GIỚI THIỆU VÀ HƯỚNG DẪN GIẢI CHI TIẾT ĐỀ THI TUYỂN SINH ĐẠI HỌC KHỐI D -...

8/10/2019 GII THIU V HNG DN GII CHI TIT THI TUYN SINH I HC KHI D - L HNG C - KIM HO - V

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ThS. L HNG C KIM HOVNH B

G(( THIU V HNG DN

GII CHI TIT THI TUYN SINH I HC

KHI D( T i b i n c s a c h a, b s u n s J

r i H X U T B N I H C Q U C Q I A H N I

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W.DAYKEMQUYNHON.UCOZ.COM

g gp PDF bi GV. Nguyn Thanh T


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NH XUT BN I HC Q u c GIA H Ni16 Hng Chui - Hai B Trng - H Ni

in toi: Bin tp - Ch bn: (04) .9714896;Hnh chnh:(04) 39714899; Tng bin tp:(04) 39714897

Fax: (04) 9714899

Chu trch nhim xu t bn:

Gim c: P H N G Qc BO

Tng bin tp: PHM TH TRM

Bin tp: TH TRANG - KIM CH I

Ch bn: N h sch HNG N

Trnh by ba: N h sch HNG N

Thc hin l in kt:Nh s ch HNG N

SCH LIN KT

ClTHIU VHNG DN Git CHI TIT THI TUYNs i m >fl HC - KHI DM s: 2L- 233H2011In 1.0.00 cun, kh 16 X 24cm ti Cng ti In Vn Lang.Giy php xut bn s: 285-2011/CXB/27-31/HQGHN, ngy 21/3/2011.Quyt nh xut bn s: 223LK-XH//Q-NXBHQGHN.

In xong v np u chiu qu III nm 2011.



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Li ni u

Cc em hc sinh thn mn

Nhm gip cc em c thm ti liu b ch trong vic n luyn th i vo cc

trng H, C, chng ti bin son cun sch-"Gii thiu v hng dn

gii ch tit thi tuyn sinh i h c khi D".

Ni dung sch gm :

Phn : Gii thiu thi mn Ton cc khi A, B, D t nm 2006 n nay.

Phn II: Gii thiu thi mn Vn cc khi c, D t nm 2006 n nay.

Phn III: Gii thiu thi mn Ting Anh khi D t nm 2006 n nay.

Gii thiu th nm 2011

Trong ln ti bn ny, chng ti b sung thi chnh thc nm 201 , cc

thi nhng nm trc cng c chun ha theo cu trc mi ca

B GD & T. Hi vng cun sch s tht s c ch cho cc em trong vic n

luyn chun b cho cc k thi sp ti.

D c gng nhiu trong bin son, song chc khng trnh khi nhng

thiu st. Xin chn thnh cm n v mong nhn c cc kin ng gp ca

cc em hc sinh v qu ng nghip cun sch ngy mt hon thin hon.



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A. PHN TON T H I M N T O N K H I A N M 2 00 6

PHN CHUNG CHO TT C CC TH SINHCu I: (2 im)

1. Kho st s bin thin v v th hm s y = 2x3 - 9x2 + 12x - 4.2. Tim m phng trnh 2 X13 - 9x2 + 12 XI = m c 6 nghim phn bit.Cu II: (2 im

1. Gii phng trnh:2(cos6 X + sin x) -sin x .co sx

= 0 .

2. Gii h phng trnh:

-s/z-2sinx

x + y - f i y =3

yjx + 1 +yjy+ 1= 4

Cu III: (2 im): Trong khng gian vi h to Oxyz, cho hnh lp phcmg

ABCD.ABCD vi A(0; 0; 0), B(l; 0; 0), D(0; i; 0), A(0; 0; 1). Gi M v N lnlt l trung im ca AB v CD.1. Tnh khong cch gia hai ng thng AC v MN.2. Vit phng trnh mt phng cha AC v to vi mt phng Oxy mt gc a

_ bit cos a = V6

Cu IV: (2 im)

sin2x.x1. Tnh tch

71/2

phn: I - I ,----

0V

COSICOS2X+ 4 sin2 X2. Cho hai s thc X. *0, y ^ 0 thay i v tho mn iu kin(x + y) xy = X2 + y 2 - xy.

Tm gi tri ln nht ca biu thc A =X y

PHN T CHNTh sinh chn cu v .a hoc cu v.bCu v.a Theo chng trnh THPT khng phn ban (2 im)1. Trong mt phng vi h to Oxy, cho cc ng thng:

(dj): X+ y + 3 = 0; (d2): X- y - 4 = 0, (d ): X- 2 = 0.Tm im M nm trn ng thng (dO sao cho khong cch t M n ngthng (d,) bng hai ln khong cch t M n ng thng (2).

/ 1 7V2. Tnh h s ca X26 trong khai trin nh thc Niutn - + X I , bit rng:

c 1 + + + c n_ = 220 1^2n+l ^2n+I ^2n+i ~(n nguyn dng, l t hp chp k ca n piin t).



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"u v.b Theo chng trnh THPT phn ban (2 im)

. Gii phng trnh 3.8X+ 4.12* - 18* 2.27* = 0.Cho hnh tr c cc y l hai hnh trn tm o v O, bn knh y bng chiu ca v

bng a. Trn ng trn y tm o ly im A, trn ng trn y tm O ly imB sao cho AB = 2a. Tnh th tch khi t din OO AB.

Phng trnh ban u c 6 nghim phn bit khi (1) c 6 nghim phn bit, tc khi ng thng y = m - 4 ct th hm s'

th hm s y = 2 I X I? - 9x2 + 12 IX - 4 c suy ra t th hm s trong;u (k hiu l (C)) bng cch:

Gi nguyn phn th ca (C) bn phi O. Ly i xng phn th trn qua Oy.T th suy ra iu kin phng trnh c 6 nghim phn bit l:

^u I.. Bn c t lm V/ v rh.

. Bin i phng trnh v dng:

2 IX13 - 9x2 + 1 2 1XI - 4 = m - 4. ( 1 )

- 9x2 + 121X1 - 4 ti 6 im phn bit.

0 < m - 4 < l 4


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t t =yx (t > 0 ), khi :

Bin i phng trnh th nht ca h v dng:

x + y - t = 3 o x + y = t + 3. Bin i phng trnh th hai ca h v dng:

x + 1+ y + I+ 2 >/(x + l)(y + l) = 16

X+ y + 2 + 2yjxy +X+ y --1 = 16

C3-i + 3 + 2 + 2\/t2 +1 + 3+ 1 = 16 2>/t2 + 1 + 4 =11 t

| l - t > 0o S

14(t2 + t + 4) = (11 - 1)2

t < 11

3t2 + 2 6 t - 105 = 0 t - 3.

H phng trnh c dng:X+ y = 6

xy = 9

tc X, y l nghim ca phng trnh:u2- 6 u + 9 = 0c3 -u = 3= > x = y = 3 tho mn iu kin (*).Vy, h c nghim (3; 3).

Cu III* T gi thit suy ra:

C(l; 1 ; 0), 0; o j , N ^ - ; 1; 0 j.

1. Khong cch gia hai ng thng AC v MNc cho bi:

d(AC, MN) = [a C ,m n ] .a 'M

|[ a ' c , m n ]

trong :

Tccl; 1; - 1 ) , MN(0; 1; 0 ), A m - ; 0; - 1j , [ a 7c ,M N ] = (1; 0; 1)

suy ra:

d(AC, MN) =

1. + 0.0 + 1.(1)2 12J ------ = ~7=~(yjl 2 +0 + 12

(vd).

2. Gi s mt phng (P) c phng trnh:(P): Ax + By + Cz + D = 0, vi A2+ B2 + c 2 > 0.

V (P) i qua hai im A \ c nn:

C + D = 0

(I)

c = D = A 4- B.

7



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Gi rip , k theo th t vtpt ca cc mt phng (P), (Oxy), ta c:n^(A;B; C), k (0; 0; 1)

T gi thit, ta c:

nP.kcos a =

ta c:

J _ |%-k| _ 1 c _ 1-l

np.k~~y/6.

Va2+ b 2

+ c2't

+- B)2 = A2+ B2 + (A + B)2

__, _ _ _ "B = -2A+5AB + 2B2 = 0 (2A + B)(A + 2B) = 0

A = -2 B

o 6(A + B)2= A + Bz + (A + B)

o 2A2+ 5AB + 2B2 = 0 (2A + B)(A + 2B) = 0

Khi :

Vi B = - 2 A th c = -A , = A, thay vo (1) ta c:(P ): A x - 2y - Az + A = 0 o (P,): X - 2y - z + 1 = 0.

Vi A = -2 B th c = -B , D - B, thay vo (1) ta c:

(P2): -2Bx + B y -B z + B = 0 (P2): 2x - y + z 1 = 0.Vy, tn ti hai mt phng (Pt) v (P2) tho mn iu kin u bi.

Cu IV.1. t t = cos2x + 4sin2x, suy ra

dt = (-2s inx.cosx + 8cosx.sinx)dx = 3sin2x.dx sin2x .dx = dt.

i cn: Vi X= 0 th t = 1.

Vi X= th t =4.2

T. ' r 1 dt . 2 rKh : = p- = vt

3 S x 3

^ . . . .2 .2 ^ 1 1 1 1 12. T gi thit: (x + y)xy =X +y ^ xy += ------X y X2 y . xy

Khi d, bng cch t a = , b = ta nhn c iu kin:X y

a + b a + b = a2+ b2- ab a + b = (a + b)2- 3ab > (a + b) - 3

o ( a + b)2 - 4(a + b ) S 0 o 0 a + b 4.

T : A = - r- + -4r- = a3+ b? = (a + b)(a2- ab + b2) = (a + b)2^ 16.X .y

Vy, ta c AMax = 16 t c khi:



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Cu v.a.1. V M thuc (do nn M(2t; t), khi khong cch t M n ng thng (d)

bng hai ln khong cch t M n ng thng (d2) iu kin l:

|2t + t + 3| 2 \ 2 t ~ t - 4 \ , ,===== - = = 3t + 3 = 2 t 4 2 + ( - i ) 2

3t + 3 = 2 ( t - 4 ) I t = 11

3t + 3 = 2(4 - 1) ^ L* =1

M j(-22; -11)

'_M2 (2; 1)

Vy, tn ti hai im Mj, M2tho mn iu kin u bi.

2. Ta c: c | n+1 =C2_k, Vk, 0 < k < 2n + 1

suy ra: 220 = 1 + C 2n+] + C2n+1 +...+ C5n+)

~ C 2n+1 + c 2n+l + + C 2n+i = T ( C 2n+1 + C 2n+'1 + + c in + ! )

s dng khai trin: c ^ n+1 + q , n+1 +... + C |^ = (1 + 1)

ta c: 22 0) ta c phng trnh:

3t* + 4t2 - t - 2 = 0 (t + l)(3t2+ t - 2) = 0 t = 1j -

X = 1.

Vy phng trnh c nghim X = 1.2. Bn c t v hnh.

Ta ln lt thc hin: K ng sinh AV Gi D l im i xng vi A qua O v H l hnh chiu ca B trn AD.



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9/394

Ta c: j BH J- A D => BH -L (AOO>.[BH_LA'A

Khi : V00 A( = BH.Sa a q q - = BH.AO.OO .3 o

rong : AO = OO = a.

