Bài giảng Kiểm tra 1 tiết Đại số 10 chương 2

Tran Sú Tuứng kim tra ẹaùi soỏ 10
CHNG II: HM S BC NHT BC HAI
KIM TRA 1 TIT
S 1
a) Trc nghim khỏch quan
Cõu 1: (0,5) Tp xỏc nh ca hm s
3
1
y 1 x
x 1
= +
+
l:
a) D = (1; 1) b) D = (1; 1]
c) D = (; 1] \ {1} d) D = (; 1] (1; + )
Cõu 2: (0,5) Cho hm s (P) : y = ax
2
+ bx + c. Tỡm a, b, c bit (P) qua 3 im
A(1; 0), B( 0; 1), C(1; 0).
a) a = 1; b = 2; c = 1. b) a = 1; b = 2; c = 1.
c) a = 1; b = 0; c = 1. d) a = 1; b = 0; c= 1.
Cõu 3: (0,5) Cho hm s y = x
2
+ mx + n cú th l parabol (P). Tỡm m, n
parabol cú nh l S(1; 2).
a) m = 2; n = 1. b) m = 2; n = 3.
c) m = 2; n = 2. d) m= 2; n = 3.
Cõu 4: (0,5) Cho hm s y = 2x
2
4x + 3 cú th l parabol (P). Mnh no
sau õy sai?

a) (P) i qua im M(1; 9). b) (P) cú nh l S(1; 1).
c) (P) cú trc i xng l .thng y = 1.
d) (P) khụng cú giao im vi trc honh.
b) T lun
Cõu 5: (8 im) Cho hm s y = (m 1)x
2
+ 2x 3 (P
m
)
a) Kho sỏt v v th hm s vi m = 2 (tng ng l (P
2
)). Bng th,
tỡm x y 0, y 0.
b) Dựng th, hóy bin lun theo k s nghim ca phng trỡnh:
2
| x 2x 3 | 2k 1.+ =
c) Vit phng trỡnh ng thng i qua nh ca (P
2
) v giao im ca (P
2
)
vi trc tung.
d) Xỏc nh m (P
m
) l parabol. Tỡm to qu tớch nh ca parabol (P
m
)
khi m thay i.
e) Chng minh rng (P
m

) luụn i qua mt im c nh, tỡm to im c
nh ú.
================
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kim tra ẹaùi soỏ 10 Tran Sú Tuứng
CHNG II: HM S BC NHT BC HAI
KIM TRA 1 TIT
S 2
a) Trc nghim khỏch quan ( 3 ) :
Cõu 1 : Tp xỏc nh ca hm s
1
y f(x) x 1
3 x
= = +

l:
a) (1;3) , b) [1;3] , c) (1;3] , c) [1;3)
Cõu 2: nh ca Parabol y = x
2
2x +2 l :
a) I(1;1) b) I(1;1) c) I(1;1) c) I(1;2)
Cõu 3 : Hm s y = 2x
2
4x + 1
a) ng bin trờn khong ( ; 1 ) b) ng bin trờn khong ( 1 ;+ )
c) Nghch bin trờn khong ( 1 ;+ ) d) ng bin trờn khong ( 4 ;2 )
b) T lun : ( 7 )
Cõu 5 ( 2 ) : Tỡm min xỏc nh v xột tớnh chn l ca hm s sau :

2

y
x 1 x 1
=
+ +
Cõu 6 ( 1,5 ): Xột s bin thiờn ca hm s :
3
y
2 x
=

trờn ( 2 ; + )
Cõu 7 :
a) (1,5 ) Tỡm Parabol y = ax
2
+ bx + 2 bit rng Parabol ú i qua im
A(3 ; 4) v cú trc i xng
3
x
2
=
.
b) ( 2 ) Kho sỏt v v th hm s va tỡm c cõu a).
=================
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Tran Sú Tuứng kim tra ẹaùi soỏ 10
CHNG II: HM S BC NHT BC HAI
KIM TRA 1 TIT
S 3
I. Phn trc nghim : ( 3 im )
Cõu 1: Hm s

2
4
x 1
y f(x)
x . 1 x
+
= =

cú tp xỏc nh l :
a)
(
]
;1
b)
( )
;1
c)
(
]
{ }
;1 \ 0
d)
( ) { }
;1 \ 0
Cõu 2: Hm s no l hm s chn :
a)
2
y 4x 2x= +
b)
y x 1 x 1= +

c)
( )
2
y x 1=
d)
y x 2 x 2= + +
Cõu 3 : im ng qui ca 3 ng thng
y 3 x; y = x+1; y = 2=
l :
a) ( 1; 2) b) ( 1; 2) c) (1; 2) d) (1; 2)
Cõu 4 : th ca hm s no i qua im A ( 1; 3 ) v ct trc honh ti im
cú x = 4 :
a)
3 12
y x
5 5
= +
b)
3 12
y x
5 5
= +
c)
3 12
y x
5 5
=
d)
3 12
y x

