De Thi Thu Cua Math.vn 1-9 Nam 2011

DIN N MATH.VN

http://math.vn thi s: 01 Cu I. (2 im)

THI TH I HC 2011 Mn thi: Ton Thi gian lm bi: 180 pht

2x + 3 (C) x+1 1) Kho st s bin thin v v th (C) ca hm s. 2) Lp phng trnh tip tuyn ca th (C) ti nhng im thuc th c khong cch n ng thng 3x + 4y 2 = 0 bng 2. Cho hm s y = Cu II. (2 im) 1) Gii phng trnh: 2) Gii phng trnh: Cu III. (1 im) Tnh gii hnx0

( ) 2 cos 2x + + 3 tan x = 1 + 3 tan x sin2 x. 3 3x3 6x2 3x 17 = 3 3 9(3x2 + 21x + 5) 3 2 cos 2x + 1 2esin x ln(1 + x2 )

lim

Cu IV. (1 im) Cho hnh chp S.ABCD c y l hnh thang vung ti A, v D, AB = AD = a,CD = 2a. Cnh bn SD vung gc vi mt phng ABCD v SD = a. Gi E l trung im ca CD. Xc nh tm v tnh bn knh mt cu ngoi tip hnh chp S.BCE. Cu V. (1 im) Cho tam gic ABC c ba cnh a, b, c tha mn iu kin 1 1 1 + 2 + 2 =2 2 +1 b +1 c +1 a 3 SABC Chng minh rng . 8 Cu VI. (2 im) 1)Trong mt phng vi h ta vung gc Oxy cho ba im I(1; 1), J(2; 2), K(2; 2). Tm ta cc nh ca hnh vung ABCD sao cho I l tm hnh vung, J thuc cnh AB, v K thuc cnh CD. 2) Trong khng gian vi h ta vung gc Oxyz cho ba im A(2; 3; 1), B(1; 2; 0),C(1; 1; 2). Tm ta trc tm H v tm ng trn ngoi tip I ca tam gic ABC. Cu VII. (1 im) Gii h phng trnh { 2 A3 54Cx + x = 29 x 2 log(x6) y = y log(3x64) 2

.

1

DIN N MATH.VN

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LI GII THI TH I HC 2011 Mn thi : Ton s: 01

x = 1 l phng trnh tim cn dc lim y = 2, lim y = 2x x+

x1

Cu I. 1) (1 im) 2x + 3 Cho hm s y = (C). Kho st s bin thin v v th (C) ca hm s . x+1 Li gii: 2x + 3 Hm s y = c tp xc nh D = R\{1}. 3 x+1 th ct Ox ti ;0 1 2 o hm y = (x + 1)2 v ct Oy ti (0; 3) y < 0, x D Hm s nghch bin trn (; 1); (1; +) lim y = ; lim y = + 4x1+

3 2 1

y = 2 l phng trnh tim cn ngang Bng bin thin x f (x) f (x) 2 1 + 2 +2 1

0

Cu I. 2) (1 im) Lp phng trnh tip tuyn ca th (C) ti nhng im thuc th c khong cch n ng thng 3x + 4y 2 = 0 bng 2. Li gii: Cch 1 lonely_abba 1 M a; 2 + (C) (a = 1) a+1 1 3a + 4 2 + 2 3a2 + 9a + 10 a+1 d(M; d) = = = 2 3a2 + 9a + 10 = 10|a + 1| () 5 5|a + 1| 1 1 11 Khi a > 1 ta c: () 3a2 a = 0 a = 0; a = M1 (0; 3); M2 ; 3 3 4 4 7 4 ; M4 ; 1 Khi a < 1 ta c: () 3a2 + 19a + 20 = 0 a = 5; a = M3 5; 3 4 3 1 Tip tuyn ca (C) ti M(xo ; yo ) c dng: y = f (xo )(x xo ) + yo f (x) = (x + 1)2 9 47 PT tip tuyn ti: M1 : l (d1 ) : y = x + 3 M2 : l (d2 ) : y = x + 16 16 1 23 M3 : l (d3 ) : y = x + M4 : l (d4 ) : y = 9x 13 16 16 Cch 2 toannh Phng trnh tip tuyn (d) ti M(xo ; yo ) thuc (C) v c khong cch ti ng thng (d ) : 3x + 4y 2 = 0 2xo + 3 1 bng 2, c dng: y yo = f (xo )(x xo ) y = (x xo ) xo + 1 (xo + 1)2 Do khong cch t M ti ng thng (d ) bng 2, |3xo + 4yo 2| 3xo + 4yo 12 = 0 nn ta c: = 2 |3xo + 4yo 2| = 10 2 + 42 3xo + 4yo + 8 = 0 3

1

Trng hp 1: Vi 3xo + 4yo 12 = 0 3xo + 4

2xo + 3 2 12 = 0 3xo xo = 0 xo + 1

Phng trnh tip tuyn (d1 ) ti M1 (0; 3) l: y = x + 3 1 11 9 47 Phng trnh tip tuyn (d2 ) ti M2 ; l: y = x + 3 4 16 16

xo = 0 1 xo = 3

2xo + 3 2 + 8 = 0 3xo + 19xo + 20 = 0 Trng hp 2: Vi 3xo + 4yo + 8 = 0 3xo + 4 xo + 1

xo =

4 Phng trnh tip tuyn (d3 ) ti im M3 ; 1 l: y = 9x 13 3 7 1 23 l: y = x + . Phng trnh tip tuyn (d4 ) ti im M4 5; 4 16 16 Cu II. 1) (1 im) Gii phng trnh: 2 cos 2x + + 3 tan x = 1 + 3 tan x sin2 x. 3 Li gii: lonely_abba K: cos x = 0 3 PT cos 2x 3 sin 2x + 3 tan x cos2 x = 1 2 sin2 x + 3 sin 2x = 0 2 x = k sin x = 0 sin x = 0 3 3 3 (k Z) (nhn) tan x = x = arctan 3 cos x 3 3 + k sin x = 2 2 2 Vy pt c 2 h nghim trn Cu II. 2) (1 im) Gii phng trnh: 3x3 6x2 3x 17 = 3 3 9(3x2 + 21x + 5) Li gii: 17 5 Phng trnh x3 2x2 x = 3 3 x2 + 7x + 3 3 5 5 5 (x 1)3 + 3(x 1) = x2 + 7x + + 3 3 x2 + 7x + f (x 1) = f 3 x2 + 7x + 3 3 3 Vi f (t) = t 3 + 3t c f (t) = 3t 2 + 3 > 0, t R , suy ra hm f (t) tng trn R. 5 Nn phng trnh x 1 = 3 x2 + 7x + 3x3 6x2 12x 8 = 0 (x + 2)3 = 4x3 x + 2 = x 3 4 3 2 Vy phng trnh c mt nghim l x = 3 41 Cu III. (1 im) 3 2 cos 2x + 1 2esin x Tnh gii hn lim x0 ln(1 + x2 ) Li gii: Mercury 3 3 2 2 cos 2x + 1 2esin x x2 cos 2x + 1 2esin x lim = lim x0 x0 ln(1 + x2 ) ln(1 + x2 ) x2 3 2 cos 2x 1 + 1 2esin x + 1 x2 = lim lim x0 ln(1 + x2 ) x0 x2 3 2 cos 2x 1 + 1 2esin x + 1 = lim x0 x2 3 2 cos 2x 1 1 2esin x + 1 = lim + lim x0 x0 x2 x2

4 3 xo = 5

2

cos 2x 1 + lim = lim x0 x2 ( cos 2x + 1) x0

2(1 esin x ) sin2 x x2 sin2 x2x 3

2

(1 2esin x )2 + 1 3 2

2

3

1 2esin 1

2x

(1 2esin x )2 + 1 1 2esin x Cu IV. (1 im) Cho hnh chp S.ABCD c y l hnh thang vung ti A, v D, AB = AD = a,CD = 2a. Cnh bn SD vung gc vi mt phng ABCD v SD = a. Gi E l trung im ca CD. Xc nh tm v tnh bn knh mt cu ngoi tip hnh chp S.BCE. Li gii: D thy BE DC. EBC vung ti E nn trc ca ng trn ngoi tip tam gic l ng thng d (BCE) ti trung im I ca BC. Gi O d l tm mt cu ngoi tip S.BEC, bn knh mt l R cu Ta c: ABD vung ti A = a2 + = a 2 DB a2 EBC vung ti E BC = a2 + a2 = a 2 S O R BCD c BD2 + BC2 = DC2 BCD vung ti B 5 BDI vung ti B: DI = BD2 + BI 2 = a 2 7 SI = SD2 + DI 2 = a a R 2 SD 2 E a a D cos SIO = sin SID = = SI 7 C Trong SIO: SO2 = OI 2 + SI 2 2OI SI cos SIO 7 7 2 a R2 = OI 2 + a2 2IO a I 2 2 7 7 R2 = OI 2 + a2 2IO a 2 2 a B A 2 = OB2 = OI 2 + BI 2 = OI 2 + a Trong BIO: R 2 7 a2 3 a2 2IO a = IO = a 2 2 2 Vy tm O c xc nh a2 9 2 a2 a 11 Bn knh R = OI 2 + = a + = 2 4 2 2 Cu V. (1 im) Cho tam gic ABC c ba cnh a, b, c tha mn iu kin 1 1 1 + 2 + 2 =2 2+1 a b +1 c +1 3 Chng minh rng SABC . 8 Li gii: Cch 1 buon_qua 1 T gi thit: 2 = 2 1 = a2 b2 + 22 b2 c2 . a +1 1 1 1 Theo cng thc tnh din tch tam gic ta c: S = ab sinC = bc sin A = ca sin B. 2 2 2 1 Suy ra: a2 b2 = 4S2 2 sin A 1 9 p dng bt ng thc Cauchy Schwarz ta c: 2 sin A sin2 A 3

1 esin sin2 x = lim 2 + 2 lim x0 sin2 x x0 x

lim

sin2 x lim x0 x2 x0

3

2

=

1 3

M vi mi tam gic ABC th: sin2 A Tng t ta c: 2a2 b2 c2 =

sin A + sin B + sinC 3 p dng bt ng thc AM GM ta c: sin A sin B sinC 3 3 3 8 16S3 Vi mi tam gic ABC ta cng c: sin A . Suy ra: 2a2 b2 c2 2 3 3 8 16S3 Kt hp cc iu trn ta c: 1 = a2 b2 + 22 b2 c2 16S2 + 3 3 3 Tng ng vi: (8S 3)(16S2 + 8 3S + 3) 0 S . 8 3 1 Vy SABC . Du ng thc xy ra khi v ch khi a = b = c = 8 2 Cch 2 can_hang2007 a2 b2 c2 T gi thit, ta suy ra 1 = 2 + 2 + 2 . a +1 b +1 c +1 S dng bt ng thc Cauchy-Schwarz, ta c a2 b2 c2 (a + b + c)2 + 2 + 2 2 . a2 + 1 b + 1 c + 1 a + b2 + c2 + 3 3 Kt hp vi trn, ta c a2 + b2 + c2 + 3 (a + b + c)2 , tc ab + bc + ca . 2 By gi, s dng cng thc Herong, ta c 1 SABC = (a + b + c)(b + c a)(c + a b)(a + b c). 4 Do bt ng thc cn chng minh tng ng vi 3 3(a + b + c)(b + c a)(c + a b)(a + b c) . 2 V chng minh kt qu ny, ta s chng minh kt qu mnh hn l 3(a + b + c)(b + c a)(c + a b)(a + b c) ab + bc + ca. Khng mt tnh tng qut, gi s b l s nm gia a v c. S dng bt ng thc AM-GM, ta c (a + b + c)(c + a b) + 3(b + c a)(a + b c) 3(a + b + c)(b + c a)(c + a b)(a + b c) . 2 Vy ta cn chng minh (a + b + c)(c + a b) + 3(b + c a)(a + b c) 2(ab + bc + ca). Bt ng thc ny tng ng vi 2(a c)2 + 2(b a)(b c) 0 (hin nhin ng). Cch 3 ltq2408 1 1 a2 b2 c2 1 T + 2 + 2 =2 + + =1 a2 + 1 b + 1 c + 1 1 + a2 1 + b2 1 + c2 (a + b + c)2 3 1 2 1 ab + bc + ca a2 b2 c2 2 + c2 + 3 a +b 8 2 3 3 Li c: Vi mi tam gic ABC th: sin A sin B sinC 8 1 2 2 2 3 3 = a b c sin A sin B sinC 1 1 3 3 S Nn: S 8 8 8 8 8 Cch 4 CSS 3 Chng ta bit ab + bc + ca . 2 Vy s dng cng thc Heron, bt ng thc Schur v AM-GM ta c 4SABC = (a + b + c)(a + b c)(b + c a)(c + a b) abc(a + b + c)

16S3 sin A sin B sinC

9 Suy ra: a2 b2 16S2 . 4

ab + bc + ca 3 1 3 = . 2 2 3 3 4

T y ta suy ra ngay iu cn chng minh. Cu VI. 1) (1 im) Trong mt phng vi h ta vung gc Oxy cho ba im I(1; 1), J(2; 2), K(2; 2). Tm ta cc nh ca hnh vung ABCD sao cho I l tm hnh vung, J thuc cnh AB, v K thuc cnh CD. Li gii: lonely_abba a Nhn xt: I(1; 1) l tm hnh vung ABCD cnh a d(I; AB) = d(I;CD) = 2 TH1: AB//CD//Oy D thy d(I; AB) = 2 = d(I;CD) = 1 (loi) TH2: AB c h s gc k AB : y = k(x + 2) + 2 CD//AB v i qua K nn CD : y = k(x 2) 2 4|k + 1| Ta c: a = AC = d(J;CD) = k2 + 1 |3k + 1| 2|k + 1| a 3 d(I; AB) = = = 4(k + 1)2 = (3k + 1)2 5k2 2k 3 = 0 k = 1; k = 2+1 2+1 2 5 k k D thy AB, CD phi c h s gc k > 0 k = 1 Vy AB : y = x + 4; CD : y = x 4 ng thng d qua I v vung gc vi AB c pt: (d) : y = (x 1) + 1 = x + 2 y = x+4 x = 1 Trung im M ca AB l nghim h y = x + 2 y=3 Trung im N ca CD l nghim h Ta c: A(x; x + 4) AB, a = 4 2 AM 2 = (x + 1)2 + (x + 1)2 y = x4 y = x + 2 x=3 y = 1

a2 = =8 4 |x + 1| = 2 x = 1; x = 3 nh A, B : (1; 5); (3; 1) Ta c: C(x; x 4) CD a2 =8 AM 2 = (x 3)2 + (x 3)2 = 2(x + 1)2 = 4 |x 3| = 2 x = 5; x = 1 nh C, D : (5; 1); (1; 3) Vy to 4 nh: (1; 5); (3; 1); (5; 1); (1; 3) Cu VI. 2) (1 im) Trong khng gian vi h ta vung gc Oxyz cho ba im A(2; 3; 1), B(1; 2; 0),C(1; 1; 2). Tm ta trc tm H v tm ng trn ngoi tip I ca tam gic ABC. Li gii: H l trc tm tam gic ABC khi v ch khi BH AC,CH AB, H mp(ABC). BH AC = 0 (x + 1) + 2(y 2) + 3z = 0 CH AB = 0 3(x 1) + (y 1) + (z + 2) = 0 AH AB, AC = 0 (x 2) 8(y 3) + 5(z 1) = 0 x = 2 x + 2y + 3z = 3 15 29 3x + y + z = 2 y= 15 1 x 8y + 5z = 17 z = 3 2 29 1 Vy H ; ; 15 15 3 I l ng trn ngoi tip tam gic ABC khi v ch khi AI = BI = CI, I mp(ABC). tm AI 2 = BI 2 (x 2)2 + (y 3)2 + (z 1)2 = (x + 1)2 + (y 2)2 + z2 CI 2 = BI 2 (x 1)2 + (y 1)2 + (z + 2)2 = (x + 1)2 + (y 2)2 + z2 AB, AC = 0 AI (x 2) 8(y 3) + 5(z 1) = 0 = 2(x + 1)2

