PHAN RIENG Thi sinh chi dugc chpn lam mpt trong hai phan phan A hoac B A... b Tim tpa dp cac diem tren duong thing y = -4 ma tu do c6 the ke den do thj C diing hai tiep tuyen.. PHAN RI
Trang 1Zau 5: Ta c6 AC = BD = 2a Gpi SO la duong cao va SO = h
Trpn h$ true tga dp Oxyz sao cho:
O(0;0;0), A(a;0;0), B(0;a;0), C(-a;0;0), D(0;-a;0), S(0;0;h)
Xac djnh tarn va ban kinh m|it cau ngoai tiep hinh chop
Do hinh chop S.ABCD la tii giac deu
nen I € OS =:i> l(0;0;Zo)
( S ) : x ^ + y ^ + z ^ - 2 z o Z + d = 0
' a 2 + d = 0
h2-2znh + d = 0
A,S€(S):
d = -a^
Zn =• h ^ - a ^
2h
Mat khac:
cos a =
• R =
SA.SB
SA.SB a^+h^
a
r ,Ol =
1 -cosa
a ( 2 c o s a - l )
2^cosa(l-cosa) 2^cosa(l-cosa)
Xac dinh tam va ban kinh mgt cau npi tiep hinh chop
Ta c6: J e OS j(0;0;r), OJ = r
2a\
^S.ABCD - gSTp/VgABCD
SxQ =4S^AB =4.isA.SB.sina = 2(a^ +h^)sina
=>STP =SABCD +SxQ =2(a^ +h^)sina + 2a^
a./cosa(l-cosa) aJcosa(l-cosa)
1 + sina-cosa 1 + sina-cosa
Tim a de mat cau ngoai tiep va npi tiep triing nhau
a^cosa(l - cosa) a ( 2 c o s a - l )
I = J o O I = O J o
-1 + sina-cosa 2^cosa(l-cosa)
<::> (sina - cos a)(sin a - cosa +1) o
Vay I = J o a = 45°
sma = cosa sina-cosa + l > 0 <» a = 45°
242
Cty TNHH MTV DWH Khang Vift
Cau 6: Khong mat tinh tong quat ta c6 the gia su: x > y > z Xethamso: f(x) =
Taco: f'(x) = (x + y + z)
1 1 1
— H + — (x + y + z) xyz + ^2x^ - yzj(y + z)
x^yz > 0, do x^ ^ y z ,
suyra f(x) dongbiehva f ( x ) < f ( l ) = 1 1
1 + - + -
(i+y+z)^-X e t f ( y ) = 1 1 1
1 + —+ -y zj
'"> Chung minh dup-c: f ( y ) < f ( l ) =
\2
Xet f(z) =
(l + y z ) ^
2 + i l ( 2 + z ) '
(2 + z)'^ ,chu'ng minh dupe f (z) < 27
I I PHAN RIENG Thi sinh chi dugc chpn lam mpt trong hai phan (phan A hoac B )
A Theo chUorng trinh chuan Cau 7.a: (C): (x -3)^ + (y +1)^ = 4 c6 tam l ( 3 ; - l ) va ban kinh R = 2 Gia su duong thSng qua P c6 vec to phap tuyen
n ( a ; b ) = > d : a ( x - l ) + b ( y - 3 ) = 0
De d la tiep tuye'n ciia (C) thi khoang each tu tam I den d bang ban kinh:
3a - b - a - 3b
- 2 < » 2a-4b
= 2 o ( a - 2 b ) ^ - a 2 + b 2 o 4 a b - 3 b 2 = 0
a^+b^ 777b^
b = 0=>d: a ( x - l ) = 0 3 > d : x - l = 0
•b(4a-3b) = 0:
b = - i a : ^ d : a ( x - l ) + - a ( y - 3 ) = 0=>d:3x + 4 y - 6 = 0
3 3
T a c 6 : P I = 2V5, PE = PF = V P I ^ - = V 2 0 - 4 = 4 Tam giac lEP dong dang voi IHF suy ra: — - — = — = ^ = Vs
IH EH IE 2
IF 2 EP 4 r- 2 8
= > I H - - ^ = -==,EH = - ^ = - L ^ P H = P I - I H = 2 7 5 - - ^ = - ^
243
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Trang 2fuyen ch<?n & Gi&i thi(u de thi Todn hgc - Nguyen Phu Khanh , Nguyen lat IHu
Cau 8.a: (S) c6 tarn l ( l ; 2; 3) va ban kinh R = 9
Giasu(P) coVTPT n = (A;B;C), (A^ + + > 0)
(P)//BC nen n l B C = (-!;!;4) =^n.BC = 0 o A = B + 4 C ^ n = (B + 4C;B;q
(P) diqua A(13;-l; 0)=^ phuong trinh (P):
(B + 4C)x + By + C z - 1 2 B - 5 2 C - 0
B + 4C + 2B + 3C-12B-52C (P) tiep xiic (S) <^ d[I,(P)] = R « , , , = ^
V ( B + 4 C ) 2 + B 2 + C 2
O B 2- 2 B C - 8 C = 0 O ( B + 2 C ) ( B - 4 C ) = 0C:>B + 2C = 0 ho^c B - 4 C = 0
Voi B + 2C = 0 chgn B = 2 , C - - 1 ,
ta duoc phuong trinh (P);-2x + 2 y - z + 28 = 0