5 5
=
Cõu 5 : Cho parabol ( P ) :
2
y x mx 2m= +
.Giỏ tr ca m tung ca nh
( P ) bng 4 l :
a) 3 b) 4 c) 5 d) 6
Cõu 6 : Hm s
2
y f(x) x 2x 5= = +
:
a) Gim trờn
( )
; 1
b) Tng trờn
( )
2;+
c) Gim trờn
( )
;2
d) Tng trờn
( )
1;+
II. Phn t lun : ( 7 im )
Bi 1 : ( 3 im )
a) V ba th ca ba hm s sau trờn cựng mt h trc ta Oxy :
1
(d ) : y 2x 2= +
2

(d ) : y x 2= +
3
(d ) : y x=
b) Gi A,B,C l giao im cỏc th hm s ó cho. Chng t ABC vuụng.
c) Vit ph.trỡnh .thng song song vi
1
(d )
v i qua giao im ca
2 3
(d ),(d )
Bi 2 : ( 2 im ) Lp bng bin thiờn v v th ca cỏc hm s sau :
a)
2
x
y
2
=
b)
2
y 2x 4x 2= +
Bi 3 : ( 2 im ) Xỏc nh a, b, c bit parabol
2
y ax bx c= + +
a) i qua im A (8; 0) v cú nh I (6, 12 )
b) i qua A( 0 ; 1) , B(1 ; 1) , C (1 ; 1 ) .
==================
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kim tra ẹaùi soỏ 10 Tran Sú Tuứng
CHNG II: HM S BC NHT BC HAI
KIM TRA 1 TIT

S 4
I. PHN TRC NGHIM (3 im)
Cõu 1 : Tp xỏc nh ca hm s y =
x 5 4 2x+
l:
a) D =
( ; 5] [2 ; ) +
b) D = [5 ; 2]
c) D =

d) D = R
Cõu 2 : Cho hm s f (x) =
2
16 x
x 2

+
. Kt qu no sau õy ỳng:
a) f(0) = 2 ; f(1) =
15
3
b) f(1) =
15
; f(0) = 8
c) f(3) = 0 ; f(1) =
8
d) f(2) =
14
4
; f(3) =

7
Cõu 3 : Trong cỏc parabol sau õy, parabol no i qua gc ta :
a) y = 3x
2
4x + 3 b) y = 2x
2
5x
c) y = x
2
+ 1 d) y = x
2
+ 2x + 3
Cõu 4 : Hm s y = x
2
+ 4x 3
a) ng bin trờn
( ; 2)
b) ng bin trờn
(2 ; )+
c) Nghch bin trờn
( ; 2)
d) Nghch bin trờn (0 ; 3)
Cõu 5 : Parabol y = 3x
2
2x + 1 cú trc i xng l:
a) x =
1
3
b) x =
2

3
c) x =
1
3
d) y =
1
3
Cõu 6 : Ta giao im ca .thng y = x + 3 v parabol y = x
2
4x + 1 l:
a)
1
;1
3


ữ

b) (0 ; 3)
c) (1 ; 4) v (2 ; 5) d) (0 ; 1) v (2 ; 2)
II. PHN T LUN (7 im)
Bi 1: Vit phng trỡnh ng thng qua A(2 ; 3) v song song vi ng
thng y = x + 1
Bi 2: Tỡm parabol y = ax
2
+ bx + 1, bit parabol ú:
a) i qua 2 im M(1 ; 5) v N(2 ; 1)
b) i qua A(1 ; 3) v cú trc i xng x =
5
2

c) cú nh I(2 ; 3)
d) i qua B(1 ; 6), nh cú tung l 3.
===================
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Tran Sú Tuứng kim tra ẹaùi soỏ 10
CHNG II: HM S BC NHT BC HAI
KIM TRA 1 TIT
S 5
I. Phn trc nghim :
Cõu 1 (0,5 im): Tp xỏc nh ca hm s
2
x 1
y
x 1
+
=

l :
a) R b) R\ {1; 1} c) R\ {1} c) (1; 1)
Cõu 2 (0,5 im): Hm s y= ( 2 +m )x + 3m ng bin khi :
a) m =2 b) m ? 2 c) m > 2 c) m < 2
Cõu 3 (0,5 im): Hm s y = f(x) = x ( x4 +3x2 + 5) l :
a) Hm s chn b) Hm s l
c) Hm s khụng chn, khụng l c) C 3 kt lun trờn u sai
Cõu 4 (0,5 im): Cho hm s
2x 1 ;x 1
y
x 7
;x 1
2

+


=
+

>



Bit f(x
0
) = 5. thỡ x
0
khụng õm tng ng l:
a) 2 b) 0 c) 1 c) 3
Cõu 5 (0,5 im): nh ca parabol y = ax
2
+ bx + c l
a)
b
;
a 4a



