dapanthithulan2vnmath2012
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THI TH VNMATH.COM P N THI TH I HC NM 2012Mn: Ton; Khi: A, B
CHNH THC p n thang im gm 4 trang.
P N - THANG IM - BNH LUN CA VNMATH.COM
Cu p n im
I 1.(1 im)
(2,0 im) Tp xc nh: D = R. S bin thin:
Chiu bin thin: y = 4x3 8x; y(x) = 0 x = 0 hoc x = 2.
0,25
Hm s ng bin trn cc khong (,2) v (0,2); nghch bintrn khong (2, 0), (2,).
Cc tr: hm s t cc i ti x = 0; yC = 0, t cc tiu ti x = 2;yCT = 4. Gii hn: lim
xy = +; lim
x+y = +.
0,25
Bng bin thin
x
y
y
2 0 +2 + 0 + 0 0 +
++44
00
44++
0,25
th: 0,25
2.(1 im)
th (Cm) ct Ox ti 4 im phn bit x1 < x2 < x3 < x4 khi v ch khi phngtrnh t2 4t + m = 0 c 2 nghim dng 0 < t1 < t2 khi v ch khi 0 < m < 4.
0,25
Din tch phn nm trn Ox l S1 =x3x2
(x4 4x2 + m)dx. Din tch phn nm diOx l S2 =
x2x1
(x4 4x2 + m)dx x4x3
(x4 4x2 + m)dx. Theo gi thit S1 = S2hay
x4x1
(x4 4x2 + m)dx = 0.
0,25
Ni cch khc,t2
t2(x4 4x2 + m)dx = 0 vi t2 = 2 +
4m. 0,25
Rt gn ta c phng trnh 24m = 3m 4. Gii ra ta c m = 0 (loi) vm =
20
9(tho).
0,25
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II 1.(1 im)
(2,0 im) Phng trnh cho tng ng vi (sin x + cos x + 1)( sin x + cos x + 2) = 0 0,25 cos x sin x + 2 = 0 hoc cos x + sin x + 1 = 0 0,25
Phng trnh cos x sin x + 2 = 0 v nghim. 0,25cos x + sin x + 1 = 0 x =
2+ k2 hoc x = + k2, k Z. 0,25
2.(1 im)
iu kin: x [
2
2,
2
2] v
1
2 x1 x2 = 1
2|x1 x2|. 0,25
Phng trnh cho tng ng vi
1 x2 x = 2(1 x2 x)(1 x2 + x). 0,25Gii phng trnh
1 x2 x = 0 c nghim x =
2
2. 0,25
Gii phng trnh
2(
1 x2 + x) = 1 c nghim x =
264
. 0,25
III Tch ra hai tch phn I =
2
6
1sin2 x
dx +
2
6
cos3xsin2 x
dx 0,25
(1,0 im) Tnh tch phn 1 0,25
Tnh tch phn 2 bng i bin. 0,25
Vy I =
3 1. 0,25IV Gi M l trung im ca AB. Khi SM (ABCD) 0,25
(1,0 im) t AB = x = 2SM. Ta c SD2 = SM2 + M C2 + CD2 nn x =a
2
2. Th tch khi
chp S.ABCD bnga3
6
48
.
0,25
Gi G l tm tam gic ACD , qua G dng ng thng d song song vi SM. DngN sao cho SMAN l hnh bnh hnh. Khi (M CN) l mt phng trung trc caSA. Dng I thuc d sao cho IN song song vi AG. I l tm mt cu.
0,25
R = OA =7a2
24, S =
7a2
6. 0,25
V P = (bc +a
b + c)(ca +
b
c + a)(ab +
c
a + b). Ta c (ca +
b
c + a)(ab +
c
a + b) = (b +
c)2 + bc[a2 1 + bc(a+b)(a+c)
]. Ta s chng minh (bc +a
b + c)[(b + c)2 + Abc] 1
4vi
A = a2 1 + bc(a+b)(a+c)
.
0,25
(1,0 im) D thy A 0. Bt ng thc cn chng minh tng ng vi a(b + c) + Ab2c2 +
bc[aA
b + c +(b+c)2] 1
4 . Ta c1
4 a(b+c) =(b + c
a)2
4 2bc(b+ca)2 and Ab2c2 0nn ta cn chng minh bc[
aA
b + c+ (b + c)2] 2bc(b + c a)2 hay 2(b + c a)2
aA
b + c+ (b + c)2.
0,25
t t =b + c
2 1
3ta c
bc
(a + b)(a + c) t
2
(a + t)2nn ta ch cn bt ng thc
mnh hn sau y 2(2t a)2 a2t
[a2 1 + t2
(a + t)2] + 4t2. Thay a = 1 2t, bt
ng thc ny tr thnh 2(4t 1)2 1 2t2t
[(1 2t)2 1 + t2
(1 t)2 ] + 4t2. Rt gn
ta c 2(16t2
11t + 2)
t(1 2t)2(1 t)2
. Ta c 4(1
t)2
4(1
1
3)2 =
16
9> 1 v
16t2 11t + 2 t(1 2t) = 2(1 3t)2 0 nn bt ng thc cui cng hin nhinng.