Voi B - 4 C = 0 chon B = 4,C = 1 ,
ta dugc phuong trinh (P): 8x + 4y + z-100 = 0
Cau 9.a: Ta c6 Z j Z j = Z2Z2 = Z3Z3 = 1 nen
Z 1 + Z 2 + Z 3
Z j Z2 Z3
1 1 1
— + — + —
1 1 1
— + — + —
Z l Z2 Z3
Z i Z2 Z3 Z1Z2 + Z2Z3 + Z3Z1
Z I Z 2 Z 3
Z i Z 2 + Z 2 Z 3 + Z 3 Z i | _
I Z 1 Z 2 Z 3
B Theo chUorng trinh nang cao
Cau 7.b: B thu^c d suy ra B :
Z1Z2 + Z 2 Z 3 + Z 3 Z 1
x = t [x = 7 - 2 m
•! , C thuQC d' cho nen C: \
[y = - 5 - t i y = m ( t - 2 m + 9) ^ m - t - 2 ^
Theo tinh chat trgng tarn: => XQ = = 2,yG = ^ = ^
^ [ t - 2 m = -3 [t = - l = > B ( - l ; - 4 )
Duong thMng (BG) qua G(2; 0) c6 vec to chi phuong u = (3; 4), cho nen ( B G )
— = ^ « > 4 x - 3 y - 8 = 0=>d(C;BG) = 2 0 - 1 5 - 8
5
244
Vay duang tron c6 tarn C(5;l) va c6 ban kinh
CauS.b: M 6 ( d ) n r > M ( l + 2t; 3 + 2t;-t)
MA = 7(2 + 2t)2+(l + 2t)2+t2 = s/gt^ +]2t + 5 = sj{3t + 2f+\
MB = V4t^+(l + 2 t ) 2 + ( - t + 5)2 = V 9 t 2 - 6 t + 26 = ^/(3t-1)^+25
Trong mp(Oxy) xet cac vecto u =(3t + 2; 1), v =:(-3t + l ; 5)
Ta c6: u + V •3S, |u| = V(3t +2)2+1, V =v'(-3t + l)2 + 25
Ta luon c6 bat dSng thuc dung: u + V
3>/5 < V(3t + 2f+\ yji^t +1)2 + 25 hay M A + MB>3>/5
D5ng thuc chi xay ra khi u va v cung huong <=> - ^ i i ^ = 1 <=> t = - i
^ ^ -3t + l 5 2 Vay, (MA + MB)^i„ =3^5 d?t duoc khi M
^au 9.b: Dieu ki^n:
x^ - 4x + 3>0 x^ -4x + 35^1
x- 3 > 0
x- 3 ^ 1
X >3
T H I : Neu x > 4 thi log4 Vx^ -4x + 3 > log41 = 0 va log4(x -3) > log41 = 0
Do do bat phuang trinh tuong duang: log4(x - 3) < log4 -/x^ -4x + 3
< » x - 3 < V x 2 - 4 x + 3<=>Vx-3< >/x^ (dung V x > 4 )
TH2: Neu 2 + N/2 < x <4 thi log4 Vx^ -4x + 3 > log4 1 = 0
va Iog4 (x - 3) < iog4 1=0 Suy ra bat phuang trinh v6 nghi^m
TH3: Neu 3< x <2 + x/2 thi log4 \lx^-4x + 3 < log^ 1 = 0
va l o g 4 ( x - 3 ) < l o g 4 l = 0
Do do bat phuang trinh tuong duong: log4(x - 3) < log4 Vx^ -4x + 3
O X- 3 < V X 2 - 4 X + 3 O N / X - 3 < V X ^ ( d i i n g Vx€(2; 2 + V 2 ) )
245
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Trang 3DETHITHl]fS638
I PHAN CHUNG CHO 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) Tim tpa dp cac diem tren duong thing y = -4 ma tu do c6 the ke den do
thj (C) diing hai tiep tuyen
Cau 2: Giai phuong trinh:
s
sin^x + sin^ X + sin^ — + X = 273 sin X + — 7t cosx
Cau 3: Giai phuong trinh: l o g 2 x + logg (2x +1) = 2
Cau 4: Tinh tich phan : I = j - cot X - tan x
I sin 2 X cos
8
^ x
Cau 5: Cho hinh hop chu nhat ABCD.A'B'C'D' c6 AB = 2, AD = 4, A A ' = 6
Gpi I, J la trung diem AB, C' D'
Gpi M, N thoa A M = mAD, BN = mBB' (O < m < l )
Tinh khoang each tu A deh(BDA') Xac djnh ban kinh r cua duong tron
giao cua mat cau (S) ngoai tiep A B D A ' va ( B D A ' )
Cau 6: Cho 0 < c < b < a < l Tim gia tr| Ion nhat cua bieu thuc:
P = a 2 ( b - c ) + b2(c-b) + c 2 ( l - c )
I I PHAN RIENG Thi sinh chi dug^c chpn lam mpt trong hai phan (phan A
hole B)
A Theo chUcrng trinh chuan
Cau 7.a: Trong mat phang Oxy, cho hai duong tron: (Cj): x^+y^ =13 va
( C j ) : (x - 6)^ + y^ = 25 cat nhau tai A ( 2 ; 3) Viet phuong trinh duong thing di
qua A va cat (Cj),(C2) theo hai day cung eo dp dai bang nhau