ữ

b)
b

;
a 4a



ữ

c)
b
;
2a 4a



ữ

c)
b
;
a 4a



ữ

Cõu 6 (0,5 im): th ca hm s y = 3x
2
+2 c suy ra t th ca hm s
y = 3x
2

nh phộp tnh tin song song vi trc Oy
a) lờn trờn 3 n v b) lờn trờn 2 n v
c) xung di 3 n v c) xung di 2 n v
II : T LUN
Cõu 1 (2 im): Tỡm tp xỏc nh cỏc hm s sau :
a)
2
x 1
y
x 5x 6

=
+ +
b)
1
y 2 3x
x 1
= +
+
Cõu 2 (3 im): Lp bng bin thiờn v v th hm s y = x
2
+ x + 2
Cõu 3 (2 im): Xỏc nh hm s bc hai bit th ca nú l mt parabol cú tung
nh l
13
4

, trc i xng l ng thng x =
3
2

, i qua im M (1 ; 3)
=============
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kim tra ẹaùi soỏ 10 Tran Sú Tuứng
CHNG II: HM S BC NHT BC HAI
KIM TRA 1 TIT
S 6
Phn 1:Trc nghim khỏch quan (3 im)
Cõu 1: Tp xỏc nh ca hm s
2
x 2
y
x 4x 3

=
+
l:
a)
{ }
D \ 1; 2; 3= Ă
b)
{ }
D \ 1; 3= Ă
c)
{ }
D \ 2= Ă
d)
] [
D ( ; 1 3; )= +

Cõu 2: Hm s y = x
2
4x + 1
a) ng bin trờn khong (; 0) v nghch bin trờn khong (0; + ).
b) Nghch bin trờn khong (; 0) v ng bin trờn khong (0; + ).
c) ng bin trờn khong (; 2) v nghch bin trờn khong (2; + ).
c) Nghch bin trờn khong (; 2) v ng bin trờn khong (2; + ).
Cõu 3: Tp xỏc nh v tớnh chn, l ca hm s
2
2
x
y
x 1
=

l:
a)
D = Ă
; hm s chn. b)
{ }
D \ 1= Ă
; hm s chn.
c)
{ }
D \ 1= Ă
; hm s chn.
c)
{ }
D \ 1= Ă
; hm s khụng chn, khụng l.

Cõu 4: Cho hm s f(x) = 3x cú tp xỏc nh l tp Q . Tỡm x f(x) = 1.
a) x = 1 b) x = 3 c) x =
1
3
c) Tt c u sai.
Cõu 5: Giao im ca th hai hm s y = x + 3 v y = x
2
4x + 1 l:
a) (4; 1) v (5; 2) b) (1; 4) v (2; 5)
c) (1; 4) v (2; 5) c) (4; 1) v (5; 2)
Cõu 6: Ph.trỡnh .thng i qua A(0; 2) v song song vi ng thng y = x l:
a) y = x + 2 b) y = 2x c) y =
1
x
2
c) y = 2x + 2
Phn II: T lun (7 im)
Cõu 7: (2 im) Tỡm tp xỏc nh ca cỏc hm s sau:
a)
1
y x 4
2 x
= + +

b)
2
y
(x 2) x 1
=
+ +

Cõu 8: (1 im) Xột tớnh chn, l ca hm s f(x) = 3x.x
Cõu 9: (2 im) Lp bng bin thiờn v v th hm s y = x
2
+ 2x + 3
Cõu 10:(2 im) Xỏc nh hm s y = ax
2
+ bx + c (a 0), bit th hm s i
qua cỏc im: A(0; 3); B(1; 4); C(1; 6).
================
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Tran Sú Tuứng kim tra ẹaùi soỏ 10
CHNG II: HM S BC NHT BC HAI
KIM TRA 1 TIT
S 7
I/ Phn trc nghim (4 im)
Bi 1: Hm s y=
x
x 1
l:
a) hm s chn b) hm s l
c) hm s khụng chn, khụng l d) hm s va chn, va l
Bi 2: Hm s y= x
2
+2x +1 ng bin trong khong :
a) (

;1) b. (

;1) c. (1;+

) d. 1 kt qu khỏc
Bi 3: Tp xỏc nh ca hm s y=
6 3x+
l :
a) (

;2) b. (

;2) c. (2;+

) d. [2;+

)
Bi 4 : th hm s :y= x
2
+2x+3 cú nh l :
a) I(1;4) b. I(1;3) c. (1;4) d. 1 kt qu khỏc
II/ Phn t lun (6im)
Bi1: Tỡm tp xỏc nh ca hm s : y=
1
2
x 2x 1 +
Bi 2: Xột tớnh chn l ca hm s : y= x
2

2x
+3
Bi 3: Xột s bin thiờn v v th ca hm s : y=x
2
+4x+3

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