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Επαναληπτικές ασκήσεις με λύσεις για τη Γ Λυκείου

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Επιμέλεια: ΕΜΕ Ημαθίας αποκλειστικά για το lisari.blogspot.com

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Επαναληπτικές ασκήσεις με λύσεις για τη Γ Λυκείου

  1. 1. - µ 2019 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 1 of 17
  2. 2. 1 f : R ! R , µ : f 0 (x)° f (x) f (x) = x°1 ex ° x , 8x 2 R (1) f (0) = 1 ∑Æ∂ f (x) ∏ 1, 8x 2 R (2) 1. f (x). 2. : f (x)+ f (°x) ∏ 2. " = " 3. µ µ µ . 4. : ex = x+ a µ a 2 R. 5. µ , µ µ f (x). 6. µ (") f (x) . 7. µ µ - f , µ (") x = 0. 8. ex ∏ ex, x 2 R. 1 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 2 of 17
  3. 3. 2 1. 1 µ : f 0 (x)°f (x) f (x) = x°1 ex°x ) f 0 (x) f (x) °1 = x°1 ex°x ) f 0 (x) f (x) = 1+ x°1 ex°x ) f 0 (x) f (x) = ex °x+x°1 ex°x ) ) f 0 (x) f (x) = ex °1 ex°x ) (ln f (x))0 = (ln(ex ° x))0 ln f (x) = ln(ex ° x)+ c. x = 0: ln f (0) = ln(e0 °0)+ c ) ln1 = ln1+ c ) c = 0 ln f (x) = ln(ex ° x). µ y = lnx . "1 ° 1" µ f (x) = ex ° x, x 2 R. 2. µ µ : f (x)+ f (°x) ∏ 2. µ µ : ex ° x+ e°x + x ∏ 2 , ex + e°x ∏ 2 , ex + 1 ex ∏ 2 , (ex )2 +1 ∏ 2ex , , (ex )2 °2ex +1 ∏ 0 , (ex °1)2 ∏ 0, . µ . " = " ex °1 = 0 , x = 0 3. f 0 (x) = (ex ° x)0 = ex °1 f 0 (x) ∏ 0 , ex ∏ 1 , ex ∏ e0 , x ∏ 0 µ µ : , µ (°1,0] f , µ [0,+1) f . 2 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 3 of 17
  4. 4. µ x0 = 0 f µ µ min f (x) = f (0) = 1 lim x!°1 f (x) = lim x!°1 (ex ° x) = +1 lim x!+1 f (x) = lim x!+1 (ex ° x) (1°1) = lim x!+1 £ ex (1° x ex ) § lim x!+1 ex = +1 lim x!+1 x ex °0 0 ¢ = DLH lim x!1 1 ex = 0 lim x!+1 ° 1° x ex ¢ = 1 µ lim x!+1 f (x) = +1. , f (A) = [1,+1). 4. µ ex = x+ a , ex ° x = a , f (x) = a i) Æ 2 (°1,1), ii) a = 1, µ iii) Æ 2 (1,+1), 5. f 0 (x) = ex °1 f 00 (x) = (ex °1)0 = ex > 0, x 2 R f µ µ . : 3 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 4 of 17
  5. 5. 6. M(x1, f (x1)) µ µ (") µ f (x), . (") : y° f (x1) = f 0 (x1)(x° x1) ) y°(ex1 ° x1) = (ex1 °1)(x° x1) µ x = 0 y = 0 : 0° ex1 + x1 = (ex1 °1)(0° x1) ) °ex1 + x1 = °x1ex1 + x1 ) x1ex1 ° ex1 = 0 ) ex1 (x1 °1) = 0 ) x1 = 1 µ M (1, e °1). µ : y°(e °1) = (e °1)(x°1) ) y° e +1 = (e °1)x° e +1 ) y = (e °1)x 4 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 5 of 17