Cau 8.a: Trong khong gian voi h? tpa dp Oxyz, cho A(3;5;4), B(3;l;4).Hay
tim tpa dp diem C thupc m5t phing ( P ) : x - y - z - l = 0 sao cho tam giac ABC
can tgi C va c6 di?n tich bing 2>/T7
246
t Cau 9.a: Tim so'phuc z c6 modun bang 1, dong thoi so'phuc w = + 2z - 1 ca
modun Ion nhat
B Theo chUorng trinh nang cao
Cau 7.b: Trong mat phing Oxy, cho ABC npi tiep trong duong tron tam I va
A(3; 3) Diem M(3;-l) nam tren duong tron I va thupc cung BC khong chua
diem A Gpi D, E Ian lupt la hinh chieu ciia diem M len cac duong thang BC, AC
Tim tpa dp cac dinh B, C biet ring tryc tam tam giac ABC la diem H(3;1),
duang thang DE c6 phuong trinh la x + 2y - 3 = 0 va hoanh dp ciia B nho hem 2
Cau 8.b: Trong khong gian voi h^ tpa dp Oxyz, cho diem A (3;-2;-2) va mat
phing ( P ) : x - y - z + l = 0 Viet phuong trinh mat phing ( Q ) di qua A, vuong goc voi mat phing (?) biet rang mat phing ( Q ) cat hai tryc Oy, Oz Ian lupt
tai diem phan bi^t M va N sao cho OM = ON
Cau 9.b: Cho so'phuc z thoa man z = 1
Tim gia trj Ion nhat ciia M = z'' - z + 2
HMGDANGIAI
I PHAN CHUNG CHO TAT CA CAC THI SINH
Cau 1:
a) Danh cho ban dpc
b) Gpi A la diem n i m tren duong thing y = -4 nen A(a;-4)
Duong thang A qua A voi hf so'goc k c6 phuong trinh y = k(x - a) - 4 Duong thang A tiep xiic voi do thj (C) khi va chi khi h? phuong trinh sau
CO nghif m:
x 3 - 3 x - 2 = 3 ( x 2 - l ) ( x - a )
~3x2+3 = k
= 0 (1)
(2)
1
I
-x3 + 3 x - 2 = k ( x - a ) - 4
-3x^ + 3 = k (x + l)r2x2 -(3a + 2)x + 3a + 2 [-3x2 ^3^1^
I Phuong trinh (1) tuong duong voi: ^g(,)^2x2 _ ( 3 , , 2 ) x ^ 3 a 2 = 0
1 Qua A ke dupe hai Hep tuyeh deh (C) khi va chi khi (2) c6 2 gia trj k
Wiac nhau, khi do ( l ) c6 diing 2 nghi^m phan bi^t X i , X 2 , dong thoi thoa
= -3xi +3, k j = -3x2 + 3 CO 2 gia trj k khac nhau
x = l
247
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Trang 4chou & Gi&i titieu ae pan noc - nguyen i-nu r<nann -,
- Truang hap 1:
g(x) phai thoa man c6 mpt nghi^m bang - 1 va nghi§m khac - 1
fg(-^) = o
hay 6a + 6 = 0 • a = - 1 kiem tra (2) thay thoa
- Truang hap 2:
g(x) phai thoa man c6 mpt nghifm kep khac - 1
\2
hay
( 3 a + 2 ) -8(3a + 2) = 0 r3(3a+ 2 ) ( a - 2 ) = 0
3a+ 2
a = — hoac a = 2, kiem tra (2) thay thoa
3
Vay, cac diem can tim la A ( - 1 ; - 4 ) , A (2;-4) hoac A
Cau 2: Ta c6 sin^ x + sin — X
v3 J
+ sin^ - + X 71
3
= 2V3sin X + — n
6 j
C O S X
-1 -cos2x + -1 - cos In
[ 3
2x +1 + cos
3
sm X + cos X C O S X
-2K
3 - c o s 2 x - 2 c o s - cos2x ^
— • = 3sinxcosx + VScos^ x
-2 -2
<» 3 = 3sin2x + yf^ilcos^ x - 1 ] o >/3sin2x + cos2x = N/3
, >/3 ^ 1 N/3
<» s m 2 x — + cos2x.- = — « sin
n
- sm —
3
2x + i i = i + k 2 7 i
6 3
2x + —= 7t — + k27i
6 3
<=>
x = — + krc
12
X = — + kn
4 ( k e Z )
Cau 3: Dieu kien: x > 0
Nhan thay x = 2 la nghiem cua phuong trinh cho v i log2 2 + log5(2.2 + l ) = 2
Ta chung minh x = 2 la nghif m duy nhat
That vay, ham so y = logj x,y = logj (2x +1) deu c6 cac co so Ion hon 1 nen
cac ham so do dong bien
248
Voi x > 2 , t a c 6 : log2 x > log2 2 = 1, log^ (2x + ] ) > logg(2.2 +1) = 1
Voi 0 < X < 2, ta c6: logj x < logj 2 = 1, logg (2x +1) < log, (2.2 +1) = 1
