dap an de thi hk1 toan 12 dong thap (day kem quy nhon 1000b tran hung dao)

7/30/2019 Dap an de Thi Hk1 Toan 12 Dong Thap (day kem quy nhon 1000B tran hung dao)

1/4

SGI O DC V O TONG THP

KIM TRA CHT LNG HC K INm hc: 2012 - 2013

Mn thi: TON - Lp 12

HNG DN CHM(Hng dn chm gm c 04 trang)

Cu Mc Ni dung im

I(3)

I.1

(2)

TX: D = Ry' = 4x

3- 4x

0.25

y' = 0 4x3

- 4x = 0

21

10

21

yx

yx

yx

0.25

x

ylim , y

xlim 0.25

Bng bin thin:x - -1 0 1 +

y' - 0 + 0 - 0 +

y+ -1 +

-2 -2

0.5

Hm s nghch bin trn tng khong (-; -1), (0; 1).Hm sng bin trn tng khong (-1; 0), (1; +).Hm st cc i ti x = 0, yC = -1.Hm st cc tiu ti x = 1, yCT = -2.

0.25

th:

-2

-1

1-1 O x

y

0.5

I.2

(1)

Ta c: x4

- 2x2

- m = 0 (*) x4

- 2x2

- 1 = m -1 0.25S nghim phng trnh (*) bng sgiao im ca hai ng:

(C): y = x4

- 2x2

- 1 v d: y = m - 10.25

ycbt

11

21

m

m

0

1

m

m

0.25

0.25

II

(2)

II.1

(1)

Ta c: 2244)( aaaa 0.25

2325 . aa = 2325 a = a2. 0.25

log5125 = 3 0.25

Vy M = 432

2

a

a 0.25


2/4

II.2

(1)

Xt hm s y = f(x) = x + 3x - 9x - 7 xc nh v lin tc trn [-2; 2].f'(x) = 3x

2+ 6x - 9

0.25

f'(x) = 0 3x2

+ 6x - 9 = 0

)2;2(3

)2;2(1

x

x 0.25

f(-2) = 15, f(2) = -5, f(1) = -12 0.25Vy: 15max]2;2[

y ti x = -2, 12min

]2;2[

y ti x = 1. 0.25

III(2)

III.a

(1)

600

O

DC

B

S

Gi O l tm hnh vung ABCD.V S.ABCD l hnh chp u nn SO (ABCD), suy ra: SO l ngcao hnh chp v OA l hnh chiu ca SA trn mp(ABCD) gc gia SA v mp(ABCD) l gc SAO bng 600.

0.25

Din tch hnh vung ABCD: SABCD = a

0.25

Xt tam gic SAO vung ti O: SO = OA.tan600

= 2

6a

0.25

Th tch khi chp: VS.ABCD =3

1SABCD.SO =

2

63

a(vtt) 0.25

III.b

(1)

Hnh nn ngoi tip hnh chp S.ABCD c:

Bn knh ng trn y: r = OA =2

2a

Chiu cao: h = SO =2

6a

0.25

di ng sinh: l = SA = 2a 0.25Din tch xung quan hnh nn: Sxq = 2rl = 2a2(vdt) 0.25

Th tch hnh nn: Vn =3

1r

2h =

12

63

a(vtt) 0.25

IV.a(1)

TX: D = R\{-1}.

t y = f(x) =1

15

x

x

f'(x) =2)1(

4

x

0.25

Gi M(1; y0) l tip im. Ta c: y0 = 311

15

0.25

Tip tuyn ti M(1; 3) c: f'(1) = 1 0.25Phng trnh: y - 3 = 1(x - 1) y = x + 2 0.25


3/4

V.a(2)

V.a.1(1)

iu kin: x > 5 0.253)2(log)5(log 22 xx 3)]2)(5[(log2 xx 0.25

(x - 5)(x + 2) = 8 x2

- 3x - 18 = 0

6

3

x

x

0.25

Kt hp iu kin, phng trnh cho c nghim x = 6 0.25

V.a.2(1)

iu kin: x 0.

t t = x1

)3

1( (t > 0 ) ta c: t2 + t - 12 > 0

3

4

t

t

0.25

Kt hp iu kin t > 0 ta c: t > 3 3)3

1(

1

x 0.25

11

x

01

x

x 0.25

x

Tp nghim bt phng trnh cho l: 0.25

IV.b(1)

TX: D = R\{1}

t y = f(x) =1

222

x

xx f'(x) =

2

2

)1(

2

x

xx

0.25

Gi M(3; y0) l tip im. Ta c: y0 =13

269

=2

5 0.25

Tip tuyn ti M(3;2

5) c: f'(3) =

4

3 0.25

Phng trnh: y - 2

5

= 4

3

(x - 3) y = 4

3

x + 4

1

0.25

V.b(2)

V.b.1(1)

TX: D = R

Ta c: y' =1

)'1(

2

2

xx

xx

0.25

=1

12

)'1(1

2

2

2

xx

x

x

0.25

=1

1

1

11

22

2

xxx

x

x

0.25

1'

1 2 xy

(cmx) 0.25

V.b.2(1)

Phng trnh honh giao im ca d v (C):

1

1

x

x

= 2x + m 2x2

+ (m - 3)x - m - 1 = 0 (x 1) (*)

t f(x) = 2x2 + (m - 3)x - m - 1d ct (C) ti hai im phn bit khi v chi khi (*) c 2 nghim phn bit

khc 1.

0.25


4/4

0132)1(

0)1(8)3( 2

mmf

mm

020

06)2( 2

m

m m R.

Vy d lun ct (C) ti hai im phn bit A, B vi mi m R.

0.25

Gi xA, xB l hai nghim phng trnh (*), ta c:

A(xA; 2xA + m), B(xB; 2xB + m) AB2 = 20]16)1[(45 2 m

Du "=" xy ra khi m = -1.

0.25

Vy khi m = -1 th AB ngn nht. 0.25Ghi ch:

1. Nu th sinh lm bi khng theo cch nu trong p n nhng ng th cho sim tng phn nh hng dn quy nh.

2. Vic chi tit ha (nu c) thang im trong hng dn chm th phi m bokhng lm sai lch hng dn chm v phi c thng nht thc hin trong ton t

chm thi.

dap an de thi hk1 toan 12 dong thap (day kem quy nhon 1000b tran hung dao)

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