Getal & ruimte: Tweede fase [11 ed.] 900184247X, 9789001842475 [PDF]

Zitiervorschau

Uitwerkingen vwo B deel 1 ELFDE EDITIE, 2015

J.H. Dijkhuis C.J. Admiraal J.A. Verbeek G. de Jong H.J. Houwing J.D. Kuis F. ten Klooster S.K.A. de Waal J. van Braak H. Liesting M. Wieringa M.L.M. van Maarseveen R.D. Hiele J.E. Romkes M. Haneveld S. Voets I. Cornelisse

Noordhoff Uitgevers Groningen

1

Inhoud 4

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Functies en grafieken

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De afgeleide functie 39

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Vergelijkingen en herleidingen 65

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Meetkunde

102

Wiskunde Olympiade 130 Gemengde opgaven 136 Voor sommige (computer)opgaven is geen uitwerking opgenomen. Deze zijn aangegeven met een *.

© Noordhoff Uitgevers bv

1 Functies en grafieken 1

Voorkennis Lineaire vergelijkingen en ongelijkheden Bladzijde 8

a1 a 10 í 3(x + 1) = 5x í (2x í 1) 10 í 3x í 3 = 5x í 2x + 1 í3x í 5x + 2x = 1 í 10 + 3 í6x = í6 í6 x= =1 í6 b 45 x í 113 = 213 x í 3 12x í 20 = 35x í 45 í23x = í25 í25 2 x= = 123 í23 c

2t í 3 = t í 113 4 6t í 9 = 12t í 16 í6t = í7 í7 t= = 116 í6

d 1,6(2x í 1) = 1,4x í 2 3,2x í 1,6 = 1,4x í 2 3,2x í 1,4x = í2 + 1,6 1,8x = í0,4 í0,4 í4 x= = = í 29 1,8 18 e

2 7 (4x

í 1) = 34 (1 í 5x) 8(4x í 1) = 21(1 í 5x) 32x í 8 = 21 í 105x 137x = 29 29 x = 137

3t í 1 5t + 1 2t + 3 í = 6 4 3 60 í 2(3t í 1) = 3(5t + 1) í 4(2t + 3) 60 í 6t + 2 = 15t + 3 í 8t í 12 í6t í 15t + 8t = 3 í 12 í 2 í 60 í13t = í71 í71 6 t= = 513 í13

f 5í

Bladzijde 9 2

a 3x > 5x í2x > 0 x 15 3p í 4 c ” 2p í 116 3 2(3p í 4) ” 12p í 7 6p í 8 ” 12p í 7 í6p ” 1 p • í 16

d 1,5(1,6x í 2) < 2,5(1,4x í 3) 2,4x í 3 < 3,5x í 7,5 í1,1x < í4,5 í11x < í45 í45 x> í11 1 x > 411

e

3 8 (5x

í 2) > 14 (2x í 5) 3(5x í 2) > 2(2x í 5) 15x í 6 > 4x í 10 11x > í4 4 x > í 11 2a í 3 4a í 1 3a + 2 í • 5 6 15 300 í 5(2a í 3) • 6(4a í 1) í 2(3a + 2) 300 í 10a + 15 • 24a í 6 í 6a í 4 í10a í 24a + 6a • í6 í 4 í 300 í 15 í28a • í325 a ” 1117 28

f 10 í

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1.1 Lineaire functies 1

Bladzijde 10

a1 a Ga je 1 naar rechts, dan ga je 2 omhoog. Het getal 3 geeft aan dat de lijn de y-as in het punt (0, 3) snijdt. b Stel l: y = ax + b. Het snijpunt met de y-as is (0, 2), dus b = 2. verticaal í1 l gaat door (0, 2) en (2, 1), dus a = = = í 12 . horizontaal 2 Dus l: y = í 12 x + 2. c Stel m: y = ax + b. Het snijpunt met de y-as is (0, í1), dus b = í1. m is evenwijdig met l dus a = í 12 . Dus m: y = í 12 x í 1. Bladzijde 11 2

a Stel k: y = ax + b. k // l, dus a = rcl = í 12 . y = í 12 x + b 1 f í2  4 + b = 3 door A(4, 3) í2 + b = 3 b=5 Dus k: y = í 12 x + 5. b m: y = ax + 3 r a  í4 + 3 = 2 door B(í4, 2) í4a + 3 = 2 í4a = í1 a = 14 c p snijden met de x-as, dus y = 0, geeft í112 x + 6 = 0 í112 x = í6 x=4 Dus p snijdt de x-as in (4, 0). n: y = 212 x + b 21 f 2 Â4 + b = 0 door (4, 0) 10 + b = 0 b = í10

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a Stel l: y = ax + b. a = rcl = í2 y = í2x + b f í2 Â í2 + b = 3 door A(í2, 3) 4+b=3 b = í1 Dus l: y = í2x í 1.

b Stel k: y = ax + b. k // m, dus a = rcm = 4. k: y = 4x + b f 4 Â í5 + b = 21 door B(í5, 21) í20 + b = 21 b = 41 Dus k: y = 4x + 41.

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a Stel p: y = ax + b. p // q, dus a = rcq = í 13.

b Snijpunt met de x-as: y = 0 geeft í 13 x + 24 = 0

í 13 x

+b y= 1 f í Â í18 + b = 30 door C(í18, 30) 3 6 + b = 30 b = 24 Dus p: y = í 13 x + 24.

© Noordhoff Uitgevers bv

í 13 x = í24 í24 x = 1 = 72 í3 Dus het snijpunt met de x-as is (72, 0). Snijpunt met de y-as: x = 0 geeft y = í 13 Â 0 + 24 = 24. Dus het snijpunt met de y-as is (0, 24).

Functies en grafieken

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a k: y = ax + 10 f a  í20 + 10 = 0 door P(í20, 0) í20a = í10 a = 12

c k: y = ax + 10 f a  0 + 10 = 0 door O(0, 0) 10 = 0 dit kan niet Er is dus geen a mogelijk.

b k: y = ax + 10 f a  2 + 10 = í4 door Q(2, í4) 2a = í14 a = í7 Bladzijde 12

a6 a m: y = í2x + b f í2 Â í8 + b = 0 door P(í8, 0) 16 + b = 0 b = í16 b m // l indien rcl = rcm, dus a = í2. m: y = í2x + b f í2 Â 10 + b = 7 door Q(10, 7) í20 + b = 7 b = 27 c k gaat door R(8, 6) want 6 = 12 Â 8 + 2. l: y = ax í 4 f aÂ8 í 4 = 6 door R(8, 6) 8a í 4 = 6 8a = 10 10 a= = 114 8 m: y = í2x + b f í2 Â 8 + b = 6 door R(8, 6) í16 + b = 6 b = 22 7

d k: y = 12 x + 2 1 f x+2=0 snijpunt met x-as: y = 0 12 2 x = í2 x = í4 Dus het snijpunt met de x-as is (í4, 0). l: y = ax í 4 f a  í4 í 4 = 0 door (í4,0) í4a í 4 = 0 í4a = 4 a = í1 m: y = í2x + b f í2  í4 + b = 0 door (í4, 0) 8+b=0 b = í8

a Kies x zo dat de formule geen a meer bevat. Voor k: y = ax + 1 geldt dat x = 0 geeft y = a  0 + 1 = 1, dus A(0, 1). Voor l: y = 2ax í 2a geldt dat y = 2ax í 2a = 2a(x í 1) x = 1 geeft y = 2a(1 í 1) = 0, dus B(1, 0). b k en l snijden in A dus l: y = 2ax í 2a f 2a  0 í 2a = 1 door A(0, 1) í2a = 1 a = í 12 k en l snijden in B dus k: y = ax + 1 f aÂ1 + 1 = 0 door B(1, 0) a+1=0 a = í1 c k: y = ax + 1 f y = a  10 + 1 = 10a + 1 xC = 10 k: y = 2ax í 2a f y = 2a  10 í 2a = 18a xC = 10 Formules aan elkaar gelijkstellen geeft 10a + 1 = 18a í8a = í1 í1 1 a= = í8 8 1 1 Dus yC = 18  10 + 1 = 10 8 + 1 = 14 + 1 = 24 .

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a Ga je 4 naar rechts, dan ga je 3 omhoog, dus ga je 1 naar rechts, dan ga je 34 omhoog. Dus rcl = 34 . yB í yA b Deel yB í yA door xB í xA. Dus rcl = x í x = 34 . B A

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Bladzijde 14

a Stel k: y = ax + b. )y 11 í 8 3 1 a= = = = )x 20 í 8 12 4 y = 14 x + b 1 f Â8 + b = 8 door (8, 8) 4 2+b=8 b=6 Dus k: y = 14 x + 6.

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Stel l: y = ax + b. ¨y í10 í 14 í24 a= = = = í 12 ¨x 50 í 2 48 y = í 12 x + b 1 f í 2 Â 2 + b = 14 door (2, 14) í1 + b = 14 b = 15 1 Dus l: y = í 2 x + 15. k en l snijden geeft + 6 = í 12 x + 15

1 4x 1 4x

+ 12 x = 15 í 6

3 4x

=9 3x = 36 x = 12 y = 14 x + 6 1 f y = 4 Â 12 + 6 = 3 + 6 = 9 x = 12 Dus E(12, 9).

10

a Stel l: y = ax + b. ¨y 4 í 1 3 a= = = = 112 ¨x 1 í í1 2 y = 112 x + b f 1+b=4 door (1, 4) 112 + b = 4 b = 212 Dus l: y = 112 x + 212 . b Stel k: y = ax + b. ¨y 0 í 5 í5 a= = í1 = = ¨x 2 í í3 5 y = íx + b f í2 + b = 0 door (2, 0) b=2 Dus k: y = íx + 2.

11

b Stel N = aM + b. M = 5 en N = 62 ¨N 86 í 62 24 = = = 135 fa= M = 20 en N = 86 ¨M 20 í 5 15 N = 135 M + b 3 f 15  5 + b = 62 M = 5 en N = 62 8 + b = 62 b = 54 Dus N = 135 M + 54. Stel M = aN + b. N = 62 en M = 5 )M 20 í 5 15 5 = = = fa= N = 86 en M = 20 )N 86 í 62 24 8 M = 58 N + b 5 f  62 + b = 5 N = 62 en N = 5 8 3834 + b = 5 b = í3334 Dus M = 58 N í 3334 . Alternatieve oplossing Omwerken van de formule geeft M uitgedrukt in N. N = 135M + 54 135 M + 54 = N 8M + 270 = 5N 8M = 5N í 270 M = 58 N í 3334

c Stel m: y = ax + b. ¨y 3 í 3 0 a= = = =0 ¨x í7 í 5 í12 y=b f b=3 door (5, 3) Dus m: y = 3. d Stel n: y = ax + b. ¨y 250 í 360 í110 a= = = = 512 ¨x 160 í 180 í20 y = 512 x + b 1 f 52 Â 180 + b = 360 door (180, 360) 990 + b = 360 b = í630 Dus n: y = 512 x í 630.

a Stel K = am + b. m = 5 en K = 10 )K 115 í 10 105 = = 15 = f a= m = 12 en K = 115 7 12 í 5 )m K = 15m + b f 15 Â 5 + b = 10 m = 5 en K = 10 75 + b = 10 b = í65 Dus K = 15m í 65.

© Noordhoff Uitgevers bv

Functies en grafieken

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b Stel F = aR + b. R = 15 en F = 300 ¨F 138 í 300 í162 = í6 = = f a= R = 42 en F = 138 ¨R 42 í 15 27 F = í6R + b f í6 Â 15 + b = 300 R = 15 en F = 300 í90 + b = 300 b = 390 Dus F = í6R + 390. c Stel g = an + b. ¨g 49 í 35 14 n = 6 en g = 35 = = = 312 fa= n = 10 en g = 49 ¨n 10 í 6 4 g = 312 n + b 1 f 32 Â 6 + b = 35 n = 6 en g = 35 21 + b = 35 b = 14 Dus g = 312 n + 14.

1

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a Stel p = aq + b. ¨p 2,25 í 7,75 í5,5 q = 150 en p = 7,75 = = = í0,02 fa= q = 425 en p = 2,25 ¨q 425 í 150 275 p = í0,02q + b f í0,02 Â 150 + b = 7,75 q = 150 en p = 7,75 í3 + b = 7,75 b = 10,75 Dus p = í0,02q + 10,75. Stel q = ap + b. ¨q 425 í 150 p = 7,75 en q = 150 275 = = í50 = f a= p = 2,25 en q = 425 ¨p 2,25 í 7,75 í5,5 p = í50q + b f í50 Â 7,75 + b = 150 p = 7,75 en q = 150 í387,5 + b = 150 b = 537,5 Dus q = í50p + 537,5. Alternatieve oplossing Omwerken van de formule geeft q uitgedrukt in p. p = í0,02q + 10,75 0,02q = íp + 10,75 q = í50p + 537,5 b p = í0,02q + 10,75 f p = í0,02 Â 250 + 10,75 = 5,75 q = 250 q = í50p + 537,5 f q = í50 Â 4,25 + 537,5 = 325 p = 4,25 Bladzijde 15

13

a Stel h = at + b. t = 12 en h = 164,0 ¨h 152,8 í 164,0 = = í113 f a= 15 t = 18 en h = 152,8 ¨t 18 í 12 13 h = í115 t + b 13 f í115 Â 12 + b = 164,0 t = 12 en h = 164,0 í22,4 + b = 164,0 b = 186,4 Dus h = í113 15 t + 186,4. b h = 156,7 geeft í113 15 t + 186,4 = 156,7 í113 15 t = í29,7 t=

í29,7 § 15,9 í113 15

De auto passeert hectometerpaal 156,7 om ongeveer 14:16 uur.

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© Noordhoff Uitgevers bv

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Stel y = ax + b. )y p + 2 í (p + 1) 1 a= = = 2p í p p )x

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y = 1p x + b 1 f p Âp + b = p + 1 door (p, p + 1) 1+b=p+1 b=p Dus y = p1 x + p. Snijden met de x-as: y = 0 geeft 1p x + p = 0 1 px

= íp x = íp2 Voor p  0 geldt dat íp2 < 0, dus x < 0. Dus elke lijn snijdt de negatieve x-as. 15

a

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Bladzijde 16

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a f (x) = 0 12 x í 1 0 = 12 x í 1 als 12 x í 1 • 0, dus als x • 2. f (x) = 0 12 x í 1 0 = í 12 x + 1 als 12 x í 1 < 0, dus als x < 2. y

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© Noordhoff Uitgevers bv

Functies en grafieken

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b g(x) = í 0 2x í 6 0 = í2x + 6 als 2x í 6 • 0, dus als x • 3. g(x) = í 0 2x í 6 0 = 2x í 6 als 2x í 6 < 0, dus als x < 3. y

1

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c h(x) = 2 í 0 13 x í 2 0 = 4 í 13 x als 13 x í 2 • 0, dus als x • 6. h(x) = 2 í 0 13 x í 2 0 = 13 x als 13 x í 2 < 0, dus als x < 6. y

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d k(x) = 5 í 0 6 í 112 x 0 = í1 + 112 x als 6 í 112 x • 0, dus als x ” 4. k(x) = 5 í 0 6 í 112 x 0 = 11 í 112 x als 6 í 112 x < 0, dus als x > 4.

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a f (x) = x + 2 í 0 2x í 1 0 = íx + 3 als 2x í 1 • 0, dus als x • 12 . f (x) = x + 2 í 0 2x í 1 0 = 3x + 1 als 2x í 1 < 0, dus als x < 12 . y

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Functies en grafieken

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y

b 1 1

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O

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22

A − 13

B

x

3

xC = 12 geeft y = í 12 + 3 = 212 , dus C (12 , 212 ) . yA = 0 geeft 3x + 1 = 0 3x = í1 x = í 13 Dus A (0, í 13 ) . yC = 0 geeft íx + 3 = 0 íx = í3 x=3 Dus B(0, 3). De oppervlakte van het ingesloten vlakdeel is 12 Â 313 Â 212 = 416 . 18

a f (x) = ax í 1 í 0 3x í 4 0 = (a í 3)x + 3 als 3x í 4 • 0, dus als x • 113 . f (x) = ax í 1 í 0 3x í 4 0 = (a + 3)x í 5 als 3x í 4 < 0, dus als x < 113 . f (x) = (a + 3)x í 5 f (a + 3)  1 í 5 = 0 door (1, 0) a+3í5=0 a=2 y

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f (x) = 5x í 5 f 5x í 5 = x y=x 4x = 5 x = 114 1 1 Dus A (14 , 14 ) . f (x) = íx + 3 f íx + 3 = x y=x í2x = í3 í3 x= = 112 í2 Dus B (112 , 112 ) . c

y y = px 2 C 1

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© Noordhoff Uitgevers bv

Functies en grafieken

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x = 113 geeft y = í113 + 3 = 123 , dus C (113 , 123 ) . y = px 1 2 f p  13 = 13 door C (113 , 123 ) 113 p = 123 4p = 5 p = 114

1

Voor p = 114 gaat de grafiek van y = px door het punt C. Voor p = 5 is de lijn y = px evenwijdig met f. Dus de lijn y = px heeft geen enkel punt met f gemeen voor 114 < p ” 5. 19

f (x) = 4 í 0 3 í 0 2x í 6 0 0 = 4 í 0 í2x + 9 0 als 2x í 6 • 0, dus als x • 3. Dit geeft f (x) = 4 í 0 í2x + 9 0 = 2x í 5 als 9 í 2x • 0, dus als 3 ” x ” 412 en f (x) = 4 í 0 í2x + 9 0 = í2x + 13 als 9 í 2x < 0, dus als x > 412 . f (x) = 4 í 0 3 í 0 2x í 6 0 0 = 4 í 0 2x í 3 0 als 2x í 6 < 0, dus als x < 3. Dit geeft f (x) = 4 í 0 2x í 3 0 = í2x + 7 als 2x í 3 • 0, dus als 112 ” x < 3 en f (x) = 4 í 0 2x í 3 0 = 2x + 1 als 2x í 3 < 0, dus als x < 112 . y

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1.2 Tweedegraadsvergelijkingen Bladzijde 18 20

14

a Bij 4x2 í 25x = 0 breng je x buiten de haakjes en bij 4x2 í 25 = 0 schrijf je de vergelijking eerst in de vorm x2 = c. b 4x2 + 25 = 0 4x2 = í25 x2 = í6,25 geen oplossingen c x(x + 2) = 0 los je op met behulp van de regel: A  B = 0 geeft A = 0  B = 0. Dit geldt niet voor x(x + 2) = 8, omdat het rechterlid niet gelijk is aan 0.

Hoofdstuk 1

© Noordhoff Uitgevers bv

Bladzijde 20 21

c b a x2 + a x + a = 0 1 b 1 b 2 x+ Â í Â 2 a 2 a

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b2 í 4ac 4a2

冑b2 í 4ac b 冑b2 í 4ac b = =í  x+ 2a 2a 2a 2a

x=í x=

) ( ) ) ( ) ) ) ) )

1 2

b + 2a

冑b2 í 4ac 2a

x=í

b 冑b2 í 4ac í 2a 2a

íb + 冑 í 4ac íb í 冑b2 í 4ac  x= 2a 2a b2

a 2x2 í 13x = 3(x í 10) 2x2 í 13x = 3x í 30 2x2 í 16x + 30 = 0 x2 í 8x + 15 = 0 (x í 3)(x í 5) = 0 x=3  x=5 b 3x2 + 2x + 7 = 7(x + 1) 3x2 + 2x + 7 = 7x + 7 3x2 í 5x = 0 x(3x í 5) = 0 x = 0  3x í 5 = 0 x = 0  3x = 5 x = 0  x = 123 c 100(x2 í 1) = 525 100x2 í 100 = 525 100x2 = 625 x2 = 614 x = 212  x = í212

d 2(x í 3)2 = 3x í 10 2(x2 í 6x + 9) = 3x í 10 2x2 í 12x + 18 = 3x í 10 2x2 í 15x + 28 = 0 D = (í15)2 í 4  2  28 = 1 15 í 1 15 + 1 x= = 312  x = =4 4 4 e 5(4x í 1)(6x í 5) = 0 4x í 1 = 0  6x í 5 = 0 4x = 1  6x = 5 x = 14  x = 56

a (3x í 2)2 = 36 3x í 2 = 6  3x í 2 = í6 3x = 8  3x = í4 x = 223  x = í113

d x(x í 1) = 12 x2 í x = 12 x2 í x í 12 = 0 (x + 3)(x í 4) = 0 x = í3  x = 4 e 2x2 = 5x 2x2 í 5x = 0 x(2x í 5) = 0 x = 0  2x í 5 = 0 x = 0  2x = 5 x = 0  x = 212 f x2 + 4 = 1 x2 = í3 geen oplossingen

b (4 í 12 x )2 = 9 4 í 12 x = 3  4 í 12 x = í3 í 12 x = í1  í 12 x = í7 x = 2  x = 14 c x2 + 6 = 5x x2 í 5x + 6 = 0 (x í 2)(x í 3) = 0 x=2  x=3

© Noordhoff Uitgevers bv

f

1 4 (2x

í 3)2 í 3 = 1

1 4 (2x

í 3)2 = 4 (2x í 3)2 = 16 2x í 3 = 4  2x í 3 = í4 2x = 7  2x = í1 x = 312  x = í 12

Functies en grafieken

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24

a 3x2 í 6x = 24 3x2 í 6x í 24 = 0 x2 í 2x í 8 = 0 (x + 2)(x í 4) = 0 x = í2  x = 4 b 3x2 í 6x = í3(x í 6) 3x2 í 6x = í3x + 18 3x2 í 3x í 18 = 0 x2 í x í 6 = 0 (x + 2)(x í 3) = 0 x = í2  x = 3 c 2x2 í 3x = 2 2x2 í 3x í 2 = 0 D = (í3)2 í 4  2  í2 = 25 3í5 3+5 x= = í 12  x = =2 4 4

d 12 x2 í 2x í 6 = 0 x2 í 4x í 12 = 0 (x + 2)(x í 6) = 0 x = í2  x = 6 e x2 í 3x = 5(x í 3) x2 í 3x = 5x í 15 x2 í 8x + 15 = 0 (x í 3)(x í 5) = 0 x=3  x=5 f 2x2 í 5x = 3x 2x2 í 8x = 0 2x(x í 4) = 0 x=0  x=4

25

a 6 í x2 = í2 íx2 = í8 x2 = 8 x = 冑8 = 2冑2  x = í冑8 = í2冑2 b 2x2 = 9x + 5 2x2 í 9x í 5 = 0 D = (í9)2 í 4  2  í5 = 121 9 í 11 9 + 11 x= = í 12  x = =5 4 4 c 3(x + 2)2 + 5 = 80 3(x + 2)2 = 75 (x + 2)2 = 25 x + 2 = 5  x + 2 = í5 x = 3  x = í7

d 12 (x í 3)2 í 3 = 5

a x2 í 5x = 0 x(x í 5) = 0 x=0  xí5=0 x=0  x=5 b x2 í 5x = 14 x2 í 5x í 14 = 0 (x + 2)(x í 7) = 0 x = í2  x = 7 c x2 í 5 = 14 x2 = 19 x = 冑19  x = í冑19 d x2 í 5 = 14x x2 í 14x í 5 = 0 (x í 7)2 í 49 í 5 = 0 (x í 7)2 = 54 x í 7 = 冑54  x í 7 = í冑54

f (2x í 1)(3x + 6) = 9x 6x2 + 12x í 3x í 6 = 9x 6x2 = 6 x2 = 1 x = 1  x = í1 g 3x(2x í 1) = 6 6x2 í 3x = 6 6x2 í 3x í 6 = 0 2x2 í x í 2 = 0 D = (í1)2 í 4  2  í2 = 17

1

26

x í 7 = 3冑6  x í 7 = í3冑6 x = 7 + 3冑6  x = 7 í 3冑6 e (2x í 1)(3x + 6) = 0 2x í 1 = 0  3x + 6 = 0 2x = 1  3x = í6 x = 12  x = í2

16

Hoofdstuk 1

1 2 (x

í 3)2 = 8 (x í 3)2 = 16 x í 3 = 4  x í 3 = í4 x = 7  x = í1 e í(2x í 1)2 + 5 = 1 í(2x í 1)2 = í4 (2x í 1)2 = 4 2x í 1 = 2  2x í 1 = í2 2x = 3  2x = í1 x = 112  x = í 12 f 8 í 3(4x í 5)2 = 5 í3(4x í 5)2 = í3 (4x í 5)2 = 1 4x í 5 = 1  4x í 5 = í1 4x = 6  4x = 4 x = 112  x = 1

1 í 冑17 1 + 冑17  x= 4 4 x = 14 í 14 冑17  x = 14 + 14 冑17 h 3x(2x í 1) = 6 í 9x 6x2 í 3x = 6 í 9x 6x2 + 6x í 6 = 0 x2 + x í 1 = 0 D = 12 í 4  1  í1 = 5 í1 í 冑5 í1 + 冑5 x=  x= 2 2 x = í 12 í 12 冑5  x = í 12 + 12 冑5 x=

© Noordhoff Uitgevers bv

27

a (x + 3)2 = 16x x2 + 6x + 9 = 16x x2 í 10x + 9 = 0 (x í 1)(x í 9) = 0 x=1  x=9 b (2x + 3)2 = í16 geen oplossingen c 2(x + 3)2 = í4x 2(x2 + 6x + 9) = í4x 2x2 + 16x + 18 = 0 x2 + 8x + 9 = 0 (x + 4)2 í 16 + 9 = 0 (x + 4)2 = 7 x + 4 = 冑7  x + 4 = í冑7 x = í4 + 冑7  x = í4 í 冑7 d (2x + 3)(4 í x) = 9 8x í 2x2 + 12 í 3x = 9 í2x2 + 5x + 3 = 0 D = 52 í 4  í2  3 = 49 í5 í 7 í5 + 7 x= =3  x= = í 12 í4 í4 e (í4x + 3)2 = 36 í4x + 3 = 6  í4x + 3 = í6 í4x = 3  í4x = í9 x = í 34  x = 214

f í4(x + 3)2 = 4x x2 + 6x + 9 = íx x2 + 7x + 9 = 0 D = 72 í 4  1  9 = 13 í7 í 冑13 í7 + 冑13 x=  x= 2 2 x = í312 í 12 冑13  x = í312 + 12 冑13

1

g x2 í (x + 1)2 = (x + 3)2 x2 í (x2 + 2x + 1) = x2 + 6x + 9 x2 í x2 í 2x í 1 = x2 + 6x + 9 íx2 í 8x í 10 = 0 x2 + 8x + 10 = 0 (x + 4)2 í 16 + 10 = 0 (x + 4)2 = 6 x + 4 = 冑6  x + 4 = í冑6 x = í4 + 冑6  x = í4 í 冑6 h (x + 3)2 + (x + 2)2 = 25 x2 + 6x + 9 + x2 + 4x + 4 = 25 2x2 + 10x í 12 = 0 x2 + 5x í 6 = 0 (x í 1)(x + 6) = 0 x = 1  x = í6

Bladzijde 21 28

a x2 + 8x = 20 x(x + 8) = 20 (x + 4)2 = 20 + 42 (x + 4)2 = 36 x+4=6 x=2 x2 + 18x = 20 x(x + 18) = 20 (x + 9)2 = 20 + 92 (x + 9)2 = 101 101 x + 9 § 12 10 + 10

(

b x2 + bx = c x(x + b) = c ( x + 12 b )2 = c +

( 12 b)2 x + 12 b = 冑c + ( 12 b ) 2

冑

x= c+

c x2 + bx = c ( x + 12 b )2 í

( 12 b)2 = c ( x + 12 b)2 = c + ( 12 b ) 2 x + 12 b = 冑c + ( 12 b ) 2

)

x + 9 § 12 Â 201 10 x§

201 20

í9=

1 1020

( 12 b ) 2 í 12 b

í9=

1 120

冑

x= c+

( 12 b ) 2 í 12 b

29

a p = í1 geeft x2 í x í 6 = 0 (x + 2)(x í 3) = 0 x = í2  x = 3 b p = 2 geeft x2 + 2x í 6 = 0 D = 22 í 4  1  í6 = 28 D > 0, dus de vergelijking x2 + 2x í 6 = 0 heeft twee oplossingen. 2 c D = p í 4  1  í6 = p2 + 24 Omdat D = p2 + 24 groter is dan nul voor elke waarde van p, heeft de vergelijking x2 + px í 6 = 0 voor elke p twee oplossingen.

30

a D = (í7)2 í 4 Â 1 Â p = 49 í 4p f 49 í 4p > 0 twee oplossingen als D > 0 í4p > í49 p < 1214

Bladzijde 22

© Noordhoff Uitgevers bv

Functies en grafieken

17

b D = (í5)2 í 4  2  íp = 25 + 8p f 25 + 8p > 0 twee oplossingen als D > 0 8p > í25 p > í318 c D = 42 í 4  í3  íp = 16 í 12p f 16 í 12p > 0 twee oplossingen als D > 0 í12p > í16 p < 113 1 d D = (í3)2 í 4  4  p = 9 í p f9íp>0 twee oplossingen als D > 0 íp > í9 p 0 twee oplossingen als D > 0 p2 > 100 p < í10  p > 10 2 2 b D = p í 4  1  4 = p í 16 f p2 í 16 < 0 geen oplossingen als D < 0 p2 < 16 í4 < p < 4 c D = p2 í 4  í2  3 = p2 + 24 f p2 + 24 > 0 voor elke p. twee oplossingen als D > 0 De vergelijking heeft dus twee oplossingen voor elke p.

32

a x = 1 invullen geeft 12 + 2  1 + p = 0 1+2+p=0 p = í3 p = í3 geeft x2 + 2x í 3 = 0 (x í 1)(x + 3) = 0 x = 1  x = í3 Dus p = í3 en de andere oplossing is x = í3. b x = 2 invullen geeft p  22 í 11  2 + 10 = 0 4p í 22 + 10 = 0 4p = 12 p=3 p = 3 geeft 3x2 í 11x + 10 = 0 D = (í11)2 í 4  3  10 = 1 11 + 1 11 í 1 x= = 123  x = =2 6 6 2 Dus p = 3 en de andere oplosing is x = 13 .

33

a í3 < p < 0  p > 0 b í4 < p < 0  0 < p < 4

34

a p = 0 geeft 3x + 1 = 0 Dit is een lineaire vergelijking, dus de vergelijking heeft één oplossing. b De discriminant geldt alleen voor het bepalen van het aantal oplossingen van een kwadratische vergelijking. Er is alleen sprake van een kwadratische vergelijking onder de voorwaarde dat p  0. c Voor p  0 is D = 32 í 4  p  1 = 9 í 4p f 9 í 4p > 0 twee oplossingen als D > 0 í4p > í9 p < 214 Voor p = 0 heeft de vergelijking één oplossing. Dus de vergelijking heeft twee oplossingen voor p < 0  0 < p < 214 .

18

Hoofdstuk 1

© Noordhoff Uitgevers bv

35

a p = 0 geeft 5x + 2 = 0, dus één oplossing voor p = 0. Voor p  0 is D = 52 í 4  p  2 = 25 í 8p f 25 í 8p > 0 twee oplossingen als D > 0 í8p > í25 p < 318 Dus de vergelijking heeft twee oplossingen voor p < 0  0 < p < 318 . b p = 0 geeft í3x í 4 = 0, dus één oplossing voor p = 0. Voor p  0 is D = (í3)2 í 4  p  í4 = 9 + 16p f 9 + 16p > 0 twee oplossingen als D > 0 16p > í9 9 p > í 16

1

9 Dus de vergelijking heeft twee oplossingen voor í 16 < p < 0  p > 0.

36

a D = 12 í 4  2  p = 1 í 8p f 1 í 8p < 0 geen oplossingen als D < 0 í8p < í1 p > 18 b p = 0 geeft x = 0, dus één oplossing voor p = 0. Voor p  0 is D = 12 í 4  p  p = 1 í 4p2 f 1 í 4p2 > 0 twee oplossingen als D > 0 í4p2 > í1 p2 < 14 í 12 < p < 12 Dus de vergelijking heeft twee oplossingen voor í 12 < p < 0  0 < p < 12 . c D = p2 í 4  2  1 = p2 í 8 f p2 í 8 > 0 twee oplossingen als D > 0 p2 > 8 p < í冑8  p > 冑8 p < í2冑2  p > 2冑2

37

a p = 0 geeft 6x + 9 = 0 6x = í9 x = í112 Voor p  0 is D = 62 í 4  p  9 = 36 í 36p f 36 í 36p = 0 één oplossing als D = 0 í36p = í36 p=1 p = 1 geeft x2 + 6x + 9 = 0 (x + 3)(x + 3) = 0 x = í3 Voor p = 0 is de oplossing x = í112 en voor p = 1 is de oplossing x = í3. b D = p2 í 4  1  1 = p2 í 4 f p2 í 4 = 0 één oplossing als D = 0 p2 = 4 p = 2  p = í2 p = 2 geeft x2 + 2x + 1 = 0 (x + 1)(x + 1) = 0 x = í1 p = í2 geeft x2 í 2x + 1 = 0 (x í 1)(x í 1) = 0 x=1 Voor p = 2 is de oplossing x = í1 en voor p = í2 is de oplossing x = 1.

© Noordhoff Uitgevers bv

Functies en grafieken

19

38 1

Stel de vergelijking is ax2 + bx + c = 0. Er moet gelden dat • de vergelijking geen twee oplossingen heeft voor p = 0 • voor p  0 er twee oplossingen zijn voor p > í4. Voor a = p zijn er geen twee oplossingen voor p = 0, dus px2 + bx + c = 0. Voor p  0 is D = b2 í 4  p  c = b2 í 4pc f b2 í 4pc > 0 twee oplossingen als D > 0 pc < 14 b2 Neem bijvoorbeeld b = 4 en c = í1, dan is p  í1 < 14  42 íp < 4 ௘ p > í4 Dit geeft de tweedegraadsvergelijking px2 + 4x í 1 = 0.

1.3 Extreme waarden en inverse functies Bladzijde 25 39

a f (x) = 0 geeft ax2 + bx = 0 x(ax + b) = 0 x = 0  ax + b = 0 x = 0  ax = íb b x = 0  x = ía b xtop = is het gemiddelde van 0 en í a . b 0 + ía b Dus xtop = =í . 2 2a b De parabool y = ax2 + bx + c ontstaat uit de grafiek van f door de verschuiving c omhoog. Bij deze verschuiving verandert de xtop niet.

40

max. is g(1) = 4 min. is h(2) = 1

41

a xtop = í

42

a xtop = í

20

Hoofdstuk 1

Bladzijde 26

b í4 =í = 2 geeft ytop = f (2) = 22 í 4  2 + 1 = í3 2a 2Â1 1 > 0, dus dalparabool en min. is f (2) = í3. b 6 b xtop = í = í = í112 geeft ytop = g ( í112 ) = 2  (í112 )2 + 6  í112 + 3 = í112 2a 2Â2 2 > 0, dus dalparabool en min. is g (í112 ) = í112 . b 6 c xtop = í = í = 10 geeft ytop = h(10) = í0,3  102 + 6  10 í 2 = 28 2a 2  í0,3 í0,3 < 0 dus bergparabool en max. is h(10) = 28. b 14 d xtop = í = í = í134 geeft ytop = k (í134 ) = 4  (í134 ) 2 + 14  í134 = í1214 2a 2Â4 4 > 0 dus dalparabool en min. is k (í134 ) = í1214 . 0,4 b =í = 20 geeft htop = í0,01  202 + 0,4  20 + 5 = 9 2a 2  í0,01 í0,01 < 0, dus bergparabool. Dus de bal komt maximaal 9 meter hoog. b h = 0 geeft í0,01x2 + 0,4x + 5 = 0 x2 í 40x í 500 = 0 (x + 10)(x í 50) = 0 x = í10  x = 50 De bal komt 50 meter verderop op de grond.

© Noordhoff Uitgevers bv

Bladzijde 28 43

a f (í1) = í(í1)2 + 6 Â í1 í 3 = í10 en f (6) = í62 + 6 Â 6 í 3 = í3 b 6 xtop = í = í = 3 en ytop = f (3) = í32 + 6 Â 3 í 3 = 6 2a 2 Â í1

1

y (3, 6)

6

ƒ –1

x

6

O

(6, –3)

(–1, –10) –10

Dus B f = 3 í10, 6 4 . b f (4) = í42 + 6 Â 4 í 3 = 5 en f (8) = í82 + 6 Â 8 í 3 = í19 xtop = 3 (zie vraag a) y (4, 5)

5

ƒ 4

O

x

8

(8, –19)

–19

Dus B f = 3 í19, 5 4 . 44

f (í3) = í 12 Â í3 + 3 = 412 en f (4) = í 12 Â 4 + 3 = 1 1

(−3, 4 2 )

y

1

42 ƒ (4, 1)

1 −3

O

1

4

x

Dus B f = 3 1, 412 4 . 45

a h = í0,004x2 + 0,62x f í0,004x2 + 0,62x = 3 h=3 í0,004x2 + 0,62x í 3 = 0 x2 í 155x + 750 = 0 (x í 150)(x í 5) = 0 x = 150  x = 5 De golfbal komt neer na 150 m, dus D golfbal = 3 0, 150 4 .

© Noordhoff Uitgevers bv

Functies en grafieken

21

0,62 b =í = 77,5 en htop = í0,004 Â 77,52 + 0,62 Â 77,5 = 24,025 2a 2 Â í0,004 De golfbal komt 24,025 m hoog, dus B golfbal = 3 0; 24,025 4 .

b xtop = í 1 46

y

4

3

x

2

y

=

y = –x2 + 4x

1

O

1

2

x

4

3

Bladzijde 30 47

xtop = í

b 6 =í = 3, dus a = 3. 2a 2 Â í1

x

3

4

5

6

f(x)

6

5

2

í3

x

6

5

2

í3

f inv(x)

3

4

5

6

y

7

6 ƒinv 5

4

3

x

2

y

=

ƒ

1

−1

O

1

2

3

4

5

6

7

x

−1

22

Hoofdstuk 1

© Noordhoff Uitgevers bv

48

a

x f (x)

í5 5

í4 0

í3 í3

í2 í4

x

5

0

í3

í4

f inv(x)

í5

í4

í3

í2

1

y

5

4

3 ƒ

y

=

x

2

1

−5

−4

−3

−2

−1

O

1

2

3

4

5

x

−1

−2

−3

−4 ƒinv −5

b f inv(5) betekent f (x) = 5, dus x2 + 4x = 5 x2 + 4x í 5 = 0 (x + 5)(x í 1) = 0 x = í5  x = 1 vold. niet vold. Dus f inv(5) = 1. c Voor elke f en f inv geldt vanwege symmetrie dat de gemeenschappelijke punten op de lijn y = x liggen. f (x) = x2 + 4x f x2 + 4x = x y=x x2 + 3x = 0 x(x + 3) = 0 x = 0  x = í3 vold. vold. niet Voor f geldt dat alleen het punt (0, 0) op de lijn y = x ligt. Er is dus precies één gemeenschappelijk punt. d Op D f = 8 k , a 4 bestaat f inv alleen voor a ” í2. Het gemeenschappelijke punt op D f = 8 k , í2 4 is (í3, í3). Dus í3 ” a ” í2.