BD2= AD2- AB2= AD2- (AB2- AA2) = 4a2 - (4a2- a2) = a2a

BD = a => AOBD u =?> BH =

( 1 )

(2)

(3)

Thay (2), (3) vo (1), ta c: V00-AB1 ax/3 a3V6 2

,a.a = 12

(vtt)

T H I M N T O N K H I B N M 2 0 06

PHN CHUNG CHO TT C CC TH SINHCu I: (2 im):Cho hm s:

( C ) : y - i L p A .X+ 2

1. Kho st s bin thin v v th (C) ca hm s.2. Vit phng trnh tip tuyn ca th (C), bit tip tuyn vung gc vi

im cn xin ca (C).Cu 11:(2 (m)

1. Gii phng trnh: cot X + sin X 1+ tan X. tan ~ j = 4.

2. Tm m phng trnh \ l \2+ mx 4- 2 = 2x + 1 c hai nghim phn bit.Cu III: (2 m)\Trong khng gian vi h to Oxyz, cho im A(0; 1; 2) v haiio thng:

fx = 1+ 1M . \ = T z } l - _ V i V = - 1 ( = K( d |) : 2 1 (d2): y = - l - 2 t , t s R .

z = 2 +t

1. Vit phng trnh mt phng (p) qua A, ng thi song song vi , (2X

2. Tim io cc im M thuc (j), N thuc (d2) sao cho ba im2. Tim io cc im M thuc (j), N thuc (d2) sao cho ba im A, M, Nthng hng.Cu IV: (2 im)

n 5

. Tnh tch phn 1= Jdx.

In 3 e + z e - 3

2. Cho hai s thc X, y thay i. Tim gi tr nh nht ca biu thc:

A = %/ ( x - l ) 2 +y 2 +yj(x + l)2+ y2 + |y - 2 |.

10



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PHN T CHNTh sinh chn cu v .a hoc cu v.b

Cu v .a Theo chng trnh THPT khng phn ban (2 im)

1. Trong mt phng vi h to Oxy, cho ng trn:

(C): X2 + y2- 2x - 6y + 6 = 0v im M (-3 ; 1). Gi T1v T2l cc tip im ca cc tip tuyn k t M n

(C). Vit phng trnh ng thng TjT2.2. Cho tp hp A gm n phn t (n > 4). Bit rng, s tp con gm 4 phn t ca

A bng 20 ln s tp con gm 2 phn t ca A. Tm k


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yVy, vi m>-~ tho mn iu kin u bi.

2

Cu III. Chuyn phng trnh ng thng (d|) v dng tham s:

X = 2u

( |) :


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Khi :

1=t - 2

t - 2= ln .

2

2. Trong mt phng \'i h o Oxy, xt cc im M(x - 1 ; -y ), N(x + 1 ; y), ta c:

7 (x - )2 + y 2 + V(x + l)2 + y2 = OM + ON > MN= y T

T , s u y ra: A >y j4 + 4y 2 + y-2 .

Xt hm s f (y) = V4 + 4y2 + y - 2| theo hai trng hp:

Trn %h-> I\Vi y < 2 th:

f(y) = 2^/l + y2 - y + 2 , f (y) = -f=^ = = -1 7V I + y 2

2yf(y) = 0 - 1 = 0

V1+ yJ 2y >0o ^ l + y =2y o 4X+ 144 < 20.2* + 80. - (*)t t - 2X(t > 0), ta c:

t2 - 20t + 64 < 0 4 < t < 16 4 < 2X< 16 2 < X< 4.

Vy, bt phng trnh c nghim 2 < X< 4.2. Bn c t v hnh. '

a. Chng minh mt phng (SAC) vung gc vi mt phng (SMB).

Xt hai tam gic vung AABM v ABCA, ta c:

ADAM 2 1 BA K ' . ,

= - = => AABM v ABCA ng dngAB AB 4 BC *

=> BM = BCA => BM + BC = BCA + BAC = 90

=> B = 90 => MB AC. (1)Mt khc, ta c:

SA JL (ABCD) => SA MB. (2)

15



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T (1) v (2)'suy ra:MB (SAC) => (SMB) _L(SAC), pcm.

b. Tnh th tch ca khi t din ANIB.Gi H l trung im AC, ta c:

NH / / SA (tb) => NH _L(ABI) v NH = - SA = -

do : VAN]B = ^ N H S J | = NH .A1.BI.5 oTrong AABM, ta c:

1 1 1 = J_

AI 1>/'= AI =

a ^3

(3)

Trong AABI, ta c:

BI2= AB2 - AI2 = a2 -

2a2

3 BI =

Thay (4), (5) vo (3), ta c: VANIB = .1 a aV5 aV _ a34.

(4)

(5)

6 2 3(vtt)

THI MN TON KHI D NM 2006

PHN CHUNG CHO TT CA CC TH SINHCu I: (2 im)'.Cho hm s:

(C): y = X-' - 3x + 2.L. Kho st s bin thin v v th (C) ca hm s.2. Gi (d) l ng thng i qua im A(3; 20) v c h s gc l m. Tm m

ng thng (d) ct th (C) ti ba im phn bit.Cu II: (2 in )1. Gii phng nh:

cos3x + cos2x - cosx - 1 = 0 .2. Gii phng trnh:

>/2x - 1 + X2 - 3x +1 = 0, X IR.Cu ni : (2 im):Trong khng gian vi h to Oxyz, cho im A(l; 2; 3) v hng thng:

X- 2 y + 2 _ z - 3 ^ x - 1 _ y - 1 _ z + 1~2 1 2 -12 -1 1 ' *' -1 2 1

1. Tim to im A i xng vi im A qua ng thng (d,).2. Vit phng trnh ng thng (d) i qua A, vung gc vi C|) v ct (d2).

16



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16/394

Cu IV: (2 im)]

1. Tnh tch phn I = | (x - 2)e2x .dx.0

2. Chng minh rng vi mi a > 0, h phng trnh sau c nghim duy nht:

e* - ey ~ ln(l + x) - ln(l + y)

y - x = a

PHN T CHNTh sinh chn cu v.a hoc cu v .bCu v.a Theo chng trnh THPT khng phn ban (2 im)

1. Trong mt phng vi h to Oxy, cho ng trn (Q v ng thng (d):

(C): X2 + y2 - 2x - 2y + 1 = 0, (d): X - y + 3 = 0.Tm to im M trn (d) sao cho ng trn tm M, c bn knh gp i bnknh ng trn (C) v tip xc ngoi vi ng trn (C).

2. Mt i thanh nin xung kch ca mt trng ph thng c 2 hc sinh, gm 5em hc sinh lp A, 4 em hc sinh p B v 3 em hc sinh lp c. Cn chn 4 hcsinh i m nhim v, sao cho 4 hc snh ny thuc khng qu 2 trong 3 lptrn. Hi c bao nhiu cch chn nh vy ?

Cu v.b Theo chng trnh THPT phn ban (2 im)

1. Gii phng trnh 2X +x - 4.2*2~x - 22* + 4 = 0.2. Cho hnh chp tam gic S.ABC c y ABC l tam gic u cnh a, SA = 2a v

vung gc vi mt phng (ABC). Gi M v N ln lt hnh chiu vung scca A trn cc ng thng SB v SG Tnh th tch ca khi chp A.BCMN.

Cu I.1. Bn c t lm.

2. ng thng (d) c phng trnh (d): y = m(x - 3) + 20.Phng trnh honh giao im ca (d) vi th (C) l:

x? - 3x + 2 = m(x - 3) + 20 (x - 3)(x2 + 3x - m + 6) = 0

X = 3

_f(x) = X + 3x - m + 6 = 0 (*)

ng thng () ct th (C) ti ba im phn bit khi:Phng nh (*) c hai nghim phn bit khc 3

0 0

32 + 3.3 m + 6 &0

4m -1 5 > 0

24 - m 0

17



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17/394

Vy, vi < m &24 tha mn iu kin u bi.4

Cu II.1. Bin i phng trnh v dng:

(cos3x cosx) - (1 - cos2x) = 0C3* -2sin2x .sinx - 2sin2x - 0 2sinx(sin2x + sinx) = 0

2sinx(2sinx.cosx + sinx) = 0 2sin2x(2cosx + 1) = 0

sin X = 0

2 c o s x + 1 = 0

sin X = 01 o

cosx = ----2

X = k n

2n _k ^ z -X = - + 2C7C

3Vy, phng trnh c ba h nghim.

2. Ta c th trnh by theo cc cch sau:Cch I :Bin i phng trnh v dng:

- X 2 + 3x - 1 > 0J2x -1 = -X 2 + 3x - 1

'2x -1 = ( - X 2 + 3 x - l ) 2

3x + 1< 0

X4 - 6x3 + 1 Ix2 ~ 8x + 2 = 0

X2 - 3x +1 < 0

i ( x - l ) ( x 3 - 5 x 2 + 6 x - 2 ) = 0

c=>

X2 - 3x + .1 < 0

( x - l ) 2( x2 - 4 x + 2 ) ~ 0

'x = lX = 2 - \2

X2 - 3x +1 < 0

X1 0 i

X2 - 4x + 2 = 0

X2 - 3x +1 < 0

X= 1

X = 2 V 2

Vy, phng trnh c hai nghim X = 1 v X - 2 - V2 .

Cch 2:t t =y /2 \~ - (t > 0), suy ra X =

Khi , phng trnh c chuyn v dng:

t r t + ) _ 3_ 1 + 1 = 0 => t4 - 4t2 + 4t - 1 = 0

c * ( t - l )2(t2+ 2t l) = 0 o

J 2 x - 1 = 1

j 2 x - { = 7 2 - 1

t -1 =0

t2 + 2t - 1= 0

X= 1

x ~ 2 y / '

t0

t = 1

t = y / 2 - l

Vy, phng trnh c hai nghim X= 1 v X- 2 - \2 .

18



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18/394

Cu III. Hai ng thng (d), (d2) theo th t c vtcp l

7 (2; -1; 1), ^(-l;2; 1).

1- Ta c th trnh by theo cc cch sau:Cch : Chuyn phng trnh ng thng (di) v dng tham s:

X = 2 + 2t

(d j): H(0; -1 ; 2).V H trung im ca AA nn A( - l ; -4 ; 1).Cch 2:Gi (P) l mt phng tho mn:

Qua A Qua A(l; 2; 20(P): 4 (P): - (P): 2 x - y + z - 3 = 0.

[(P)(dj) IvtpfUjCfc-l;!).

Gi H l hnh chiu vung gc ca A ln ng thng (di), ta c {H} = (d) n (P)nn to H l nghim h phng trnh:

X- 2 y + 2 Z -3 x + 2y = -2 o

IIX

2 -1 1 - y = -1 => H(; -1; 2)

2 x - y + z - 3 - 0 2 x - y + z = 3 z = 2

V H l trung im ca AA nn A (1; -4 ; 1).Cch 3:Gi s im A(x; y; z), suy ra:

Trung im H ca AA thuc (d| )

[AAJL(d,) 0 ,

(I + x)(l + x + a)

=> Hm s y = f(x) nghch bin trn D.Mt khc, ta ln lt c:

1im f(x) = e_I - ea- - (-co) + lna = +00,X 1+

lim f(x) = limX > + c o X >+00

ex(l - ed) + In1 + X+ a

1 + x

= e+00( l - e a) + lnl =- oo .