  6. 6. 7. f [0,1] f (x) ∏ (e °1)x, : E = 1R 0 [f (x)°(e °1)x]dx = 1R 0 (ex ° x°(e °1)x)dx = 1R 0 ex dx° 1R 0 xdx°(e °1) 1R 0 xdx = = [ex ]1 0 ° h x2 2 i1 0 °(e °1) h x2 2 i1 0 = e °1° °1 2 °0 ¢ °(e °1) °1 2 °0 ¢ = e °1° 1 2 ° e°1 2 = = e °1° 1 2 ° e 2 + 1 2 = e 2 °1 = e°2 2 ø.µ. 8. µ µ ex ∏ ex, x 2 R µ : ex ∏ ex , ex ° x ∏ ex , f (x) ∏ (e °1)x, µ f µ . 5 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 6 of 17
  7. 7. 3 f (x) = ln ≥ x+ p x2 +1 ¥ . 1. µ f . 2. f µµ . 3. µ f µ . 4. µ f µ - µ . 5. f f °1 (x) µ - µ , µ µ µ . 6. µ (") Cf µ µ µ µ x0 = 0 µ Cf , µ (") x = 1. 7. E1 µ Cf , x0 x x = 0,x = 1 E2 µ Cf °1 , y0 y y = 0,y = 1 : E1 + E2 = 2 ≥ 1° p 2+ln(1+ p 2) ¥ 8. : lim x!0 µ f °1 (x)· ex x ∂ 6 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 7 of 17
  8. 8. 4 1. x+ p x2 +1 > 0 • x < 0, x+ p x2 +1 > 0 ) p x2 +1 > °x ) ≥p x2 +1 ¥2 > (°x)2 ) x2 +1 > x2 +1 ) 1 > 0, • x > 0, x+ p x2 +1 > 0 ) p x2 +1 > °x, Af = R 2. µ Af = R x 2 R °x 2 R µ : f (°x) = ln ≥ °x+ p (°x)2 +1 ¥ = ln ≥p x2 +1° x ¥ = ln ≥p x2+1°x ¥≥p x2+1+x ¥ ≥p x2+1+x ¥ = = ln ≥p x2+1 ¥2 °x2 ≥p x2+1+x ¥ = ln x2 +1°x2 ≥p x2+1+x ¥ = ln 1≥p x2+1+x ¥ = ln1°ln ≥p x2 +1+ x ¥ = = °ln ≥p x2 +1+ x ¥ = °f (x) f Cf µµ . 3. f f 0 (x) = ≥ x+ p x2+1 ¥0 x+ p x2+1 = 1+ ≥p x2+1 ¥0 x+ p x2+1 = 1+ (x2+1) 0 2 p x2+1 x+ p x2+1 = 1+ 62x 62 p x2+1 x+ p x2+1 = = x+ p x2+1p x2+1 x+ p x2+1 1 = x+ p x2+1≥ x+ p x2+1 ¥p x2+1 = 1p x2+1 f 0 (x) = 1p x2+1 > 0, x 2 R, f R . 4. f 00 (x) = ≥ 1p x2+1 ¥0 = ° ≥p x2+1 ¥0 ≥p x2+1 ¥2 = ° 62x 62 p x2+1 x2+1 = ° x≥p x2+1 ¥ (x2+1) f 00 (x) = 0 , ° x≥p x2+1 ¥ (x2+1) = 0 , x = 0 f 00 (x) > 0 , ° x≥p x2+1 ¥ (x2+1) > 0 , x < 0 f 00 (x) < 0 , ° x≥p x2+1 ¥ (x2+1) < 0 , x > 0 7 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 8 of 17
  9. 9. : µ (°1,0] Cf µ [0,+1) Cf µ x0 = 0 f µ O(0,0). 5. µ f 0 (x) = 1p x2+1 > 0 f R, "1°1", . f (x) = y , ln ≥ x+ p x2 +1 ¥ = y , x+ p x2 +1 = ey , p x2 +1 = ey ° x ≥p x2 +1 ¥2 = (ey ° x)2 , x2 +1 = e2y °2xey + x2 , 2xey = e2y °1 , x = e2y °1 2ey ) f °1 (y) = e2y °1 2ey = e2y 2ey ° 1 2ey = ey 2 ° 1 2ey = ey 2 ° e°y 2 ) f °1 (y) = 1 2 (ey ° e°y ) , f °1 (x) = 1 2 (ex ° e°x ),x 2 R g(x) = f °1 (x) = 1 2 (ex ° e°x ) g0 (x) = 1 2 (ex + e°x ) > 0 g00 (x) = 1 2 (ex ° e°x ) g00 (x) = 0 , 1 2 (ex ° e°x ) = 0 , ex ° e°x = 0 , ex = e°x , x = 0 8 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 9 of 17
