Khảo sát chất lượng môn toán - chuyên vĩnh phúc

KTHITHIHCLN1NMHC2013ư2014

TRNGTHPTCHUYấNVNHPHC

Mụn:Toỏn12.Khi A,A1,B.

chớnhthc
(thigm01trang)

1) Khosỏtsbinthiờnvvthcahmskhi m =1 .
2) Tỡmttccỏcgiỏtrcathams m ạ0 saochotiptuyncathtigiaoimcanúvi

trctungtovihaitrctomttamgiỏccúdintớchbng4.
Cõu2. (1,25 im) . Giiphngtrỡnh:

)

(

)

3 1 - 3 cos 2x + 3 1 + 3 sin 2x = 8 ( sin x + cos x )

(

)

3 sin 3 x + cos 3 x - 3 -3 3 .

x
ỡ 2 1

ù x - x = y- y
Cõu3.(1,25im) .Giihphngtrỡnh: ớ
( x, yẻ Ă).
ù 5y -1 - x y = 1
ợ
3
x + 6 - 4 7 x + 2
Cõu4. (1,0im). Tớnhgiihn: L = lim
x đ 2
x -2
Cõu5.(1,0im).Chohỡnhchúp S .ABCD cúỏylhỡnhvuụngvicnh 2a ,mtbờn ( SAB) nm

trongmtphngvuụnggúcvimtphng ( ABCD) v SA = a ,SB =a 3 .
Hóytớnh thtớchcahỡnhchúp S .ABCD vkhongcỏchgiahaingthng AC v SB theo a .
Cõu6.(1,0im).Xộtcỏcsthcdng a, b,c thomón ab + bc + ca =7abc .Tỡmgiỏtr nh nht
cabiuthc: P=

THIKHSCLLNINMHC2013 2014
HNGDNCHMTON12A,B,A1

Thigianlmbi:180phỳt(Khụngkthigiangiao)

A.PHNCHUNGCHOTTCTHSINH(8,0im)
Cõu 1.(2,5im).Chohms y = mx 3 - ( 2m + 1 )x 2 + m +1 ( Cm ) .

(

SGDưTVNHPHC
TRNGTHPTCHUYấN

8a 4 + 1 108b5 + 1 16c6 + 1
+
+
a2
b2
c 2

Hngdnchung.
ư Mimtbitoỏncúthcúnhiucỏchgii,trongHDCnychtrỡnhbyslcmtcỏch
gii.Hcsinhcúthgiitheonhiucỏchkhỏcnhau,nuývchoktquỳng,giỏmkho
vnchoimtiacaphnú.
ư Cõu(Hỡnhhckhụnggian),nuhcsinhvhỡnhsaihockhụngvhỡnhchớnhcabitoỏn,
thỡkhụngchoimcõu(Hỡnhhcgiitớch)khụngnhtthitphivhỡnh.
ư imtonbichmchititn0.25,khụnglmtrũn.
ư HDCnycú04 trang.
Cõu
Nidungtrỡnhby
im
1
1. Khi m = 1: y = x 3- 3 x +2
+TX: Ă
0.25
+Sbinthiờn: y  = 3 x 2 - 3 = 3 ( x - 1)( x + 1), y  = 0 x = 1
y  > 0 x < -1 x >1 suyrahmsngbintrờn cỏckhong ( -Ơ -1) , (1 +Ơ)
y  < 0 -1 < x <1 suyrahmsnghchbintrờn ( -11).

0.25

Hmstcciti x = -1, ycd = y ( -1)=4 hmstcctiuti x = 1, yct = y (1)=0.
3

2ử
3
2 ử
ổ
ổ
lim y = lim x3 ỗ 1 - 2 + 3 ữ = -Ơ lim y = lim x3 ỗ 1- 2 + 3 ữ = +Ơ
x đ-Ơ
x đ+Ơ
xđ+Ơ
x
x
x
x
ố
ứ
ố
ứ

x đ-Ơ

x

B.PHNRIấNG (2,0im).Thớsinhchclmmttronghaiphn(phn1 hoc2)
1.TheochngtrỡnhChun



y'

+

1

1

0

0

+

+

Cõu7A.(1,0im) .Trongmtphngvihtrcto Oxy ,chohỡnhbỡnhhnh ABCD cú A ( 20 )
,B ( 30) vdintớchbng 4 .Bitrnggiaoimcahaingchộo AC v BD nmtrờnng
1
2
3
2013
Cõu8A(1,0im).Tớnhtng: S1 = 12 .C2013
+ 2 2 .C2013
+ 3 2 .C2013
+ L+ 2013 2 .C2013



2.Theochngtrỡnhnõngcao.

