luyện thi đại học môn toán

Tuye'n chQtt & Giai thifu dethi Todn UQC - Nguyen Phu Khanh, Nguyen Tai Thu.. PHAN RIENG Thi sinh chi dupe chpn lam mpt trong hai phan phan A hole B A... TwygH chgn & Giai thifu dethi To

Tuye'n chQtt & Giai thifu dethi Todn UQC - Nguyen Phu Khanh, Nguyen Tai Thu

nen A H = D H = ^ va B C ± ( A H D )

=> ( A B C ) 1 ( A H D )

Ke D K 1 A H => D K 1 ( A B C )

=>DAK = 45°, D A H = 45°

=> ADAK vuong can tai K

ADAH vuong can tai H

=>K = H : ^ D H l ( A B C )

Di^n tich tarn giac ABC la:

SABC = •^ABACsin60° = — ^

The tich khoi tu dien ABCD la V = - D H S A B C = T " = I T

Ke HE 1 AB => DE 1 AB Vay goc giiia 2 mp (ABD) va (ABC) la goc giua

hai duang thSng DE va HE va bang DEH

Gpi CF la duong cao xuat phat tir C cua tarn giac deu ABC canh a

nen co CF = , HE = - C F =

2 2 4 nen tan DEH = DH

HE = 2 =>DEH = arctan2

Vay, goc giua hai mp ( D A B ) va (ABC) la DEH = arctan2

Cau 6: Xet ham so: f(t) = ln(t + 1 ) - Vt+T voi t > - 1

Taco: f ( t ) = ^ " , ^ va f ( t ) = 0 o t = 3

^ ' 2(t + l ) ^ '

f'(t) doi dau tu duong sang am khi qua 3 nen f (3) la gia tri Ion nhat

Suy ra f (t) < f (3) < 0, do do f(x) + f(y) + f ( y ) < 0

hay In(x +1) + In(y +1) + ln(z +1) < Vx + 1 + ^ y + 1 + Vz + 1

I I PHAN RIENG Thi sinh chi dupe chpn lam mpt trong hai phan (phan A

hole B)

A Theo chi/ang trinh chuan

Cau 7.a: Duong tron c6 tam l(l;-2),ban kinh R = 3

Vi M e ( d ) nentpa dp M ( t ; t + l )

Cty TNHH MTV DWH Khang Vift

£)e tu M CO the ke dugc hai tiep tuyen den ( C ) thi I M > R <=> 2t^ + 4t +1 > 0

„ , > ^ h o , c , < - : ^ ( )

Phuong trinh di qua hai tiep diem A , B c6 dang:

( t - l ) ( x - l ) + (t + 3)(y + 2 ) - 9 = 0

3t + l : Ta c6: d ( N ; A B ) = ,

2V2t^+4t + 10 3t + l

TwygH chgn & Giai thifu dethi Toan hpc - Nguyen Phu Khanh, Nguyht Tat Thu

M $ t p h i n g (P) d i qua M(l; 2; 1) c6 vecto phap tuyen n = U j = ( l ; 2 ; - l )

p h u o n g trinh la: l ( x - l ) + 2 ( y - 2 ) - l ( z - l ) = 0 hay x + 2 y - z - 4 = 0

M^i phan g (P) d i qua M(5; 10; - 3 ) c6 vecto pha p tuye n n = U j = ( l ; 2 ; - l )

p h u o n g trinh la: l ( x - 5 ) + 2 ( y - 1 0 ) - l ( z + 3) = 0 hay x + 2 y - z - 2 8 = 0

Vay, C O 2 mat phang (P): x + 2 y - z - 4 = : 0 hoac x + 2 y - z - 2 8 = 0

Cau 9.a: T i m tat ca cac so phuc z , biet |z - 1 - 2if + zi + z = 11 + 2i (l)

G(?i so phijrc z = a + b i (a,b e # ) thoa man de bai

B Theo chUorng trinh nang cao

Cau 7.b: D u o n g tron (C) c6 t a r n ! ( - ; 2), R = 4 va diem I thuoc d u o n g thang A

D u o n g tron ( C ) c6 tam J ban k i n h R' = l va tiep xiic ngoai v o i d u o n g tron

(C) suy ra quy tich cua d i e m I la d u o n g tron ( K ) C 6 tam I ban k i n h R + R' = 5

h a y ( K ) : (x + l f + ( y - 2 f = 2 5

Khoang each ciia I t o i A la Ion nhat k h i I la giao diem cua d u o n g t h i n g d d i

qua J va v u o n g goc v o i A v o i d u o n g tron (K)

