Kiem Tra CII DS9 Co MTD An
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Transcript of Kiem Tra CII DS9 Co MTD An
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KIM TRA I S . Nm hc: 2011 2012MN : TON . LP 9
( Thi gian lm bi : 45 pht khng k thi gian pht )
H v tn :Lp :
:A. Trc nghim:( 4 im)
Khoanh trn vo ch ci ng trc cu tr li ng:1. Hm s y =
3. 3
3
m
x
m
++
l hm s bc nht khi:
A. m 3 B. m -3 C. m > 3 D. m 32. im nm trn th hm s y = -2x + 1 l:
A. (1
2;0) B. (
1
2;1) C. (2;-4) D. (-1;-1)
3. Hm s bc nht y = (k - 3)x - 6 ng bin khi:A. k 3 B. k -3 C. k > -3 D. k > 3
4. ng thng y = 3x + b i qua im (-2 ; 2) th h s b ca n bng:A. -8 B. 8 C. 4 D. -4
5. Hai ng thng y = ( k -2)x + m + 2 v y = 2x + 3 m song song vi nhau khi:
A. k = -4 v m =1
2B. k = 4 v m =
5
2C. k = 4 v m
1
2D. k = -4 v m
5
26. Hai ng thng y = - x + 2 v y = x + 2 c v tr tng i l:
A. Song song B. Ct nhau ti mt im c tung bng 2C. Trng nhau D. Ct nhau ti mt im c honh bng 2
7. Cho hm s : y = x 1 c th l ng thng (d).ng thng no sau y i qua gc ta v ct ng thng (d)?
A. y = 2x 1 B. y = x C. y = 2x D. y = x + 18. Cho hm s y = 4x + 2 .Khng nh no sau y l sai:
A. th hm s l ng thng song song vi ng thng y = 4x + 5B. Gc to bi ng thng trn vi trc Ox l gc tC. th hm s ct trc tung ti im c tung bng 2D. Hm s nghch bin trn R
B.T LUN: (6 im)Bi 1: ( 2im) Cho ng thng y = (2 k)x + k 1 (d)
a) Vi gi tr no ca k th (d) to vi trc Ox mt gc t ?b) Tm k (d) ct trc tung ti im c tung bng 5 ?
Bi 2: ( 4im) Cho hai hm s y = 2x 4 (d) v y = x + 4 (d)a) V th hai hm s trn cng mt phng ta ?b) Gi giao im ca ng thng (d) v (d)vi trc Oy l A v B , giao im ca hai ng thng l C.Xc nh ta im C v tnh din tchABC ? Tnh cc gc caABC ?
Bi lm
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MA TRN KIM TRA MN TON LP 9Nm hc : 2011 2012
Cp
Ch
Nhn bit Thng hiuVn dung
CnCp Thp Cp CaoTNKQ TL TNKQ TL TNKQ TL TNKQ TL
Hm s bc nhtv th( 4 tit )
Nhn bit c hms bc nht ; hm sng bin, nghch
bin
Bit v th cahm s bc nhty = ax + b ( a 0) .
Bit tm ta giaoim ca hai th.
Vn dng kin thc tnh c khongcch, din tch mt
hnh,S cu hiS im%
21
10%
10,5
5%
11
10%
10,5
5%
10,5
5%
11
10%
64
45
ng thngsong song v
ng thng ctnhau
( 2 tit )
Nhn bit c v trtng i ca haing thng l th ca hm s bcnht.
Cn c vo cc hs xc nh c vtr tng i ca haing thng l th ca hm s bcnht.
Xc nh cc dngng thng linquan n ngthng ct nhau, songsong.
S cu hiS im
%
10,5
5%
10.5
5%
11
10%
3
20H s gc cang thng
( 3 tit )
Hiu c h s gcca ng thngy = ax + b ( a 0)
Xc nh c h sgc ca ngthng.
Vit c phngtrnh ng thng.
S cu hiS im%
10,5
5%
10,5
5%
11,5
10%
11
10%
43
35
Tng s cuTng s im%
42,5
25%
32
20%
43,5
35%
22
20%
13
100
P N BIU IM
TRC NGHIM( 4im)Mi cu chn ng c 0,5 im1 2 3 4 5 6 7 8D A D B C B C A
T LUN: ( 6im)Cu 1:
( 2im)a) ng thng (d) to vi trc Ox mt gc t th a < 0
Tc l : 2 k < 0 k > 2b) ng thng (d) ct trc tung ti im c tung bng 5 th b = 5
Tc l : k 1 = 5 k = 6
0.50.50.5
0.5Cu 2:
( 4im)a) Xc nh ng cc im thuc thV ng th 2 hm s
0.51
4
2
- 2
-
y
- 5 x
g x( ) = - x +
f x( ) = 2x -4
O
Q
N
M
H
42
^
>
2
4
y
x
=
4
y
x=
KE
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b) V Q l giao im ca hai ng thng (d ) v ( d) nn ta c phngtrnh honh giao im: 2x - 4 = - x + 4
3x = 8 x =8
3
y =- x + 4 = -8
3+ 4 =
4
3
Vy C(8
3;4
3)
SABC =12
AB. CH = 12
.8 . 83
= 323
c) p dng t s lng gic vo tam gic vung AOE ta c:
tanA =OE
OA=
1
2 A 26
034
Tam gic vung BOK ta c: OB = OK = 4nn l tam gic vung cn A =45
0
Tam gic ABC c A B C+ + = 1800
Suy ra C = 1800 (26034 + 450) = 108026
0.25
0.25
0.25
0.25
0.5
0.5
0.25
0.25