Cau 4: I = J - cot x - tan X
cos^ x - s i n ^ X
" sin 2 X cos
8
4
= 2 V ^ f
-J C I
2 x - ^
4
cot2x
^ x = smx.cosx
^ sin2x
8
cos2x.cos + s i n 2 x s i n
-4 -4 j
d x
sin2x(cos2x + sin2x)
<^X = 2^/^t-™*2x
1
Dat t = cot2x dt = ^ — d x => di = —\—
sin 2x 2 sin 2x
Doi can: x = — =>t = l , x = — =:>t = 0
8 4
+ cot2x sin^2x
dx
dx
I = 2V2 ' t ^
1 + t 2 = Vij_i^t = V2'
i l + t i 1
-= V 2 ( t - l n | t + l | ) ^ -= 7 2 ( l - l n 2 ) Cau 5: Chpn he true tpa dp Axyz sao cho:
A ( 0 ; 0 ; 0 ) , B(2;0;0), C(2;4;0), D ( 0 ; 4 ; 0 ) ,
A ' ( 0 ; 0 ; 6 ) , B'(2;0;6), C'(2;4;6), D'(0;4;6)
^ l ( l ; 0 ; 0 ) , J(l;4;6), M ( 0 ; 4 m ; 0 ) , N(2;0;6m)
1 Tinh khoang each t u A den ( B D A ' )
( B D A ' ) : - + ^ + - = l o 6 x + 3y + 2 z - 1 2 = 0
^ ^ 2 4 6 ^
= > d [ A , ( B D A ' ) ] = y
2 Xac djnh tam K va ban kinh R cua
mat cau (S) ngoai tie'p A B D A ' (S): x^ + y^ + z^ - 2ax - 2by - 2cz = 0
(A6(S))
4 - 4 a = 0 [a = l
1 6 - 8 b = 0 = > i b = 2 36-12c = 0 c = 3
t + 1 dt
B, D, A ' e ( S ) :
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Trang 5=>K(1;2;3), R = VTi
Xac dinh ban kinh r ciia duang tron giao ciia ( S ) va ( B D A ' )
, , 6.1 + 3.2 + 2.3-12
Gpi d la khoang each t u diem K den ( B D A ' j => d =
V36 + 9 + 4
6
7
Ta CO r = ^ R ^ - d ^ = 5V26
C a u 6 : D o b > c va a^ < l , n e n c6: P < b - c + b ^ ( c - b ) + c ^ ( l - c )
hay P < ( b - c ) ( l - b ) ( l + b) + c 2 ( l - c )
Ta c6: (b - c)(l - b)(l + b) = 2(b - c ) ^ ( l - b ) ^ ( l + b)
-|3
<2
( b - c ) 4 ^ ( i - b ) % i ( i b )
2_
27
Suy ra P < ^ ( _ c ) % ( l - c), dat f (c) = ^(Vs - cf + c^ ( l - c)
27' vol C € 0;1
Vay, maxP = - ^ ^ khi (a;b;c) = ( 1 1
I N/3 J
I I P H A N R I E N G T h i s i n h c h i d u Q c c h p n l a m m p t t r o n g h a i p h a n ( p h a n A
h o a c B )
A Theo chUtfng trinh chuan
C a u 7.a: T u gia thiet: ( C j ) c6 tarn I = (0;0),R - yjl3
( C 2 ) cotam j ( 6 ; 0 ) , R ' = 5
Gpi duang thang d qua A (2; 3) c6 vecto chi phuong
x = 2 + at
y = 3 + bt
x = 2 + at
u = ( a ; b ) = > d :
d c i t ( C j ) t^ii A , B: o y = 3 + bt o
x 2 + y 2 =13
(a2 + b2)t2+2(2a + 3 b ) t = 0
2a + 3b „
=>t = —3 5- = > B
a^+b^'
b ( 2 b - 3 a ) a ( 3 a - 2 b ) ' a^+b^ ' a^+b^
250
d cat ( C j ) tai A, C thi tpa dg cua A, C la nghiem ciia he:
2 ( 4 a - 3 b )
x = 2 + at
y = 3 + bt =>t = - a + b , 2 , 1 , 2
{x-ef+y^ =25
10a^-6ab + 2b^ 3a^+8ab-3b^l
a^+b^
a^+b^ ' Neil 2 day cung bang nhau thi A la trung diem cua A, C
Tu" do ta CO phuong trinh:
( 2 b 2 - 3 a b ) i 0 a 2 - 6 a b + 2b2
a^+b^ a^+b^
x = 2
y = 3 + t
' / / ? = ( 3 ; 2 ) = > d :
= 4 o 6 a 2 - 9 a b = 0
a = 0 = > d :
a = —b=> u =
2
y = 3 + 2t
C a u 8.a: C thuQC mat phang (P) nen C ( a ; b ; a - b - l ) AABC can tai C <=> A C = BC
o ^ ( a - 3 f + ( b - 5 f + ( 5 - a + b f = J ( a - 3 f + ( b - l f + ( 5 - a + b f ^ b
Ta CO AB = 4, trung diem AB la I(3; 3;4)
SAABC = ^ C I A B = 2V17 => CI = VT7 ^ ( 3 - a)^ + (8 - a)^ = V l 7
<=> a = 4 hoac a = 7 Vay,c6hai diem C(4 ;3;0), C(7;3;3)
C a u 9.a: Gpi z = x + yi (x, y e R ) , Theo gia thiet, ta c6: |z| = 1 o x^ + y^ = 1 y^ = 1 - , x e [ - l ; l '