Vay, khoang each t u I t o i A Ion nhat bang 5

^ I t t S b : T i m dugc tpa dp d i e m A ( 7 ; 16; 1 4 ) G(?i U j , i 3 ^ , r i p Ian l u g t la cac vecto chi p h u o n g ciia d , A va vecto phap

t^yeh ciia ( P ) Gia s u u j = (a; b; c) a^ + b^ + c^ > 0

Neu x < y t h i p h u o n g trinh cho v6 nghi^m

Neu X = y t h i p h u o n g trinh t h u 2 tro thanh: x + V2x + 1 = 1 + Vx + 2

o£THiTHiirsdi9

I PHAN CHUNG CHO TAT CA CAC THI SINK

Cau 1: Cho ham so y = - 3x^ + (3m - 3)x + 2 c6 do thi la ( C „ )

a) Khao sat su bien thien va ve do thi (C) cua ham so khi m = 1

b) T i m m de ham so c6 eye d^i, eye tieu eiing v o i diem l ( - l ; - l ) t^o thanh

tam giae vuong tai I

Cau 2: Giai phuong trinh:

sin^ X - Vscos^x - i s i n 2x 2 (sin x - cos x) - — s i n 2x

Cau 3: Giai h § phuong trinh:

x 3 + ( 2 - y ) x 2 + ( 2 - 3 y ) x - 5 ( y + l ) = 0

Cau 4 , , , , ^ r fln^x-31nx + 3

: Tmh tich phan: I = -, r— d x

J x { l n x - 2 )

Cau 5: Cho tam di?n Oxyz c6 xOy = yOz = zOx = a Tren Ox, Oy, Oz lay cac

diem A, B, C sao cho OA = OB = OC = k > 0 T i m dieu ki?n eua a de t u di^n

OABC CO the tich Ion nha't

Cau 6: Cho a,b,e>0 thoaman: (a + b - e ) ( b + c - a ) ( e + a - b ) = l

Chung minh rang: ' a + b + c a^+b^+c^

II PHAN RIENG Thi sinh chi dugrc chpn lam mpt trong hai phan (phan A

hole 6)

A Thee chUorng trinh chuan

Cau 7.a: Trong mat phSng tpa dp Oxy, cho duong tron ( K ) : x^ + y^ = 4 va hai

diem A (0; 2), B ( 0 ; - 2 ) Gpi C, D (A A, B) la hai diem thupe ( K ) va doi xung

voi nhau qua true tung Biet r i n g giao diem E eua hai duong t h i n g AC, BD

n l m tren duong tron (KJ ) : x^ + y^ + 3x - 4 = 0, hay tim tpa dp eua E

Cau 8.a: Trong khong gian tpa dp Oxyz, cho hinh thoi ABCD eo dinh B thupc

trye Ox, dinh D thupe mat p h l n g (Oyz) va duong eheo AC nam tren duong

t h i n g d : = ^ = j T i m tpa dp cac dinh A, B, C, D ciia hinh thoi ABCP

biet di?n tich hinh thoi ABCD b i n g ISyJl (dvdt)

126

3 + 51

*Su 9.a: T i m so phuc z thoa man z + —^ 5i = 0

" z Theo chi/crng trinh nang cac

Cau 7.b: Trong mat p h i n g tpa dp Oxy, cho hinh thang vuong ABCD, vuong tai

va D Phuong trinh A D : x - yyjl = 0 Trung diem M eiia BC eo tpa dp M ( l ; 0)

giet BC = CD = 2AB T i m tpa dp eiia diem A

Cau 8.b: Trong khong gian toa dp Oxyz, cho mat phang ( ? ) : 2 x + y + z - 6 = 0

va m | t cau (S): x^ + y^ + z^ + 4x + 6y - 2z J-11 = 0 T u diem M tren (P) dyng tiep tuyen M N den mat cau ( N la tiep diem) Tim M de M N ngan nhat, tinh Idioang each ngan nhat do

Cau 9.b: Giai phuong trinh sau: ^ l o g ^ (x^ + 2x j - log j (x + 3) = logg - — ^

H(/dNGDiiNGlAl

I PHAN C H U N G C H O T A T C A C A C THf S I N H

C a u l : a) Danh cho ban dpc

b) Taeo: y' = 3x^ - 6 x + 3 m - 3 = 3(x^ - 2 x + m - l ) Ham so'CO eye dai, eye tieu khi y' = 0 eo hai nghi^m phan bi?t Xj, Xj, y' trift tieu va doi dau qua moi nghi^m, nghla la phai eo:

A ' > 0 < = > l - ( m - l ) > 0 o m < 2 Voi m < 2 thi do thj ham so luon eo eye dai A(xj;y(xj)), eye tieu B(x2;y(x2))

V i l y ' ^ ' ^ ^ = ° nen: jyi'^l) = ( 2 m " 4 ) x i + m + 1 [y'(x2) = 0 y(x2) = ( 2 m - 4 ) x 2 + m + l Khi do: I A = ( x i + 1 ; ( 2 m - 4 ) x i + m + 2), iB = ( x 2 + 1 ; ( 2 m - 4 ) x 2 + m + 2) Tam giae lAB vuong tgi I nen c6: lA.IB = 0