T , suy ra vi a > 0 phcmg trnh (*) lun c nghim duy nht, tc l h c

nghim duy nht.Cu v.a.1. ng trn (C) c tm 1(1; 1) v bn knh R = 1.

im M thuc (d) nn c to M(x; X + 3).

Khi , iu kin K cabi ton l:

IM = 2R + R = 3R IM2= 9 (x - i) 2+ (x + 2)2= 9

x 2 + x - 2 = 0 X

X - -2M,(l; 4)

l M2(-2; 1)'

Vy, tn ti hai im Mi, Mztho mn iu kin u bi.

2. Ta i thc hin bi ton ngc:

a. Mi cch chn 4 hc sinh i m nhim v t ba lp A, B, c ng vi mt thp chp 4 c 12 phn t, tc s cch chn bng:

cfj=495.

b. S cch chn 4 hc sinh i lm nhim v t ba lp A, B, c m mi lp c tnht mt em c tnh nh sau:

Lp A c 2 hc sinh, cc lp B, c mi p c 1 hc sinh th s cchchn bng:

c^ .c i .d =120.

Lp B c 2 hc sinh, cc lp A, c mi p c 1 hc sinh th s cchchn bng:

c . c . c = 90.

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21/394

Lp c c 2 hc sinh, cc lp A, B mi lp c hc sinh th s cchchn bng:

C 3 .C5 .C4 - 60.

Vy, s cch chn 4 hc sinh i lm nhim v t ba lp A, B, c m mi lp c tnht mt em bng: 120 + 90 -f- 60 = 270.

Khi , s cch chn phi tm l 495 - 270 = 225.Cu v.b. . Bin i phcmg trnh v dng:

.2 ... ^ ( .. \ ..2 .. ^ ( 2,x +K 2ZX 1-1 4.2X"X- 4 1= 0 o 22x 2x x - 1 I~4 | 2X' X-1 I= 0

(22x - 4 ) | V 2~x - l j1 1= 0 22* = 4 2x = 2 "X=

, 2X "x =1 X - X= 0 X= 0

Vy, phng trinh c nghim X = 0 v X = 1.2. Bn c v hnh,Gi K l trung im BC v H l hnh chiu vung gc ca A trn SK, ta c:

BC -1AK BC (SAK) => BC AH=> AHJ_(SBC)B C 1 S A

1A.BCNM- ^AH.Sbcnmdo : V

Trong ASAK, ta c

( 1 )

g ASAK, ta c:

I _ 1 1 1 1 X9AH2 AK2 + AS2 _ ayf +4a2 ~~ 12a:

r*\AH = a.

Ta c:

Ssmn _ SM SN ( S M \

>ASBC SB ' SC VSB J

SA

VSB J

4a

(2)

2 \ 2

4a2 + a 2

16

25

_ o 9 o . 9a2^/9=>S bc nm - Sasbc- - ^ -

Thay (2), (3) vo (1), ta c:

V = i /9a 2V9 _ 3a3%/3A.BCNM 3*a^j9* 100~ 50

(3)

(vtt).

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22/394

[ t h i m n t o n k h i A N M 2 0 0 7

PHN CHNG CHO TT C CC TH SINH

Cu I: (2 im):Cho hm s:

X2 +2( m + l)x + m2 +4 m __.y = --------------- ----------------------------, m l th am s.(1)

x + 2

1. Kho st s bin thin v v .th hm s (1) khi m = -1 .2. Tm m hm s (1) c cc i v cc tiu, ng thi cc im cc tr ca

th cng vi gc to o to thnh mt tam gic vung ti o.Cu II: (2 im)1. Gii phng trnh:

2. Tm ra phng trnh 3V x 1 + mVx" + 1 = 2 C nghim thc.Cu III: (2 im): Trong khng gian vi h to Oxyz, cho im A(, 1; 2) v hai

ng thng: '

. Chng minh rng Cdx) v (d2) cho nhau.2. Vit phng trnh ng thng (d) vung gc vi mt phng(P) c phng

trnh 7x + y - 4z = 0 v ct hai ng thng (dj), (d2).Cu IV: (2 im)

1. Tnh din ch hnh phng gii hn bi cc ng y = (e + 1 )x, y = (1 + ex)x.2. Cho ba s thc dng X, y, 2thay i v tho mn iu kin xyz = 1. Tmgitr nh nht ca biu thc:

PHN T CHNT h sinh chn cu v .a hoc cu v .b

Cu v.a Theo chng trnh THPT khng phn ban (2 im)1. Trong mt phng vi h to Oxy, cho AABC c A(0; 2), B(-2; -2), C(4; -2).

Gi H l chn ng cao k t B, M v N ln lt l trung im ca cc cnh ABv BG Vit phng trnh ng trn i qua cc im H, M, N.

2. Chng minh rng:

(l + sin2x)c o sx + (l + cos2x)s in x = 1+ sin2x .

X = - 1 4 -

y = l + t2 = 3

1+ 2t

t R.

A = x 2( + z) t y2(2+ x) | z2(x + y)y \ f + 2zVz zVz + 2xV x xyfx + 2y-J

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23/394


24/394

2. iu kin X>1 .

Chia hai v ca phung trnh cho V X + 1 , ta c:

Vx+1 V x + 1Khi , phng trnh c dng: 3t2 - 2t = -m . (*)Xt hm s f(t) = 3t2- 2t trn tp D = [0; 1), ta c:

r) = 6t - 2, f(t) = 06t -2=0 t = .

Ta c bng bin thin:

0 1/30 0

0f -1/3

phng trnh ban u c nghim thc iu kin l phng trnh (*) cnghim t G [0; 1), tc ng thng y = -m ct th hm s f(t) = 3t2 - 2t trn tpD = [0; 1), ta c:

1 1 1 ~ 1- < - m


25/394

iQua M. _ Qua M j(0 ;l;-2 )(P,): r - - (P,>: - - -

[Cpvtcp n v u, [vtpt ri| =[ n, u ,3= ( -3 ;- 1 5 ;9)

(P,): X + 5y + 3z + 1 = 0.

Xc nh phng trnh mt phng (P2), ta c:Qua M2 _ [Qua M2(-l ;l ;3 )

(p2: ' - (P2): 1 -

Cpvtcp n v u2 [vtpt n2 =[n, u2] = (4;-8;5)

N(2; 0; -1) (d).

0 (P2): 4x - 8y + 5z - 3 = 0.Vy, to cc im thuc ng thng (d) tho mn:

X+ 5y + 3z + 1= 0

4x - 8y + 5z - 3 = 0

Phng trnh ng thng (d) c cho bi:

Qua N (2 ;0 ;- 1)( d ) : f *v (


26/394

;(d2) c ( P 2) iQua M2(P2): 2 2 (P2N _ 2 _ _

(P) (P2) I Gp vtcp n v u.

[vtpt n2=[n , u 23= (4 ;-8 ; 5)

Ta im F nghim ca h phng trnh:

4x -8y + 5z -3 = 0 f x =2 x _ y - l _ z + 2 =>< y= 0 F(2; 0; -1).

12 ~~ -1 1 [z = ~ l

Vy, phng trnh ng thng (d) c dng:

X = 2 + 7t

(P2): 4x - 8y + 5z - 3 = 0.

Qua F (2 ;0;-l)(d): - ' o (d):

[vtcp n(7;l; - 4 )y = t , t e l .

2 = i 4t

(dj):

Cch 4 : Gi n vtpt ca mt phng (P), ta c n (7; 1; -4).

Chuyn phng trnh ng thng (d!> v dng tham s.

X= 2u

y = l - u , u e I R .

z = - 2 + u

Gi s (d) l ng thng cn dng v (d) ct (d|) v (d2) theo th t ti ccim F, E. Khi :

im F (d,) suy ra F(2u; 1 - u; u - 2). im E e (d2) suy ra E(2t - 1; 1 + 1; 3). EF vung gc vi mt phng (p) ta c:

2 u- 2 t + l - u - t u - 5 _ fu = IE F / / n ------_ . - : = ------- \

7 1 - 4 t = -2

Khi , ng thng (d) c cho bi:

"x =2 + 7t

y = t , t e R .

z = - l - 4 t

F(2; 0; -1).

Qua F (2 ;0;-l)(): {

[vtcp n(7;l; - 4 )

Cu IV .I. Honh giao im ca hai th l nghim ca phng trnh:

X = 0(e + l)x = (1 + ex)x x(ex- e) = 0

Khi :

X = 1

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27/394

1 I - 1 1s =jjx(ex - e)| dx = Jx(e - ex)dx = e Jx.dx - Jxe* .dx.

0 0 () 0t, ' u "

(1)

Ta ln lt: Vi li th:

1,==ex

(2)

Vi I2th t:

u = X du = dx

dv = e*d x i V = e"

(3)Khi : I, = xe - fe x = e - e = I.I[> J lo0

Thay (2), (3) vo (1), ta c s= 12. Vi gi thit X, y, z dng v xyz = 1, ta c:

x2(y-f z) > x3.2^/z = 2x2^ = 2xVx

chng minh tng t, ta cng c y 2 (z + x) > 2 y j y v z 2(x + y) > 2 z 4 z .

Khi , ta nhn c nh gi;

> 2 xVx 2y>/ 2zVz

y^/y+2zVz z \ fz + 2 x j x x y x + l ^

Bng vic s dng n ph:

4c + a - 2b

a = xVx + 2 yy f

b = y j + 2 z \ fz

c = zVz + 2x>/x

xVx =

yV =4a + b - 2c

4 b + c - 2 aZyfz=

9

khi , ta nhn c bt ng thc mi:2 r 4c + a 2b 4a + b 2c 4b 4- c 2a

A > ------ -------+------- ------ + ------------9 \ b c a

2 f c a b N\ ( a b c''= 4 ++ + + + - 69 c u c

> (4.3 + 3 6) = 2.9

Vy, ta c AMin = 2, t c khi:

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28/394

H(; ).

xyz = 1x = y = z x = y = 2 = 1.

a = b = c

Cu v.a.

1. Ta ln lt c:

. M v N ln lt l trung im ca cc canh AB v BC nn M (-1; 0), N( 1; -2).

Gi s H(x; y), ta c:

H e A C 4 x +A(y - 2 ) 0 |x = l

b H X A C ^ [4(x + 2 ) - 4(y + 2) = 0 ^ \ y = 1

Gi s ng trn (C) ngoi tip AHMN c dng:

(C): X2 + y2- 2ax - 2by + c = 0, vi a2+ b2- c > 0.

im H, M, N e (C), ta c:

a = l / 2

b = -1 / 2, tho mn.

c = - 2

Vy phng trnh ng trn (C): X2+ y2 - X+ y - 2 = 0.