5

Vy I

6x + 2y + 2z = 9 4x 2y 4z = 1 x 8y + 5z = 17

14 61 1 ; ; 15 30 3 Cu VII. (1 im) 2 A3 54Cx + x = 29 x Gii h phng trnh . 2 log(x6) y = y log(3x64) 2 Li gii: trantrongtai1234 64 K: x N ; x > ; y > 0 3 2 T phng trnh A3 54Cx + x = 29 ta c ngay: x = 29 x th xung phng trnh 2 log(x6) y = y log(3x64) 2 ta c: y2 = 2y y do y > 0 , ta ly cn hai v, ri t = t > 0 2t = 2t 2t 2t = 0 2 f (t) = 2t ln 2 2, f (t) = 2t (ln 2)2 > 0, f (t) = 0 c 1 nghim duy nht f (t) c ti a 2 nghim (lp bng bin thin ) nn: t = 1v t = 2 l 2 nghim y = 2 v y = 4 Vy h PT c 2 nghim l (29; 2), (29; 4)

x = 14 15 61 y= 30 1 z = 3

6

DI N N MATH.VN

http://math.vn thi s : 02

THI TH I H C 2011 Mn thi: Ton Th i gian lm bi: 180 pht

Cu I. (2 i m) Cho hm s y = x3 3mx + 2, v i m l tham s 1) Kh o st s bi n thin v v th (C) c a hm s v i m = 1. 2) Tm cc gi tr c a m th hm s c hai i m c c tr A, B sao cho b ng 18, trong I(1; 1). Cu II. (1 i m) Gi i phng trnh Cu III. (1 i m) Gi i h phng trnh sau trn R: Cu IV. (1 i m)2

IAB c di n tch

x 3x 2 2 sin cos cos x = 2 sin 2x 3. 8 2 8 2 3x = 8y2 + 1 3y = 8x2 + 1.

Tnh tch phn

I=1

x + ln x dx. (1 + x)2

Cu V. (1 i m) Cho hnh chp S.ABCD c y ABCD l hnh thoi v AB = BD = a, SA = a 3, SA (ABCD). 2 G i M l i m trn c nh SB sao cho BM = SB, gi s N l i m di ng trn c nh AD. Tm 3 v tr c a i m N BN DM v khi tnh th tch c a kh i t di n BDMN. Cu VI. (1 i m) Cho a, b, c l di ba c nh c a tam gic nh n ABC. Ch ng minh r ng b3 c3 a3 + + 12pR2 , cos A cos B cosC trong p l n a chu vi v R l bn knh ng trn ngo i ti p ABC. Cu VII. (1 i m) Trong m t ph ng v i h t a Oxy cho tam gic ABC c phng trnh ng cao AH : 3x + 2y 1 = 0, phn gic trong CK : 2x y + 5 = 0 v trung i m M(2; 1) c a c nh AC. Tnh chu vi v di n tch c a c a tam gic ABC. Cu VIII. (1 i m) Trong khng gian v i h t a Oxyz cho m t c u (S) tm I(1; 2; 1); bn knh R = 4 v ng x y1 z+1 = . L p phng trnh m t ph ng (P) ch a (d) v c t m t c u (S) th ng (d) : = 2 2 1 theo m t ng trn c di n tch nh nh t. Cu IX. (1 i m) Cho t p A = {1, 2, 3, . . . , 2011} v n A, n 1006. G i B l t p con c a A c n ph n t v B ch a ba s t nhin lin ti p. H i c bao nhiu t p B nh v y ?

1

DI N N MATH.VN

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L I GI I THI TH I H C 2011 Mn thi : Ton s : 02

Cu I. 1) (1 i m) Cho hm s y = x3 3mx + 2, v i m l tham s . Kh o st s bi n thin v v th (C) c a hm s v i m = 1. L i gi i: m = 1 Hm s l y = x3 3x + 2 c TX l D = R th y = 3x2 3 = 3(x2 1) Giao v i tr c tung: (0; 2) x = 1 y = 4 Giao v i tr c honh: (2; 0), (1; 0) y =0 x=1 y=0 Gi i h n t i v c c lim y = +; lim y = . 4x+ x

B ng bi n thin x y + 1 0 4 1 0 + +2

+ y 0 hm s ng bi n trn (; 1) ; (1; +) hm s ngh ch bi n trn (1; 1). i m c c i (1; 4), i m c c ti u (1; 0).

2

1

0

1

2

Cu I. 2) (1 i m) Tm cc gi tr c a m th hm s c hai i m c c tr A, B sao cho IAB c di n tch b ng 18, trong I(1; 1). L i gi i: trantrongtai1234 Ta c: y = 3x2 3m = 3(x2 m). s c C v CT m > 0 hm g i A, B l 2 c c tr th: A( m; 2 + 2m m); B( m; 2 2m m) 4m m Phng trnh ng th ng qua AB l: y (2 + 2m m) = (x + m) y = 2 2mx 2 m |2m 1| Kho ng cch t I n t AB l d(I; AB) = , di o n AB = 4m + 16m3 4m2 + 1 1 |2m 1| m di n tch tam gic IAB l S = 18 4m + 16m3 = 18 2 4m2 + 1 3 )(2m 1)2 = (4m2 + 1)4 18 m(2m 1)2 = 18 (4m + 16m 4m3 4m2 + m 18 = 0 (m 2)(4m2 + 4m + 9) = 0 m = 2 Cu II.) (1 i m) x 3x Gi i phng trnh 2 2 sin cos cos x = 2 sin 2x 3. 8 2 8 2 L i gi i: Cch 1: lonely_abba PT 2 sin 2x + 2 sin x 2 2 cos x = 2 sin 2x 3 4 cos 2x sin 2x + 2 sin x 2 2 cos x = 2 sin 2x 3 cos 2x 3 sin 2x + 2 sin x 2 cos x + 3 = 0 4 t t = x a v c: sin 2t + 3 cos 2t 3 sint + cost 3 = 0 4 6 sin2 t + (2 cost 3) sint + cost = 0 () 2 + 24 cost = (2 cost + 3)2 sin x = (2 cost 3) sint = 2 1 () sint = cost 3 Cch 2: ltq2408 1

Phng trnh tng ng v i:

2x + 2 sin x 2 2 cos x = 2 sin 2x 3 2 sin 4 (cos x sin x) (cos x + sin x) + 3 (1 sin + 2 (sin x 2 cos x) = 0 2x) 2 (cos x sin x) (2 cos x sin x) 2 (2 cos x sin x) = 0 (2 cos x sin x) 2 cos x 2 sin x 2 = 0

Cu III.) (1 i m) 3x = 8y2 + 1 Gi i h phng trnh sau trn R: 3y = 8x2 + 1. L i gi i: Cch 1: canhochoi u tin ta xt hm f (t) = 3t + 8t 2 + 1 ch ng t r ng x = y. Xt g(x) = 9x 8x2 1, g (x) = 9x ln 9 16x, g (x) = 9x ln 92 16, 16 = Ta c g () 1, 44. g (x) = 0 x = log9 ln 92 T BBT ta d suy ra c 2 s z1 , z2 th a pt g (x) = 0. T d dng l p BBT c a hm s g(x). 1 T BBT c a hm s g(x), suy ra pt g(x)=0 c t i a ba nghi m, m ta l i c g(1) = g(0) = g = 0, 2 1 nn 1, 0, l ba nghi m c a PT. 2 Cch 2: vokhachuyy Ta c: 8y2 + 1 1 (3x 1) x 0. 8x2 + 1 1 (3y 1) y 0. 8t 0,t 0 f (t) ng bi n. Xt f (t) = 8t 2 + 1 f (t) = 8t 2 + 1 gi s x y 3x 3y 8y2 + 1 8x2 + 1 y x suy ra x = y Xt g(x) = 9x 8x2 1, g (x) = 9x ln 9 16x, g (x) = 9x ln 92 16, 16 g (x) = 0 x = log9 = Ta c g () 1, 44. ln 92 T BBT ta d suy ra c 2 s z1 , z2 th a pt g (x) = 0. T d dng l p BBT c a hm s g(x). 1 = 0, T BBT c a hm s g(x), suy ra pt g(x)=0 c t i a ba nghi m, m ta l i c g(1) = g(0) = g 2 1 nn 1, 0, l ba nghi m c a PT. 2 Cu IV. (1 i m) 2 x + ln x Tnh tch phn I = dx. 2 1 (1 + x) L i gi i: Cch 1: canhochoi 2 x + ln x 2 1 2 2 ln x 1 dx = dx + dx dx I= 2 2 2 (1 1 + x) 1 1+x 1 (1 + x) 1 (1 + x) 2 3 2 1 dx = ln |x + 1| = ln 1 1+x 2 1 2 2 2 ln x ln x 2 3 ln 2 2 ln 2 1 dx = + ln |x| ln |x + 1| = + ln 2 ln Ta c: . V yI= 1+x 1 3 2 1 1 1 (1 + x)2 3 6 2 1 1 2 1 dx = = 2 x+1 1 6 1 (1 + x) Cch 2: trantrongtai1234 x+1 1 1 t u = x + ln x du = dx. dv = dx v = 2 x (x + 1) (x + 1) 21 2 (x + ln x) 2 (x + ln x) 2 2 ln 2 1 Ta c: I = + dx = + ln |x| . VyI= 1+x 1+x 3 6 1 1 1 1 x Cu V. (1 i m) Cho hnh chp S.ABCD c y ABCD l hnh thoi v AB = BD = a, SA = a 3, SA (ABCD). G i M l 2

2 i m trn c nh SB sao cho BM = SB, gi s N l i m di ng trn c nh AD. Tm v tr c a i m N 3 BN DM v khi tnh th tch c a kh i t di n BDMN. L i gi i: lonely_abba a2 3 Ta c: ABD l tam gic u SBDA = V ME SA (E AB); BH AD (H AD) 4 ME MB DN 4 a2 3 SBDN = SAB : SA ME = = SBDN = SA SB SBDA DA 5 5 MB 2a 3 Th tch kh i t di n MNDB l: ME = SA = SB 3 1 1 2a 3 a2 3 2a3 Ta c: ME SA; SA AB ME AB V = ME.SBAN = = 3 3 3 5 15 DE l hnh chi u vung gc c a DM trn (ABCD) S Theo nh l 3 ng vung gc, BN DM BN DE MB 2a BE MB = BE = AB = Ta c: BA SB SB 3 2 Ta c: AE = BA BD 3 M a x x BD BN = BA a a a2 BABD = 2 D BN DE AE BN = 0 2a 2 2(a x) x ax 2 H BA + BABD BD = 0 3a 3a a a N 2a.a2 2(a x) x a2 a2 (a x) C A + =0 3a 3a a 2 a E 5 2 4a ax a2 = 0 x = 6 3 5 B Cu VI. (1 i m) Cho a, b, c l di ba c nh c a tam gic nh n ABC. Ch ng minh r ng b3 c3 a3 + + 12pR2 , cos A cos B cosC trong p l n a chu vi v R l bn knh ng trn ngo i ti p ABC. L i gi i: CSS S d ng cc cng th c quen thu c c2 + a2 b2 a2 + b2 c2 b2 + c2 a2 , cos B = , cosC = , cos A = 2bc 2ca 2ab a2 b2 c2 2p = a + b + c, R2 = , 2(a2 b2 + b2 c2 + c2 a2 ) (a4 + b4 + c4 ) ta c th vi t l i b t ng th c c n ch ng minh d i d ng a2 b2 c2 3abc(a + b + c) + 2 + 2 . b2 + c2 a2 c + a2 b2 a + b2 c2 2 a2 b2 a4 Do tam gic ABC nh n nn min{b2 + c2 , c2 + a2 , a2 + b2 } > max{a2 , b2 , c2 }, i u ny cho php ta s d ng b t ng th c Cauchy-Schwarz nh sau a2 (a2 + b2 + c2 )2 (a2 + b2 + c2 )2 . = 2 b + c2 a2 a2 (b2 + c2 a2 ) 2 a2 b2 a4 V y ta ch c n ch ng minh (a2 + b2 + c2 )2 3abc(a + b + c), hi n nhin ng v (a2 + b2 + c2 )2 (ab + bc + ca)2 3abc(a + b + c) theo AM-GM. Cu VII.) (1 i m) Trong m t ph ng v i h t a Oxy cho tam gic ABC c phng trnh ng cao AH : 3x + 2y 1 = 0, phn gic trong CK : 2x y + 5 = 0 v trung i m M(2; 1) c a c nh AC. Tnh chu vi v di n tch c a c a tam gic ABC.