M ( 02)nmtrờnngthng AB v MC = 2 ,tỡmtocỏcnhcatamgiỏc.

Cõu8 B(1,0im). Tớnhtng: S2 =

C

0
2013

C

1
2013

2
2013

C

ưưưưưưưưưưHTưưưưưưưưưư

0

+th
ư GiaoOx:

2013
2013

C
+
+

+ L+
1
2
3
2014

+

4
y

thng y = x ,hóytỡmtocacỏcnh C,D.

Cõu7B(2,0im).TrongmtphngvihtaOxychotamgiỏc ABC cúngcaokt B v
phõngiỏctrongkt A lnltcúphngtrỡnh : 3x + 4 y + 10 =0 v x - y + 1 =0 .Bitrngim

0.25

( -2 0 ) , (10 )

( 0 2)
I( 0 2)
imun:
suyra
I( 0 2)
thtxngqua

4

ư GiaoOy:

ư

2

2. th (Cm) : y = mx 3 - (2 m + 1) x + m +1 cttrctungti M (0 m +1).

y = 3mx 2 - (2m + 1) ị y ( 0 ) = - ( 2m +1)
Tú,khi m ạ0, tiptuyn tm ca (Cm) ti Mcúphngtrỡnh

0.50

0.25
0.25


y = - (2m + 1) x + m +1
Do (tm) tovihaitrctamttamgiỏccúdintớchbng4nờntacúh

5

1
ỡ
1
ỡ
ùùmạ - 2
ùmạ 2
ớ
ớ
ù m+ 1 ì m+ 1 = 8 ù( m + 1)2 = 8 2m+ 1
ợ

ù
2m + 1
ợ
2

3

0.50

M

Giih,thuc m = 7 56 v -9 72. ichiuiukinvktlun
+ýrng sin 2 x + 1 = (sin x + cos x )2 sin 3 x = -4sin 3 x +3sinx v cos 3 x = 4 cos 3 x -3cosx
nờnphngtrỡnh cvitvdng
(sin x + cos x)( 3 sin 3 x - cos 3 x) =0
p
+Giiphngtrỡnh sin x + cos x =0 tachnghim x = - + kp ,k ẻ Â
4
p
+Giiphngtrỡnh 3 sin 3 x - cos 3 x =0 tachnghim x = + lp ,l ẻ Â
6
+Ktlunnghim
1
iukin x ạ 0,y
5
Tphngtrỡnhthnhtcahsuyrahoc y =x 2 hoc xy = -1
1
+Nu xy = -1 thỡ x < 0< y vphngtrỡnhthhaitrthnh 5 y- 1 +
=1
y

Phngtrỡnhnytngngvi 5 y 2 - y =

ỡù y 1
y - 1 ớ
2
ùợ2 y = 1 - 2 y - 5y

0.25

0.25
0.25
0.25
0.25

0.5

0.5

Giiphngtrỡnh,c ( x y ) = (11), ( 2 2), ( - 7 - 41 7 - 41)
Ktlunnghim

xđ2

3

) (

x+6 -2 -

4

x-2

) = limổ

7 x + 2 - 2

4

0.25

ổ
ử
ỗ
ữ
x+6 -8
7 x + 2 - 16
L = limỗ
ữ
4
x đ 2
2
ổ
ử
7 x + 2 + 4 ữ
ỗỗ ( x - 2 ) ỗ 3 ( x + 6 ) + 2 3 x + 6 + 4 ữ ( x - 2 ) 7 x + 2 + 2
ữ
ố
ứ
ố

ứ
ổ
ử
ỗ
ữ 1 7
1
7
13
L = limỗ
=ữ=
4
x đ 2 ổ
2
ử
12
32
96
3
3
7x + 2 + 2
7 x + 2 + 4 ữ
ỗỗ ỗ ( x + 6 ) + 2 x + 6 + 4 ữ
ữ
ứ
ốố
ứ

(

(

)(

)(

)