© Noordhoff Uitgevers bv

Functies en grafieken

23

49

a

1

x

0

8

f(x)

2

0

x

2

0

f inv(x)

0

8

y

9

8

7

ƒinv

6

y

=

x

5

4

3

2

1 ƒ −1

O

1

2

3

4

5

6

7

8

9

x

−1

Stel f inv(x) = ax + b. Door (0, 8), dus b = 8. f inv(x) = ax + 8 f aÂ2 + 8 = 0 Door (2, 0) 2a = í8 a = í4 Dus f inv(x) = í4x + 8 Alternatieve oplossing Elk punt P(xP, yP) van de grafiek van f is na spiegeling in de lijn y = x het punt (yP, xP) op de grafiek van f inv. f wordt gegeven door de formule y = í 14 x + 2. f inv wordt dus gegeven door de formule x = í 14 y + 2. Vrijmaken van y geeft x = í 14 y + 2 1 4y

Dus f inv(x) = í4x + 8. b

24

x

1

3

g(x)

2

0

x

2

0

ginv(x)

1

3

Hoofdstuk 1

= íx + 2 y = í4x + 8

© Noordhoff Uitgevers bv

y

y

1

3

3

ginv

= y

y

=

x

4

x

4

2

2

1

1 g

O

1

2

3

4

x

O

1

2

3

4

x

Dus ginv(x) = 3 í x met D ginv = 8 k , 2 4 . Alternatieve oplossing g(1) = 3 í 1 = 2, dus B g = 8 k , 2 4 . Vanwege symmetrie geldt dat Bginv = Dg en Dginv = Bg, dus Dginv = 8 k , 2 4 . g wordt gegeven door de formule y = 3 í x. g inv wordt dus gegeven door de formule x = 3 í y. Vrijmaken van y geeft x = 3 í y y=3íx inv Dus g (x) = 3 í x met Dginv = 8 k , 2 4 .

1.4 Tweedegraadsfuncties met een parameter Bladzijde 32 50

a p = 1 geeft f (x) = íx2 + 6x + 1 b 6 xtop = í = í = 3 en ytop = í32 + 6 Â 3 + 1 = 10 2a 2 Â í1 De top is dus (3, 10). b De top ligt op de x-as als de vergelijking íx2 + 6x + p = 0 precies één oplossing heeft. D = 62 í 4 Â í1 Â p = 36 + 4p f 36 + 4p = 0 één oplossing als D = 0 4p = í36 p = í9 De top ligt op de x-as voor p = í9. Bladzijde 33

51

a

b ƒp

ƒp

x x

D = (í6)2 í 4 Â 2 Â p = 36 í 8p f 36 í 8p = 0 Er moet gelden D = 0 í8p = í36 p = 412

© Noordhoff Uitgevers bv

D = 36 í 8p f 36 í 8p > 0 Er moet gelden D > 0 í8p > í36 p < 412

Functies en grafieken

25

52

De gra¿ek van y = ax2 + bx + c met a < 0

1

twee snijpunten met de x-as D>0

53

a

x

x

x

één snijpunt (raakpunt) met de x-as D=0

geen snijpunt met de x-as D 0 Er moet gelden D > 0 p2 > 36 p < í6  p > 6

p b =í = í 18 p 2a 2Â4 2 1 2 1 2 p í 18 p2 + 5 = í 16 p +5 ytop = fp (í 18 p ) = 4 ( í 18 p ) + p  í 18 p + 5 = 16

xtop = í

Bladzijde 34 56

p = 1p 2  í2 4 ytop = f ( 14 p) = í2  ( 14 p ) 2 + p  14 p + 1 = í 18 p2 + 14 p2 + 1 = 18 p2 + 1

xtop = í

Het maximum is 9 geeft 18 p2 + 1 = 9 1 2 8p p2

=8 = 64 p = 8  p = í8

26

Hoofdstuk 1

© Noordhoff Uitgevers bv

57

xtop = í

p = í 12 p 2Â1

ytop = f (í 12 p) = (í 12 p) + p  (í 12 p) + 3 = 14 p2 í 12 p2 + 3 = í 14 p2 + 3 2

1

De top (í 12 p, í 14 p2 + 3) invullen in y = x + 1 geeft í 14 p2 + 3 = í 12 p + 1 í 14 p2 + 12 p + 2 = 0 p2 í 2p í 8 = 0 (p í 4)(p + 2) = 0 p = 4  p = í2 58

6

a xtop = í

=í

3 p

2Âp 3 3 ytop = f í p = p  í p

( ) ( )

3 9 9 18 + 6Âíp + 1 = p í p + 1 = íp + 1 9 De extreme waarde is í2 geeft í p + 1 = í2 9 í p = í3 í3p = í9 p=3 2 b p = 3 geeft f (x) = 3x + 6x + 1 De grafiek is een dalparabool, dus de extreme waarde is een minimum. 59

a xtop = í

2

p

= íp 2  12 ytop = f(íp) = 12  (íp)2 + p  íp + q = 12 p2 í p2 + q = í 12 p2 + q

y = x2 + x + 1 1 f í 2 p2 + q = (íp)2 + íp + 1 top (íp, í 12 p2 + q) í 12 p2 + q = p2 í p + 1 q = 112 p2 í p + 1 b xtop = íp = 2, dus p = í2 en q = 112 Â (í2)2 í í2 + 1 = 9. Bladzijde 35 60

a xtop = í

b xtop = í

c xtop = í

p

2Â1 p 2Â2

= í 12 p, dus xtop = í 12 p í 12 p = xtop p = í2xtop = í 14 p, dus xtop = í 14 p í 14 p = xtop p = í4xtop

4p = 2 p, dus xtop = 23 p 2 Â í3 3 2 3 p = xtop

d xtop = í

6 2Âp

3 3 = í p , dus xtop = í p

pxtop = í3 í3 p=x top 4p 2 2 e xtop = í = í p , dus xtop = í p 2 Â p2 pxtop = í2 í2 p=x top

p = 112 xtop Bladzijde 36 61

a xtop = í

p = 4p, dus p = 14 xtop. 2 Â í 18

ytop = í 18 xtop2 + pxtop í 6 p = 14 xtop

¶ ytop = í 18 xtop2 + 14 xtop  xtop í 6

ytop = í 18 xtop2 + 14 xtop2 í 6 ytop = 18 xtop2 í 6

Dus de formule van de kromme is y = 18 x2 í 6.

© Noordhoff Uitgevers bv

Functies en grafieken

27

b xtop = í

6 3 3 = í , dus p = í x . 2Âp p top

ytop = pxtop2 + 6xtop + p 3 3 ¶ ytop = í x  xtop2 + 6xtop + í x 3 p = íx top top top 3 ytop = í3xtop + 6xtop í x

1

top

3 ytop = 3xtop í x top 3 Dus de formule van de kromme is y = 3x í x voor p  0. í2p 1 1 c xtop = í = , dus p = x . 2 2Âp top p ytop = p2xtop2 í 2pxtop + 3 1 2 1 ¶ ytop = x 1  xtop2 í 2  x  xtop + 3 p=x top top top ytop = 1 í 2 + 3 ytop = 2 Dus de formule van de kromme is y = 2 voor p  0. íp 1 d xtop = í = 2Âp 2 ytop = pxtop2 í pxtop + 1 2 ¶ ytop = p  (12 ) í p  12 + 1 xtop = 12 ytop = 14 p í 12 p + 1

( )

ytop = í 14 p + 1 Dus de coördinaten van de toppen van fq zijn ( 12, í 41 p + 1) . De formule van de kromme is x = 12 voor p  0. 62

a xtop = í

p = 1 p, dus p = 2xtop. 2 Â í1 2

ytop = íxtop2 + pxtop + 2p f ytop = íxtop2 + 2xtop  xtop + 2  2xtop p = 2xtop ytop = íxtop2 + 2xtop2 + 4xtop ytop = xtop2 + 4xtop De formule van de kromme is y = x2 + 4x.

1.5 Grafisch-numeriek oplossen Bladzijde 38 63

a x = í1, x = 1, x = 2 en x = 3 b x = í1 geeft (í1)4 í 5 Â (í1)3 + 5 Â (í1)2 + 5 Â í1 í 6 = 0 klopt x = 1 geeft 14 í 5 Â 13 + 5 Â 12 + 5 Â 1 í 6 = 0 klopt x = 2 geeft 24 í 5 Â 23 + 5 Â 22 + 5 Â 2 í 6 = 0 klopt x = 3 geeft 34 í 5 Â 33 + 5 Â 32 + 5 Â 3 í 6 = 0 klopt

64

a x = í2, x = 2 en x = 4 b x = í2  x = 2  x = 4 Bladzijde 40

65

28

a Voer in y1 = í 13 x3 í 12 x2 + 5x + 4. De optie zero (TI) of ROOT (Casio) geeft x § í4,33  x § í0,77  x § 3,60. b De optie minimum geeft x § í2,79 en y § í6,60. De optie maximum geeft x § 1,79 en y § 9,44. Dus min. is f (í2,79) § í6,60 en max. is f (1,79) § 9,44. c Voer in y2 = íx + 3. Intersect geeft x § í4,99  x § í0,16  x § 3,65.

Hoofdstuk 1

© Noordhoff Uitgevers bv

66

a Voer in y1 = x3 í 4x2 + 3. De optie zero (TI) of ROOT (Casio) geeft x § í0,79  x = 1  x § 3,79. b Voer in y1 = x4 í 4x3 + 2x2 + x í 1. De optie zero (TI) of ROOT (Casio) geeft x § í0,58  x § 3,34. c Voer in y1 = 0,4x3 + 2x2 + x í 2 en y2 = x + 2. Intersect geeft x § í4,51  x § í1,76  x § 1,26. d Voer in y1 = 0,2x5 í x4 + 4x2 en y2 = 0,2x + 3. Intersect geeft x § í1,45  x = í1  x = 1  x = 3  x § 3,45.

67

a Voer in y1 = 0,2x4 í x3 í x2 + 8x + 2. De optie optie minimum geeft x § í1,62 en y § í7,96 en geeft x § 3,69 en y § 4,74. De optie optie maximum geeft x § 1,67 en y § 9,47. Dus min. is f (í1,62) § í7,96, max. is f (1,67) § 9,47 en min. is f (3,69) § 4,74. b Voer in y1 = í113 x3 + 3x2 + 40x í 28. De optie minimum geeft x § í2,50 en y § í88,42. De optie maximum geeft x § 4,00 en y § 94,67. Dus min. is f (í2,50) § í88,42 en max. is f (4,00) § 94,67.

68

a Voer in y1 = 0 x3 í 9x 0 en y2 = 5. Intersect geeft x § í3,25  x § í2,67  x § í0,58  x § 0,58  x § 2,67  x § 3,25. b Voer in y3 = x + 5. Intersect met y1 geeft x § í3,10  x § í2,87  x § í0,51  x § 0,66  x § 2,44  x § 3,39.

69

a Voer in y1 = 0,5x3 í 5x2 + 20. De optie zero (TI) of ROOT (Casio) geeft x § í1,84  x § 2,28  x § 9,56. b Voer in y1 = 0,1x4 + 0,1x3 í 12x2 + 50 en y2 = 25x. Intersect geeft x = í10  x § í3,53  x § 1,26  x § 11,27. c Voer in y1 = 0 x4 í x3 + x í 5 0 en y2 = x + 3. Intersect geeft x § í1,48  x § í1,26  x = 1  x = 2. d Voer in y1 = 0 x3 í 5x2 í 2x + 24 0 en y2 = 20. Intersect geeft x § í2,55  x = í1  x § 0,76  x § 5,24.

70

xtop = í

1

p2 = í 14 p2 en 2Â2 ytop = fp ( í 14 p2 ) = 2 ( í 14 p2 )2 + p2 Â í 14 p2 + p = 18 p4 í 14 p4 + p = í 18 p4 + p

( í 14 p2, í 18 p4 + p ) y = 8x + 4

¶ 8 Â í 1 p2 + 4 = í 1 p4 + p 4 8

í2p2 + 4 = í 18 p4 + p 1 4 8p

í 2p2 í p + 4 = 0

Voer in y1 = 18 x4 í 2x2 í x + 4. Optie zero (TI) of ROOT (Casio) geeft x § í3,236  x = í2  x § 1,236  x = 4, dus p § í3,236  p = í2  p § 1,236  p = 4. Voor p = í3,236 is xtop = í 14  (í3,236)2 § í2,618 en ytop = í 18  (í3,236)4 + í3,236 § í16,943, a > 0, dus fp is een dalparabool, dus min. is fí3,236(í2,618) = í16,943. Voor p = í2 is xtop = í 14  (í2)2 = í1 en ytop = í 18  (í2)4 + í2 = í4, dus min. is fí2(í1) = í4. Voor p = 1,236 is xtop = í 14  1,2362 = í0,382 en ytop = í 18  1,2364 + 1,236 = 0,944, dus min. is f1,236(í0,382) = 0,944. Voor p = 4 is xtop = í 14  42 = í4 en ytop = í 18  44 + 4 = í28, dus min. is f4(í4) = í28. 71

a Voer in y1 = íx2 + 6x en y2 = x + 4. Intersect geeft x = 1  x = 4. b Voor 1 < x < 4.

© Noordhoff Uitgevers bv

Functies en grafieken

29

Bladzijde 42 72 1

a Voer in y1 = x2 í 3x en y2 = 14. Intersect geeft x § í2,531 en x § 5,531. y

y = x 2 – 3x y = 14

– 2,531

O

x

5,531

x2 í 3x ” 14 geeft í2,531 ” x ” 5,531 b Voer in y1 = x2 + 2x en y2 = 11. Intersect geeft x § í4,464 en x § 2,464. y y = x 2 + 2x y = 11

O

– 4,464

2,464

x

x2 + 2x > 11 geeft x < í4,464  x > 2,464 c Voer in y1 = 8x2 + 6x í 35. De optie zero (TI) of ROOT (Casio) geeft x = í2,5 en x = 1,75. y y = 8x 2 + 6x – 35

O

– 2,5

x

1,75

8x2 + 6x í 35 • 0 geeft x ” í2,5  x • 1,75 d Voer in y1 = x3 + 4,5x2 en y2 = 19x + 60. Intersect geeft x = í6, x = í2,5 en x = 4. y

y = x 3 + 4,5x 2

y = 19x + 60

–6 –2,5

O

4

x

x3 + 4,5x2 < 19x + 60 geeft x < í6  í2,5 < x < 4

30

Hoofdstuk 1

© Noordhoff Uitgevers bv

a x4 > 81

1 4 2x + 1 f (x) 1 4 x +1= 2

< 9

¶

}

} f(x) x4

c

{

73

g(x)

= 81 4 4 x =冑 81 = 3  x = í冑 81 = í3

g(x)

9

1

1 4 2x x4

=8 = 16 4 4 x =冑 16 = 2  x = í冑 16 = í2

y ƒ

y

g

ƒ

g

O

–3

x

3

x

2

g(x)

x3 = í8 3 x =冑 í8 = í2

1 4 2x

d

¶

y ƒ

–2

+ 1 < 9 geeft í2 < x < 2

1 3 3 (x í 1) > 9 f(x) g(x) 1 3 (x í 1) 9 = 3 (x í 1)3 27

x

O

{

f(x)

–2 O

}

}

x4 > 81 geeft x < í3  x > 3 b x3 < í8

= 3 x í 1 =冑 27 xí1=3 x=4

y ƒ

g

g

x3 < í8 geeft x < í2 O

1 3 (x

© Noordhoff Uitgevers bv

4

x

í 1)3 > 9 geeft x > 4

Functies en grafieken

31

74

a Voer in y1 = 0,1x3 í 2x2 + 8x + 10 en y2 = íx + 15. Intersect geeft x § 0,65, x § 5,66 en x § 13,69. y

1

y = 0,1x3 – 2x2 + 8x + 10

O 0,65

5,66

x

13,69

y = –x + 15

0,1x3 í 2x2 + 8x + 10 • íx + 15 geeft 0,65 ” x ” 5,66  x • 13,69 b Voer in y1 = í0,5x4 + 3x3 í 4x2 + 8 en y2 = x + 7. Intersect geeft x § í0,52, x § 0,45, x § 2,29 en x § 3,78. y

y=x+7

y = –0,5x 4 + 3x 3 – 4x 2 + 8

– 0,52 O

0,45

2,29

x

3,78

í0,5x4 + 3x3 í 4x2 + 8 • x + 7 geeft í0,52 ” x ” 0,45  2,29 ” x ” 3,78 c Voer in y1 = 0 x3 í 10x 0 en y2 = 2x + 8. Intersect geeft x § í3,24, x § í3,06, x § í0,69, x § 1,24, x = 2 en x § 3,76. y y = 2x + 8

y = |x3 − 10x|

–3,24 –3,06

–0,69 O

1,24 2

3,76

x

0 x3 í 10x 0 ” 2x + 8 geeft í3,24 ” x ” í3,06  í0,69 ” x ” 1,24  2 ” x ” 3,76

32

Hoofdstuk 1

© Noordhoff Uitgevers bv

d Voer in y1 = 0 x4 + x2 í 5x í 10 0 en y2 = 8 í 0 2x í 4 0 . Intersect geeft x § í1,32, x § í1,10, x § 1,69 en x § 2,21. 1

y

y = | x4 + x2 − 5x − 10|

y = 8 − |2x − 4|

–1,32 –1,10 O

1,69 2,21

x

0 x4 + x2 í 5x í 10 0 ” 8 í 0 2x í 4 0 geeft í1,32 ” x ” í1,10  1,69 ” x ” 2,21

Diagnostische toets Bladzijde 44 1

a Stel k: y = ax + b met a = rck = 2. y = 2x + b f 2  í1 + b = 6 door A(í1, 6) í2 + b = 6 b=8 Dus k: y = 2x + 8. b Stel l: y = ax + b. l // m, dus a = rc m = í 12 . y = í 12 x + b 1 f í2  9 + b = 3 door B(9, 3) í412 + b = 3 b = 712 Dus l: y = í 12 x + 712 . c n: y = ax + 5 f a  í10 + 5 = 0 door (í10, 0) í10a + 5 = 0 í10a = í5 a = 12

2

a Stel k: y = ax + b met a =

¨y í2 í 2 í4 = = = í 12 . ¨x 3 í í5 8

y = í 12 x + b 1 f í 2 Â í5 + b = 2 door A(í5, 2) 212 + b = 2 Dus k: y =

í 12 x

b = í 12 í 12 .

b Stel l: y = ax + b met a =

¨y 135 í 60 75 = = = 3. ¨x 65 í 40 25

y = 3x + b f 3 Â 40 + b = 60 door P(40, 60) 120 + b = 60 b = í60 Dus l: y = 3x í 60. © Noordhoff Uitgevers bv

Functies en grafieken

33

3

a Stel W = at + b. t = 4 en W = 500 ¨W 2900 í 500 = = 300 fa= t = 12 en W = 2900 ¨t 12 í 4 W = 300t + b f 300 Â 4 + b = 500 t = 4 en W = 500 1200 + b = 500 b = í700 Dus W = 300t í 700. b t = 5,2 geeft W = 300 Â 5,2 í 700 = 860.

4

a f (x) = 2x + 1 í 0 3x í 6 0 = íx + 7 als 3x í 6 • 0, dus als x • 2.

1

f (x) = 2x + 1 í 0 3x í 6 0 = 5x í 5 als 3x í 6 < 0, dus als x < 2. y

5

4 ƒ 3

2

1

−1

O

1

2

3

4

5

6

7

8

9

x

−1

−2

−3

−4

−5

b oppervlakte vlakdeel is 12 Â (7 í 1) Â 5 = 15

34

Hoofdstuk 1

© Noordhoff Uitgevers bv

5

a 3x2 í x = 0 x(3x í 1) = 0 x = 0  3x = 1 x = 0  x = 13 b 3x2 í 9x = 12 3x2 í 9x í 12 = 0 x2 í 3x í 4 = 0 (x + 1)(x í 4) = 0 x = í1  x = 4 c 3x2 í x = 2 3x2 í x í 2 = 0 D = (í1)2 í 4  3  í2 = 25 1í5 1+5 x= = í 23  x = =1 6 6 2 d x + 14 = 16 x2 = 2 x = 冑2  x = í冑2 e (2x í 3)2 = 81 2x í 3 = 9  2x í 3 = í9 2x = 12  2x = í6 x = 6  x = í3 f (3x + 2)(x í 1) = 0 3x + 2 = 0  x í 1 = 0 3x = í2  x = 1 x = í 23  x = 1 g x2 = 7x + 13 x2 í 7x í 13 = 0 D = (í7)2 í 4  1  í13 = 101 7 í 冑101 7 + 冑101 x=  x= 2 2 x = 312 í 12 冑101  x = 312 + 12 冑101

h (3x + 2)(x í 1) = x(x + 5) 3x2 í 3x + 2x í 2 = x2 + 5x 2x2 í 6x í 2 = 0 x2 í 3x í 1 = 0 D = (í3)2 í 4 Â 1 Â í1 = 13 x=

3 í 冑13 3 + 冑13  x= 2 2

x = 112 í 12 冑13  x = 112 + 12 冑13 i (x + 2)2 = 3x + 7 x2 + 4x + 4 = 3x + 7 x2 + x í 3 = 0 ( x + 12 ) 2 í 14 í 3 = 0

(x + 12 ) 2 = 314

冑 +冑

冑 í冑

x + 12 = 314  x + 12 = í 314 x=

í 12

13 4

x=

í 12

13 4

x = í 12 + 12 冑13  x = í 12 í 12 冑13 j (x í 3)2 í (x + 1)2 = (x í 4)2 x2 í 6x + 9 í x2 í 2x í 1 = x2 í 8x + 16 íx2 = 8 x2 = í8 geen oplossingen

6

a D = 42 í 4  2  p = 16 í 8p f 16 í 8p < 0 geen oplossingen als D < 0 í8p < í16 p>2 b D = p2 í 4  3  17 = p2 í 204 f p2 í 204 > 0 twee oplossingen als D > 0 p2 > 204 p < í冑204  p > 冑204 p < í2冑51  p > 2冑51 c p = 0 geeft 2x + 5 = 0, dus één oplossing. Voor p  0 is D = 22 í 4  p  5 = 4 í 20p f 4 í 20p > 0 twee oplossingen als D > 0 í20p > í4 p < 15 De vergelijking heeft twee oplossingen voor p < 0  0 < p < 15 .

7

p = 0 geeft í6x + 12 = 0 í6x = í12 x=2 Voor p  0 is D = (í6)2 í 4  p  12 = 36 í 48p f 36 í 48p = 0 één oplossing als D = 0 í48p = í36 p = 34 3 3 2 p = 4 geeft 4 x í 6x + 12 = 0 x2 í 8x + 16 = 0 (x í 4)2 = 0 x=4 Voor p = 0 is de oplossing x = 2 en voor p = 34 is de oplossing x = 4.

© Noordhoff Uitgevers bv

1

Functies en grafieken

35

Bladzijde 45 1

b 8 =í = í2 geeft ytop = f (í2) = 2 Â (í2)2 + 8 Â í2 + 5 = í3 2a 2Â2 2 > 0, dus dalparabool en min. is f (í2) = í3. b 4 b xtop = í = í = 5 geeft ytop = g(5) = í0,4 Â 52 + 4 Â 5 í 3 = 7 2a 2 Â í0,4 í0,4 < 0, dus bergparabool en max. is f (5) = 7.

8

a xtop = í

9

a f (0) = 3 en f (5) = 0,6 Â 52 í 4,8 Â 5 + 3 = í6 í4,8 b xtop = í = í = 4 en ytop = f (4) = 0,6 Â 42 í 4,8 Â 4 + 3 = í6,6 2a 2 Â 0,6 y

3 (0, 3) O

x

5

ƒ (5, –6)

–6,6

(4; –6,6)

Dus B f = 3 í6,6; 3 4 . b f (2) = 0,6 Â 22 í 4,8 Â 2 + 3 = í4,2 en f (10) = 0,6 Â 102 í 4,8 Â 10 + 3 = 15 xtop = 4 en ytop = í6,6 (zie vraag a) y

(10, 15)

15

ƒ

O

2

10

x

(2; –4,2) –6,6

(4; –6,6)

Dus B f = 3 í6,6; 15 4 . 10

xtop = í

f

36

b í1 = í 1 = 1, dus a = 1. 2a 2Â2

x

í3

í2

í1

0

1

f(x)

3

í0,5

í3

í4,5

í5

x

3

í0,5

í3

í4,5

í5

í3

í2

í1

0

1

inv(x)

Hoofdstuk 1

© Noordhoff Uitgevers bv

y

1 5

4 ƒ 3

y

=

x

2

1

−5

−4

−3

−2

−1

O

1

2

3

4

5

x

−1

−2

−3

ƒinv

−4

−5

11

a D = p2 í 4  í1  í3 = p2 í 12 f p2 í 12 = 0 x-as raken, dus D = 0 p2 = 12 p = 冑12  p = í冑12 p = 2冑3  p = í2冑3 b positief maximum betekent twee snijpunten met de x-as, dus D > 0 f p2 í 12 > 0 D = p2 í 12 p2 > 12 p < í2冑3  p > 2冑3

12

a xtop = í

p b =í = í 12 p 2a 2Â1 ytop = fp (í 12 p) = (í 12 p)2 + p  í 12 p + 6p = 14 p2 í 12 p2 + 6p = í 14 p2 + 6p

Extreme waarde is í13 geeft í 14 p2 + 6p = í13 í 14 p2 + 6p + 13 = 0 p 2 í 24p í 52 = 0 (p + 2)(p í 26) = 0 p = í2  p = 26 b (í 12 p, í 14 p2 + 6p) f í5  í 12 p + 10 = í 14 p2 + 6p y = í5x + 10 212 p + 10 = í 14 p2 + 6p 1 2 1 4 p í 32 p + 10 2 p í 14p + 40

=0 =0 (p í 10)(p í 4) = 0 p = 10  p = 4

© Noordhoff Uitgevers bv

Functies en grafieken

37

13 1

14

2p = íp, dus p = íxtop. 2Â1 ytop = xtop2 + 2pxtop + p f ytop = xtop2 + 2  íxtop  xtop + íxtop p = íxtop ytop = xtop 2 í 2xtop 2 í xtop ytop = íxtop2 í xtop Dus de formule van de kromme is y = íx2 í x. xtop = í

Voer in y1 = í 15 x3 + x2 + 2x í 5. De optie zero (TI) of ROOT (Casio) geeft x § í2,59  x § 1,62  x § 5,97. De optie minimum geeft x § í0,81 en y § í5,86. De optie maximum geeft x § 4,14 en y § 6,23. Dus min. is f (í0,81) § í5,86 en max. is f (4,14) § 6,23.

15

a Voer in y1 = x4 í 4x2 en y2 = 0,5x í 2. Intersect geeft x § í1,75  x § í0,86  x § 0,69  x § 1,93. b Voer in y1 = 0 x3 í 3x 0 en y2 = í 12 x + 2. Intersect geeft x § í2,11  x § 0,65  x § 1,46  x § 1,89.

16

a Voer in y1 = x2 + 5x en y2 = x3 + 2x2 í 6x + 1. Intersect geeft x § í3,89, x § 0,09 en x § 2,80. y y = x 2 + 5x

y = x 3 + 2x 2 – 6x + 1 – 3,89 O

x

2,80 0,09

x2

x3

2x2

+ 5x ” + í 6x + 1 geeft í3,89 ” x ” 0,09  x • 2,80 b Voer in y1 = 10 í 0 4 í 3x 0 en y2 = 0 x3 í 4x2 + x 0 . Intersect geeft x § í0,78, x § 2,40, x § 3,35 en x § 3,90. y y = | x3 − 4x2 + x| y = 10 − |4 − 3x|

– 0,78

O

2,40

3,35 3,90

x

10 í 0 4 í 3x 0 < 0 x3 í 4x2 + x 0 geeft x < í0,78  2,40 < x < 3,35  x > 3,90

38

Hoofdstuk 1

© Noordhoff Uitgevers bv

2 De afgeleide functie Voorkennis Herleiden 2

Bladzijde 48

a1 a (2x í 5)2 = 4x2 í 20x + 25 b (3 + h)2 = h2 + 6h + 9 c ( 12 x í 1) ( 12 x + 1) = 14 x2 í 1 d 2(3x í 2)2 + 3(2x í 1)2 = 2(9x2 í 12x + 4) + 3(4x2 í 4x + 1) = 18x2 í 24x + 8 + 12x2 í 12x + 3 = 30x2 í 36x + 11 e (3x + 2)2 í (2x í 6)2 = 9x2 + 12x + 4 í (4x2 í 24x + 36) = 9x2 + 12x + 4 í 4x2 + 24x í 36 = 5x2 + 36x í 32 1 2 2 f (6x í 5) í 2 (2x í 4) = (36x2 í 60x + 25) í 12 (4x2 í 16x + 16) = 36x2 í 60x + 25 í 2x2 + 8x í 8 = 34x2 í 52x + 17 Bladzijde 49

a2 a f (2x) = 14 Â (2x)2 í 2x + 2 = 14 Â 4x2 í 2x + 2 = x2 í 2x + 2 b f (x + 5) = 14 (x + 5)2 í (x + 5) + 2 = 14 (x2 + 10x + 25) í x í 5 + 2 = 14 x2 + 212 x + 614 í x í 3 = 14 x2 + 112 x + 314 c f (3x + 1) = 14 (3x + 1)2 í (3x + 1) + 2 = 14 (9x2 + 6x + 1) í 3x í 1 + 2 = 214 x2 + 112 x + 14 í 3x + 1 = 214 x2 í 112 x + 114 d f (x + h) = 14 (x + h)2 í (x + h) + 2 = 14 (x2 + 2xh + h2 ) í x í h + 2 = 14 x2 + 12 xh + 14 h2 í x í h + 2 a3 a y =

3xh + 2h2 h(3x + 2h) = = 3x + 2h, mits h  0 h h

b y=

axh + bh2 h(ax + bh) = = ax + bh, mits h  0 h h

c y=

a(x + h)2 í ax2 a(x2 + 2xh + h2 ) í ax2 ax2 + 2axh + ah2 í ax2 2axh + ah2 = = = h h h h

=

h(2ax + ah) = 2ax + ah, mits h  0 h

© Noordhoff Uitgevers bv

De afgeleide functie

39

2.1 Snelheden Bladzijde 50

a1 Tussen A en C neemt de daling steeds meer toe, tussen C en B neemt de daling steeds meer af.

2

a2 Afnemend dalend op 8 k , 19. Toenemend stijgend op 81, 29. Afnemend stijgend op 82, 39. Toenemend dalend op 83,m 9. Bladzijde 51

a3 Afnemend dalend op 8 k , í49 en op 83, 59. Toenemend stijgend op 8í4, í29 en op 85,m 9. Afnemend stijgend op 8í2, 19. Toenemend dalend op 81, 39. a4 a Constant stijgend op 80, 39. Afnemend stijgend op 83, 49. Toenemend dalend op 84, 59. Afnemend dalend op 85, 79. Toenemend stijgend op 87, 109. b Bij de steilste klim is de snelheid het laagst. Dus na 7 minuten. a5

%s 90 í 0 = = 18 %t 5 í 0 Dus de gemiddelde snelheid gedurende de eerste vijf seconden is 18 m/s. Bladzijde 52

%s 12,5 í 5 7,5 = = = 0,375 km/minuut = 22,5 km/uur %t 40 í 20 20 3 30, 60 4 : %s = 15 í 10 = 5 = 16 km/minuut = 10 km/uur %t 60 í 30 30 b De grafiek is niet overal even steil. c Trek de lijn door de punten (0, 0) en (20, 5) door totdat hij de grafiek weer snijdt. Dat is in het punt (60, 15). Dus voor p = 60.

a6 a 3 20, 40 4 :

a7 De gemiddelde snelheid op 3 0, t 4 wordt steeds kleiner als je t steeds groter neemt. Bladzijde 53

%y 5 í 1 4 = = = 1. %x 4 í 0 4 %y 0 í 5 í5 b Op 3 2, 6 4 is = = = í114 . %x 6 í 2 4 c Trek de lijn k met rc = 35 door het punt (0, 1). k snijdt de grafiek in (5, 3), dus p = 5. d Trek de halve lijn met beginpunt (1, 4) en rc = 14 . De lijn snijdt de grafiek voor x > 1 in drie punten. Er zijn dus drie waarden voor q.

a8 a Op 3 0, 4 4 is

a9 f (0) = í3 %y = 4, dus f (1) = í3 + 4 = 1. %x %y Op 3 1, 3 4 is = 2, dus f (3) = 1 + 2 Â 2 = 5. %x %y Op 3 3, 6 4 is = í2, dus f (6) = 5 + 3 Â í2 = í1. %x %y Op 3 6, 10 4 is = í1, dus f (10) = í1 + 4 Â í1 = í5. %x Op 3 0, 1 4 is

40

Hoofdstuk 2

© Noordhoff Uitgevers bv

y 5 4 ƒ 3 2

2

1 O –1

1

2

3

4

5

6

7

8

x

9

–2 –3 –4 –5

Er zijn meerdere mogelijkheden, omdat alleen de punten (0, í3), (1, 1), (3, 5), (6, í1) en (10, í5) vastliggen. Bladzijde 54 10

a f (0) = í4 %y 1 2 = Â 1 í 2 Â 1 = í112 , dus f (1) = í4 í 112 = í512 %x 2 %y 1 2 Op 3 0, 2 4 is = Â 2 í 2 Â 2 = í2, dus f (2) = í4 + 2 Â í2 = í8 %x 2 %y 1 2 Op 3 0, 3 4 is = Â 3 í 2 Â 3 = í112 , dus f (3) = í4 + 3 Â í112 = í812 %x 2 %y 1 2 Op 3 0, 4 4 is = Â 4 í 2 Â 4 = 0, dus f (4) = í4 + 4 Â 0 = í4 %x 2 %y 1 2 Op 3 0, 5 4 is = Â 5 í 2 Â 5 = 212 , dus f (5) = í4 + 5 Â 212 = 812 %x 2 Een mogelijke grafiek is Op 3 0, 1 4 is

y

8

6

4

2

O

1

2

3

4

x

−2

−4

−6

−8

© Noordhoff Uitgevers bv

De afgeleide functie

41

%y = í 12 Â 02 + 0 + 1 = 1 en de grafiek gaat door (0, 1), dus f (í3) = 1 í 3 Â 1 = í2. %x %y Op 3 í3, í2 4 is = í 12 Â (í2)2 + í2 + 1 = í3, dus f (í2) = í2 + í3 = í5 %x %y Op 3 í3, í1 4 is = í 12 Â (í1)2 + í1 + 1 = í 12 , dus f (í1) = í2 + 2 Â í 12 = í3 %x %y Op 3 í3, 1 4 is = í 12 Â 12 + 1 + 1 = 112 , dus f (1) = í2 + 4 Â 112 = 4 %x %y Op 3 í3, 2 4 is = í 12 Â 22 + 2 + 1 = 1, dus f (2) = í2 + 5 Â 1 = 3 %x %y Op 3 í3, 3 4 is = í 12 Â 32 + 3 + 1 = í 12 , dus f (3) = í2 + 6 Â í 12 = í5 %x Een mogelijke grafiek is

b Op 3 í3, 0 4 is

2

y

5

4

3

2

1

−2

−1

O

1

2

x

−1

−2

−3

−4

−5

11 a

a yA = f (1) = 12 í 4 Â 1 + 1 = í2 yB = f (5) = 52 í 4 Â 5 + 1 = 6 b

%y 6 í í2 8 = = =2 %x 5 í 1 4

Bladzijde 55 12 a

42

a

%y f (3) í f (í1) í6 í 6 = = = í3 3 í í1 4 %x

b

%y f (4) í f (1) í4 í í4 = = =0 4í1 3 %x

c

%y f (1) í f (í5) í4 í 50 = = = í9 1 í í5 6 %x

d

%y f (4) í f (í5) í4 í 50 = = = í6 4 í í5 9 %x

Hoofdstuk 2

© Noordhoff Uitgevers bv

13 a

Bij een lineaire functie is ieder differentiequotiënt gelijk aan de richtingscoëf¿ciënt.

