Đề thi thử đại học lần I Năm 2013 – 2014 môn Toán Khối A, B và A1

... 
b,PT tt của đồ thị (C) tại điểm 0M ( 0 0,x y )là: 2 20 0 0( 1) 2 2 1 0x x y x x+ − − + − = ( ∆ ) 
 .................................................................................................................................................. 
d(I; ∆ )= 00
4
00
02 2
2
21 ( 1)
xx
xx
=− 
= ⇔ 
=+ − 
 .............................................................................................................................................. 
có 2 pt tt là y=-x+1 và y=-x+5 
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 CâuII 
 (2 đ) a, 2os 10cos( ) 3 sin 2 5 06c x x x
Π
+ + − + = 22cos 1 10cos( ) 3 sin 2 6 0
6
x x x
pi
⇔ − + + − + = 
 ....................................................................................................................................... 
cos2x - 3 sin 2 10cos( ) 6 0
6
x x
pi
+ + + = 2cos(2 ) 10cos( ) 6 0
3 6
x x
pi pi
⇔ + + + + = 
 ............................................................................................................................................ 
24cos ( ) 10cos( ) 4 0
6 6
x x
pi pi
⇔ + + + + = 22cos ( ) 5cos( ) 2 0
6 6
x x
pi pi
⇔ + + + + = 
 .................................................................................................................................................. 
cos( ) 2
6
x
pi
+ = − (loại) hoặc cos( 1)
6 2
x
pi
+ = −
52 ; 2
2 6
x k x kpi pipi pi⇒ = + = − + , k z∈ 
 ............................................................................................................................................ 
b,Giải hệ PT 
3
2 4 3
1 1 2;(1)
9 (9 );(2)
x y
x y y x y y
 + + − =

− + = + −
 đ/k y 1≤ 
(2) 3 3( )( 9) 0 9 0
y x
x y x y
x y
=
⇔ − + − = ⇔ 
+ − =
 ................................................................................................................................... 
Thay y=x vào(1) ta có pt: 3
0
1 1 2
11 6 3
x y
x x
x y
= =
+ + − = ⇔ 
= = − ±
 ..................................................................................................................................... 
Do y 1≤ ta có (1) 3 1 2 1 2 7x y x⇔ + = − − ≤ ⇒ ≤ 
 ....................................................................................................................................... 
3 9 1 0x y⇒ + − ≤ − < pt (2) vô nghiệm 
Vậy hệ pt có nghiệm là:x=y=0 và x=y= -11 6 3± 
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 CâuIII 
 (1 đ) 
I=
2 24 4 4
2 2 2
4 4 4
s inx 1 s inx 1
1 2cos 1 2cos 1 2cos
x xdx dx dx
x x x
pi pi pi
pi pi pi
− − −
+
= +
+ + +∫ ∫ ∫
 1( )I ( )2I 
 ............................................................................................................................ 
giải 1I =
0 2 24
2 2
0
4
sin sin
1 2cos 1 2cos
x x x xdx dx
x x
pi
pi
−
+
+ +∫ ∫
.xét J=
0 2
2
4
sin
1 2cos
x x dx
xpi
−
+∫
 ,Đặt t=-x 
0 2 24
2 2
0
4
s inx s inx
1 2cos 1 2cos
x xdx dx
x x
pi
pi
−
⇒ = −
+ +∫ ∫
 suy ra 1 0I = 
 ................................................................................................................................ 
4 4 4
2 2 2
2
24 4 4
(t anx)
11 2cos t an x+3
os ( 2)
cos
dx dx dI dx
x
c x
x
pi pi pi
pi pi pi
− − −
= = =
+ +
∫ ∫ ∫ .Đặt tanx=t 
x 
-
4
pi
4
pi
t -1 1 
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. 
1
2 2
1 3
dtI
t
−
=
+∫
 Đặt t= 3 tanz 2
3
os x
dtdt
c
⇒ = 
t -1 1 
z 
6
pi
− 
6
pi
6 6
2 2 2
6 6
3 1
cos (3 tan 3) 3 3 3
dxI dx
x x
pi pi
pi pi
pi
− −
= = =
+∫ ∫
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 CâuIV 
 (1 đ) 
2
2 01 3sin120
2 2ABCD
aS a= = ,
2 3
8ACI
aS∆ = 
 .......................................................................... 
SB=BO=
2
a
 ,V=
31 3
.
3 48ACI
aSB S∆ = (DVTT) 
E
O
B
A
C
S
D
I
 ........................................................................... 
Đặt O(0;0;0) ;A 3( ;0;0)
2
a
; C 3( ;0;0)
2
a
− ; 
B(0;
2
a
;0); B(0;
2
a
;0);S(0;
2
a
; 
2
a );I 3( ; ; )
4 4 4
a a a
(CI =