3

L i gi i: Cu VIII.) (1 i m) Trong khng gian v i h t a Oxyz cho m t c u (S) tm I(1; 2; 1); bn knh R = 4 v ng th ng x y1 z+1 = . L p phng trnh m t ph ng (P) ch a (d) v c t m t c u (S) theo m t ng trn (d) : = 2 2 1 c di n tch nh nh t. L i gi i: trantrongtai1234 G i H hnh chi u c a I ln ng th ng (d): G i (Q) l m t ph ng qua I vung gc v i (d) pt mp (Q) : 2x 2y z = 5 4 1 5 ; ) nh v y ta c H thu c (d) v (Q) H( ; 3 3 3 ta c IH = 10 < 4 = R (d) c t m t c u G i (G) l m t ph ng i qua H v vung gc v i IH L p phng trnh m t ph ng c t m t theo m t ng trn c di n tch nh nh t. Tng ng kho ng cch t tm n mp l l n nh t. Bi ton quay tr v gi ng (A-2008) PT mp(P) i qua H nh n IH l vct php tuy n: x + 5y 8z = 13 Th t v y g i (X) l mp b t k ch a (d) v A l hnh chi u c a A ln (X) AA nh hn ho c b ng AH. T c AH l kho ng cch l n nh t t A n mp b t k ch a (d). V y PT: x + 5y 8z = 13 chnh l pt mp(P) c n tm. Cu IX. (1 i m) Cho t p A = {1, 2, 3, . . . , 2011} v n A, n 1006. G i B l t p con c a A c n ph n t v B ch a ba s t nhin lin ti p. H i c bao nhiu t p B nh v y ? L i gi i:

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Cu I. (2 i m) Cho hm s y = x4 2mx2 + 2 (Cm) 1) Kh o st s bi n thin v v th c a hm s khi m = 1 2) Tm t t c cc gi tr c a tham s m th (Cm) c ba i m c c tr t o thnh m t tam gic 3 9 ; c ng trn ngo i ti p i qua i m D . 5 5 Cu II. (2 i m) 1) Gi i phng trnh : 2) Gi i h phng trnh: Cu III. (1 i m) Tnh tch phn I=01 2

sin x =

16cos6 x + 2cos4 x . 54 51cos2 x x2 + 2y2 3x + 2xy = 0 xy(x + y) + (x 1)2 = 3y(1 y)

.

ln(1 x) dx. 2x2 2x + 1

Cu IV. (1 i m) Cho hnh chp S.ABCD c y ABCD l hnh vung c nh a. Hnh chi u c a S trng v i tr ng tm tam gic ABD. M t bn (SAB) t o v i y m t gc 60o . Tnh theo a th tch c a kh i chp S.ABCD. Cu V. (1 i m) Cho s th c a, b, c [0; 1]. Tm gi tr l n nh t, gi tr nh nh t c a bi u th c P = a5 b5 c5 (3(ab + bc + ca) 8abc). Cu VI. (2 i m) 1) Trong m t ph ng Oxy cho i m A(1; 4) v hai ng trn (C1 ) : (x 2)2 + (y 5)2 = 13, (C2 ) : (x 1)2 + (y 2)2 = 25. Tm trn hai ng trn (C1 ), (C2 ) hai i m M, N sao cho tam gic MAN vung cn t i A. 2) Trong khng gian v i h t a Oxyz, cho M (1; 2; 3). L p phng trnh m t ph ng i qua M c t ba tia Ox t i A, Oy t i B, Oz t i C sao cho th tch t di n OABC nh nh t. Cu VII. (1 i m) Gi i b t phng trnh 4x 2x+2 x2 2x 3

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htt p:/ /m ath .vnBng bin thinx x+ 1

Cu I. 1) (1 im) Cho hm s y = x4 2mx2 + 2 (Cm). Kho st s bin thin v v th ca hm s khi m = 1 Li gii: th m = 1 hm s y = x4 2x2 + 2 c TX l D = R y = 4x3 4x = 4x x2 1 4 x=0 y=2 Nn y = 0 x = 1 y = 1 3 x=1 y=1 y > 0 1 < x < 0 hoc 1 < x < + hm s ng bin trn (1; 0) ; (1; +) 2 y < 0 < x < 1 hoc 0 < x < 1 hm s nghch bin trn (; 1) ; (0; 1). Gii hn lim y = +; lim y = +.

im cc i (0; 2), im cc tiu (1; 1) ; (1; 1). th giao vi trc tung: x = 0 y = 2. th ct trc tung ti im B(0; 2)

2

1

1

2

Cu I. 2) (1 im) Tm tt c cc gi tr ca tham s m th (Cm) c ba im cc tr to thnh mt tam gic c ng trn 3 9 ngoi tip i qua im D ; . 5 5 Li gii: Cch 1. Ta c y = 4x3 4mx = 4x(x2 m) . Hm s c cc i v cc tiu khi v ch khi y c 3 nghim phn bit v i du khi x qua 3 nghim y c 3 nghim phn bit t(x) = x2 m c 2 nghim phn bit khc 0 m > 0 (1) y = x4 2mx2 + 2 (2) Vi K hm s c cc im cc tr vi to l nghim HPT x3 mx = 0 (3) 3 mx) mx2 + 2 = mx2 + 2 (4) (do (3)) Ta c (2) y = x(x 2y x2 = (5) m T (4) c y2 = m2 x4 4mx2 + 4 = m2 x(x3 mx) + m(m2 4)x2 + 4 = m(m2 4)x2 + 4(do (3)) Hay y2 = (m2 4)(2 y) + 4 (6) (do (5)) 1 T (5)&(6) ta thu c x2 + y2 = m2 + 4 (2 y) + 4(7) m Nh vy theo suy lun trn th to cc im cc tr cng tho mn PT (7) , m (7) l PT ca ng trn . 1 Do ng trn (T ) qua cc im cc tr ca th hm s c PT x2 + y2 = m2 + 4 (2 y) + 4 m 9 81 1 1 3 9 + = m2 + 4 +4 By gi (T ) qua D( ; ) 5 5 25 25 m 5 3 2m + 1 = 0 (m 1)(m2 + m 1) = 0 m = 1; m = 1 5 m 2 1 + 5 Kt hp K m > 0 ta thu c cc gi tr cn tm l m = 1 v m = 2 Cch 2. Hm s c y = 4x3 4mx = 4x(x2 m). y l tch ca mt nh thc v mt tam thc nn hm s c cc 1

htt p:/ /m ath .vnI= /4

i v cc tiu khi v ch khi x2 m c 2 nghim phn bit khc 0 m > 0 Do h s bc 3 dng nn hm s c cc i A(0, 2), cc tiu B ( m, 2 m2 ), B( m, 2 m2 ). Tm I ca ng trn qua 4 im A, B, B , M s nm trn Oy v B, B i xng nhau qua Oy do I(0, b) 2 9 9 + b b = 1 IA = IM (2 b)2 = 25 5 2 = m + (2 m2 b)2 thay b = 1 vo IA = IB (2 b) 1 + 5 m 1 + (1 m2 )2 = 0 m(m 1)(m2 + m 1) = 0 m = 1 hay m = (v iu kin m > 0) 2 Cu II. 1) (1 im) 16cos6 x + 2cos4 x Gii phng trnh : sin x = . 54 51cos2 x Li gii: 16cos6 x + 2cos4 x >0 sin x > 0 sin x = 1 cos2 x t: t = cos2 x (0 < t < 1) 54 51cos2 x 16t 3 + 2t 2 Phng trnh ban u tr thnh: 1t = 0 54 51t (48t 2 + 4t)(54 51t) + 51(16t 3 + 2t 2 ) 1 > 0( do 0 < t < 1) f (t) = + 2 (54 51t) 2 1t f (t) l hm ng bin, m f (3/4) = 0 t = 3/4 chnh l nghim duy nht cos2 x = 3/4 sin x = 1/2 ( do sin x > 0) x = /6 + k2 hay x = 5 /6 + k2 Cu II. 2) (1 im) x2 + 2y2 3x + 2xy = 0 . Gii h phng trnh: xy(x + y) + (x 1)2 = 3y(1 y) Li gii: x2 + 2y2 3x + 2xy = 0 (1) 2 = 3y(1 y) (2) xy(x + y) + (x 1) 2 + 2y y(x 1)) = 0 (1) (2) = (x + 1)(1 y x = 1 pt v nghim. 2 3y y2 1 3y y2 1 2 + y2 3x = 0 3y 1 (1) (x + y) + y2 = 3 x= y y y 2 1 1 1 1 y+3 +2 = 0 y+3 = 1 hay y + 3 =2 3 y+3 y y y y y = 1 2 hay y = 1 + 2 x = . . . hay x = . . . y2 + 2y 1 = 0 1 + 5 x = . . . hay x = . . . 1 5 2 +y1 y =0 hay y = y= 2 2 Cu III. (1 im) 1 2 ln(1 x) Tnh tch phn I = dx. 2 0 2x 2x + 1 Li gii: /4 1 + tant dt t 1 2x = tant 2dx = (tan2 t + 1)dt Ta c: I = ln 2 0 u 1 + tan /4 /4 /4 1 4 du = t u = t ta c: I = du = ln ln ln(1+tan u)du 4 2 1 + tan u 0 0 0 ln

/4 /4 1 + tan u ln 2du = I ln 2du du 2 0 0 0 /4 1 /4 1 = ln 2 I= ln 2du = (u ln 2) 2 0 2 8 0 Cu IV. (1 im) Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a. Hnh chiu ca S trng vi trng tm tam gic

2

ABD. Mt bn (SAB) to vi y mt gc 60o . Tnh theo a th tch ca khi chp S.ABCD. Li gii:

htt p:/ /m ath .vnA E D G B C 3

Gi G l trng tm tam gic ABD E l hnh chiu ca G ln AB SGAB AB (SGE) Ta c: GEAB SEG = 600 SG = GE tan SEG = 3GE Mt khc: G l trng tm tam gic ABD 1 a GE = BC = 3 3 a3 3 1 VSABCD = SG.SABCD = 3 9

S

Cu V. (1 im) Cho s thc a, b, c [0; 1]. Tm gi tr ln nht, gi tr nh nht ca biu thc P = a5 b5 c5 (3(ab + bc + ca) 8abc). Li gii: T iu kin ban u ta d dng suy ra c 3(ab + bc + ca) 8abc 0, do vy gi tr nh nht ca P l 0, t c khi trong ba s a, b, c c t nht mt s bng 0. Tip theo ta s tm gi tr ln nht ca P. S dng bt ng thc AM-GM, ta c 3(ab + bc + ca) 8abc + 5abc 6 (ab + bc + ca abc)6 = . P 6 26 V b + c 2 bc bc nn ab + bc + ca abc = a(b + c bc) + bc b + c bc + bc = b + c 2, (ab + bc + ca abc)6 26 6 = 1. t suy ra P 26 2 Nh vy ta tm c gi tr ln nht ca P l 1, c c khi c a, b, c u bng 1. Cu VI. 1) (1 im) Trong mt phng Oxy cho im A(1; 4) v hai ng trn (C1 ) : (x 2)2 + (y 5)2 = 13, (C2 ) : (x 1)2 + (y 2)2 = 25. Tm trn hai ng trn (C1 ), (C2 ) hai im M, N sao cho tam gic MAN vung cn ti A. Li gii:

Cu VI. 2) (1 im) Trong khng gian vi h ta Oxyz, cho M (1; 2; 3). Lp phng trnh mt phng i qua M ct ba tia Ox ti A, Oy ti B, Oz ti C sao cho th tch t din OABC nh nht. Li gii: Gi A(a, 0, 0) Ox; B(0, b, 0) Oy;C(0, 0, c) Oz v M(1, 2, 3) nn a, b, c > 0 x y z 1 2 3 Pt mt phng (ABC) l : + + = 1 v M (ABC) + + = 1 a b c a b c 1 1 ta c: VOABC = OC. OA, OB = abc 6 6 1 2 3 6 1 p dng bt ng thc C-si ta c : + + 3 3 = 1 abc 27 a b c abc 6 1 2 3 = = a b c a = 3, b = 6, c = 9 ng thc xy ra khi : 1+2+3 =1 a b c x y z Vy phng trnh mt phng l : + + = 1 3 6 9 Cu VII. (1 im) Gii bt phng trnh 4x 2x+2 x2 2x 3 Li gii: PT (2x 2)2 (x 1)2 |2x 2| |x 1| ()

htt p:/ /m ath .vn4

Xt f (x) = 2x x 1 trn R. f (x) = 2x ln 2 1 f (x) = 0 x = log2 (ln(2)) 0, 53 f ( log2 (ln(2))) 0.09 BBT: + Vi x 1 () 2x 2 x 1 2x x 1 0 x [0; 1] Kt hp K ta c: x = 1 + Vi x < 1 () (2x 2) (x 1) 2x x 1 0 x (; 0] [1; +) Kt hp K ta c: x (; 0] Kt lun: Tp nghim bt pt ban u l S = (; 0] {1}

DI N N MATH.VN



Cu I. (2 i m) Cho hm s y = x4 + 6x2 5. 1 Kh o st s bi n thin v v th (C) c a hm s . 2 Tm cc gi tr c a m phng trnh (x2 5)|x2 1| = m c 6 nghi m phn bi t. Cu II. (2 i m) x3 2x =2 6 x2 1 x2 1 14x2 21y2 + 22x 39y = 0 2 Gi i h phng trnh sau trn R: 35x2 + 28y2 + 111x 10y = 0. 1 Gi i phng trnh: Cu III. (1 i m)3

Tnh tch phn

I=0

x dx. 9x

Cu IV. (1 i m) Cho kh i l p phng ABCD.A B C D c nh a. G i M l trung i m c a BC, i m N chia o n CD theo t s 2. M t ph ng (A MN) chia kh i l p phng thnh hai ph n. Tnh th tch m i ph n. Cu V. (1 i m) 1 1 1 + + = 16. a b c a2 + 2b2 Tm gi tr l n nh t v gi tr nh nh t c a bi u th c P = . ab Cho cc s dng a, b, c th a mn (a + b + c) Cu VI. (2 i m) 1 Trong m t ph ng t a Oxy cho tam gic ABC c B(4; 0), c nh AC qua O, phng trnh trung tr c AC l x + y 1 = 0, phng trnh ng cao qua C l 5x + y 12 = 0. Tnh di n tch tam gic ABC. 2 Cho t di n ABCD c A(1; 1; 6), B(3; 2; 4),C(1; 2; 1), D(2; 2; 0). Tm i m M thu c ng th ng CD sao cho chu vi tam gic MAB nh nh t. Tnh gi tr nh nh t . Cu VII. (1 i m) Gi i b t phng trnh: 1 log2 (x) 2 log2 (5x 6)2

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Cu I. 1) (1 im) Cho hm s y = x4 + 6x2 5. Kho st s bin thin v v th (C) ca hm s. Li gii: th hm s y = x4 + 6x2 5 c TX l D = R 3 + 12x = 4x x2 3 y = 4x 4 x = 0 y = 5 x= 3 y=4 Nn y = 0 x= 3 y=4 2 y < 0 3 < x < 0 hoc 3 < x + < hm s nghch bin trn 3; 0 ; 3; + 3 y > 0 < x < 3 hoc 0 < x < hm s ngh bin trn ; 3 ; 0; 3 . 2 2 Gii hn lim y = ; lim y = . Bng bin thinx x+

im cc tiu (0; 5), im 3; 4 . 3; 4 ; th ct trc honh ti ( 5; 0), (1; 0), (1; 0), ( 5; 0) th ct trc tung ti im (0; 5)

cc 4

i im

2

4

Cu I. 2) (1 im) Tm cc gi tr ca m phng trnh (x2 5)|x2 1| = m c 6 nghim phn bit. Li gii: h4

x2 1 (x2 5) = m (1) Xt th hm s y = x2 1 (x2 5) v ng thng y = m ta c y = x4 6x2 + 5 vi x > 1 v x < 1 y = x4 + 6x2 5 vi 1 < x < 1 suy ra cch v - V th hm s y = x4 + 6x2 5 - Gi nguyn phn th ng vi 1 < x < 1 - ly i xng vi phn th ng vi x > 1 v x < 1 qua Ox da vo th nhn thy phng trnh (1) c 6 nghim phn bit khi 4 < m < 0

