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