14 a

a

y

ƒ

2

O

x

%y f (3) í f (1) 23 í 3 = = = 10 3í1 2 %x %y f (4) í f (í2) 57 í 3 c = = =9 4 í í2 6 %x d Stel l: y = ax + b. %y 3 í í13 f (í3) = í13, dus A(í3, í13) en f (1) = 3, dus B(1, 3), dus a = = = 4. %x 1 í í3 y = 4x + b 4Â1 + b = 3 door B(1, 3) f b = í1 Dus l: y = 4x í 1. b

15 a

a Voer in y1 = í4,9x2 + 44,1x. h in meter

O

tijd in seconden

t

b De optie maximum geeft x § 4,5 en y = 99,225. Dus na 4,5 seconden. c De derde seconde is van t = 2 tot t = 3. Op t = 2 is h = 68,6 De steen stijgt 88,2 í 68,6 = 19,6 m. Op t = 3 is h = 88,2 f %h 68,6 í 0 d Op t = 2 is h = 68,6, dus gemiddelde snelheid is = = 34,3 m/s. 2í0 %t e De optie zero (TI) of ROOT (Casio) geeft x = 0 en x = 9. Op t = 8,5 is h = 20,825. %h 0 í 20,825 Op 3 8,5; 9 4 is = = í41,65 m/s. 9 í 8,5 %t Dus de gemiddelde snelheid gedurende de laatste seconde is 41,65 m/s. 16 a

a De gemiddelde snelheid op het interval 3 2, 3 4 is b Op 3 2; 2,14 is de gemiddelde snelheid

%s (2,13 + 2,1) í (23 + 2) = = 13,61 m/s. 2,1 í 2 %t

Op 3 2; 2,014 is de gemiddelde snelheid

%s (2,013 + 2,01) í (23 + 2) = = 13,0601 m/s. 2,01 í 2 %t

Op 3 2; 2,0014 is de gemiddelde snelheid

© Noordhoff Uitgevers bv

%s (33 + 3) í (23 + 2) 30 í 10 = = = 20 m/s. 3í2 1 %t

%s (2,0013 + 2,001) í (23 + 2) = = 13,006001 m/s. 2,001 í 2 %t

De afgeleide functie

43

c Op een kleiner wordend interval komt de gemiddelde snelheid steeds dichter bij 13 m/s. De gemiddelde snelheid op het interval 3 2; 200014 is %s (2,00013 + 2,0001) í (23 + 2) 10,00130006 í 10 § 13,0006 m/s. = = 2,0001 í 2 0,0001 %t d Voor %t = 0 is 2

%s 0 = . Dit is niet gedefinieerd. %t 0

Bladzijde 57 17 a

%s (10冑4 Â 2,01 + 1 í 10) í (10冑4 Â 2 + 1 í 10) § 6,66. = 0,01 %t De snelheid op t = 2 is bij benadering 6,66 m/s.

a Op 3 2; 2,014 is

%s (10冑4 Â 20,01 + 1 í 10) í (10冑4 Â 20 + 1 í 10) § 2,22. = 0,01 %t De snelheid op t = 20 is bij benadering 2,22 m/s. 6,66 + 2,22 b De gemiddelde snelheid van de snelheden op t = 2 en t = 20 is = 4,44 m/s. 2 Op 3 20; 20,014 is

%s (10冑4 Â 11,01 + 1 í 10) í (10冑4 Â 11 + 1 í 10) § 2,98. = 0,01 %t De snelheid op t = 11 is bij benadering 2,98 m/s. Dus de snelheid op t = 11 is niet gelijk aan het gemiddelde van de snelheden op t = 2 en t = 20. Op 3 11; 11,01 4 is

18 a

%s 0,4 Â 3,012 í 0,4 Â 32 = = 2,404. 0,01 %t De snelheid op t = 3 is bij benadering 2,40 m/s. %s 0,4 Â 5,012 í 0,4 Â 52 Op 3 5; 5,01 4 is = = 4,004. 0,01 %t De snelheid op t = 5 is bij benadering 4,00 m/s. Op 3 3; 3,01 4 is

(

) (

)

(

) (

)

5 5 8í í 8í 1,01 + 2 1+2 %s 19 a Op 3 1; 1,01 4 is § 0,55. = 0,01 %t De snelheid op t = 1 is bij benadering 0,55 m/s. 5 5 8í í 8í 2,01 + 2 2+2 %s Op 3 2; 2,01 4 is § 0,31. = 0,01 %t De snelheid op t = 2 is bij benadering 0,31 m/s. 20 a

21 a

%A 1,20 Â 1,153,001 í 1,20 Â 1,153 § 0,26. = 0,001 %t De snelheid op t = 3 is bij benadering 0,26 m 2 /dag. Op 3 3; 3,001 4 is

De gemiddelde snelheid op 3 212 ,16 4 is %s 75冑32 + 4 í 150 í (75冑5 + 4 í 150) 450 í 150 í 225 + 150 225 = = = 1 = 1623 m/s. %t 1312 1312 132 De snelheid op t = 8 is

%s 75冑16,02 + 4 í 150 í (75冑16 + 4 í 150) § 16,8 m/s. = 0,01 %t

%s 75冑18,02 + 4 í 150 í (75冑18 + 4 í 150) § 16,0 m/s. = 0,01 %t 2 Omdat 16,0 < 163 < 16,8 moet er op [8, 9] een tijdstip zijn waarop de snelheid gelijk is aan de gemiddelde De snelheid op t = 9 is snelheid op 3 212 , 16 4 .

44

Hoofdstuk 2

© Noordhoff Uitgevers bv

2.2 Raaklijnen en hellinggrafieken Bladzijde 59 22 a

a De gemiddelde snelheid op het interval 3 2, 5 4 is %s (í52 + 10 Â 5) í (í22 + 10 Â 2) 25 í 16 = = = 3 m/s. 5í2 3 %t De gemiddelde snelheid op het interval 3 2, 4 4 is %s (í42 + 10 Â 4) í (í22 + 10 Â 2) 24 í 16 = = = 4 m/s. 4í2 2 %t De gemiddelde snelheid op het interval 3 2, 3 4 is %s (í32 + 10 Â 3) í (í22 + 10 Â 2) 21 í 16 = = = 5 m/s. 3í2 1 %t De gemiddelde snelheid op het interval 3 2; 2,5 4 is %s (í2,52 + 10 Â 2,5) í (í22 + 10 Â 2) 18,75 í 16 = = = 5,5 m/s. 2,5 í 2 0,5 %t b Van de vier lijnen komt de lijn AB4 het dichtst bij de lijn die de grafiek in A raakt.

2

Bladzijde 60 23 a

2x + 4 en y2 = 冑9x í 2. xí1 Intersect geeft x = 3 en y = 5, dus S(3, 5). Stel k: y = ax + b. Voer in y1 =

De optie d y/d x (TI) of d/d x (Casio) geeft a = c

dy = í1,5. d dx x=3

y = í1,5x + b f í1,5 Â 3 + b = 5 door (3, 5) í4,5 + b = 5 b = 9,5 Dus k: y = í1,5x + 9,5. Stel l: y = ax + b. De optie d y/d x (TI) of d/d x (Casio) geeft a = c

dy d = 0,9. dx x=3

y = 0,9x + b f 0,9 Â 3 + b = 5 door (3, 5) 2,7 + b = 5 b = 2,3 Dus l: y = 0,9x + 2,3. Bladzijde 61 24 a

a Stel k: y = ax + b. Voer in y1 = x2 + x í 2. De optie d y/d x (TI) of d/d x (Casio) geeft a = c

dy = í1. d d x x = í1

y = íx + b f í í1 + b = í2 f (í1) = í2, dus A(í1,í2) 1 + b = í2 b = í3 Dus k: y = íx í 3. b Stel l: y = ax + b. 2 Voer in y1 = + 3. xí1 De optie d y/d x (TI) of d/d x (Casio) geeft a = c

dy = í0,5. d dx x=3

y = í0,5x + b f í0,5 Â 3 + b = 4 f (3) = 4, dus A(3, 4) í1,5 + b = 4 b = 5,5 Dus l: y = í0,5x + 5,5.

© Noordhoff Uitgevers bv

De afgeleide functie

45

25 a

5x2 . x2 + 1 Stel k: y = ax + b. Voer in y1 = 1 +

De optie d y/d x (TI) of d/d x (Casio) geeft a = c y = í0,8x + b f í0,8 Â í2 + b = 5 f (í2) = 5, dus A(í2, 5) 1,6 + b = 5 b = 3,4 Dus k: y = í0,8x + 3,4. Stel l: y = ax + b.

2

De optie d y/d x (TI) of d/d x (Casio) geeft a = c

dy = í0,8. d d x x = í2

dy = 2,5. d dx x=1

y = 2,5x + b f 2,5 Â 1 + b = 3,5 f (1) = 3,5, dus B(1; 3,5) b=1 Dus l: y = 2,5x + 1. Snijpunt S van k en l volgt uit í0,8x + 3,4 = 2,5x + 1 í3,3x = í2,4 x = 0,727… f y = 2,818… y = 2,5x + 1 Dus S(0,73; 2,82). 26 a

a Voer in y1 = 0,6x2. Gebruik de optie d y/d x (TI) of d/d x (Casio). ds De snelheid op t = 3 is c d = 3,6 m/s. dt t=3 De snelheid op t = 5 is c

ds d = 6 m/s. dt t=5

b Na 5 seconden is er 0,6 Â 52 = 15 m afgelegd. Tussen t = 5 en t = 10 wordt 5 Â 6 = 30 m afgelegd. Dus na 10 seconden is 15 + 30 = 45 m afgelegd. 27 a

a Voer in y1 = 37 +

3 . 2x + 1

De optie d y/d x (TI) of d/d x (Casio) geeft c Dus op t = 2 is de snelheid í0,24 °C/uur. b De optie d y/d x (TI) of d/d x (Casio) geeft c

dT d = í0,24. dt t=2

dT d § í0,12. dt t=3

Op t = 3 is de lichaamstemperatuur T = 37,43 °C. 0,43 Het duurt dan nog § 3,5 uur voordat de temperatuur van 37 °C is 0,12 bereikt. Dus op t § 6,5. Bladzijde 62 28 a

a Voer in y1 =

600x . 4x

De optie d y/d x (TI) of d/d x (Casio) geeft c

dH d § í57,9. dt t=1

Dus op t = 1 is de snelheid í 57,9 mg/uur. b De optie d y/d x (TI) of d/d x (Casio) geeft c

dH d = í66,47... dt t=2

Op t = 2 is de hoeveelheid verdovend middel H = 75 mg 75 Het duurt dan nog § 1,1 uur voordat het verdovend middel uit het 66,47... lichaam verdwenen is. Dus op t § 3,1.

46

Hoofdstuk 2

© Noordhoff Uitgevers bv

29 a

200x2 + 1200x + 450 . 4x2 + 9 De optie maximum geeft x = 1,5 en y = 150. De inspanning duurde 1,5 minuut en de maximale hartslagfrequentie is 150 slagen per minuut. b Voer in y2 = 120. Intersect geeft x = 3,67... Het duurt 3,67 í 1,5 = 2,17 minuten § 130 seconden vanaf het moment van beëindigen van de inspanning. dF De optie d y/d x (TI) of d/d x (Casio) geeft c § í13,6. d d t t = 3,67... Op dat moment neemt de hartslag af met ongeveer 14 slagen per minuut. 240x2 + 1440x + 540 . c Voer in y1 = 4x2 + 9 De optie maximum geeft x = 1,5 en y = 180. Dus de maximale hartslag van een wielrenner is 180 slagen per minuut. De maximale hartslag van een hardloper is 150 slagen per minuut. 180 k= = 1,2 150 d 120k = 120 Â 1,2 = 144 slagen per minuut Voer in y2 = 144. Intersect geeft x = 3,67... dF De optie d y/d x (TI) of d/d x (Casio) geeft c = í16,3 d d t t = 3,67... a Voer in y1 =

2

í16,3 § 1,2, dus de snelheid waarmee de hartslag van de wielrenner afneemt is í13,6 k keer zo groot als de snelheid waarmee de hartslag van een hardloper afneemt. Bladzijde 64 30 a

a positief negatief b In de top is de helling nul. c Voer in y1 = íx2 + 4x en gebruik de optie d y/d x (TI) of d/d x (Casio).

d

x-coördinaat

í1

0

1

2

3

4

helling

6

4

2

0

í2

í4

helling 6 5 4 3 2 1 O –1

1

2

3

4

x

–2 –3 –4

e De lijn geeft voor elke x de helling van de grafiek van f.

© Noordhoff Uitgevers bv

De afgeleide functie

47

Bladzijde 65 20 a

y

y

2

g

ƒ

O

1

x

5

helling

O

O

1

x

5

helling

1

3

x

5

O

1

3

5

8

x

Bladzijde 66 32 a

a De hellinggrafiek ligt op het interval 8a, b9 boven de x-as en is daar stijgend. b De hellinggrafiek ligt op het interval 8c, d 9 onder de x-as en is daar stijgend. c De hellinggrafiek snijdt de x-as in (p, 0) en gaat daar over van stijgend (boven de x-as) naar dalend (onder de x-as). d De hellinggrafiek heeft een negatief minimum bij x = q.

33 a

a

hellinggra¿ek van f

gra¿ek van f

onder de x-as snijdt de x-as boven de x-as snijdt de x-as onder de x-as snijdt de x-as boven de x-as

dalend top stijgend top dalend top stijgend

8í4, í39 x = í3 8í3, 09 x=0 80, 29 x=2 82, 49 b

helling 2 1 – 3 – 2 –1 O –1

1

2

3

x

–2 y

–3

48

Hoofdstuk 2

O

2

x

© Noordhoff Uitgevers bv

34 a

helling

helling

O

1

2

3

x

O

2

3

1

2

3

2

x

y

y

O

1

1

2

3

x O

x

Bladzijde 67 35

a Voer in y1 = 3x4 + 4x3 í 12x2 + 2. De optie minimum geeft x = í2 en y = í30 en x = 1 en y = í3. De optie maximum geeft x = 0 en y = 2. De toppen zijn (í2,í30), (0, 2) en (1,í3). y b ƒ –2

O

1

x

helling

–2

O

1

x

c De optie d y/d x (TI) of d/d x (Casio) geeft a = c d De optie zero (TI) of ROOT (Casio) geeft x § í2,8  x § í0,4  x § 0,5  x § 1,4.

© Noordhoff Uitgevers bv

dy = 24. d d x x = í1

De afgeleide functie

49

e

y

ƒ –0,4 – 2,8

0,5

O

x

1,4

2

y g

–0,4 – 2,8

36

O

0,5

x

1,4

a Voer in y1 = 0,1x3 + x2 í 6. De optie maximum geeft x § í6,7 en y § 8,8. De optie minimum geeft x = 0 en y = í6. y

ƒ

–6,7

x

O

helling

–6,7

x

O

b De optie zero (TI) of ROOT (Casio) geeft x § í9,3, x § í2,9 en x § 2,2. y

g

– 9,3

50

Hoofdstuk 2

–2,9

O

2,2

x

© Noordhoff Uitgevers bv

37

a

y

helling

2 O

x

De grafiek van g ontstaat uit de grafiek van f door een verticale verschuiving. Dus hebben f en g voor elke waarde van x dezelfde helling. De hellinggrafieken van f en g vallen dus samen. b Stel h(x) = x2 í 4x + c f 32 í 4 Â 3 + c = 4 door (3, 4) 9 í 12 + c = 4 í3 + c = 4 c=7 Dus h(x) = x2 í 4x + 7.

2.3 Limiet en afgeleide Bladzijde 69 38

a

b f (1) = 1, f (1,9) = 3,61, f (1,99) = 3,9601, f (2,01) = 4,0401 23 í 2 Â 22 0 c f (2) = = en dat is onbepaald. 2í2 0 39

a

6(x + 2) 6 = mits x  í2 x(x + 2) x

c

(x í 1)(x í 4) = x í 4 mits x  1 xí1

b

(x + 2)(x + 3) x + 2 1 = = 2 x + 1 mits x  í3 2 2(x + 3)

d

x2 í x x(x í 1) = = x mits x  1 xí1 xí1

Bladzijde 71 40

a f (2) =

22 + 2 í 6 0 = = 0, dus de functie f is continu in 2 en heeft dus geen perforatie. 2í3 í1

x2 + x í 6 (x + 3)(x í 2) = = x + 3 mits x  2. xí2 xí2 De grafiek van g heeft een perforatie voor x = 2. (x + 3)(x í 2) x2 + x í 6 lim = lim = lim (x + 3) = 5 xí2 xí2 xm2 x m2 xm2 De perforatie is dus (2, 5).

b g(x) =

© Noordhoff Uitgevers bv

De afgeleide functie

51

41

c lim

(x í 2)(x í 3) x2 í 5x + 6 = lim = lim (x í 3) = í1 xí2 xí2 xm2 xm2 xm2

d lim

2x2 2 =3 x m 0 3x2

c lim

b lim

x(2x í 1) 2x2 í x 2x í 1 = lim = lim = í 13 3x 3x 3 xm0 xm0 xm0

d lim

(x í 1)(x + 1) x2 í 1 = lim = lim (x + 1) = 2 xí1 xm1 x í 1 xm1 xm1 De perforatie is dus (1, 2). xí2 xí2 1 b lim 2 = lim = lim = 14 x + 2 (x í 2)(x + 2) x í 4 m m m x 2 x 2 x 2

(x í 4)(x + 1) x2 í 3x í 4 = lim = lim (x + 1) = 5 xí4 xí4 xm4 xm4 xm4 De perforatie is dus (4, 5). x+1 x+1 1 d lim 2 = lim = lim = í 15 x í 4 (x í 4)(x + 1) x í 3x í 4 m m m x í1 x í1 x í1

b lim

2

42

43

a lim

a lim

De perforatie is dus (2, 14 ) .

44

(x í 4)(x + 4) x2 í 16 = lim = lim (x + 4) = 8 m m x í 4 xí4 x 4 x 4 xm4

x2 í 5x + 6 22 í 5 Â 2 + 6 0 = = =0 xí4 2í4 í2 xm2

a lim

x2 í 10x + 25 12 í 10 Â 1 + 25 16 = = = í4 xí5 1í5 í4 xm1 x(3x + a) 3x2 + ax = lim = lim (3x + a) = a x x xm0 xm0 xm0 h(4h + 2a) 4h2 + 2ah = lim = lim (4h + 2a) = 2a h h hm0 hm0 hm0

c lim

De perforatie is dus ( í1, í 15 ) .

(x í a)(x + a) x2 í a2 lim x í a = lim = lim (x + a) = 2a xía xma xma xma De perforatie is dus (a, 2a). y=xí1 f 2a í 1 = a (a, 2a) a=1 Dus voor a = 1 ligt de perforatie op y = x í 1. Bladzijde 72

45

a Voor elke x is de helling gelijk aan 3, omdat de grafiek een rechte lijn is, dus de helling is de lijn y = 3. b De formule van de helling is y = 0. Bladzijde 73

46

47

f (4 + h) í f (4) h hm0 1 12 (4 + h)2 í 112 Â 42 = lim h hm0 1 12 (16 + 8h + h2) í 112 Â 16 = lim h hm0 24 + 12h + 112 h2 í 24 = lim h hm0 1 2 12h + 12 h = lim h hm0 = lim (12 + 112 h) = 12

b f '(x) = lim

hm0

hm0

f (3 + h) í f (3) h hm0

a f '(3) = lim

f (x + h) í f (x) h hm0

b f '(x) = lim

(3 + h)2 í 4(3 + h) í (32 í 4 Â 3) h hm0

= lim

9 + 6h + h2 í 12 í 4h í í3 h hm0

= lim

= lim = lim

2h + h2 h hm0 = lim (2 + h) = 2

52

f (x + h) í f (x) h hm0 1 12 (x + h)2 í 112 x2 = lim h hm0 1 2 12 (x + 2xh + h2) í 112 x2 = lim h hm0 1 2 12 x + 3xh + 112 h2 í 112 x2 = lim h hm0 1 2 3xh + 12 h = lim h hm0 = lim (3x + 112 h) = 3x

a f '(4) = lim

(x + h)2 í 4(x + h) í (x2 í 4x) h hm0 x2 + 2xh + h2 í 4x í 4h í x2 + 4x h hm0

2xh + h2 í 4h h hm0 = lim (2x + h í 4) = 2x í 4

= lim

= lim

hm0

hm0

Hoofdstuk 2

© Noordhoff Uitgevers bv

Bladzijde 74 48

f (x + h) í f (x) h hm0 a(x + h) í ax = lim h hm0 ax + ah í ax = lim h hm0 ah = lim hm0 h = lim (a) = a

f (x + h) í f (x) h hm0 aía = lim hm0 h 0 = lim hm0 h = lim (0) = 0

a f '(x) = lim

b f '(x) = lim

2

hm0

hm0

49

f (x + h) í f (x) h hm0

a (x + h)3 = (x + h)(x + h)2 = (x + h)(x2 + 2xh + h2 ) = x3 + 2x2h + xh2 + x2h + 2xh2 + h3 = x3 + 3x2h + 3xh2 + h3

c f '(x) = lim

a(x + h)2 + b(x + h) + c í (ax2 + bx + c) h hm0

= lim

f (x + h) í f (x) h hm0

= lim

a(x + h)3 í ax3 h hm0

= lim

a(x2 + 2xh + h2) + bx + bh + c í ax2 í bx í c h hm0

b f '(x) = lim

ax2 + 2axh + ah2 + bx + bh + c í ax2 í bx í c h hm0 2axh + ah2 + bh = lim h hm0 = lim (2ax + ah + b) = 2ax + b

= lim

a(x3 + 3x2h + 3xh2 + h3) í ax3 h hm0 ax3 + 3ax2h + 3axh2 + ah3 í ax3 = lim h hm0 3ax2h + 3axh2 + ah3 = lim h hm0 = lim (3ax2 + 3axh + ah2) = 3ax2 = lim

hm0

hm0

50 a

a f (x) = c  g(x) geeft f (x + h) í f (x) c  g(x + h) í c  g(x) c  (g(x + h) í g(x)) f࣠࣠ƍ(x) = lim = lim = lim h h h hm0 hm0 hm0 g(x + h) í g(x) = c  g'(x) h hm0 b s(x) = f (x) + g(x) geeft s(x + h) í s(x) f (x + h) + g(x + h) í (f (x) + g(x)) sƍ(x) = lim = lim h h hm0 hm0 = c  lim

f (x + h) + g(x + h) í f (x) í g(x) f (x + h) í f (x) + g(x + h) í g(x) = lim h h hm0 hm0

= lim = lim

hm0

(

)

f (x + h) í f (x) g(x + h) í g(x) f (x + h) í f (x) g(x + h) í g(x) + + lim = f࣠࣠ƍ(x) + gƍ(x) = lim h h h h hm0 hm0

Bladzijde 75 51 a

a b c d

f (x) = 5x6 í 3x5 + 2x í 7 geeft f '(x) = 30x5 í 15x4 + 2 g(x) = í2x8 í 4x4 + 7,2 geeft g'(x) = í 16x7 í 16x3 h(x) = í 13 x3 í 12 x2 í x í 1 geeft h'(x) = íx2 í x í 1 k(q) = 1 + 3q í 3q2 í 5q7 geeft k'(q) = 3 í 6q í 35q6

Bladzijde 76 52 a

a f (x) = (5x + 7)(4 í 3x) = 20x í 15x2 + 28 í 21x = í15x2 í x + 28 geeft f '(x) = í30x í 1 b g(x) = (3x + 6)2 í 8x = 9x2 + 36x + 36 í 8x = 9x2 + 28x + 36 geeft g'(x) = 18x + 28 c h(x) = 5(x í 3)2 + 5(2x í 1) = 5(x2 í 6x + 9) + 10x í 5 = 5x2 í 30x + 45 + 10x í 5 = 5x2 í 20x + 40 geeft h'(x) = 10x í 20 d k(x) = í3(x í 1)(5 í 9x) í 8(x í 7) = í3(5x í 9x2 í 5 + 9x) í 8x + 56 = í3(í9x2 + 14x í 5) í 8x + 56 = 27x2 í 42x + 15 í 8x + 56 = 27x2 í 50x + 71 geeft k'(x) = 54x í 50

© Noordhoff Uitgevers bv

De afgeleide functie

53

f (x) = (3x í 1)(x2 + 5x) = 3x3 + 15x2 í x2 í 5x = 3x3 + 14x2 í 5x geeft f '(x) = 9x2 + 28x í 5 g(x) = (3x3 í 1)2 = 9x6 í 6x3 + 1 geeft g'(x) = 54x5 í 18x2 h(x) = (5x5 í 3)(3x í 2) = 15x6 í 10x5 í 9x + 6 geeft h'(x) = 90x5 í 50x4 í 9 k(x) = 5 í 3(x4 í x)(x + 1) = 5 í 3(x5 + x4 í x2 í x) = 5 í 3x5 í 3x4 + 3x2 + 3x geeft k'(x) = í 15x4 í 12x3 + 6x + 3 e l(t) = (5t3 í t)(3t5 + t) = 15t8 + 5t4 í 3t6 í t2 = 15t8 í 3t6 + 5t4 í t2 geeft l' (t) = 120t7 í 18t5 + 20t3 í 2t f m(q) = 1 í (3q2 í 2)2 = 1 í (9q4 í 12q2 + 4) = 1 í 9q4 + 12q2 í 4 = í 9q4 + 12q2 í 3 geeft m'(q) = í36q3 + 24q

53 a

a b c d

54 a

a Een tweedegraadsfunctie is van de vorm f (x) = ax2 + bx + c met a  0. Differentiëren geeft f࣠࣠ƍ(x) = 2ax + b. Dit is een lineaire functie. b Stel h(x) = ax2 + bx + c. Differentiëren geeft hƍ(x) = 2ax + b. hƍ(x) = 2ax + b f b = 12 en a = 3 g(x) = 6x + 12 Dus h(x) = 3x2 + 12x + c. 12 xtop = í = í2 2Â3 h(x) = 3x2 + 12x + c f 3  (í2)2 + 12  í2 + c = í7 h(í2) = í7 12 í 24 + c = í7 í12 + c = í7 c=5 Dus h(x) = 3x2 + 12x + 5.

2

2.4 Toepassingen van de afgeleide Bladzijde 78 55 a

a p(x) = f (x)  g(x) = x2  (3x í 7) = 3x3 í 7x2 geeft p'(x) = 9x2 í 14x f (x) = x2 geeft f '(x) = 2x g(x) = 3x í 7 geeft g'(x) = 3 b p'(x) = 9x2 í 14x f is niet gelijk, dus pƍ(x) is niet f࣠࣠ƍ(x). g࣠ƍ(x) f '(x)  g'(x) = 2x  3 = 6x c p'(x) = 9x2 í 14x f '(x)  g(x) + f (x)  g'(x) = 2x(3x í 7) + x2  3 = 6x2 í 14x + 3x2 = 9x2 í 14x Dus p'(x) = f '(x)  g(x) + f (x)  g'(x). Bladzijde 79

56 a

a f (x) = (2 í 3x2 )(2 + 7x) geeft f '(x) = 3 2 í 3x2 4'  (2 + 7x) + (2 í 3x2)  3 2 + 7x 4' = í6x  (2 + 7x) + (2 í 3x2)  7 b g(x) = (2x í 5)2 = (2x í 5)(2x í 5) geeft g'(x) = 3 2x í 5 4'  (2x í 5) + (2x í 5)  3 2x í 5 4 ' = 2  (2x í 5) + (2x í 5)  2 = 4(2x í 5) c h(x) = (x2 í 3x)(x3 + x2 + x) geeft h'(x) = 3 x2 í 3x 4'  (x3 + x2 + x) + (x2 í 3x)  3 x3 + x2 + x 4 ' ௘ = (2x í 3)  (x3 + x2 + x) + (x2 í 3x)  (3x2 + 2x + 1) d j(x) = (3x2 í 4)2 = (3x2 í 4)(3x2 í 4) geeft j'(x) = 3 3x2 í 4 4'  (3x2 í 4) + (3x2 í 4)  3 3x2 í 4 4' = 6x  (3x2 í 4) + (3x4 í 4)  6x = 12x(3x4 í 4)

57 a

冑x  冑x = x Neem van het linker- en rechterlid de afgeleide. 3冑x  冑x 4' = 3 x 4' 3冑x 4'  冑x + 冑x  3冑x 4' = 1 2冑x  3冑x 4' = 1 : 2冑x 3冑x 4' = 1 2冑x Dus f '(x) =

54

Hoofdstuk 2

1 . 2冑x

© Noordhoff Uitgevers bv

58 a

59 a

a p = fgh = fg  h = jh geeft p' = j'h + jh' j = fg geeft j' = f 'g + fg' Dus p' = ( f 'g + fg' )h + fgh' = f 'gh + fg'h + fgh' b p = fghj geeft pƍ = 3 fgh 4 '  j + fgh  3 j 4 ' = ( f 'gh + fg'h + fgh')  j + fgh  j' = f 'ghj + fg'hj + fgh'j + fghj'

2

a q(x) Â n(x) = t(x) Differentiëren van beide leden geeft q'(x) Â n(x) + q(x) Â n'(x) = t'(x). b q'(x) Â n(x) + q(x) Â n'(x) = t'(x) q'(x) Â n(x) = t'(x) í q(x) Â n'(x) t'(x) í q(x) Â n'(x) n(x)

q'(x) =

t(x) t'(x) í q(x) Â n'(x) c Invullen van q(x) = in q'(x) = geeft q'(x) = n(x) n(x) Teller en noemer vermenigvuldigen met n(x)geeft q'(x) =

t'(x) í

t(x) Â n'(x) n(x) n(x)

n(x) Â t'(x) í t(x) Â n'(x) . (n(x))2

Bladzijde 81 60 a

a f (x) =

(x + 5) Â 1 í (x í 2) Â 1 x + 5 í x + 2 7 xí2 geeft f '(x) = = = x+5 (x + 5)2 (x + 5)2 (x + 5)2

b f (x) =

(2x í 1) Â 0 í 2 Â 2 2 í4 geeft f '(x) = = 2x í 1 (2x í 1)2 (2x í 1)2

c f (x) =

(2x2 + 1) Â 3x2 í x3 Â 4x 6x4 + 3x2 í 4x4 2x4 + 3x2 x3 geeft f '(x) = = = 2x2 + 1 (2x2 + 1)2 (2x2 + 1)2 (2x2 + 1)2

d f (x) =

(3 í x2 ) Â 1 í (x í 2) Â í2x 3 í x2 + 2x2 í 4x x2 í 4x + 3 xí2 geeft f '(x) = = = 3 í x2 (3 í x2 )2 (3 í x2 )2 (3 í x2 )2

e f (x) =

3 í x2 + x3 geeft xí2

f '(x) =

(x í 2) Â í2x í (3 í x2 ) Â 1 í2x2 + 4x í 3 + x2 íx2 + 4x í 3 + 3x2 = + 3x2 = + 3x2 2 2 (x í 2) (x í 2) (x í 2)2

f f (x) = x í 61 a

(x + 4) Â 0 í 2 Â 1 2 2 í2 geeft f '(x) = 1 í =1í =1+ x+4 (x + 4)2 (x + 4)2 (x + 4)2

a f (x) = x2 í 3x í 1 geeft f '(x) = 2x í 3 f '(4) = 2 Â 4 í 3 = 5 f (4) = 42 í 3 Â 4 í 1 = 3 b yA = f (4) c De helling in A is f '(4). Bladzijde 82

62 a

a f (x) = 14 x3 í 3x2 + 6x + 5 geeft f '(x) = 34 x2 í 6x + 6 Snijden met y-as, dus x = 0 geeft f (0) = 5, dus A(0, 5). Stel k: y = ax + b. a = f '(0) = 34 Â 02 í 6 Â 0 + 6 = 6 k: y = 6x + b f b=5 door A(0, 5) Dus k: y = 6x + 5.

© Noordhoff Uitgevers bv

De afgeleide functie

55

b g(x) = 13 x2(x2 í 9) + 4 = 13 x4 í 3x2 + 4 geeft g'(x) = 113 x3 í 6x Stel l: y = ax + b. a = g'(2) = 113 Â 23 í 6 Â 2 = í113 y = í113 x + b r 11 Â 2 + b = í223 g(2) = í223 , dus B ( 2, í223 ) í 3 í223 + b = í223 b=0 1 Dus l: y = í13 x. x2 + 4x geeft c h(x) = 2 x +1

2

h'(x) =

(x2 + 1) Â (2x + 4) í (x2 + 4x) Â 2x 2x3 + 4x2 + 2x + 4 í 2x3 í 8x2 í4x2 + 2x + 4 = = (x2 + 1)2 (x2 + 1)2 (x2 + 1)2

Snijden met de x-as dus h(x) = 0 geeft

x2 + 4x =0 x2 + 1 x2 + 4x = 0 x(x + 4) = 0 x = 0  x = í4 vold. vold.

Stel m: y = ax + b. í4 Â 02 + 2 Â 0 + 4 a = h'(0) = =4 (02 + 1)2 y = 4x + b fb = 0 door O(0, 0) Dus m: y = 4x. Stel n: y = ax + b. í4 Â (í4)2 + 2 Â í4 + 4 4 a = h'(í4) = = í 17 ((í4)2 + 1)2 4 y = í 17 x+b r í 4 Â í4 + b = 0 door C(í4, 0) 17 16 17 + b = 0

b = í 16 17 4 Dus n: y = í 17 x í 16 17 .

63 a

a f (x) = 12 x3 í 2x2 + 2 geeft f'(x) = 112 x2 í 4x Stel k: y = ax + b. a = f'(4) = 112 Â 42 í 4 Â 4 = 8 y = 8x + b f 8Â4 + b = 2 f (4) = 2, dus A(4, 2) 32 + b = 2 b = í30 Dus k: y = 8x í 30. b Stel m: y = ax + b. a = f '(í2) = 112 Â (í2)2 í 4 Â í2 = 14 y = 14x + b f 14 Â í2 + b = í10 f(í2) = í10, dus B(í2, í10) í28 + b = í10 b = 18 Dus m: y = 14x + 18.

64 a

a g(x) = 2x2 í 6x geeft g'(x) = 4x í 6 Stel l: y = ax + b. a = g'(í3) = 4 Â í3 í 6 = í18 y = í18x + b f í18 Â í3 + b = 36 g(í3) = 36, dus A(í3, 36) 54 + b = 36 b = í18 Dus l: y = í18x í 18.

56

Hoofdstuk 2

© Noordhoff Uitgevers bv

b g(x) = 0 geeft 2x2 í 6x = 0 x(2x í 6) = 0 x=0  x=3 Dus P(3, 0). Stel n: y = ax + b. a = g'(3) = 4  3 í 6 = 6 y = 6x + b f 6Â3 + b = 0 door P(3, 0) 18 + b = 0 b = í18 Dus n: y = 6x í 18.

2

Bladzijde 83 65 a

(2x + 5) Â 2x í (x2 í 4) Â 2 4x2 + 10x í 2x2 + 8 2x2 + 10x + 8 x2 í 4 geeft f '(x) = = = 2x + 5 (2x + 5)2 (2x + 5)2 (2x + 5)2 Stel k: y = ax + b. 2 Â (í2)2 + 10 Â í2 + 8 a = f '(í2) = = í4 (2 Â í2 + 5)2 y = í4x + b f í4 Â í2 + b = 0 A(í2, 0) 8+b=0 b = í8 Dus k: y = í4x í 8.

a f (x) =

Stel l: y = ax + b. 2 Â 22 + 10 Â 2 + 8 4 a = f '(2) = =9 (2 Â 2 + 5)2 4 y = 9x + b 4 r 9 Â2 + b = 0 B(2, 0) b = í 89 Dus l: y = 49 x í 89 . b f (0) = í 45 dus C ( 0, í 45 ) f '(0) =

8 25

s

8 m: y = 25 x í 45

Bladzijde 84 66 a

f (x) =

2x í 5 geeft x2 í 4

(x2 í 4) Â 2 í (2x í 5) Â 2x 2x2 í 8 í 4x2 + 10x í2x2 + 10x í 8 = = (x2 í 4)2 (x2 í 4)2 (x2 í 4)2 Stel k: y = ax + b. í8 a = f '(0) = = í 12 16 y = í 12 x + b r b = 114 f (0) = 114 dus A ( 0, 114 ) Dus k: y = í 12 x + 114 . 2x í 5 yB = 0 geeft 2 =0 x í4 2x í 5 = 0 2x = 5 x = 212 voldoet f '(x) =

k: y = í 12 x + 114

(

2 12 , 0

)

s

í 12 Â 212 + 114 = 0 klopt!

Hieruit volgt dat B ook op de lijn k ligt, dus de bewering is waar.

© Noordhoff Uitgevers bv

De afgeleide functie

57

67 a

a f (x) = (x2 í 4)(x + 1) = x3 + x2 í 4x í 4 geeft f '(x) = 3x2 + 2x í 4 Stel k: y = ax + b. a = f '(í3) = 3  (í3)2 + 2  í3 í 4 = 17 y = 17x + b f 17  í3 + b = í10 f (í3) = í10, dus A(í3, í10) í51 + b = í10 b = 41 Dus k: y = 17x + 41. b B is het snijpunt met de y-as, dus xB = 0. Stel l: y = ax + b. a = f '(0) = í4 y = í4x + b f b = í4 f (0) = í4, dus B(0, í4) Dus l: y = í4x í 4. c f (x) = 0 geeft (x2 í 4)(x + 1) = 0 x2 í 4 = 0  x + 1 = 0 x2 = 4  x = í1 x = 2  x = í2  x = í1 Dus C(2, 0). Stel m: y = ax + b. a = f '(2) = 3  22 + 2  2 í 4 = 12 y = 12x + b f 12  2 + b = 0 door C(2, 0) 24 + b = 0 b = í24 Dus m: y = 12x í 24.

68 a

f (x) = x3 + ax2 + a2x geeft f '(x) = 3x2 + 2ax + a2 Stel k: y = mx + n. m = f '(a) = 3  a2 + 2a  a + a2 = 6a2

2

y = 6a2x + n 6a2 Â a + n = 3a3 f (a) = 3a3, dus A(a, 3a3 ) f 3 6a + n = 3a3 n = í3a3 2 3 Dus k: y = 6a x í 3a . 69 a

a

b De lijn y = 4x í 14 raakt de grafiek van f . c Er geldt f '(xR ) = 4. Bladzijde 85 70 a

a f (x) = íx2 + 2x + 3 geeft f '(x) = í2x + 2 f '(x) = 4 geeft í2x + 2 = 4 í2x = 2 x = í1 xA = í1 en yA = f (í1) = 0, dus A(í1, 0). b k is evenwijdig met l, dus rck = rcl = í6 en f '(xB ) = rck = rcl = í6. f '(xB ) = í6 geeft í2x + 2 = í6 í2x = í8 x=4 xB = 4 en yB = f (4) = í5, dus B(4, í5).

58

Hoofdstuk 2

© Noordhoff Uitgevers bv

71 a

a f (x) = 13 x3 + 12 x2 í 10x + 6 geeft f '(x) = x2 + x í 10 f '(x) = 2 geeft x2 + x í 10 = 2 x2 + x í 12 = 0 (x + 4)(x í 3) = 0 x = í4  x = 3 xA = í4 en yA = f (í4) = 3223 , dus A ( í4, 3223 ) .

2

xB = 3 en yB = f (3) = í1012 , dus B ( 3, í1012 ) . b raaklijn is evenwijdig met k, dus f '(xC ) = f '(xD ) = rck = 10. f'(x) = 10 geeft x2 + x í 10 = 10 x2 + x í 20 = 0 (x + 5)(x í 4) = 0 x = í5  x = 4 Dus xC = í5 en xD = 4. 72 a

a f (x) = 13 x3 í x2 í 1 geeft f '(x) = x2 í 2x Stel k: y = ax + b. a = f '(4) = 42 í 2 Â 4 = 8 y = 8x + b 1 f 8 Â 4 + b = 43 f (4) = 413 , dus P ( 4, 413 ) 32 + b = 413 í 2723 . x2 í 2x

Dus k: y = 8x b f '(x) = 3 geeft

b = í2723

=3 x2 í 2x í 3 = 0 (x + 1)(x í 3) = 0 x = í1  x = 3 xQ = í1 en yQ = f (í1) = í213 , dus Q ( í1, í213 ) . xR = 3 en yR = f (3) = í1, dus R(3, í1).