D
O
C

a 3
(HlhỡnhchiucaAtrờnAB).
2
0.25
1
2a3
Tú,do ( SAB ) ^( ABCD ) nờn VS .ABCD = SH ì AB ì AD =
(.v.t.t)
3
3
1
+DoABCDlhỡnhvuụng,nờn S ABC = S ADC = S ABCD suyra
2
1
a3
VS . ABC = VS .ABCD =
(.v.t.t)
2
3

0.25
1
M VS .ABC = ì AC ì SB ì d ( AC SB ) ìsin (ã
AC SB ) nờn
6
2 a3 3
d ( AC SB) =
AC ì SB ìsin ã
AC SB

+Tgithitsuyratamgiỏc SAB vuụngtiSv SH =

)

0.25

0.5

)

+Gi O,M theothtltrungim AC , SD. Khiú (ã
AC SB ) =(ã
OAOM )
p dng nh lý cụưsin cho tam giỏc AOM

x +6 -2
7 x + 2 - 2ử
ỗ
ữữ
x đ 2ỗ

x-2
ố x-2
ứ
3

B

A

(

+Nu y =x 2, thayvophngtrỡnhthhai,tac 5 x 2 - 1 = 1 +x | x |.

(
L = lim

H

0.5

Do y 1 nờnhphngtrỡnhnyvụnghim.

4

S

6
tớnh c cosã
AOM =
4

suy ra

0.25

10
sin (ã
AC SB ) = sinã
AOM =
4
2a
d
AC

SB
=
L
=
Vy (
(.v..d)
)
0.25
5
Chỳý: Vibitoỏnny(phntớnhkhongcỏch),cúnhiucỏchgii,chnghnhcsinhcúthsdngvect,
tahaydngonvuụnggúcchung.Nucỏchgiiỳngvchoktquỳng,giỏmkhovnchoimti
acaphnú.CỏchgiitrongbitoỏnnysdngktqucaBitp6(tr.26)SGKHỡnhhc12(CCT)
6
1 1 1
Vitligithitvdng + + =7
0.25

a b c
pdngbtngthcAMưGM,tacú
1
1
A = 8a 2 + 2 4," = " a=
2a
2
2
2
2
1
3
3
B = 54b + 54b + 2 + 2 + 2 10," = " b=
0.5
9b 9b 9b
3
1
1
1
C = 16c 4 + 2 + 2 3," = " c=
4c 4c
2


Tú,vi D=

1
1
1

+
+
,theobtngthcCauchy Bunhiacopsky ưSchwarz,thỡ
2 a 2 3b 2 2c 2
2

P = A + B + C + D 4 + 10 + 3 +

7a

1
1
1
ổ 1 1 1ử
ỗ + + ữ = 24," = " a = c = ,b=
2 + 3 + 2 ố a b c ứ
2
3

KL
Gi Ilgiaoimhaingchộocahỡnhbỡnhhnh,ththỡ I ( aa) vialsthcnoú.
Suyra C ( 2a - 2 2a ) , D ( 2a -3 2a ).

8a

0.25
0.25

k
k

ak = k.( k - 1) C2013
+ kC2013
= k.( k - 1)

ak = 2012 ì 2013C

k -2
2011

+ 2013C

k -1
2012

2013!
2013!
+ k.
"k = 1,2,...,2013
k ! ( 2013 - k ) !
k ! ( 2013 -k )!

"k =1,2,...,2013

0
1
2011
0
1
2012
S1 = 2012 ì 2013 ( C2011

+ C2011
+ L + C2011
+ C2012
+ L + C2012
) + 2013 ( C2012
)

S1 = 2012 ì 2013 ì ( 1 + 1)

2011

2012

+ 2013 ì ( 1 + 1)

= 2012 ì 2013 ì 2 2011 + 2013 ì 2 2012 = 2013 ì 2014 ì2 2011

hb : 3 x + 4 y + 10 = 0, la : x - y + 1 = 0
+Do M ( 0 2) ẻ( AB ) nờnim N(11) ixngvi Mqua la nmtrờn AC.
+SuyraAlgiaoimcangthngdquaN,vuụnggúcvi hb vngthng la. Tú
A( 45 ).