I o ( x i +l)(x2 + l ) + [ ( 2 m - 4 ) x j + m + 2 ] [ ( 2 m - 4 ) x 2 + m + 2] = 0

o ( 4 m ^ - 1 6 m + 17)xiX2 +(2m^ - 7 ) ( x i +X2) + m^ +4m + 5 = 0 (•) Theo dinh ly V i - et: xj + X2 = 2, Xj Xj = m - 1

K h i do (*) tro thanh:

127

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|4m^ -16m + 1 7 J ( m - l ) + (2m^ - 7J2 + m^ + 4 m + 5 = 0

<=> 4m^ - 15m^ + 37m -26 = 0<=>m = l

Do'i chieu dieu ki$n ta c6 m = 1 la gia trj can tim

Cau 2: Phuong trinh cho tuong duong vai (s inx - cosx)^sin x + Vscosx - 2j = 0

sinx - cosx = 0 <=> tanx = 1 o x = — + krt, k e Z

Phuong trinh thu nhat tuong duong vai: x^ + 2x^ + 2x - 5 = y ^x^ + 3x + sj

o ( x - l ) ( x ^ +3x + 5) = y(x2+3x + 5) <^(x^ + 3x + 5)(x - y -1) = 0

THI: x^ + 3x + 5 = 0 ta thay v6 nghi?m

TH2: X - y -1 = 0 hay y = x -1 thay vao phuong trinh thu 2, ta dugfc:

Tuong tif, ta cung c6: BC = CA = AB = 2ksin j nen AABC deu

Trong hinh chop O.ABC ta c6 cac m|it ben deu la tam giac can t^i O day la

am giac deu ABC, nen A.ABC la hinh chop tam giac deu

Gpi G1^ trpng tam A A B C , the thi OG 1 ( A B C ) , khi do VO.ABC =-^&ABC-^

AN la duofng cao AABC deu

Taco: A N = — B C = k V S s i n

-2 -2

^ A G = ^ A N = ^ s i n P

3 3 2 Xet AAOG vuong t^i O , ta c6:

O G 2 = A 0 2 A G 2 = k 2 l s m ^

Dodo, V o A B C f - ^ l - - s i n 2 - (dvtt)

3 2 ^ ' DSt t = sin^ —, do 0 < a < — nen 0 < t < —

2 3 4 Khido: V o A B C = ^ k ^ , , t^ t^ voi te f 3^ 0;-

129

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Tuyen CUQU & Gi&i thifu dethi Toan htn !sJ<^uii,'-ii I'hii Kh,ii:h \^u\ihi Iiil lUu.^

Dieu can chiing minh tro thanh: X + y • + z ^ + + + xy + yz + zx

Dat ^ + y + ^^ t > l K h i d 6 (•) tro thanh t^-|t2+i>() vai t > l

3 1 Xethamso: f ( t ) = t^ — t ^ + - voi t > l V ; 2 2

Taco: f (t) = S t ' ' - 3 t , voi V t > l thi f ' ( t ) > 0 dodo f(t) dongbie'n voi t > l

Tud6,tadu(?c f ( t ) > f ( l ) = 0

I I PHAN R I E N G Thi sinh chi dugic chpn lam mpt trong hai phan (phan A

hoac B)

A Theo chUorng trinh chuan

Cau 7.a: Vi C, D thupc duong tron (K) ma lai doi xiing voi nhau qua true tung

nen toa do 2 diem CO dang la: C(a;b), D(-a;b) (a,b ^O)

Taco: a^+h^-^^ ( l )

Plurong trinh duong thang:

A C : ( b - 2 ) x - a ( y - 2 ) = 0, BD:(b + 2)x + a{y + 2) = 0

{ 2a ( b - 2 ) x - a ( y - 2 ) = 0

Tu (l) va (2) suy ra 8a^ - 6ab = 0 o 4a = 3b

Cau8.a:Goi B(b;0;0), D(0;di;d2) la toa do thoadebai

Goi (Q) la mat phSng di qua B(b;0;0)va vuong goc vai duong thSng AC fien nhan u ^ ^ = ( 2 ; - l ; l ) lam vecto phap tuyen, nen c6 phuang trinh:

2 x - y + z - 2 b = 0 Goi I la trung diem AC nen I € d => I (-3 + 2 t ; - t ; t ) Han nOa, I cung la giao diem cua d va (Q), nen toa do diem I thoa phuang trinh:

YB + YD = 2 y i , tim du(?c:

^B ^^D

Vi AC nam tren duang thSng d nen goi t j , t2 e ^ sao cho:

A ( - 3 + 2 t i ; - t i ; t i ) , C ( - 3 + 2t2;-t2;t2) => A C = (2t2 - 2 t i ; t i - t j ; t j - t j ) Theo de bai ta c6 di^n tich hinh thoi bang 18N/2 , suy ra i A C B D = IsVi o^/6(v-tO^-3V3 = 18N/2 hay t 2 - t i = 2 ( l )

A B C D la hinh thoi nen c6: A B = C B => t j +1^ - 6 (2)

Tu (l) va (2) tadupc t j = 2 : ^ A ( 1 ; - 2 ; 2 ) , t2 = 4 =^C(5;-4;4)

Cau 9.a: Gia su z = x + yi, (x, y e ^ ) suy ra z = x - yi

Theo gia thiet, ta c6 : z.z + 3 + 5i - 5i.z = 0 <=> x^ + y^ + 3 + 5i - 5xi - 5y = 0

GQi I la tam mat cau, tam giac IMN vuong tai N, ta c6: IN^ + MN^ = IM^

Ma IN khong doi nen MN ngan nhat khi IM ngan nhat, tiic la M la hinh chieu

cua I len mat phang (?)

