De Kiem Tra Chuong 4 Dai So 9 (toan6789.wordpress.com)

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KiÓm tra ®¹i sè 9 ch−¬ng IV A/khoanh trßn vµo ch÷ c¸i tr−íc kh¼ng ®Þnh ®óng

C©u 1:Hµm sè y=f(x)=(3- 2 2 )x 2 cã: A.f(-10)
B.f(-5) > f(-6)

C. f(5)
D. f(2) > f(3)

C©u 2 : Parapol y = x 2 vµ ®−êng th¼ng y = -2 Cã sè ®iÓm chung lµ : A.1

B .2

C. 3

D.0

C©u 3 : Cho ph−¬ng tr×nh : x 2 - 7 x -8 = 0 (*) th× : A . PT(*) v« nghiÖm B . PT(*) cã tæng 2nghiÖm lµ 7 , tÝch 2nghiÖm lµ -8 C . PT(*) cã 2 nghiÖm lµ -1 vµ -8 D . PT(*) cã 2 tæng nghiÖm lµ -8 vµ tÝch 2 nghiÖm lµ 7.

C©u4 : A. PT x 2 + 5 x -1 = 0 v« nghiÖm . B . PT x 2 - 4 x = 4 cã nghiÖm kÐp C. PT x 2 -x + 5 =0 cã tæng 2 nghiÖm lµ 1 , tÝch 2nghiÖm lµ 5 D . PT x 2 + 6 x = 9 cã 2 nghiÖm ph©n biÖt B. PhÇn tù luËn :

Bai Bai 1 : 1 a)VÏ ®åthÞ hµm sè y= - x 2 3 1 b) T×m to¹ ®é ®iÓm chung cña ®å thÞ hµm sè y = - x 2 víi ®å thÞ hµm sè 3 y=2x-

7 3

1 c) T×m m ®Ó pa ra pol y = - x 2 vµ ®−êng th¼ng y = x + m – 1 tiÕp xóc nhau. 3 3 Bai Bai 2 : T×m 2 sè biÕt tæng vµ tÝch cña chóng lÇn l−ît lµ vµ -1 . 2

Bai Bai 3 : Cho PT : x 2 - 2 ( m-1 ) x + m -5 = 0 ( x-lµ Èn , m lµ tham sè ) a) CMR : ph−¬ng tr×nh cã 2 nghiÖm ph©n biÖt víi moÞ m b) T×m m ®Ó PT cã 2 nghiÖm x 1,, , x 2 tho¶ mTn (3 x 1 -1 ).(3 x 2 - 1 ) = 4 HÕt

§¸p ¸n :

I.Tr¾c nghiÖm (2 ®iÓm ): C©u

1

2

3

4

®¸p ¸n

C

D

B

D

II.T− luËn : Bµi 1 (3 ®iÓm ) a. VÏ ®å thÞ hµm sè

y=-

1 2 x 3

1 b)Hoµnh ®é ®iÓm chung cña cña ®å thÞ hµm sè y = - x 2 víi ®å thÞ hµm sè 3 7 1 7 y = 2 x - lµ nghiÖm cña pt - x 2 = 2 x 3 3 3

Gi¶I pt ta ®−îc 2 nghiÖm lµ 1vµ 7 . TÝnh ®−îc tung ®é cña 2 ®iÓm chung -1/3 Vµ -49/3 VËy 2 ®iÓm chung lµ M(1, -1/3) vµ N (7;-49/3). 1 c) pa ra pol y = - x 2 vµ ®−êng th¼ng y = x + m – 1 tiÕp xóc nhau.Khi vµ chØ 3 1 2 khi pt - x = x+m -1 Cã nghiÖm kÐp 21-12m = 0 m = 7/4 3 3 Bµi 2 (2 ®iÕm): V× 2 sè cã tæng vµ tÝch cña chóng lÇn l−ît lµ vµ -1 . 2 3 x −1 = 0 2 Ta cã 2 nghiÖm lµ 2 vµ -1/2

VËy 2 sè ®ã lµ nhgiÖm cña pt : x 2 −

Bµi 3 (3 ®iÓm ) a) ∆ = m 2 - 3m + 6 3 = (m - ) 2 +15/4 > 0 ∀ m 2 VËy ph−¬ng tr×nh cã 2 nghiÖm ph©n biÖt b)theo viÐt ta cã x 1 + x 2 =2(m – 1) vµ x 1 .x 2 = m - 5 mµ (3x 1 -1)(3x 2 -1)=4 gØai hÖ ph−¬ng tr×nh ta cã m = 15 vËy víi m = 15 th× tho¶ mTn ®Çu bµi

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