2. Ta c:

(1 +x )2 = c + C nX+ C jnX2+ ... + c x2*. (1)

(1 - X)2" = c - X+ c ' X2- ... + c lx2n. (2)

Tr theo v (1) v (2), ta c;

(1 + X)2- - (1 - X)2" = 2(cix + cx5+... + C r 'x 2n- ')

l + l - 2 a - 2 b + c = 0 2a + 2b - c = 2


29/394

Cu v.b.

f 4x - 3 > 0 31. iu kin: x> . (*)

[2x + 3 > 0 4

Bin i bt phng trnh v dng:

log3(4x - 3)2- log*(2x. + 3 ) ^ 2 0 Iog3(4x - 3)2^ log?(2x + 3) + 2log?(4x - 3)2 og-,9(2x + 3) (4x - 3)22 9(2x + 3)

3 16x2- 42X- 18 < 0 - - < x SHAD.Theo gi thit:

(SAD) JL (ABCD) => SH (ABCD) => SH X BP. (1)Mt khc, trong hnh vung ABCD, ta c:

ACDH = ABCP => CH JL BP. (2)

T (I) v (2) suy ra BP JL (SCH).

Ta li c:

M N / / S C ~ =>(AMN) // (SHC) => BP _L(AMN) AN //.CH

=> BP _LAM, pcm.b. Tnh th tch ca khi t din CMNP.

K MK vung gc vi mt phng(ABCD), ta c:

CMNP M K .S iCNP, (3)

ironR :

M K=- -SH = , (4)2 4

SCNP= ~ C N .C P = .

Thay (4), (5) vo (3), ta c:

v _1 aj3 a2_* ,JVcNp - 3 - 4 ; g - 96 ()

(5)

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30/394

TH MN TON KHI B NM 2007

PHN CHUNG CHO TT C CC TH SINH

Cu I: (2 im):Ch hm s:y = - x ' + 3x2 + 3(m2 - l)x - 3m2 - 1, m i tham s. (1)

1. Kho st s bin thin v v th hm s (I ) khi m = 1.2. Tm m hm s' (1) c cc i v cc tiu, ng thi cc im cc tr ca

th cch u gc to o.

Cu II: (2 im)1. Gii phng trnh 2sin22x + sin7x - = sinx.2. Chng minh rng vi mi gi tr ng ca tham s m, phng trnh sau c hai

nghim thc phn bit:

X2 + 2x - 8 - y/m (x - 2 ) .

Cu III: (2 im):Trong khng gian vi h to Qxyz, cho mt cu (S) v mtphng (P) c phng trnh:

(S): X2+ 2 + z2- 2x + 4y + 2z - 3 = 0,(P): 2x - y + 2z - 14 = 0.

1. Vit phng trnh mt phng (Q) cha trc Ox v ct (S) theo mt ng trnc bn knh bng 3.

2. Tm to im M thuc mt cu (S) sao cho khong cch t M ti mt phng(P) l n nht.

Cu IV : (2 im)1. Cho hnh phng H gii hn bi cc ng y =x.lnx, y = 0, X= e. Tnh th tch

ca khi trn xoay to thnh kh quay hnh H quanh c Ox.2. Cho ba s thc dng X, y, z thay i. Tm gi tr nh nht ca biu thc:

A = Xr \ / 1 N r 1z 1

+ -2 XV2 y z) \ 2 zx

PHN T CHNTh sinh chn cu v .a hoc cu v.b

Cu v .a Theo chng trnh THPT khng phn baa (2 im)

1. Tnh h s ca XK1ong khi trin nh thc Niutcm (2 + x ) \ bit rng:

3"C -3" 'cl + 3 -C; - , . .+ ( - l )" C ; = 2048.(n nguyn duong, l t hp chp k ca n phn t).

2. Trong mt phng vi h to Oxy, cho im A(2; 2) v cc cmg thng:

(d,) X+ y - 2 = 0, (d2):'x + y - 8 = 0.

Tm to cc im B v c ln lt thuc (d), (d2) sao cho AABC vung cnti A.

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31/394

Cu v.b Theo chng trnh THPT phn ban (2 im)

1. Gii phng trnh ^>/21j + ^.+ l j 2 V2= 0.

2. Cho hnh chp t gic u S.BCD c y ABCD l hnh vung cnh a. Gl im i xng ca D qua trung im ca SA, M v N ln lt l ung ica cc cnh AE v BC. Chng minh rang MN vung gc vi BD v tkhong cch gia hai ng thng MN v AC.

Cu I.1. Bn c t lm.2. o hm:

y' = -3 x 2 + 6x 4- 3(m2 - 1),y* = 0 -3x2 + 6x + 3(m2 - 1) = 0

a. -Hm s c cc i v cc tiu khi:(1) c hai nghim phn bit

o A > 0 o m 2 > 0 o m ^ 0 .Khi , th hm s c hai im cc tr l:

A(1 - m ; -2 - 2m3) v B(1 + m ; -2 + 21^).b. cc im cc tr ca th cch u gc to o , iu kin l:

OA = OB OA2= OB2

1. Bin i phng trnh v dng:2sin22x - 1 + sin7x - sinx = 0 -cos4x + 2cos4x.sin3x = 0

cos4x(2sin3x - 1) = 0

f(x) = X2 - 2x - m2 + 1= 0. (1)

(1 - m)2 + (-2 - 2m3)2 = (1 + m)2 + (-2 + 2m3)2

111*0 ] 8m 2m = 0 m = .2

Vy, vi m = tho mn iu kin u bi.2

Cu II.

3x = 7 t-+2k7t6

571 2knX = 4

18 3Vy, phng trnh c ba h nghim.

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32/394

2. Vi m > 0 th iu kin ca phng trnh l X > 2.Phng trnh c bin i tng ng v dng:

c > (x 2)[(x - 2)(x + 4)2 - m] = 0

(x - 2)(x + 4) = ^ / m ( x - 2 ) (x - 2)2(x + 4)2= m(x - 2)

X = 2

X3 + 6 x 2 - 3 2 - m = 0 (*)

Ta ch cn i chng minh phng trnh (*) lun c ng mt nghim rhuc

khong (2;+co)^Tht vy, xt hm s y = x + 6x2- 32 trn tp D = (2; +co), ta c:

y = 3x2+ 12 x> 0, Vx eD => Hm s ng bin trn D.Ta c bng bin thin:

X 2 +00

y +

y 0+O0

T bng bin thin ta thy vi mi m > 0 phng trnh (*) lun c ng mtnghim thuc khong (2; +00).

Vy, vi mi m > 0 phng trnh ban u lun c hai nghim thc phn bit.

Cu III. Mt cu (S) c tm 1(1; -2; -1) v bn knh R = 3.. Mt phng (Q) ct (S) theo mt ng trn c bn knh bng 3 th phi i qua

I. T , ta c:

Qua _ Qua I(Q):j o ( Q ) : _ r __

[O xc i( Q) [Cp vtcp i vOI

^ ( Q ) :;Qua I( l ; -2; - i )

Ivtpt n = [T, ] = ( 0 ;l ; -2 )(Q): y - 2z = 0.

2. Nhn xt rng ng thng (d) i qua tm I ca mt cu (S) v vung gc vi mtphng (P) th (d) s ct (S) ti hai im A, B. V khi , nu d(A, (P)) > d(B. (P)) th(M, (p)) ln nht khi M = A. Ta c:

X = 1 + 2t

y = -2 - 1 , t M.

z = -1 + 2t

Qua 1(1; - 2 ;-1)

^ (d): A(- l ; -1; -3) v B(3; -3 ; 1).

Ta c:

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33/394

2.(1) (1) + 2 .(3 )14d(A , (P)) = 1 T T t To - 2------ = 7

^ + ( - \ ) 2 + 2 2

| 2 . 3 - ( -3 ) + 2 .1 -141d(B, (P )) = 1 1 = 1, => d(A, (P)) > d(B, (P)).

yj22 + ( - l ) 2+ 22

Vy, vi im M(-l; -1; -3) thuc mt cu (S) th khong cch t M ti mtphng (P) l n nht.

Cu IV.1. Honh giao im ca hai th y = x.lnx v y = 0 l nghim ca phng trnh:

x.Inx = 0 Inx = 0 X= 1.Khi , th tch ca khi trn xoay to thnh khi quay hnh H quanh trc Ox l:

V = TC(x.Inx)2 dx = 71Jx2.ln2 xdx1 I

2Inx.dxu = In X

t: < [dv =. x2.dx

Khi :

V = 71 .ln2x3

---- fx .lnxdx = ---------- fx -lnxdx. (1)3 J 3 3J j J J I

I

[ u = n XVi I, ta t: <

dv = x .dx

du -

V=

x

Khi :

1= .Inx3

1 c 3 31 r 2 J e X- - X .dx = - -----3- .3 9

2e +1(2)

Thay (2) vo (1), ta c V =27

2. Bin i A v dng:2 2X 4-y + z

+

Nhn xt rng: X7+ y 2 +X = V-----------b

xyz2 _2 7 2 > 22 _ X + y y + 2 z + X

> xy + yz + zx

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34/394

do , ta c bt ng thc:

A >f 2 _ 2 _2 \X yz I xy + yz + ZX y2 , Z21 1 1 1

' H-----1----- 1----xyz z X y

2 1X 1+

1+

2 y

\ 2+

2 1 f _

2 z

X2 1 1~---1----------2 2x 2x

\ ' L - i . _L' f 2


35/394

SUV ra X2 v - y 2l nghim ca phng trnh:

V - -1 fx2 = 4V2- 3v - 4 = 0

X = 2

_v = 4 [ - y - = - 1 " [y = 1

Bi X, y cng u nn ta ln lt c:

f x - 2 - f t - 1 - 2 - 3 - 3 ^ , ( 3 ; - ! )

[ y= = u - 4 = 1 \u = 5 [0 , (5; 3) ' .

X - - 2 ft 1-2 _ ft = ~1- . B2(-1; 3)o < < /2+1 j = - , iu kin t > 0.

Phng nh c bin i v dng;; t- + 2 V2 = 0 t22 t y f + 1= 0

"t = V2 + 1

>-s( V 2 - ! ) * = y / + X = -1

_t = 2 - l F -0w

* = ^ 2 - 1 . X = 1

Vy, phng trnh c hai nghim l X= 1 .

36



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36/394

2. Bn c T v hnh.

a. Chng minh rng MN vung gc vi BD.Gi p l trung im ca SA, ta c ngay:

MNCP l hnh bnh hnh => MN // CP => MN // (SAC). (1)

Mt khc, ta c BD -L (SAC). (2)

T (1) v (2) suy ra BD _LMN.

b. Tnh khong cch sia hai ng thng MN v C.Ta c nh gi:

MN (SAC) d(MN, AC) = d(MN, (SAC)) = d(N, (SAC))

1 1 F?=d(B , (SAC)) = BD = (vd).

2 4 4

T HI MN TON KHI D NM 2007

PHN CHUNG CHO TT C CC TH SINH2x

Cu I: (2 im)'.Cho hm s: y =X + 1

1. Kho st s bin thin v v th (C) ca hm s.2. Tm to im M thuc (C), bit tip tuyn ca (C) ti M ct hai trc Ox, Oy

1ti A, B v AOAB c din tch bng .