73 a

fp = x3 + px2 + 2x í 3 geeft fp '(x) = 3x2 + 2px + 2 fp '(x) = í1 geeft 3x2 + 2px + 2 = í1 3x2 + 2px + 3 = 0 Twee raaklijnen betekent twee oplossingen, dus D > 0 f 4p2 í 36 > 0 D = (2p)2 í 4  3  3 = 4p2 í 36 4p2 > 36 p2 > 9 p < í3  p > 3 Bladzijde 86

74 a

a h = í5t2 + 25t geeft v = í10t + 25 t = 0 geeft v = 25 m/s b t = 3 geeft v = í10  3 + 25 = í5, dus de snelheid is 5 m/s omlaag. c Op het hoogste punt is de snelheid nul. v = 0 geeft í10t + 25 = 0 í10t = í25 t = 2,5 Dat is na 2,5 seconden het geval. d h = 0 geeft í5t2 + 25t = 0 í5t(t í 5) = 0 t=0t=5 Dus na 5 seconden is de bal weer op de grond. t = 5 geeft v = í10  5 + 25 = í25 De bal komt met een snelheid van 25 m/s weer op de grond.

75 a

a s = 0,8t2 geeft v = 1,6t Op t = 3 is v = 1,6 Â 3 = 4,8 m/s. Op t = 6 is v = 1,6 Â 6 = 9,6 m/s.

© Noordhoff Uitgevers bv

De afgeleide functie

59

30 m/s 3,6 30 30 v= geeft 1,6t = 3,6 3,6 t § 5,21 Dus na 5,21 seconden. c In de eerste zes seconden legt de auto 0,8 Â 62 = 28,8 m af . Tussen t = 6 en t = 10 legt de auto 4 Â 9,6 = 38,4 m af . Dus in de eerste tien seconden 28,8 + 38,4 = 67,2 m. b 30 km/uur =

2

2.5 Hellingen en raaklijnen met GeoGebra Bladzijde 88 76 a

a b c d

* * * *

Bladzijde 89 77 a

a b c d e f g

* P(1, í5) * Het differentiequotiënt is gelijk aan de richtingscoëf¿ciënt van lijn l. l: y = í2x í 2 en B(2, í6). * l: y = 4x í 18

78 a

a l: y = 4x + 1 b De hellingen zijn 6, 2, 0, í2, í4. c In (6, 0) is de helling í6. In (7,í7) is de helling í8. Bladzijde 90

79 a

a b c d e

* * y = 2x í 6 ( 5, 412 ) Door de punten die boven de parabool liggen gaat geen raaklijn.

80 a

a b c d

* * y = íx + 4 Ja, het antwoord klopt.

Bladzijde 91 81 a

a formule hellinggrafiek is y = 2x í 2 b formule hellinggrafiek is y = x2 í 2 c De hellinggrafiek heeft twee toppen.

82 a

a b c d e

83 a

a f'(x) = x í 6 b g'(x) = íx + 4 c h'(x) = íx + 4

60

Hoofdstuk 2

* in (2, 2) en in (í2,í2). (0, 0) Door (5, 0) gaan twee raaklijnen. Door (1, 0) gaat één raaklijn.

© Noordhoff Uitgevers bv

84 a

De grafieken van g en h hebben dezelfde helling. Dit is geen toeval omdat de grafiek van h ontstaat uit de grafiek van g door deze 4 omhoog te verschuiven. De hellingen van de beide grafieken zijn dus gelijk.

Diagnostische toets 2

Bladzijde 92 1

a Afnemend dalend op 8k , 19. Toenemend stijgend op 81, 212 9. Afnemend stijgend op 8212 , 49. Toenemend dalend 84, 69. Afnemend dalend 86, 712 9. Toenemend stijgend op 8712 ,m 9. %y 4 í í512 912 = = = 316 4í1 3 %x %y í4 í í2 í2 c Op 3 2, 6 4 is = = = í 12 6í2 4 %x d Trek de halve lijn met beginpunt O(0, 0) en rc = 1. Deze lijn gaat ook door (4, 4) en (8, 8). De lijn snijdt de grafiek in drie punten, dus er zijn drie waarden van p.

b Op 3 1, 4 4 is

2

a Op 3 1, 4 4 is

%y f (4) í f (1) 8 í í212 = = = 312 . 4í1 3 %x

%y f (1) í f (í1) í212 í 12 = = = í112 . 1 í í1 2 %x c Stel l: y = ax + b. b Op 3 í1, 1 4 is

%y f (3) í f (í2) í112 í í4 1 = =2 = 5 3 í í2 %x 1 y = 2x + b 1 f  í2 + b = í4 door A(í2, í4) 2 í1 + b = í4 b = í3 Dus l: y = 12 x í 3. a=

3

%s 9 í 5 4 = = = 0,5 km/minuut = 30 km/uur %t 9 í 1 8 b De gemiddelde snelheid op [0, 9] is de richtingcoëf¿ciënt van de lijn door (0, 0) en (9, 9), dus rc = 1. De halve lijn met beginpunt (1, 5) en rc = 1 snijdt de grafiek in (4, 8), dus p = 4. a Op 3 1, 9 4 is

(

) (

)

10 10 10 í í 10 í 1,01 + 1 1+1 %s c Op 3 1; 1,01 4 is = = 2,48... km/min 0,01 %t De snelheid op t = 1 is bij benadering 2,48. . . Â 60 § 149 km/uur. 4

a Voer in y1 = 冑2x í 3. De optie d y/d x (TI) of d/d x (Casio) geeft helling = c

dy = 1. d dx x=2

dy b De optie d y/d x (TI) of d/d x (Casio) geeft snelheid = c = 12 . d 1 d x x = 32 c Stel k: y = ax + b. dy De optie d y/d x (TI) of d/d x (Casio) geeft a = c = 1. d dx x=6 3 y = 13 x + b f f (6) = 3, dus B(6, 3) Dus k: y = 13 x + 1.

© Noordhoff Uitgevers bv

1 3

Â6 + b = 3 2+b=3 b=1

De afgeleide functie

61

5

2

a Voer in y1 = í 14 x3 + 3x2. Gebruik de optie d y/d x (TI) of d/d x (Casio). ds De snelheid op t = 2 is c d = 9 m/s. dt t=2 b t = 6 geeft s = 54 ds De snelheid op t = 6 is c d = 9 m/s. dt t=6 Tussen t = 6 en t = 10 wordt 4 Â 9 = 36 m afgelegd. Dus na 10 seconden is 54 + 36 = 90 m afgelegd. Bladzijde 93

6

a

y

b

y ƒ

2

1

1

–2

–1

O

1

2

3

x

–2

–1

7

–1

O

–1

–1

–2

–2

–3

–3

O

1

2

3

1

2

3

x

y

helling

–2

ƒ

2

1

2

3

x

–2

–1

O

x

a Voer in y1 = í0,2x3 + x2 í 2. De optie minimum geeft x = 0. De optie maximum geeft x = 313 . helling

O

62

Hoofdstuk 2

1

33

x

© Noordhoff Uitgevers bv

b De optie zero (TI) of ROOT (Casio) geeft x § í1,3, x § 1,8 en x § 4,5. y

ƒ

2 – 1,3

O

1,8

x

4,5

y

g

– 1,3

O

1,8

4,5

x

x2 í 2x í 8 (í2)2 í 2 Â í2 í 8 0 = = =0 xí2 í2 í 2 í4 x m í2

a lim

8

(x + 4)(x í 2) x2 + 2x í 8 = lim = lim (x + 4) = 6 xí2 xí2 xm2 xm2 xm2

b lim

(x + 2)(x í 2) x2 í 4 = lim = lim (x + 2) = 4 xí2 xm2 x í 2 xm2 xm2

c lim lim

9

x m í1 x2

x+1 x+1 1 = lim = lim =1 + 6x + 5 x m í1 (x + 1)(x + 5) x m í1 x + 5 4

De perforatie is dus (í1, 14 ) . 10 a

f (3 + h) í f (3) h hm0

a f '(3) = lim

5(3 + h)2 + 4 í (5 Â 32 + 4) h hm0

= lim

5(9 + 6h + h2) + 4 í 49 h hm0

= lim

45 + 30h + 5h2 + 4 í 49 = lim h hm0 30h + 5h2 h hm0

= lim

= lim (30 + 5h) = 30

f (x + h) í f (x) h hm0 5(x + h)2 + 4 í (5x2 + 4) = lim h hm0 2 5(x + 2xh + h2) + 4 í 5x2 í 4 = lim h hm0

b f ƍ(x) = lim

5x2 + 10xh + 5h2 + 4 í 5x2 í 4 h hm0 10xh + 5h2 = lim h hm0 = lim (10x + 5h) = 10x = lim

hm0

hm0

11 a

a f (x) = 0,6x3 í 1,3x2 + 7 geeft f'(x) = 1,8x2 í 2,6x b g(p) = 4p3 + p2 í 11p + 20 geeft g'(p) = 12p2 + 2p í 11

12 a

a b c d

f (x) = (3 í x)(5 + 2x) = 15 + 6x í 5x í 2x2 = í2x2 + x + 15 geeft f '(x) = í4x + 1 g(x) = (3x + 1)2 = 9x2 + 6x + 1 geeft g'(x) = 18x + 6 h(x) = x(2x í 1)2 = x(4x2 í 4x + 1) = 4x3 í 4x2 + x geeft h'(x) = 12x2 í 8x + 1 k(x) = 13 x3 + 2x2(x í 4) + 6 = 13 x3 + 2x3 í 8x2 + 6 = 213 x3 í 8x2 + 6 geeft k'(x) = 7x2 í 16x

© Noordhoff Uitgevers bv

De afgeleide functie

63

13 a

a f (x) = b g(x) =

(3x í 1) Â 2 í (2x í 3) Â 3 6x í 2 í 6x + 9 2x í 3 7 geeft f'(x) = = = 2 2 3x í 1 (3x í 1) (3x í 1) (3x í 1)2 (x2 + 1) Â 4x3 í x4 Â 2x 4x5 + 4x3 í 2x5 2x5 + 4x3 x4 geeft g'(x) = = = 2 +1 (x2 + 1)2 (x2 + 1)2 (x + 1)2

x2

c h(x) = x2 í 2 14 a

(x + 1) Â 0 í 3 Â 1 3 3 í3 geeft h'(x) = 2x í = 2x í = 2x + x+1 (x + 1)2 (x + 1)2 (x + 1)2

a f (x) = í 16 x3 + 12 x2 + 4x + 1 geeft f '(x) = í 12 x2 + x + 4 Stel k: y = ax + b. a = f '(2) = í 12 Â 22 + 2 + 4 = 4 y = 4x + b 2 f 4 Â 2 + b = 93 f (2) = 923 , dus A ( 2, 923 ) 8 + b = 923 123 .

b = 123

Dus k: y = 4x + b Stel l: y = ax + b. C is het snijpunt met de y-as, dus xC = 0. a = f ƍ(0) = 4 y = 4x + b f b=1 f (0) = 1, dus C(0, 1) Dus l: y = 4x + 1. 15 a

f (x) = 13 x3 í x2 í x + 2 geeft f '(x) = x2 í 2x í 1 Raaklijn evenwijdig met k, dus f '(xA ) = f '(xB ) = rck = 2. f '(x) = 2 geeft x2 í 2x í 1 = 2 x2 í 2x í 3 = 0 (x + 1)(x í 3) = 0 x = í1  x = 3 xA = í1 en yA = f (í1) = 123 , dus A ( í1, 123 ) . xB = 3 en yB = f (3) = í1, dus B(3, í1).

64

Hoofdstuk 2

© Noordhoff Uitgevers bv

3 Vergelijkingen en herleidingen Voorkennis Stelsels lineaire vergelijkingen en kwadratische ongelijkheden Bladzijde 96 1

a l:

x

0

2

y

í6

0

m:

x

0

1

y

1

0

n:

x

0

3

y

0

3

p:

x

0

4

y

2

0

3

of l: 3x í y = 6 geeft l: y = 3x í 6, dus rcl = 3 en l snijdt de x-as in (0, í6). m: x + y = 1 geeft m: y = íx + 1, dus rcm = í1 en m snijdt de x-as in (0, 1). n: x í y = 0 geeft n: y = x, dus rcn = 1 en n snijdt de x-as in (0, 0). p: x + 2y = 4 geeft l: y = í 12 x + 2, dus rcp = í 12 en p snijdt de x-as in (0, 2). y

4

3 m

2 l 1 p

−3

−2

−1

O

1

2

3

4

5

x

−1

−2

−3 n −4

−5

−6

b rcl = 3, rcm = í1, rcn = 1 en rcp = í 12 . a2 a Snijpunt met de x-as, dus y = 0 geeft 4x í 3 Â 0 = 24 4x = 24 x=6 Dus het snijpunt met de x-as is (6, 0). Snijpunt met de y-as, dus x = 0 geeft 4 Â 0 í 3y = 24 í3y = 24 y = í8 Dus het snijpunt met de y-as is (0, í8).

© Noordhoff Uitgevers bv

Vergelijkingen en herleidingen

65

b Invullen van A(8, 3) geeft 4 Â 8 í 3 Â 3 = 24 32 í 9 = 24 23 = 24 Dit klopt niet, dus A ligt niet op l. Invullen van B(18, 16) geeft 4 Â 18 í 3 Â 16 = 24 72 í 48 = 24 24 = 24 Dit klopt, dus B ligt op l. Invullen van C(í30, í48) geeft 4 Â í30 í 3 Â í48 = 24 í120 + 144 = 24 24 = 24 Dit klopt, dus C ligt op l. c 4x í 3y = 24 r 4 Â 16 í 3 Â p = 24 (16, p) 64 í 3p = 24 í3p = í40 í40 p= = 1313 í3 d 4x í 3y = 24 r 4 Â q í 3 Â 48 = 24 (q, 48) 4q í 144 = 24 4q = 168 q = 42

3

Bladzijde 97

a3

a Uit 3x + y = 7 volgt y = í3x + 7. y = í3x + 7 en x í 4y = 11 geeft x í 4(í3x + 7) = 11 x + 12x í 28 = 11 13x = 39 x=3 x = 3 en y = í3x + 7 geeft y = í3 Â 3 + 7 = í2. De oplossing is (x, y) = (3, í2). b Uit x + 2y = í3 volgt y = í 12 x í 112 . y = í 12 x í 112 en 4x í 3y = 10 geeft 4x í 3 (í 12 x í 112 ) = 10 4x + 112 x + 412 = 10 512 x = 512 x=1 x = 1 en y = í 12 x í 112 geeft y = í 12 Â 1 í 112 = í2. De oplossing is (x, y) = (1, í2). c Uit 5x + 4y = 56 volgt y = í114 x + 14. y = í114 x + 14 en x í 6y = 1 geeft x í 6 (í114 x + 14) = 1 x + 712 x í 84 = 1 812 x = 85 x = 10 x = 10 en y = í114 x + 14 geeft y = í114 Â 10 + 14 = 112 . De oplossing is (x, y) = (10, 112 ) .

4

a k:

x

0

6

y

í4

0

l:

x

0

5

y

1

0

of k: 2x í 3y = 12 geeft k: y = 23 x í 4, dus rck = 23 en k snijdt de x-as in (0, í4). l: x + 5y = 5 geeft l: y = í 15 x + 1, dus rcl = í 15 en l snijdt de x-as in (0, 1).

66

Hoofdstuk 3

© Noordhoff Uitgevers bv

y

2 l 1

−1

O

1

2

3

4

5

6

x

7

−1

3

−2 k −3

−4

−5

b Uit x + 5y = 5 volgt y = í 15 x + 1. y = í 15 x + 1 en 2x í 3y = 12 geeft 2x í 3 (í 15 x + 1) = 12 2x + 35 x í 3 = 12 235 x = 15 x = 510 13 1 1 10 2 x = 510 13 en y = í 5 x + 1 geeft y = í 5 Â 513 + 1 = í 13 . 2 De oplossing is (x, y) = (510 13 , í 13 ) .

Bladzijde 98

a5 a Voor geen enkele x. b Voor elke x. c Voor x  112 .

ƒ

b 2x2 í 3x í 2 • 0 f f(x) f (x) = 0 geeft 2x2 í 3x í 2 = 0 D = (í3)2 í 4  2  í2 = 25 3í5 3+5 x= x= 4 4 x = í 12  x = 2 t

t

a6 a x2 í 5x í 14 < 0 f f(x) f (x) = 0 geeft x2 í 5x í 14 = 0 (x í 7)(x + 2) = 0 xí7=0 x+2=0 x = 7  x = í2

ƒ −2

7

x

f (x) < 0 geeft í2 < x < 7

− 21

2

x

f (x) • 0 geeft x ” í 12  x • 2

© Noordhoff Uitgevers bv

Vergelijkingen en herleidingen

67

•

c 2x2 + x í 6 ” 0 f f(x)

ƒ

3 1 21

–2

f (x) ” 0 geeft í2 ” x ”

u

f (x) = 0 geeft 2x2 + x í 6 = 0 D = 1 í 4  2  í6 = 49 í1 í 7 í1 + 7 x= x= 4 4 x = í2  x = 112

d (x í 1)2 + 3x + 14 í 5(x í 2)2 • 0 x2 í 2x + 1 + 3x + 14 í 5(x2 í 4x + 4) • 0 x2 + x + 15 í 5x2 + 20x í 20 • 0 í4x2 + 21x í 5 • 0 f f(x) f (x) = 0 geeft í4x2 + 21x í 5 = 0 D = 441 í 4  í4  í5 = 361 í21 í 19 í21 + 19 x= x= í8 í8 x = 5  x = 14

x 1 4

112

5

x

ƒ

f (x) • 0 geeft a7 a x2 í 2x冑5 + 5 > 0 f f(x) f (x) = 0 geeft x2 í 2x冑5 + 5 = 0 (x í 冑5 )2 = 0 x í 冑5 = 0 x = 冑5

1 4

”x”5

d

u

c 9x2 í 24x + 16 ” 0 f f(x) f (x) = 0 geeft 9x2 í 24x + 16 = 0 (3x í 4)2 = 0 3x í 4 = 0 3x = 4 x = 113

ƒ

ƒ

x

5

1

f (x) > 0 geeft x < 冑5  x > 冑5 ofwel x  冑5. b x2 í 40 ” 0 x2 ” 40 geeft í冑40 ” x ” 冑40, ofwel í2冑10 ” x ” 2冑10.

13

x

f (x) ” 0 geeft x = 113 d (2x í 5)2 í (x í 7)2 í 2x(x í 3) • 0 4x2 í 20x + 25 í (x2 í 14x + 49) í 2x2 + 6x • 0 4x2 í 20x + 25 í x2 + 14x í 49 í 2x2 + 6x • 0 x2 í 24 • 0 x2 • 24 geeft x ” í冑24  x • 冑24 x ” í2冑6  x • 2冑6

Bladzijde 99

c

a8 a D = p2 í 4  1  3p = p2 í 12p r p2 í 12p > 0 twee oplossingen als D > 0 f f(p) f (p) = 0 geeft p2 í 12p = 0 p(p í 12) = 0 p = 0  p = 12 ƒ

0

12

p

De vergelijking heeft twee oplossingen voor p < 0  p > 12.

68

Hoofdstuk 3

© Noordhoff Uitgevers bv

c

b D = (í2p)2 í 4  1  p = 4p2 í 4p r 4p2 í 4p < 0 geen oplossingen als D < 0 f f(p) f (p) = 0 geeft 4p2 í 4p = 0 4p(p í 1) = 0 p=0p=1 ƒ 0

p

1

De vergelijking heeft geen oplossingen voor 0 < p < 1. c D = p2 í 4 Â 1 Â í3 = p2 + 12 r p2 + 12 > 0 twee oplossingen als D > 0 Voor elke waarde van p is p2 + 12 > 0. De vergelijking heeft twee oplossingen voor elke waarde van p. a9 a

3

x

ƒp

s

Er moet gelden D < 0 p2 + 8p < 0 D = p2 í 4  í1  2p = p2 + 8p r g (p) f g (p) = 0 geeft p2 + 8p = 0 p(p + 8) = 0 p = 0  p = í8 g −8

b

p

0

fp heeft een negatief maximum voor í8 < p < 0. y = 12

x ƒp

t

Er moet gelden, dat de vergelijking fp(x) = 12 twee oplossingen heeft. fp(x) = 12 geeft íx2 + px + 2p í 12 = 0 D = p2 í 4  í1  (2p í 12) = p2 + 8p í 48 r p2 + 8p í 48 > 0 twee oplossingen als D > 0 f g (p) g (p) = 0 geeft p2 + 8p í 48 = 0 (p + 12)(p í 4) = 0 p = í12  p = 4 g −12

4

p

Het maximum van fp is groter dan 12 voor p < í12  p > 4.

© Noordhoff Uitgevers bv

Vergelijkingen en herleidingen

69

3.1 Hogeregraadsvergelijkingen Bladzijde 100

3

a1 a x3 = 10 heeft één oplossing, omdat de gra¿ek van f en de lijn y = 10 één snijpunt hebben. x3 = í10 heeft één oplossing, omdat de gra¿ek van f en de lijn y = í10 één snijpunt hebben. b x4 = 10 heeft twee oplossingen, omdat de grafiek van g en de lijn y = 10 twee snijpunten hebben. x4 = í10 heeft geen oplossingen, omdat de grafiek van g en de lijn y = í10 geen snijpunten hebben. Bladzijde 101

a2 a x

x2

x3

x4

x5

x6

1

1

1

1

1

1

2

4

8

16

32

64

3

9

27

81

243

729

4

16

64

256

1024

5

25

125

625

6

36

216

b * Bladzijde 102

a3 a

1 3 4x

+ 60 = 6

1 3 4x x3

= í54 = í216 x = í6 b 100 í 3x4 = 55 í3x4 = í45 x4 = 15 4 15 4 15 x =冑  x = í冑

c

1 2 (4x

í 1)5 + 3 = 19

1 2 (4x

í 1)5 = 16 (4x í 1)5 = 32 4x í 1 = 2 4x = 3 x = 34 d 212 (1 í 2x)6 í 6 = 14 212 (1 í 2x)6 = 20 (1 í 2x)6 = 8 6 6 8  1 í 2x = í冑 8 1 í 2x = 冑 68 68 í2x = í1 + 冑  í2x = í1 í 冑 6 6 x = 12 í 12  冑 8  x = 12 + 12  冑 8

a4 a x6 = 20 6 20 6 20 x =冑  x = í冑 3 b 5x = 135 x3 = 27 x=3

c 1 í 3x5 = 97 í3x5 = 96 x5 = í32 x = í2 d 14 x8 + 3 = 10 1 8 4x x8

=7 = 28 8 28 8 28 x =冑  x = í冑

a5 a 5x4 í 1 = 4 5x4 = 5 x4 = 1 x = 1  x = í1

70

Hoofdstuk 3

b 5x3 í 1 = 9 5x3 = 10 x3 = 2 32 x =冑

© Noordhoff Uitgevers bv

c 8x3 + 2 = 1 8x3 = í1 x3 = í 18 x = í 12 a6 a 3(x í 2)4 + 7 = 37 3(x í 2)4 = 30 (x í 2)4 = 10 4 4 x í 2 =冑 10  x í 2 = í冑 10 4 4 x = 2 +冑 10  x = 2 í 冑 10 1 5 b 100 í 3 (4x í 3) = 19 í 13 (4x í 3)5 = í81 (4x í 3)5 = 243 4x í 3 = 3 4x = 6 x = 112

a7 a 5x4 í 3 = 17 5x4 = 20 x4 = 4 4 4 x =冑 4  x = í冑 4 3 b 4x í 5 = 1367 4x3 = 1372 x3 = 343 x=7

d 5x6 + 7 = 97 5x6 = 90 x6 = 18 6 6 x =冑 18  x = í冑 18 1 2 (3x

í 1)4 = 8 (3x í 1)4 = 16 3x í 1 = 2  3x í 1 = í2 3x = 3  3x = í1 x = 1  x = í 13 d 6 í (2x í 1)3 = 1 í(2x í 1)3 = í5 (2x í 1)3 = 5 3 5 2x í 1 = 冑 c

3

3 2x = 1 + 冑 5 35 x = 12 + 12 Â 冑

c 3(4x í 5)3 = 15 (4x í 5)3 = 5 35 4x í 5 = 冑 3 4x = 5 + 冑 5 35 x = 114 + 14  冑 d 17 í 2(1 í 3x)4 = 5 í2(1 í 3x)4 = í12 (1 í 3x)4 = 6 46 46 1 í 3x = 冑  1 í 3x = í冑 4 4 6  3x = í1 í 冑 6 í3x = í1 + 冑 4 4 x = 13 í 13  冑 6  x = 13 + 13  冑 6

a8 a x3 í x2 í 2x = 0 x(x2 í x í 2) = 0

b x(x2 í x í 2) = 0 x(x + 1)(x í 2) = 0 x = 0  x = í1  x = 2

Bladzijde 103

a9 a x3 í 5x2 + 6x = 0 x(x2 í 5x + 6) = 0 x(x í 2)(x í 3) = 0 x=0x=2x=3 b x3 í 5x2 = 6x x3 í 5x2 í 6x = 0 x(x2 í 5x í 6) = 0 x(x + 1)(x í 6) = 0 x = 0  x = í1  x = 6

10 a

a x4 í 10x2 + 9 = 0 Stel x2 = u. u2 í 10u + 9 = 0 (u í 1)(u í 9) = 0 u=1u=9 x2 = 1  x2 = 9 x = 1  x = í1  x = 3  x = í3

© Noordhoff Uitgevers bv

c x3 = 4x2 + 12x x3 í 4x2 í 12x = 0 x(x2 í 4x í 12) = 0 x(x + 2)(x í 6) = 0 x = 0  x = í2  x = 6 d x4 í 13x2 + 36 = 0 Stel x2 = u. u2 í 13u + 36 = 0 (u í 4)(u í 9) = 0 u=4u=9 x2 = 4  x2 = 9 x = 2  x = í2  x = 3  x = í3 b x4 í 8x2 í 9 = 0 Stel x2 = u. u2 í 8u í 9 = 0 (u í 9)(u + 1) = 0 u = 9  u = í1 x2 = 9  x2 = í1 x = 3  x = í3

Vergelijkingen en herleidingen

71

c x4 + 16 = 10x2 x4 í 10x2 + 16 = 0 Stel x2 = u. u2 í 10u + 16 = 0 (u í 2)(u í 8) = 0 u=2u=8 x2 = 2  x2 = 8 x = 冑2  x = í冑2  x = 2冑2  x = í2冑2 3

11 a

f (x) =

(x2 + 2)  í10 í í10x  2x í10x2 í 20 + 20x2 10x2 í 20 í10x (x) geeft f ‫މ‬ = = 2 = x2 + 2 (x2 + 2)2 (x2 + 2)2 (x + 2)2

f '(x) = 59 geeft

12 a

d x3 + 25x = 10x2 x3 í 10x2 + 25x = 0 x(x2 í 10x + 25) = 0 x(x í 5)2 = 0 x=0x=5

10x2 í 20 5 = (x2 + 2)2 9 2 9(10x í 20) = 5(x2 + 2)2 90x2 í 180 = 5(x4 + 4x2 + 4) 90x2 í 180 = 5x4 + 20x2 + 20 5x4 í 70x2 + 200 = 0 x4 í 14x2 + 40 = 0 Stel x2 = u. u2 í 14u + 40 = 0 (u í 10)(u í 4) = 0 u = 10  u = 4 x2 = 10  x2 = 4 x = 冑10  x = í冑10  x = 2  x = í2

2

a 2x4 í 11x + 12 = 0 Stel x2 = u. 2u2 í 11u + 12 = 0 D = (í11)2 í 4  2  12 = 25 11 + 5 11 í 5 u= = 121  u = =4 4 4 b 2x4 í 11x2 + 12 = 0 x2 = 112  x2 = 4

冑

冑

x = 112  x = í 112  x = 2  x = í2 13 a

a 6x4 + 2 = 7x2 6x4 í 7x2 + 2 = 0 Stel x2 = u. 6u2 í 7u + 2 = 0 D = (í7)2 í 4  6  2 = 1 7+1 2 7í1 1 u= =2 u= =3 12 12 x2 = 12  x2 = 23 x=

冑

1 2

x=í

冑

1 2

x=

冑

2 3

x=í

b 2x4 = x2 + 3 2x4 í x2 í 3 = 0 Stel x2 = u. 2u2 í u í 3 = 0 D = (í1)2 í 4  2  í3 = 25 1+5 1í5 u= = í1  u = = 112 4 4 x2 = í1  x2 = 112

冑

冑

x = 112  x = í 112

72

Hoofdstuk 3

冑

2 3

c 4x4 + 7x2 = 2 4x4 + 7x2 í 2 = 0 Stel x2 = u. 4u2 + 7u í 2 = 0 D = 72 í 4  4  í2 = 81 í7 í 9 í7 + 9 1 u= = í2  u = =4 8 8 x2 = í2  x2 = 14 x = 12  x = í 12 d 16x4 + 225 = 136x2 16x4 í 136x2 + 225 = 0 Stel x2 = u. 16u2 í 136u + 225 = 0 D = (í136)2 í 4  16  225 = 4096 136 + 64 136 í 64 u= = 214  u = = 614 32 32 x2 = 214  x2 = 614 x = 112  x = í112  x = 212  x = í212

© Noordhoff Uitgevers bv

Bladzijde 105 14 a

a 4x4 + 153 = 53x2 4x4 í 53x2 + 153 = 0 Stel x2 = u. 4u2 í 53u + 153 = 0 D = (í53)2 í 4  4  153 = 361 53 + 19 53 í 19 u= = 414  u = =9 8 8 x2 = 414  x2 = 9

冑

冑

x = 414  x = í 414  x = 3  x = í3 4x4

3

21x2

+ = 148 4x4 + 21x2 í 148 = 0 Stel x2 = u. 4u2 + 21u í 148 = 0 D = 212 í 4  4  í148 = 2809 í21 í 53 í21 + 53 u= = í914  u = =4 8 8 x2 = í914  x2 = 4 x = 2  x = í2 c 4x6 + 35 = 24x3 4x6 í 24x3 + 35 = 0 Stel x3 = u. 4u2 í 24u + 35 = 0 D = (í24)2 í 4  4  35 = 16 24 + 4 24 í 4 u= = 212  u = = 312 8 8 x3 = 212  x3 = 312 b

3 1 3 1 x =冑 22  x = 冑 32

d 64x5 + 27x = 224x3 64x5 í 224x3 + 27x = 0 x(64x4 í 224x2 + 27) = 0 x = 0  64x4 í 224x2 + 27 = 0 Stel x2 = u. 64u2 í 224u + 27 = 0 D = (í224)2 í 4  64  27 = 43264 224 í 208 1 224 + 208 u= = 338  u = =8 128 128 3 1 2 2 x = 38  x = 8

冑

冑

x = 338  x = í 338  x = 15 a

冑

1 8

 x=í

冑

1 8

a * b “E = “ B = 90° + ABC ऴ + FEC “C (in +ABC) = “ C (in +FEC) r CE EF x 1 geeft = = CB AB x + 1 1 + AD x(1 + AD) = x + 1 x+1 1 + AD = x x+1 1 1 AD = x í 1 = 1 + x í 1 = x

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73

c Uit de stelling van Pythagoras in +ABC volgt BC2 + AB2 = AC2

( )

1 (1 + x)2 + 1 + x

1 + 2x + x2 + 1 +

2

= 16

2 1 + = 16 x x2

x2 + 2x + 2 +

2 1 + = 16 x x2

d x2 + 2x + 2 +

2 1 + = 16 x x2

3

x2 + 2 +

1 2 + 2x + = 16 2 x x

x2 + 2 +

1 1 +2 x+ = 16 x x2

( ) ( )

1 1 Stel x + = u, dus x + x x

2

1 = x2 + 2 + 2 = u2 x

u

u2 + 2u = 16

e u2 + 2u = 16 u2 + 2u í 16 = 0 D = 22 í 4  1  í16 = 68 í2 + 冑68 í2 í 冑68 u= = 冑17 í 1  u = = í冑17 í 1 2 2 vold. vold. niet 1 x + x = 冑17 í 1 x2 + 1 = (冑17 í 1) x x2 í (冑17 í 1) x + 1 = 0 2 D = (冑17 í 1) í 4  1  1 = 14 í 2冑17 x=

冑17 í 1 + 冑14 í 2冑17

 x=

冑17 í 1 í 冑14 í 2冑17

2 2 x = 2,760...  x = 0,362... Dus de ladder staat 100 + 276 = 376 cm of 100 + 36 = 136 cm hoog tegen het huis. 16

a x3 í 12x = x2 x3 í x2 í 12x = 0 x(x2 í x í 12) = 0 x(x + 3)(x í 4) = 0 x = 0  x = í3  x = 4 y b g

–3

O

4

x

ƒ

c x3 í 12x < x2 geeft x < í3  0 < x < 4

74

Hoofdstuk 3

© Noordhoff Uitgevers bv

Bladzijde 106 17

a Stel f (x) = x3 en g(x) = 2x2 + 8x. f (x) = g(x) geeft x3 = 2x2 + 8x x3 í 2x2 í 8x = 0 x(x2 í 2x í 8) = 0 x(x + 2)(x í 4) = 0 x = 0  x = í2  x = 4 y

ƒ

g

c Stel f (x) = x4 en g(x) = 3x2 + 10. f (x) = g(x) geeft x4 = 3x2 + 10 x4 í 3x2 í 10 = 0 Stel x2 = u. u2 í 3u í 10 = 0 (u + 2)(u í 5) = 0 u = í2  u = 5 x2 = í2  x2 = 5 x = 冑5  x = í冑5 y

−2 O

4

ƒ

–3 O

1

3

g

x

x3 • 2x2 + 8x geeft í2 ” x ” 0  x • 4 b Stel f (x) = x3 + 2x2 en g(x) = 3x. f (x) = g(x) geeft x3 + 2x2 = 3x x3 + 2x2 í 3x = 0 x(x2 + 2x í 3) = 0 x(x + 3)(x í 1) = 0 x = 0  x = í3  x = 1 y

ƒ

g

x

− 5

x

5

x4 • 3x2 + 10 geeft x ” í冑5  x • 冑5 d Stel f (x) = 23 x4 + 9 en g(x) = 7x2. f (x) = g(x) geeft 23 x4 + 9 = 7x2 2x4 + 27 = 21x2 2x4 í 21x2 + 27 = 0 Stel x2 = u. 2u2 í 21u + 27 = 0 D = (í21)2 í 4  2  27 = 225 21 + 15 21 í 15 u= = 112  u = =9 4 4 x2 = 112  x2 = 9

冑

冑

x = 112  x = í 112  x = 3  x = í3 g

y

ƒ

x3 + 2x2 < 3x geeft x < í3  0 < x < 1

−3

2 4 3x

© Noordhoff Uitgevers bv

− 1 12

1 12

3

冑

x

+ 9 ” 7x2 geeft í3 ” x ” í 112 

冑1

1 2

”x”3

Vergelijkingen en herleidingen

75

18

Een negatief minimum, dus de gra¿ek heeft twee snijpunten met de x-as. ƒp

x

( )

5 p + p x2 + p2x + 4p = 0

Er moet gelden D > 0. 5 p4 í 16p2 í 80 > 0 D = (p2)2 í 4  p + p  4p = p4 í 16p2 í 80 ¶ f f (p) u

( )

3

f (p) = 0 geeft p4 í 16p2 í 80 = 0 Stel p2 = u. u2 í 16u í 80 = 0 (u í 20)(u + 4) = 0 u = 20  u = í4 p2 = 20  p2 = í4 p = 冑20 = 2冑5  p = í冑20 = í = 2冑5 ƒ

–2 5

O

2 5

p

Gegeven was fp met p > 0, dus fp heeft een negatief minimum voor p > 2冑5. Bladzijde 107 19

x = í1 geeft 0 4 Â í1 í 5 0 = 9 0 í4 í 5 0 = 9 0 í9 0 = 9 klopt! x = 312 geeft 0 4 Â 312 í 5 0 = 9 0 14 í 5 0 = 9 090 = 9 klopt! Dus x = í1 en x = 312 zijn beiden oplossing van de vergelijking.

