Hfoir s6

rf

tra.dranli y = x' -

4x2

+3

0,25

XCt sybitin thi€n:

lirn v-+@
lim u=+o:'x-+<-

t-#t

y'= 4x3 -8x

lx=0

'v'=0<)ll,
Hnm

=+J2
s0 d6ng bi6n tcn {-f;0)v,i(J2;+"o)

Hnm s6 nghich bi6n tr€n ("*;-Ji)ua(o;f)
Hem S detctrc d4i ter.x = 0 -> lcs =3

Him

datcuc ti6u tdi "x = +Ji
bi6n*riOn:

s6

-*

--:n

D6 thi:
Dths cit Oy tei di6m

4 l* ='l

o

+0-

A(B 3) vnnhsn Oy hm

J,

trrrc d6i xrmg-

Xdt pt h"Atth tlQ grao tli€m crla (C) ve O:: xo'4x2 + m+2 = 0(2)
(2) c6 4 ngliQrn phdn bi$t
0c6 2 nghiQm phin biQt duong
o(=) pt t2 -4t +m+2=

/>
+0,25


f

a'=4-m-z>o

<+{S=2>A
I

e-2
0,25

lr=n+z>a
Ggi cto nghi$m cua (2).ln +a vi +b (0 < b.< a)
Do tinh tldi xfrng cria tl3 thi qua Oy r€n de tho6 m6n ycbt thi:

Jtr'

o

- 4xz + m+Zft= -ttr'

-4xz

+m+z)&
0,25

b

firo -4x2 +m+2)dr=A

0

*453-y+(m+2)b=o
cr

3Da

-20b2 +6(m+ 2) = 913;

0,25

Me b h nghipm cria (2) n€n ba -4b2 + n+2=0(4)

ru

lo

1,,=T

(3) vn (4)=+

lo

1=,

l^ ie(1;z>

Dap s6

2.1
2.1

lil

0.2s

* =?.

td
Dk: sin2x * 0
Phuons

-

'cos2x.cosx\21160
L-*\.1asin 2x.sin.r ,/ sin' x cos' r
| 160

cosr 2
trlnh:

<> |

f-

2sin2.r.cos.r'sin2x' cosox

ll160
sinlx costr

-

9

g

0,25

A....'......-f-=-

A-

9

l-2sin2.r.cost.r
(sin

160

x.cosx)"

9

D$t (sinx.cosx)z = r(r > 0)
Ta tlugc:

l-2t

160

t2g
-=el60t2
<+ (l6r

*

0,25
+ l8r

-9

=0
3Xl0r + 3) = 0

e.t=-3

0,25

l6

Vgy (sin-r.cosx)2
't6= 3

e

sin2r =

tf 2 e

-f,

o
=

4srn2

2x

=3

frkr
+-+-

0,25

62

1.t

2.2

/>

c{ nglria v6i mqi x

Bpt

3ts,:e3'"${o
DAt:

3' =

lu+v>

*ali

>Jmi,3-fr

*6

= v ta duqc:

0

0,25

>mzr'

fr*u

{u+v>0

^< o }4=Y
[(u -v)'

<+{

:+ 3' =

.,ffi € 32' --3' -6= 0 :+ (3' - 3X3' +T) = 0

mEiaAi€t'=>
->

r.{,C

'0,?5

0,25

aABC.dou o4nh a

=a

vi BD.
I
I
BD - (ABCD) n@ nDl|

+ Gqi O ld giao'cua AC

AotBD
A,OLBD
+ A()A' *

+

CD' lt
=

0,25

=90")
d(CD';Qt BD))

6ae.QDoA'eo

A'B

d(C;(A'

+

BD]I' =11tr;(A' BD.})'

(vi AO * CO) '
+ lrtrl: J{H vudng 96o v6i

A'O

0,25

(1)

c6
AA'L BD\+

AH

(z)
BD L
BD)
Tir(il) v$ (2) a AE L(A'BD',
v0y:
d(CD";(A'BD)) = AII: AO-sin AOH
AO

t

a

= jsrnou"=

a

I

0,25

0,25

"Jj
"Ji
Z.T=T
I.I*L

!v'-z*qJx'-t
D$ t =\[7a 4 f = x' -l>

t-dt =

x&

0,25

t}OI CAn:

xlll.D
t l0l1

,:1sff*.='!;Hr;J[*-*]
=

tt) =
i[U' -rl-]rnpr. t; ](t"t-Jr"o)

4,25

0,25
0,25

/>

I

ry'

!t=
(xz

Xdt

+3y.\x*Jffi/y

y=0=>M=0
0,25

Xdt y

*0+M = ry'U-Jxz +t2y2)=

-(rTtryit}

0,25

EItr=4nui(t>0)

M=f(t)=#

f'(t)

=

2-t +z.rlt+t
6(t

+4)2.J1+t

f'(t)=0<+r=8
Bing bi6n thi€n:

=} M dat GTLN.

Gi& sri:

lzy,
fzt'=3r'
x' =g c)i

** =*

[rt0

0,25

.-ld

A(a;0)

BeOx

I

B e BC :4x +3y- 16 = gJ

MBC vu6ngtei A;

=

4,25

rt+'01
la

C e BC

1l

+c6;We1
0,25

4,25

=1b-4y =iv-q{r-+.;)
=l
ela-41=
I I 3 <) la

ia=7

/>

FF66t(-t;

t;-t); bin kinh R = 5
tld) ii6p tuy6na cln cm c6 vtcp il =(a;b"c\;.* 6)
+A IOx o T.T = 0.Voi i = (l$;O}'= a = 0

t?5

+ U(5;-l|);u(0,b,c)
t- -l

V4,;l=(-c -b.-sc.sb)

+

I

tifu xuc (S) <=> d(Ii

A) = R

lF,;l
(+3=)

0,23

lrl

+a:
b2

€s[=
,*l-1*

a;25

x==4
V$y A c6 pf:

yttl

4,25

=*l
6
(1d)

Gt

* z*{{*rr*isine)

+ i =j (."r* -,'in
c6

e)= |tead-

*t*

isin(-

p[

0,25

-r*,=.tr(rer?.t'*?)

vay:*=#[{-o-f).,'*[

0,25

3* 5t
Gt=-e-{=-++2kn

+

e=

)

7

0,25

!+kzn

=]trorf *i*il

Hay

0,25

t.

/>