2

2 2

2

4

Cu II. 1) (1 im) x3 2x Gii phng trnh: =2 6 x2 1 x2 1 Li gii: K: |x| > 1

1

x( x2 1 1)( x2 1 + 1) =2 6 PT x2 1( x2 1 1) x x4 x2 x4 2x2 x+ = 2 6; (x > 0) x2 + 2 = 24 2 + +2 = 24 x 1 x 1 x2 1 x2 1 x2 1 x4 = 4 x4 16x2 + 16 = 0 x = 8 + 48 hay x = 8 48 x2 1 Cu II. 2) (1 im) 14x2 21y2 + 22x 39 = 0 Gii h phng trnh sau trn R: 35x2 + 28y2 + 111x 10y = 0. Li gii: Cu III. (1 im) 3 x Tnh tch phn I = dx. 9x 0 Li gii: 6 9 6 3 9t 9 t: 9 x = t dx = dt, x t I = dt = 1 dt t t 9 0 9 6 9 0 1 9 dt = 18 sin u cos u du, t u vi 9 cos2 = 6 t: = t cos2 u 6 0 1 9 sin2 u du = 9 du+9 cos 2u du = 9u + sin 2u I = 18 1sin ucos u du = 18 cos2 u 2 0 0 0 0 0 Cu IV. (1 im) Cho khi lp phng ABCD.A B C D cnh a. Gi M l trung im ca BC, im N chia on CD theo t s 2. Mt phng (A MN) chia khi lp phng thnh hai phn. Tnh th tch mi phn. Li gii: Gi th tch phn cha A l V1 v phn cn li l V2 hnh v Gi E, F ln lt l giao im ca MN vi AB, AD. A D Gi P l giao im ca A F vi DD v Q l giao im ca A E vi BB . 2a MC a a = BE, DF = = . Ta c DN = , NC = 3 3 2 4 5a 2a a 5a B suy ra AE = , AF = , BQ = , DP = . C P 3 4 5 5 3 25a 1 1 D F A VE.AA F = AA AF AE = 3 2 72 Q 1 1 a3 N VE.BMQ = BM BQ BE = 3 2 45 a3 1 1 VF.PDN = DP DN DF = C 3 2 360 M B 3 29a V1 = VE.AA F VE.BMQ VF.PDN = 90 E 61a3 3 V = V2 = a 1 90 Cu V. (1 im) 1 1 1 + + = 16. Cho cc s dng a, b, c tha mn (a + b + c) a b c a2 + 2b2 Tm gi tr ln nht v gi tr nh nht ca biu thc P = . ab Li gii: 2 a2 + 2b2 = x+ a = xb, b = yc, x > 0, y > 0 P = ab x 2

2 73 5 Cu VI. 1) (1 im) Trong mt phng ta Oxy cho tam gic ABC c B(4; 0), cnh AC qua O, phng trnh trung trc AC l x + y 1 = 0, phng trnh ng cao qua C l 5x + y 12 = 0. Tnh din tch tam gic ABC. Li gii: Ta c ng trung trc AC : x + y 1 = 0. Suy ra ptr t AC qua O(0; 0) : x y = 0 1 1 Gi H l giao im ca AC v ng trung trc ca n: H ; 2 2 PT AB i qua B(0; 4) v vung gc vi ng cao qua C c dng: x5y4 = 0 D dng Suy ra A(1; 1) 12 M H l trung im ca AC Suy ra C(2; 2); AB = 26; CH = d(C/AB) = Suy ra SABC = 6 26 Cu VI. 2) (1 im) Cho t din ABCD c A(1; 1; 6), B(3; 2; 4),C(1; 2; 1), D(2; 2; 0). Tm im M thuc ng thng CD sao cho chu vi tam gic MAB nh nht. Tnh gi tr nh nht . Li gii: AB = (2; 3; 10), CD = (1; 4; 1) nn AB.CD = 0 AB CD. Gi l mp qua AB v vung gc CD, phng trnh l (x + 1) 4(y 1) + (z 6) = 0 x 4y + z 1 = 0 (1) Ta c ngay gi tr ln nht ca P l 7 +

1 1 1 1 1 + + + + 1 = 16 = 16 (xy + y + 1) a b c xy y 1 1 y 1 1 x + xy + + y + + = 13 xy + xy2 + + y2 + + 1 = 13y x xy y x x 1 1 (x + 1) y2 + x + 13 y + + 1 = 0 () x x Xem y l ph tr bc 2 n y > 0 Gi 2 nghim l x1 , x2 phng trnh ny c t nht 1 nghim dng 1 +1 1 x1 x2 > 0 x nn ta phi c x1 x2 < 0 hoc Ta lun c x1 x2 = = >0 x1 + x2 > 0 x+1 x nn loi trng hp x1 x2 < 0. Vy pt () c t nht 1 nghim dng khi v ch khi 1 x + 13 1 x x1 + x2 = 0 x + 13 x+1 x 2 1 1 v x + 13 4 (x + 1) +1 0 x x 1 30 1 2 1 x2 + 163 30x + 2 0 x+ 30 x + + 161 0 t 2 30t + 161 0 x x x x 1 7 3 5 7 3 5 x + x 23 x + 1 2 2 2 2 x+ 7 x 2 1 1 1 2 = 7+ 13.85 + 7+ P = x+ = x+ x x x 7 3 5 73 5 2 2 7 3 5 Du = xy ra khi x = 0.145898 2 2 7 3 5 73 5 b 0.381966 Vy ta ch cn chn c = 1; b = 2.618; a = 2 2 2 94 5 (a + b + c)

3

x = 1 + t ng thng CD c pt tham s l y = 2 4t (2). z = 1 + t CD ct mp ti im c ta l nghim ca h (1), (2) . Giao im ny l hnh chiu vung gc ca A ln CD v cng l hnh chiu vung gc ca B ln CD nn cc on vung gc ny l on ngn nht, vy giao im ny l im M cn tm cho chu vi tam gic ABC nh nht. 3 1 1 1 3 ; 0; Gii h (1), (2) ta c t = , x = , y = 0, z = . Nn M 2 2 2 2 2 Chu vi tam gic l 5 2 9 2 13 2 7 2 AB + AM + BM = 22 + 32 + 102 + + 12 + + + 22 + 2 2 2 2 198 146 + = 113 + 2 2 Cu VII. (1 im) 1 2 Gii bt phng trnh: (x) log 2 log2 (5x 6)2 Li gii: Trc ht ta c nhn xt sau: Vi iu kin xc nh s tn ti th log2 X log2 Y lun cng du hoc cng trit tiu vi X Y Quay li bi ton ta c: x > 0 (1) 1 2 (2) 6x 5 = 0 2 2 (x) log2 (5x 6) 4 log2 x log 2 log2 (5x 6) 0 () 2 log2 (x) log2 (5x 6)2 Cc rng buc (1); (2) kt hp li di iu kin chung l: 6 := D x (0; +) \ 5 log2 (5x 6)2 log2 x4 0 Khi y: () 2 log2 (x) log2 (5x 6)2 Theo nhn xt th trn D ta c: log2 (5x 6)2 log2 x4 lun cng du hoc cng trit tiu vi (5x 6)2 x4 = (x + 6)(x 1)(x 2)(x 3) log2 (x) lun cng du hoc cng trit tiu vi x 1 log2 (5x 6)2 lun cng du hoc cng trit tiu vi (5x 6)2 1 = 5(x 1)(5x 7) (x 2)(x 3) Vy tc l trn D th: () 0 (x 1)(5x 7) Sau khi v trc an du ra gii v kt hp vi x D 7 6 Ta c tp nghim ca bt phng trnh ra l: 1; [2; 3] \ 5 5

4

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PH N CHUNG (7,0 i m) Cho t t c th sinh Cu I. (2 i m) Cho hm s y = x3 + 6x2 + 9x + 3 1 Kh o st s bi n thin v v th (C) c a hm s . 2 Tm cc gi tr c a k t n t i 2 ti p tuy n v i (C) phn bi t nhau v c cng h s gc k , ng th i ng th ng i qua cc ti p i m (c a 2 ti p tuy n v i (C)) c t cc tr c Ox, Oy tng ng t i A v B sao cho OB = 2011.OA Cu II. (2 i m) 1 Gi i phng trnh : 1 x 2 sin2 x = tan2 cos 2x + 4 cos x + 3 2 2 x3 + 2y3 = x2 y + 2xy 2 Gi i h phng trnh : 2 x2 2y 1 + 3 y3 14 = x 23

(x, y R)

Cu III. (1 i m) Tnh tch phn I =1

(x2 2x 2

2010

x 1)2011 + 2012 sin4

x dx 2

Cu IV. (1 i m) Cho hnh chp S.ABC c y l tam gic vung t i A , BC = a v ABC = 300 . M t ph ng (SBC) vung gc v i y, hai m t ph ng (SAB) v (SAC) cng t o v i m t ph ng y gc 60o . Tnh th tch kh i chp S.ABC theo a. Cu V. (1 i m) Cho cc s dng x, y, z tho mn x + y + 1 = z Tm gi tr l n nh t c a bi u th c x 3 y3 F= (x + yz)(y + zx)(z + xy)2 PH N RING (3,0 i m) Th sinh ch lm m t trong hai ph n A ho c B Ph n A theo chng trnh chu n Cu VIa. (2 i m) 1 Trong m t ph ng v i h to Oxy cho tam gic ABC bi t 3 chn ng phn gic trong ng v i cc nh A, B,C l n l t l A (1; 1), B (3; 2), C (2; 3) . Vi t phng trnh cc ng th ng ch a 3 c nh c a tam gic ABC. 2 Trong khng gian v i h to Oxyz cho hnh chp tam gic S.ABC c A; B thu c tr c honh v phng trnh x+1 y z+3 x1 y2 z3 hai ng phn gic ngoi c a hai gc BSC; CSA l n l t l: (la ) : = = , (lb ) : = = 2 3 4 2 2 6 Hy vi t phng trnh ng phn gic trong (lc ) c a gc ASB Cu VIIa. (1 i m) Tm t p h p cc i m bi u di n s ph c 2z + 3 i bi t |3z + i|2 zz + 9 Ph n B theo chng trnh nng cao Cu VIb. (2 i m) 1 Trong m t ph ng v i h to Oxy cho i m A ch y trn Ox , i m B ch y trn Oy sao cho o n AB lun b ng a khng i . Tm t p h p cc i m M trn o n AB sao cho MB = 2MA 2 Trong khng gian v i h to Oxyz cho t gic ABCD c A(1; 2; 1), C(2; 4; 1) . Hai nh B, D thu c ng x1 y2 z th ng = = sao cho BD = 4. G i I l giao i m hai ng cho c a t gic v bi t r ng 1 2 3 dt(ABCD) = 2011dt(IAD). Tnh kho ng cch t D t i ng th ng AC. Cu VIIb. (1 i m) Cho 2 phng trnh z2 + mz + 2 = 0 v z2 + 2z + m = 0 . Tm cc gi tr th c c a m 2 phng trnh c t nh t m t nghi m ph c chung.

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Cu I. 1) (1 im) Cho hm s y = x3 + 6x2 + 9x + 3. Kho st s bin thin v v th (C) ca hm s . Li gii: Hm s y = x3 + 6x2 + 9x + 3 c TX l D = R th giao vi trc tung ti im B(0; 3) y = 3x2 + 12x + 9 = 3(x + 1)(x + 3) th giao vi ng thng: y = 1 x = 1 y = 1 Nn y = 0 ti hai im (4; 1), (1; 1) x = 3 y = 3 th y > 0 < x < 3 hoc 1 < x < + hm s ng bin trn (; 3) ; (1; +) y < 0 3 < x < 1 3 hm s nghch bin trn (3; 1). Gii hn lim y = +; lim y = . 2 Bng bin thin x y + 3 0 3 x+ x

1 0 +

+ +4 3 2 1

1

y 1 im cc i (3; 3), im cc tiu (1; 1).

1 2

Cu I. 2) (1 im) Tm cc gi tr ca k tn ti 2 tip tuyn vi (C) phn bit nhau v c cng h s gc k , ng thi ng thng i qua cc tip im (ca 2 tip tuyn vi (C)) ct cc trc Ox, Oy tng ng ti A v B sao cho OB = 2011.OA Li gii: i vi th hm s bc 3 th mi tip tuyn ta c mt v ch mt tip im tng ng . V honh tip im x ca tip tuyn dng y = kx + m vi (C) l nghim PT f (x) = k 3x2 + 12x + 9 k = 0 (1) tn ti hai tip tuyn vi (C) phn bit nhau c cng h s gc k th cn v l (1) c 2 nghim phn bit = 9 + 3k > 0 k > 3 (2) Khi to cc tip im (x; y) ca 2 tip tuyn vi (C) l nghim HPT y = 1 x + 2 (3x2 + 12x + 9) 2x 3 3 + 6x2 + 9x + 3 y=x 3 3 2 3x2 + 12x + 9 = k 3x + 12x + 9 = k k6 2k 9 y = k 1 x + 2 2x 3 x+ y= 3 3 3 3 3x2 + 12x + 9 = k 2 3x + 12x + 9 = k k6 2k 9 Suy ra to cc tip im cng tho mn PT y = x+ . 3 3 k6 2k 9 Vy ng thng qua cc tip im l d : y = x+ 3 3 Do d ct Ox, Oy tng ng ti A v B sao cho OB = 2011.OA nn c th xy ra cc trng hp sau : 9 +) Nu A O th B O, TH ny c tho mn nu v ch nu d qua O k = 2 OB +) Nu A O, khi trong tam gic vung OAB cho tan OAB = = 2011 m tan OAB bng hoc bng OA k6 vi i ca h s gc ca d , tc l ta c = 2011 k = 6039; k = 6027, 3 i chiu vi (2) , ta c kt qu ca TH ny l : k = 6039 9 Tm li cc gi tr cn tm ca k l k = v k = 6039 2 1