Cau 9.b: Dieu ki#n: x > 0

?huong trinh cho tuong duong vol

o log3 (x^ + 2x) + logg (x + 3) = log3 3 - logg (x +1)

log3[x(x + 2)(x + 3)(x +1)] = log3 3 o [ x ( x + 2)(x + 3)(x +1)] = 3

<=>]

o [ x ( x + 3)(x + 2)(x + l)] = 3 « (x2+3x)(x2+3x + 2) =3 (*)

D | t t = x^ + 3x, phuong trinh (•) tro thanh: t^ + 2t - 3 = 0 o t = -3, t = 1

V6i t = l tuc x^ + 3x = 1 <:> x^ + 3x - 1 = 0 o x = ~^ thoa dieu ki?n

Vol t = -3 tuc x^ + 3x = -3 o x^ + 3x + 3 = 0 phuong trinh nay v6 nghifm

Vly, phuong trinh cho c6 nghi?m x = -3 + M

Cty TNHHMIV I )VVI rKhangVi$t

DETHITH(jfSd20

I PHAN CHUNG CHO TAT CA CAC THI SINH

Cau 1: Cho ham so y = x"* - 2mx^ - 1 c6 do thj la (C^), m la tham so

a) Khao sat su bien thien va ve do thi (C) cua ham so khi m = 1

b) Tim m de ham so c6 ba eye trj tao thanh mot tam giac c6 dp dai c^nh day gap doi ban kinh duong tron ngogi tiep tam giac do

n (67rx - 371^ jcosx.sin^ x + 47i(l + sin^ x)

Cau 4: Tinh tich phan: I = '- = _ — 5 ^ ^ x

0 Vl + sin-'x

Cau 5: Cho hinh chop S.ABCD, day la hinh chu nhat c6 AB = 3, BC = 6, mSt

phJing ( SAB) vuong goc voi mat phMng day, cac mat phSng (SBC) va (SCD) cimg tao voi m | t phSng (ABCD) cae goc bang nhau Biet khoang each giua hai duong thSng SA va BD bSng S Tinh the tich khoi chop S.ABCD va cosin goc

giiia hai duong thing SA va BD

Cau 6: Cho a, b, c la cac so thyc duong thoa man a + b + c = 3

Chung minh rang : 8\/abc < 9

11 PHAN RIENG Thi sinh chi duQC chpn lam mpt trong hai phan (phan A

hoac B)

A Theo chiTorng trinh chuan

Cau 7.a: Trong mat phang tpa dp Oxy, cho tam giac ABC vuong can tgi A,

phuong trinh BA: 2x - y - 7 = 0, duong thang AC di qua diem M ( - l ; 1) diem A nSm tren duong th3ng A: x - 4y + 6 = 0 Tim tpa dp cae dinh cua tam giac ABC biet rang dinh A eo hoanh dp duong

Cau 8.a: Trong khong gian voi hq tpa dp Oxyz, cho hai duong thang

x _ y + 1 _ z x - 1 _ y + 1 _ z - 4

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Tuyen chqn & Gi&i thifu de thi Todn hqc - Nguyen Phu Khdnh , Nguyen Tat Thu

Viet phuong trinh duong t h i n g A cat ca hai duong t h i n g d ^ d j dong thoi

vuong goc v6i mat phang (P): x + 4 y - 2 z + 5 = 0

z - 1 = 5 va 17 z + z = 5zz

V >

B Theo chi/tfng trinh nang cao

Cau 7.b: Trong mat p h i n g toa do Oxy, cho 2 duong t h i n g Ian lugt c6 phuong

trinh la ( d j ) : 2x - 3y - 3 = 0 va ( d j ) : 5x + 2y - 1 7 = 0 Viet phuong trinh duong

t h i n g d i qua giao diem cua ( d j ) , ( d j ) Ian lugt cat cac tia Ox, Oy tai A va B

2

sao cho AB

' A O A B J

dat gia trj nho nha't

Cau 8.b: Trong khong gian voi h§ tga do Oxyz, cho diem A(3; - 2 ; - 2 ) va mat

p h i n g (P): x - y - z + l = 0 Viet phuong trinh mat p h i n g (Q) d i qua A, vuong

goc voi mat p h i n g (P) biet rang mat p h i n g ( Q ) cat hai tryc Oy, Oz Ian lugt

tai diem phan biet M va N sao cho O M = O N

Cau9.b: Giaibat phuong trinh: logj Vx^ - 5 x + 6 + log^ V x - 2 > ^ l o g j (x + 3)

