άλγεβρα προαγωγικών εξετάσεων με λύσεις 2011 ευαγγελική...

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Transcript of άλγεβρα προαγωγικών εξετάσεων με λύσεις 2011 ευαγγελική...

  • 1. P (x) = x + 1 x1 + + 1 x + 0 P (x) x= P (x) P (x) x x= P (x) = x3 + 2x2 7x 8 1 P (x) P (x) : (x + 1) P (x) P (x)0

2. x 1 (x) =3 (1) , (2) , (3) () + ( + 1) = 36 =2 = 13 1 f (x) = (ln x)2 + ln x f f f (x) = 0 [0, 2011] f () = f ()= = e 3. - 2011, 13 2011 : : 1 x3 +2x2 7x8 x+1 x3x2x2 +x 8x2 7x8P (x) = x + 1 x1 + + 1 x + 0 x2x 8x8 .8x +801. P (x) x = ; Horner - 2. P (x); 1, 1, 8 x2 + x 8.3. P (x) x 3. x = . P (x) = x3 + 2x2 7x 8 = (x + 1) x2 + x 8 :1. 62. 63. 13 : P (x) = 0 (x + 1) x2 + x 8 = 0 1. 62 x + 1 = 0 x2 + x 8 = 0 2. 62 1 + 331 33 x = 1 x =x=223. 674. (x + 1) x2 + x 80. x + 1 x1 2 x2 + x 8 : P (x) = x3 + 2x2 7x 8 1 331+ 33 x 21 2 1. 1 P (x) x+1 + +2. P (x) : (x + 1).3. P (x). x2 + x 8+ +4. P (x)0.: 1. 62. 6 3. 6 4. 7P (x) + + - 1. : P (1) = (1)3 + x 2 1 33 x 11 2 2 2 (1) 7 (1)8 = 0 1 -x 1 + 2 33 12 . 11 , 2 2 33 1, 2 + 1 33 .12 Horner: 3 12 -7-8-1 x *-1-181 (x) = 11 -80 3 1. - x = 1 . 1 . 2. (1) , (2) , (3) -2. : . 1 4. 3. 4 () + ( + 1) = 36(1)2 1 f (x) = (ln x) + ln () = 2 .x1. f . () = 13 .:1.5 2. 43.162. f .3. 1. ax - f (x) = 0 [0, 2011]. a1 0a1. a = 1 34. f () = f () = = e.1 14: 1. 62. 6 3. 6 4. 731 0 1143 1. - (1, 4) . (4, +) . 1 x0 x0. x0 2 3 f (0, +).2. (1) = 1 , (2) = 1 , (3) = 1 . 3 3 3 2. f y y . 2 ( (2)) = (1) (3) 0 . f : y y. 22 3 x x f (x) = 0. 1 11212=f (x) = 0 (ln x) + ln x = 0 (ln x) ln x = 3 33 0 ln x (ln x 1) = 0 ln x = 0 ln x = 1 x = .1 x = e. f x x A (1, 0) , B (e, 0).3. () = 2 (1) 3. f (x) = 0 x (2) + (3) = 36 f x = 1 x = e. e = 2, 71 x 1 f (x) = 023 1 1 x = 1. += 36 (2)3 3x = 1 x = 0 x = 0 + 2k x = y = 1 . y0, y = 1. 0 2k (k Z) x = 2k . - 3(2) f (x) = 0 . y 3 + y 2 36 = 0[0, 2011] 0, 2, 4 , . . . , 2010y 3 + y 2 36 S = 2 + 4 + ... + 2010 = 2 (1 + 2 + ... + 1005) 36 1, 2, 3, 4, 6, 9, 12, 18, 36 1 + 2 + ... + 1005 . 3 . - 1005 - y 3 + y 2 36 y 3 1, 1 1005 (1+1005)1005 = 505 515. S = 2 505 515 = y 3 + y 2 36 = (y 3) y 2 + 4y + 121011 030. 2 y 2 + 4y + 12 -4. f (x) = (ln x)2 ln x : . y 3 + y 2 36 = 0 y = 3 f () = f () 2 2 (ln a) ln = (ln ) ln 1 22= 3 = 10(ln a) (ln ) (ln ln ) = 0 3(ln a ln ) ((ln a + ln )) (ln ln ) = 0 () = 13 (x) = 4x (1) (ln a ln ) (ln a + ln 1) = 04 + 4+1 = 36(3) : ln aln = 0 ln a = ln 36(3) 4 + 4 4 = 36 5 4 = 36 4 =5 = 36lnln 4 = ln 36 ln 4 = ln 36 = ln 45 55 ln a+ln 1 = 0 ln a+ln = ln - 1 ln () = ln e = e.2