Cu II. 1) (1 im) 2 sin2 x 1 x Gii phng trnh : = tan2 cos 2x + 4 cos x + 3 2 2 Li gii: x K: cos = 0; cos 2x + 4 cos x + 3 = 0. Vi k y pt tng ng: 2 1 + cos2 x 1 cos x 1 + cos2 x 1 cos x = = 2 x + 2 cos x + 1) 2 2(cos 2(1 + cos x) (cos x + 1) 1 + cos x 2 x = (1 cos x)(1 + cos x) 1 + cos2 x = 1 cos2 x cos x = 0 x = + k (tho) 1 + cos 2 Vy pt c h nghim l x = + k 2 Cu II. 2) (1 im) x3 + 2y2 = x2 y + 2xy Gii h phng trnh : (x, y R) 2 x2 2y 1 + 3 y3 14 = x 2 Li gii: KX ca HPT l x2 2y 1 0 (1) PT u ca h c vit li l x(x2 2y) y(x2 2y) = 0 (x y)(x2 2y) = 0 x = y (do (1) nn x2 2y 1 > 0) 3 Thay vo PT th 2 ca h ta c : 14 x3 = 2 x2 2x 1 + 2 x (2) i vi PT (2) c 2 cch gii quyt ta : 3 +) Cch 1: Do x2 2x 1 0 nn 14 x3 2 x 14 x3 8 12x + 6x2 x3 x2 2x 1 0 Mt khc khi x = y th (1) cho ta x2 2x 1 0 . Do x2 2x 1 = 0 x = 1 2. Vy x = y = 1 2 tho mn HPT cho v l 2 nghim ca HPT ban u. +) Cch 2: t u = 2 x , v = x2 2x 1(v 0) , ta c u3 6v2 = 14 x3 . Khi (2) c dng : 3 3 u 6v2 = u + 2v u3 6v2 = (u + 2v)3 v[v2 + 3(u + v)2 + 3v] = 0 v = 0 ; u = v = 0 (do v 0) v = 0 x2 2x 1 = 0 x = 1 2. Do vy x = y = 1 2 (tho mn (1)) 2x = 0 u=v=0 2 . H ny v nghim. x 2x 1 = 0 Tm li HPT u c 2 nghim l x = y = 1 2 Cu III. (1 im) 3 x 2010 Tnh tch phn I = (x2 2x 2 x 1)2011 + 2012 sin4 dx 2 1 Li gii: Cch 1: 3 3 x x [(x 1)3 3(x 1)]2010 (x 1) sin4 dx + 2012 sin4 dx = (I1 ) + (I2 ) I= 2 2 1 1 t x 1 = t dx = dt; x = 1 t = 2; x = 3 t = 2 2 t Vy Ta thy I1 = (t 3 3t)2010t cos4 dt 2 2 3 3t)2010t cos4 t l hm l I = 0 m f (t) = (t 1 2 2 2 2 t t Nhim v by gi ch tnh I2 = 2012 cos4 dt = 4024 cos4 dt = 1006 (1 + cos t)2 dt 2 2 2 0 0 M ci tch phn ny khai trin ra ri h bc l xong. Kt qu l 3018 Cch 2: 3 3 x 2010 (x2 2x 2 x 1)2011 dx + 2012 Ta c I = sin4 dx 2 1 1 +) Xt tch phn J =1 3

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t x = 2 t ta c dx = dt v : x = 1 t = 3; x = 3 t = 1 Do J = 3

1

(x2 2x 2

2010

x 1)2011 dx

((2 t)2 2(2 t) 2

2010

2 t 1)2011 dt = 2

3 1

(t 2 + 2t + 2

2010

1 t)2011 dt

=

http://www.VNMATH.com 3 20101

(t 2 2t 2

t 1)2011 dt = J. Suy ra J = 03

+) Xt tch phn K = 2012 Ta c J = 20123 1

sin42

1 cos x 2

1

x dx 2

3

dx = 503

3 503 3 503 1 = = 3018 (3 4 cos 2 x + cos 4 x)dx = 3x 2 sin 2 x + sin 4 x 2 1 2 4 1 Vy I = J + K = 0 + 3018 = 3018 Cu IV. (1 im) Cho hnh chp S.ABC c y l tam gic vung ti A , BC = a v ABC = 300 . Mt phng (SBC) vung gc vi y, hai mt phng (SAB) v (SAC) cng to vi mt phng y gc 60o . Tnh th tch khi chp S.ABC theo a. Li gii: Cch 1: V SH BC (H BC); HE AB (E AB); HF AC (F AC) V (SBC) (ABC) vi giao tuyn l BC SH (ABC) Ta c: SHE = SHF HE = HF m AEHF l hnh ch nht AEHF l hnh vung; cnh l b TH1: H nm trong on BC

1

(1 2 cos 2 x + cos2 x)dx

S

A

F

C

E B

H

a 3 a Ta c: AB = ; AC = 2 2 a 3 a Ta c: BE = AB AE = b; FC = AC AF = b 2 2 EH a 3 a BE BEH HFC = BE.FC = b2 b b = b2 HF FC 2 2 a2 3 ab( 3 + 1) a2 3 ab a(3 3) + b2 = b2 = b= 4 2 4 4 31 2 3 a(3 3) 3 a Ta c: SH = b 3 = SABC = 4 8 1 1 a(3 3) 3 a2 3 a3 (3 3) VS.ABC = SH.SABC = = 3 3 4 8 32 TH2: H nm ngoi on BC v pha B HE AC BEH HFC = < 1 HE < EB < AE V l, v vy ko c TH ny EB AB TH3: H nm ngoi on BC v pha C

3


S

F

H

A C

E

CA BA a a 3 a 3 a(3 + 3) Ta c: BAC BEH = CA.BE = AB.HE (b + )= bb= HE BE 2 2 2 4 a(3 + 3) 3 a2 3 Ta c: SH = b 3 = SABC = 4 8 1 1 a(3 + 3) 3 a2 3 a3 (3 + 3) VS.ABC = SH.SABC = = 3 3 4 8 32 Cch 2: Dng ng thng Az(ABC) v thit lp h ta Axyz nh hnh v di y:z

B

S

A

C

y

B x

a 3 a Chng ta t b = ; c = v c c: A (0; 0; 0) , B (b; 0; 0) , C (0; c; 0) 2 2 ta cng gi th lun l: S (x; y; z) v thy l: SABC .d (S; (ABC)) bc.d (S; (Axy)) bc|z| VS.ABC = = = 3 6 6 Mt khc: AS = (x; y; z) , BS = (x b; y; z) , CS = (x; y c; z) Vy nn mt phng (SBC) c vector php tuyn l: BS CS = (cz; bz; bc by cx) na rng php tuyn ca (ABC) chnh l k (0; 0; 1) nn s kin (SBC)(ABC) dn n: . = 0 bc = by + cx na k () Hai vector php tuyn ln lt ca (SCA) v (SAB) l: CS AS = (cz; 0; cx), AS BS = (0; bz; by) nb nc 4

Th nn vic to gc 60 gia (ABC) vi cc mt phng (SCA) v (SAB) s cho ta: 1 |by| |cx| = = 2 (bz)2 + (by)2 (cz)2 + (cx)2 |z| c: |x| = |y| = 3 T () ta dn n hai kh nng sau: bc 3 b2 c2 3 a3 (3 3) Nu x; y cng du th th |z| = |x| 3 = nn th tch cn tnh l: VS.ABC = = . b+c 6(b c) + 32 b2 c2 3 a3 (3 + 3) bc 3 nn th tch cn tnh l: VS.ABC = = . Nu x; y tri du th th |z| = |x| 3 = bc 6(b c) 32 a3 (3 + 3) a3 (3 3) Vy c hai kt qu cho th tch cn tnh l: V1 = v V2 = 32 32 Cu V. (1 im) Cho cc s dng x, y, z tho mn x + y + 1 = z Tm gi tr ln nht ca biu thc x 3 y3 F= (x + yz)(y + zx)(z + xy)2 Li gii: Cch 1: t t = xy. T gi thit s dng bt ng thc AM-GM, ta d thy z = x + y + 1 2 xy + 1 = 2t + 1. By gi, s dng bt ng thc Cauchy-Schwarz, ta c 2 x y + yz zx = xy(z + 1)2 t 2 (2t + 1 + 1)2 = 4t 2 (t + 1)2 . (x + yz)(y + zx) Li c (z + xy)2 (2t + 1 + t 2 )2 = (t + 1)4 . Kt hp cc nh gi ny li, ta suy ra t6 t4 F 2 = . [4t (t + 1)2 ] [(t + 1)4 ] 4(t + 1)6 Mt khc, s dng bt ng thc AM-GM, ta li c 2 t t 3 t . t +1 = + +1 3 2 2 4 t4 4 V nh th, F . = 6 729 2 3 t 4 3 4 ng thc xy ra khi v ch khi x = y = t = 2 v z = 2t + 1 = 5. Cch 2: x 1 y xy x x y y xy Ta c : x + y + 1 = z + + = 1 . + . + . =1 z z z z yz yz zx zx z xy x y t tan = , tan = , tan = z yz zx xy x y ta ch cn chn 0 < , , < v hm tang tun hon chu k v do , , >0 2 z yz zx xy x x y y xy Khi t . + . + . =1 z yz yz zx zx z ta c tan tan + tan tan + tan tan = 1 (1) . Ta tm gi s + = 2 1 tan tan Vy (1) tan = tan = cot( + ) + + = (2) tan + tan 2 (Ch l hai gc nhn bng nhau khi v ch khi tang ca chng bng nhau) A B C By gi ta chn = , = , = vi 0 < A, B,C < (3) ( m bo 0 < , , < ) 2 2 2 2 Khi (2) cho ta A + B +C = (4) 5


V vi cch chon A, B nh vy cng m bo rng tho ci hi trc ta tm gi s + = Hai iu (3)&(4) chng t A, B,C l ba gc ca mt tam gic xy x x y y xy Tm li t . + . + . =1 z yz yz zx zx z xy = tan A z 2 x B = tan ng thc ny chng t tn ti ABC sao cho 2 yz y C = tan zx 2 2 xy 2 2A tan 1 1 2 Ta c F = z xy = A yz zx C B 1+ 1 + tan2 1+ 1+ 1 + cot2 1 + cot2 z 2 x y 2 2 2 2 A 1 2A B C 1 2A A B C B +C = sin2 sin sin = cos 1 sin sin cos sin 2 2 2 2 2 2 2 2 2 2 3 2 A A A 2 2 1 sin 2 + sin 2 + 2 2 sin 2 1 A A A = 2 sin . sin 2 2 sin . 4 2 2 2 4 3 27 B = C Du bng xy ra A 2 sin = 2 3 tan B = tan C 2 2 A 2 tan = 2 5 x y = yz zx xy = 4 z 5 x+y+1 = z x=y=2 z=5


2

2

=

Vy max F =

2 2 27 Cu VIa. 1) (1 im) Trong mt phng vi h to Oxy cho tam gic ABC bit 3 chn ng phn gic trong ng vi cc nh A, B,C ln lt l A (1; 1), B (3; 2), C (2; 3) . Vit phng trnh cc ng thng cha 3 cnh ca tam gic ABC. Li gii: Cu VIa. 2) (1 im) Trong khng gian vi h to Oxyz cho hnh chp tam gic S.ABC c A; B thuc trc honh v phng x1 y2 z3 trnh hai ng phn gic ngoi ca hai gc BSC; CSA ln lt l: (la ) : = = , 2 3 4 x+1 y z+3 (lb ) : = = Hy vit phng trnh ng phn gic trong (lc ) ca gc ASB 2 2 6 Li gii: = SA , = SB , = SC Chng ta k hiu: ea eb ec SA SB SC Khi ba vector n v , , s ln lt l ba vector ch phng ca SA, SB, SC. ea eb ec Theo quy tc hnh bnh hnh th + , + , + s ln lt l cc vector ch phng ca cc ng ea eb eb ec ec ea phn gic trong ca cc gc: ASB; BSC; CSA rng: ( + ) ( ) = ( + ) ( ) = ( + ) ( ) = 12 12 = 0 ea eb ea eb eb ec eb ec ec ea ec ea , , ln lt l cc vector ch phng ca l , l , l , ba ng phn gic ngoi ny Vy ea eb eb ec ec ea a b c u i qua S thm na: = 1. ( ) + (1). ( ) ea eb eb ec ec ea Vy la , lb , lc ng phng. Theo bi th (2; 3; 4), (2; 2; 6) cng l cc vector ch phng ca la , lb ua ub 6

( ua ub ) = n (5; 2; 1) s c phng vung gc vi phng ca lc . 2 Cng rng lc nm trong (SAB) m A, B (Ox) nn lc vung gc vi OS. 1 Ta c ngay S(1; 2; 3) nn OS = (1; 4; 3) l mt vector ch phng ca lc ( l = 0 ). n uc uc 4 By gi nhn thy rng lc lc 1 OS = (1; 1; 1) chnh l vector ch phng ca l (hy l: = ) uc u u 0 nn c c c 6 Vy phng trnh chnh tc ca ng phn gic ngoi cn tm l: x1 y2 z3 (lc ) : = = 1 1 1 Cu VIIa. (1 im) Tm tp hp cc im biu din s phc 2z + 3 i bit |3z + i|2 zz + 9 Li gii: Cch 1: t z = a + bi v z = 2z + 3 i = a + b i suy ra a = 2a + 3; b = 2b 1 a 3 b +1 suy ra a= ;b = (1) 2 2 2 + (3b + 1)2 a2 + b2 + 9 tng ng 4a2 + 4b2 + 3b 4 0 T phng trnh () ta c: 9a b +1 em (1) th vo phng trnh ny ta c: (a 3)2 + (b + 1)2 + 3 4 0 2 25 7 (2) (a 3)2 + (b + )2 tng ng: 4 16 7 5 Vy tp hp cc im M biu din cho s phc z l hnh trn tm I 3; , bn knh R = 4 4 Cch 2: t: z1 = (2a + 3) + (2b 1)i. Gi M biu din s phc c to : M(2a + 3; 2b 1) Ta c: |3z + i|2 z + 9 2(2a + 3)2 12(2a + 3) + (2b 1)2 + 7(2b 1) + 15 8 z 7 7 (2a + 3)2 6(2a + 3) + (2b 1)2 + (2b 1) + 0 2 2 7 5 Vy tp hp cc im M biu din cho s phc z1 l hnh trn tm I 3; , bn knh R = 4 4 Cu VIb. 1) (1 im) Trong mt phng vi h to Oxy cho im A chy trn Ox , im B chy trn Oy sao cho on AB lun bng a khng i . Tm tp hp cc im M trn on AB sao cho MB = 2MA Li gii: Gi A(x; 0)thuc Ox,im B(0; y)thuc Oy. Ta c AB = a a2 = x2 + y2 2x y ; Ta c MB = 2AM MB = 2AM M 3 3 2x y 4x2 2 y2 t = c; = d c2 = ;d = 3 3 9 9 4a2 4 2 m c2 + 4d 2 = (x + y2 ) = 9 9 4a2 Vy PT qu tch im M : c2 + 4d 2 = (c; d l honh v tung ca im M) 9 Cu VIb. 2) (1 im) Trong khng gian vi h to Oxyz cho t gic ABCD c A(1; 2; 1), C(2; 4; 1) . Hai nh B, D thuc x1 y2 z ng thng = = sao cho BD = 4. Gi I l giao im hai ng cho ca t gic v bit rng 1 2 3 dt(ABCD) = 2011dt(IAD). Tnh khong cch t D ti ng thng AC. Li gii: bi ny ta cn tnh c: 125 - Gc gia hai ng thng AC v BD c sin bng: 126 7

v th

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1 4 5 - Khong cch t A n BD v t C n BD ln lt bng: ; 14 14 1 1 16 5 - T dt(ABCD) = (d(A; BD) + d(C; BD)).BD = dt(IAD) = d(A; BD).ID suy ra ID = 2 2 2011 14 - Gi hnh chiu ca D xung AC l H. 200 T tam gic DHI vung ti H c DH = ID. sin(AC, BD) = 21.2011 Cu VIIb. (1 im) Cho 2 phng trnh z2 + mz + 2 = 0 v z2 + 2z + m = 0 . Tm cc gi tr thc ca m 2 phng trnh c t nht mt nghim phc chung. Li gii: Gi s 2 phng trnh c nghim phc chung z. Khi , z l nghim ca h (1) v (2). Cng theo v 2 phng trnh ca h c (m + 2)(z + 1) = 0 suy ra m = 2 hoc z = 1 Vi m = 2 thay trc tip vo 2 phng trnh thy chng trng nhau nn c nghim phc chung. Vi z = 1 thay vo (1) tm c m = 3 Kt lun: m = 2; m = 3