  10. 10. : µ (°1,0] Cg µ [0,+1) Cg µ x0 = 0 g µ O(0,0). 6. µ (") Cf x0 = 0 : 9 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 10 of 17
  11. 11. (" :) y° f (0) = f 0 (0)(x°0) ) y°0 = 1·(x°0) ) y = x Cf [0,+1) f (x) ∑ x, x 2 [0,+1) " = " µ x = 0 f (x)° x ∑ 0. : E = 1R 0 |f (x)° x|dx = 1R 0 (x° f (x))dx = 1R 0 xdx° 1R 0 f (x)dx = = h x2 2 i1 0 ° 1R 0 ln ≥ x+ p x2 +1 ¥ dx = 1 2 ° 1R 0 ln ≥ x+ p x2 +1 ¥ dx I1 = 1R 0 ln ≥ x+ p x2 +1 ¥ dx = 1R 0 (x)0 ·ln ≥ x+ p x2 +1 ¥ dx = = h x·ln ≥ x+ p x2 +1 ¥i1 0 ° 1R 0 x· xp x2+1 dx = h x·ln ≥ x+ p x2 +1 ¥i1 0 ° 1R 0 x· xp x2+1 dx = = ln(1+ p 2)° 1R 0 x· xp x2+1 dx p x2 +1 = t ) x2 +1 = t2 ) 2xdx = 2tdt ) xdx = tdt x = 0 ! t = 1 x = 1 ! t = p 2 I1 = ln(1+ p 2)° p 2R 1 t t dt = ln(1+ p 2)°[t] p 2 1 = ln(1+ p 2)° p 2+1 E = 1 2 °ln(1+ p 2)+ p 2°1 = p 2° 1 2 °ln(1+ p 2). 7. µµ Cf Cg y = x : E1 = E2 : E1 + E2 = (ABO°)°2E = (OB)·(AB)°2 °p 2° 1 2 °ln(1+ p 2) ¢ = = 1·1°2 p 2+1+2ln(1+ p 2) = 2°2 p 2+2ln(1+ p 2) = 2· ° 1° p 2+ln(1+ p 2) ¢ 8. lim x!0 ≥ f °1 (x)· ex x ¥ = lim x!0 ≥ ex °e°x 2 · ex x ¥ = lim x!0 e2x °1 2x °0 0 ¢ = DLH lim x!0 (e2x °1) 0 (2x)0 = = lim x!0 62e2x 62 = lim x!0 e2x = 1 10 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 11 of 17
  12. 12. 5 f : R ! R µ : ° x2 +1 ¢ · h f (x)°ln p x2 +1 i = x· £ x° ° x2 +1 ¢ f 0 (x) § , 8x 2 R (3) 1. f (x). 2. . 3. µ µ . 4. µ µ µ . 5. µ . 6. 2f (x) = x+ln 2 e , x 2 [1,+1) 7. E2 E1 = ln4 4°º °1, : • E1 µ f x = °1, x = 1 y = ln p 2 2 • E2 µ f x = °1, x = 1 y = 0. 8. lim x!+1 f (x)+x f (x)+ex . 11 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 12 of 17
  13. 13. 6 1. 3 µ : ° x2 +1 ¢ · f (x)° ° x2 +1 ¢ ·ln p x2 +1 = x2 ° x ° x2 +1 ¢ f 0 (x) , ° x2 +1 ¢ · f (x)+ x ° x2 +1 ¢ f 0 (x) = ° x2 +1 ¢ ·ln p x2 +1+ x2 :(x2 +1) , f (x)+ xf 0 (x) = ln p x2 +1+ x· x x2+1 , (x· f (x))0 = (x)0 ·ln p x2 +1+ x· ≥ ln p x2 +1 ¥0 , (x· f (x))0 = ≥ x·ln p x2 +1 ¥0 x· f (x) = x·ln p x2 +1+ c x = 0: 0 = 0+ c , c = 0 µ , x· f (x) = x·ln p x2 +1+ c : x6=0 , f (x) = ln p x2 +1, x 6= 0 f (x) = ln p x2 +1, x 6= 0 f µ µ , µ x0 = 0, µ : f (0) = lim x!0 f (x) = lim x!0 ≥ ln p x2 +1 ¥ = ln1 = 0 , f (0) = 0, f (x) = ln p x2 +1. , f (x) = ln p x2 +1, x 2 R . 2. Af = R, x 2 R °x 2 R µ f (°x) = ln p (°x)2 +1 = ln p x2 +1 = f (x), f . 3. x 2 R : f 0 (x) = ≥ ln p x2 +1 ¥0 = ≥p x2+1 ¥0 p x2+1 = 2x 2 p x2+1 p x2+1 = x x2+1 f 0 (x) = 0 , x x2+1 = 0 , x = 0 f 0 (x) > 0 , x x2+1 > 0 x2 +1>0 , x > 0 µ : 12 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 13 of 17