20

a 0 2x í 1 0 = 8 2x í 1 = 8  2x í 1 = í8 2x = 9 2x = í7 x = 412  x = í312 b 0 x2 í 3 0 = 1 x2 í 3 = 1  x2 í 3 = í1 x2 = 4  x2 = 2 x = 2  x = í2  x = 冑2  x = í冑2

76

Hoofdstuk 3

c 0 2x2 í 5 0 = 11 2x2 í 5 = 11  2x2 í 5 = í11 2x2 = 16  2x2 = í6 x2 = 8  x2 = í3 x = 2冑2  x = í2冑2 d 0 5 í x2 0 = 11 5 í x2 = 11  5 í x2 = í11 íx2 = 6  íx2 = í16 x2 = í6  x2 = 16 x = 4  x = í4

© Noordhoff Uitgevers bv

21

a 0 2x4 í 5 0 = 15 2x4 í 5 = 15  2x4 í 5 = í15 2x4 = 20  2x4 = í10 x4 = 10  x4 = í5 4 10 4 10 x =冑  x = í冑 3 b 0 2x í 5 0 = 15 2x3 í 5 = 15  2x3 í 5 = í15 2x3 = 20  2x3 = í10 x3 = 10  x3 = í5 3 10 35 x =冑  x = í冑 4 2 c 0 x í 5x 0 = 6 x4 í 5x2 = 6  x4 í 5x2 = í6 x4 í 5x2 í 6 = 0  x4 í 5x2 + 6 = 0 Stel x2 = u. u2 í 5u í 6 = 0  u2 í 5u + 6 = 0 (u í 6)(u + 1) = 0  (u í 2)(u í 3) = 0 u = 6  u = í1  u = 2  u = 3 x2 = 6  x2 = í1  x2 = 2  x2 = 3 x = 冑6  x = í冑6  x = 冑2  x = í冑2  x = 冑3  x = í冑3 d 0 x6 í 10x3 0 = 24 x6 í 10x3 = 24  x6 í 10x3 = í24 x6 í 10x3 í 24 = 0  x6 í 10x3 + 24 = 0 Stel x3 = u. u2 í 10u í 24 = 0  u2 í 10u + 24 = 0 (u í 12)(u + 2) = 0  (u í 4)(u í 6) = 0 u = 12  u = í2  u = 4  u = 6 x3 = 12  x3 = í2  x3 = 4  x3 = 6 3 12 3 34 36 x =冑 í2  x = 冑  x =冑  x =冑

3

3.2 Stelsels vergelijkingen Bladzijde 109 22

Uit 7x + 4y = 1 volgt y = í134 x + 14 . y = í134 x + 14 en 5x í 4y = 13 geeft 5x í 4 (í134 x + 14 ) = 13 5x + 7x í 1 = 13 12x = 14 x = 116 x = 116 en y = í134 x + 14 geeft y = í134 Â 116 + 14 = í119 24 . De oplossing is (x, y) = (116 , í119 24 ) . Bladzijde 110

23

5x í 4y = í8 íx + 4y = í12 + 4x = í20 x = í5 r 5 + 4y = í12 íx + 4y = í12 4y = í17 y = í414 1 Dus (x, y) = (í5,í44 ) .

a b

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77

í2x + y = 7 í2x + 3y = í1 í í2y = 8 y = í4 r í2x í 4 = 7 í2x + y = 7 í2x = 11 x = í512 1 Dus (x, y) = (í52 , í4) . ௘í x í 3y = í8 c b í2x + 3y = í1 + í3x = í9 x=3 r í2 Â 3 + 3y = í1 í2x + 3y = í1 í6 + 3y = í1 3y = 5 y = 123 2 Dus (x, y) = (3, 13 ) . b b

3

24 a

3x í 4y = 7 2x + 3y = 16 + 5x í y = 23 Nee, er is geen variabele geëlimineerd. 3x í 4y = 7 b b 2x + 3y = 16 í x í 7y = í 9 Nee, er is geen variabele geëlimineerd.

a b

Bladzijde 111 25 a

4x í y = 13 3 12x í 3y = 39 geeft e 2x + 3y = í11 1 2x + 3y = í11 + 14x = 28 x=2 2 Â 2 + 3y = í11 2x + 3y = í11 r 4 + 3y = í11 3y = í15 y = í5 Dus (x, y) = (2, í5). 3x í 2y = 7 ௘ 2 6x í 4y = 14 b b geeft b 5x í 4y = 10 1 5x í 4y = 10 í x = 4 f 3 Â 4 í 2y = 7 3x í 2y = 7 12 í 2y = 7 í2y = í5 y = 212 Het snijpunt is (4, 212 ) . c Stel de aannemer bouwt x huizen van type A en y huizen van type B. Hieruit volgt dat 325x + 175y = 8800 1 325x + 175y = 8800 b geeft b x + y = 40 175 175x + 175y = 7000 í 150x = 1800 x = 12 De aannemer gaat 12 huizen van type A bouwen. a b

0 0

0 0

0 0

78

Hoofdstuk 3

© Noordhoff Uitgevers bv

Bladzijde 112 26 a

a e

3x + 5y = í7 2x + y = 0

0 51 0 geeft

e

௘3x + 5y = í7 10x + 5y = 0 í í7x = í7 x=1 2Â1 + y = 0 2x + y = 0 r 2+y=0 y = í2

Dus (x, y) = (1, í2). 2x í 4y = 6 1 ௘2x í 4y = 6 b e geeft e 3x í y = 19 4 12x í 4y = 76 í í10x = í70 x=7 2 Â 7 í 4y = 6 2x í 4y = 6 r 14 í 4y = 6 í4y = í8 y=2 Dus (x, y) = (7, 2). 4x + y = 13 2 8x + 2y = 26 c e geeft e x í 2y = 1 1 x í 2y = 1 + 9x = 27 x=3 4 Â 3 + y = 13 4x + y = 13 r 12 + y = 13 y=1 Dus (x, y) = (3, 1).

0 0

3

0 0

27 a

3x í 2y = í12 2 6x í 4y = í 24 geeft e x + 4y = 38 1 x + 4y = 38 + 7x = 14 x=2 2 + 4y = 38 x + 4y = 38 r 4y = 36 y=9 Het snijpunt is (2, 9). 2x + 5y = 26 3 6x + 15y = 78 b e geeft e 3x í 2y = 1 2 6x í 4y = 2 í 19y = 76 y=4 2x + 5 Â 4 = 26 2x + 5y = 26 r 2x + 20 = 26 2x = 6 x=3 Het snijpunt is (3, 4). a e

0 0

0 0

28 a

a Stel er zitten x muntstukken van 1 euro en y muntstukken van 2 euro in zijn spaarpot. Hieruit volgt dat x + y = 50 en x + 2y = 87. x + 2y = 87 b e x + y = 50 í y = 37 x = 13 x + y = 50 r Daan heeft 13 munten van 1 euro.

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79

29 a

Stel de groenteman verkoopt x appels en y peren. Hieruit volgt dat 1,4x + 1,7y = 452 10 14x + 17y = 4520 geeft e e x + y = 295 17 17x + 17y = 5015 í í3x = í495 x = 165 Hij verkoopt die dag 165 kg appels.

0 0

30 a 3

5x + 2y = 69 1 5x + 2y = 69 geeft e x + 3y = í7 5 5x + 15y = í35 í í13y = 104 y = í8 x + 3 Â í8 = í7 x + 3y = í7 r x í 24 = í7 x = 17 Dus (x, y) = (17, í8). 2x í 5y = í19 4 8x í 20y = í76 geeft e b e 5x + 4y = 35 5 25x + 20y = 175 + 33x = 99 x=3 2 Â 3 í 5y = í19 2x í 5y = í19 r 6 í 5y = í19 í5y = í25 y=5 Dus (x, y) = (3, 5). 0,8x + 0,2y = 1 ௘3 2,4x + 0,6y = 3 geeft e c e 0,3x í 0,3y = 1,5 2 0,6x í 0,6y = 3 + 3x =6 x=2 0,8 Â 2 + 0,2y = 1 0,8x + 0,2y = 1 r 1,6 + 0,2y = 1 0,2y = í0,6 y = í3 Dus (x, y) = (2, í3). a e

0 0

0 0

0 0

3x í 2y = í12 2 6x í 4y = í24 geeft e x + 4y = 38 1 x + 4y = 38 + 7x = 14 x=2 2 + 4y = 38 x + 4y = 38 r 4y = 36 y=9 Het snijpunt is (2, 9).

0 0

31 a

a e

32 a

Stel er zijn x jongens en y meisjes. De totale leeftijd van de jongens is dan 15,6x en van de meisjes is 16,8y. De totale leeftijd van alle personen samen is 15 Â 16,4 = 246. Hieruit volgt dat 168x + 168y = 2520 168 x + y = 15 geeft e e 15,6x + 16,8y = 246 10 156x + 168y = 2460 í 12x = 60 x=5 Er zijn dus vijf jongens op de verjaardag.

0 0

80

Hoofdstuk 3

© Noordhoff Uitgevers bv

33 a

34 a

Stel de rechthoek heeft een lengte van x en een breedte van y. Voor omtrek rechthoek geldt 2x + 2y = 26. Voor omtrek vijf rechthoeken geldt 2x + 10y = 50. 2x + 10y = 50 e 2x + 2y = 26 í 8y = 24 y = 3 2x + 2 Â 3 = 26 2x + 2y = 26 r 2x + 6 = 26 2x = 20 x = 10 De eerste rechthoek heeft een lengte van 10 en een breedte van 3.

3

3x + 2y = 18 geeft y = í112 x + 9 en 6x + 4y = 15 geeft y = í112 x + 334 . Bij deze vergelijkingen horen twee evenwijdige lijnen. Deze lijnen hebben geen snijpunten, dus heeft het stelsel geen oplossingen. Bladzijde 113

35 a

a y = x2 + bx + c r 12 + b  1 + c = í2 door (1, í2) 1 + b + c = í2 b + c = í3 b y = x2 + bx + c r 22 + b  2 + c = 3 door (2, 3) 4 + 2b + c = 3 2b + c = í1 b + c = í3 c e 2b + c = í1 í íb = í2 b=2 2 + c = í3 b + c = í3 r c = í5 Dus b = 2 œ c = í5.

36 a

(1, 8) invullen geeft a  12 + c = 8 dus a + c = 8. (2, 17) invullen geeft a  22 + c = 17 dus 4a + c = 17. a+c = 8 e 4a + c = 17 í í3a = í9 a =3 3+c=8 a + c = 8r c=5 Dus y = 3x2 + 5.

37 a

(3, 9) invullen bij k geeft a  3 + b = 9 dus 3a + b = 9. (3, 9) invullen bij l geeft íb  3 + 9a = 9 dus 9a í 3b = 9. 3a + b = 9 3 9a + 3b = 27 geeft e e 9a í 3b = 9 9a í 3b = 9 1 + 18a = 36 a=2 3Â2 + b = 9 3a + b = 9 r 6+b=9 b=3 Dus a = 2 œ b = 3.

0 0

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81

38 a

3

a (2, í1) invullen in y = x2 + px + q geeft 22 + p  2 + q = í1 dus 2p + q = í5. (2, í1) invullen in y = 2px í q geeft 2p  2 í q = í1 dus 4p í q = í1. 2p + q = í5 e 4p í q = í1 + 6p = í6 p = í1 2  í1 + q = í5 2p + q = í5 r í2 + q = í5 q = í3 Dus p = í1 œ q = í3. b y = x2 í x í 3 2 x í x í 3 = í2x + 3 y = í2x + 3 r 2 x +xí6=0 (x + 3)(x í 2) = 0 x = í3  x = 2 x = í3 invullen geeft y = í2  í3 + 3 = 9. Het andere snijpunt is (í3, 9). Bladzijde 114

39 a

(í2, í10) invullen geeft a  (í2)2 + b  í2 + c = í10 dus 4a í 2b + c = í10. (0, 4) invullen geeft a  02 + b  0 + c = 4 dus c = 4. (3, 5) invullen geeft a  32 + b  3 + c = 5 dus 9a + 3b + c = 5. c=4 4a í 2b = í14 4a í 2b + c = í10 r c=4 9a + 3b = 1 9a + 3b + c = 5 r 4a í 2b = í14 3 12a í 6b = í 42 geeft e 2 9a + 3b = 1 18a + 6b = 2 + 30a = í40 1 a = í13 r 9  í113 + 3b = 1 9a + 3b = 1 í12 + 3b = 1 3b = 13 b = 413 1 2 1 Dus y = í13 x + 43 x + 4.

e

0 0

40 a

f (x) = ax2 + bx + c geeft f ‫(މ‬x) = 2ax + b. De lijn k met rck = 1 raakt de grafiek van f in (2, 6), dus f ‫(މ‬2) = 2a  2 + b = 4a + b = 1. De lijn l met rcl = í2 raakt de grafiek van f in (8, 3), dus f ‫(މ‬8) = 2a  8 + b = 16a + b = í2. 16a + b = í2 e 4a + b = 1 í 12a = í3 1 a = í4 4  í 14 + b = 1 4a + b = 1 r í1 + b = 1 b=2 y = í 14 x2 + 2x + c r 1 2 í4  2 + 2  2 + c = 6 door (2, 6) í1 + 4 + c = 6 3+c=6 c=3 f (8) = í 14  82 + 2  8 + 3 = 3, dus f gaat ook door (8, 3). Dus a = í 14 , b = 2 en c = 3.

41 a

Nee.

82

Hoofdstuk 3

© Noordhoff Uitgevers bv

42 a

a Substitutie van y = x2 í 18 in x í y = 2 geeft x í (x2 í 18) = 2 x í x2 + 18 = 2 íx2 + x + 16 = 0 x2 í x í 16 = 0 D = (í1)2 í 4  1  í16 = 65 1 í 冑65 1 1 1 + 冑65 1 1 x= = 2 í 2 冑65  x = = 2 + 2 冑65 2 2 1 1 1 1 1 1 x = 2 í 2 冑65 geeft y = 2 í 2 冑65 í 2 = í12 í 2 冑65 x = 12 + 12 冑65 geeft y = 12 + 12 冑65 í 2 = í112 + 12 冑65 Dus (x, y) =

( 12 í 12 冑65, í112 í 12 冑65 )

 (x, y) =

( 12 + 12 冑65, í112 + 12 冑65 )

3

b Substitutie van y = x2 í 3 in x í y = í3 geeft x í (x2 í 3) = í3 x í x2 + 3 = í3 íx2 + x + 6 = 0 x2 í x í 6 = 0 (x í 3)(x + 2) = 0 x = 3  x = í2 x = 3 geeft y = 32 í 3 = 6 x = í2 geeft y = (í2)2 í 3 = 1 Dus (x, y) = (3, 6)  (x, y) = (í2, 1). c 3x + y = 5 geeft y = í3x + 5 Substitutie van y = í3x + 5 in x2 + y2 = 25 geeft x2 + (í3x + 5)2 = 25 x2 + 9x2 í 30x + 25 = 25 10x2 í 30x = 0 10x(x í 3) = 0 x=0x=3 x = 0 geeft y = í3  0 + 5 = 5 x = 3 geeft y = í3  3 + 5 = í4 Dus (x, y) = (0, 5)  (x, y) = (3, í4). 43 a

a 3x + 2y = 10 geeft 2y = 10 í 3x 4y2 = (10 í 3x)2 2 Substitutie van 4y = (10 í 3x)2 in x2 + 4y2 = 100 geeft x2 + (10 í 3x)2 = 100 x2 + 100 í 60x + 9x2 = 100 10x2 í 60x = 0 10x(x í 6) = 0 x=0x=6 x = 0 geeft 2y = 10 í 3  0 = 10 ofwel y = 5. x = 6 geeft 2y = 10 í 3  6 = í8 ofwel y = í4. Dus (x, y) = (0, 5)  (x, y) = (6, í4). b 2x + y = 4 geeft y = 4 í 2x Substitutie van y = 4 í 2x in (x í 3)2 + y2 = 8 geeft (x í 3)2 + (4 í 2x)2 = 8 x2 í 6x + 9 + 16 í 16x + 4x2 = 8 5x2 í 22x + 17 = 0 D = (í22)2 í 4  5  17 = 144 22 í 12 22 + 12 x= =1x= = 325 10 10 x = 1 geeft y = 4 í 2  1 = 2 x = 325 geeft y = 4 í 2  325 = í245 Dus (x, y) = (1, 2)  (x, y) = (325 , í245 ) .

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Vergelijkingen en herleidingen

83

c x + 4y = 9 geeft x = 9 í 4y Substitutie van x = 9 í 4y in (x + 2)2 + (y í 3)2 = 10 geeft (9 í 4y + 2)2 + (y í 3)2 = 10 (11 í 4y)2 + (y í 3)2 = 10 121 í 88y + 16y2 + y2 í 6y + 9 = 10 17y2 í 94y + 120 = 0 D = (í94)2 í 4  17  120 = 676 94 + 26 94 í 26 9 y= =2y= = 317 34 34 y = 2 geeft x = 9 í 4  2 = 1 9 9 2 geeft x = 9 í 4  317 y = 317 = í517

3

2 9 ). Dus (x, y) = (1, 2)  (x, y) = (í517 , 317

3.3 Regels voor het oplossen van vergelijkingen Bladzijde 116 44 a

a 5x(x2 í 4) = 15(x2 í 4) 5x3 í 20x = 15x2 í 60 5x3 í 15x2 í 20x + 60 = 0 Je kent geen methode om deze derdegraadsvergelijking algebraïsch op te lossen. b 5x = 15 geeft als oplossing x = 3 en je kunt zien dat bijvoorbeeld x = 2 ook een oplossing is van de gegeven vergelijking. Dus door links en rechts te delen door x2 í 4 gaan er oplossingen verloren, omdat x2 í 4 nul kan zijn. c 5x(x2 í 4) = 15(x2 í 4) x2 í 4 = 0  5x = 15 x2 = 4  x = 3 x = 2  x = í2  x = 3 Bladzijde 118

45 a

a (4x í 1)2 = (3x í 2)2 4x í 1 = 3x í 2  4x í 1 = í3x + 2 x = í1  7x = 3 x = í1  x = 37 2 2 2 b (3x í 5) = 4x 3x2 í 5 = 2x  3x2 í 5 = í2x 3x2 í 2x í 5 = 0  3x2 + 2x í 5 = 0 2 D = 22 í 4  3  í5 = 64 D = (í2) í 4  3  í5 = 64 2+8 2í8 í2 í 8 í2 + 8 x= x= x= x= 6 6 6 6 x = í1  x = 123  x = í123  x = 1 c (x2 í 4x)(x2 í 8) = 0 x2 í 4x = 0  x2 í 8 = 0 x(x í 4) = 0  x2 = 8 x = 0  x = 4  x = 冑8  x = í冑8 x = 0  x = 4  x = 2冑2  x = í2冑2

84

Hoofdstuk 3

d x3(x í 3) = 8(x í 3) x í 3 = 0  x3 = 8 x=3 x=2 e 2x(x2 í 4) = 6(x í 2) 2x(x + 2)(x í 2) = 6(x í 2) x í 2 = 0  2x(x + 2) = 6 x = 2  2x2 + 4x í 6 = 0 x = 2  x2 + 2x í 3 = 0 x = 2  (x + 3)(x í 1) = 0 x = 2  x = í3  x = 1 f x(x í 2)(x2 í 3) = 0 x = 0  x = 2  x2 = 3 x = 0  x = 2  x = 冑3  x = í冑3

© Noordhoff Uitgevers bv

46 a

a Voer in y1 = (3x + 4)(x í 2)3. y

ƒ

O

x

3

Optie minimum geeft x § í0,5 en y § í39,1, dus B f = Cí 39,1; m 9. b (3x + 4)(x í 2)3 = 0 3x + 4 = 0  x í 2 = 0 3x = í4  x = 2 x = í113  x = 2 Dus de nulpunten zijn x = í113 en x = 2. c f (x) = (3x + 4)(x í 2)3 r (3x + 4)(x í 2)3 = 3x + 4 y = 3x + 4 3x + 4 = 0  (x í 2)3 = 1 x = í113  x í 2 = 1 x = í113  x = 3 x = í113 geeft y = 3  í113 + 4 = 0, dus A (í113 , 0) . x = 3 geeft y = 3  3 + 4 = 13, dus B(3, 13). d f (x) = (3x + 4)(x í 2)3 r (3x + 4)(x í 2)3 = (3x + 4)(x í 2) y = (3x + 4)(x í 2) 3x + 4 = 0  x í 2 = 0  (x í 2)2 = 1 3x = í4  x = 2  x í 2 = 1  x í 2 = í1 x = í113  x = 2  x = 3  x = 1 x = í113 geeft y = (3  í113 + 4) (í113 í 2) = 0 x = 1 geeft y = (3  1 + 4)(1 í 2) = í7 x = 2 geeft y = (3  2 + 4)(2 í 2) = 0 x = 3 geeft y = (3  3 + 4)(3 í 2) = 13 Dus de snijpunten zijn (í113 , 0) , (1, í7), (2, 0) en (3, 13). 47 a

a Kruiselings vermenigvuldigen geeft x  x = 2(x + 4) x2 = 2x + 8 x2 í 2x í 8 = 0 b x2 í 2x í 8 = 0 (x í 4)(x + 2) = 0 x = 4  x = í2

© Noordhoff Uitgevers bv

Vergelijkingen en herleidingen

85

Bladzijde 120 48 a

a

xí3 = 112 x+1

xí3 3 = x+1 2 (x í 3) Â 2 = (x + 1) Â 3 2x í 6 = 3x + 3 íx = 9 x = í9 voldoet xí1 b x +1=3 xí1 x =2 xí1 2 x =1 x í 1 = 2x íx = 1 x = í1 voldoet

3

49 a

5x2 í 15 =0 x2 + 5 5x2 í 15 = 0 5x2 = 15 x2 = 3 x = 冑3  x = í冑3 vold. vold. 2 x í3 xí1 b 2 = x + 1 x2 + 1 x2 í 3 = x í 1 x2 í x í 2 = 0 (x í 2)(x + 1) = 0 x = 2  x = í1 vold. vold. a

3x2 í 10 =2 x2 + 1 3x2 í 10 = 2(x2 + 1) 3x2 í 10 = 2x2 + 2 x2 = 12 x = 冑12 = 2冑3  x = í冑12 = í2冑3 voldoet voldoet

50 a

a

86

Hoofdstuk 3

3x + 4 x + 18 = xí1 x (3x + 4)  x = (x í 1)(x + 18) 3x2 + 4x = x2 + 18x í x í 18 2x2 í 13x + 18 = 0 D = (í13)2 í 4  2  18 = 25 13 í 5 13 + 5 x= =2x= = 412 4 4 voldoet voldoet 2x í 5 x + 2 d = 4 í x 3x í 4 (2x í 5)(3x í 4) = (4 í x)(x + 2) 6x2 í 8x í 15x + 20 = 4x + 8 í x2 í 2x 7x2 í 25x + 12 = 0 D = (í25)2 í 4  7  12 = 289 25 í 17 4 25 + 17 x= =7 x= =3 14 14 voldoet voldoet c

x2 í 4 x2 í 4 = 2x + 5 x + 4 x2 í 4 = 0  2x + 5 = x + 4 x2 = 4  x = í1 x = 2  x = í2  x = í1 vold. vold. vold. x2 + 1 x + 3 d = x+1 x+1 x2 + 1 = x + 3 x2 í x í 2 = 0 (x í 2)(x + 1) = 0 x = 2  x = í1 vold. vold. niet c

b

x3 í 8 x3 í 8 = x2 + 2 x + 8 x3 í 8 = 0  x2 + 2 = x + 8 x3 = 8  x2 í x í 6 = 0 x = 2  (x í 3)(x + 2) = 0 x = 2  x = 3  x = í2 vold. vold. vold.

© Noordhoff Uitgevers bv

c

3x2 í 10 2 = (x2 + 1)2 25 25(3x2 í 10) = 2(x2 + 1)2 75x2 í 250 = 2(x4 + 2x2 + 1) 75x2 í 250 = 2x4 + 4x2 + 2 í2x4 + 71x2 í 252 = 0 2x4 í 71x2 + 252 = 0 Stel x2 = u. 2u2 í 71u + 252 = 0 D = (í71)2 í 4  2  252 = 3025 71 + 55 71 í 55 u= =4  u= = 3112 4 4 x2 = 4  x2 = 3112

冑

冑

x = 2  x = í2  x = 3112  x = í 3112 x = 2  x = í2  x = 112 冑14  x = í112 冑14 vold. vold. vold. vold. 51 a

6x2 í 12 = 113 (x2 í 1)2 6x2 í 12 4 = (x2 í 1)2 3 2 3(6x í 12) = 4(x2 í 1)2 18x2 í 36 = 4(x4 í 2x2 + 1) 18x2 í 36 = 4x4 í 8x2 + 4 í4x4 + 26x2 í 40 = 0 2x4 í 13x2 + 20 = 0 Stel x2 = u. 2u2 í 13u + 20 = 0 D = (í13)2 í 4  2  20 = 9 13 + 3 13 í 3 u= = 212  u = =4 4 4 1 2 2 x = 22  x = 4

冑

3

冑

x = 212  x = í 212  x = 2  x = í2 x = 冑10  x = í 12 冑10  x = 2  x = í2 vold. vold. vold. vold. 1 2

a Kwadrateren geeft 2x í 5 = 9 2x = 4 x=2 b Een wortel kan niet negatief zijn. Bladzijde 121

52 a

d

a x = 冑5x + 14 kwadrateren geeft x2 = 5x + 14 x2 í 5x í 14 = 0 (x í 7)(x + 2) = 0 x = 7  x = í2 x = 7 geeft 7 = 冑49 voldoet x = í2 geeft í2 = 冑4 voldoet niet b 3x = 冑8x + 20 kwadrateren geeft 9x2 = 8x + 20 9x2 í 8x í 20 = 0 D = (í8)2 í 4  9  í20 = 784 8 í 28 8 + 28 x= = í119  x = =2 18 18

冑

x = í119 geeft í313 = 1119 voldoet niet x = 2 geeft 6 = 冑36 voldoet

© Noordhoff Uitgevers bv

c 5冑x = x kwadrateren geeft 25x = x2 x2 í 25x = 0 x(x í 25) = 0 x = 0  x = 25 x = 0 geeft 0 = 0 voldoet x = 25 geeft 25 = 25 voldoet d 3x = 冑18x + 72 kwadrateren geeft 9x2 = 18x + 72 9x2 í 18x í 72 = 0 x2 í 2x í 8 = 0 (x í 4)(x + 2) = 0 x = 4  x = í2 x = 4 geeft 12 = 冑144 voldoet x = í2 geeft í6 = 冑36 voldoet niet

Vergelijkingen en herleidingen

87

53 a

a 4 í 3冑x = 2 í3冑x = í2

冑x = 23

kwadrateren geeft x = 49 x = 49 geeft 4 í 3 Â 23 = 2 voldoet

b 5冑x í 2x = 0 5冑x = 2x kwadrateren geeft 25x = 4x2 4x2 í 25x = 0 x(4x í 25) = 0 x = 0  4x = 25 x = 0  x = 614 x = 0 geeft 0 í 0 = 0 voldoet x = 614 geeft 5  212 í 2  614 = 0 voldoet

3

Bladzijde 122 54 a

a 2x + 冑x = 10

冑x = 10 í 2x

kwadrateren geeft x = (10 í 2x)2 x = 100 í 40x + 4x2 4x2 í 41x + 100 = 0 D = (í41)2 í 4  4  100 = 81 41 + 9 41 í 9 x= =4 = 614 8 8 x = 4 geeft 8 + 2 = 10 voldoet x = 614 geeft 1212 + 212 = 10 voldoet niet

b

冑x + 12 = x

kwadrateren geeft x + 12 = x2 x2 í x í 12 = 0 (x í 4)(x + 3) = 0 x = 4 geeft 冑16 = 4 voldoet x = í3 geeft 冑9 = í3 voldoet niet

55 a

c 2x í 5冑x = 3 2x í 3 = 5冑x kwadrateren geeft (2x í 3)2 = 25x 4x2 í 12x + 9 = 25x 4x2 í 37x + 9 = 0 D = (í37)2 í 4  4  9 = 1225 37 í 35 1 37 + 35 x= =4  =9 8 8 x = 14 geeft 12 í 212 = 3 voldoet niet x = 9 geeft 18 í 15 = 3 voldoet d 5 í 2冑x = 3 í2冑x = í2

冑x = 1

kwadrateren geeft x=1 x = 1 geeft 5 í 2 = 3 voldoet

c 2x + 冑x = 6 冑x = 6 í 2x kwadrateren geeft x = (6 í 2x)2 x = 36 í 24x + 4x2 4x2 í 25x + 36 = 0 D = (í25)2 í 4  4  36 = 49 25 í 7 25 + 7 x= = 214  x = =4 8 8 x = 214 geeft 412 + 112 = 6 voldoet x = 4 geeft 8 + 2 = 6 voldoet niet d 10 í x冑x = 2 8 = x冑x kwadrateren geeft 64 = x3 x3 = 64 x=4 x = 4 geeft 10 í 4冑4 = 2 voldoet

a Je krijgt u2 + u í 6 = 0 u2 + u í 6 = 0 (u + 3)(u í 2) = 0 u = í3  u = 2 冑 b x x = í3 geeft geen oplossingen voor x omdat x冑x niet negatief kan zijn. Bladzijde 123

56 a

Je controleert de oplossing in een vergelijking voorafgaand aan het kwadrateren, dus ook de vergelijking x2 Â 冑x = 2 is toegestaan om te controleren.

88

Hoofdstuk 3

© Noordhoff Uitgevers bv

57 a

a x3 í 9x冑x + 8 = 0 Stel x冑x = u. u2 í 9u + 8 = 0 (u í 1)(u í 8) = 0 u=1  u=8 x冑x = 1  x冑x = 8 kwadrateren geeft x3 = 1  x3 = 64 x=1x=4 x = 1 geeft 1冑1 = 1 voldoet x = 4 geeft 4冑4 = 8 voldoet b x3 + 27 = 28x冑x x3

í 28x冑x + 27 = 0

Stel x冑x = u. u2 í 28u + 27 = 0 (u í 1)(u í 27) = 0 u = 1  u = 27 x冑x = 1  x冑x = 27 kwadrateren geeft x3 = 1  x3 = 729 x=1x=9 x = 1 geeft 1冑1 = 1 voldoet x = 9 geeft 9冑9 = 27 voldoet

c 8x3 + 8 = 65x冑x 8x3 í 65x冑x + 8 = 0 Stel x冑x = u. 8u2 í 65u + 8 = 0 D = (í65)2 í 4  8  8 = 3969 65 + 63 65 í 63 1 u= =8 u= =8 16 16 x冑x = 18  x冑x = 8 kwadrateren geeft 1 x3 = 64  x3 = 64

3

x = 14  x = 4 x = 14 geeft 14

冑

1 4

= 18 voldoet

x = 4 geeft 4冑4 = 8 voldoet d x5 í 33x2  冑x + 32 = 0 Stel x2  冑x = u. u2 í 33u + 32 = 0 (u í 1)(u í 32) = 0 u = 1  u = 32 x2  冑x = 1  x2  冑x = 32 kwadrateren geeft x5 = 1  x5 = 1024 x=1x=4 x = 1 geeft 12  冑1 = 1 voldoet x = 4 geeft 42  冑4 = 32 voldoet

58 a

a x3 + 30 = 11x冑x

c x5 + 10 = 7x2 Â 冑x

x3 í 11x冑x + 30 = 0

x5 í 7x2 Â 冑x + 10 = 0

Stel x冑x = u. u2 í 11u + 30 = 0 (u í 5)(u í 6) = 0 u=5u=6 x冑x = 5  x冑x = 6 kwadrateren geeft x3 = 25  x3 = 36 3 25 3 36 x =冑  x =冑

Stel x2  冑x = u. u2 í 7u + 10 = 0 (u í 2)(u í 5) = 0 u=2u=5 x2  冑x = 2  x2  冑x = 5 kwadrateren geeft x5 = 4  x5 = 25 5 5 x =冑 4  x =冑 25

3 25 geeft 冑 3 25 冑 3 x =冑 Â 冑25 = 5 voldoet

5 4 geeft x =冑

3 3 3 x =冑 36 geeft 冑 36 Â 冑冑 36 = 6 voldoet

b x3 + 125 = 126x冑x

(冑5 4 )2 Â 冑冑5 4 = 2 voldoet 2 5 5 5 x =冑 25 geeft (冑 25 ) Â 冑冑 25 = 5 voldoet d 32x5 + 32 = 1025x2 Â 冑x

x3 í 126x冑x + 125 = 0

32x5 í 1025x2 Â 冑x + 32 = 0

Stel x冑x = u. u2 í 126u + 125 = 0 (u í 1)(u í 125) = 0 u = 1  u = 125 x冑x = 1  x冑x = 125 kwadrateren geeft x3 = 1  x3 = 15625 x = 1 x = 25 x = 1 geeft 1冑1 = 1 voldoet x = 25 geeft 25冑25 = 125 voldoet

Stel x2  冑x = u. 32u2 í 1025u + 32 = 0 D = (í1025)2 í 4  32  32 = 1046529 1025 + 1023 1025 í 1023 1 u= = 32  u = = 32 64 64 1 x2  冑x = 32  x2  冑x = 32 kwadrateren geeft 1 x5 = 1024  x5 = 1024 x = 14  x = 4 x = 14 geeft

( 14 )2 Â 冑14 = 321 voldoet

x = 4 geeft (4)2 Â 冑4 = 32 voldoet

© Noordhoff Uitgevers bv

Vergelijkingen en herleidingen

89

59 a

3

Met isoleren, kwadrateren, controleren: x í 冑x = 12 x í 12 = 冑x kwadrateren geeft (x í 12)2 = x x2 í 24x + 144 = x x2 í 25x + 144 = 0 (x í 9)(x í 16) = 0 x = 9  x = 16 x = 9 geeft 9 í 3 = 12 voldoet niet x = 16 geeft 16 í 4 = 12 voldoet

Met de substitutie 冑x = u: x í 冑x = 12 x í 冑x í 12 = 0 Stel 冑x = u. u2 í u í 12 = 0 (u í 4)(u + 3) = 0 u = 4  u = í3 冑x = 4  冑x = í3 x = 16

3.4 Herleidingen Bladzijde 125 60 a

a (3x í 2)2 = 9x2 í 12x + 4 b (4x + 3)(4x í 3) = 16x2 í 9 c (x + 2)3 = (x + 2)(x + 2)2 = (x + 2)(x2 + 4x + 4) = x3 + 4x2 + 4x + 2x2 + 8x + 8 = x3 + 6x2 + 12x + 8

61 a

a (2x í 1)3 = (2x í 1)(2x í 1)2 = (2x í 1)(4x2 í 4x + 1) = 8x3 í 8x2 + 2x í 4x2 + 4x í 1 = 8x3 í 12x2 + 6x í 1 b (2x2 + 1)3 = (2x2 + 1)(2x2 + 1)2 = (2x2 + 1)(4x4 + 4x2 + 1) = 8x6 + 8x4 + 2x2 + 4x4 + 4x2 + 1 = 8x6 + 12x4 + 6x2 + 1 c ((x2 í 1)(x2 + 1))2 = (x4 í 1)2 = x8 í 2x4 + 1

62 a

a

2x5 í 32x 2x(x4 í 16) 2x(x2 + 4)(x2 í 4) 2x(x2 + 4)(x2 í 4) = = = = 2x(x2 + 4) mits x  2 œ x  í2 x2 í 4 x2 í 4 x2 í 4 x2 í 4

b

x4 + 4x2 + 4 (x2 + 2)(x2 + 2) x2 + 2 = 2 = x4 í 4 (x í 2)(x2 + 2) x2 í 2

c

x4 í 9x2 x2(x2 í 9) x2(x í 3)(x + 3) = = x(x + 3) mits x  0 œ x  3 = x(x í 3) x(x í 3) x2 í 3x

Bladzijde 126 63 a

64 a

xP = p geeft yP = p2, dus P(p, p2 ) en xQ = q geeft yQ = q2, dus Q(q, q2 ). Stel k: y = ax + b. %y yQ í yP q2 í p2 (q + p)(q í p) a= = =q+p = qíp = qíp %x xQ í xP y = (p + q)x + b (p + q)  p + b = p2 Door P(p, p2 ) r 2 p + pq + b = p2 b = ípq Dus k: y = (p + q)x í pq. 1 2x2 1 2x2 í 1 a 2x í x = x í x = x x ( x + 1) x ( x + 2) x2 + 2x + x2 + x 2x2 + 3x x x + + = = = x + 1 x + 2 (x + 1)(x + 2) (x + 1)(x + 2) (x + 1)(x + 2) (x + 1)(x + 2) x + 2 (x + 1)(x + 2) x2 + 3x + 2 = = x+3 x+3 x+3 x x2 1 2 x = x  = = 2 x mits x  0 2 2 2 x

b (x + 1) Â c

()

Dus de herleiding is niet juist voor x = 0.