4Cu II: (2 im)

1. Gii phngtrnh: ^ sin + COSj + Vj c o s x = 2.

2. Tm gi trca tham s m h phng trnh sauc nghim thc:

1 1 _ X + +V + = 5X y

X I 4 + y ' H = 15m ] 0

x y

Cu XI: (2 im):Trong khng gian vi h to Oxyz, cho hai im A(l; 4; 2),B(-l; 2; 4) v ng thng (A) c phng trnh:

( A ) : i i r = y i = .-1 1 2

1. Vit phcng trnh ng thng (d) i qua trng tm G ca AOAB v vung gcvi mt phng (OAB).

2. Tm to im M thuc ng thng (A) sao cho MA2+ MB2nh nht.

37



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37/394

Cu IV: (2 im)e

1 . Tnh tch phn I ~ Jx3.ln2 x.dx.1

2 . Cho 0< b < a, chng minh rng 2a + < I 2 +2a ) \ 2M . I

, c n u n g 11111111 r a n g I Z, -t- -V :

PHN T CHNTh snh chn cu v.a hoc cu v.b

Cu v.a Theo chng trnh THPT khng phn ban (2 im)1. Tnh h s ca Xs trong khai trin a thc x( l - 2x)5+ x2(l + 3x)\2- Trong mt phng vi h to Oxy, cho ng trn (C) v ng thng () c

phng trnh:(C): (x - 1)2+ (y + 2)2= 9, (d): 3x - 4y + m = (X

Tm m trn (d) c duy nht mt im p m t k c hai tip tuyn PA,PB ti (C) (A, B l cc tip im) sao cho APAB u.

Cu v .b Theo chng trnh THPT phn ban (2 im)1. Gii phng trnh log2(4* + 15.2X+ 2 7 ) + 2 log2 -----------= 0.

v 1 4.2X- 32. Cho hnh chp S.ABCD c y l hnh thang vung ti A v B, BA - BC a, AD =

2a. Cnh bn SA vung gc vi y v SA = \2. Gi H l hnh chiu vung gcca A trn SB. Chng minh rng ASCD vung v tnh khong cch t H n mt

phng (SCD).

Cu I.1. Bn c t lm.2

2. o hm: y' = -----.(x + 1)2

im M(a; y(a))e(C), khi phng trnh tip tuyn ti M c dng:

2 2a2(): y = y(a)(x - a) + y(a) (d ): y = ^-x +

(a + l)2 (a + 1)2 'To giao im A ca tip tuyn () vi Ox l nghim ca h:

y = 0 r

fy = 2 2a ~ A( - a 2; 0).y ----------7x + ------ ~T y = - a

L (a + !) (a + ) u

To giao im B ca p tuyn (d) vi Oy l nghim ca h:x = 0 x = 0 ,

2 .. 2a2 1 2a2 0;y = L K+ y = ~- --y ^ (a +1)

(a + (a + \) 1 . ( + \)

2a2

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38/394

Ta c: Sq A g o OA..OB = o OA..OB OAB 4 2 4 2

- a2a"

(a + i= 4a4 = (a + l)2

2

2a = a +1

2a2 = a - 1

2a - a - I =0

2a2 + a +1 = 0

a = 1

1a = -2

Mid; 1)

M2 ~ ; - 2

Vy, tn ti hai im M ,, M2 tho mn iu kin u bi.C uI I .I. Bin i phng trnh v dng:

sin2 + cos2 + 2sin -COS +y3cosx = 22 2 2 2

1 + sin X + V 3 cosx = 2 o s i n X+ -v/3 COSX = 1

* : >/3 ' 1 _ . 7^ 1 ^ - s i n X + cosx = - r o s i n X + - = - '2 , 2 2 I . 3J 2

X + =+ 2kt

3 6

X + = 7C- + 2k7t3 6

X = + 2kn

6

X = - + 2kr2

Vy, phng trnh c hai h nghim.2. Trc tin, ta c:

a. t u - X + iu kin u > 2 th:X

x3+,? = ( x + x ) ~ 3( x + x ) =u',_ 3 u '

b. Tng tu, t y = y + iu kin VI > 2 thy3+ - = V3- 3 v .. y y

Khi , h phng trnh c dng:

u + V = 5 I u + V = 5

j u 3+ V3 - 3 ( u +v) = 15m - 10 . [ uv = 8 - m

suy ra u, V l nghim ca phng trnh:

t2 5t + 8 m = 0 t2 5t + 8 = m. (1)H cho c nghim khi v ch. khi phng trnh (1)C hai nghim thc tho

mn 11 > 2.

Xt hm s y = t2 - 5t + 8 trn tp D = ( - 00; -2 ]o [2 ; +oo), ta c:

39



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39/394

y = 2t - 5, y = 0 2t - 5 = 0 t = .2

T bng bin thin ca hm s suy r h c nghim khi v ch khi:7

< m < 2 hoc m > 22 .4

7Vy, vi < m < 2 hoc m > 22 tho mn iu kin u bi.4

C u m .1- Tam gic OAB c trng tm G(0; 2; 2).

Khi , phng trnh ng thng (d) c cho bi:

():

(d): Hm s nghch bin trn D

Cu v.a.

1. Ta ln lt thc hin:

a. Trong khai trin:5 5

x ( l - 2 x ) = x ^ c . ( - 2 x ) k = x c ( - 2 ) k xk.k=0 k=(>

H S ca Xs trong khai in trn l c(~2)4.

b. Trong khai trin:10 10

X2(1 +3x)m = X2 cf(3x)k =X2 cf3kxkk=0 k=0

H S ca Xs trong khai trin trn l c*()33.

Vy, h s ca Xs l: c(~ 2) 4 +C '^ 3 = 3320.



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41/394

2. ng trn (C) c tm I( ; -2) v bn knh R = 3.

Ta c nhn xt:

APAB u APB = 60 API = 30 IP = 21A = 2R - khng i

p thuc ng trn (C) tm I bn knh 2R.

Nh vy p l giao im ca ( C ) v ().

Khi , c duy nht mt im p iu kin (d) tip xc vi (C), suy .ra:_ 3.1 - 4 . ( 2) + m| , , ,

(I, (d)) = 2R j---- :> Z.L...... 1 = 6 m +11 = 30V 32+ ( -4 )2

m = 19

m = 41

Vy, vi m = 19 hoc m = -41 tho mn iu kin u bi.

Cu v.b.

1. iu kin 4.2* - 3 > 0. (*)

Bin i tng ng phng trnh v dng:

. log2(4x +15.2x +2 7)- 21 og 2(4.2x - 3 ) = 0

og2 (4 X+ 15.2X+ 27 ) = log2 (4 .2 X- 3 ) 2

^ 22x +15.2X+2 7 = 16.22x - 24 .2* + 9

15.22* - 3 9 .2 X-1 8 = 0. (1)t t = 2 \ t > 0 ta c:

t>05t2 - 13t - 6 = 0 t = 3 o 2X= 3 X= log23.

Vy, phng trnh c nghim Xlog23.

2. Bn c t \ ' hnh.

a. Chng minh rng ASCD vung.

Gi I l trung im ca AD, ta c:

A = B = IC = a :=> AACD vung ti c => CD -L AC.Mt khc, ta c:

CD SA => CD 1 (SAC) => CD 1 s c => ASCD vung ti cb. Tnh khong cch t H n mt phng (SCD).

Nhn xt rng BH ct.mt phng (SCD) ti s, do :

d(H, (SGD)) = .d(B, (SCD)).d(B, (SC D SB SB

m + 11 = 30

m + 11 - -30

42



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42/394

^ ^ . SH SA2 SA2 2a2 2Ta ln t c: 1 = -~ =- = -~~ = - = .

SB SB2 sA 2 + AB2 2a2 4- a 2 3

VB.SCD = d(B, (SCD))S^cd c=> d(B, (SC)) = 3^ BSCD = ^ -BCD-> ASCD SSCD

SiBCD = |A B .B C = i a 2.

S ASCD = -SC.CD = -VsA2 + AB2 + BC2 . V i c 2 + ID2 = a 2y f l .

T , suy ra d(H, (SCD)) = .

THI MN TON KHI A NM 2008

PHN CHUNG CHO TT C CC TH SINHCu I: (2 im)'. Cho hm s:

mx2 + (3m2 2)x - 2y = , m l tham s.

X + 3m1. Kho st ;S bin thin v v th hm s (1) khi m = 1.2. Tm m gc gia hai ng tim cn ca th hm s (1) bng 45.Cu II: (2 im)

1. Gii phng lnh:

(1 )

sinx1h

s i n X

1 2

= 4sin| -X 37^ l 4

2. Gii h phng trnh:

X2 + y + x3y + xy2 + xy = - 4

X 4 + y 2 + xy(I + 2x) =4

, (x ,ye R) .

Cu III: (2 im): Trong khng gian vi h to Oxyz, cho im A(2; 5; 3) vng thng:

X -1 y _ 2 - 27 ~ '

(d):2 1 2

1. Tm to hnh chiu vung gc ca A trn ng thng (d).2. Vit phng trnh mt phng (P) cha (d) sao cho khong cch t A n (P) ln nht.Cu IV: (2 im)

n/61. Tnh tch phn I = J

tan x.dxcos2x

43



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43/394

2. Tim cc gi tr ca tham s m phng trnh sau c ng hai nghim phn bit:

x . + \ 2 x - 2 ^ 6 - X 4-2 y j 6 - X = m , ( m G R ) .

PHN T CHNTh sinh, chn cu v.a hoc cuy.b

Cu v .a Theo chng trnh THPT khng phn ban (2 im)1. Trong mt phng vi h to Oxy, hy vit phng trnh chnh tc ca Eip

(E), bit rng (E) c tm si bng v hnh ch nht c s ca (E) c chu vi

bng 20.

2. Cho khai trin (1 + 2x)n = a0 + ajX + ...+ trong n e N* v cc h s ao

a J ^ tho mn h thc a0 = 4096 . Tm s ln nht ong cc^ 2

s a l 5 an.

Cu v .b Theo chng trnh THPT phn ban (2 im)

1. Gii phng trnh log2x-j (2x 2 + X- ) + logx + j(2x - l) 2 = 4.

2. Cho lng tr ABC-ABC c di cnh bn bng 2a, y ABC l tam gic yung

ti A, AB = a, AC = a a/s v hnh chiu vng gc ca nh A trn mt phng(ABC) l trung im ca cnh BC. Tnh theo th tch ca khi chp A.ABC vtnh csin ca gc gia hai ng thng AA\ BC\

Cu I.1. Vi m = 1 hm s c dng:

(H): y =X + X - 2

C> (H): y = X - 2 +x + 3 X + 3

Ta ln lt c:a. Hm s xc nh trn D = \ {-3}.

b. S bin thin ca hm s: Gii hn ca hmsti v cc, gii hn v cc v cc ng tim cn:

lim y = - c o ; lim y = +CO .X c X

lim y = 00 nn X = - 3 l ng

tim cn ng.

l m [y -( x- 2 )] = 0 nn y = X 2x->oo

l ng tim cn xin. Bng bin thin:

44



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44/394


45/394

sin X+ cos X= 0 j tanx = -1

1 [ 1 sin2x = Y= sin2x = 7=

4 L V2

Vy, phng trnh c ba h nghim.