8

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PH N CHUNG (7,0 i m) Cho t t c th sinh Cu I. (2 i m) x+3 . x1 1 Kh o st s bi n thin v v th (C) c a hm s cho. 2 Tm i m A trn ng th ng x = 5 sao cho t A ta c th v n (C) hai ti p tuy n m hai ti p i m cng v i i m B(1; 3) th ng hng. Cho hm s : y = Cu II. (2 i m) x x x 2 3x 2 cos 6 sin = 2 sin + 2 sin + . 5 12 5 12 5 3 5 6 1 3 x + x2 8x 2 + 3 x3 20. 2 Gi i phng trnh sau trn t p s th c: x = 1 + 2 5

1 Gi i phng trnh :

Cu III. (1 i m) Tnh tch phn: I =

dx (9 x2 )3

0

Cu IV. (1 i m) Cho hnh chp S.ABCD c y ABCD l hnh vung c nh a, ng cao SA = a, M l i m thay i trn c nh SB. M t ph ng (ADM) c t SC t i i m N. Ta k hi u V1 ,V2 l n l t l th tch cc kh i a di n SADMN v MNADCB. 5 V1 = . Tm v tr c a i m M trn c nh SB V2 4 Cu V. (1 i m) Cho ba s th c dng a, b, c c tch b ng 1. Ch ng minh r ng: (a + b) (b + c) (c + a) 7 3 a+b+c+ . 3 7

PH N RING (3,0 i m) Th sinh ch lm m t trong hai ph n A ho c B Ph n A theo chng trnh chu n Cu VIa. (2 i m) 1 Trong m t ph ng v i h t a Oxy cho tam gic ABC v i i m A(2; 7), ng th ng AB c t tr c Oy t i E sao 13 cho AE = 2EB. Bi t r ng tam gic AEC cn t i A v c tr ng tm l G 2; . Vi t phng trnh c nh BC. 3 x5 y6 z+3 x2 y3 z+3 2 Trong khng gian v i h t a Oxyz cho hai ng th ng: : = = , : = = . 13 1 4 13 1 4 G i () l m t ph ng ch a hai ng th ng trn. Tm t a hnh chi u vung gc c a i m C(3; 4; 2) trn (). Cu VIIa. (1 i m) Gi i phng trnh z4 + 4 = 0 trn t p s ph c. Ph n B theo chng trnh nng cao Cu VIb. (2 i m) 1 Trong m t ph ng v i h t a Oxy g i d l ng th ng i qua i m A(0; 1) v t o v i ng th ng d : x + 2y + 3 = 0 m t gc 45o . 7 Vi t phng trnh ng trn c tm n m trn d , ti p xc v i d v c bn knh b ng . 5 2 Trong khng gian v i h t a Oxyz cho tam gic ABC v i A(1; 2; 1), B(2; 1; 3) v C(4; 7; 5). G i H l tr c tm c a tam gic ni trn. Vi t phng trnh ng th ng i qua H v vung gc v i m t ph ng (ABC). Cu VIIb. (1 i m) Tm m phng trnh: 2 log2 (x 1) = 1 + log2 (5 mx) c ng m t nghi m.

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Cu I. 1) (1 im) Cho hm s y = x3 3x2 + (m 6)x + m 2 (m l tham s). Kho st v v th khi m = 9 Li gii: th10 8

Hm s y = x3 3x2 + 3x + 7 = (x 1)3 + 8 Bng bin thin

6 4 2

2

2

Cu I. 2) (1 im) 3 11 Tm m th hm s c hai im cc tr v khong cch t im A ; n ng thng i qua hai 2 4 im cc tr ln nht. Li gii: Hm s c o hm: y = 3x2 6x + m 6 th hm s a thc bc 3 c 2 cc tr khi: y = 0 phi c 2 nghim phn bit: = 32 3(m 6) > 0 m < 9 1 1 4m 2 Ta c : y = y x m6 x+ 4 + 3 3 3 3 v im cc tr c hong l nghim ca y = 0 nn ng thng (d) qua 2 cc tr c pt l: 2 4m y= m6 x+ 4 (2m 18)x 3y + 4m 12 = 0 2kx 3y + 4k + 24 = 0 vi k = m 9 < 0 3 3 11 3 2k 3 + 4k + 24 7 |4k + 9| 2 4 Khong cch t A n d l : l = = > 0 nn 4k + 9 = 0 4 4k2 + 9 4k2 + 9 (4k + 9)2 72(4k + 9)(k 4) Xt f (k) = f (k) = 2+9 4k 4k2 + 9 Suy ra khng tn ti k f (k) t gi tr ln nht. Do khng tn ti m khong cch t A n d ln nht. Cu II. 1) (1 im) Gii phng trnh 4 sin2 x + tan x + 2(1 + tan x) sin 3x = 1 Li gii: PT 4 sin2 x 2 + 1 + tan x + 2 sin + tan x) = 0 3x(1 2 sin2 x 2 cos2 x + (1 + tan x)(1 + sin 3x) = 0 2 2 x sin2 x) + (1 + tan x)(1 + 2 sin 3x) = 0 2(cos 1 + 2 sin 3x 2(cos x sin x)(cos x + sin x) + (cos x + sin x) =0 cos x 1 + 2 sin 3x (cos x + sin x) 2 cos x + 2 sin x + =0 cos x TH 1. cos x + sin x = 0 tan x = 1 x = + k 4 1

1 + 2 sin 3x = 0 2 cos2 x + sin 2x + 1 + 2 sin 3x = 0 TH 2. 2 cos x + 2 sin x + cos x cos 2x sin 2x = 2 sin 3x sin( 2x) = sin 3x x = 20 + k2 hay x = 34 + k2 4 5 Cu II. 2) (1 im) 2 x + y2 + y + 3 3 y = x + 2 Gii h phng trnh y3 + y2 3y 5 = 3x 3 3 x + 2 Li gii: Ta c: 2 x + y2 + y + 3 (1 + 3)(x + 2 + 3y) x + 2 + 3 y ng thc xy ra khi x = 1; y = 1. Thay vo phng trnh di, thy tha mn. p s: (x; y) = (1; 1). Cu III. (1 im) 3 ln(3 + x2 ) Tnh tch phn I= dx 1 x(4 x) 2 Li gii: Cu IV. (1 im) Cho hnh chp S.ABC c SA = SB = ASB = ASC = BSC = ni tip trong mt cu bn knh bng R, SC, 8 3 3 bit th tch khi chp S.ABC bng R . Tnh 27 Li gii: Gi M l trung im ca SA; O l tm ng trn ngoi tip tam gic ABC; I l tm mt cu ngoi tip t din SABC. t x = SA, a = AB, r = OA. Ta c tam ABC u. gic 2 3 2 3 3 2 2 2 1 a = 2x sin , r = a= x sin , SABC = a = 3x sin , SI SO = SM SA = x2 , 2 3 2 3 2 4 2 2 x2 4 2 SO = , SO2 = SA2 OA2 , x2 = 4R2 1 sin ; 2R 3 2 4 2 2 1 8 3 3 R 1 sin2 sin VSABC = SO SABC = 3 3 3 2 2 2 8 3 3 4 1 4 1 Vy VSABC = R 1 sin2 sin = sin2 = 1 sin2 27 3 2 2 9 3 2 2 3 4 3 = 1 = + k . Vy = . sin 2 3 3 3 Cu V. (1 im) a2 1 b2 1 c2 1 Cho cc s thc a, b, c tha mn 0 < a b c v + + = 0. a b c Tm gi tr nh nht ca biu thc P = a + b2011 + c2012 Li gii: T gi thit ta c a 1 v c 1 b2 1 c2 1 a2 1 1 khi + = 0 (b + c)(1 ) 0 bc 1. (1) b c a bc c2 1 a2 1 b2 1 1 1 1 1 Li c = + = + (a + b)= (a + b) 1 2 ab( 1) = c a b a b ab ab 1 1 1 ab 1 = 2 ab ab c 1+ 0 c abc2 1. (2) c ab ab ab ab 3 2011 c2010 = 3 3 abc2 (bc)2010 3. Kt hp (1), (2) v Theo BT AM-GM ta c P 3 ab ng thc xy ra khi a = b = c = 1. Vy min P = 3. Cu VIa. 1) (1 im) Trong h ta Oxy cho ng trn (C) : (x 1)2 + (y 2)2 = 4 v hai ng thng d1 : mx + y m 1 = 0, d2 : x my + m 1 = 0. Tm m mi ng thng d1 , d2 ct (C) ti hai im phn bit sao cho bn giao im to thnh mt t gic c din tch ln nht. Li gii: 2

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www.VNMATH.comCch 1: ng trn (C) c tm I(1; 2) v c bn knh R = 2. Gi A, B l giao im ca d1 vi (C); C, D l giao im ca d2 vi (C)(A, B, C, D theo th t trn ng trn); h1 , h2 ln lt l khong cc t I n d1 , d2 . 1 |m| Ta c h1 = < R, h2 = < R nn d1 , d2 lun ct (C) ti 2 im phn bit. 2+1 m m2 + 1 Ta c AB = 2 R2 h2 , CD = 2 R2 h2 . R rng d1 d2 nn AB CD . 1 2

1 (4m2 + 3)(3m2 + 4) (4m2 + 3) + (3m2 + 4) Nn SABCD = AB CD = 2 R2 h2 R2 h2 = 2 =7 1 2 2 m2 + 1 m2 + 1 Du bng xy ra khi v ch khi m = 1. Cch 2: R rng d1 v d2 i qua P(1; 1) nm trong ng trn vi mi m nn mi ng thng ct ng trn ti hai im phn bit. Vect php tuyn ca d1 v d2 ln lt l = (m; 1) v = (1; m) = 0 d1 d2 . n1 n2 n1 n2 Gi A,C v B, D ln lt l cc giao im ca d1 , d2 vi ng trn (C), H, K ln lt l hnh chiu ca tm I(1; 2) trn d1 v d2 th 1 1 2 AC .BD2 = 2 (R2 IH 2 )(R2 IK 2 ) = 2 R4 R2 .IP2 + IH 2 .IK 2 SABCD = AC.BD = 2 2 2 IH 2 + IK 2 2 2 R4 R2 .IP2 + = (2R2 IP2 ) = 2R2 IP2 = 7 (Do IH 2 + IK 2 = IP2 ) 2 2 ng thc xy ra khi IH = IK IHPK l hnh vung IP = IH 2 = 1 m = 1 m2 + 1 Cu VIa. 2) (1 im) 16 v im Trong khng gian vi h ta Oxyz, cho mt cu (S) : (x + 1)2 + (y 1)2 + (z + 1)2 = 9 1 A 0; 0; . Vit phng trnh ng thng i qua A vung gc vi ng thng cha trc Oz v tip xc 3 vi mt cu (S) Li gii:

Cu VIIa. (1 im) Cho s phc z tha mn |z|2 2(z + z) 2(z z)i 9 = 0. Tm gi tr ln nht v gi tr nh nht ca |z| Li gii: t: z = a + bi |z|2 2(z + z) 2(z z)i 9 = 0 a2 + b2 4a + 4b = 9 a2 + b2 4(a b) = 9 2 theo BT: a b 2(a2 + b2 ) 9 = a2 + b 4(a b) a2 + b2 4 2(a2 + b2 ) 2) a2 + b2 4 2(a2 + b 9 0 2 2 + 17 (a2 + b2 ) 2 2 17 2 2 + 17 |z| 2 2 17 Cu VIb. 1) (1 im) Trong h ta Oxy cho hai ng trn (C1 ) : x2 + y2 2x 4y + 3 = 0, (C2 ) : x2 + y2 6x 8y + 20 = 0 v A(2; 2). Vit phng trnh ng thng i qua A v ct mi ng trn (C1 ), (C2 ) ti hai im phn 2 2 bit v 2 d1 + 5 d2 = 13 (d1 , d2 l khong cch t tm ca cc ng trn (C1 ), (C2 )n ) Li gii: Cu VIb. 2) (1 im) Trong khng gian vi h ta Oxyz, cho mt cu (S) : (x 1)2 + (y 1)2 + z2 = 1. Gi A l mt im ty x1 y1 z1 trn ng thng : = = . T A v cc tip tuyn AT1 , AT2 , AT3 n mt cu (S). Tm ta 1 2 1 im A bit mp(T1 T2 T3 ) to vi mt gc 30o . Li gii: Cu VIIb. (1 im)

3

www.VNMATH.comCho s phc z = 0 tha z z3

z + z

3

+ |z| +

3

1 |z|3

2

=6

Li gii: z = r(cos + i. sin ) ng thc tr thnh 1 1 1 2 cos 6 + r6 + 6 = 2 = z3 + 3 = z + . z z r (P + 1)2 (P 2) 0 P 2 max P = 2

z+

1 z

2

3 P3 3P

4

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PH N CHUNG (7 i m) Cho t t c th sinh Cu I. (2 i m) Cho hm s y = x3 3x2 + (m 6)x + m 2 (m l tham s ) 1 Kh o st v v th khi m = 9 2 Tm m th hm s c hai i m c c tr v kho ng cch t i m A i m c c tr l n nh t. Cu II. (2 i m) 1 Gi i phng trnh 2 Gi i h phng trnh Cu III. (1 i m)3

3 11 ; 2 4

n ng th ng i qua hai

4 sin2 x + tan x + 2(1 + tan x) sin 3x = 1 2 x + y2 + y + 3 3 y = x + 2 y3 + y2 3y 5 = 3x 3 3 x + 2 ln(3 + x2 ) dx x(4 x) 2