8b

0.25
0.25

1
2
3

2013
Tớnhtng: S1 = 12 .C2013
+ 2 2 .C2013
+ 3 2 .C2013
+ L+ 2013 2 .C2013

1ử
ổ
+BlgiaoimcangthngAMvi hb. Tú B ỗ -3- ữ
4ứ
ố
+Do MC = 2 nờn C lgiaoimcangtrũntõmMbỏnkớnh 2 vingthngd.
ổ 33 31ử
Suyra C(11) hoc C ỗ ữ
ố 25 25ứ
0
C2013
C1
C2
C2013
Tớnhtng: S2 =
+ 2013 + 2013 + L+ 2013
1
2
3
2014
k
C2013
Shngtngquỏtcatngl ak =
" k = 0,1,2,...,2013

k +1
Ck
2013!
1
2014 !
ak = 2013 =
=
ì
" k = 0,1,2,...,2013
k + 1 ( k + 1) ì k ! ( 2013 - k ) ! 2014 ( k + 1) ! ( 2013 -k )!

Vytac
S2 =

ak =

k + 1
C2014
2014

" k =0,1,2,...,2013

1
1 ộ
2 2014 - 1
2014
1
2
2014
0 ự

ì ( C2014
+ C2014
+ L+ C2014
=
ì ( 1 + 1) - C2014
=
)
ở
ỷ
2014
2014
2014

CmnthyNguynDuyLiờn()giti
www.laisac.page.tl

Mụn:Toỏn12.Khi D.

chớnhthc
(thigm01trang)

Thigianlmbi:180phỳt(Khụngkthigiangiao)

A.PHNCHUNGCHOTTCTHSINH(7,0im)
Cõu I(2,0im).Chohms y = - x 3 + ( 2m + 1 )x 2 - m -1 ( Cm ) .

Vi a =2 : C ( 2 4 ) , D (1 4) vi a = -2 : C ( -6 -4 ) , D ( -7 -4)

k
k

Shngtngquỏtcatngl ak = k 2 C2013
= k .( k - 1 + 1)C2013
"k =1,2,...,2013

7b

0.25

Tú,dodintớchcahỡnhbỡnhhnhbng4nờn 2a = 4 a = 2.
Ktlun

KTHITHIHCLN1NMHC2013ư2014

TRNGTHPTCHUYấNVNHPHC

0.25
0.25
0.25
0.25

0.25
0.25

1) Khosỏtsbinthiờnvvthcahmskhi m =1 .
2) Tỡm m ngthng y = 2mx - m -1 ctctthhms ( Cm ) tibaimphõnbitcú

honhlpthnhmtcpscng.
3
2
CõuII(2,0im)1)Giiphngtrỡnh: 2 sin x - 3 = 3 sin x + 2 sin x -3 tan x .

(

)

4
ỡ
2
2
= 13
2
ù9 x + y + 2xy +
x
( y)
ù
2)Giihphngtrỡnh: ớ
.
ù 2x + 1 = 3
ùợ
x- y

(

)

3

3x + 2 - 3x - 2
x -2
Cõu IV (1,0 im). Cho hỡnh chúp S .ABCD cú ỏy l hỡnh bỡnh hnh vi AB =2a , BC =a 2 ,

BD =a 6 .Hỡnhchiuvuụnggúcca S lờnmtphng ABCD ltrngtõm G catamgiỏc BCD ,
bit SG =2a .
Tớnhthtớch V cahỡnhchúp S .ABCD vkhongcỏchgiahaingthng AC v SB theo a .
1 1 1
CõuV(1,0im).Cho x,y lcỏcsdngthomón + + =3.Tỡmgiỏtrlnnhtcabiu
xy x y
3y
3x
1
1
1
+
+
- thc: M =
x( y + 1) y ( x + 1) x + y x 2 y 2
CõuIII(1,0im). Tớnhgiihn: L = lim
x đ 2

0.25

B.PHNRIấNG (3im).Thớsinhchclmmttronghaiphn(phn1 hoc2)
1.TheochngtrỡnhChun

0.25

CõuVIA(2,0im)1)Trongmtphngvihtrcto Oxy ,chohỡnhthangcõn ABCD cúhai
ỏyl AB , CD haingchộo AC, BD vuụnggúcvinhau.Bit A ( 03), B ( 34) v C nmtrờn
trchonh.Xỏcnhtonh D cahỡnhthang ABCD .

0.25

2 ử
ổ
2)Tỡmshngkhụngcha x trongkhaitrin: p ( x )= ỗ 3 x +
ữ .Bitrngsnguyờndng n
x ứ
ố
thomón Cn6 + 3Cn7 + 3Cn8 + Cn9 =2Cn8+ 2

n

CõuVIIA(1,0im).Xỏcnh m hm s: y = ( m2 - 3m ) x + 2 ( m -3 )cos x luụnnghchbintrờn Ă
0.25

2.Theochngtrỡnhnõngcao.