H o n n i i a : w = z^ + 2 z - l = x^ - y ^ +2xyi + 2 ( x - y i ) - l
= ( x 2 - y ^ + 2 x - l j + 2 y ( x - l ) i w ' 2 - y 2 + 2 x - l ) + 4 y ^ ( x - l )
= 2 x ^ + x - l ) + ( l - x 2 ) ( x - l ) ^ = 2 \ / 4 x ^ - x 2 - 4 x + 2
Xethamso f(x) = 4 x ^ - x ^ - 4 x + 2 , x € [ - l ; l "
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Trang 6f {x) = 12x2 - 2 x - 4 , f ( x ) = 0<=>x = - ^ , x = |
Lai C O f ( - l ) = l , i
[ 2
2
I3j 27 , f ( l ) = l
Dodo max w = 2 /maxfTx) = Vl3 datdug-ckhi x = — , y - ± —
2 2
1 >/3 1 V3
Vay, CO hai so phuc can tim la Zj = ' ^2 ~ ~2 ~"2~' '
B Theo chUofng trinh nang cao
Cau 7.b: Theo gia thiet ta c6 A, H , M nam tren duong thang x - 3 = 0 suy ra D
la trung diem cua H M => D(3;0)
Duong thang BC di qua D va c6 vecto phap tuyen HD = (0;l) => BC: y = 0
Goi F la hinh chieu cua M len duong thiing AB suy ra F thuQC duong th^ng
DE suy ra F ( 3 - 2 t ; t )
Taco FA.FM = 0 » 5 t 2 - 2 t - 3 = 0 « t = — hoac t = l
5
THl: Voi t = — ^ ¥
va A F nen tpa dp B la nghif m cua h$
F A : 3x + y -12 = 0 Do B la giao cvia BC 3x + y - 1 2 = 0 [x = 4
y = 0 <=> -i loai do X n < 2 y = 0
TH2: Voi t = 1 => F(1;1) => A F : x - y = 0 Do B la giao cua BC va A F nen tpa
' x - y = 0 [x = 0
do B la n g h i e m cua he
y = 0 y^O •B(0;0)
Do C nam tren duong thSng BC nen C(u;0) Mat khac H la true tam non
HB vuong goc voi AC, khi do: AC.HB = 0 c ^ 3 u - 9 - 3 = 0 » u = 4=>C(4;0)
Cau 8.b: Gia sir ng la mot vecto phap tuyen ciia ( Q )
Khi do n ^ l n j ^ ( l ; - l ; - l )
M|t phSng (Q) cat hai true Oy, Oz tai M(0;a;0),N(0;0;b) phan bi^t sa^
fa = b 9t 0 cho O M = ON nen a = b «•
a = -h^O
Neu a = b thi M N = (0;-a;a)//u(0;-l;l) va n g l u
= (2;1;1)
nen UQ =
252
u,ni
cry iNHHMl V UVVH J^nang vifi
i Khi do mat phang (Q) :2x + y + z - 2 = 0 va ( Q ) catOy, Qztai M(0;2;0) va
>j(0;0;2) (thoa man) Neu a = - b t h i M N = (0;-a;-a)//u(0;l;l) va n ^ 1 u
= (0;1;-1)
nen n ^ = u , n p
Khi do mat phSng (Q): y - z = 0 va (Q) di Oy, Oz tai M(0; 0; 0) va N(0; 0; 0)
V?y, ( Q ) :2x + y + z - 2 = 0 la m$t phSng can tim
|;au 9.b: Cdch /.-Gpi z = x + yi (x,y € R )
I Theo gia thiet, ta c6 |z| = 1 ^ yjx^ + = 1 y^ = 1 - x^ x 6 [ - 1 ; l ] ( l )
Khi do: M - z ^ - z + 2 (x^ - 3xy2 - x + 2) + (3xV - - y ) i
= V ( x ' - 3 x y 2 - x + 2) + ( 3 x 2 y - y 3 - y )
= ^ ( x 3 - 3 x y 2- x + 2 f + ( 3 x 2 - y 2 - i ) ' y 2
= y(4x^-4x + 2) + ( 4 x ^ - 2 ) ( l - x 2 ) = V 4 x 3 - x 2 - 4 x + 2
Xet ham so f(x) = 4x^ - x ^ - 4 X + 2 , X G [ - 1 ; 1
Taco: f'(x) = 12x2 2 x 4 va f'(x) = O o x = i hoac x =
-L 9 i e 6 f ( - l ) = f ( l ) = l , f
I ' 2 j
'1^
.3
8
3
I Vay maxM = 2 /maxf(x) = 7 l 3 , dat dupe khi x = - - y = ±
Vi-i.-i] ^' • 1
Gfch2.GQi z = eos(p +isin(p,ta CO z'^ =(eoscp + isin(p)^ =eos3(p + isin3(p
Do do M = z-*-z + 2 cos 39 + i s in39-cos(p-isin(p+2
cos 3(p - cos (p + 2 + i (s in3(p - sin cp)
= ^^(cos3{p-cos(p + 2)^ +(sin3(p-sin<p)^
= ^6 - 2(cos(pcos3(p + sincp + sin3(p) + 4(cos3(p - cos(p)
= V6 - 2 cos 2{p - 8 s in2(p sin (p = ^ 6 - 2 ( 2 c o s ^ ( p - l ) - 16(l -cos^ (p)cos(p
= 2 A / 4 C O S ^ P - COS^ (p - 4cos(p + 2