Neu m < 0 thi y ' = 0 c6 1 nghiem x = 0 va doi da'u t u ( - ) sang ( + ) nen c6

1 cue trj ( khong thoa bai toan )

Neu m > 0 thi y' = 0 c6 3 nghi?m x = 0, x = - V m , x = \/m va doi da'u qua

moi nghiem nen ham so da cho c6 3 cue tri

Giasu: A ( 0 ; - l ) , B ( - V m ; - m ^ - l ) , c ( N / m ; - m ^ - l )

Tarn giac ABC can tai A nen canh day la BC vol:

BC = 2 V m , AB = AC = Vm + m *

Ta CO d i ^ n tich tam giac S ^ ^ B C = ^BC.d(A,BC) = m^%/m (dvdt)

Cty TNHHMTVjyWH Khang Vie

T U u - - - ' o AB.ACBC BC +

Theo bai toan, ta co: R = —— = —r- <=>

4S \ A B C 4m^%/rn = 4m

<=> m ' ^ - 2 m N / m + 1 = 0 <=> m = 1 (thoa man) Vay, m = 1 thoa man yeu cau bai toan

Cau 2: Dieu k i ^ n : x 5^ m - , m e Z

4

„, u u A sinx cos2x 2 { c o s x - l )

Phuong trinh cho tuong duong: + = — '—

cosx sin2x 2sin2xcos2x

1 cos X - 1

sin2x sin2xcos2x <=> cos2x = cos X - 1 o 2 cos^ x = cos x cos X = 0

1 « cos X = —

X e t h a m so: f { t ) = t^ + 2\/t voi t > 0 , ta c6: f ' ( t ) = 2t + - ^ > 0 i : i f ( t ) dong

v t

bien t > 0 , khi do phuong trinh: f(x + l) = f ( y ) o y ^ x + 1

Thay vao p h u o n g trinh t h u hai ta dugc: Vx + 1 + 7 x - 2 = 3

ruye'n chgn & Giai thifu dethi Todn hQC - Nguyen Phu Khdnh, Nguyen Tat Thu

Khi do I = (67rx - 3n^ j^Vl + sin' = 471^

Zau 5: Ha SH 1 A B => S H 1 ( A B C D ) (do (SAB) 1 ( A B C D ) = A B )

Ke H K 1 CD => t u giac H B C K la hinh chu nhat

Gpi P la goc giua hai duang thiing B D va S A => p = ( B D , S A ) = ( A K , S A )

Ta C O S K = 6V2,SA = A K = 3N/5 Trong tam giac S A K

A S ^ + A K ^ - S K ^ 45 + 45 - 72 1

cos S A K =

2AS.AK 2 3 V 5 3 V 5 5

Vgy p = SAK =

arccos-Cau 6: Ta c6: a^ + + c^ = (a + b + c)^ - 3{a + b)(b + c)(c + a) < 27 -24abc

Khi do: ?

3 V 3

V

D|t t = abc, ta C O 0 < t < a + b + c' = > 0 < t < l

Xet ham so: f (t) = ^ 9 - 8 t + 8 ^ voi t € (0;1^

Cty TNHH MTV P W H khang Vifi

Taco: f ( t ) = |

Nh$n thay, ( 9 - 8 t ) ^ - t ^ = 9 ( l - t ) ( 9 - 7 t ) > 0 =>f'(t)<0 vol mpi t 6 ( 0 ; l )

guy ra f (t) nghich bien voi mpi t e (0;l] va f (t) < f (l) = 9 Dang thiic xay ra khi a = b = c = 1

JI PHAN RIENG Thi sinh chi du-grc chpn lam mpt trong hai phan (phan A holcB)

A Theo chi/orng trinh chuan

Cau 7.a: Ggi diem A e A => A(4yo - 6; yg) nAC = (yo -1;5 - 4yo)

Tam giac ABC vuong can tai A, nen c6:

6 y o - 7 1 cosACB =

^ 5 ( l 7 y 2-42yo+26) ^

« 1 3 y 2 - 4 2 y o + 3 2 = 0<=>yo=2=>Xo=2 hoac yj, = ^ = > XQ = ( l o a i )

Vay,A(2;2), B(3;-1), C(5;3) la tpa dp can tim

Cau S.a: Phuong trinh tham so ciia d j :

Cau 9.a: Gia su z = x + y i (x, y e M^^ => z = x - y i

B Theo chUtfng trinh nang cao

Cau 7.b: Giao diem ciia ( d , ) va ( d 2 ) l a M ( 3 ; l )

Cdch 1: S,^oAB = ^ A B O H v o i H la chan d u o n g cao ha tir O len A B

AB N2

J O H ^ V i O H < O M OHmax = O M thi nho nhat O H '