0.25

Cõu VI B (2,0 im) 1) Trong mt phng vi h ta Oxy ,lp phng trỡnh chớnh tc ca elip
( E)bitrngcúmtnhvhaitiờuimca ( E)tothnhmttamgiỏcuvchuvihỡnhchnht

(

)

csca ( E) l 12 2 + 3 .
0.25

2
3

2013
2)Tớnhtng : S = 1.2.C2013
+ 2.3.C2013
+ L+ 2012.2013.C2013

CõuVII B (1,0 im).Xỏc nh m hm s: y = ( m 2 + m + 1) x + ( m 2 - m + 1)sin x +2m luụn ng
bintrờn Ă
ưưưưưưưưưưHTưưưưưưưưưư


Bagiaoiml: A ( 0 - m -1) B ( 1m -1 ) C ( 2m4m 2 - m -1)
TRNGTHPTCHUYấNVNHPHC

1
(*)
2
Spspcỏchonhtheothttngdntacúcỏcdóyssau
ã 0 1 2m lpthnhcpscng 0 + 2m = 2.1 m =1 thomón(*)
1
ã 0 2m 1lpthnhcpscng 0 + 1 = 2.2m m= thomón(*)
4
1
ã 2m 0 1 lpthnhcpscng 2m + 1 = 2.0 m= - thomón(*)
2
1 1
Ktlun:m= - 1
2 4

KTHITHIHCLN1NMHC2013ư2014

Tacú: A , B , C phõnbit m ạ 0mạ

Mụn:Toỏn12.Khi D.

chớnhthc
(thigm01trang)

Thigianlmbi:180phỳt(Khụngkthigiangiao)
HNGDNCHMTHI
(Vnbnnygm05trang)

I)Hngdnchung:
1)Nuthớsinhlmbikhụngtheocỏchnờutrongỏpỏnnhngvnỳngthỡchosimtng
phnnhthangimquynh.
2)Vicchitithoỏthangim(nucú)tronghngdnchmphimbokhụnglmsailch
hngdnchmvphicthngnhtthchintrongcỏcgiỏoviờnchmthi.
3)imtonbitớnhn0,25im.Saukhicngimtonbi,ginguyờnktqu.
II)ỏpỏnvthangim:
Cõu
ỏpỏn
im
Chohms y = - x 3 + ( 2m + 1)x 2 - m -1 ( Cm ) .
1,0
1)Khosỏtsbinthiờnvvthcahmskhi m =1 .
Khi m =1 hmstrthnh y = - x 3 + 3x 2 -2
CõuI Tpxỏcnh:Rhmsliờntctrờn R.
0,25
Sbinthiờn:lim y = +Ơ lim y = -Ơ .thhmskhụngcútimcn.
xđ-Ơ

2,0

xđ+Ơ

Bngbinthiờn:
x à
01
y
+0

y +à



2+à
0+
2

0.25

yU =0
2
thcahmscúdngnhhỡnhdiõy:

à

0.25

3

(

0.25

0.25

)

2

1)Giiphngtrỡnh: 2 sin x - 3 = 3 sin x + 2 sin x -3 tan x .(1)
iukin: cos x ạ0
Phngtrỡnh óchotngngvi:
CõuII 2 sin 3 x.cos x - 3 cos x = 3 sin 2 x + 2 sin x -3 sin x

(

)

0.25

2 sin3 x.cos x - 3cos x = -3cos 2 x.sin x +2 sin 2 x
2 sin 2 x ( sin x.cos x - 1) + 3 cos x ( sin x.cos x - 1)= 0
2,0

( sin x.cos x - 1) ( 2 sin 2 x + 3 cos x )= 0

0.25

ổ 1

ử
ỗ sin 2x - 1 ữ ( 2 - 2 cos 2 x + 3 cos x )= 0
ố2
ứ
ộ cos x = 2 (VN)
2 cos 2 x - 3 cos x - 2 =0 (do sin 2x - 2 ạ 0,"x ) ờ
ờ cos x= - 1
ờở
2
1
2p
cos x = - x =
+ k 2 p ,kẻ Â ( thomón iukin)
2
3
2p
Vyphngtrỡnhcúhaihnghim: x =
+ k 2 p ,kẻ Â
3
4
ỡ
2
2
= 13
2
ù9 ( x + y ) + 2xy +
x
y)
(
ù