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Trang 7Tuyen chQti & Giai thifu dethi Toan hgc - TQguyen Phu Khanh , Nguyen Tat Thu
Dat t = cos(p(-l < t < l ) ,ta c6 M = 2^41^-1^-41 + 2
Xet ham so f (t) = 4t^ - - 4t + 2, t e [ - 1 ; f
1 2 Taco: f (t) = 12t^ 2 t 4 va f (t) = O o t = hoac t =
-Laico f ( - l ) = f ( l ) , f I 2
_13 / 2 >
" 4 '
8
3
Vay maxM = 2^maxf(t) = N/I3 ,dat du(?c khi
t = cosffi = - - o (D = ± — + k27t,k e Z
DETHITH(lrsd39
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 = - +1, c6 do thj la (C)
a) Khao sat sy bien thien va ve do thj (C) ciia ham so
b) Tim tren do thi (C) nhung diem A sao cho tiep tijyen tai A cat (C) tai
hai diem B, C khac A va B, C nam ve 2 phia doi voi A
sin3x-cos3x Cau 2: Giai phuong trinh: cos2x + •
2 s i n 2 x - l • = sinx(l + tanx)
Cau 3: Giai phuong trinh: Vsx^ +14x + 9 -\/x^ - x - 2 0 = sVx + l
Cau 4: Tinh tich phan: I = cos* x sin^ xdx
0
Cau 5: Cho hinh lap phuong ABCD.A'B'C'D' canh a Tren canh AB lay dier
M dat A M = m (0 < m < a) Mat phing ( A ' M C ) cat C D ' tai N Chimg min
A ' M C N la hinh binh hanh Tim vi tri M de di$n tich A ' M C N nho nhat kM
do tinh goc giCra ( C M N ) , ( M N D )
Cau 6: Cho x, y, z la cac so duong thoa man dieu ki?n x + y + z = xyz
Chung minh rSng: 4(x + y + z)^ - ( x y + yz + zx +1)^ > xy + yz + zx
254
I
CtyTNHHMTV DWH Khang Vift
I P H A N R I E N G Thi sinh chi durgc chpn lam mpt trong hai phan (phan A o|c B)
Theo chucrng trinh chuan
Cau 7,a: Trong mat phiing tga dp Oxy, cho 2 duong thSng dj : 2x - y - 1 = 0,
d2 : 2x + y - 3 = 0 Gpi I la giao diem ciia dj va'dj, A la diem thupc dj va A
CO hoanh dp duong khac 1 (O < < l ) Lap phuong trinh duong thSng A di qua A, cat dj tai B sao cho di$n tich AIAB bang 6 va IB = 3IA
Cau 8.a: Trong khong gian voi h? tryc tpa dp vuong goc Oxyz, cho mat ph^ng (P): X + y + z - 3 = 0 va duong thing A : ^ ^ = ^ = — Lap phuong trinh duong thing d, nam trong mat phang (P), vuong goc voi duong thing A va each
8 duong thang A mpt khoang bang - =
•\i66
Cau9.a: Timsophuc z thoa man ( z - l ) ( z + 2ij lasothycva |z|nh6nhat
B Theo chUomg trinh nang cao
x^ v^
Cau 7.b: Trong m | t phang tpa dp Oxy, cho Elip — + ^ = 1 ( a > b > 0 ) bie't
V a 2 - b 2 1
= -7= hinh chu nhat co so cat Ox tai A, A' va cat Oy tai B, B' Lap
phuong trinh Elip biet difn tich hinh tron npi tiep hinh thoi ABA'B' c6 di?n h'ch bang 47t
Cau S.b: Trong khong gian tpa dp Oxyz cho cac diem A(l;3;4),B(l;2;-3), C(6; - 1 ; 1) va mSt phang (a): x + 2y + 2z - 1 = 0 Lap phuong trinh mat cau (S)
CO tam nim tren mat phing (a) va di qua ba diem A, B, C Tim difn tich hinh chieu ciia tam giac ABC tren mat phang (a)
Cau 9.b: Trong cac so phuc thoa man dieu k i | n |z + 1 + 2i| = 1 ,so phuc z nao
CO modun nho nhat
" P H A N C H U N G C H O T A T CA C A C T H I S I N H
CSu 1:
a) Danh cho ban dpc
Hi b) Gpi A (a; a"* - +1J la diem thoa man de bai
Ta c6: y' = 4x^ - 2x Phuong trinh tiep tiiyeh (d) ciia (C) tai A la:
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Trang 8y = ( 4 a 3 - 2 a ) ( x - a ) + a ^ - a 2 + l
Phuong trinh hoanh dp giao diem ciia (d) va (C) la:
x4 _ x2 +1 = (4a^ - 2a)(x - a) + a^ - a H l « (x - af (x^ + 2ax + Sa^ - a) = 0
o x = a hoac g(x) = x^+2ax + 3a^-a = 0
Theo bai toan thi g(x) = 0 c6 2 nghi^m phan biet x^,x^ sao cho: Xj < a <
A' = -2a^+a>0 ^ 1
( x i - a ) ( x 2 - a ) < 0 6
Cau 2: Dieu kien: sin 2x ^ ^, cos X ^ 0
Phuong trinh da cho tuong duong:
eos2x + 3sinx-4sin^x-4cos3x + 3cosx ^^.^^^^
2 s i n 2 x - l
, 2 (sinx + cosx)(2sin2x-l) sin x(sin x + cos x)
2 s i n 2 x - l cosx sin X + cos X = 0
^ sinx cosx-smx + 1 =
cosx
o cos X - s i n x +
THI: sinx + cosx = 0 <=> tanx = - 1 o x = - - - + k 7 t , k e Z
4 TH2: cosx-sinx + l = - ^ ^ o ( c o s x - s i n x ) ( l + cosx) = 0
cosx cosx-sinx = 0
1 + cos x = 0 <=>
tan x = 1 cos x = - 1
X = — + K 7 t , ^
4 ( k e Z )
X = 7l + k27t
Doi chieu dieu ki|n, suy ra cac hp nghi^m ciia phuong trinh da cho la:
x = ± — + k7t, X = 7t + k27t,k6
4
Cau 3: Dieu ki^n:
5x^ +14X + 9S0
x + l > 0 o
x 2- x - 2 0 > 0
X €
X €
9 u [ - l ; + o o )
x e ( - o o ; - 4 ] u [ 5 ; + o o )
o x e 5;+co)
Phuong trinh cho viet lai: V5x^+14x + 9 = sVx+T + Vx^ - x - 20
<=>5x^ +14x + 9 = x2 +24x + 5 + 10^(x + l)(x^ - x - 2 0 Sx^ + 14x + 9 = x^ + 24x + 5 + 10^(x + l)(x2 - x - 2 o )
o 2x^ - Sx + 2 = 5^(x + l)(x + 4 ) ( x - 5 )
o 2(x2 - 4x - 5) + 3(x + 4) = 5^(x2 -4x - 5)(x + 4) (*)
Chia hai ve ciia (•) cho ^{y} - 4 x - 5 J ( x + 4) ta dugc:
x - - 4 x - 5 ^ 3 =5
x + 4 'x^ - 4 x - 5
/X - 4 x - 5
E>at t = J———^— dupe phuong trinh
2t + = 5 » 2 t 2 5 t + 3 = 0 « t = l , t =
-t 2
THI: t = 1 o Ix-" - 4 x - 5 2 r ^ . 5-761
x + 4 = l < : > X ^ - 5 x - 9 = 0c:>X = - (khong thoa)
hoac X = ^ " ^ y ^ (thoa man de bai)
TH2: t = - o x ^ - 4 x - 5 3
2 ^ V — x T i — = 2'^^^ ^'^^^'^^^^^^^'4 (khong thoa)
hoac X = 8 (thoa man de bai)
Vay, nghifm ciia phuong trinh la: x = llJ^^ x = 8,
Cau 4:
1= Jcos'*xsin^ xdx = i Jcos^xsin^ 2xdx = — J(l-cos4x)dx + i Jcos2xsin^2xdx
= — J ( l - cos 4x)dx + - fsin^ 2xd(sin 2x) =
Cau 5:
Chon h§ trijc tpa dp Axyz sao cho:
||| A{0;0;0), B(a;0;0), C(a;a;0), D(0;a;0),
sin^2x
X 1 , sm4x +
16 64 24
n
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Trang 9A'(0;0;a), B'(a;0;a), C'(a;a;a), D'(0;a;a) => M(m;0;0)
1 Chung minh A'MCN la hinh binh hanh z
= -a(a;m-a;m) ^
Taco: MA',MC
=>n(A'MC) =(a;m-a;m)
=i> (A'MC): ax + (m - a)y + mz - am = 0
Phuong trinh tham so ciia CD:
x = t
y = a{teM) ^
z = a
N = C'D'n(A'MC)=>N(a-m;a;a)
V?y A • MCN la hinh binh hanh X
2 Tim vj tri M de difn rich A'MCN nho nhat, rinh goc giiia (CMN),(MND)
Ta c6: S^'MCN = [MA',MC]| = 2aVm2-am + a2 = 2a^
^(SAMCNLn=a2Vi«m = |
V$y M la trung diem ciia AB => ACMN can tai C
fCIlMN GQI I la trung diem cua MN
a
m —
V 2,
\ 3a'
>cos IC;ID) = - = COS(CMN),(MND)
' 3 Cau6:D|it T = 4(x + y+ z)^-(xy + yz + zx + l f - x y - y z - z x
T = 4(x + y + zf-(xy + yz + zx)^-3(xy + yz + zx)
Taco: 3(xy+ yz + zx)<(x + y + zf nen T<^^^-^^^^ + 3(x + y + zf
Tir X + y + z = xyz ta CO xyz > 3V3 , vi the neu t = (x + y + z)^ thi t > 27
Xet ham so: f (t) = ^ + 3t voi t ^ 27
258
II PHAN RIENG Thi sinh chi dugfc chpn lam mpt trong hai phan (phan A
ho9C B)
A Theo chUarng trinh chuan
Cau 7.a: I = dj n => tao dp ciia I la nghi^m cua h?