K h i do A B nhan O M lam vec to phap tuyen Ta vie't duoc p h u o n g trinh A B

Cach 2: Phuong trinh d u o n g thang d c6 dang: a(x - 3) + b ( y - l ) = 0 ,(a,b > O)

Theo bai toan, ta t i m du^c:

Phuong trinh d u o n g thang can t i m la: 3x + y - 1 0 = 0

Cau 8.b: Gia su n g la m p t vecto phap tuyen ciia ( Q )

K h i do n Q l n p ( l ; - l ; - l )

M a t p h ^ n g ( Q ) cat hai tryc Oy va Oz tai M ( 0 ; a ; 0 ) , N ( 0 ; 0 ; b ) phan bi?t sao

cho O M = O N nen a = b o a = b^O hoac a = - b * 0

= { 0 ; 1 ; - 1 ) nen n ^ = u,np

K h i do mat p h l n g ( Q ) : y - z = 0

( Q ) c i t O y , Oz tai M ( 0 ; 0 ; 0 ) va N ( 0 ; 0 ; 0 ) (loai)

Vay, ( Q ) : 2x + y + z - 2 = 0

Cau 9.b: Dieu kien: x > 3

Phuong trinh da cho tuong duong:

Tuyen chgn & Giai thifu dethi Todn hgc - Nguyen Phu Khdnh, Nxuyht Tai Thu

OETHITHUfSdzi

I PHAN C H U N G C H O T A T C A C A C T H I S I N H

Cau 1: Cho ham so : y = x'^ - 3x^ + mx +1 c6 do thi la (C^^)

a) Khao sat sy bien thien va ve do thi (C) cua ham so khi m = 0

b) Tim m de ham so c6 cue dai, cxfc tieu Gpi (A) la duong thang di qua hai

diem eye dai, cue tieu Tim gia trj ion nhat khoang each tir diem I - ; — den

Cau 5: Cho hinh hpp dung ABCD.A'B'C'D' eo day la hinh thoi e^nh a, BAD=a

voi cosa=-, canh ben AA' = 2a Gpi M la diem thoa man DM = k.DA va N la

4

trung diem cua canh A'B' Tinh the tich khoi tu dien C'MD'N theo a va tim

kde C ' M I D ' N

Cau 6: Cho 3 so thuc khong am a, b, c thoa man a + b + c = 3 Tim gia tri nho

nhai cua bieu thuc: P = a + b^ + c^

II PHAN R I E N G Thi sinh chi dupe chpn lam mpt trong hai phan (phan A

hoac B)

A Theo chUorng trinh chuan

Cau 7.a: Trong mat phang tpa dp Oxy, cho tam giac ABC vuong tai A va diem

B(1;1) Phuong trinh duong thSng AC: 4x + 3y - 32 = 0 Tia BC lay M sao cho

5>/2 BM.BC = 75 Tim C biet ban kinh duong tron ngoai tiep tam giac AMC la — ^

Cau 8.a: Trong khong gian Oxyz, cho hai duong thSng (dj): = =

(d;): ^ 2 ~^ \ \g (P): x + y-2z + 5 = 0 Lap phuong

trinh duong thSng (d) song song voi m^t phSng (P) va cat (d^), ( d j ) Ian lupt

tai A, B sao cho dp dai doan AB nho nhat

Cau 9.^: Tinh modun cua so'phiic z, biet: z = (2 - i)"^ + ( l + i)'* - ^ — i -

0 Theo chUomg trinh nang cao

Cau 7.b: Trong mSt phSng tpa dp Oxy, cho hinh vuong ABCD eo phuong trinh duong thing AB: 2x + y - 1 = 0, va C, D Ian lupt thupc 2 dupng thing

d j : 3x - y - 4 = 0, dj : x + y - 6 = 0 Tinh di|n tich hinh vuong

x = - t Cau 8.b: Trong khong gian Oxyz, cho duong thang (d): y = 1 + 2t va mSt cau

[z = -3-2t (S): x^ + y^ + z^ - 2x - 6y + 4z -11 = 0 Viet phuong trinh mat phing (p) vuong goc duong thang (d), cat mat cau (S) theo giao tuyen la mpt duong tron c6 ban kinh r = 4

Cau 9.b: Tim so phue z thoa man (l - 3i) z la so thuc va z - 2 + 5i = 1

x _ l _

3 3 2m -2 x + — + 1 m , Chia da thiic y cho y', ta dupe: y = y'

Gia sir ham so c6 eye d^i, eye tieu t^ii cae diem (xi;yi),(x2;y2) •

Vi y'(xj) = 0,y'(x2) = 0 nen phuong trinh duong thing (A)qua hai diem

eye dgi, eye tieu la: y = r2m_2^ + ^ + 1 hay y = —(2x + l)-2x + l

( \

Ta thay, duong thang (A) luon di qua diem co'djnh A —;2 so'goc

|*a duong thing lA la k = | Ke IH 1 ( A ) ta thay d(l;A) = I H ^ I A = |

I Ding thuc xay ra khi I A ± ( A ) o ^ - 2 = - i = - - < » m = l V^y, max d ( l ; A ) = | k h i m = 1