0,25
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Vy gi tr ln nht ca P bng1
4. Du bng xy ra khi x = y =
1
2, z = 0 v cc
hon v tng ng..
0,25
VI.a 1.(1 im)
(2,0 im) Trng hp 1: Phng trnh AC c dng kx
y + 1
11k = 0. 0,25
cos B = cos C hay1
5.
10=
|3k + 1|k2 + 1
10
. 0,25
Gii ra k =12
(loi) hoc k =211
. 0,25
Trng hp 2: AC c dng x = k. Loi. 0,25
2.(1 im)
A(2 t, 3 + t, 4 + t), B(1+2t,2 + t, 22t). AB song song vi (P) nn nP.AB = 0.Do t + t = 2.
0,25
Kt hp vi AB = 3
6 ta c t = 1 hoc t = 2. 0,25Vi t = 1 ta c A(1,3, 4) loi v A thuc (P). 0,25
Vi t = 2 ta c phng trnh AB : x 51 = y 01 = z + 22 . 0,25VII.a 1.(1 im)
(1,0 im) Gi z = a + bi vi a, b R. V z2+z+1z2z+1 l s thc nn n c phn o bng 0. 0,25
Tnh ton a n b(a2 + b2 1) = 0. 0,25V m un ca z khc 1 nn b = 0. 0,25
Vy z l s thc. 0,25
VI.b 1.(1 im)
(2,0 im) A, B chia on ni tm I, J theo t s k = 3 hoc k = 3. 0,25
Vit cng thc tnh A, B. 0,25Tnh ra A(11,10), B(5,4). 0,25Phng trnh ng trn cn tm l x2 + y2 16x + 14y + 95 = 0. 0,252.(1 im)
Gi C(x,y, 4 2x y) thuc (P). Gi I l trung im ca AC ta cI(
x + 4
2,
y + 1
2,
3 2x y2
). V I thuc ng thng BD nn thay vo phng trnh
ca d ta suy ra c to ca C.
0,25
Gi B(2 + t, + 2t, 2 t) thuc d. V ABCD l hnh ch nht nn IA = IB vi Isuy ra t to ca C.
0,25
T suy ra phng trnh n t. Gii phng trnh ny c 2 gi tr ca t. 0,25Suy ra to ca B v D. 0,25
VII.b 1.(1 im)
(1,0 im) Gi z = a + bi vi a, b R. V z2+z+1z2z+1 l s thc nn n c phn o bng 0. 0,25
Tnh ton a n b(a2 + b2 1) = 0. 0,25V z khng l s thc nn b khc 0. 0,25
Vy m un ca z bng 1. 0,25
. . . . . . . . . . . . . . . . . . . . . . . .Ht . . . . . . . . . . . . . . . . . . . . . . . .
LU V MT S CU C NHIU CCH GII DNH CHO TH SINH D THITRN VNMATH.COM (XEM TRANG SAU):
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Nu th sinh gii ng bng cch khc p n th vn cho nguyn s im. Nu th sinh c cch gii khc v hay th ban gim kho c th xut im thng.
im thng khng qu 1 im.
p n ny lt b mt s chi tit so vi hng dn chm dnh cho gim kho
ca VNMATH trnh hin tng hc sinh ch c p n m khng c gng tgii nhng bi ton hay t ra l cc tnh ton.
MT S LI SAI THNG GP TRONG T THI TH HAI M TH SINHCN RT KINH NGHIM
KHNG loi nghim x = 2 trong cc cu II.2, cu VIa, VI.b.
H nghim ca phng trnh lng gic KHNG c k Z.
Sau khi t n ph th KHNG c vit phng trnh hay h phng trnh c tng ng
vi phng trnh hoc h phng trnh mi.MT S CU C NHIU CCH GII
CU II.b Ngoi cch gii trong p n, mt cch gii khc l t n ph. C th t hai nph u = x, v =
1 x2. Cng c th t 1 n ph (bn c t suy ngh).
Cu IV. Bn c th to ho bi ton ny vi h trc to thch hp. Vic cn li l tnhton cho chnh xc. Hy th sc.
Cu III. tnh tch phn th hai bn cnh vic i bin theo cch thng thng t = cos x tacng c th dng php i bin t =
1
sin x.
C t nht l ba cch gii cho cu VI.b. Mt cch khc p n l z2+z+1z2z+1
l s thc nn
1 +2z
z2 z + 1 l s thc hayz
z2 z + 1 l s thc. Do ngch o ca n cng l s thc
(z khc 0). Suy ra z +1
zl s thc. iu ny c ngha l z +
1
z= z +
1
z. Thu gn ta c
(z z)(|z|2 1) = 0. Kt hp vi gi thit ta suy ra z l s thc.Thm mt cch khc l gi s z2+z+1
z2z+1
= k, k l s thc. Khi z l nghim ca phng trnh(1 k)z2 + (1 + k)z + 1 k = 0. Nu k = 1 th z = 0. Nu k = 1 th ta bin lun theo . Nu khng m th phng trnh bc hai trn c nghim thc. Nu m th phng trnh bc haitrn c hai nghim phc lin hp v tch ca chng bng |z|2 = 1 (tri vi gi thit). Vy z l
s thc.
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