J 2 x 2 x 4 -y-l=:0 fx = l - 3 = 0=^ jy = l-^(^^ ^)
Tu gia thuyet dj c6 VTPT n^ =(2;-l), dz c6 VTPT =(2;1) Gpi cp la goc
4-1 cuadj vadj =>cos(p = ^ — ^ = |i :>sin(p = | =>S.,AB = - I A 3 I A - i = ^ ^
5 5 5 ^^^^ 2 5 5
Gia thuye't: S^j^g = 6 => lA^ = 5 => IB^ = 45
VA € dj => A(a,2a -1) voi a > a*\
/ a = 2=>A(2;3)
VBedj =>B(a,3-2b)
"b = 4=>B(4;-5)
b = -2=>B(-2;7)
"di A(2;3), B(4;5) phuong trinh can tim la = o 4x + y -11 = 0
IB^ =(b-l)2 +(2-2b)2 =5(b-l)^ IB^ =45o(b-l)2 =9<»
^ ,<x-2 y - 3 can tim la - - ^ i
Voi A (2; 3); B(-2;7) phuong trinh can tim la ^ ^ ,«x-2 y - 3 4-2 -5-3
•an rim la = ± - 2 - 2 7-3 <:>x + y - 5 = 0
'u8.a:Tac6 (P) c6 VTPT np =(l;l;l),
4 coVTCP i^ = (l;3;-l), M(1;0;0)6A
d l A u J 1 np r < J "1 = (-4;2;2) GQI (Q) la m|t phang chua d va song song voi A Khi do ta chpn
= -2(4;l;7) suy ra (Q) codang 4x + y + 7z + d = 0
4 + d vTaco d(A;d) = d(A;(p)) = d(M;(P)) = ^ Ket hpp voi gia thiet ta dupe: 4 + d >/66 V66 4 + d =8<:*d = 4 ho$c d = -12
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Trang 10THl: Neu d = 4 => ( Q ) : 4x + y + 7z + 4 = 0
Ch(?n d i e m N f - i ; ^ ; - l l e (P)n(Q) = d
V 3 3 J
1 13 suy ra phuong trmh: d : ^ = ^ = — ^
TH2; Neu d =-12 =^ (Q) : 4x + y + 7z -12 = 0
Chondiem N(l;l;l) e (P)n(Q) = d
suy ra phuong trmh: d : — ^ = —— = ——
Cau9.a: Goi z = a + bi (a,b€R)
Theo gia thiet, ta c6: (z - l)(z + 2i) = [(a -1) + bi][a - (b - 2)i
= a(a-l) + b(b-2) + [ab-(a-l)(b-2)]i (z-l)(z + 2i) lasothirc o a b - ( a - l ) ( b - 2 ) = 0»2a + b - 2 = 0 o b = 2-2a
Khi do z = a+(2-2a)i
Ta CO |z| = ^a^ +(2-2a)^ = Vsa^ -8a+ 4 = ^ 4
mm 5 5 5 - ^ 5 5 , khi a = -=>b = - Vay z = - + —i 4 , 2 4 2
B Theo chUtfng trinh nang cao
Cau 7.b: Di?n tich hinh tron npi tiep hinh thoi ABA'B' bang 47i ban kinh
duong tron r = 2
O la tarn hinh tron, ke OK ± AB'=> r = OK = 2
1
Xet tarn giac vuong OAB', ta c6: 1 1 1 1 1 • + — ^ 0 - = -.- + - ^ (1)
OK^ OA^ OB^ 4 b^
Tir gia thuyet:
^ ^ i ^ = 4-oa = V ^ V 7 ^ « a 2 = 2 a 2 - 2 b 2 « a 2 = 2 b 2 (2)
a V2
+ — = 4
a^ b^
a2=12 x2 y2
b 2 = 6 12 6 = 1
260
Cau 8.b: Goi I(a; b; c) la tarn mat cau ta c6:
(1 - a)2 + (3 - b)2 + (4 - cf = (1 - a)2 + (2 - b)2 + (-3 - c)^ (1 - a)2 + (3 - bf + (4 - cf = (6 - af + (-1 - b)^ + (1 - c)^
IA = IB
lA = IC <»
Ie(a) a + 2b + 2c-1 = 0
<=> b + 7c = 6 5a - 4b - 3c = 6 <=>
a + 2b + 2c -1 = 0
a = l
b = -l=>I(l;-l;l)
c = l
R2 = I A 2 =25 ^(S):(x-1)2 +(y+ 1)^ +(z-l)2 ^25 Tam giac ABC deu canh bang 5V2 nen S^gc =
AB = (0;-l;-7), AC = (5;-4;-3) =:> n = cos ((a), (ABC)) = |cos(.^,p)| =
Goi S' la dien tich hinh chieu cua tam giac ABC len mat phang (a) Taco S' = SABc-':os((a),(ABC)) = : ^ ^ = ^ (dvdt)
;au 9.b: Cach 1: Gpi z = x + yi (x,y e R )
Theo gia thiet, ta c6:
z +1 + 2i| = 1 |x +1 + (y + 2)i| = 1 <» (x +1)^ + (y + 2)^ = 1 Tap hop cac diem M(x; y) bieu dien so phiic z la duong tron (C) c6 tam l(-l;-2), ban kinh R = l
Diem M(x;y) bieu dien so phuc zco modun nho nliat la mot trong cac giao diem ciia duong thang OI voi duong tron (C) (sao cho dp dai doan OM nho nhat) Phuong trinh duong th^ng OI la y = 2x
Toa dp diem M thoa h^:
(x + l f ^ ( y + 2f =1
y = 2x (x + l f = i y = 2x y = x = -l + -2 + hay x = l
l-i _2-Al
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