Ố Thời gian: 120 phút (không kể phát đề) !"#"$%& 3 2 1 2 3 1 3 y x x x= − + − ' (")#%*+,$,-.+"/C01'"$%&+234 5 64"784+60""94":";4<<=">45?C,$0*0@A4<+";4< 0, 2, 3y x x= = = B ! 64"0*0+60":"4%' 3 2 0 x I dx 1 x = + ò B ( ) 5 1 ln e I x x x dx= + ∫ ! 9C401'%&:"D0z5E+2F4< ( ) 1 2 (4 5 ) 1 3i z i i− + − = + GH! I0%4""I00"@J4<+294"4$#+"90"K@L0M$:"N47$4"234<0"#0"@J4<+294"O:"N4"#P0:"N4B "Q#0"@J4<+294""R4 S 'B 2#4<T"C4<<'4,="8+I'UVWX0"#"' A(1;2;0) B(3;4; 2)- ,$P+:";4< Y x y z 4 0- + - = ZE+:"@J4<+294"P+:";4<[\'"']^,$,C4<<O0,=P+:";4< B IM$+"_'`4 IA IB 0+ = uur uur r `W,E+:"@J4<+294"P+0N+,$+E:Va0,= P+:";4< b ' Tìm x (0; )Î +¥ thỏa mãn : ( ) x 2 0 2sin t 1 dt 0- = ò B "Q#0"@J4<+294"4<0'# S 5B 2#4<T"C4<<'4,="8+I'UVWX0"#"' A(1;2;0) B(3;4; 2)- ,$P+:";4< Y x y z 4 0- + - = ZE+:"@J4<+294"P+:";4<[\'"']^,$,C4<<O0,=P+:";4< B IM$+"_'`4 3IA 2IB 0- = uur uur r `W,E+:"@J4<+294"P+0N+,$+E:Va0,= P+:";4< b 5 cd+%&:"D0 ( ) z x yi x,y R= + Î 9VW%'#0"# ( ) 2 x yi 8 6i+ = + Hết HƯỚNG DẨN ĐỀ 6 I. PHẦN CHUNG;(7 điểm) e'fcY ¡ f 2 ' 4 3y x x= − + gh 2 1 ' 0 4 3 0 3 x y x x x = = ⇔ − + = ⇔ = f=">4 lim x y →+∞ = +∞ ,$ lim x y →−∞ = −∞ • ^)4<5E4+"34 - $%&.4<5E4+2340*0T"#)4< ( ) ;1−∞ ,$ ( ) 3;+∞ - $%&4<"/0"5E4+234 ( ) 1;3 - 0i0> 1 1; 3 ÷ - 0i0+ ( ) 3; 1− f.+"/ .+"/"$%&0j++2k0"#$4"+>!:"458+ lU+%&+"U0.+"/ x B ! S y 1− 1 3 1 3 − 1− 1 3 Y5784+60""94":";4<<=">45?C,$0*0@A4<+";4< 0, 2, 3y x x= = = M$ 3 3 2 2 1 2 3 1 3 S x x x dx= − + − ∫ 3 3 2 2 1 2 3 1 3 x x x dx = − − + − ÷ ∫ 3 4 3 2 1 2 3 12 3 2 x x x x = − − + − ÷ 3 4 = B'64"0*0+60":"4%' 3 2 0 x I dx 1 x = + ò P+ 2 u 1 x du 2xdx= + Þ = m0n4Y u 4 x 3 u 1 x 0 = = Þ = = o#OY 4 1 4 1 I du u 1 1 2 u = = = ò ZnW I 1= B5 ( ) 5 5 6 1 1 1 ln ln e e e I x x x dx x xdx x dx= + = + ∫ ∫ ∫ 64" 5 1 1 ln e I x xdx= ∫ P+ 5 6 1 ln 6 du dx u x x dv x dx x v = = ⇒ = = 6 5 6 6 6 1 1 1 1 1 ln ln 5 1 6 6 6 36 36 e e e e x x x x x x e I dx + = − = − = ∫ f64" 7 7 6 2 1 1 1 7 7 e e x e I x dx − = = = ∫ ZnW 6 7 5 1 1 36 7 e e I + − = + !'0O ( ) ( ) ( ) ( ) ( ) ( ) ( ) + = + = + = + + + + + + + = = = = = + + + 2 2 2 1 2 (4 5 ) 1 3 1 2 1 3 (4 5 ) 1 2 3 8 3 8 1 2 3 8 3 6 8 16 19 2 19 2 1 2 1 2 1 2 1 2 5 5 5 i z i i i z i i i z i i i i i i i i z z z i i i i o#O 2 2 19 2 19 2 73 365 5 5 5 5 5 5 z i = + = + = = ữ ữ GH! Hc sinh hc chng trỡnh no thỡ ch c lm phn dnh riờng cho chng trỡnh ú (phn 1 hoc phn 2) "Q#0"@J4<+294""R4 SB 1. Vit phng trỡnh mt phng (Q) i qua hai im A, B v vuụng gúc vi mt phng (P). Mt phng (P) cú vect phỏp tuyn l : P n (1; 1;1)= - uur , AB (2;2; 2)= - uuur Vỡ (Q) qua A,B v vuụng gúc vi (P) nờn (Q) cú mt vect phỏp tuyn l: ( ) Q P 1 1 1 1 1 1 n n ;AB ; ; 0;4;4 2 2 2 2 2 2 ổ ử - - ữ ỗ ộ ự ữ ỗ = = = ữ ỗ ờ ỳ ữ ở ỷ ỗ - - ữ ỗ ố ứ uuur uur uur Do ú phng trỡnh mt phng (Q) l 4(y 2) 4(z 0) 0 y z 2 0 - + - = + - = Vy phng trỡnh (Q): y z 2 0+ - = 2. Gi I l trung im ca AB. Hóy vit phng trỡnh mt cu tõm I v tip xỳc vi mt phng (P). Do I tha món IA IB 0+ = uur uur r nờn I l trung im ca AB Ta trung im I ca AB l: I(2;3; 1)- Gi (S) l mt cu cú tõm I v tip xỳc vi (P) Bỏn kớnh ca mt cu (S) l: R d(I,(P)) 2 3 1 4 6 2 3 3 3 = - - - - = = = Vy phng trỡnh mt cu (S) l 2 2 2 (x 2) (y 3) (z 1) 12- + - + + = b ' Tỡm x (0; )ẻ +Ơ tha món : ( ) x 2 0 2sin t 1 dt 0- = ũ (1) Ta cú: ( ) x x 2 0 0 x 1 1 2sin t 1 dt cos2tdt sin2t sin2x 0 2 2 - = - = - = - ũ ũ Do ú: 1 (1) sin2x=0 sin2x=0 2 2x k k x 2 - = p p = Do x (0; )ẻ +Ơ nờn ta chn k x 2 p = vi k Z + ẻ 2 "Q#0"@J4<+294"4<0'# S5B# 1. Vit phng trỡnh mt phng (Q) i qua hai im A, B v vuụng gúc vi mt phng (P). Mt phng (P) cú vect phỏp tuyn l : P n (1; 1;1)= - uur , AB (2;2; 2)= - uuur Vỡ (Q) qua A,B v vuụng gúc vi (P) nờn (Q) cú mt vect phỏp tuyn l: ( ) Q P 1 1 1 1 1 1 n n ;AB ; ; 0;4;4 2 2 2 2 2 2 ổ ử - - ữ ỗ ộ ự ữ ỗ = = = ữ ỗ ờ ỳ ữ ở ỷ ỗ - - ữ ỗ ố ứ uuur uur uur Do ú phng trỡnh mt phng (Q) l 4(y 2) 4(z 0) 0 y z 2 0 - + - = + - = Vy phng trỡnh (Q): y z 2 0+ - = 2. Gi I l im tha món 3IA 2IB 0- = uur uur r . Hóy vit phng trỡnh mt cu tõm I v tip xỳc vi mt phng. Gi I(x;y) l im tha món 3IA 2IB= uur uur , ta cú: ( ) ( ) ( ) ( ) ( ) ( ) 3 1 x 2 3 x x 3 3IA 2IB 3 2 y 2 4 y y 2 z 4 3 0 z 2 2 z ỡ ỡ ù ù - = - = - ù ù ù ù ù ù ù ù = - = - = - ớ ớ ù ù ù ù ù ù = - = - - ù ù ù ợ ù ợ uur uur . Suy ra: I( 3; 2;4)- - Gi (S) l mt cu cú tõm I v tip xỳc vi (P) Bỏn kớnh ca mt cu (S) l: 3 2 4 4 1 3 R d(I,(P)) 3 3 3 - + + - - = = = = Vy phng trỡnh mt cu (S) l 2 2 2 1 (x 3) (y 2) (z 4) 3 + + + + - = b 5 Xột s phc ( ) z x yi x,y R= + ẻ . Tỡm x, y sao cho ( ) 2 x yi 8 6i+ = + Ta cú: ( ) 2 2 2 4 2 2 2 2 2 2 x yi 8 6i x y 2xyi 8 6i x 3 9 x 8x 9 0 x 9 x 8 y 1 x y 8 x 3 3 3 xy 3 x 3 y y y x x x y 1 + = + - + = + ộ ỡ = ù ù ờ ỡ ù ớ ỡ ỡ ờ - - = = ù ù ù - = ỡ ù = ù ù - =ù ù ờ ù ù ù ù ợ ù ù ù ù ù ờ ớ ớ ớ ớ ờù ù ù ù ỡ= = - ù = = ù ù ù ù = ù ợ ù ờ ù ù ù ù ù ợ ợ ớ ù ờ ù ợ ù = - ờ ù ợ ở Vy giỏ tr x, y cn tỡm l x 3 y 1 ỡ = ù ù ớ ù = ù ợ hoc x 3 y 1 ỡ = - ù ù ớ ù = - ù ợ f . 6 6 1 1 1 1 1 ln ln 5 1 6 6 6 36 36 e e e e x x x x x x e I dx + = − = − = ∫ f 64 " 7 7 6 2 1 1 1 7 7 e e x e I x dx − = = = ∫ ZnW 6 7 5 1 1 36 7 e e I + − = + !'0O (. 1= B5 ( ) 5 5 6 1 1 1 ln ln e e e I x x x dx x xdx x dx= + = + ∫ ∫ ∫ 64 " 5 1 1 ln e I x xdx= ∫ P+ 5 6 1 ln 6 du dx u x x dv x dx x v = = ⇒ = = 6 5 6 6 6 1 1 1 1. (")#%*+,$,-.+"/C01'"$%&+234 5 64 "784 +60 ""94":";4<<=">45?C,$0*0@A4<+";4< 0, 2, 3y x x= = = B ! 64 "0*0 +60 ":"4%' 3 2 0 x I