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Tuye'n chgn b Giai thi?u dethi Todu hgc - Nguyett Phu Khanh , Nguyen Tat Thu

Cau 2: Phuang trinh cho tuong duong voi phuang trinh:

sin3x + 3sinx = 4sin^ x.cosx + 2cosx + sinx-cosx

<=> sin 3x + s inx = 2 sin x.sin 2x + cos x - s inx

o 2 sin 2x.cos X - 2 sin 2x.s inx = cos x - s inx

<=>(cosx-sinx)(2sin2x-l) = 0 o c o s x = sinx hoac 2 s i n 2 x - l = 0

Voi cosx = sinx <=> x = — +

Cau 3: Dieu ki#n: x > 3

Phuong trinh cho tuong duong: 5(2Vx-3 - N/2X + I ) = 2X -13

Nhan 2 ve voi bieu thiic lien hg-p va dat thua so chung:

( 2 X- 1 3 ) ( 5 - 2 N / X ^ - N / 2 X + I ) = 0 c^x = y hoac 2N/)r^ + N/2X +1 = 5

, f x < 6 2^/x^ + ^ / 2 ^ = 5<=>27(x-3)(2x + l ) = 18-3x<^• X^ - 8 8 X + 336 = 0

z:>X = 4

13 Vgy, phuong trinh cho c6 nghiem la: x = - y , x = 4

D^t P(b) = 3 - b - c + b2+c^ voi be[U,3'

Taco: P'(b) = 2 b - 1 va P'(b) = O o b = i

Tirdo, ta duoc P > m i n P = c''-c + —

4 Xet P(c) = c 3 - c + J voi ce[0;3"

Theo chi/orng trinh chuan

7.a: Gpi I la tarn duong tron ngoai tiep tarn giac AMC

Vi B n i m ngoai duong tron ( l )

nen ta c6: BM.BC = BM.BC ( l )

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V^y, phuong trinh duong t h i n g (d) la: = =

Phuong trinh duong trung tryc I N cua A C => A C n I N = N C ( 8 ; 0 ) hoac Ijifc^^/ ^ 4^ Dfb 6

C(2;8) ^ ^' ^ ' ^ ' ^ ^

Cdch2.Tu M d\mg M K 1 B C , ( K € A B )

G<?i I la trung diem K C => I la tarn duong tron ngogi tiep tam giac A M C

(Do t u giac A K C M npi tiep)

Cau 9.a: Ta c6: (2 - i) = 3 - 4i va ( l + i)'' = (2i)^ = -4

Ta C O A A B C dong d^ng A M B K nen: — = — A B B K = M B B C = 75

M B B K

Phuong trinh duong t h i n g A B qua diem B(1;1) va c6 VTPT (3; - 4 ) :

3 x - 4 y + l = 0

V i A la giao diem cua A B va A C nen A ( 5 ; 4 )

V i ABCD la hinh vuong nen j C D l n

Cau8.a:Dat A ( - l + a;-2 + 2a;a), B ( 2 + 2 b ; l + b ; l + b)

Mat p h i n g (P) vuong goc duong thang (d) nen phuong trinh mat p h i n g (P) bdang: - x + 2 y - 2 z + D = 0

I

=>AB = (-a + 2b + 3;-2a + b + 3;-a + b + l )

Do AB song song v o i (P) nen AB 1 np = ( l ; l ; - 2 ) <=> b = a - 4

d(l,(P),) = V R = - r 2 D + 9

4 ^ o D + 9 = 9<=> D = 0

D = -18

Vay, CO hai mat phang can tim la: - x + 2y - 2z = 0, - x + 2y - 2z - 1 8 = 0

Cau 9.b: Gia su z = x + y i , khi do ( l - 3i)z = ( l - 3i)(a + hi) = a + 3b + (b - 3a)i

I PHAN C H U N G C H O TAT CA CAC THI SINH

Cau 1: Cho ham so: y = x'' - 3x + 2 c6 do thj la (C)

a) Khao sat su bien thien va ve do thj (C) cua ham so

b) T i m toa do diem M thuoc duong thang (d) c6 phuong trinh y = -3x + 2

sao cho t u M ke duoc hai tiep tuyen toi do thi (C) va hai tiep tuyen do vuong

goc voi nhau

Cau 2: Giai phuong trinh: sinxcos2x + cos^ x|tan^ x - 1 j + 2sin"' x = 0

Cau 3: Giai h# phuong trinh:

A B C = 60" Hai mat p h i n g ( S A D ) va ( S B C ) la hai tam giac vuong Ian lugt tai

A va C Dong thai cac mat phMng nay ciing hop voi m^t day mpt goc a Tinh

the tich khoi chop S A B C D theo a va a

T-ty MTV DWH Khung Viei Cau 6: Cho a, b, c l a 3 sothuc duong thoa man: (a + c)(b + c) = 4c^