 3 3
; ; )
4 4 4
a a a
= (3 3;1;1); (0; ; ) (0;2;1)
4 2 2
a a aSD a= − − = −




; 
;n CI SD = = 
 






2
( 1; 3 3;6 3)
8
a
− − − ptmp(α ) chứa CI // SD là 
3( ) 3 3( 0) 6 3( 0) 0
2
a
x y z− + − − + − = 33 3 6 3 0
2
a
x y z⇔ + − + = 
 ........................................................................................................................................ 
2 2
3 3 3
2 2( ; ( ))
1 (3 3) (6 3 )
a a
d D α
− +
=
+ +
=
3
136
a
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 CâuV 
 (1 đ) ( )2 2 2 2 2 23 ( )( )a b c a b c a b c+ + = + + + + = 3 3 3 2 2 2 2 2 2a b c a b b c c a ab bc ca+ + + + + + + + . 
 ............................................................................................................................................ 
3 2 22a ab a b+ ≥ ; 3 2 22b bc b c+ ≥ ; 3 2 22c ca c a+ ≥ 3(⇒ 22 2 2 2 2) 3( ) 0a b c a b b c c a+ + ≥ + + > 
 ..................................................................................................................................... 
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VT 2 2 2 2 2 2
ab bc ac
a b c
a b c
+ +≥ + + + =
+ +
2 2 2
2 2 2
2 2 2
9 ( )
2( )
a b c
a b c
a b c
− + +
+ + +
+ +
; Đặt t= 2 2 2a b c+ + 
 ............................................................................................................................................ 
VT
9 9 1 3 13
2 2 2 2 2 2 2
t t t
t
t t
−≥ + = + + − ≥ + − =4 dấu bằng xẩy ra khi a=b=c=1 
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 Tựchọn 
 cơ bản 
 CâuVI 
 (1 đ) 
a,Ta cóA(2;1) B∈BH⇒B(b;7-2b) 
 ........................................................................................................................................... 
M là trung điểm của BC⇒C(2-b;2b-5) 
 ........................................................................................................................................... 
( ;2 6);AC b b BH= − −




⊥ AC 
. 0BHU AC =




 


 12
5
b⇒ = 12 11 2 1( ; ); ( ; )
5 5 5 5
B C⇒ − − 
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 CâuVII 
 (1 đ) 
a, Véc tơ chỉ phương đt d: (2;1;1); 1;0;1) , (1; 1; 1)d dU AB U U AB∆  = = ⇒ = = − − 


 


 

 



 ............................................................................................................................................ 
Pt đt d: 1 1
1 1 1
x y z− −
= =
− −
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 CâuVIII 
a,
1 7
15
k
n
K
n
C
C
−
= 
! ! 71 :( 1)!( 1)! !( )! 15
n nk n
k n k k n k
≤ ≤ ⇔ =
− − + −
 .............................................................................................................................................. 
15.⇔ 15. ! 7. !( 1)!( 1)! !( )!
n n
k n k k n k
⇔ = ⇔
− − + −
15 7 15 7 7 7
1
k n k
n k k
= ⇔ = − +
− +
 ............................................................................................................................................... 
7n=22k-7
22 1 7 21
7
k
n k n⇔ = − ⇒ = ⇒ = 
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 Tự chọn 
 nâng cao 
 CâuVI 
b,R= ( , )
6
13o
d ∆ = .Gọi I( 0 0; )x y là tâm đường tròn (C)
2 2
0 0 1
9 4
x y
⇒ + = (1) 
 ............................................................................................................................................... 
0 0
( ; )
2 3 6 6
13 13I
x y
d R∆
− +
= ⇔ = ⇔ 0 0
0 0
2 3 12 0;(2)
2 3 0;(3)
x y
x y
− + =

− =
Từ (1) và (2)suy ra: 2 20 0 0 0( 2) 1 2 12 27 09 3
x x
x x+ + = ⇔ + + = vô nghiệm 
 ............................................................................................................................................... 
Từ(1)và(3)suyra:
2 2
0 0
0
3 21
9 9 2
x x
x+ = ⇔ = ± 
Khi 2 20 0
3 2 3 2 362 ( ) : ( ) ( 2)
2 2 13
x y C x y= ⇒ = ⇒ − + − = 
 ............................................................................................................................................ 
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Khi 0 0
3 2 2 ( )
2
x y c= − ⇒ = − ⇒ : 2 23 2 36( ) ( 2)
2 13
x y+ + + = 
 CâuVII 
b,Đặt 
1 2
1 1
2 1 1
1
x t
x y z
t y t
z t
= +
− − 
= = = ⇔ =

= +
 (1) 
................................................................................................................................................. 
dt d cắt (p) ta có 1+2t-2t+1+t=0 2 ( 3; 2; 1)t A⇔ = − ⇒ − − − 
................................................................................................................................................ 
(1; 2;1); (1 3;3) , ( 3; 2; 1)p Q p Qn n U n n∆  = − = − ⇒ = = − − − 


 

 

 

 


................................................................................................................................................ 
PTđường thẳng 
3 2 1
:
3 2 1
x y z+ + +∆ = = 
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 CâuVIII 
 (1 đ) 
b, ( )2014 0 1 2 2 2014 20142014 2014 2014 20141 ...x c c x c x c x+ = + + + + (1) 
( )2014 0 1 2 2 2014 20142014 2014 2014 20141 ...x c c x c x c x− = − + − + (2) 
Lấy (1)+(2) Ta có f(x)= 2014 2014 0 2 2 2014 20142014 2014 2014(1 ) (1 ) 2 2 ... 2x x c x c x c x+ + − = + + + 
......................................................................................................................................... 
Lấy đạo hàm 2 vế ta được 
f’(x)=2014 2013 2013(1 ) 2014(1 )x x+ − − = 2 4 3 2014 20132014 2014 20144 8 ... 4028c c x c x+ + + 
........................................................................................................................................... 
Thay x=1 ta được f’(1)= 2013 2 4 20142014 2014 20142014.2 4 8 ... 4028c c c= + + + 2013
1007
.2
2
A⇒ = 
Chú ý: ( Học sinh giải cách khác đúng vẫn đạt điểm tối đa) 
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