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

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

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.


+ 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:

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


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


)(

)(

)

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

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

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

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

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

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

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

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


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


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


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

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

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

(

(

0.25

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

+
+
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

0.25

1,0

2

3

3x + 2 - 3x - 2

GH=CJm

0.25
0.25

)

0.25

0.25

0.25

0,25
0.25
0.25
0.25

)

0,25

1,0


th

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

ỡù f ( -1) Ê 0

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

cú mt nh v hai tiờu im ca

(

0.25

(

k
( k - 1) .k.C2013

C

= 2C

8
n+ 2

C

8

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
0.25

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

0,25

1,0

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

0,25

ohm: y  = m - 3m - 2 ( m -3 )sin x

0,25
0,25

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


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

0,25

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

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 .
0.25

0,25

( 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 )