Tim gia tri Ion nhat cua bieu thiic: Q = — - — + —^— + ———

b + 3c a + 3c bc + ca

II PHAN RIENG Thi sinh chi du<?c chpn lam mgt trong hai phan (phan A hole B)

A Theo chUcrng trinh chuan

Cau 7.a: Trong mat phang tpa do Oxy, cho duong tron ( C ) : ( x - l ) ^ +(y+2)^ =4

M la diem di dpng tren duong thang d: x - y + 1 = 0 Chung minh rSng t u M ke dup-c hai tiep tuyen M T j , M T j toi (C) ( T j , Tj la tiep diem) va tim toa do diem M , bie't duong thang TjTj di qua diem A ( 1 ; - 1 )

Cau S.a: Trong khong gian toa do Oxyz, cho diem M(0; - I ; 2), N ( - l ; I ; 3)

Viet phuong trinh mat phang (R) d i qua M , N va tao voi mat phang (P):

2 x - y - 2 z - 2 = 0 mot goc nho nhat

Tinh mo dun ciia so'phuc w = z

B Theo chUtfng trinh nang cao

Cau 7,b: Trong mat phang toa do Oxy, cho hinh thoi ABCD c6 phuong trinh

hai canh AB va A D theo t h u t u la x + 2y - 2 = 0 va 2x + y +1 = 0 Canh BD chua diem M ( l ; 2) Tim toa dp cac dinh cua hinh thoi

Cau 8.b: Trong khong gian toa dp Oxyz, cho mat cau (S): (x - 2)^ + (y + 2)^ + (z - 1 f = 1

Tim toa dp diem M thupc tryc Oz sao cho t u Mice dupe ba tiep tuyen M A , MB,

MC toi mat cau (S) va diem D ( l ; 2; 5) thupc mat phSng (ABC)

a) Khao sat su bien thien va ve do thj (C) ciia ham so

b) Goi M ( a ; b ) la didm cSn t i m M e (d) => b =-3a + 2 Tiep tuyen cua do thj (C) t?i diem (xo;yo) la y=(3x^-3J(x-Xo)+x^-3xo+^ Cau 9.b: Giai he phuong trinh:

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Tuyin chgn & Giai thifu dethi Toi'ui hoc Nguyen I'lui Khdnh , Nguyen Tat Thu

Tiep tuyen d i qua M ( a ; b )

V|y CO hai diem thoa m a n de bai la: M

Cau 2: Dieu ki?n c o s x * 0

sinxcos2x + cos^ x t a n ^ x - l j + 2sin^x = 0

s i n x ^ l - 2 s i n ^ xj + 2sin^ x - 1 + 2sin^ x = 0 o 2sin^ x + s i n x - 1 = 0

Xet i{t) = t^+/t voi t ^ O T a c o : f ( t ) = 2t + ^ , f ' ( t ) > 0 v o l V t > 0

ham so' f ( t ) d o n g bien tren nua khoang [0;+oo)

f ^ > / x - 2 J = £ ( 3 - y ) o x - 2 = 3- y hay x = 5 - y , thay vao p h u o n g t r i n h t h u

Cty TNIIII M T V m'X'H KhangVi?

Cau 4: Dat: t = cos x = > d t = - s i n x d x

d u = - d t

t _ _ 1

Tuyi'n chgn b Giai thifu dethi Todn UQC - Nguyen Phii Khdnh , Nguyen Tilt Thu

Do tinh doi xurigcua x, y ta dat S = x + y, P = xy =>S + P = 3

=i> P = 3 - S > 0 hay S < 3 De ton tai S, P ta c6:

S 2 - 4 P > 0 < o S 2 - 4 ( 3 - S ) > O o S > 2

S ^ - 2( 3 - S ) + 3S ^ 3 - S ^ S ^ 3 3 3S + (3-S) + 9 S 2 S 2

Tom lai, voi 2 < S < 3, luon c6 Q :

A Theo chi/crng trinh chuan

Cau 7.a: Duong tron (C) c6 tam l ( l ; - 2 ) ban kinh r = 2, M nam tren d nen

M ( m ; m + 1) =:>IM = ^ ( m - l f + ( m + 3 f = ^ 2 ( m + l f + 8

Vi I M > 2 nen M nam ngoai (C), do do qua M ke Avtqc 2 tiep tuyeh toi (C)

Gpi J la trung diem I M nen toa dp diem J

Truong hop ( d j ) : x - y + 3 = 0

Duong thSng ( B D ) di qua M va vuong goc voi ( d , ) nen ( B D ) : X + y - 3 = 0 Suyra B - A B n B D = > B ( 4 ; - l ) , D = A D n BD => D ( - 4 ; 7 )

Gpi I = BD n { d ^ ) 1(0;3) Vi C doi xung voi A qua I nen C

Truong hop ( d 2 ) : 3x + 3y - 1 = 0 Ban doc lam tuong tu

4 13

3 ' 3 j

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