Tnh tch phn

I=1

Cu IV. (1 i m) Cho hnh chp S.ABC c SA SB = SC, ASB = ASC = BSC = n i ti p trong m t c u bn knh b ng R, bi t th = 8 3 3 R . Tnh tch kh i chp S.ABC b ng 27 Cu V. (1 i m) Cho cc s th c a, b, c th a mn 0 < a b c v Tm gi tr nh nh t c a bi u th c a2 1 b2 1 c2 1 + + = 0. a b c 2011 + c2012 P = a+b

PH N RING (3 i m) Th sinh ch lm m t trong hai ph n A ho c B Ph n A theo chng trnh chu n Cu VIa. (2 i m) 1 Trong h t a Oxy cho ng trn (C) : (x 1)2 + (y 2)2 = 4 v hai ng th ng d1 : mx + y m 1 = 0, d2 : x my + m 1 = 0. Tm m m i ng th ng d1 , d2 c t (C) t i hai i m phn bi t sao cho b n giao i m t o thnh m t t gic c di n tch l n nh t. 16 1 2 Trong khng gian v i h t a Oxyz, cho m t c u (S) : (x + 1)2 + (y 1)2 + (z + 1)2 = v i m A 0; 0; . 9 3 Vi t phng trnh ng th ng i qua A vung gc v i ng th ng ch a tr c Oz v ti p xc v i m t c u (S) Cu VIIa. (1 i m) Cho s ph c z th a mn |z|2 2(z + z) 2(z z)i 9 = 0 Tm gi tr l n nh t v gi tr nh nh t c a |z| Ph n B theo chng trnh nng cao Cu VIb. (2 i m) 1 Trong h t a Oxy cho hai ng trn (C1 ) : x2 + y2 2x 4y + 3 = 0, (C2 ) : x2 + y2 6x 8y + 20 = 0 v A(2; 2). Vi t phng trnh ng th ng i qua A v c t m i ng trn (C1 ), (C2 ) t i hai i m phn bi t v 2 2 2 d1 + 5 d2 = 13 (d1 , d2 l kho ng cch t tm c a cc ng trn (C1 ), (C2 ) n ) 2 Trong khng gian v i h t a Oxyz, cho m t c u (S) : (x 1)2 + (y 1)2 + z2 = 1. G i A l m t i m ty x1 y1 z1 trn ng th ng : = = . T A v cc ti p tuy n AT1 , AT2 , AT3 n m t c u (S). Tm t a i m 1 2 1 o. A bi t mp(T1 T2 T3 ) t o v i m t gc 30 Cu VIIb. (1 i m) Cho s ph c z = 0 th a z z3

+

z z

3

+ |z|3 +

1 |z|3

2

= 6 Tm gi tr l n nh t c a P = z +

1 z

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PH N CHUNG (7 i m) Cho t t c th sinh Cu I. (2 i m) Cho hm s y = x4 2(m + 1)x2 + 2m + 1, (Cm ) (m l tham s ). 1 Kh o st s bi n thin v v th c a hm s khi m = 1. 2 Xc nh m th hm s cho c t tr c honh t i 4 i m phn bi t A, B,C, D l n l t c honh x1 , x2 , x3 , x4 , (x1 < x2 < x3 < x4 ) sao cho tam gic ACK c di n tch b ng 4, v i K(3; 2). Cu II. (2 i m) 1 Gi i phng trnh: 2 Gi i h phng trnh: Cu III. (1 i m) Tnh tch phn I=0 2

2

1 1 sin 2x = 4 sin x 1 . sin x 6 2 sin x (x 2)(2y 1) = x3 + 20y 28 . 2( x + 2y + y) = x2 + x

5 cos x 4 sin x dx (sin x + cos x)7

Cu IV. (1 i m) Cho hnh l p phng ABCD.A B C D c nh a. Trn cc o n AD , BD l n l t l y cc i m M, N sao cho AM = DN = x, (0 < x < a 2). Tm x MN l o n vung gc chung c a AD v BD. Cu V. (1 i m) Cho 3 s a, b, c [0; 2] tho mn : a + b + c = 3. Tm gi tr l n nh t c a M = a2 + b2 + c2 . ab + bc + ca

PH N RING (3,0 i m) Th sinh ch lm m t trong hai ph n A ho c B Ph n A theo chng trnh chu n Cu VIa. (2 i m) 1 Cho ABC c phng trnh c a trung tuy n xu t pht t A v ng cao k t B l n l t l: 2x 5y 1 = 0, x + 3y 4 = 0. ng th ng BC i qua i m K(4; 9). L p phng trnh ng trn ngo i ti p ABC, bi t r ng nh C n m trn ng th ng d : x y 6 = 0. x2 y1 z1 2 Trong khng gian v i h tr c Oxyz, cho (P) : x + y z + 1 = 0, d : = = . G i I l giao i m 1 1 3 c a d v (P). Vi t phng trnh c a ng th ng n m trong (P), vung gc v i d v cch i m I m t kho ng b ng 3 2. Cu VIIa. (1 i m) Cho s ph c z sao cho: z+i = 1. Tm cc s ph c z tho mn i u ki n: |z + 3i 2| = 4 z 3i

Ph n B theo chng trnh nng cao Cu VIb. (2 i m) 1 Trong m t ph ng v i h t a Oxy, cho tam gic ABC bi t ng cao v trung tuy n xu t pht t A l n l t c phng trnh: 6x 5y 7 = 0; x 4y + 2 = 0. Tnh di n tch ABC, bi t r ng tr ng tm c a tam gic thu c tr c honh v ng cao xu t pht t nh B i qua i m E(1; 4). x2 y2 z1 2 Trong khng gian to Oxyz, cho i m M(2; 2; 1), ng th ng d : = = v m t c u 2 1 2 2 + y2 + z2 + 4x 6y + m = 0. Xc nh cc gi tr c a m ng th ng d c t m t c u (S) t i 2 i m phn (S) : x bi t A, B sao cho MA = 5MB. Cu VIIb. (1 i m) Cho s ph c z tho mn: zi = 1. Tm s ph c z sao cho z + 1 c m t acgumen b ng . z + 3i 6


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Cu I. 1) (1 im) Cho hm s y = x4 2(m + 1)x2 + 2m + 1, (Cm ) (m l tham s). Kho st s bin thin v v th ca hm s khi m = 1. Li gii: th3 2 1

Hm s y = x4 4x2 + 3 Bng bin thin

2

1 1

1

2

Cu I. 2) (1 im) Xc nh m th hm s cho ct trc honh ti 4 im phn bit A, B,C, D ln lt c honh x1 , x2 , x3 , x4 , (x1 < x2 < x3 < x4 ) sao cho tam gic ACK c din tch bng 4, vi K(3; 2). Li gii: x4 2(m + 1)x2 + 2m + 1 = 0 (1). t t = x2 ,t 0, ta c t 2 2(m + 1)t + 2m + 1 = 0 (2) th hm s ct trc honh ti 4 im phn bit th pt (2) c 2 nghim phn bit t > 0 = (m + 1)2 (2m + 1) > 0 m = 0 S = 2(m + 1) > 0 () Vi k () th th hm s ct trc honh 1 m > P = 2m + 1 > 0 2 ti 4 im phn bit c honh theo th t t1 , t2 , t2 , t1 , vi t1 > t2 1 Theo gt: SACK = AC.|yk | = 4 AC = t2 + t1 = 4 t1 + t2 + 2 t1t2 = 16 2 p dng nh l Vi-et cho phng trnh (2) ta c: m7 0 2(m + 1) + 2 2m + 1 = 16 m 7 = 2m + 1 m=4 m2 16m + 48 = 0 Cu II. 1) (1 im) 1 1 Gii phng trnh: 2 sin 2x = 4 sin x 1 . sin x 6 2 sin x Li gii: iu kin: sin x = 0. PT (4 sin x 2) sin 2x = 8 sin2 x 2 sin x 1 6 2 sin x 1 = 0 (1) 2(2 sin x 1) sin 2x = (2 sin x 1)(4 sin x + 1) 2 sin 2x = 4 sin x + 1 (2) 6 6 1 5 + k2 (1) sin x = x = + k2 hoc x = 2 6 6 (2) cos 2x 3 sin 2x = 4 sin x + 1 4 sin x + 2 sin2 x + 2 3 sin x cos x = 0 sin x + 3 cos x = 2 7 cos x = 1 x = + k2 6 6 Cu II. 2) (1 im) (x 2)(2y 1) = x3 + 20y 28 Gii h phng trnh: . 2( x + 2y + y) = x2 + x Li gii: 1

www.VNMATH.com PT hai ca h x + 2y + 2 x + 2y + 1 = x2 + 2x + 1 ( x + 2y + 1) = (x + 1)2 th x + 2y = x hoc x + 2y = x 2 x0 TH 1: x + 2y = x thay vo phng trnh th nht ta c 13x2 11x 30 = 0 2y = x2 x x + 2y = x 2

x+2 0 thay vo phng trnh th nht ta c pt bc 2 theo x 2y = x2 + x + 1 Cu III. (1 im) 2 5 cos x 4 sin x dx Tnh tch phn I= 7 0 (sin x + cos x) Li gii: 2 5 sin x 4 cos x t x = t ta c I = dx 7 2 0 (sin x + cos x) 2 5 cos x 4 sin x 2 5 sin x 4 cos x 2 dx dx + dx = Suy ra 2I = 7 7 6 0 (sin x + cos x) 0 (sin x + cos x) 0 (sin x + cos x) dx 1 2 1 2 2 1 + tan2 x d tan x = = 6 8 0 8 0 4 4 cos x 4 1 2 = 1 + 2 tan2 x + tan4 x d tan x 8 0 4 4 4 2 1 2 1 7 = tan x + tan3 x + tan5 x = 8 4 3 4 5 4 15 0 7 Vy I = 30 Cu IV. (1 im) Cho hnh lp phng ABCD.A B C D cnh a. Trn cc on AD , BD ln lt ly cc im M, N sao cho AM = DN = x, (0 < x < a 2). Tm x MN l on vung gc chung ca AD v BD. Li gii: Gi hnh v TH2: Cu V. (1 im) a2 + b2 + c2 . Cho 3 s a, b, c [0; 2] tho mn : a + b + c = 3. Tm gi tr ln nht ca M = ab + bc + ca Li gii: Cch 1: a2 + b2 + c2 9 +M = = 2 V th ta ch cn tm min ca A = ab + bc + ca ab + bc + ca ab + bc + ca 3 +ab + bc + ca = a2 + 3a + bc = f (a), f (a) = 0 a = 2 Vi a [0, 2] th a2 + 3a + bc t min khi a = 0 hoc a = 2 + Vi a = 0 b + c = 3, A = bc = b(3 b) = b2 + 3b khi b [1, 2], f (1) = 2, f (2) = 2 A 2 +Vi a = 2 A = 2 + bc, b + c = 1 A = 2 + b(1 b) = b2 + b + 2 = f (b) 9 5 b [0, 1] m f (0) = f (1) = 2 A 2 M 2 = 2 2 Du = t c khi a, b, c l cc hon v ca (0, 1, 2) Cch 2: Ta c: (a 2)(b 2)(c 2) 0 abc 2(ab + bc + ca) + 4(a + b + c) 8 0 abc + 4(a + b + c) 8 12 8 ab + bc + ca =2 2 2 9 9 5 A= 2 2 = ab + bc + ca 2 2 Du = xy ra (a; b; c) = (0; 1; 2) v cc hon v 2

www.VNMATH.comCu VIa. 1) (1 im) Cho ABC c phng trnh ca trung tuyn xut pht t A v ng cao k t B ln lt l: 2x 5y 1 = 0, x + 3y 4 = 0. ng thng BC i qua im K(4; 9). Lp phng trnh ng trn ngoi tip ABC, bit rng nh C nm trn ng thng d : x y 6 = 0. Li gii: Gi B(4 3b; b),C(c; c 6) ta c KB(3b; b + 9); KC(c 4; c + 3). 7k 9 27 5k K, B,C thng hng nn KB = kKC. T ta tnh c b = ,c = 4 4k 21k2 + 38k + 27 7k2 38k + 27 Gi M l trung im BC ta tnh c M ; 8k 8k V M thuc ng trung tuyn AM nn ta c ta M tha mn phng trnh AM. 77k2 + 258k 81 = 0 27 Gii ra ta c k = 3 hoc k = 77 Vit phng trnh AC tm A theo hai trng hp. Phn cn li l n gin cc bn t gii Cu VIa. 2) (1 im) x2 y1 z1 Trong khng gian vi h trc Oxyz, cho (P) : x + y z + 1 = 0, d : = = . Gi I l giao 1 1 3 im ca d v (P). Vit phng trnh ca ng thng nm trong (P), vung gc vi d v cch im I mt khong bng 3 2. Li gii: Cu VIIa. (1 im) z+i Cho s phc z sao cho: = 1. Tm cc s phc z tho mn iu kin: |z + 3i 2| = 4 z 3i Li gii: Cch 1: t z = a + bi vi a; b R; (z = 3i). Ta c: z+i |z + i| = = 1 |z + i|2 = |z 3i|2 a2 + (b + 1)2 = a2 + (b 3)2 b + 1 = (b 3) b = 1 z 3i |z 3i| V iu kin: |z + 3i 2| = 4 |z + 3i 2|2 = 16 (a 2)2 + 42 = 16 a = 2 Vy z = 2 + i (tho) Cch 2: Gi A, B, M, I biu din ba s phc z, i, 3i, 2 3i trn mt phng phc. 1 + 3 Khi y A, B thuc trc o v MA = MB nn M nm trn trung trc AB yM = = 1. 2 Mt khc IM = 4 nn M thuc ng trn (I; 4). Do khong cch t I n ng thng (d) : y = 1 bng 4 nn (d) tip xc vi (I) ti M. Suy ra IM Oy hay xM = xI = 2. Kt lun M(2; 1) hay z = 2 + i l s cn tm. Cu VIb. 1) (1 im) Trong mt phng vi h ta Oxy, cho tam gic ABC bit ng cao v trung tuyn xut pht t A ln lt c phng trnh: 6x 5y 7 = 0; x 4y + 2 = 0. Tnh din tch ABC, bit rng trng tm ca tam gic thuc trc honh v ng cao xut pht t nh B i qua im E(1; 4). Li gii: Ta c A(2; 1) Gi G(a; 0) v G thuc trung tuyn nn suy ra G(2; 0). Gi M l trung im BC ta c AG = 2GM. 1 suy ra M 4; Vit c BC : 5x + 6y + 23 = 0 suy ra B(1 + 6t; 3 5t);C(7 6t; 5t + 2). 2 19 V BE vung gc vi AC ta c iu kin l 61t 2 + 42t 19 = 0 t = 1 hoc t = 61 n y chia hai trng hp gii Cu VIb. 2) (1 im) 3

www.VNMATH.comx2 y2 z1 = = v mt cu 2 1 2 (S) : x2 + y2 + z2 + 4x 6y + m = 0. Xc nh cc gi tr ca m ng thng d ct mt cu (S) ti 2 im phn bit A, B sao cho MA = 5MB. Li gii: Trong khng gian to Oxyz, cho im M(2; 2; 1), ng thng d : Cu VIIb. (1 im) zi Cho s phc z tho mn: = 1. Tm s phc z sao cho z + 1 c mt acgumen bng . z + 3i 6 Li gii: Cch 1: t z = a + bi vi a; b R; (z = 3i) Ta c: zi |z i| = = 1 |z i|2 = |z + 3i|2 a2 + (b 1)2 = a2 + (b + 3)2 b 1 = (b + 3) b = 1 z + 3i |z + 3i| a+1 i + = (a + 1)2 + 1 cos + i sin z + 1 = (a + 1)2 + 1 6 6 (a + 1)2 + 1 (a + 1)2 + 1 a+1 > 0 a+1 > 0 a+1 3 = = cos a = 31 6 2 4(a + 1)2 = 3(a + 1)2 + 3 (a + 1)2 = 3 (a + 1)2 + 1 Vy z = 3 1 i (tho) Cch 2: Gi M, A, B biu din ba s phc z + 1, 1 + i, 1 3i trn mp phc. Khi y A(1; 1) v B(1; 3) v MA = MB nn M (d), vi (d) l trung trc ca AB, song song vi Oy nn M(x0 ; 1). 1 Li c tan = x0 = 3. x0 6 Suy ra M( 3; 1) z + 1 = 3 i z = 3 1 i.