  14. 14. , µ (°1,0] f , µ [0,+1) f . µ x0 = 0 f µ µ min f (x) = f (0) = 1 4. f 00 (x) = ≥ x x2+1 ¥0 = (x)0 ·x2 +1°x(x2 +1) 0 (x2+1)2 = x2 +1°2x2 (x2+1)2 = °x2 +1 (x2+1)2 f 00 (x) = 0 , °x2 +1 (x2+1)2 = 0 , x = °1 x = 1 µ : µ (°1, ° 1], [1 , +1) f , µ [°1,1] f . µ A ° °1,ln p 2 ¢ B ° 1,ln p 2 ¢ - µ µ . 5. [Af = R µ . µ : ∏ = lim x!+1 f (x) x = lim x!+1 ln p x2+1 x (1 1 ) = lim x!+1 x x2+1 1 = lim x!+1 x x2+1 = 0 Ø = lim x!+1 f (x) = lim x!+1 ≥ ln p x2 +1 ¥ = +1 µ µ +1, µ °1. µ : 13 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 14 of 17
  15. 15. 6. x 2 [1,+1) : 2f (x) = x+ln 2 e , 2f (x) = x+ln2°ln e , f (x) = 1 2 x+ 1 2 ln2° 1 2 , f (x) = 1 2 x° 1 2 + 1 2 ln2 µ f B ° 1,ln p 2 ¢ - : (") : y° 1 2 ln2 = f 0 (1)(x°1) , y° 1 2 ln2 = 1 2(x°1) , y = 1 2 x° 1 2 + 1 2 ln2 f x 2 [1,+1) : f (x) ∑ 1 2 x ° 1 2 + 1 2 ln2 " = " µ µ µ x = 1. : f (x) = 1 2 x° 1 2 + 1 2 ln2 , x = 1 14 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 15 of 17
  16. 16. 7. µ E2 x 2 [°1,1] f (x) ∏ 0 : E2 = R1 °1 f (x)dx = R1 °1 f (x)dx = R1 °1 ln p x2 +1dx f :ÆΩø∂Æ = 2· R1 0 ln p x2 +1dx = = 2· R1 0 (x)0 ln p x2 +1dx =2 µh x·ln p x2 +1 i1 0 ° R1 0 x· ≥ ln p x2 +1 ¥0 dx ∂ = = 2 ≥ ln p 2°0° R1 0 x· x x2+1 dx ¥ = 2ln p 2°2· R1 0 x2 x2+1 dx = = 2· 1 2 ln2°2 R1 0 x2 +1°1 x2+1 dx = ln2°2 ≥R1 0 x2 +1 x2+1 dx° R1 0 1 x2+1 dx ¥ = = ln2°2+2 R1 0 1 x2+1 dx = ln2°2+2I1 I1 = R1 0 1 x2+1 dx x = "'t ) dx = 1 æ¿∫2t dt = (1+"'2 t)dt x = 0 : "'t = 0 , t = 0 x = 1 : "'t = 1 , t = º 4 I1 = Rº 4 0 1 1+"'2t (1+"'2 t)dt = Rº 4 0 dt = º 4 µ , E2 = ln2°2+2· º 4 = ln2°2+ º 2 ø.µ. µ E1 : E1 = (AB°¢)°E2 = 2· 1 2 ·ln2° ° ln2°2+ º 2 ¢ = 2° º 2 ø.µ. , E2 E1 = ln2°2+º 2 2°º 2 = ln2°(2°º 2 ) 2°º 2 = ln2 2°º 2 °1 = 2ln2 4°º °1 = ln4 4°º °1 . 15 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 16 of 17
  17. 17. 8. lim x!+1 f (x)+x f (x)+ex = lim x!+1 ln p x2+1+x ln p x2+1+ex (1 1 ) = lim x!+1 ≥ ln p x2+1+x ¥0 ≥ ln p x2+1+ex ¥0 = = lim x!+1 x x2+1 +1 x x2+1 +ex = lim x!+1 x+x2+1 x2+1 x+(x2+1)ex x2+1 = lim x!+1 x2 +x+1 x+(x2+1)ex (1 1 ) = = lim x!+1 2x+1 2xex+(x2+1)ex+1 (1 1 ) = lim x!+1 2 ex (x2+1+4x+2) = lim x!+1 2 ex (x2+4x+3) = 0 16 23.05.2019 Αποκλειστικά στο lisari.blogspot.gr Page 17 of 17

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