90

Hoofdstuk 3

© Noordhoff Uitgevers bv

Bladzijde 127 65 a

(

)

( )

66 a

(

)

20 20 4(x í 1) 20 4x í 4 í 2 20(4x í 6) 80x í 120 2 2 4í í = = = = Â xí1 xí1 xí1 xí1 xí1 xí1 xí1 (x í 1)2 (x í 1)2 4x 4 b y= en x = 1 geeft y = , dus delen door 0. x+1 2 xí1 0 4x(x í 1) 4Â0 0 y= en x = 1 geeft y = = = 0. x+1 2 2 a y=

a y= b y= c y=

d y= e y=

()

20 5 40 5 35 í í = = x 2x 2x 2x 2x x2(x í 1) 10 í x2(x í 1) 10 í x3 + x2 íx3 + x2 + 10 10 10 í x2 = í = = = xí1 xí1 xí1 xí1 xí1 xí1 x í 1 2x2(x í 1) 2x3 í 2x2 2x2 2 mits x  1 = 2x  = = x+1 x+1 x+1 x+1 xí1 x(x í 1) x x x x2 í x + 1 x3 í x2 + x 1 1 x+ + = = =  xí1 xí1 xí1 xí1 xí1 xí1 xí1 (x í 1)2 6 5 30 =  x í 2 x + 2 (x í 2)(x + 2)

( )

(

)

(

3

)

( )

x+1 2x x+1 1 x+1 f y= = = Â xí1 2x x í 1 2x(x í 1) 67 a

4x 4 = x+1 xí1 4x(x í 1) = 4(x + 1) 4x2 í 4x = 4x + 4 4x2 í 8x í 4 = 0 x2 í 2x í 1 = 0 D = (í2)2 í 4  1  í1 = 8 2 + 冑8 2 + 2冑2 2 í 冑8 2 í 2冑2 x= = = 1 + 冑2  x = = = 1 í 冑2 2 2 2 2 voldoet voldoet 4x 4 b f (x)  g(x) = 6 geeft =6  x+1 xí1 16x =6 x2 í 1 16x = 6(x2 í 1) 16x = 6x2 í 6 6x2 í 16x í 6 = 0 3x2 í 8x í 3 = 0 D = (í8)2 í 4  3  í3 = 100 8 + 10 8 í 10 x= =3  x= = í 13 6 6 voldoet voldoet a f (x) = g(x) geeft

© Noordhoff Uitgevers bv

Vergelijkingen en herleidingen

91

c f (x) í g(x) = 7 geeft

4x 4 í =7 x+1 xí1 4(x + 1) 4x(x í 1) í =7 (x + 1)(x í 1) (x + 1)(x í 1) 4x2 í 4x í 4x í 4 =7 (x + 1)(x í 1) 4x2 í 8x í 4 =7 x2 í 1 4x2 í 8x í 4 = 7(x2 í 1) 4x2 í 8x í 4 = 7x2 í 7 í3x2 í 8x + 3 = 0 D = (í8)2 í 4  í3  3 = 100 8 + 10 8 í 10 1 x= = í3  x = =3 í6 í6 voldoet voldoet

3

d f (x) =

(x + 1)  4 í 4x  1 4x + 4 í 4x 4x 4 geeft f ‫(މ‬x) = = = 2 2 x+1 (x + 1) (x + 1) (x + 1)2

(x í 1) Â 0 í 4 Â 1 4 í4 geeft g ‫(މ‬x) = = xí1 (x í 1)2 (x í 1)2 4 í4 f ‫(މ‬x) + 2 Â g ‫(މ‬x) = 0 geeft + 2Â =0 2 (x + 1) (x í 1)2

g(x) =

4(x í 1)2 í8(x + 1)2 + =0 (x í 1)2 Â (x + 1)2 (x í 1)2 Â (x + 1)2 4(x2 í 2x + 1) í 8(x2 + 2x + 1) =0 (x í 1)2 Â (x + 1)2 4x2 í 8x + 4 í 8x2 í 16x í 8 =0 (x í 1)2 Â (x + 1)2 í4x2 í 24x í 4 =0 (x í 1)2 Â (x + 1)2 í4x2 í 24x í 4 = 0 x2 + 6x + 1 = 0 D = 62 í 4 Â 1 Â 1 = 32 x=

í6 + 冑32 í6 í 冑32  x= 2 2

í6 + 4冑2 í6 í 4冑2 = í3 + 2冑2  x = = í3 í 2冑2 2 2 voldoet voldoet x=

68 a

Vermenigvuldig de teller en de noemer van

Bladzijde 128

x+ 69 a

a y=

92

Hoofdstuk 3

(

2x + 13 met 3. x+1

)

3 3 x+  (x + 1) x+1 x ( x + 1) + 3 x 2 + x + 3 x+1 = = = x x  (x + 1) x(x + 1) x(x + 1)

© Noordhoff Uitgevers bv

( (

) )

5 5  (x í 1) 10 + 10(x í 1) + 5 10x í 5 xí1 xí1 b y= mits x  1 = = = 6x í 9 3 6(x í 1) í 3 3 6í 6í  (x í 1) xí1 xí1 10 +

c y=

70 a

a N=

10x  2p 20px 10x mits p  0 = = 2 2p2 + x2 x x2 p+ p+  2p 2p 2p

(

)

2400ab 600a 600a  4b mits b  0 = = 12b2 í a2 a2 a2 3b í 3b í  4b 4b 4b

(

)

3

50 50  (x2 + 1) x2 + 1 x2 + 1 50 1000 b A = 25x + 20  = 25x + 20  2 = 25x + = 25x + 20  2 x  (x + 1) x(x + 1) x(x2 + 1) x 150p  q 150p 150pq 50p(p + 5q) 150pq 150 c K = 50 + p p = 50p + p + = 50p + = 50p + = =  p ° ¢ p + 5q p + 5q p + 5q + 5 + 5 q q q + 5 Âq

( )

50p2 + 250pq + 150pq 50p2 + 400pq mits q  0 = p + 5q p + 5q 71 a

4p í 1 3x substitueren in N = geeft x+5 2p + 3 12x 3x í 1  (x + 5) í1 4 12x í (x + 5) 12x í x í 5 11x í 5 x+5 x+5 N= mits x  í5 = = = = 3x 6x + 3(x + 5) 6x + 3x + 15 9x + 15 6x 2 +3 + 3  (x + 5) x+5 x+5

a p=

( (

b N = 9 geeft

72 a

) )

11x í 5 =9 9x + 15 11x í 5 = 9(9x + 15) 11x í 5 = 81x + 135 í70x = 140 x = í2

1000 5x2 + 1000 5x2 1000 = x + x = 5x + x x 6t2 + 12t + 1500 6t2 12t 1500 500 b K= + + = = 2t + 4 + 3t 3t 3t 3t t 5a2 + 8a 5a2 8a 4 1 c F= = 2 + 2 = 22 + a 2a2 2a 2a a A=

d N=

6p2 í 3p í 1 6p2 3p 1 1 í í = = 3p í 112 í 2p 2p 2p 2p 2p

© Noordhoff Uitgevers bv

Vergelijkingen en herleidingen

93

Bladzijde 129 73 a

a Tijd heenreis =

d d d d uur en tijd terugreis = uur, dus t = + . 12 48 12 48

afstand 2d en afstand is 2d, dus v = . tijd d d + 12 48 2 . Teller en noemer delen door d geeft v = 1 1 + 12 48 48 2 2 2 v= = = = 2Â = 19,2 km/uur 5 5 1 4 1 1 + + 12 48 48 48 48 2 Â ab 2ab 2ab 2 2 c h= = = = = 1 1 a b b+a a+b b+a + + Â ab a b ab ab ab 6a 18a 3 3 3 7 7 d h= a, dus p = 111 . = = 3Â = = = 111 1 1 6 11 11 11 1 3 2 + + + + a 2a 3a 6a 6a 6a 6a b Gemiddelde snelheid is

3

( )

Bladzijde 130 74 a

75 a

2 y= x xy = 2 2 x= y B B+2 A(B + 2) = B AB + 2A = B AB í B = í2A B(A í 1) = í2A 2A B=í Aí1

a A=

Qí5 Q PQ = Q í 5 PQ í Q = í5 Q(P í 1) = í5 5 Q=í Pí1

b P=

94

Hoofdstuk 3

Fí2 Fí1 R(F í 1) = F í 2 RF í R = F í 2 RF í F = R í 2 F(R í 1) = R í 2 Rí2 F= Rí1 18 d L = 320 í qí1 c R=

18 = 320 í L qí1 320 í L =

18 qí1

qí1=

18 320 í L

q=1+

18 320 í L

© Noordhoff Uitgevers bv

Bladzijde 131 76 a

77 a

a

1 1 =2+ a b

b

1 1 =2+ a b

1 2b 1 + = a b b

1 1 í2= a b

1 2b + 1 = a b Neem van beide leden het omgekeerde. b a= 2b + 1

1 1 = í2 b a 1 1 2a = í b a a

3

1 1 í 2a = b a a b= 1 í 2a

a p uitdrukken in q 2 1 p=5íq 1 5q 2 p= q íq 1 5q í 2 p= q q p= 5q í 2

q uitdrukken in p 2 1 p=5íq 1 2 q=5íp 2 5p 1 q= p íp 2 5p í 1 q= p q  (5p í 1) = 2p 2p q= 5p í 1

b m schrijven als functie van n 1 1 3 = í m 2 n n 6 1 í = m 2n 2n 1 ní6 = 2n m 2n m= ní6

n schrijven als functie van m 1 1 3 = í m 2 n 3 1 1 = í n 2 m 3 m 2 í = n 2m 2m 3 mí2 = 2m n n(m í 2) = 6m 6m n= mí2

Bladzijde 132 78 a

a

b

tí2 t ÂP = tí3 tí1 P=

t tí3 Â tí1 tí2

P=

t(t í 3) (t í 1)(t í 2)

P=

t2 í 3t (t í 1)(t í 2)

3x =5íy x+y (x + y)(5 í y) = 3x 5x í xy + 5y í y2 = 3x 2x í xy = y2 í 5y x(2 í y) = y2 í 5y y2 í 5y x= 2íy

© Noordhoff Uitgevers bv

c K = 90 í

2N N + 0,2

2N = 90 í K N + 0,2 2N = (N + 0,2)(90 í K) 2N = 90N í KN + 18 í 0,2K KN í 88N = 18 í 0,2K N(K í 88) = 18 í 0,2K 18 í 0,2K N= K í 88

Vergelijkingen en herleidingen

95

79 a

a F=

1 1 + K 2K 2 1 + 2K 2K

1 2R + 2 í3 = N 5R + 2

F=

3 2K

1 2R + 2 3(5R + 2) í = 5R + 2 N 5R + 2

K= b

1 2R + 2 +3= N 5R + 2

F=

2K = 3

c

3 F

3 2F

1 2 = 10 í T S

1 2R + 2 í 15R í 6 = N 5R + 2 1 í13R í 4 = N 5R + 2 N=

5R + 2 í13R í 4

1 10S 2 í = T S S 1 10S í 2 = T S T=

S 10S í 2

Diagnostische toets Bladzijde 134 1

a 3x3 + 5 = 86 3x3 = 81 x3 = 27 x=3 b 5x4 í 6 = 9 5x4 = 15 x4 = 3 4 4 x =冑 3  x = í冑 3 3 c 2x + 19 = 5 2x3 = í14 x3 = í7 3 x =冑 í7

1 d 12 (x + 2)4 = 32 1 (x + 2)4 = 16

x + 2 = 12  x + 2 = í 12 x = í112  x = í212 e 100 í (2x + 1)5 = 68 í(2x + 1)5 = í32 (2x + 1)5 = 32 2x + 1 = 2 2x = 1 x = 12 f (2x + 4)3 = 10 3 10 2x + 4 = 冑 3 10 2x = í4 + 冑 3 x = í2 + 12  冑 10

2

96

a x3 = x2 + 20x x3 í x2 í 20x = 0 x(x2 í x í 20) = 0 x(x í 5)(x + 4) = 0 x = 0  x = 5  x = í4 b x4 í 6x2 + 5 = 0 Stel x2 = u. u2 í 6u + 5 = 0 (u í 1)(u í 5) = 0 u=1u=5 x2 = 1  x2 = 5 x = 1  x = í1  x = 冑5  x = í冑5

Hoofdstuk 3

c x4 í 6x3 + 5x2 = 0 x2(x2 í 6x + 5) = 0 x2(x í 1)(x í 5) = 0 x=0 x=1x=5 d x8 + x4 = 42 x8 + x4 í 42 = 0 Stel x4 = u. u2 + u í 42 = 0 (u + 7)(u í 6) = 0 u = í7  u = 6 x4 = í7  x4 = 6 4 4 x =冑 6  x = í冑 6

© Noordhoff Uitgevers bv

3

a 5x4 í 6x2 + 1 = 0 Stel x2 = u. 5u2 í 6u + 1 = 0 D = (í6)2 í 4  5  1 = 16 6+4 6í4 1 u= =5 u= =1 10 10 1 u=5 u=1

b 3x6 + 3 = 10x3 3x6 í 10x3 + 3 = 0 Stel x3 = u. 3u2 í 10u + 3 = 0 D = (í10)2 í 4  3  3 = 64 10 + 8 10 í 8 1 u= =3 u= =3 6 6 1 x3 = 3  x3 = 3

x2 = 15  x2 = 1 x=

冑

1 5

x=í

冑

1 5

 x = 1  x = í1

31 3 x =冑 3  x = 冑3

3

x = 15 冑5  x = í 15 冑5  x = 1  x = í1 4

a Stel f (x) = 2x2 + 3x en g(x) = x3. f (x) = g(x) geeft 2x2 + 3x = x3 x3 í 2x2 í 3x = 0 x(x2 í 2x í 3) = 0 x(x + 1)(x í 3) = 0 x = 0  x = í1  x = 3 y

b Stel f (x) = 23 x2 + 1 en g(x) = 13 x4 í 4 . f (x) = g(x) geeft 23 x2 + 1 = 13 x4 í 4 2x2 + 3 = x4 í 12 x4 í 2x2 í 15 = 0 Stel x2 = u. u2 í 2u í 15 = 0 (u + 3)(u í 5) = 0 u = í3  u = 5 x2 = í3  x2 = 5 x = í冑5  x = 冑5 y

ƒ

g ƒ

− 5 –1 3

5

x

x

g

2x2 + 3x > x3 geeft x < í1  0 < x < 3

5

a 0 x2 í 4 0 = 21 x2 í 4 = 21  x2 í 4 = í21 x2 = 25  x2 = í17 x = 5  x = í5

2 2 3x

+ 1 > 13 x4 í 4 geeft x < í冑5  x > 冑5

b 0 4x3 í 5 0 = 19 4x3 í 5 = 19  4x3 í 5 = í19 4x3 = 24  4x3 = í14 x3 = 6  x3 = í312 3 1 3 x =冑 6  x = í冑 32

6

4x + 5y = 27 ௘1 4x + 5y = 27 geeft e í2x + 3y = 25 2 í4x + 6y = 50 + 11y = 77 y=7 r í2x + 3 Â 7 = 25 í2x + 3y = 25 í2x + 21 = 25 í2x = 4 x = í2 Dus (x, y) = (í2, 7).

a e

© Noordhoff Uitgevers bv

0 0

Vergelijkingen en herleidingen

97

2x + 3y = 7 2 ௘4x + 6y = 14 geeft e 15x í 6y = 24 5x í 2y = 8 3 + 19x = 38 x=2 r 5 Â 2 í 2y = 8 5x í 2y = 8 10 í 2y = 8 í2y = í2 y=1 Dus (x, y) = (2, 1).

b e

3

0 0

2x í 5y = 1 ௘6x í 15y = 3 ௘3 geeft e 6x + 15y = 39 6x + 15y = 39 1 + 12x = 42 1 x = 32 1 r 2 Â 32 í 5y = 1 2x í 5y = 1 7 í 5y = 1 í5y = í6 y = 115 1 1 Het snijpunt is (32 ,15 ) .

0 0

7

e

8

(2, 18) invullen geeft a  22 + b  2 = 18, dus 4a + 2b = 18. (í4, 0) invullen geeft a  (í4)2 + b  í4 = 0, dus 16a í 4b = 0. ௘ 8a + 4b = 36 4a + 2b = 18 ௘2 geeft e 16a í 4b = 0 1 16a í 4b = 0 + 24a = 36 1 a = 12 1 r 4  12 + 2b = 18 4a + 2b = 18 6 + 2b = 18 2b = 12 b=6 1 2 Dus y = 12 x + 6x.

0 0

e

9

a Substitutie van y = x2 í 4x + 6 in 2x + 3y = 10 geeft 2x + 3(x2 í 4x + 6) = 10 2x + 3x2 í 12x + 18 = 10 3x2 í 10x + 8 = 0 D = (í10)2 í 4  3  8 = 4 10 í 2 10 + 2 x= = 113  x = =2 6 6 1 1 2 1 4 x = 13 geeft y = (13 ) í 4  13 + 6 = 29 x = 2 geeft y = 22 í 4  2 + 6 = 2 Dus (x, y) = (131 , 249 )  (x, y) = (2, 2). b 3x í y = 7 geeft y = 3x í 7. Substitutie van y = 3x í 7 in x2 + (y í 4)2 = 13 geeft x2 + (3x í 7 í 4)2 = 13 x2 + (3x í 11)2 = 13 x2 + 9x2 í 66x + 121 = 13 10x2 í 66x + 108 = 0 D = (í66)2 í 4  10  108 = 36 66 í 6 66 + 6 x= =3x= = 335 20 20 x = 3 geeft y = 3  3 í 7 = 2 x = 335 geeft y = 3  335 í 7 = 345 Dus (x, y) = (3, 2)  (x, y) = (335 , 345 ) .

98

Hoofdstuk 3

© Noordhoff Uitgevers bv

10

a (x2 í 6)(x2 í 2x) = 0 x2 í 6 = 0  x2 í 2x = 0 x2 = 6  x(x í 2) = 0 x = 冑6  x = í冑6  x = 0  x = 2 b (2x2 í 1)2 = (6x + 1)2 2x2 í 1 = 6x + 1  2x2 í 1 = í(6x + 1) 2x2 í 6x í 2 = 0  2x2 í 1 = í6x í 1 2 x í 3x í 1 = 0  2x2 + 6x = 0 2 D = (í3) í 4  1  í1 = 13 2x(x + 3) = 0 3 í 冑13 3 + 冑13 x=  x=  x = 0  x = í3 2 2 x = 112 í 12 冑13  x = 112 + 12 冑13  x = 0  x = í3

c x(x2 í 1) = 4(x2 í 1) x = 4  x2 í 1 = 0 x = 4  x2 = 1 x = 4  x = 1  x = í1 d (x3 í 9x)(x2 í 3) + 9x = x3 (x3 í 9x)(x2 í 3) = x3 í 9x x3 í 9x = 0  x2 í 3 = 1 x(x2 í 9) = 0  x2 = 4 x = 0  x2 = 9  x = 2  x = í2 x = 0  x = 3  x = í3  x = 2  x = í2 3

Bladzijde 135 11

12

x2 í 5x + 6 =0 2x + 4 x2 í 5x + 6 = 0 (x í 2)(x í 3) = 0 x=2  x=3 vold. vold. x2 í 4 x2 í 4 b = 2x + 1 x í 4 2x + 1 = x í 4  x2 í 4 = 0 x = í5  x2 = 4 x = í5  x = 2  x = í2 vold. vold. vold. a

a

冑3x + 5 + 1 = 5 冑3x + 5 = 4

kwadrateren geeft 3x + 5 = 16 3x = 11 x = 323 x = 323 geeft 冑11 + 5 + 1 = 5 voldoet

2x í 1 4x + 1 = x + 1 5x í 1 (2x í 1)(5x í 1) = (x + 1)(4x + 1) 10x2 í 2x í 5x + 1 = 4x2 + x + 4x + 1 6x2 í 12x = 0 6x(x í 2) = 0 x=0  x=2 vold. vold. 2x2 í 4 d = 134 x+5 c

2x2 í 4 7 = x+5 4 2 4(2x í 4) = 7(x + 5) 8x2 í 16 = 7x + 35 8x2 í 7x í 51 = 0 D = (í7)2 í 4  8  í51 = 1681 7 í 41 7 + 41 x= = í218  x = =3 16 16 voldoet voldoet b 3x = 5冑x + 4 kwadrateren geeft 9x2 = 25(x + 4) 9x2 = 25x + 100 9x2 í 25x í 100 = 0 D = (í25)2 í 4  9  í100 = 4225 25 í 65 25 + 65 x= = í229  x = =5 18 18

冑

x = í229 geeft í623 = 5 í229 + 4 voldoet niet x = 5 geeft 15 = 5冑9 voldoet

© Noordhoff Uitgevers bv

Vergelijkingen en herleidingen

99

c x = 冑x + 6 x í 6 = 冑x kwadrateren geeft (x í 6)2 = x x2 í 12x + 36 = x x2 í 13x + 36 = 0 (x í 4)(x í 9) = 0 x=4x=9 x = 4 geeft 4 = 冑4 + 6 voldoet niet x = 9 geeft 9 = 冑9 + 6 voldoet

3

d 2x + 3冑x = 2 3冑x = 2 í 2x kwadrateren geeft 9x = (2 í 2x)2 9x = 4 í 8x + 4x2 í4x2 + 17x í 4 = 0 D = 172 í 4  í4  í4 = 225 í17 í 15 í17 + 15 1 x= =4x= =4 í8 í8 x = 4 geeft 8 + 3冑4 = 2 voldoet niet

冑

x = 14 geeft 12 + 3 13

a x3 í 189 = 20x冑x x3 í 20x冑x í 189 = 0 Stel x冑x = u. u2 í 20u í 189 = 0 (u + 7)(u í 27) = 0 u = í7  u = 27 x冑x = í7  x冑x = 27

1 4

= 2 voldoet

b x5 + 12 = 8x2  冑x x5 í 8x2  冑x + 12 = 0 Stel x2  冑x = u. u2 í 8u + 12 = 0 (u í 2)(u í 6) = 0 u=2u=6 x2  冑x = 2  x2  冑x = 6 kwadrateren geeft x5 = 4  x5 = 36 54 5 36 x =冑  x =冑

x冑x = í7 heeft geen oplossing x冑x = 27 kwadrateren geeft x2 = 729 x=9 冑 x = 9 geeft 9 9 = 27 voldoet

5 x =冑 4 geeft

(冑5 4 )2 Â 冑冑5 4 = 2 voldoet 2 5 5 5 x =冑 36 geeft (冑冑 36 ) Â 冑冑 36 = 6 voldoet

14

a (2x + 3)3 = (2x + 3)(2x + 3)2 = (2x + 3)(4x2 + 12x + 9) = 8x3 + 24x2 + 18x + 12x2 + 36x + 27 = 8x3 + 36x2 + 54x + 27 x4 í 16 (x2 í 4)(x2 + 4) b 2 = = x2 + 4 mits x  2 œ x  í2 x í4 x2 í 4

15

a y = 2x í b y=

(

x í 1 2x(x í 2) x í 1 2x2 í 4x í x + 1 2x2 í 5x + 1 í = = = xí2 xí2 xí2 xí2 xí2

) (

) ( )

(

)

x x 2x + Â (x í 1) 2x(x í 1) + x xí1 xí1 2x2 í 2x + x 2x2 í x = = = 2 = 2 x+1 (x + 1) Â (x í 1) x í1 x2 í 1 x í1

2x + 16

) (

3 x 3 2(x í 1) 3 2x í 2 í x 3 xí2 3x í 6 x 2í í = = = = x x xí1 x x xí1 xí1 xí1 xí1 x(x í 1)

a y=

( (

) )

2 2 í 3  (x + 1) í3 2 í 3(x + 1) 2 í 3x í 3 í3x í 1 x+1 x+1 b y= mits x  í1 = = = x = 4(x + 1) í x 4x + 4 í x 3x + 4 x 4í 4í  (x + 1) x+1 x+1

17

a N=

4x2 í 50 4x2 50 25 í = = 2x í 2x 2x 2x x

b B=

6p2 í 3p + 4 6p2 3p 4 4 í + = = 2p í 1 + 3p 3p 3p 3p 3p

100 Hoofdstuk 3

© Noordhoff Uitgevers bv

18

3P í 2 2P í 3 V(2P í 3) = 3P í 2 2PV í 3V = 3P í 2 2PV í 3P = 3V í 2 P(2V í 3) = 3V í 2 3V í 2 P= 2V í 3

a V=

b R = 40 í

8 aí1

8 = 40 í R aí1 8 =aí1 40 í R aí1= a=

3 4 c p+q=6 3 4 p=6íq 3 6q 4 p= q íq 3 6q í 4 p= q p q = 3 6q í 4 p=

3

3q 6q í 4

8 40 í R

8 +1 40 í R

© Noordhoff Uitgevers bv

Vergelijkingen en herleidingen 101

4 Meetkunde Voorkennis Rekenen met wortels Bladzijde 138

1

a 2冑3 Â 3冑5 = 6冑15 b

5冑10 = 5冑2 冑5

c 3a冑2 Â a冑7 = 3a2 Â 冑14 d

4

2冑14 2 = 冑2 3冑7 3

冑2 Â 12 a冑3 = 14 a2 Â 冑6 6 6 冑2 6冑2 3 = 冑2 f = Â = 5冑2 5冑2 冑2 10 5 2

e

1 2a

a

1 1 冑3 冑3 1 = = 冑3 Â = 冑3 冑3 冑3 3 3

b

冑

c d e f

冑1 1 冑2 冑2 1 = = 冑2 Â = 冑2 冑2 冑2 2 2 冑9 3 3 冑2 3冑2 冑412 = 冑92 = 冑2 = 冑2 = 冑2 Â 冑2 = 2 = 112 冑2 ( 12 冑5 )2 = ( 12 )2 Â (冑5 )2 = 14 Â 5 = 114 ( 12 a冑2 )2 = ( 12 a )2 Â (冑2 )2 = 14 a2 Â 2 = 12 a2 ( 23 a冑3 )2 = ( 23 a )2 Â (冑3 )2 = 49 a2 Â 3 = 113 a2 1 2

=

Bladzijde 139 3

a b c d e f

4

a b c d e f

冑24 + 冑6 = 冑4  6 + 冑6 = 2冑6 + 冑6 = 3冑6 10 10 冑5 10冑5 冑80 í = 冑16  5 í  = 4冑5 í = 4冑5 í 2冑5 = 2冑5 冑5 冑5 冑5 5 冑18a í 冑8a = 冑9  2a í 冑4  2a = 3冑2a í 2冑2a = 冑2a 冑3a 冑3a 冑12a + 冑34 a = 冑4  3a + = 2冑3a + = 2冑3a + 12 冑3a = 212 冑3a 冑4 2 9冑2 9 9 冑2 í 冑2 = í 冑2 = 118 冑2 í 冑2 = 18 冑2  í 冑2 = 8 4冑2 4冑2 冑2 冑1 冑3 4 冑3 1 1 1 1 = 13 冑3 í 13 冑3 = 冑3  = 3 冑3 í 3 冑48 í 冑3 = 3 冑16  3 í 冑3 冑3 3 冑25 冑49 5 7 12 12 冑2 12冑2 + + = = = = 6冑2  = 冑2 冑2 冑2 冑2 冑2 冑2 冑2 2 a冑8 í a冑2 = a冑4  2 í a冑2 = 2a冑2 í a冑2 = a冑2 冑a 冑a 冑2 冑2a 冑2a + 冑12 a = 冑2a + a = 冑2a + = 冑2a +  = 冑2a + = 冑2a + 12 冑2a = 112 冑2a Äb 冑2 冑2 冑2 2 冑64 1 2 2 8 1 2 2 2 ( 23 a冑3 ) 2 + a2  冑719 = ( 23 )2  a2 (冑3 ) 2 + a2  冑649 = 49 a2  3 + a2  = 13 a + a  = 13 a + 23 a = 4a2 冑9 3 ( 14 a冑2 ) 2 + ( 34 a冑2 ) 2 = ( 14 ) 2  a2  ( 冑2 )2 + ( 34 ) 2  a2  ( 冑2 ) 2 = 161 a2  2 + 169 a2  2 = 18 a2 + 118 a2 = 114 a2 5a 3a 10a 9a a a 冑2 a冑2 1 í = í = = = a冑2  = 3冑2 2冑2 6冑2 6冑2 6冑2 6冑2 冑2 12 12

冑12 + 冑24 = 冑 1 2

102 Hoofdstuk 4

1 2

25 2

+

冑

49 2

=

© Noordhoff Uitgevers bv

Bladzijde 140 5

a b c d e f

6

a b c d e f

7

( 3冑2 í 冑5 ) 2 = 18 í 6冑10 + 5 = 23 í 6冑10 ( 2冑2 + 3冑3 ) 2 = 8 + 12冑6 + 27 = 35 + 12冑6 (5冑3 + 2) (5冑3 í 2) = 75 í 4 = 71 (a í 冑3 ) 2 = a2 í 2a冑3 + 3 (a í a冑2 ) 2 = a2 í 2a2 Â 冑2 + 2a2 = 3a2 í 2a2 Â 冑2 (4 í 12 a冑2 ) 2 = 16 í 4a冑2 + 12 a2 (2a冑2 í a冑3 ) 2 = 8a2 í 4a2 Â 冑6 + 3a2 = 11a2 í 4a2 Â 冑6 3 3 ( 12 冑2 + 34 冑3 ) 2 = 12 + 34 冑6 + 27 16 = 216 + 4 冑6 (2 í 冑2 )2 = 4 í 4冑2 + 2 = 6 í 4冑2 (112 冑2 í 12 冑3 )2 = 412 í 112 冑6 + 34 = 514 í 112 冑6 (3冑2 + 5) (3冑2 í 5) = 18 í 25 = í7

(冑 1

2 2

+

3 冑2

) (冑 2

=

1

2 2

+

冑 ) ( 冑 ) 6

2

2 2

=

7

2 2

2

=

4

49 = 618 8

BC2 = AB2 + AC2 BC 2 = (冑3 í 冑2 )2 + (冑3 + 冑2 )2 BC2 = 3 í 2冑6 + 2 + 3 + 2冑6 + 2 BC2 = 10 BC = 冑10

C

3+ 2

A

3− 2

B

4.1 Goniometrische verhoudingen en gelijkvormigheid Bladzijde 141 1

BC BC geeft tan(30°) = AB 6 BC = 6 tan(30°) § 3,464 3,464... BC geeft tan(40°) = b tan“ (BDC) = BD BD 3,464... § 4,13. Dus BD = tan(40°)

a tan(“ A) =

Bladzijde 142

2

BC a In +BCD is sin(70°) = 5 BC = 5 sin(70°) = 4,69... BC 4,69... = 10 AC “ BAC § 28,0° b “ ADC = 180° í 70° = 110° “ACD § 180° í 110° í 28,0° = 42,0° In +ABC is sin(“ BAC) =

3

In +ABC is tan(28°) = AB =

4 AB

4 = 7,52... tan(28°)

BM = 12 AB = 3,76...

BC 4 = BM 3,76... “BMC § 46,8°

In +BCM is tan(“BMC) =

© Noordhoff Uitgevers bv

Meetkunde 103

4

BC In +ABC is sin(40°) = 10 BC = 10 sin(40°) § 6,42... AB cos(40°) = 10 AB = 10 cos(40°) § 7,66... BM = 12 BC § 3,21... BM 3,21... In +ABM is tan(“ BAM ) = = AB 7,66... “ BAM § 22,8° Bladzijde 143

BM 6 BM = 6 cos(50°) = 3,85... BC sin(50°) = 6 BC = 6 sin(50°) = 4,59... AB = 2BM § 7,71... BC 4,59... In +ABC is tan(“ BAC) = = AB 7,71... “ BAC § 30,8° “ AMC = 180° í 50° = 130° “ACM § 180° í 130° í 30,8° = 19,2°

5

In +BCM is cos(50°) =

6

In +ABD is tan(20°) =

4

BD 10 BD = 10 tan(20°) = 3,63... BE = 2BD = 7,27... BE 7,27... In +ABE is tan(“ EAB) = = 10 10 “EAB = 36,05...°

BC = 2BE = 14,55... BC 14,55... In +ABC is tan(“ CAB) = = 10 10 “ CAB = 55,51...° “CAE = “CAB í “EAB = 55,51...° í 36,05...° § 19,5° 7

a In +AHF is tan(“HFA) =

AH 2 = =1 FH 2

E

“HFA = 45° EG 2 = =2 FG 1 “EFG § 63,4° “ F = “HFA + “ EFG § 45° + 63,4° = 108,4°

In +FGE is tan(“EFG) =

DQ 5 = AQ 2 “ DAQ = 68,1...°

b In +AQD is tan(“ DAQ) =

2 F

1

1 G

H 2 A

EP 4 = BP 5 “PBE = 38,6...° “SAB = “DAQ en “ABS = “PBE “ASB = 180° í “ABS í “ BAS = 180° í 38,6...° + 68,1...° § 73,1°’

D E

In +PBE is tan(“ PBE) =

C S F

P

104 Hoofdstuk 4

A

R

Q

B

© Noordhoff Uitgevers bv

a +ABC ऴ +DEC b In +ABC is AC2 = AB2 + BC2 = 64 + 36 = 100, dus AC = 10.

8

AC

AB

CD

DE

Dus DE =

geeft

10

8

5

DE

5Â8 = 4. 10

Bladzijde 146

“CAB = “EAD f +ABC ऴ +AED “ABC = “AED

9

AB

BC

AC

AE

DE

AD

geeft

9

7

AC

5

DE

3

4

5Â7 9Â3 = 525 , dus CE = AC í AE = 525 í 5 = 25 . DE = = 389 en AC = 9 5 10

a “ BAE = “ AED (Z-hoeken) f +EDA ऴ +AEB “AEB = “ADE = 90° “ABE = “ BEC (Z-hoeken) f +AEB ऴ +BCE “AEB = “BCE = 90° +EDA ऴ +AEB f +EDA ऴ +BCE +AEB ऴ +BCE b In +ABE is BE2 + AE2 = AB2 BE2 + 36 = 100 BE2 = 64 BE = 8 c Uit +EDA ऴ +AEB volgt DE

AD

AE

AE

BE

AB

AD = 11

geeft

DE

AD

6

6

8

10

6Â8 6Â6 = 445 en DE = = 335 . 10 10

“DAS = “ SCE (Z-hoeken) f +ADS ऴ +CES “ADS = “SEC (Z-hoeken) In +CDE is DE2 = CD2 + CE2 DE2 = 144 + 25 DE2 = 169 DE = 13 Stel DS = x, dan is ES = 13 í x. AD

DS

CE

ES

geeft

7

x

5

13 – x

Dus 5x = 7(13 í x) 5x = 91 í 7x 12x = 91 7 7 x = 712 , dus DS = 712 12

“ABS = “ SDC (Z-hoeken) f +ABS ऴ +CDS “BAS = “ SCD (Z-hoeken) In +BCD is BD2 + BC2 = CD2 BD2 + 225 = 625 BD2 = 400 BD = 20

© Noordhoff Uitgevers bv

Meetkunde 105

Stel BS = x, dan is DS = 20 í x. 10 AB BS geeft 25 CD DS

x 20 – x

Dus 25x = 10(20 í x) 25x = 200 í 10x 35x = 200 x = 557 , dus BS = 557 13

4

“ABC = “ADE = 90° f +ABC ऴ +ADE “BAC = “DAC AB

BC

AD

DE

geeft

20

15

8

DE

8  15 = 6. 20 “ DES = “ SBC (Z-hoeken) f +EDS ऴ +BCS “EDS = “SCB (Z-hoeken) In +BCD is CD2 = BC2 + BD2 CD2 = 144 + 225 CD2 = 369 CD = 冑369 = 冑9  41 = 3冑41 Stel DS = x, dan is CS = 3冑41 í x. Dus DE =

DE

DS

BC

CS

geeft

6

x

15

3冑41 í x

Dus 15x = 6(3冑41 í x) 15x = 18冑41 í 6x 21x = 18冑41 x = 67 冑41, dus DS = 67 冑41 14

“AEB = “CFB = 90° f +ABE ऴ +CBF “ABE = “FBC AB

BE

BC

BF

geeft

10

BE

15

5

C

10  5 = 313 , dus CE = 15 í 313 = 1123 . 15 “ ABC = “ DEC (F-hoeken) f +ABC ऴ +DEC “BAC = “EDC (F-hoeken)

15

Dus BE =

AB DE Dus DE =

BC

geeft

CE

1123

10 Â 15

10

15

DE

1123

D

A

= 779 .

15

E

G

5

F

5

B

Bladzijde 147 15

“ACB = “ DEB (F-hoeken) f +ABC ऴ +DBE “CAB = “ EDB (F-hoeken) Stel BD = x, dan is AB = 7 + x. AB

BC

BD

BE

geeft

7+x

10

x

4

Dus 10x = 4(7 + x) 10x = 28 + 4x 6x = 28 x = 423 , dus BD = 423 en AB = 7 + 423 = 1123 .

106 Hoofdstuk 4

© Noordhoff Uitgevers bv

Indien “ACB = 90°, dan geldt in +ABC dat AB2 = AC2 + BC2 AB2 = 36 + 100 AB2 = 136 AB = 冑136 1123 > 冑136, dus “ ACB > 90°, dus bewering III is waar. b Uit +ABC ऴ +DBE volgt AC

BC

DE

BE

geeft

6

10

DE

4

6Â4 = 225 . 10 “ CAS = “ DES (Z-hoeken) f +ASC ऴ +ESD “ ACS = “ EDS (Z-hoeken)

DE =

AS

AC

ES

DE

geeft

AS

6

AE – AS

225

4

225 AS = 6(AE í AS) 225 AS = 6AE í 6AS 825 AS = 6AE 6AE AS = 2 = 57 AE 85 16

a De som van de hoeken van een driehoek is altijd 180°, dus “ A + “ B + “ C1 + “ C2 = 180°. b +ACM is een gelijkbenige driehoek met “A = “C1 (basishoeken). +BCM is een gelijkbenige driehoek met “B = “C2 (basishoeken). c “A + “B + “C12 = 180° ¶ “C1 + “C2 + “C12 = 180° “A = “C1 “B = “C2 “C12 + “C12 = 180° 2  “C12 = 180° “C12 = 90° Bladzijde 149

17

a “ACB = “MNB = 90° f +ABC ऴ +MBN “ABC = “MBN b AB BC 2BM BC geeft BM BN BM BN 2BM  BN = BM  BC 2BN = BC BN = 12 BC, dus BN = CN. c In +CMN is CN 2 + MN 2 = CM 2 MN 2 = CM 2 í CN 2 In +BNM is BN 2 + MN 2 = BM 2 MN 2 = BM 2 í BN 2 2 2 MN = BM í BN 2 MN 2 = CM 2 í CN 2 ¶ BM 2 í CN 2 = CM 2 í CN 2 BN = CN BM 2 = CM 2 BM = CM d BM = CM f AM = BM = CM AM = BM Dus A, B en C liggen op een cirkel met middelpunt M met straal AM en middellijn AB. Hiermee is de stelling bewezen.

18

a De de¿nitie zegt dat een raaklijn één punt met de cirkel gemeen heeft. Dus elk ander punt op de raaklijn ligt niet op de cirkel. Dit kan alleen als voor elk punt P van de raaklijn geldt MP > MA.

© Noordhoff Uitgevers bv

Meetkunde 107

b Voor elk punt P waarvoor P  A geldt dat MP > MA. Dus MA is de korste verbinding van punt M tot raaklijn l. Uit de definitie van de afstand een een punt tot een lijn volgt dat MA C l. Bladzijde 150 19

l is raaklijn aan cirkel met middelpunt M, dus AM C l f “MAN = 180°, dus MN C l l is raaklijn aan cirkel met middelpunt N, dus AN C l

20

MQ C l en MP C k, dus “ AQM = “ APM = 90°. In +APM is AP2 + MP2 = MA2 In +AQM is AQ2 + MQ2 = MA2 ¶ AP2 + MP2 = AQ2 + MQ2 MP = MQ AP2 + MQ2 = AQ2 + MQ2 AP2 = AQ2 AP = AQ

l Q

A

M

4 P k

21

a “ ACB = 90° en “ ADB = 90° (stelling van Thales) In +ADB is BD2 + AD2 = AB2 BD2 + 16 = 36 BD2 = 20 BD = 2冑5 In +ABC is BC2 + AC2 = AB2 BC2 + 9 = 36 BC2 = 27 BC = 3冑3 b In +ABC is sin(“ABC) = “ABC = 30°

AC 3 = AB 6

AD 4 = AB 6 “CAB § 41,8° “CBD = “ABC + “CAB § 30° + 41,8° = 71,8°  90°, waaruit volgt dat B niet op de cirkel met middellijn CD ligt. (omgekeerde stelling van Thales) In +ADB is sin(“ABD) =

22

Zie de figuur hiernaast. Uit opgave 20 weet je dat AQ = PQ = BQ. Teken DM loodrecht op NB. In +DMN is DM = 冑(3 + 5)2 í 22 = 冑60 = 2冑15, dus AB = 2冑15. Dus AQ = PQ = BQ = 冑15.

c1 3

M

A

3 P

Q

5

c2 N

l

2

D

B

3

k

108 Hoofdstuk 4

© Noordhoff Uitgevers bv

23

In +ADC is CD2 + AD2 = AC2 CD2 + 1 = 9 CD2 = 8 CD = 冑8 § 2,83 dm De hoogte van het bankje is 28,3 cm..