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PH N CHUNG (7 i m) Cho t t c th sinh Cu I. (2 i m) Cho hm s y = x3 + (1 2m)x2 + (2 m)x + m + 2(1) m l tham s . 1 Kh o st s bi n thin v v th (C) c a hm s (1) v i m = 2. 2 Tm tham s m th c a hm s (1) c ti p tuy n t o v i ng th ng d : x + y + 7 = 0 gc , bi t 1 cos = . 26 Cu II. (2 i m) 1 Gi i h phng trnh: 2 Gi i phng trnh: x3 + 7y = (x + y)2 + x2 y + 7x + 4 3x2 + y2 + 8y + 4 = 8x 2cos2 x + 2 cos x 3 + 4 3 sin x = 0 x sin2 2 .

Cu III. (1 i m) 3

Tm tch phn

I= 6

dx sin3 x.cos5 x

Cu IV. (1 i m) Cho hnh chp S.ABC c y ABC l tam gic vung cn nh A, AB = a 2. G i I l trung i m c a BC, hnh chi u vung gc H c a S ln m t y (ABC) th a mn: IA = 2IH, gc gi a SC v m t y (ABC) b ng 60o . Hy tnh th tch kh i chp S.ABC v kho ng cch t trung i m K c a SB t i (SAH). Cu V. (1 i m) Cho x, y, zl ba s th c dng thay i v th mn: x2 + y2 + z2 = 1. a xy yz zx Tm gi tr nh nh t c a bi u th c: P= . + + 4 xy 4 yz 4 zx

PH N RING (3,0 i m) Th sinh ch lm m t trong hai ph n A ho c B Ph n A theo chng trnh chu n Cu VIa. (2 i m) 1 Trong h tr c t a Oxy cho hnh bnh hnh ABCD tm I. Bi t A(0; 1) v B(3; 4) thu c Parabol (P) : y = x2 2x + 1, I n m trn cung AB c a (P) sao cho tam gic IAB c di n tch l n nh t. Tnh to hai nh C v D. 2 Trong h to Oxyz cho tam gic ABC c B(1; 4; 3), phng trnh cc ng th ng ch a ng trung tuy n k x y1 z7 x1 y3 z4 t A v ng cao k t C l n l t l: (d1 ) : = = ; (d2 ) : = = . Tnh chu vi tam gic 1 1 2 2 1 1 ABC. Cu VIIa. (1 i m) Tm ph n th c v ph n o c a s ph c z bi t r ng |z|2 12 = 2i(3 z)

Ph n B theo chng trnh nng cao Cu VIb. (2 i m) 1 Trong h tr c t a Oxy cho tam gic ABC. Bi t A(4; 6), C 4 ; 2 v tm ng trn n i ti p tam gic ABC 3

2 8 l K ; . Tnh to nh B c a tam gic. 3 3 2 Trong h to Oxyz, cho tam gic ABC bi t phng trnh ng phn gic AD, trung tuy n AM l: x+1 y1 z3 x y1 z+3 (d1 ) : = = ; (d2 ) : = = v C(2; 0; 1). Tnh di n tch tam gic ABC 3 2 2 1 12 2 Cu VIIb. (1 i m) Trong t t c cc s ph c z = 6 th a mn w = z + 8i l m t s z6 o th s no c modun l n nh t ?

Tnh gi tr l n nh t ?


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Cu I. 1) (1 im) Cho hm s y = x3 + (1 2m)x2 + (2 m)x + m + 2 (1), m l tham s.Kho st s bin thin v v th (C) ca hm s (1) vi m = 2. Li gii: th4 3 2 1

Hm s y = x3 3x2 + 4 Bng bin thin

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2

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Cu I. 2) (1 im) Tm tham s m th ca hm s (1) c tip tuyn to vi ng thng d : x + y + 7 = 0 gc , 1 bit cos = . 26 Li gii: Cch 1: 2 *gi h s gc ca tip tuyn vi th (C) ti M(xo , yo ) dng k = f (xo ) = 3xo + 2xo (1 2m) + 2 m u1 .u2 1 = 6k2 13k + 6 = 0 k = 3/2 hay k = 2/3 *Tip tuyn to vi d gc |u1 ||u2 | 26 2 + 2x (1 2m) + 2 m = 3/2 3x2 + 2x (1 2m) + 1/2 m = 0 * Vi k = 3/2 th: 3xo o o o 0 m 1/2 hay m 1/4 *tng t: k = 2/3 m 1 hay m 3/4 Vy nhng gi tr ca m tho l: m 1/2 hay m 1/4 Cch 2: Gi vct php tuyn l = (a, b) iu kin l : a2 + b2 = 0 Theo bi ra ta c : n |a + b| 1 3a + 2b = 0(1) = 6a2 + 13a + 6b2 = 0 (3a + 2b)(2a + 3b) = 0 2a + 3b = 0(2) 26 a2 + b2 2 Ta xt cc trng hp sau : 2 (1) : a = 2, b = 3. Phng trnh tip tuyn c dng : y = x + 3 2 Thay o hm vi h s gc ny ta c : 3x2 + 2(1 2m)x + 2 m = 3 3 m Cn 0 4m2 m 3 0 4 m1 Lm tng t vi trng hp sau ta c : 3 (2) : a = 3, b = 2. Phng trnh tip tuyn c dng : y = x + 2 2 + 2(1 2m)x + 2 m = 3 Thay o hm vi h s gc ny ta c : 3x 2 1 m 4 Cn 0 8m2 2m 1 0 1 m 2 1

www.VNMATH.com1 m 4 Kt hp bng cch ly hp ca hai h ta c nhng gi tr ca m tho l: 1 m 2 Cu II. 1) (1 im) x3 + 7y = (x + y)2 + x2 y + 7x + 4 Gii h phng trnh: . 3x2 + y2 + 8y + 4 = 8x Li gii: Cch 1: x2 (x y) = (x + y)2 + 7(x y) + 4 (1) Bin i h phng trnh mt cht nh sau : 4 = 3x2 y2 + 8(x y) (2) 2 (x y) = (x + y)2 + 7(x y) 3x2 y2 + 8(x y) Thc hin php th (2) vo (1) ta c : x x2 (x y) = 2x2 + 2xy + 15(x y) x2 (x y) = 2x(x y) + 15(x y) (x y)(x2 + 2x 15) = 0 Trng hp 1: x = y Thay vo phng trnh (2) c ngay : 4x2 + 4 = 0 . Phng trnh ny v nghim ! y = 1 x=3 y2 + 8y + 7 = 0 y = 7 Trng hp 2: x2 + 2x 15 = 0 2 + 8y + 119 = 0 (v nghim) x = 5 y Vy h cho c cc nghim sau : (3, 1), (3, 7) Cch 2: (x2 7)(x y) x2 2xy (y2 + 4) = 0 (1) HPT 3x2 + (y2 + 4) 8(x y) = 0 (2) 2 7)(x y) + 2x(x y) 8(x y) = 0 (x y)(x2 + 2x 15) = 0 Thc hin php th (x x=y tng t cch 1. x2 + 2x 15 = 0 Cu II. 2) (1 im) 2cos2 x + 2 cos x 3 + 4 3 sin x = 0 Gii phng trnh: x sin2 2 Li gii: x K: sin = 0 x = k2 . PT 2 cos2 x + 2 cos x 3 = 2 3 sin x(1 cos x) = 0 2 cos2 x 3 sin x. cos x 3 sin2 + 2 cos x + 2 3 sin x = 2 x 0 (cos x + 3 sin x)2 + 2(cos x + 2 3 sin x) = 0 (cos x + 3 sin x)(cos x + 3 sin x 2) = 0 cos x + 3 sin x = 0 cos(x ) = 0 x = + k x = 23 + k 6 6 2 cos(x ) = 1 x = k2 x = + k2 cos x + 3 sin x = 2 6 6 6 KT k ta nhn c cc nghim ny. Cu III. (1 im) 3

Tm tch phn Li gii: I= 6 6 6

I= 6

dx sin x.cos5 x3 6

dx = 3 x. cos8 x tan3

6

(tan2 x + 1)3 d(tan x) = tan3 x

6 6

1 tan x + tan x

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d(tan x)

6 1 1 1 4 1 3 = tan x + 3 + 3 tan x + 3 d(tan x) = tan x + tan2 x + 3 ln | tan x| = 2x tan x tan x 4 2 tan 2 6 6 tnh tip nhe. Cu IV. (1 im) Cho hnh chp S.ABC c y ABC l tam gic vung cn nh A, AB = a 2. Gi I l trung im ca BC, hnh chiu vung gc H ca S ln mt y (ABC) tha mn: IA = 2IH, gc gia SC v mt y (ABC) bng 60o . Hy tnh th tch khi chp S.ABC v khong cch t trung im K ca SB ti (SAH). Li gii:

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AB IA a Ta c: IB = IA = IC = = a; IH = = 2 2 2 a 5 HC = IH 2 + IC2 = 2 a 15 SH = HC. tan 60o = 2 1 1 VS.ABC = SH.SABC = SH.AB2 3 6 1 a 15 2 a3 15 = . .2a = 6 2 6 Gi E trung im SI; KE l ng trung bnh SBI BI a KE BI; KE = = 2 2 BI AH; BI SH BI (SAH) a KE (SAH) d(K; (SAH)) = KE = 2

K

E

H I B

60o C

A

Cu V. (1 im) Cho x, y, z l ba s thc dng thay i v tha mn: x2 + y2 + z2 3. = xy yz zx Tm gi tr ln nht ca biu thc: P= . + + 4 xy 4 yz 4 zx Li gii: Cu VIa. 1) (1 im) Trong h trc ta Oxy cho hnh bnh hnh ABCD tm I. Bit A(0; 1) v B(3; 4) thuc Parabol (P) : y = x2 2x + 1, I nm trn cung AB ca (P) sao cho tam gic IAB c din tch ln nht. Tnh to hai nh C v D. Li gii: Cu VIa. 2) (1 im) Trong h to Oxyz cho tam gic ABC c B(1; 4; 3), phng trnh cc ng thng cha ng trung tuyn x1 y3 z4 x y1 z7 = ; (d2 ) : = = . k t A v ng cao k t C ln lt l: (d1 ) : = 1 1 2 2 1 1 Tnh chu vi tam gic ABC. Li gii: Cu VIIa. (1 im) Tm phn thc v phn o ca s phc z bit rng |z|2 12 = 2i(3 z) Li gii: Gi z = a + bi vi a; b R Ta c: a2 + b2 12 = 2i(3 a bi) = 2(3 a)i + 2b a=3 a=3 a=3 2 + b2 12 = 2b 2 2b 3 = 0 a b b = 1hay b = 3 Vy c 2 s phc z = 3 i, z = 3 + 3i tha bi. S z = 3 i c phn thc l 3, phn o l 1. S z = 3 + 3i c phn thc l 3, phn o l 3 Cu VIb. 1) (1 im) 4 Trong h trc ta Oxy cho tam gic ABC. Bit A(4; 6), C ; 2 v tm ng trn ni tip tam gic 3 2 8 ABC l K ; . Tnh to nh B ca tam gic. 3 3 Li gii: 3

www.VNMATH.comCu VIb. 2) (1 im) Trong h to Oxyz, cho tam gic ABC bit phng trnh ng phn gic AD, trung tuyn AM l: x+1 y1 z3 x y1 z+3 (d1 ) : = = ; (d2 ) : = = v C(2; 0; 1). Tnh din tch tam gic ABC 3 2 2 1 1 2 Li gii: Cu VIIb. (1 im) z + 8i Trong tt c cc s phc z = 6 tha mn w = l mt s o th s no c modun ln nht ? Tnh gi tr z6 ln nht ? Li gii: Cch 1: t: z = x + iy Khi : x + (y + 8)i (x + (y + 8)i)((x 6) yi) x(x 6) + (y + 8) (y + 8)(x y 6)i = = + = (x 6) + yi (x 6)2 + y2 (x 6)2 + y2 (x 6)2 + y2 x(x 6) + (y + 8)y l s o = 0 x(x 6) + (y + 8)y = 0 (x 3)2 + (y + 4)2 = 25 () 2 + y2 (x 6) T () suy ra cc im M biu din cc s phc cho nm trn ng trn tm I = (3; 4) bn knh R = 5. 4 x Ta c |z|max M () sao cho OMmax ng thng ni hai im 0 I c phng trnh y = 3 y = 4 x (1) Vy tm im M ta xt h phng trnh sau: 3 (x 3)2 + (y + 4)2 = 25 () D thy h ny c hai nghim: Vi x = 0 y = 0 z = 0 Vi x = 6 y = 8 z = 6 8i Vy z = 6 8i Cch 2: z + 8i z 8i Gi thit w + w = 0 + = 0. |z|2 3(z + z) + 4i(z z) = 0 |z + 4i 3| = 5. z6 z6 Mt khc 5 = |z + 4i 3| |z| |3 4i| hay |z| 10. ng thc xy ra khi v ch khi z + 4i 3 = k(3 4i) |(k + 1)(3 4i)| = 10 k = 1 z = 6 8i. |z| = 10 Th (1) vo () ta c (x 3)2 + 16 (x 3)2 = 25 (x 3)2 = 9 x = 6 hoc x = 0 9

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