(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e ) e OE z 1 1 e E xy e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0

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Download "(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e ) e OE z 1 1 e E xy e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0"

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4 (1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e ) e OE z 1 1 e E xy e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0 Q y P y k 2 M N M( 1 0 0) N(1 0 0) 4 P Q M N C EP EQ = EM EN = 1 3 = 3 k EP 2 = 3 P (p 1 p 2 0) ( 1 < p < 1) EP 2 = (p + 2) 2 + ( 1 p 2 ) 2 = 4p + 5 k = 3 4p + 5 Q y 3 1 p 2 4p + 5 (3) F (cos θ sin θ 0) (0 < θ < π) BG = t AF t ( ) 2 OG = OB + taf = t ( cos θ 1 sin θ 1 ) 5 ( ) 2 = 5 + t(cos θ 1) t sin θ 4 5 t G z 0 t = 4 5 OG = ( 4 5 cos θ sin θ 0 ) G C ( 4 5 cos θ 2 ) 2 ( 4 ) sin θ = 1

5 F x cos θ = 5 16 (4) R (X R Y R 0). 3 A P U α OU = OA + α AP ( = 1 + α(p 1) α ) 1 p 2 1 α β OU = OB + β BR ( 2 = (X 5 + β R 2 ) βy R ) 5 β 1 + α(p 1) = 2 ( 5 + β X R 2 ) 5 α 1 p 2 = βy R 1 α = β 3 α = 4β β p2 = βy R 5 β = 0 β 0 Y R = (4β + 1) 1 p 2 5β 1 X R = (4p 2)β + p + 2 5β R C X 2 R + Y 2 R = 1 { } 2 { } (4p 2)β + p + 2 (4β + 1) 1 p = 1 5β 5β ()β 2 12pβ (4p + 5) = 0 4p(4β 2 3β 1) + 5(β 2 1) = 0 4p(β 1)(4β + 1) + 5(β + 1)(β 1) = 0 (β 1){()β + 4p + 5} = 0 P (3) F p 5 16 β = 1 4p 5 β = 1 α = 1 3 U P R

6 β = 4p 5 α = 3 R y Y R Y R = 3 1 p 2 4p + 5 U y Y U Y U = 3 1 p 2 5 A B P R U (1) E AB 6 A B P R U E ( π ) π C π 2 P R C π C 2 P R. (2) π EP C P Q Q = R R y 3 1 p 2 4p A B P R (= Q) U E U y Y A B U xy A B U 1 < p < 1 2 Q P U E B A (2) EQ : EP = k : 1 = 3 4p + 5 : 1 EQ QP = 3 2(2p + 1)

7 A U P 0 Y 1 p 2 y PU U A = PU 1 UA = p2 Y Y AB : BE = A B : B E = 1 : 4 PU U A A B B E EQ PU UA AB BE EQ 1 p2 Y 1 { } Y 4 3 = 1 2(2p + 1) Y = 3 1 p < p < 5 16 U Q P E B A PU U A A B B E EQ PU UA AB BE EQ

8 1 p2 Y Y (2p + 1) = 1 Y = 3 1 p < p < 1 P Q E A B U PU U A A B B E EQ PU UA AB BE EQ Y 1 p 2 Y (2p + 1) = 1 Y = 3 1 p 2 U y 3 1 p 2

9 É d Ê n 9 1 f(x) = x(log x) n (0 < x < 1) (1) 0 < x < 1 log x < 0 f 0 (x) = log x + n x 2 (log x) n+1 f 0 (x) = 0 log x = n Ú x = e n x a n a n = e n f(a n ) = f(e n ) = n 1 e n ( n) n = # e n ;n n x (0) Ý a n Ý (1) f 0 (x) + 0 f(x) % & x (0) Ý a n Ý (1) f 0 (x) 0 + f(x) & % f(a n ) U n n (2) f 00 (x) = 2(log x)2 + 3n(log x) + n(n + 1) x 3 (log x) n+2 f 00 (x) = 0 2(log x) 2 + 3n(log x) + n(n + 1) = 0 t = log x ÝÝ1 0 < x < 1 t < 0 2t 2 + 3nt + n(n + 1) = 0 ÝÝ2 D D = 9n 2 8n(n + 1) = n(n 8) > 0 (Û n 9) = 3n 2 < 0 n(n + 1) = 2 > 0 < 0 < t 1 0 < x < 1 2 x x f 00 (x) f(x) 2 C 1 2 log x = 3n n 2 8n 4 p x b n c n (b n < c n ) b n = e 3n n 2 8n 4

10 C (3) log b n = 3n + n 2 8n log a n 4n n 9 #1 1 n ; log b n = n 1 log a n n 3n + C n 2 8n 4n C = n 4 n 2 8n 4n = log b n #1 1 log a n n 5 n 2 ; = 3n + 16C > 0 4n"n 4 + n 2 8n : C n 2 8n 4n C = n n 2 8n (n 2 4n 20) 4n 2 n2 n 5 n 2 2n(3n 20) 100 = C > 0 n 2 Qn n 2 8n + (n 2 4n 20)i (Û n 9 2n(3n 20) 2 9(3 9 20) = 126) 1 1 n 5 n 2 log b n 1 1 ÝÝ3 log a n n Z Z 1 (4) f(x) dx = x (log (log x)1 n x) n dx = + C (C ) 1 n (5) a n x b n f(x) S n = Z bn a n f(x) dx = = (log a n) 1 n 1 n (log x)1 n 1 n T# log b n log a n ; 1 n 1l b n = (log b n) 1 n (log a n ) 1 n a n 1 n = ( n)1 n 1 n T# log b n log a n ; 1 n 1l (Û log a n = n) = n1 n n 1 T# log b n log a n ; 1 n 1l n n S n = n n 1 T# log b n ; 1 n 1l ÝÝ4 log a n n lim n!1 n 1 = lim 1 n!1 1 1 = 1 n 3 (1 n) (< 0) #1 1 n ;1 n # log b n ; 1 n #1 1 log a n n 5 n 2 ;1 n lim #1 1 n!1 n ;1 n = lim #1 1 n!1 n ; n #1 1 n ; = e lim log #1 1 n!1 n 5 n 2 ;1 n = lim(1 n) n!1 lim #1 1 n!1 n 5 n 2 ;1 n = e n + 5 n 2 = lim #1 1 n!1 n ;#1 + 5 n = log e ÝÝ5 n 2 n + 5 log #1 n + 5 n 2 ; ; log #1 n + 5 n 2 ; n 2 n+5

11 5 lim # log b n ; 1 n = e n!1 log a n 4 lim n n S n = e 1 n!1

1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1

1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1 ABCD ABD AC BD E E BD : () AB = AD =, AB AD = () AE = AB + () A F AD AE = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD AB + AD AB + 7 9 AD AB + AD AB + 9 7 4 9 AD () AB sin π = AB = ABD AD

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1 26 ( ) ( ) 1 4 I II III A B C (120 ) ( ) 1, 5 7 I II III A B C (120 ) 1 (1) 0 x π 0 y π 3 sin x sin y = 3, 3 cos x + cos y = 1 (2) a b c a + 6 ( ) 6 5 ( ) 4 I II III A B C ( ) ( ), 5 7 I II III A B C ( ) () x π y π sin x sin y =, cos x + cos y = () b c + b + c = + b + = b c c () 4 5 6 n ( ) ( ) ( ) n ( ) n m n + m = 555 n OAB P k m n k PO +

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