2)Giihphngtrỡnh: ớ
.
ù 2x + 1 = 3
ùợ
x- y
ỡ
ộ
1 ự
2
2
ù5 ( x + y ) + 4 ờ( x - y ) +
ỳ = 13
2
ù
( x - y) ỳỷ
ờở
Vitlihphngtrỡnh: ớ
/K x - y ạ0
1
ù
ù( x + y ) + ( x - y ) + ( x - y ) = 3
ợ

2)Tỡm m ngthng y = 2mx - m -1 ct( Cm ) tibaimphõnbitcúhonh
lpthnhmtcpscng
Xộtphngtrỡnhhonh giaoim:
- x 3 + ( 2m + 1 )x 2 - m - 1 = 2mx - m -1 x 3 - ( 2m + 1 )x 2 + 2mx = 0
ộ x = 0
x ( x 2 - ( 2m + 1 )x + 2m )=0 ờờ x = 1
ờởx = 2m

0.25

0.25

0.25

0.25

1
iukin b 2 .
x - y

1,0

t a = x + y b = x - y+

0.25

5
ỡ
2
2
2
ùỡ5a + 4 ( b - 2 )= 13 ỡ9a - 24a + 15 = 0 ùa = 1 a=
Hóchotrthnh: ớ
ớ
ớ
3
ợb = 3 - a

ùợb = 3 - a
ợùa + b = 3

0.25


ỡ x + y = 1
ỡa = 1
ỡ x + y = 1 ỡ x = 1
ù
ã ớ
ớ
ớ
ớ
1
x
y
+
=
2
ợb = 2
ợx - y = 1
ợ y = 1
ù
x
y
ợ
5
ỡ
ùùa= 3

ã ớ
Loi
ùb = 3 - a = 3- 5 = 4
ùợ
3 3
Vyhphngtrỡnhcúmtnghimduynht ( x y ) =( 11 )

0.25

0.25

CõuIII L = lim

(

3

) (

) = lim ổ

3x + 2 - 2 + 2 - 3x - 2

ỗ
x đ 2 ỗ
ố

x-2

xđ 2

3x + 2 - 2
3x - 2 - 2ử
= L1 - L2
ữữ
x-2
x-2
ứ

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3

1,0

3x + 2 - 2
3x + 2 - 8
L1 = lim
= lim
xđ 2
x đ 2
2
x - 2
ổ
3
( x - 2 ) ỗ ( 3x + 2 ) + 2 3 3x + 2 + 4ửữ
ố
ứ
3
1

L1 = lim
=
2
x đ 2 3
( 3x + 2 ) + 2 3 3x + 2 +4 4

L2 = lim
xđ 2

3x - 2 - 2
3x - 2 - 4
= lim
x đ 2
x - 2
( x - 2 ) 3x - 2 + 2

(

)

3
3
=
x đ 2 3x - 2 +2
4
1 3
1
L = L1 - L2 = - = -
4 4
2

Cho hỡnh chúp S .ABCD cú ỏy l hỡnh bỡnh hnh vi AB =2a , BC =a 2 ,
BD =a 6 .Hỡnhchiu vuụnggúcca S lờn mtphng ABCD ltrngtõm G ca
CõuIV tamgiỏc BCD ,bit SG =2a .
Tớnhthtớch V cahỡnhchúp S .ABCD vkhongcỏchgiahaingthng AC v
SB theo a .

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L2 = lim

1,0

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1,0

NhnxộtABCDlhỡnhchnht(do AB 2 + AD 2 =BD 2 )
1
4 2 3
SG.S ABCD =
a
3
3
K l im i xng vi D qua C, H l hỡnh chiu vuụng gúc ca G lờn BK suy ra
BK ^( SHG ) .GiIlhỡnhchiuvuụnggúccaGlờnSHsuyraGI=d(AC,SB)

t a =

(

(

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VS .ABCD =

( a + b)
1
1
> 0, b= >0,theobitacú 3- ( a + b ) = ab Ê
(BTCauchy),
x
y
4
kthpvi a + b >0 suyra a + b 2
3a
3b
ab
Tatỡmgiỏtrlnnhtca M =
+
+
- a 2 - b2
b + 1 a + 1 a +b
(a + b) 2 - 2ab + a + b
ab
=3
+
- (a + b)2 + 2ab