2. Bin i h phng trnh v ng:

2 5X + y + x y (x + y ) + x y = - -

2 2 5(x + y) + xy = - - -

4

x = - + ki4

X= + k t , k e z .8

571 . _X = -HK7

8

t u = X2 + y v V = xy , h phng trnh tr thnh:

5u + uv + V =

, 5 2 \ 5 2 5u + u - u - - u

4 ) 4 4

5 . 2V = u4

= - 4

V4

2x3 + X - 3 = 0

46



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46/394

x =

Vy, h phng trnh c ba cp nghim ' 31Z - J 5

Cu ra . ng thng (d) c vtcp u(2; 1; 2).. Gi H l hnh chiu vung gc ca A ln ng thng (d), ta c th trnh byheo hai cch sau:

Cch 1: Chuyn phng trinh ng thng (d) v dng tham s:

x = l + 2t

(d): iy = t , t e R .

z = 2 + 2t^ *

V im H () nn H(1 + 2t; t; 2 + 2t), suy ra AH (-1 + 2t; t - 5; -1 + 2t). H l hnh chiu vung gc ca A ln (d) iu kin l:

AH(d) H-L H. = 0

2-(-l + 2t) + l( t - 5) + 2(1 + 2t) = 0 t = 1 => H(3; 1; 4).

Cch2: Gi (P) l mt phng tho mn:

Qua A Qua A(2; 5; 3) '

: 1 (p>: - (P): 2x + y + 22 - 15 = 0.

[(P) JL (d) [vtpt u(2; 1; 2)

V H} - (d) n (p) nn to H l nghim h phng trnh:X - z - 2

2 1 22x + y + 2 z- 1 5 = 0

X = 3

y = l:

z = 4

H(3; ; 4).

. Gi K l hnh chiu vung gc ca A ln mt phng (P), ta c:

d(A, (P)) - AK < AH - tnh cht ng vung gc v ng xin.

Do , khong cch t A n (P) Icrn nht khi v ch khi K = H- Suy ra, mthng (P) cn dng s i qua H, do :.

Qua A(2;5;3)(P) : - (P): X4y + 2 + 1 5 0.

[vtpt AH(1;-4;1)

Cu IV.1. Bin i tch phn v dng:

7T./6tan x.dx

cos2 x - s n 2 x

n/6

- ftan x.x

( ( 1 - t a n 2 x)cos2 X

47



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47/394

t t = tanx, suy ra dt =

i cn:* Vi X = 0 th t = 0.

Vi X = th t -6 s

Khi :

dx

COS2 X

ao:

l ^ t4 At

>= 0 1 1 0

1/V3I-

1/^3/dt = - J t2 +1 +

0 ^ t 2 - ldt

-/[(Kte-ril

T + t + ( lnlt _ 1 l_ i n l + 1l)

dt

I / SI V----h t H---ll

v 3 2 Sl)

1/V5

= i ln (2 + ) -10

9^3

2. iu kin: < , x ^ < = > 0 < x < 6 .[ 6 - X > 0

.t VT ca phng trnh l f(x), chng ta i xt hm s ny trn tpD = [; 6], ta c:

1 1 1 1~ 2\/2x * y/x 26~X

,2%/2x 2tl6-Z)u(x) v(x)

Nhn xt rng:

u(x) = 0 o -- / = 0 i l 2 x =3 / 6 - X & X = 2.2^/2x 2%/6-x

-----= 0 o V2x = V - X X = 2-v(x) = 0 1\/2x \ 6 - K

T , suy ra:

f (x ) = O o X = 2

v vi nh gi rng:

Khi X e [0; 2) th u(x), v(x) cng m, do f (x) 0.

48



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48/394

Ta c bng bin thin:

X 0 2 6

y ' - 0 + 0

y 0 1

f(2) ^

phng trnh ban u c hai nghim thc phn bit iu kin l ng thngy = m ct th hm s y = f(x) trn tp D = [0; 1) ti hai im phn bit, ta c:

Max {f(0), f(6)} < m < f (2) 2-J+ 2$6 - = 4a2 = 9b2 2a = 3b = 3(5 - a)3 a 9 a2

a = 3 => b = 2.2 2

Vy, Elp (E) c phng trnh + = 1.2. t:

12 nf(x) = (1 + 2x)n = a


49/394

r

Cu v.b.]. Bin i phng trnh v dng: log2x -j(2x - l)(x +1) + logx + j(2x - 1)2 = 4.

o < 2x -1 * 1 1 , ...iu kin: g2*- i(x 4- 1) = 1 X+ = 2x 1 X= 2.

- Vi t = 2 th: log2x- i(x +. 1) = 2X '+ 1 = (2x - l)2cs> 4x2- 5x = 0X= 0 (loi)

X = 5 /4

Vy, phng trnh c hai nghim X= 2 v X= .

2. Bn c t v hnh.a. Tnh th tch ca khi chp A\ABC.

Gi H l trung im ca BC, suy ra AH J- (ABC) nn:

V, = - A' H.S/ = A H.AB.AC.6

Trong , ta ln lt c:

AB = a, AC = aV3 .

()

(2)

(3)

AH = -B C = ^ a 2 +3a2 =a.2 2

AH2= AA2 - AH2 = 3a2 => A H =

a3Thay (2), (3) vo (1), ta c VA. ABC = .

b. Tnh csin ca gc gia hai ng thng A A \ BC \

Trong AABH vung ti A \ ta c:

BH2 = A B2+ AH2 = 4a2 => BH .= 2a => ABBH cn t B \

Gi a l gc gia hai ng thng A A \ BC \ suy ra:

, ~ s a ~ B BH v cos a = - = .

2.2a 4

50



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50/394

T H I MN TON KHI B NM 2008


Cu I: (2 im)'.Cho hm s:y = 4x3 - 6x2 + 1. ( )

1. Kho st s bin thin v v th hm s (1).

2. Vit phng trnh tip tuyn ca th hm s (1), bit rng tip tuyn iqua im M (-l ;-9).

Cu II.I (2 im)

1. Gii phng trnh sin* X - yfsCOS3 X= sin X.COS2 X - sin2 X, COS X.2. Gii h phng trnh:

Cu: (2 im): Trong khng gian vi h to Oxyz, cho ba im A(0; ; 2),B(2; 2; 1), Q 2; 0; 1). ,1. Vit phng trnh mt phng i qa ba im A, B, c.2. Tm to i m M thuc mt phng 2x + 2y + z - 3 = 0sao cho MA = MB = M C

Cu IV: (2 im)

1. Tnh tch phn I = ------- .MU T T.Jill yyT COSx)

2. Cho hai s thc X, y thay i tho mn h thc X2 + 2 = 1. Tm gi tr ln nhtv gi t nh nht ca biu thc:

A _ 2(x2 + 6xy)

1+2 xy + 2y2"

PHN T CHNTh sinh chn c v .a hoc cu v .b

Cn v.a Theo chng trnh THPT khng phn ban (2 im)

(n nguyn ng, C* t hp chp k ca n. phn t).

2. Trong mt phng vi h o Ox, xc nh to nh c ca AABC, bitrng hnh chiu vung gc ca c trn ng thng AB im H (-; ->,ng phn gic trong ca gc A c phng trnh X - y + 2 = 0 v ng cao kt B c6 phng trnh 4x + 3y - = 0.

X4 + 2x:,y + x2y2 = 2x + 9

X2 + 2xy = 6x + 6, (x,y e R).

5.1



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51/394

Cu v .b Theo chng trnh THPT phn ban (2 im)

f x2 +x)1. Gii bt phng trnh log07 log6----- < 0.

' I X + 4 J2. Cho hnh chp S.ABCD c y ABD l hnh vung cnh 2a, SA a

SB = aV3 v mt phng (SAB) vung gc vi mt phng y. Gi M, N nlt l trung im ca AB, BC. Tnh theo a th tch khi chp S.BMDN v csin

ca gc gia hai ng thng SM, DN.

Cu I.1. Ta ln lt c:

a. Hm s xc nh trn D = M .b. S bin thin ca hm s:

Gii hn ca hm s ti v cc:

4 - +- lim y = lim X X

+00 khi X - +CO

-00 khi X > -co

Bng bin thin:y = 12x2 - 12x, y' = 0 12xa - 12x = 0 X = 0 hoc X = 1.

0 1 +000

"+00

im un:y" = 24x -1 2 , y" = 24x - 12 = 0 o X = 1/2.

V y" i du khi X qua im 1/2 nn th hm s c mt im un 1(1/2; 0).

c. th ca hm s.2. Ta c th trnh by theo cc cch sau: ;Cch 1.Gi s honh tip im l X= Xo, khi phng trnh tip tuyn c dng:

(d): y = y(x)(x - x0) +- y(x0)

(d): y =.(12x - 12x,j)(x x0)-+'4x 6x +1. - (1)

im -9 )e (d ), suy ra>

. -9 = (12 x - I2x


52/394

Ta ln lt: Vi X(, = - , thay vo (1) ta c tip tuyn (i): y = 24x +15.

Vi X(, = , thay vo (1) ta c tip tuyn (d2): y = X - .4 4 4

Vy, tn ti hai tip tuyn (|), (d2) ca th tho mn iu kin.Cch 2 :Phng trnh ng thng (d) qua M (-l; -9) vi h s gc k, c dng:

y = k(x + 1) - 9. (1)

ng thng () tip xc vi th hm s khi h sau c nghim:4x3 - 6 x 2 +1 = k(x + 1 ) - 9 (x + l)2(4x 5) = 0

12x2 - 12x = k

X = 1

5X =

4

12x - 2 x = k

k = 2 4

k . i -

1 2 x 2 - 1 2 x = k

Ta ln lt:

Vi k = 24, thay vo (1) ta c tip tuyn ( j): y = 24x + 15.

Vi k = , thay vo (1) ta c tip tuyn (d2): y - X- .

Vy, tn ti hai tip tuyn (dj), (2) ca th tho mn iu kin.Cu ri.1. Ta c th trnh by theo bai cch sau:Cch :Bin phng trnh v dng:

\ /3 c o s 3 X - V 3 si n 2 X. COS X + sin X. COS2 X - si n 3 X = 0

-s/3 co sx (cos2 X- sin2 x) + sinx(cos2 X- sin2 x) = 0

[Z - \ o -v/3 COSX + sinX = 0(V3COSX + si nx jcos2x = 0

cos2x = 0

r~ta n X = - V 3

cos2x = 0

X = + kn

3 X = - - 7- + k i

371 , K

X = + k -4 2

,k z .

Vy, phng trnh c hai h nghim.

Cch2: V cosx = 0 khng phi l nghim ca phng trnh nn chia c hai v caphng nh cho c o s \ *0Tta c:

t a n 3 X - = t an X - \ / 3 t a n 2 X.

t t = tanx, ta uc:

t3 + 73 t2 - i - n/3 = 0 t(t2 -1 ) + V3(t2 -1 ) = 0

(t2 - ) t + ^ = 0

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53/394

t = 1

t = s

tanx = 1

tanx = -7 3

X = + k r c4

71 i_X= - - f + kTt3

, k e z

Vy, phng trnh c ba h nghim.

2. H phng trnh c bin i tng ng v dng:r(x2 + xy)2 = 2x 4-9 / 2 *\2

_ x2- + 3x + 3xy = + 3x + 3

= 2x + 9

X4 + 12x3 + 48x2+ 64x = 0 x(x + 4 - 0 o

Ta ln lt: V i X = 0, th:

o . y = 3 , v nghim.

Vi X - -4 , th:17

4y = -8 + 3 (-4) + 3 y .

17

X = 0

X = -

Vy, h c nghim -4 ; |.

Cu III.

1. Ta c th trnh by theo hai cch sau:Cch : Mt phng (Q) i qua ba im A, B, c c cho bi:

Qua A _ Qua A(0;l;2)(Q) : \ _ . _ < = > ( Q ) :

Cp vtcp AB, AC vtpt n = I^AB, AC J =

(Q): X + 2y - 4z + 6 = 0.

Cch 2:Gi s mt phng (Q) c phng trnh:

(Q): Ax + By + Cz + D = 0 vi A2+ B2 +C2> 0.

V A, B, c thuc


54/394

2. Ta c th trnh by theo hai cch sau:

Ck h h MA = MB = MC th M thuc trc ng trn () ca AABC (l ngthng i qua tm ng trn ngoi tip ABC v vung gc vi mt phng (ABQ).

Nhn xt rng:

AB.AC = 0 AABC vung ti A

=> Trung im 1(0; -1 ; 1) ca BC l tm ng trn ngoi tip AABC.T , suy ra M thuc ng thng () tho mn:

x = t

y = - l + 2 t , t E .

2 = 1 - 4t

T , suy ra to im M l nghim ca h phng trnh:

Qua I Qua 1(0;1;1)

[(d) -l(Q ) [vtcp n( l; 2; - 4 )

X - 1

y = + 2t

z = 1 - 4t2 x + 2 y + z - 3 = 0

X = t

y = -1 + 2t

z ~ l ~ 4 l2t + 2(2t i) + (1 4t) - 3 = 0

X = 2

y = 3 .

z = -7

Vy, vi im M(2; 3; -7 ) tho mn iu kin u bi.

Cch 2: Gi s M(x; y; z), M thuc mt phng (P): 2x + 2y + z - 3 = 0 sao choMA = MB = MC iu kin l:

M (p)

MA2 = MB2 M(2; 3; -7 ).

2 = -7

2 x + 2 y + z = 3


55/394

t t = sinx + cosx + I, suy ra:dt = (cosx - sinx)dx (sinx - cosx)dx = -t.

i cn: V ix = 0 t h t = 2.

4 - J

V X= th t - y / + 1.4

F) ^+lHt ^ +1Khi : 1 = - * = 2 t2 2t2. Bin i A v dng:

2(x2 + 6xy) _ 2(x2 + 6xy)A =

xz + y2 + 2xy + 2y2 X2 + 2xy + 3y2

Nhn xt rng:

Vi y = 0 th X2= 1, suy ra A = 2.

* V y 5* 0 th ta c bin i tip cho A bng cch chia c TS v MS cho y2:\

A =

f X , X7 +6-

^ - + 2 - + 3y y

t X - ty, ta c:

* 2(2+6t )r\ -t + 2t + 3

Ta xt hai trng hp:

(A - 2)t2 + 2(A - 6>t + 3A = 0. (*)

7 rng hp : Vi A = 2 th (*) c nghim t = 4

Trng hp2: Vi A 5*2 th (*) c nghim khi:

A > 0 (A - 6)2 - 3A(A -2 )> 0< = > A 2 + 3 A - 1 8 < 0 -6 < A < 3.

Vy, ta c kt lun:

AMin= 6, t c khi: t = ----2

X2 + y 2 =1

3X= y

2

i y 2 + y 2 = i4

3X= - y

2

,2 _ Ay = 133

X = y2

_ _ _ 3 _ 2

3 2

x Vi5 y ~ J 13

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56/394

AMa*= 3, t c khi:

X2 -hy2 =1t = 3

X = 3y

9y + y = 1

X= 3y

2 ___y l0 oX= 3y

X= ~=r& y = 1

X=

J 3

7

Cu v.a.I. Ta c:

n + k!(n + l - k ) (k + l ) (n -k )

n + 2 { (n+1)! (n + 1)!

1 k!(n - k ) ! / . . . A _ k ! ( n - J k ) _ 1 ,= - .---- -------(n + 1 - k + k + l) = - = - , pcm.

n + 2 n! } n! C;

2. Bn c t v hnh.

Ki hiu:

a. ng phn gic ong ca gc A (d (), c vtcp Uj (1;1).

b. ng cao k t B l (d2) c vtcp u2 (3;- 4).

Gi H(a; b) l im i xng ca H qua (j) th H thuc AC. Ta c:

Hr _Lu7

[Trungim ca HHthuc (d)

l.(a + 1)+ l.(b +1) = 0

1 a - i b -I+ 2= 0

a = -3b = 1

1).

2 2Phng trnh ng thng AC c cho bi:

(Qua H Qua H '(-3; 1)( A C ) : \ , (AC): 11

\ ( C U (d 2) vtpt u 2 (3; - 4 )

(AQ: 3 x -4 y + 13 = 0.

V (A Q n(d |) = {A} nn to ca A l nghim ca h:

3 x -4 y +13 = 0 J X -5X - y + 2 = 0 ^ y = 7

A(5; 7),

Phng trnh ng thng CH c cho bi:

[Qua H [Qa H (-l; -1 )(CH): ^ (CH): < 1

CH _LHA [vtpt HA(6; 8)

(CH): 3x + 4y + 7 = 0.

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57/394

V (AC)o(CH) = {C} nn to ca c l nghim ca h:3x - 4y + 13 = 0 x = -1 0 /3

3x + 4y + 7 = 0 ^ |y = 3/4C|

3 4

Cu v.b.1. Bin i tng ng bt phng trnh v dng:

X+ 4 X+ 4(*)

-4 < X< -3

X> 8X+ 4 X+ 4

Vy, nghim ca bt phng trnh l tp (-4 ; -3 ) u (8; +co).

tRCh : Mt s em hc sinh mc phi sai lm khi gii (*)bng cch nhn cho hai v.Hy nh rng iu ch c th c thc hin khi MS un m hoc lun dng.

2. Bn c t ii.

a* V a - * S.BMDN ~

b. Csin ca gc gia hai ng thng SM, ND bng .

TH I M N TON KH I D 'NM 2008


Cu I: (2 im)'.Cho hm s:y = x? - 3x2 + 4. (1)

1- Kho st s bin thin v v th hm s (1).2. Chng minh rng mi ng thng i quaim1(1;2) vi h s gc k

(k > -3 ) u ct th hm s (1) ti ba im phn bit I, A, B ngthi t ltrung im ca on thng AB.

Cu II: (2 im)1. Gii phng trnh 2sinx( 1 + cos2x) + sin2x = 1 + 2cosx.

2. Gii h phng trnh:xy + X + y = X2 - 2 y 2

X y = 2x -2y, ( x , y e i ) .

Cu III: (2 im):Trong khng gian vi h to Oxyz, cho bn im A(3; 3; 0),B(3; 0; 3), C(0; 3; 3), D(3; 3; 3).1. Vit phng tr mt cu i qua bn im A., B, c, D.2. Tim to tm ng trn ngoi tip AABC.

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58/394

Cu IV: (2 im)

1. Tnh tch phn I = J lnx;dx .! x

2. Cho hai s thc X, khng m thay i. Tm gi tr ln nht v gi tr nh nhtca biu thc:

(n nguyn dng, C l t hp chp k ca n phn t).

2. Trong mt phng vi h to Oxy, cho.parabol (P): y2 = 16x v im A (l 4).Hai im phn bit B, c (B v c khc A) di ng trn (P) sao cho BAC = 90.Chng minh rng ng thng BC lun i qua mt im c nh.

Cu v.b Theo chng trnh TH PT phn ban (2 im)

X2 3x + 21. Gii bt phng trnh log 1------------> 0.

2 x

2. Cho ng tr ng ABGABC c y ABC l tam gic vung AB = AC = a, cnhbn AA= a,y2 .Gi M l trung im .ca BC. Tnh theo a th tch khi lng tr

ABC.ABC v khong cch gia hai ng thng AM, BC

A (x - 1)(1 - xy)

(l + x)2(l + y)2 '

PHN T CHNTh sinh chn cu v .a hoc cu v .b

Cu v.a Theo chng trnh TK PT khng phn ban (2 im)I. Tm s nguyn n tho mn h thc:

Cu I.1. Ta ln lt c

a. Hm s >b. S bin tl

Gii

X-00 khi X>-00

* Bng bin thin:

X= 0y' = 3x2- 6 x , y = 0 c > 3x2- 6 x = 0

X= 2

59



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59/394

im un:y" = 6x - 6, y" = 0


60/394

(2y + l)7 2 y - yV 2y+ 1-1 = 2(2y +1) - 2y

(y + l)yj2y - 2(y +1) y = 2 2y = 4y= 2= > x = 5.

Vy, h phng trnh c nghim (5 ; 2).

Cu III.1. Ta c th trnh by theo cc cch sau:Cch :Gi s mt cu (S) c dng:

(S): X2+ y2 + z2- 2ax - 2by - 2cz + = 0, iu kin a2+ b2+ c2- d > 0.im A, B, c, D (S), ta c:

6a + 6 b - d = 18

a = b = c = , _ ,c$ < 2 ,thoa mn iu kin.

[d - 06a + 6b + 6c - d = 27

Vy, phng trnh mt cu (S) c dng:

(S): X2 + y2 + z2 - 3x - 3y - 3z = 0.Cch 2:Gi s mt cu (S) c tm I(a; b; c), kh ta ciu kin:

6a + 6c d = 18

6b + c -d = 18

IA = IB

A - C ^

1A = ID

I2 - IB2

IA2 = IC2

IA2 =ID7

(a - 3)2 + (b - 3)2 + c2 = (a - 3)2 + b2 + (c - 3)2

(a - 3)2 + (b - 3)2 + c2 - a2 + (b - 3)2 + (c -3 ) 2

( a - 3 ) 2 +(b - 3 ) 2 + c2 = (a - 3)2 + ( b - 3 ) 2 + (c -3 )2b = c -

_ _ _ u _ _ 3 3 3a = c a = b = c = - = > I 2 u 2 2

3c = 2

Vy, phng trnh mt cu (S) c cho bi:

(S):

Tm II 2 ; 2 ; 2 J 2 2

/27 2 ] +iy '2j + l z 2Bn knh R = A = J -27

___

2. Mt phng (Q) i qua ba im A, B, c c cho bi:[Qua A

(Q):[Cp vtcp AB, AC

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61/394

[Qua A(3;3;0)

^ iQ): jv fp t n =[B, c] = ( -9 ; -9 ; -9 ) , chn n(l;l;l)

(Q): x + y + z - 6 = 0.ng trn (C) ngoi tip AABC c tm H l hnh chiu vung gc ca I ln (Q). Gi () l ng thng qua I v vung gc vi (Q), ta c:

(d): Qual12 2 - 2 ; o (d ):

vtcpn(l;l;l)

Bng cch thay phng trnh ca () vo (Q), ta c:

r 1 ( I I t ) T ( | 1 1 ) - 6 r 0

X= 3/2 + 1

y = 3/2 + 1, te K .

z = 3/2 + 1

o t = => H(2; 2; 2).2

Vy, ng trn ngoi tip AABC c tm H(2; 2; 2).lu V.

u = lnx1. t: 1 0. (*)

Bin i tng ng h phng trnh v dng:

[og2(x2 + y 2)^ io g 2(2xy) fx 2 + y2 =2xy [x = y

.-.'..*..2 ,4 ~ 1V * 1 x - V + x - 43 X -xy+y - 3-+ ^ x ^ - x y + y "

X = y 2

X = y ~ -2

Vy, h phng trnh c hai cp nghim (2; 2) v (-2; -2).