C

2

2

1 A

1 1

D

1

B

4.2 De sinusregel en de cosinusregel Bladzijde 152

C

24

4

D 76°

A

12

B

AD , dus AD = 12 sin(76° ) = 11,64... 12 “C = 180° í 48° í 76° = 56° 11,64... 11,64... In +ACD is sin(56°) = § 14,04. , dus AC = AC sin(56°) In +ABD is sin(76°) =

25

a

C

b

A

b

6,8

50°

75° c

6,8 b = sin(50°) sin(75°) 6,8 Â sin(75°) b= § 8,6 sin(50°)

6,8 c = sin(50°) sin(55°) 6,8 Â sin(55°) c= § 7,3 sin(50°)

B

Ȗ = 180° í 50° í 75° = 55° Bladzijde 153 26

CD , dus CD = b sin(Į). b CD In +BCD is sin( ȕ) = a , dus CD = a sin(ȕ).

a In +ACD is sin(Į) =

b Uit a volgt a sin(ȕ) = bsin(Į) b sin(Į) sin(ȕ) b a = sin(Į) sin(ȕ)

a=

AE c In +ABE is sin( ȕ) = c , dus AE = c sin(ȕ). AE , dus AE = b sin(Ȗ). In +ACE is sin(Ȗ) = b

© Noordhoff Uitgevers bv

Meetkunde 109

d Uit c volgt b sin(Ȗ) = c sin(ȕ) c sin(ȕ) b= sin(Ȗ) c b = sin(ȕ) sin(Ȗ) e

27

a b = sin(Į) sin(ȕ)

a b c b c  sin(Į) = sin(ȕ) = sin(Ȗ) = sin(ȕ) sin(Ȗ)

a

C 11°

4

88° A

81° 420 m

B

“ACB = 180° í 88° í 81° = 11° 420  sin(81°) 420 AC geeft AC = § 2174 m = sin(11°) sin(81°) sin(11°) 28

a “DAC = 180° í “BAC = 180° í Į CD , dus CD = b sin(180° í Į). b In +ACD is sin(180° í Į) = b CD c In +BCD is sin(ȕ) = a , dus CD = a sin(ȕ). d Uit b en c volgt a sin(ȕ) = b sin(180° í Į) b sin(180° í Į) sin(ȕ) b a = sin(180° í Į) sin(ȕ)

a=

Bladzijde 154 M

29 50°

5,3 K

20°

110°

“M = 180° í 20° í 110° = 50° 5,3 KM = sin(20°) sin(110°) 5,3  sin(110°) KM = § 14,6 sin(20°)

110 Hoofdstuk 4

L

KL 5,3 = sin(20°) sin(50°) 5,3 Â sin(50°) KL = § 11,9 sin(20°)

© Noordhoff Uitgevers bv

30

a

C

C

6

6

5

5

A

50°

A

B

50° B driehoek 2

driehoek 1

5 6 c b = = sin(50°) sin(ȕ) sin(Ȗ)

4

6 Â sin(50°) = 0,91... 5 ȕ § 66,8° Ȗ = 180° í 50° í 66,8...° § 63,2° 5 c = sin(50°) sin(63,1...°) sin(ȕ) =

5 Â sin(63,1...°) § 5,8 sin(50°) 6 c 5 c = = sin(50°) sin(ȕ) sin(Ȗ) c=

6 Â sin(50°) = 0,91... 5 ȕ = 180° í 66,8...° § 113,2° Ȗ = 180° í 50° í 113,1...° § 16,8° c 5 = sin(50°) sin(16,8...°) sin(ȕ) =

c=

31

5 Â sin(16,8...°) § 1,9 sin(50°)

a

C

10

B

46°

C

10

8

A

B

46°

8

A

10  sin(46°) 10 8 geeft sin(“A) = dus “A = 64,0...°. = 8 sin(“A) sin(46°) In de stomphoekige driehoek is “A = 180° í 64,0...° § 116,0°. b In de scherphoekige driehoek is “C = 180° í 46° í 64,0...° = 69,9...° 8  sin(69,9...°) 8 AB geeft AB = § 10,4 = sin(46°) sin(69,9...°) sin(46°) In de stomphoekige driehoek is “C = 180° í 46° í 115,9...° = 18,0...°. 8  sin(18,0...°) 8 AB geeft AB = § 3,4 = sin(46°) sin(18,0...°) sin(46°)

© Noordhoff Uitgevers bv

Meetkunde 111

c

C

10

B

7

46°

De cirkel met middelpunt C en straal 7 snijdt het andere been van hoek B niet. 32

a

C

4

6

A

60° B

BC sin(60°) = , dus BC = 6 sin(60°) § 5,20. 6 Er is geen driehoek mogelijk voor a < 5,20. b Er is precies één driehoek mogelijk voor a = 5,20  a • 6. c Er zijn twee driehoeken mogelijk voor 5,20 < a < 6. 33

a

D 152,7° 16,1°

A

10,3°

B

E 71,8° C

“CAD = 10,3° + 16,1° = 26,4° 235  sin(71,8°) 235 AD In +ACD is , dus AD = = = 502,08... sin(26,4°) sin(71,8°) sin(26,4°) In +ABD is “ABD = 180° í 16,1° í 152,7° = 11,2°. In +ABD is

502,08... Â sin(152,7°) 502,08... AB , dus AB = § 1186 m. = sin(11,2°) sin(152,7°) sin(11,2°)

Bladzijde 155 34

a

M

C

37,72°

A 48° B

11,03°

Amsterdam ligt op 52,5° NB en Benin City op 4,5° NB, dus Ȗ = 52,5° í 4,5° = 48°. MA = MB = 6378 km “MAB is gelijkbenig met tophoek M, dus 180° í 48° “MAB = “MBA = = 66°. 2 “BAC = 180° í 66° í 37,72° = 76,28° “ABC = 180° í 66° í 11,03° = 102,97° “ACB = 180° í 76,28° í 101,97° = 0,75° 112 Hoofdstuk 4

© Noordhoff Uitgevers bv

In +ABM is

AB AM = sin(“AMB) sin(“ABM)

6378 AB = sin(48°) sin(66°) 6378  sin(48°) AB = = 5188,3... km sin(66°) AC AB In +ABC is = sin(“ ABC) sin(“ ACB) 5188,3... AC = sin(102,97°) sin(0,75°) 5188,3...  sin(102,97°) § 386 258 km sin(0,75°) Dus de gevraagde afstand is 386 300 km. AC =

35

a De sinusregel geeft

4 5 6 = = . sin(Į) sin(ȕ) sin(Ȗ)

4

Bij elke combinatie van twee breuken zijn er twee onbekenden. Je kunt dus niet de sinusregel gebruiken. QR 5 6 . b De sinusregel geeft = = sin(50°) sin(“Q) sin(“R) Bij elke combinatie van twee breuken zijn er twee onbekenden Je kunt dus niet de sinusregel gebruiken. Bladzijde 156 36

a In +ACD is AD2 + CD2 = AC2 x2 + h2 = b2 b In +BCD is BC2 = BD2 + CD2 a2 = (c í x)2 + h2 a2 = c2 í 2cx + x2 + h2 2 2 c a = c í 2cx + x2 + h2 2 f a2 = c2 í 2cx + b x2 + h2 = b2 2 2 2 a = b + c í 2cx x d In +ACD is cos(Į) = , dus x = b cos(Į). b e a2 = b2 + c2 í 2cx 2 f a = b2 + c2 í 2cb cos(Į) x = b cos(Į) a2 = b2 + c2 í 2bc cos(Į)

37

a In +ACD is AD2 + CD2 = AC 2 x2 + h2 = b2 b In +BCD is BC2 = BD2 + CD2 a2 = (c + x)2 + h2 a2 = c2 + 2cx + x2 + h2 2 2 2 x +h =b f a2 = c2 + 2cx + b2 a2 = c2 + 2cx + x2 + h2 a2 = b2 + c2 + 2cx c In +ACD is “CAD = 180° í Į. AD Verder is cos(“CAD) = . AC x Dus cos(180° í Į) = , ofwel x = b cos(180° í Į). b d cos(180° í Į) = ícos(Į) f x = íb cos(Į) x = b cos(180° í Į) a2 = b2 + c2 + 2cx 2 f a = b2 + c2 + 2c  íb cos(Į) x = íb cos(Į) a2 = b2 + c2 í 2bc cos(Į) Dus de cosinusregel geldt ook voor stomphoekige driehoeken.

Bladzijde 157

© Noordhoff Uitgevers bv

Meetkunde 113

38

a In +ABD is AD2 = AB2 + BD2 í 2  AD  BD  cos(“B) 62 = 82 + 32 í 2  8  3  cos(“B) 36 = 73 í 48 cos(“B) 48 cos(“B) = 37 37 cos(“B) = 48 “B = 39,5...° In +ABC is AC2 = AB2 + BC2 í 2  AB  BC  cos(“B) 37 AC2 = 82 + 72 í 2  8  7  48 AC2 = 2623

冑

AC = 2623 § 5,164

b

C

4

7

A

39,5...° E

B

8

CE , dus CE = 7 sin(39,5...°) = 4,45... 7 opp +ABC = 12 Â 8 Â 4,45... § 17,84

In +BCE is sin(39,5...°) =

39

a2 = b2 + c2 í 2bc cos(Į) 52 = 62 + 72 í 2 Â 6 Â 7 Â cos(Į) 25 = 85 í 84 cos(Į) 84 cos(Į) = 60 60 cos(Į) = 84 Į § 44,4° Ȗ = 180° í 44,41...° í 57,12...° § 78,5° Dus Į § 44,4°, ȕ § 57,1° en Ȗ § 78,5°.

40

EF2 = DF2 + DE2 í 2  DF  DE  cos(“D) 42 = 72 + 52 í 2  7  5  cos(“D) 16 = 74 í 70 cos(“D) 70 cos(“D) = 58 58 cos(“D) = 70 “D § 34,0° DF2 = EF2 + DE2 í 2  EF  DE  cos(“E) 72 = 42 + 52 í 2  4  5  cos(“E) 49 = 41 í 40 cos(“E) 40 cos(“E) = í8 í8 cos(“E) = 40 “E § 101,5° “F = 180° í 34,04...° í 101,53...° § 44,4°

114 Hoofdstuk 4

b2 = a2 + c2 í 2ac cos(ȕ) 62 = 52 + 72 í 2 Â 5 Â 7 Â cos(ȕ) 36 = 74 í 70 cos(ȕ) 70 cos(ȕ) = 38 38 cos(ȕ) = 70 ȕ § 57,1°

F

7

D

5

4

E

© Noordhoff Uitgevers bv

41

42

a a2 = b2 + c2 í 2bc cos(Į) a2 = 52 + 62 í 2 Â 5 Â 6 Â cos(50°) a2 = 22,43... a § 4,74 b b2 = a2 + c2 í 2ac cos(ȕ) 25 = 22,43... + 36 í 2 Â 4,73... Â 6 Â cos(ȕ) 25 = 58,43... í 56,83... Â cos(ȕ) 56,83... Â cos(ȕ) = 33,43... 33,43... cos(ȕ) = 56,83... ȕ § 54,0°

Alternatieve oplossing a b = sin(Į) sin(ȕ) 4,73... 5 = sin(50°) sin(ȕ) 5 Â sin(50°) 4,73... ȕ § 54,0°

sin(ȕ) =

a BC2 = AC2 + AB2 í 2  AC  AB  cos(“A) 82 = 72 + 102 í 2  7  10  cos(“A) 64 = 149 í 140 cos(“A) 140 cos(“A) = 85 85 cos(“A) = 140 “A = 52,61...° CD In +ACD is sin(52,61...°) = , dus CD = 7 sin(52,61...°) = 5,56.... 7 1 O(+ABC) = 2  10  5,56... § 27,8

4

C

8

7

A

B

D 10

Bladzijde 158 43

a In +ABS is BS2 = AB2 + AS2 í 2  AB  AS  cos(“A) 4,52 = 102 + 72 í 2  10  7  cos(“A) 20,25 = 149 í 140 cos(“A) 128,75 cos(“A) = 140 “A § 23,1° In +ABC is BC2 = AC2 + AB2 í 2  AC  AB  cos(“A) 128,75 BC2 = 142 + 102 í 2  14  10  140 BC2 = 38,5 BC § 6,2 CE b In +ACE is sin(23,12...°) = , dus CE = 14 sin(23,12...°) § 5,49. 14 O(ABCD) = AB  CE § 10  5,49 = 54,9

D

C 7

4,5 S 4,5

7 A

B

10

C

D

14

A

44

23,1° 10

a Zie de figuur hiernaast. “B1 = “A1 = 40° (Z-hoeken) “B2 = 180° í “B3 = 180° í 110° = 70° (gestrekte hoek) “ABC = “B1 + “B2 = 40° + 70° = 110°

B

E

N

B N

40°

1

40°

3

110°

2

70°

C

1

A

© Noordhoff Uitgevers bv

Meetkunde 115

b AC2 = AB2 + BC2 í 2  AB  BC  cos(“B) AC2 = 2002 + 3002 í 2  200  300  cos(110°) AC2 = 171042,4... AC § 414 m Dus Harm had 414 meter moeten lopen.

B 110°

C

BC AC c = A sin(“ BAC) sin(“ ABC) 413,57... 300 = sin(“ BAC) sin(110°) 300  sin(110°) sin(“ BAC) = 413,57... “BAC § 43° Dus de koers is 40° + 43° = 83°. d “C1 = “B2 = 70° (Z-hoeken) “ C4 = 180° í 110° í 42,97...° = 27,02...° (hoekensom driehoek) “ACD = “C3 = 360° í 70° í 230° í 27,02...° = 32,97...° (volle hoek) In +ADC is AD2 = AC2 + CD2 í 2  AC  CD  cos(“ACD) AD2 § 413,57...2 + 4002 í 2  413,57...  400  cos(33,0°) AD2 § 53 473,17... AD § 231 m 231,24... CD 400 AD geeft = = sin(“ DAC) sin(“ ACD) sin(“ DAC) sin(32,97...°)

4

Dus sin(“ DAC) =

300 m

200 m

N

B

N

1

40°

A

3

110°

N

2

70°

40° 43°

27°

414 m

400 Â sin(32,97...°) 231,43...

70° C1

4

33°

3

2

230°

400 m

“ DAC § 70° Harm had 231 meter met koers 40° + 43° + 70° = 153° moeten lopen D

4.3 Lengten en oppervlakten Bladzijde 160 45

De oppervlakte is gelijk aan de oppervlakte van een rechthoek met zijden van 3 en 2 plus de oppervlakte van een halve cirkel met straal 1 cm, dus O = 3 Â 2 + 12 Â ʌ Â 12 = 6 + 12 ʌ § 7,57 cm 2 = 757 mm 2.

2 1

1

1 2

Bladzijde 161 46

a In +ABC is

10 sin(70°) 10 AB geeft AB = = = 12,26... sin(50°) sin(70°) sin(50°)

O(+ABC) = 12 Â 8,66... Â 12,26... = 53,11... De oppervlakte van het gevraagde gebied is 53,11 % í 47 a

70 Â ʌ Â 8,66 %2 § 7,30. 360

CD , dus CD = 5 sin(30°) = 2,5. 5 1 O(+ABC) = 2 Â 6 Â 2,5 = 7,5 De oppervlakte van het gevraagde vlakdeel is 30 7,5 í Â ʌ Â 42 = 7,5 í 113 ʌ. 360

In +ADC is sin(30°) =

1

4

A

116 Hoofdstuk 4

C

30° 4

D

B

© Noordhoff Uitgevers bv

49

DE , geeft DE = 4 sin(60°) = 3,46... 4 1 O(ABCD) = 2 Â (3 + 6) Â 3,46... = 15,58... 60 De oppervlakte van het gevraagde vlakdeel is 15,58... í Â ʌ Â 42 § 7,21. 360

In +AED is sin(60°) =

C

4

60°

A

49

3

D

E

AM = BM = CM = 5 In +ADM is DM = 冑52 í 42 = 3. O(+ABC) = 12 Â 8 Â (5 + 3) = 32 De oppervlakte van het gevraagde gebied is ʌ Â 52 í 32 = 25ʌ í 32.

B

2 C

4

5

M

3

5 A

50

In +ABC is BC = 冑102 í 42 = 冑84 = 2冑21. O(blauwe vlakdelen) = O(halve cirkel ) í O(+ABC) = 12 Â ʌ Â 52 í 12 Â 4 Â 2冑21 = 1212 ʌ í 4冑21 2 In +DEM is DM = DE2 + EM2 52 = DE2 + 32 DE2 = 25 í 9 = 16, dus DE = 4. O(gele vlakdelen) = O(halve cirkel ) í O(+ABD) = 1212 ʌ í 12 Â 10 Â 4

4

A

5

5

B

4

D

M

3

E

B

5

D

= 1212 ʌ í 20 Het verschil tussen de oppervlakten van het blauwe en het gele gebied is 1212 ʌ í 4冑21 í (1212 ʌ í 20) = 20 í 4冑21. Alternatieve oplossing In +ABC is BC = 冑102 í 42 = 冑84 = 2冑21. In +DEM is DM 2 = DE 2 + EM 2 52 = DE2 + 32 DE2 = 25 í 9 = 16, dus DE = 4. O(blauwe vlakdelen) í O(gele vlakdelen) = O(+ABD) í O(+ABC) = 12 Â 10 Â 4 í 12 Â 4 Â 2冑21 = 20 í 4冑21

© Noordhoff Uitgevers bv

Meetkunde 117

Bladzijde 162 51

4

In +ABC is 52 = 62 + 102 í 2  6  10  cos(“ BAC) 25 = 136 í 120 cos(“ BAC) 111 cos(“BAC) = 120 “BAC = 22,33...° CF In +AFC is sin(22,33...°) = 10 CF = 10 sin(22,33...°) = 3,79... O(ABCD) = 6  3,79... = 22,79... In +ACD is 62 = 52 + 102 í 2  5  10  cos(“CAD) 36 = 125 í 100 cos(“ CAD) 89 cos(“CAD) = 100 “CAD = 27,12...° “BAD = “BAC + “CAD = 22,33...° + 27,12...° = 49,45...° De oppervlakte van het gevraagde vlakdeel is 22,79... í 2 Â

52

“C = 180° í “A í “B = 180° í 70° í 45° = 65° 10 sin(45°) AC 10 geeft AC = = = 7,80... sin(45°) sin(65°) sin(65°) CD In +ADC is sin(70°) = 7,80... CD = 7,80...  sin(70°) = 7,33... O(+ABC) = 12  10  7,33... = 36,65... De oppervlakte van het gevraagde gebied is 36,65... í

53

D

1

5

5

E

5

C

5

5

A

1

5

B

F

49,45... Â ʌ Â 52 § 1,22. 360 C

7,80...

70 + 45 Â ʌ Â 52 § 11,57. 360

a Binnen het vierkant bevinden zich één cirkel en vier kwartcirkels. De oppervlakte van het blauwe gedeelte binnen de cirkel is dus 2ʌr2. In +ABC is AB = BC en AB2 + BC2 = AC2 2 Â AB2 = (4r)2 2 Â AB2 = 16r2 AB2 = 8r2 AB = r冑8 De oppervlakte van het vierkant is (r冑8)2 = 8r2. 2ʌr2 ʌ Het deel van het vierkant dat blauw is, is 2 = . 4 8r Alternatieve oplossing Door het vierkant met diagonalen in kwarten te delen, is binnen elk kwart de verhouding gelijk aan de verhouding binnen het gehele vierkant. Door twee kwarten tegen elkaar te leggen ontstaat een kleiner vierkant. Zie de figuur hiernaast. ʌr2 ʌr2 ʌ Het deel van het vierkant dat blauw is, is = 2= . 2 4 (2r) 4r

A

70°

B

D

C

D r

r

r

r A

B

r

r

r

r

b O(vierkant) = a, dus zijden vierkant zijn 冑a. O(cirkel) = b, dus ʌr2 = b b r2 = ʌ r=

b ʌ Å

b = 2冑bʌ. Åʌ De lengte van de rode lijn = 4 Â zijde vierkant + 2 Â halve omtrek cirkel. = 4冑a + 2冑bʌ Omtrek cirkel = 2ʌr = 2ʌ

118 Hoofdstuk 4

© Noordhoff Uitgevers bv

c Het blauwe gebied bestaat uit een halve cirkel en deel I en deel II Door deel I en deel II te verplaatsen, vormen deze samen met de halve cirkel een rechthoek. De oppervlakte van het blauwe gebied is dus 2 Â 4 = 8 cm 2.

2

2 II

I 2

2 II

I

Bladzijde 163 54

55

360° = 60°. 6 b AM = BM en “AMB = 60°, dus +ABM is gelijkzijdig en dus is AM = BM = AB = 4. In +ANM is AN 2 + MN 2 = AM 2 22 + MN2 = 42 MN2 = 16 í 4 = 12, dus MN = 冑12 O(+ABM) = 12  4  冑12 § 6,93 c O(ABCDEF) = 6  O(+ABM) = 6  6,92... § 41,6

M

a In +ABM is “AMB =

a

4

4

4

2

A

2

B

N

C

A

70°

B

D

CD b In +ADC is sin(70°) = , dus CD = AC Â sin(70°). AC O(+ABC) = 12 Â AB Â CD = 12 Â AB Â AC Â sin(70°) Bladzijde 164

56

M

O(+AMB) = 12  AM  BM  sin(“AMB) 360° = 12  6  6  sin 8 = 12,72... O(ABCDEFGH ) § 8  12,72... § 101,82

( )

6

6

A

57

360° 360° , dus “AMK = . 7 14 1,5 360° In +AKM is sin , dus AM = = 14 AM

B

“AMB =

( )

M

1,5 = 3,45... 360° sin 14

( )

O(+ABM) = 12 Â 3,45... Â 3,45... Â sin

( )

360° = 4,67... 7

O(ABCDEFG) = 7 Â 4,67... § 32,71

A

© Noordhoff Uitgevers bv

1,5

K

1,5

B

Meetkunde 119

Bladzijde 165 58

360° = 72°, dus “AMC = 36°. 5 140 140 In +ACM is sin(36°) = , dus AM = = 238,18... m. AM sin(36°) 1 O(+ABM) = 2  238,18...  238,18...  sin(72°) = 26977,08... m 2 O(Pentagon) = 5  26977,08... = 134885,42... m 2 § 13,5 ha “AMB =

M

36° 36°

A

4

59

60

360° = 40° en “AMC = 20°. 9 Neem AM = BM = a, dan geldt O(negenhoek) = 9  12  a  a  sin(40°) geeft 412 a2 sin(40°) = 180 40 a2 = sin(40°) 40 a = sin(40°) = 7,88% Å AC , dus AC = 7,88...  sin(20°) = 2,69... In +ACM is sin(20°) = 7,88... De omtrek van de negenhoek is 18  2,69... § 48,56.

C

140 m

B

“AMB =

M

a

A

O(cN ) = 10, dus ʌ Â AN 2 = 10 10 AN 2 = ʌ AN = Zie de figuur hiernaast. In +ASN is sin(40°) =

In +AMS is sin(25°) =

10 Åʌ AS

10 Åʌ

20° 20°

a

B

C

A

, dus AS =

M

10 Â sin(40°) Åʌ

u AM Â sin(25°) =

AS , dus AS = AM Â sin(25°) AM AM =

1 2

O(+ABM) = Â 2,71... Â 2,71... Â sin(50°) § 2,82 61

140 m

25°

40° S

25°

10 Â sin(40°) Åʌ

10 Â sin(40°) Åʌ sin(25°)

40°

N

B

§ 2,71...

a O(+ABC) = 12 Â AB Â CE b O(+ABC) = 12 Â BC Â AD c O(+ABC) = 12 Â AB Â CE 1 2

O(+ABC) = Â BC Â AD

1

 AB  CE = 12  BC  AD AB  CE = BC  AD

¶ 2

C E

Bladzijde 167

62

CD = 冑62 + 32 = 冑45 = 冑9 Â 冑5 = 3冑5 De zijde × hoogte-methode in +BCD geeft CD × BE = BC × DF 3冑5 Â BE = 4 Â 6 8 冑5 8 4Â6 24 BE = = = Â = 冑5 = 135 冑5 3冑5 3冑5 冑5 冑5 5

120 Hoofdstuk 4

3

D

F

1

1

A

6

B

© Noordhoff Uitgevers bv

63

64

65

BC = 冑42 + 22 = 冑20 = 冑4 Â 冑5 = 2冑5 De zijde × hoogte-methode in +ABC geeft BC × AD = AB × AC 2冑5 Â AD = 4 Â 2 4Â2 4 冑5 4 AD = = Â = 冑5 冑 冑 2 5 5 冑5 5 a CD = 冑132 í 52 = 冑144 = 12 b De zijde × hoogte-methode in +ABC geeft AC × BE = AB × CD 13 Â BE = 10 Â 12 10 Â 12 120 3 = 13 = 913 BE = 13 BE = 冑42 + 22 = 冑20 = 冑4 Â 冑5 = 2冑5 De zijde × hoogte-methode in +BCE geeft BE × CF = BC × EG 2冑5 Â CF = 4 Â 4 8 冑5 8 4Â4 = CF = Â = 冑5 = 135 冑5 2冑5 冑5 冑5 5

D

C

2

E

G

2 F A

66

O(ABCD) = AB Â CF = 8 Â CF = 32, dus CF = 4. BF = 冑52 í 42 = 3 AC = 冑112 + 42 = 冑137 De zijde × hoogte-methode in +ABC geeft AC × BE = AB × CF 冑137 Â BE = 8 Â 4 32 冑137 32 8Â4 = = 冑137 BE = Â 冑137 冑137 冑137 137

B

4

D

C E

5 5

A

8

B

3

4

F

4.4. Vergelijkingen in de meetkunde Bladzijde 169 67

a In +ACD is AC 2 = AD2 + CD2 (2a)2 = a2 + CD2 CD2 = 4a2 í a2 CD2 = 3a2 CD = a冑3 b CD = 7冑3 geeft a冑3 = 7冑3, dus a = 7. AC = 2a = 2 Â 7 = 14 Bladzijde 170

68

AB = 8 geeft AC =

8 8 冑2 8冑2 = = 4冑2 Â = 冑2 冑2 冑2 2

EF = 6冑3 geeft DF =

6冑3 = 3冑3 en DE = 冑3 Â DF = 冑3 Â 3冑3 = 9 2

KM = 10 geeft MN =

10 = 5 en KN = 冑3 Â MN = 5冑3 ¶ KL = KN + NL = 5冑3 + 5 2

MN = 5 geeft NL = 5

RS = 3 geeft QS = 3冑2 dus PQ = 3冑2 Â 冑3 = 3冑6

© Noordhoff Uitgevers bv

Meetkunde 121

4

69

Noem het middelpunt van de omgeschreven cirkel G. 360° 180° í 60° +ABG is gelijkbenig en “AGB = = 60°, dus “A = “B = = 60°. 6 2 Hieruit volgt dat +ABG een gelijkzijdige driehoek is. Uit AK = 4 en “A = 60° volgt GK = 4冑3. O(ABCDEF) = 6  O(+ABG) = 6  12  8  4冑3 = 96冑3

G

A

+KLG is een gelijkzijdige driehoek met zijde 4冑3, dus KL = 4冑3, KH = 2冑3 en GH = 2冑3 Â 冑3 = 6. O(KLMNOP) = 6 Â O(+KLG) = 6 Â 12 Â 4冑3 Â 6 = 72冑3 De oppervlakte van het gekleurde gebied is O(ABCDEF) í O(KLMNOP) = 96冑3 í 72冑3 = 24冑3.

4

60°

60° 4

K

4

B

G

L

4 3

H

Alternatieve oplossing Noem het middelpunt van de omgeschreven cirkel G. 360° +ABG is gelijkbenig en “AGB = = 60°, 6 180° í 60° dus “A = “B = = 60°. 2 Hieruit volgt dat +ABG een gelijkzijdige driehoek is. Op dezelfde manier is aan te tonen dat +PKG een gelijkzijdige driehoek is. Hieruit volgt dat “S = 90° en dat “AKS = 180° í 90° í 60° = 30°. AK = 4 geeft AS = 12  4 = 2 en KS = 2冑3. De oppervlakte van het gekleurde gebied is 12  O(+AKS) = 12  12  2  2冑3 = 24冑3.

K G 30°

P

60°

S

4

A

70

30° 30°

60°

30° 4

60°

60° K

4

B

a In +ABE is AB = 冑2 en AE = 1. In +BDE is DE = 冑3 en BD = 2. AD = AE + DE = 1 + 冑3 AD 1 + 冑3 1 + 冑3 冑2 冑2 + 冑6 1 In +ACD is AC = = = = 2 冑2 + 12 冑6. Â = 冑2 冑2 冑2 冑2 2 CD = AC = 12 冑2 + 12 冑6 BC = AC í AB = 12 冑2 + 12 冑6 í 冑2 = í 12 冑2 + 12 冑6 b sin(15°) =

1 1 BC í 2 冑2 + 2 冑6 = = í 14 冑2 + 14 冑6 BD 2

c cos(15°) =

CD = BD

122 Hoofdstuk 4

1 2

冑2 + 12 冑6 2

= 14 冑2 + 14 冑6

© Noordhoff Uitgevers bv

Bladzijde 171 71

Stel de hoogte CD = x, dan is x x 冑3 x冑3 1 AD = = = x冑3 en BD = x  = 冑3 冑3 冑3 3 3

C

Uit AD + DB = AB volgt 13 x冑3 + x = 12 x(

1 3

冑3 + 1 ) = 12

x=

1 3

36 12 = 冑3 + 1 冑3 + 3

O(+ABC) = 12 Â AB Â CD = 12 Â 12 Â 72

x

A

45° D

B

36 216 = 冑3 + 3 冑3 + 3

Stel de zijde van de regelmatige achthoek x, dan is AP = BQ =

60°

6

D

C

x 冑2 x冑2 1 x = = x冑2. Â = 冑2 冑2 冑2 2 2

4

Uit AP + PQ + BQ = AB volgt 12 x冑2 + x + 12 x冑2 = 6 x冑2 + x = 6 x(冑2 + 1) = 6

W x

6 x= 冑2 + 1

x x

A

P

Q

B

Bladzijde 172

73

74

In de rechthoekige +AEM is “MAE = 30° en ME = r, dus AE = r冑3. Uit AE + ED = AD volgt r冑3 + r = 6 r(冑3 + 1) = 6 6 r= 冑3 + 1 C

6 In +AEM is ME = r dus AM = 2r met r = (zie opgave 73). 冑3 + 1 Stel de straal van de kleine cirkel x, dan is AN = 3r + x. In +ADN is AD = 6 dus AN =

6 12 冑3 12冑3 = 4冑3 Â2 = Â = 冑3 冑3 冑3 3

3r + x = 4冑3 geeft x = 4冑3 í 3r x = 4冑3 í 3 Â x = 4冑3 í

6 冑3 + 1

18 冑3 + 1

r

P

N

M

r A

30° E

B

D 12

75

D

a Stel de straal van het cirkeltje x. 2 冑2 2 = In +ASD is AD = 2, dus DS = Â = 冑2 冑2 冑2 冑2 ¶ 1 + x = 冑2 DS = 1 + x x = 冑2 í 1 Dus de straal van het cirkeltje is 冑2 í 1.

1

1

1 x S

A

C

x 1 1 1 B

© Noordhoff Uitgevers bv

Meetkunde 123

b Stel DE = x. Dan is CE = AE = 24 í x. In +ACD is AD2 + DE2 = AE2 122 + x2 = (24 í x )2 144 + x2 = 576 í 48x + x2 48x = 432 x=9 Dus de zijde van de ruit is 24 í 9 = 15.

F

D

24 − x

E

x

C

24 − x

12

A

4

Bladzijde 173 76

77

Noem de straal van de grote gele cirkel a en van de kleine gele cirkel b, 2a + 2b = a + b, dus EM = a + b. dan is de straal van de buitenste cirkel 2 De oppervlakte van het blauwe gedeelte is ʌ(a + b)2 í ʌa2 í ʌb2 = 2abʌ, dus 2abʌ = 2ʌ geeft ab = 1. Zie de figuur hiernaast met M het middelpunt van de buitenste cirkel. Stel AD = x. DM = EM í DE = a + b í 2b = a í b In +ADM is AD2 + DM 2 = AM 2 x2 + (a í b)2 = (a + b)2 x2 + a2 í 2ab + b2 = a2 + 2ab + b2 x2 = 4ab x = 2冑ab x = 2冑ab f x = 2冑1 = 2, dus AB = 4. ab = 1

a

M

A

x

B

D b

E

a CD = 6 í 2  AE 3 3 冑3 3冑3 = = 冑3. In +AED is “A = 60° en DE = 3, dus AE =  = 冑 冑 3 3 冑3 3 Dus CD = 6 í 2冑3. b O(ABCD) = 12 (AB + CD)  DE = 12 (6 + 6 í 2冑3 )  3 = 112 (12 í 2冑3 ) = 18 í 3冑3 Bladzijde 174

78

Stel CD = x, dan is AB = AE + x.

O(ABCD) = 12 (AB + CD) Â DE = 12 ( x + 3冑3 + x ) Â 9

9

= 412 (2x + 3冑3 ) = 9x + 1312 冑3

x

D

9 冑3 9冑3 9 = = 3冑3. In +AED is “A = 60° en DE = 9, dus AE =  = 冑3 冑3 冑3 3

A

60° E

C

9

B

O(ABCD) = 54 geeft 9x + 1312 冑3 = 54 9x = 54 í 1312 冑3 x= Dus CD = 6 í 112 冑3.

124 Hoofdstuk 4

54 í 1312 冑3 = 6 í 112 冑3 9

© Noordhoff Uitgevers bv

79

a h = 4 geeft AE =

4 4 冑3 4冑3 = 113 冑3 en BF = 4. = Â = 冑3 冑3 冑3 3

D

DC = EF = 10 í 4 í 113 冑3 = 6 í 113 冑3 O(ABCD) = 12 (10 + 6 í 113 冑3 ) Â 4 = 32 í 223 冑3

C

4

A

60°

45° E

h h 冑3 h冑3 1 = h冑3 en BF = h. b DE = h geeft AE = = Â = 冑3 冑3 冑3 3 3

F

D

h ( 13 冑3 + 1 ) = 8 24 8 = h=1 冑 冑 3 + 1 3 +3 3 144 24 = 冑3 + 3 冑3 + 3

B

10

AE + EF + BF = 10 geeft 13 h冑3 + 2 + h = 10

O(ABCD) = 12 (10 + 2) Â

4

C

h

A

60°

45° E

F 10

B

4

c AE = 13 h冑3 en BF = h. Stel CD = x. Uit AB = 10 volgt x = 10 í 13 h冑3 í h 1 3 h冑3 + h = 10 í x h ( 13 冑3 + 1 ) = 10 í x 10 í x h=1 3 冑3 + 1 Uit O(ABCD) = 25 volgt 12  (10 + x)  h = 25 50 h= 10 + x 50 10 í x = . Dus 1 10 + x 3 冑3 + 1 10 í x 50 en y2 = . 冑3 + 1 10 + x Intersect geeft x § í4,60  x § 4,60 vold. niet vold. Dus CD § 4,60. Voer in y1 =

1 3

Alternatieve oplossing AE = 13 h冑3, BF = h en CD = EF = 10 í 13 h冑3 í h. O(ABCD) = 12 (10 + 10 í 13 h冑3 í h)  h = 12 h ( 20 í 13 h冑3 í h ) = 10h í 16 h2  冑3 í 12 h2 O(ABCD) = 25 geeft 10h í 16 h2  冑3 í 12 h2 = 25 Voer in y1 = 10x í 16 x2  冑3 í 12 x2 en y2 = 25. Intersect geeft x = 3,425...  x = 9,254.... h = 3,425... geeft CD = 10 í 13  3,425...  冑3 í 3,425... § 4,60 h = 9,254... geeft CD = 10 í 13  9,254...  冑3 í 9,254... § í4,60 voldoet niet Dus CD § 4,60.

© Noordhoff Uitgevers bv

Meetkunde 125

80

a Uit “A = 60° en AM = a volgt AN = 12 a en MN = 12 a冑3.

M

O(ABCDEF) = 6  O(+ABM) = 6  12  a  12 a冑3 = 112 a2  冑3 b O(ingeschreven cirkel) = ʌ  MN 2 = ʌ  ( 12 a冑3 )2 = ʌ  14 a2  3 = 34 a2ʌ

a

c O(binnen zeshoek en buiten cirkel) = O(zeshoek) í O(cirkel) = 112 a2 Â 冑3 í 34 a2ʌ O(binnen zeshoek en buiten cirkel) = 10 geeft 112 a2 Â 冑3 í 34 a2ʌ = 10

60°

A

B

N

( 冑3 í ) = 10

a2 112

3 4ʌ

10 = 41,34... 112 冑3 í 34 ʌ a § 6,43 a2

81 4

=

O(ABCDEF) = 6 Â 12 Â 1 Â 12 冑3 = 112 冑3

M

O(PBCQ) = 13 Â O(ABCDEF) = 13 Â 112 冑3 = 12 冑3 Stel PR = x, dan is BR = x冑3. O(PBCQ) = 12 (PQ + BC) Â BR = 12 (2x + 1 + 1) Â x冑3 = (x + 1) Â x冑3 = (x2 + x)冑3 1 Dus 2 冑3 = (x2 + x)冑3

1

1 2

3

1 2

= x2 + x 1 = 2x2 + 2x 2x2 + 2x í 1 = 0 D = 22 í 4  2  í1 = 12 í2 í 冑12 í2 í 2冑3 í2 + 2冑3 x= = = í 12 í 12 冑3  x = = í 12 + 12 冑3 4 4 4 voldoet niet voldoet PQ = 2x + 1 = 2 (í 12 + 12 冑3) + 1 = í1 + 冑3 + 1 = 冑3

A

B

1 2

E

D Q x S

F

C

1

1

R x 3

x A

B

2x

P

Diagnostische toets Bladzijde 176 1

BC , dus BC = 20 sin(38°) = 12,31... 20 12,31... geeft “BDC § 64,0°. In +BCD is tan(“ BDC) = 6 “ADC § 180° í 64,0° = 116,0°

C

In +ABC is sin(38°) =

20

“ACD § 180° í 116,0° í 38° = 26,0° 2

a “ABC = “DBE f +ABC ऴ +DBE “CAB = “EDB = 90° CA

AB

DE

BD

DE =

geeft

9

12

DE

8

A

38° D

6

B

9Â8 =6 12

BE = 冑62 + 82 = 10

126 Hoofdstuk 4

© Noordhoff Uitgevers bv

b “ACS = “SDE (Z-hoeken) f +ACS ऴ +EDS “CAS = “SED (Z-hoeken) AE = 冑42 + 62 = 冑52 = 2冑13 Stel AS = x, dus ES = 2冑13 í x. AC

AS

DE

ES

geeft

9

x

6

2冑13 í x

Dus 6x = 9(2冑13 í x ) 6x = 18冑13 í 9x 15x = 18冑13 x = 115 冑13 Dus AS = 115 冑13. 3

“APM = “BPN f +AMP ऴ +BNP “MAP = “NBP = 90° Stel NP = x, dus MP = x + 8. AM MP 5 x+8 geeft BN NP 3 x

4 A B 5 3 M

Dus 5x = 3(x + 8) 5x = 3x + 24 2x = 24 x = 12 In +AMP is AP2 + AM 2 = MP 2 AP2 = 202 í 52 = 375 AP = 冑375 = 5冑15 4

a

6

40°

A

55° D

C

5

P

x

C

“ACB = 180° í 40° í 55° = 85° 6 sin(85°) 6 AB geeft AB = = = 9,29... sin(40°) sin(85°) sin(40°) CD In +BCD is sin(55°) = , dus CD = 6 sin(55°) = 4,91... 6 1 O(+ABC) = 2  9,29...  4,91... § 22,85. A

5

N

3

5

B

C

5

3

3

35° B

A

35° B

5 sin(35°) 3 5 geeft sin(ȕ) = = 3 sin(35°) sin(ȕ) ȕ § 72,9° In de stomphoekige driehoek is ȕ § 180° í 72,9° = 107,1°. Dus Ȗ § 180° í 72,9° í 35° = 72,1° of Ȗ § 180° í 107,1° í 35° = 37,9°.