ab + a + b + 1
a +b
1ộ
12
ự
= ờ -(a + b)2 + a + b+
+ 2ỳ (do ab = 3 - (a +b))
4ở
a+b
ỷ
12
t t = a + b 2 xộthms: g (t ) = -t 2 + t+ +2 trờn [ 2 +Ơ)
t
12
g Â(t ) = -2t - 2 + 1 < 0, "t2 suyra g (t) nghchbintrờn (2, +Ơ)
t
3
Do ú max g (t ) = g (2) =6 suy ra giỏ tr ln nht ca M bng
t c khi
[ 2,+Ơ )
2
a = b = 1 x = y =1.
1
1
3a
3b
ab
+
+
- a 2 - b2

Cỏch 2 t a = > 0, b= >0,theobitacú M =
x
y
b + 1 a + 1 a +b
( a + ab + b ) a ( a + ab + b )b ab 2 2
M=
+
+
- a - b .
b +1
a + 1
a +b
ab
ab
ab
ab
ab
ab
1
M=
+
+
Ê
+
+
= a b + b a + ab (BTAMưGM)
b + 1 a + 1 a +b 2 b 2 a 2 ab 2
1
1 ộ a ( b + 1) b ( a+ 1) a + bự 3
M Ê a b + b a + ab Ê ờ

+
+
ỳ = ,(BTAMưGM)
2
2ở 2
2
2 ỷ 2
dubngkhi a = b =1
3
Vygiỏtrlnnhtca M bng tckhi a = b = 1 x = y =1.
2
1)Trong mtphng vi htrcto Oxy ,chohỡnh thangcõn ABCD cúhaiỏy l
Cõu
AB , CD haingchộo AC, BD vuụnggúcvinhau.Bit A ( 03), B ( 34) v C
VIA
nmtrờntrchonh.Xỏcnhtonh D cahỡnhthang ABCD .
Cỏch1

1,0
3

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1,0

2

3

3x + 2 - 3x - 2

Tớnhgiihn : L = lim
x đ 2
x -2

1
1
1
2a
2a
=
+
ị CJ =
ị GH =
CJ 2 BC 2 CK2
3
3
TamgiỏcSHGvuụngGsuyraGI=a.
Vy:d(AC,SB)=a
1 1 1
Cho x,y lcỏcsdngthomón + + =3.Tỡmgiỏtrlnnhtcabiuthc:
xy x y
3y
3x
1
1
1
+
+
- CUV M =
x( y + 1) y ( x + 1) x +y x 2 y 2

GH=CJm

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)

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)

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1,0


th

f ( t ) = -2 ( m - 3 )t + m 2 -3m trờnon [ -11] lmtonthng

ỡù f ( -1) Ê 0

f ( t ) Ê 0 "t ẻ [ -11] ớ
ùợf ( 1)Ê 0

2,0

2
ùỡ2 ( m - 3 ) + m - 3m Ê 0
ùỡ( m - 3 )( m + 2 ) Ê 0 ỡ-2 Ê m Ê 3
ớ
ớ
2 Ê m Ê 3
ớ
2
2
m
3
+
m
3m
Ê
0
)
ù (
ợù( m - 3 )( m - 2 )Ê 0 ợ2 Ê m Ê 3
ợ
Vyhmsnghchbintrờn Ă thỡ 2 Ê m Ê3

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Cõu
VIB

1)TrongmtphngvihtaOxy ,lpphngtrỡnhchớnhtccaelip ( E)bitrng

2,0

chnhtcsca ( E) l 12 2 + 3 .

C ẻ Ox ị C ( c0)

AC ^ BD ị 3dc + c 2 - 3c - 3d + 12 = 0( 1 )
3 7
IltrungimAB ị I( )
2 2
8 - 3c
ổ 3d + 2c dử
JltrungimDC ị J ỗ
ữ ,t IJ ^ AB ị d =
( 2 )
2 ứ
5
ố 2
ộ c = 6
Thay(2)vo(1)cú: 2c 2 - 9c - 18 = 0 ờ
-3
ờ c=
ở
2
c = 6 ị d = -2 ị D( 0 -2 )( tm )