_ x = yi ^

X = 2

TH I MN TON KHI B NM 2009

PHN CHNG CHO TT C CC TH SINH (7-0 im)

Cu I: (2 im):Cho hm s:

y = 2x4- 4 x 2. (1)1. Kho st s bin thin v v th (C) ca hm s (1).

2. Vi gi no ca m phng trnh X2 1X2 2 - m c ng 6 nghim phn bit.Cu II: (2 im)

1. Gii phng trnh: s inX+ COSX.sin2x + V3 cos.3x = 2(cos4x + sin3 x).

x y + X + = 7 y2. Gii h phng lnh: < ' - -.

[x y + xy +1 = 13y

Cu III: ( im):Tnh lch phn I = ^ -dx.i (* + 0

73



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73/394

Cu IV: ( im):Cho hnh lng tr tam gic ABC.ABC c BJET = a, gc gia

ng thng BB v mt phng (ABC) bng 60, AABC vung ti c v BAC = 60.Hnh chiu vung gc ca im B trn mt phng (ABC) ng vi trng tm caABC. Tnh th tch khi t din A ABC theo a.

Cu V: (1 im):Cho cc s thc X, y thay i tho mail (x + y) + 4xy > 2.Tm gi tr nh nht ca biu thc:

PHN RING (3.0 im)Th snh c lm mt trong hai phn (phn A hoc B)

Cu VI. a (2 im)

1. Trong mt phng vi h to Oxy, cho ng trn (C) v hai ng thng (Aj),(A2) c phng trnh:

Xc nh to tm K v bn knh ca ng trn (Q), bit ng trn (Ci) tipxc vi cc ng thng (Aj), (A2) v tm K thuc ng trn (Q .

2. Trong khng gian vi h to Oxyz, cho t din ABC c cc nhA(l; 2; 1), B(-2; 1; 3), C(2; - I ; 1) v D(0; 3; 1). Vit phng trnh mt phng(P) i qua A, B sao cho khong cch t c ti (P) bng khong cch t D ti (P).

Cu VI. b (2 im)

1. Trong mt phng vi h to Oxy, cho AABC cn ti A c nh A(1; 4) v ccnh B, c thuc ng thng (A): X - y - 4 = 0. Xc nh to im B v c,bit din tch AABC bng 18.

2. Trong khng gian vi h to Oxyz, cho hai im A(-3; 0; 1), B(l; -1 ; 3) v mtphng (P) c phng trnh:

Trong cc ng thng i qua A v song song vi (P), hy vit phng trnhng thng m khong cch t B n ng thng l nh nht.

Cu VILb ( im):Tim gi t ca tham s m ng thng y = -X + m ct

A = 3(x4 + y4+ x2y2) - 2(x2+ y2) + 1.

(C):(x - 2)2 + y 2 =J , (A,): X - y = 0, (2); X - 7y = 0

Cu VII. a ( im):Tm s phc tho mn:

(P): X - 2y + 2z - 5 = 0 .

th hm s' y = ti hai im phn bit A, B sao cho AB = 4.X

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74/394

N H G I V N H H NG T H C H I 3

Cu I.1. Vi hm s:

y = f(x) = ax4+ bx2+ c?vi a 960ta in lt c:

Min xc nh D = R-o hin: o hm cp mt:

y' = 4ax + 2bx = 2x(2ax2+ b).Phng trnh y= 0 hoc c mt nghim (a.b > 0) hoc c ba nghim phn bit.

Do hm s hoc ch c mt cc tr hoc c ba cc tr. o hm cp hai:

y" = 12ax2 + 2b.Do hm s hoc c hai im un hoc khng c im un.

Gii hn:

im y = lim ax4( l +ax ax

) =+00khi a > 00 0 khi a < 0

Bng bin thin: Du ca y' ph thuc vo du ca a (a > 0 hay a < 0) v duca a.b, do ta c bn trng hp bin thin khc nhau.

th ca hm s: o c bn trng hp khc nhau v chiu bin thin nn th ca hm trng phng c bn dng sau y:

Vi a > 0

C mt cc trL

0 X

V i a c O

C ba cc tr

y\i .

/ )

1 1 X

Cu I.1! Ta ln lt c:

a. Hm s xc inh trn D = .b. S bin thin ca hm s:

Gii hn ca hm s ti v cc:

lim y = lim [2x4( 1 - r ) ] - +.

75



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75/394

Bng bin thin:

y - 8x:' - 8x, y' = 0 8x3- 8x = 0 o

- 0

X= 0

X= 1

X -00 4-00

0 0 0

H"CO +00

im un: y = 24x2- 8, " = 0 o 24x2- 8 = 0x =

un l u, 101 . t [ 1 .. 10 .v U J -7=r; - -T -9 ) 2{ s 9

T th, yu cu ca bi ton c tho mn khi v ch khi:0 < 2m < 2 < = > 0 < m < l .

Cu II.1. Bin i phng trnh v dng:

(s n x - 2 s in 3x) + cosx.s in2x + -v/ cos3x = 2cos4 x

( l - 2 s i n 2 x )s in x + co sx .sin2 x + %/3 cos3x = 2c os 4x

co s2 x.sinx + cosx .sin2x + V3 cos3x = 2co s4 x

VI y" i du khi X qua cc im nn th hm s c hai imV3

c. th ca hm s: Ta tm thm vi im trn th A(-2; 0), B(2; 0).2. Vit li phng trnh di dng: 12x4- 4x21= 2m

Phng trnh c ng 6 nghim phn bit khi v ch khi ng thng y = 2m ct th hm s y= |2x4- 4 x 2 ti 6 im phn bit.

th y = 12x4- 4x21 gm:

+ Phn t trc honh tr in ca th (C): y = 2x4 - 4x2.+ i xng phn th (C) pha di trc honh qua trc honh.

76



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/ 1 yfjsin 3 x 4- V3 cos3x = 2c os4 x -sin3x + ~ ~ c o s3 x = cos4x

2 2

cosl 3x | = cos4x

n

X = + 2k7T6

271 , k e z .A _ 71 71 * j7 4x = -3 x + + 2kti X - -+ k 16 L 42 7

Vy, phng trnh c hai h nghim.>. V y = 0 khng phi l nghim ca phng trnh nn ta bin i:

X 1X = 7

y yX

+ -y y'

(V1 X

X + + = 7 y

x 2 + - + = 13 r x _j_i _ 2i = 13

l y y 11 y1 X

t u = X + v V- , h phng trnh c bin i v dng:y y

u + V = 7 u + V = 7

u2 - V = 13 | u 2 + u - 2 0 = 0

u + V = 7

u - -5

u = 4

u = -5 & V = 12

u = 4 & V = 3

Ta ln lt: Vi u = -5 v V= 12 th:

X + - - 5y

* =12

112y + -" = -5 12y2+ 5 y + 1= 0

y -S , v nghim._ n [x = 12y

X = 12y

* Vi u = 4 v V= 3 thi:1

X += 4y

x _ ->

.yX = 3 & y = 1

X = 1 & y = 1 / 3

13y + = 4 3y2- 4 y + l = 0

. y ' i . -

l x =3yX = 3y **

y = l

_y -1 /3

X = 3y

Vy, h phng trnh c hai cp nghim (3; 1) v 1; .

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'u III. t:

u = 3 + Inx

dxdv =

( x + i

du = X

1

x + 1

Khi :

o ! (3 33 + lnx rI = - I +

dx 3 ln3 dx

X + 1 '1 ] x(x + 1)

B _ (A + B)x + A

Vi r , ta c:

__ A

x ( x + 1 ) X X + 1 x ( x + 1)

T , tch phn r c vit li di dng:

x(x + 1) V ^

V

(1)

x + 1r = ) ( x " x + r ) dx = ( l n lx | - l n lx + l | ) l = in

Thay (2) vo (I ), ta c I = ^ ^ + in .

Cu IV.Bn c t v hnh.

Gi D l trung im ca AC v G l trng tm AABC, ta c:

BG 1 (ABC) =>Tb G = 60,

= In. (2)2

nrBG = BB.sinB'BG = , BG =

2 2BD =

3a

T'rong AABC, ta C:

BC = 5 ^ , A C = 2 2

CD =AB

,^ 2 _ 3AB2 AB2 9a2BC + CD2= BD2 c=>

16 16

13 26 104(vtt ) .

I __ 9a 3Khi : VA. ABC = VB. ABC = ^ B 'G .S bc =

Cu V. Kt hp bt ng thc (x + y)2 > 4xy vi bt ng thc iu kin, ta c:

(X + y + (X + y)2> 2 o X + y > 1.

Khi :

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| a - b 2V2V t , bn knh ng trn (Q) c cho bi: R j = =~=

8 4 Nj ' , 2V2Vy, ng trn (C() c tm K-; I v bn knh R= -.

2. mt phng (p) cch u hai im c , D ch c th xy ra hai trng hp:

(p) song song vi CD.

(P) i qua trung im ca CD.T nhn xt trn, ta ln lt:

(P) song song vi CD, suy ra:

f Qua A(p) i

ICp vtcp ABv CD

(P) -'Qua A(l;2;l)

,vtpt n = [B, CD] = (-8; - 4 ; - 14 )

(P): 4x + 2y + 7z - 1 5 = 0

(P) i qua trung im I( 1; 1; 1) ca CD, suy ra:

- Qua A _ Qua A(Z;2;J)(P)H __


80/394

Xy 4 = 0

2 r 97 (x +1) + (y - 4) =

Vy, to hai im b ; J v c^; - j hoc ngc li.

2. Gi () l ng thng cn tm.

Ta c: (d) nm trorig mt phng (Q) qua A v song song vi (P), suy ra:

^ Qua A ^(Q) :{ , (Q): X2y + 2z + = 0.

Gi K, H theo th t l hnh chiu vung gc ca B trn (d) v (Q), ta c:

BK > BH => AH l ng thng cn tm.

T , suy ra to ca H tho mn h phng trnh:

X1 _ y +1 _ z - 3 IX1/91 -2 = T ~ ^ y = 11/9 => H

_ - 1 9 9 9x -2 y + 2z + l = 0 2= 7/9

Khi , phng trnh ng thng (d) c cho bi:

Qua A(-3; 0;1)

(d):

X= -3 + 26t (d ): {y = 1lt , t e K .

z = 1- 2t

Cu VILb. Phng trnh honh giao im ca (d) vi th hm s l:

x 2 - l- -X + m

C5- f(x) = 2x2 - mx - 1 = 0 vi X* 0.

ng thng (d) ct th hm s ti hai im phn bit khi:

Af > 0 m2+ 8 > 0, lun ng.

Khi , (1) c hai nghim phn bit X,, x2 honh ca A. B tho mn:

(1)

mX, 4- X, =

GIỚI THIỆU VÀ HƯỚNG DẪN GIẢI CHI TIẾT ĐỀ THI TUYỂN SINH ĐẠI HỌC KHỐI D -...

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Transcript of GIỚI THIỆU VÀ HƯỚNG DẪN GIẢI CHI TIẾT ĐỀ THI TUYỂN SINH ĐẠI HỌC KHỐI D -...