-3

5
5
ị d = ịD( 6 )( ktm )
2
2
2
(HcsinhphikimtraiukinthụngquavộctABvvộctDCcựngchiu)
Ktlun: D( 0 -2 )

cú mt nh v hai tiờu im ca

(

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(

k
( k - 1) .k.C2013

C

= 2C

8
n+ 2

C

8

n+ 2

+C

15

= 2C
15

8
n+ 2

2 ử
ổ
k
Khiú p ( x ) = ỗ 3 x +
ữ = ồ C15
xứ
k =0
ố

C

9
n+2

=C

8
n + 2

k

15

30 - 5 k
6

ổ 2 ử
k
k
ỗ
ữ = ồC152 x
k =0
ố xứ
30 - 5k
Shngkhụngcha x tngngvi
= 0 k =6
6
6
6
Shngkhụngcha x phitỡml C15.2 =320320

Cõu
VIIA

S = 2012.2013.( 1 + 1)

1,0
Cõu
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n =15

15 - k

( x )
3

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2

2

iukinhmsluụnnghchbin trờn Ă y  0"x ẻ Ă

iukinhmsluụnnghchbintrờn Ă y Â Ê 0"x ẻ Ă
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( m2 + m + 1) + ( m2 - m + 1)cos x 0 "x ẻ Ă
(m

2

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+ m + 1) + ( m 2 - m + 1) t 0 "t ẻ [ -11] vi t =cos x

f ( t ) = ( m 2 + m + 1) + ( m 2 - m + 1) t , "t ẻ [ -11]

ỡù f ( 1) 0
onthng f ( t ) 0 "t ẻ [ -11] ớ
ùợf ( -1) 0

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1,0

ohm y  = ( m + m + 1) + ( m - m +1)cos x
2

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ohm: y  = m - 3m - 2 ( m -3 )sin x

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Xỏcnh m hms: y = ( m + m + 1) x + ( m - m + 1)sin x +2m ngbintrờn Ă
2

th

1,0

m 2 - 3m - 2 ( m - 3 ) sin x Ê 0 "x ẻ Ă m 2 - 3m - 2 ( m - 3 ) t Ê 0 "t ẻ [ -11] ,t = sin x

1,0

=2012.2013.2

1,0
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2011

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Xỏcnh m hms: y = ( m2 - 3m ) x + 2 ( m -3 )cos x luụnnghchbintrờn Ă
2

"k =2,3,...,2013.

2013!
k - 2
= ( k - 1) .k.
= 2012.2013.C2011
"k = 2,3,...,2013
k ! ( 2013 -k )!

2011

7B
9
n+ 2

k
2013

0
1
2
2011
Vy S = 2012.2013.( C2011
+ C2011
+ C2011
+ L+ C2011
)

iukin:n ẻ Ơ* ,n 9
9
n +3

)

Xộtshngtngquỏt: ( k - 1).k .C

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n

2 ử
ổ
2)Tỡmshngkhụngcha x trongkhaitrin: p ( x )= ỗ 3 x +
ữ .Bitrngs
x ứ
ố
nguyờndng n thomón Cn6 + 3Cn7 + 3Cn8 + Cn9 =2Cn8+ 2

0,5

2
3
2013
2)Tớnhtng: S = 1.2.C2013
+ 2.3.C2013
+ L+ 2012.2013.C2013

c=

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)

ỡc 2 = a 2 - b2
ỡ a= 6
ù
ù
ù
3
x 2 y2
b
=
2
c

= 1

ớ
ớb = 3 3 ( E): +
2
36 27
ù
ùc= 3
ợ
ù4 ( a + b ) = 12 2 + 3
ợ

(

1,0

)

vchuvihỡnhchnhtcsca ( E) l 12 2 + 3 .
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( E)to thnh mt tam giỏc u v chu vi hỡnh

x 2 y2
( E ) : 2 + 2 = 1( a > b>0) vi2tiờuim F1 ( -c0 ) F2 ( c0 ) ( c 2 = a 2 - b 2, c >0)
a
b
2nhtrờntrcnhl B1 ( 0 -b ) , B2 ( 0b ) theogt:tamgiỏc B1F1F2 ( DB1 F1F )u

( DC ): x - 3 y - c = 0 ị D( 3d +cd )

uuur
uuur
AC( 0 -3 ) BD( 3d + c - 3d - 4 )

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trờn on [ -11] l mt

ỡ 2m 2 + 2 0 "mẻ Ă
ớ
ị m 0.Vy m 0 thomónyờucubitoỏn
ợ 2m 0

CmnthyNguynDuyLiờn()giti
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