© Noordhoff Uitgevers bv

Meetkunde 127

b

C

5

A

35° B

BC sin(35°) = geeft BC = 5 sin(35°) § 2,87 5 Er is slechts één driehoek mogelijk voor a § 2,87  a • 5. 6

4

a 52 = 62 + 42 í 2  6  4  cos(“Q) 48 cos(“ Q) = 52 í 25 27 cos(“Q) = 48 “Q § 55,8° b PM 2 = 42 + 32 í 2  4  3  cos(55,77...°) = 11,5 geeft PM § 3,39.

R

3 M

5

3

P

7

a In +ABD is 82 = 102 + 152 í 2  10  15  cos(“ABD) 300 cos(“ ABD) = 325 í 64 261 cos(“ABD) = 300 “ABD = 29,54...° DE In +BDE is sin(29,54...°) = 10 DE = 10 sin(29,54...°) = 4,93... O(ABCD) = 15  4,93... § 73,96

15

D

8

A

Q

4

C

10

8

E

B

F

15

b BF = 冑82 í 4,93...2 = 6,3 In +AFC is AC 2 = AF 2 + CF2 AC = 冑21,32 + 4,93...2 AC § 21,9 Bladzijde 177 8

82 = 92 + 62 í 2  9  6  cos(“CAB) 108 cos(“CAB) = 117 í 64 53 cos(“CAB) = 108 “CAB = 60,61...° CD In +ADC is sin(60,61...°) = , dus CD = 6 sin(60,61...°) = 5,22... 6 O(+ABC) = 12  9  5,22... = 23,52... “ ABC + “BCA + “ CAB = 180°, dus de oppervlakte van het vlakdeel dat binnen de driehoek maar buiten de cirkels ligt is 180 23,52... í  ʌ  32 § 9,39. 360

128 Hoofdstuk 4

C

8

6

A

B

D 9

© Noordhoff Uitgevers bv

O(omgeschreven cirkel) = 25ʌ geeft ʌr2 = 25ʌ, dus r = 5. 360° “ AMB = = 45°, dus “AMK = “BMK = 22,5°. 8 AK , dus AK = 5 sin(22,5°) = 1,91... In +AKM is sin(22,5°) = 5 De omtrek van de achthoek is 16  1,91... § 30,6. De oppervlakte van de achthoek is 8  12  5  5  sin(45°) § 70,7.

9

E

F

G

D

M

H

5

C

5 22,5° 22,5°

A

B

K

4 10

11

BM = 冑62 + 22 = 冑40 = 冑4 Â 冑10 = 2冑10 De zijde × hoogte-methode in +ABM geeft BM × AE = AB × AM 2冑10 Â AE = 6 Â 2 6Â2 6 冑10 6 AE = = = 冑10 = 35 冑10 Â 2冑10 冑10 冑10 10 Zie de figuur hiernaast. De zijde × hoogte-methode in +BCM geeft BM × CF = BC × GM 2冑10 Â CF = 4 Â 6 4Â6 12 冑10 12 CF = = = 冑10 = 115 冑10 Â 2冑10 冑10 冑10 10

D

2 M

G E 2

2 F

A

In +ABC is AC = 12 en “ACB = 45°, dus AB = BC = In +ABD is AB = 6冑2 en “ BAD = 30°, dus BD =

C

B

6

12 12 冑2 12 = Â = 冑2 = 6冑2. 冑2 冑2 冑2 2

6冑2 6冑2 冑3 6 = Â = 冑6 = 2冑6. 冑3 冑3 冑3 3

O(+ACD) = 12 Â CD Â AB = 12 Â (6冑2 í 2冑6 ) Â 6冑2 = 3冑2 Â (6冑2 í 2冑6 ) = 36 í 6冑12 = 36 í 6冑4 Â 冑3 = 36 í 12冑3 12

Stel AD = x. Dan is BC = CD = x. x x 冑2 1 AE = BF = = Â = x冑2. 冑2 冑2 冑2 2 AB = AE + EF + BF = 冑2 + x + AB = 18 geeft x冑2 + x = 18 x(冑2 + 1) = 18 18 x= 冑2 + 1 18 Dus AD = . 冑2 + 1 1 2x

13

D

x

C

x

x

冑2 = x冑2 + x

1 2x

A

In +BDF is “ B = 60° en BD = 6, dus BF = 12  6 = 3 en DF = 3冑3. In +AEC is “A = 30° en CE = DF = 3冑3, dus AE = 冑3  3冑3 = 9. Stel CD = EF = x geeft O(ABCD) = 12  (9 + x + 3 + x )  3冑3 = (6 + x )  3冑3 O(ABCD) = 36, dus (6 + x)  3冑3 = 36 36 6+x= 3冑3 12 12 冑3 x= í6=  í 6 = 123 冑3 í 6 = 4冑3 í 6 冑3 冑3 冑3

E

B

F 18

C x D

6

A

30°

60° E xF

B

Dus AB = 9 + 4冑3 í 6 + 3 = 6 + 4冑3.

© Noordhoff Uitgevers bv

Meetkunde 129

Wiskunde Olympiade 2006 A-vragen Bladzijde 180

1

4 Ali heeft om te beginnen 11 24 deel van al het geld. Aan het eind heeft Ali 9 . Om de verhoudingen te kunnen vergelijken zorgen we ervoor dat de breuken dezelfde noemer krijgen, in dit geval is dat 72. Ali ging van 33 32 24 15 16 72 naar 72 van het totaal. Bente bleef op 72 van het totaal en Chris ging van 72 naar 72 . Dus antwoord D is het goede antwoord.

2

In de linker driehoek is de derde hoek gelijk aan 180 í 7x graden. In de rechter driehoek is de derde hoek gelijk aan 180 í 13x graden. De beide basishoeken van de middelste driehoek zijn elk gelijk aan een van die hoeken. Dus moet gelden: 5x + (180 í 7x) + (180 í 13x) = 180 dus 15x = 180 en dus x = 12. Dus antwoord D is het goede antwoord.

3

De getallen 1 t/m 9 achter elkaar zijn negen cijfers. De getallen 10 t/m 99 achter elkaar zijn 90 getallen van twee cijfers, dus in totaal 180 cijfers. Samen zijn dat 189 cijfers. 1788 í 189 = 1599. 1599 Er zijn dus = 533 getallen van drie cijfers opgeschreven, te beginnen met 100. 3 Dus n = 99 + 533 = 632. Dus antwoord B is het goede antwoord.

4

6=0+0+6=0+1+5=0+2+4=0+3+3=1+1+4=1+2+3=2+2+2 Met 0, 0 en 6 kunnen we drie getallen maken, 600, 60 en 6, waarbij nullen aan het begin weggelaten zijn. Zo ook drie getallen met 0, 3 en 3 en ook met 1, 1 en 4 . Met 0, 1 en 5 kunnen we zes getallen maken; ook met 0, 2 en 4 en ook met 1, 2 en 3. Met 2, 2 en 2 kunnen we alleen het getal 222 maken. In totaal dus 3 Â 3 + 3 Â 6 + 1 = 28. Dus antwoord D is het goede antwoord.

5

Vergelijk de linker kolom met de diagonaal waarin 10 staat. In het middelste vakje (onder de N ) moet 13 staan omdat moet gelden 13 + 10 = 12 + 11. Elke rij, kolom of diagonaal is dus in totaal 11 + 13 + 15 = 39. Midden onder moet 17 staan omdat 39 í (12 + 10) = 17. Dus N = 39 í 13 í 17 = 9. Dus antwoord B is het goede antwoord.

6

Door de cijfers 1, 2, 3 en 4 precies één keer te gebruiken kun je 24 verschillende getallen maken. Als je die onder elkaar zet dan heb je op de plaats van de eenheden elk van de vier cijfers 1, 2, 3 en 4 precies zes keer. Dat levert opgeteld 60. Op de plaats van de 10-tallen komen de 1, 2, 3 en 4 ook precies zes keer voor. Dat levert een bijdrage van 600 aan de som. Evenzo voor de plaats van de 100- en 1000-tallen. De cijfers op de plaats van de 100-tallen dragen dus 6000 bij en de cijfers op de plaats van de 1000-tallen 60000. Totaal: 66660. Dus antwoord E is het goede antwoord. Bladzijde 181

7

Om het verschil van twee breuken – weer als breuk geschreven – zo klein mogelijk te maken moet de noemer zo groot mogelijk en de teller zo klein mogelijk zijn. De noemer kan maximaal 90 worden, de noemer van de ene breuk is dan 10 en de andere noemer is 9. De teller is minimaal 1. Die waarden worden 1 9 1 en bij 10 í 89 en is dus 90 . aangenomen bij 19 í 10 Dus antwoord C is het goede antwoord.

130 Wiskunde Olympiade

© Noordhoff Uitgevers bv

8

“PRB = 90° í 12 “B en “QRA = 90° í 12 “A “PRQ = 180° í “PRB í “QRA = 180° í (90° í 12 “B) í (90° í 12 “A) = 12 “B + 12 “A

A

R

= 12 (“B + “A) = 12  90° = 45° Dus antwoord B is het goede antwoord.

Q

B

C

P

B-vragen 1

Vanuit A heb je twee mogelijkheden: òf naar B òf naar D. Vanuit B (en ook vanuit D) heb je voor de volgende beurt weer twee mogelijkheden òf naar A òf naar C (ook vanuit D weer naar A of naar C). Vanuit A (en ook vanuit C) heb je voor de derde beurt weer twee mogelijkheden: naar B of naar D (ook weer naar B of D vanuit C). Bij elke stap dus twee mogelijkheden. Na 9 stappen ben je in B of D. Voor de tiende stap heb je dan maar één mogelijkheid naar A òf vanuit B òf vanuit D. In totaal dus 29 = 512 verschillende wandelingen.

2

De vier cijfers van het getal moeten een rekenkundige rij vormen. We tellen het aantal systematisch: - eerst de getallen 1111, 2222, … , 9999. Dat zijn er negen. - dan de getallen 1234, 2345, 3456, … , 6789 en 9876, 8765, … , 4321, 3210. Dat zijn er dertien. - vervolgens de getallen 1357, 2468, 3579 en 9753, 8642, 7531, 6420. Dat zijn er zeven. - vervolgens alleen nog het getal 9630. In totaal dus 9 + 13 + 7 + 1 = 30.

3

Q is de loodrechte projectie van P op de zijde AD. Dan AQ2 + PQ2 = AP2 = 100. ABCD is een vierkant, dus 2PQ = AQ + 10 ofwel AQ = 2PQ í 10. Dus (2PQ í 10)2 + PQ2 = 5PQ2 í 40PQ + 100 = 100. Dus PQ = 8, de zijde van het vierkant is 16 en de oppervlakte 256.

E

D

C

10

P

Q

10

10

A

4

B

ab is te schrijven als 10a + b, dus ab = K × (a + b) 10a + b = Ka + Kb 10a í Ka = Kb í b (10 í K)  a = (K í 1)  b Omdat K positief en geheel moet zijn, hoeven we alleen de waarden 1 t/m 10 voor K te onderzoeken. Verder moet natuurlijk gelden dat 1 ” a ” 9 en 0 ” b ” 9. Als K = 1 dan is a = 0, en dat mag niet. Dus K = 1 voldoet niet. Als K = 2 dan is 8  a = b en dan a = 1, b = 8 en ab = 18, dus deelbaar door 9. Dus K = 2 voldoet niet. Als K = 3 dan is 7  a = 2  b dus a = 2, b = 7 en ab = 27, dus deelbaar door 9. Dus K = 3 voldoet niet. Als K = 4 dan is 6  a = 3  b dus a = 1, b = 2 en ab = 12 of a = 2, b = 4 en ab = 24. 12 en 24 zijn niet deelbaar door 9. Dus K = 4 voldoet. Als K = 5 dan is 5  a = 4  b dus a = 4, b = 5 en ab = 45 en dat is deelbaar door 9. Dus K = 5 voldoet niet. Als K = 6 dan is 4  a = 5  b dus a = 5, b = 4 en ab = 54 en dat is deelbaar door 9. Dus K = 6 voldoet niet. Als K = 7 dan is 3  a = 6  b dus a = 2, b = 1 en ab = 21 of a = 4, b = 2 en ab = 42. 21 en 42 zijn niet deelbaar door 9. Dus K = 7 voldoet. Als K = 8 dan alleen ab = 72 en dat is deelbaar door 9. Dus 8 voldoet niet. Als K = 9 dan alleen ab = 81 en dat is deelbaar door 9. Dus 9 voldoet niet. Als K = 10 dan is b = 0. Dat geeft wel oplossingen, bijvoorbeeld 30 = 10  (3 + 0). Dus K = 10 voldoet. Dus 4, 7 en 10 zijn mogelijke waarden voor K.

© Noordhoff Uitgevers bv

Wiskunde Olympiade 131

2007 A-vragen Bladzijde 182 1

Als M uit vier enen bestaat dan is 2007 Â M = 2229777, dus de som van de cijfers is 3 Â (2 + 7) + 9 = 4 Â 9. Als M uit vijf enen bestaat dan is 2007 Â M = 22299777; de som van de cijfers is dan 3 Â (2 + 7) + 9 + 9 = 5 Â 9. Als M uit 2007 enen bestaat dan is de som van de cijfers 2007 Â 9 = 18063. Dus antwoord C is het goede antwoord.

2

Je hoeft hier de noemers niet gelijknamig te maken om de breuken te vergelijken. Je kunt de noemers 16 1 14 5 15 allemaal dicht bij de 100 kiezen: 0,16 = , = , = . 100 7 98 33 99 ník k k+1 13 14 15 16 k+1 k 17 Omdat bij k < n geldt < > 0, vinden we < < , immers í = < < . n + 1 n n(n + 1) 97 98 99 100 101 n n+1 Dus antwoord E is het goede antwoord.

3

Uit de negen punten kun je op

4

a+

5

6

9Â8Â7 = 84 manieren drie punten kiezen. Maar als drie punten op één lijn 3Â2Â1 liggen, dan levert dat geen driehoek op. Dat komt acht keer voor, drie keer horizontaal, drie keer verticaal en twee keer diagonaal. Dus 84 í 8 = 76 mogelijkheden. Dus antwoord A is het goede antwoord.

( )

ba + 1 ab + 1 1 1 = 13 Â b + a geeft = 13 Â a dus a = 13b en dus zeven paren (13, 1), (26, 2), … , (91, 7) . b b Dus antwoord B is het goede antwoord. Route A is langer dan route B want 冑5 > 2. Dus A valt af. Route E is langer dan route A, want 冑5 > 1, dus E valt af. Route D is langer dan route B want 冑13 > 冑10. Dus D valt af. Route C en route B hebben een stuk met lengte 冑5 en een stuk met lengte 2 gemeenschappelijk. Blijft over te vergelijken de rest van C, 2冑5 en de rest van B, 1 + 冑10. 冑10 + 1 < 2冑5, dus route B is het kortst. Dus antwoord B is het goede antwoord. We vervolgen de rij met alleen de laatste cijfers van de getallen: we vinden 2, 2, 4, 8, ..2, ..6, ..2, ..2, ..4, ..8, ..2, ..6, ..2, enz. We zien dat een groep van zes cijfers zich telkens herhaalt. Omdat 2007 = 334 Â 6 + 3 is het laatste cijfer van het 2007e getal gelijk aan dat van het derde getal, dus 4. Dus antwoord C is het goede antwoord. Bladzijde 183

7

8

9n + 9n + 9n = 3 Â 9n = 3 Â 32n = 32n + 1. Dus n = 1003. Dus antwoord C is het goede antwoord. Laat in gedachten eerst de zes leerlingen een stoel pakken en op een rij gaan zitten. Vervolgens kiezen de leraren met hun stoel een plaats tussen de leerlingen. De heer Aap kan dan zijn stoel op vijf plaatsen tussen de zes leerlingen neerzetten en gaan zitten. De heer Noot kan daarna zijn stoel op vier plaatsen neerzetten en gaan zitten en voor mevrouw Mies zijn er dan nog drie plaatsen over. Het aantal mogelijkheden is dus 5 Â 4 Â 3 = 60. Dus antwoord C is het goede antwoord. B-vragen

1

Ga uit van het ongunstigste geval. Dan pak je alle kaartjes met een 1 t/m een 9 en dat zijn er 45. Van alle andere kaartjes, met nummers 10 t/m 50, pak je er telkens negen met hetzelfde nummer. Dat zijn er 41 Â 9 = 369. Je hebt dan in totaal al 45 + 369 = 414 kaartjes gepakt. Als je 414 + 1 = 415 kaartjes pakt, dan moeten er dus ten minste tien bij zijn met hetzelfde nummer.

132 Wiskunde Olympiade

© Noordhoff Uitgevers bv

2

Zie de ¿guur hiernaast. De drie vierhoeken ABSR, RSQP en PQCD zijn gelijkvormig. PQ RS DC PQ RS 2 Daarom geldt: = = geeft = = dus PQ RS AB PQ RS 16 PQ2 = 2 Â RS en RS2 = 16 Â PQ geeft PQ3 = 64, dus PQ = 4 en RS = 8. De zijden van vierhoek RSQP zijn dus twee maal zo lang als de overeenkomstige zijden van vierhoek PQCD en de zijden van vierhoek ABSR zijn weer twee maal zo lang als de overeenkomstige zijden van vierhoek RSQP. PR = 2 Â DP en AR = 2 Â PR = 4 Â DP, dus AD = 7 Â DP, dus DP = 4. Evenzo vind je CQ = 3. De gevraagde omtrek van vierhoek PQCD is 4 + 4 + 2 + 3 = 13.

4

C 3

P

Q

4

28

21 R

S

8

A

3

2

D

B

16

8 # 5 = (5 + 3) # 5,

5 # (5 + 3) 5 + 3 8 = = en 8 # 5 = 83 Â (5 # 3) 3 3 5#3

5 # 3 = (3 + 2) # 3,

3 # (3 + 2) 3 + 2 5 = = en 5 # 3 = 52 Â (3 # 2) 2 2 3#2

3 # 2 = (2 + 1) # 2,

2 # (2 + 1) 2 + 1 = = 3 en 3 # 2 = 3 Â (2 # 1) 1 2#1

2 # 1 = (1 + 1) # 1,

1 # (1 + 1) 1 + 1 = = 2 en 2 # 1 = 2 Â (1 # 1). 1 # 1 = 1 + 2 = 3, dus 1 1#1

2 # 1 = 2 Â 3 = 6, dus 3 # 2 = 3 Â 6 = 18, dus 5 # 3 = 52 Â 18 = 45, dus 8 # 5 = 83 Â 45 = 120. 4

Noem de lengte van de zijden van de gelijkzijdige driehoek x. Zie de ¿guur hiernaast. x2 = BD2 + 1 BD = EA + AF ¶ x2 = (冑x2 í 16 + 冑x2 í 9 )2 + 1 x2 = EA2 + 42 geeft EA = 冑x2 í 16 x2 = 2x2 í 25 + 2冑x2 í 16 Â 冑x2 í 9 + 1 x2 = AF2 + 32 geeft AF = 冑x2 í 9 íx2 + 24 = 2冑x2 í 16 Â 冑x2 í 9 kwadrateren geeft x4 í 48x2 + 576 = 4(x4 í 25x2 + 144) 3x4 í 52x2 = 0

C x

1 D

3

E

B

x

3

x

A

F

52 2 = 冑39 Å3 3 vold. niet vold. Dus de lengte van de zijde van de vlag is 23 冑39. x=0

© Noordhoff Uitgevers bv

 x=

Wiskunde Olympiade 133

2008 A-vragen Bladzijde 184 1

2

Als we de gegevens in een tabel zetten, zien we dat de lootjes van Birgit en Cedric nog over zijn. Nu trekt Ersin niet het lootje van Cedric, dus wel dat van Birgit. Dus trekt Alex juist het lootje van Cedric. Dus antwoord C is het goede antwoord. Zie de tabel. Uit F + 10 + 3 = F + D + 7 volgt D = 6. Uit 7 + E + 3 = C + D + E = C + 6 + E volgt C = 7 + 3 í 6 = 4. Dus antwoord B is het goede antwoord.

A

B

C

D

E

?

A

D

E

?

A

B

7

C

D

E

F

10

3

3

Het getal 720 bevat alleen de priemfactoren 2, 3 en 5. Het priemgetal 2 komt viermaal voor (eenmaal in 2, tweemaal in 4 en eenmaal in 6), het priemgetal 3 tweemaal (in 3 en in 6) en 5 eenmaal. De delers zonder factor 3 en 5 zijn 1, 2, 4, 8 en 16. De delers met één factor 3 en geen factor 5 zijn 3, 6, 12, 24 en 48. Met factor 9 en geen factor 5: 9, 18, 36, 72 en 144. Zonder de factor 5 zijn er dus 15 delers. Deze 15 delers geven elk vermenigvuldigd met 5 de resterende 15 delers. In totaal 30 delers. Dus antwoord D is het goede antwoord.

4

Volgens de stelling van Pythagoras is 0 AC 0 = 5. Driehoek ACD is dus gelijkbenig met top C. In deze driehoek verdeelt de hoogtelijn uit C de driehoek in twee driehoeken met zijden 3, 4 en 5. Vierhoek ABCD is dus op te delen in drie driehoeken met zijden 3, 4 en 5. De gevraagde oppervlakte is dus 3 Â 12 Â 3 Â 4 = 18. Dus antwoord B is het goede antwoord.

D 5 C

3

4 3

5

Als x een positief veelvoud van 6 is dat eindigt op een 4 (zoals 24), dan eindigen de eerstvolgende veelvouden van 6 op een 0 (x + 6), een 6 (x + 12), een 2 (x + 18), een 8 (x + 24), een 4 (x + 30), dus het eerstvolgende veelvoud van 6 dat weer op een 4 eindigt is x + 30. Bij elke groep van dertig opeenvolgende getallen zit dus precies één veelvoud van 6 dat op een 4 eindigt. Omdat in totaal 90 000 getallen liggen tussen 10 000 en 99 999, zijn er onder deze getallen 90 000 : 30 = 3000 zulke zesvouden eindigend op een 4. Dus antwoord C is het goede antwoord.

6

Trek een lijn door H evenwijdig aan AD. Noem de snijpunten van deze lijn met AB en CD respectievelijk P en Q. Nu geldt

0 HP 0 0 HQ 0

=

D

A

3

Q

B

C

0 AE 0

1 = , dus 0 HQ 0 = 34 Â 0 PQ 0 = 34 Â 3 = 94 . 0 CD 0 3

De oppervlakte van driehoek CDH is dus 12 Â 3 Â 94 = 27 8 . Dus antwoord E is het goede antwoord.

H A

B

P E

Bladzijde 185 7

Voor de twee G-blokjes zijn er 6 + 5 + 4 + 3 + 2 + 1 = 21 mogelijkheden om op de 7 plaatsen gelegd te worden (zie ¿guur). Bij elke keuze zijn er voor de twee N-blokjes 4 + 3 + 2 + 1 = 10 mogelijkheden; daarna liggen de drie E-blokjes vast. In totaal dus 21 Â 10 = 210. Dus antwoord A is het goede antwoord.

G G G

G G

G G G

8

2

2)2

5)2

í =1 ((x í (x2 í 2)2 í 5 = 1  (x2 í 2)2 í 5 = í1 (x2 í 2)2 = 6  (x2 í 2)2 = 4 x2 í 2 = 冑6  x2 í 2 = í冑6  x2 í 2 = 2  x2 í 2 = í2 twee opl. geen opl. twee opl. één opl. Dus in totaal 2 + 0 + 2 + 1 = 5 oplossingen. Dus antwoord B is het goede antwoord.

134 Wiskunde Olympiade

G

G

G

G G G G

G

© Noordhoff Uitgevers bv

B-vragen 1

In de eerste kolom komen van boven naar beneden achtereenvolgens te liggen 1 Â 1, 1 Â 2, 1 Â 3, ...1 Â 8 graankorrels. In totaal dus in de eerste kolom: 1 Â (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8). In de tweede kolom liggen er: 2 Â (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8). In de derde kolom liggen er: 3 Â (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8).

· · ·

In de achtste kolom ten slotte: 8 Â (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8). Totaal dus: (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) Â (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8). Omdat 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 zijn er in totaal dus 362 = 1296 graankorrels.

2

Als je alle 50 oneven getallen uit de verzameling {1, 2, 3, ...,100} pakt dan zijn die bij elkaar opgeteld  50  (1 + 99) = 2500. Dat is 400 te weinig. Vervang dus de kleinste oneven getallen door de grootst mogelijke even getallen, telkens per twee omdat 400 even is. Als we 1 en 3 vervangen door 100 en 98 krijgen we 2694. Daarna 2694 í 5 í 7 + 96 + 94 = 2872. We moeten dus nog ten minste één zo’n stap doen. En die lukt: 2872 í 9 í 11 = 2852, dus plaats bijvoorbeeld 20 en 28 terug; dan hebben we som 2900 gevonden met 6 even getallen en bovendien hebben we laten zien dat het met minder niet lukt. Het antwoord is dus 6.

3

Uit x +

1 2

( ) () () () ( ) ( ) Â

x3 + 4

1 x

3

= x3 + 3x2

1 1 = x+ x x3

3

1 1 + 3x x x

í3 x+

2

+

1 x

3

= x3 + 3x +

( )

3 1 1 1 + 3 = x3 + 3 + 3 x + volgt dat x x x x

1 = 53 í 3 5 = 110, dus n = 110. x

Zie de ¿guur hiernaast. 0 AQ 0 2 + 0 QP 0 2 = 36 en 0 BQ 0 2 + 0 SP 0 2 = 25 (want 0 BQ 0 = 0 CS 0 ), Dus 0 AQ 0 2 + 0 QP 0 2 + 0 BQ 0 2 + 0 SP 0 2 = 61. Verder 0 BQ 0 2 + 0 QP 0 2 = 49. Dus 0 DS 0 2 + 0 SP 0 2 = 0 AQ 0 2 + 0 SP 0 2 = 61 í 49 = 12 en 0 DP 0 = 冑12 = 2冑3.

D

C

P T

A

© Noordhoff Uitgevers bv

S

R

Q

B

Wiskunde Olympiade 135

Gemengde opgaven 1 Functies en grafieken Bladzijde 186 1

Stel k: y = ax + b. %y 4 í 2 2 a= = = %x 7 í 2 5 y = 25 x + b f 2 Â2 + b = 2 door A(2, 2) 5 4 5 +b=2 b = 115 , dus k: y = 25 x + 115.

l: y = ax + 11 f a  7 + 11 = 4 door B(7, 4) 7a = í7 a = í1, dus l: y = íx + 11. m: y = 2x + b f 2Â2 + b = 2 door A(2, 2) 4+b=2 b = í2, dus m: y = 2x í 2. Snijpunt berekenen van l en m. íx + 11 = 2x í 2 í3x = í13 í13 x= = 413 geeft y = í413 + 11 = 623 , dus C (413 , 623 ) . í3 m snijden met de x-as, y = 0 geeft 2x í 2 = 0 2x = 2 x = 1, dus D(1, 0). l snijden met de x-as, y = 0 geeft íx + 11 = 0 íx = í11 x = 11, dus E(11, 0). P is het midden van DE dus P(6, 0). Stel n: y = ax + b. a=

¨y 623 í 0 623 = = = í4 ¨x 413 í 6 í123

y = í4x + b f í4 Â 6 + b = 0 door P(6, 0) í24 + b = 0 b = 24, dus n: y = í4x + 24. Snijpunt berekenen van k en n. 2 1 5 x + 15 = í4x + 24 425 x = 2245 2 2 3 2 3 ). x = 511 , geeft y = í4 Â 511 + 24 = 311 , dus (511 , 311

2

a Stel H = at + b. t = í5 en H = 485 %H 200 í 485 = = í11,4 f a= t = 20 en H = 200 20 í í5 %t H = í11,4t + b f í11,4 Â 20 + b = 200 door (20, 200) í228 + b = 200 b = 428 Dus H = í11,4t + 428.

136 Gemengde opgaven

© Noordhoff Uitgevers bv

b Stel K = at + b. t = í2 en K = 253 %K 220 í 253 = = í1,5 f a= t = 20 en K = 220 20 í í2 %t K = í1,5t + b f í1,5 Â 20 + b = 220 door (20, 220) í30 + b = 220 b = 250 Dus K = í1,5t + 250. c Stel N = at + b. t = 0 en N = 15,0 %N 16,7 í 15,0 = = 0,085 f a= t = 20 en N = 16,7 20 %t N = 0,085t + b f b = 15 door (0; 15,0) Dus N = 0,085t + 15. d í1,5t + 250 = í11,4t + 428 9,9t = 178 t § 18 Dus in het jaar 2008. e N is in miljoenen, dus 10N is in honderdduizendtallen. A = H Â 10N = (í11,4t + 428) Â 10(0,085t + 15) = (í11,4t + 428) Â (0,85t + 150) = í9,69t2 í 1710t + 363,8t + 64 200 = í9,69t2 í 1346,2t + 64 200 Dus a = í9,69, b = í1346,2 en c = 64 200. Bladzijde 187

3

a f (x) = 5 í 0 112 x í 3 0 = 5 í 112 x + 3 = í112 x + 8 als 112 x í 3 • 0, dus als x • 2 en f (x) = 5 í 0 112 x í 3 0 = 5 + 112 x í 3 = 112 x + 2 als 112 x í 3 < 0, dus als x < 2. g(x) = x í 1 + 0 2x í 3 0 = x í 1 + 2x í 3 = 3x í 4 als 2x í 3 • 0, dus als x • 112 en g(x) = x í 1 + 0 2x í 3 0 = x í 1 í 2x + 3 = íx + 2 als 2x í 3 < 0, dus als x < 112 . y g 5

4

3

ƒ

2

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−1

O

1

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3

4

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6

7

x

−1

−2

b íx + 2 = 112 x + 2 í212 x = 0 x=0 x = 0 geeft f (0) = 112 Â 0 + 2 = 2

© Noordhoff Uitgevers bv

Gemengde opgaven 137

í112 x + 8 = 3x í 4 í412 x = í12 í12 x = 1 = 223 í42 x = 223 geeft g (223 ) = 3 Â 223 í 4 = 4 Dus de snijpunten zijn (0, 2) en (223 , 4) . c g(x) = íx + 2 f íxA + 2 = 3 A(xA, 3) íxA = 1 xA = í1 112 x

f (x) = B(xB, 3)

+2

y g 5

f 112 xB + 2 = 3 112 xB = 1 3xB = 2 xB = 23

4

A

g(x) = 3x í 4 f 3xC í 4 = 3 C(xC, 3) 3xC = 7 xC = 213 f (x) = í112 x + 8 1 f í12 xD + 8 = 3 D(xD, 3) í112 xD = í5 3xD = 10 xD = 313

3

B

D C

2 ƒ 1

−1

AB = 23 í í1 = 123 , BC = 213 í 23 = 123 en

O

1

2

3

4

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6

7

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−1

−2

CD = 313 í 213 = 1. Dus CD heeft de kleinste lengte. d h(x) = (112 x + 2) (íx + 2) = í112 x2 + x + 4 als x < 112 1 xtop = í = 1 geeft ytop = h ( 13 ) = 416 . 2  í112 3 y = í112 x2 + x + 4 met D = 8k , 112 9 en B = 8k , 416 4 . h(x) = (112 x + 2) (3x í 4) = 412 x2 í 8 als 112 ” x < 2 h (112 ) = 218 en h(2) = 10. y = 412 x2 í 8 met D = 3112 , 29 en B = 3 218 , 109. h(x) = (í112 x + 8) (3x í 4) = í412 x2 + 30x í 32 als x • 2 xtop = í

30 = 313 geeft ytop = h (313 ) = 18. 2 Â í412

y = í412 x2 + 30x í 32 met D = 3 2, m 9 en B = 8k , 18 4 . 4

a 7x2 = 5x 7x2 í 5x = 0 x(7x í 5) = 0 x = 0  7x = 5 x = 0  x = 57 b 2x2 + x = 3 2x2 + x í 3 = 0 D = 12 í 4  2  í3 = 25 í1 í 5 í1 + 5 x= = í112  x = =1 4 4

138 Gemengde opgaven

c (x + 2)(x í 6) = 9 x2 í 6x + 2x í 12 = 9 x2 í 4x í 21 = 0 (x í 7)(x + 3) = 0 x = 7  x = í3

© Noordhoff Uitgevers bv

d (x í 3)2 í (x + 1)2 = x2 í 1 x2 í 6x + 9 í (x2 + 2x + 1) = x2 í 1 x2 í 6x + 9 í x2 í 2x í 1 = x2 í 1 íx2 í 8x + 9 = 0 x2 + 8x í 9 = 0 (x + 9)(x í 1) = 0 x = í9  x = 1 e (2x í 3)2 = 36 2x í 3 = 6  2x í 3 = í6 2x = 9  2x = í3 x = 412  x = í112 5

6

f 4 í (x í 2)2 = 7x í 3 4 í (x2 í 4x + 4) = 7x í 3 4 í x2 + 4x í 4 = 7x í 3 íx2 í 3x + 3 = 0 x2 + 3x í 3 = 0 D = 32 í 4  1  í3 = 21 í3 í 冑21 í3 + 冑21 x=  x= 2 2 x = í112 í 12 冑21  x = í112 + 12 冑21

a p = 0 geeft 6x = 0, dus één oplossing. Voor p  0 is D = 62 í 4  p  3p = 36 í 12p2 f 36 í 12p2 > 0 twee oplossingen als D > 0 í12p2 > í36 p2 < 3 í冑3 < p < 冑3 De vergelijking heeft twee oplossingen voor í冑3 < p < 0  0 < p < 冑3. b x = 6 geeft 62 + p  6 í 6p2 = 0 36 + 6p í 6p2 = 0 í6p2 + 6p + 36 = 0 p2 í p í 6 = 0 (p í 3)(p + 2) = 0 p = 3  p = í2 p = 3 geeft x2 + 3x í 6  32 = 0 p = í2 geeft x2 í 2x í 6  (í2)2 = 0 2 x + 3x í 54 = 0 x2 í 2x í 24 = 0 (x í 6)(x + 9) = 0 (x í 6)(x + 4) = 0 x = 6  x = í9 x = 6  x = í4 Voor p = 3 is de andere oplossing x = í9 en voor p = í2 is de andere oplossing x = í4. c Voor p  0 is D = (í2p)2 í 4  p  4 = 4p2 í 16p f 4p2 í 16p = 0 één oplossing als D = 0 p(4p í 16) = 0 p = 0  4p = 16 p=0  p=4 p = 0 geeft 4 = 0, geen oplossingen p = 4 geeft 4x2 í 2  4x + 4 = 0 4x2 í 8x + 4 = 0 x2 í 2x + 1 = 0 (x í 1)(x í 1) = 0 x=1 Dus voor p = 4 is de oplossing x = 1. a Voer in y1 = x3 í 3x2 + 4. De optie maximum geeft x = 0 en y = 4. De optie minimum geeft x = 2 en y = 0. Dus min. is f (2) = 0 en max. is f (0) = 4.

b

y

4

(0, 4)

(3, 4)

ƒ

O

(2, 0)

3

x

f (0) = f (3) = 4 is de grootste functiewaarde en f (2) = 0 is de kleinste functiewaarde. Dus B f = 3 0, 4 4 .

© Noordhoff Uitgevers bv

Gemengde opgaven 139

7

a Voer in y1 = 14 x4 + 23 x3 í 312 x2 í 12x. Optie minimum geeft x = í3 en y = 6,75 en geeft x § 2,56 en y § í31,74. Optie maximum geeft x § í1,56 en y § 9,15. Dus min. is f (í3) = 6,75 en min. is f (2,56) § í31,74 en max. is f (í1,56) § 9,15. b 26 ” a ” 31 Bladzijde 188

8

a

x

0

í1

í6 í212

f (x)

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112

2

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í212

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ƒ 1 B

A –1

O

1

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–1

–2 C –3

–4

–5

b xtop = í

3 = 3 en ytop = f (3) = 2, dus max. is f (3) = 2. 2 Â í 12

B f = 8k, 2 4 c T(3, 2) en yc = f (0) = í212 , dus C (0, í212 ) . ¨y 2 í í212 = = 112 . Stel y = ax + b met a = ¨x 3í0 y = 112 x + b f b = í212 door C (0, í212 ) Dus y = 112 x í 212 . d f (x) = 0 geeft í 12 x2 + 3x í 212 = 0 x2 í 6x + 5 = 0 (x í 1)(x í 5) = 0 x=1x=5 O(+ABT ) = 12  (5 í 1)  2 = 4 e f (x) = í4 geeft í 12 x2 + 3x í 212 = í4 x2 í 6x + 5 = 8 x2 í 6x í 3 = 0 (x í 3)2 í 9 í 3 = 0 (x í 3)3 = 12 x í 3 = 冑12  x í 3 = í冑12 x = 3 + 冑12  x = 3 í 冑12 x = 3 + 2冑3  x = 3 í 2冑3 voldoet niet voldoet Dus a = 3 í 2冑3. 140 Gemengde opgaven

© Noordhoff Uitgevers bv

a Voer in y1 = 0 x2 í x í 6 0 . x f (x)

í3 í2

í1

0

1

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3

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6

4

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y

8 ƒ 7

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b Optie maximum geeft x = 0,5 en y = 6,25, dus B f = 3 0; 6,25 4 . c a=3 x

0

6

f inv(x)

3

4

y

8 ƒ

=

x

7

y

9

6

5 ƒinv

4

3

2

1

O

© Noordhoff Uitgevers bv

1

2

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x

Gemengde opgaven 141

d b=3 x

4

0

f inv(x)

2

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=

x

ƒ

3 ƒinv 2

1

O

10

1

2

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a ƒp

x

Er moet gelden D < 0 f p2 í 8 < 0 D = p2 í 4  12  4 = p2 í 8 p2 < 8 í冑8 < p < 冑8 í2冑2 < p < 2冑2 p b xtop = í = íp 1 ytop = fp(xtop) = 12  (íp)2 + p  íp + 4 = 12 p2 í p2 + 4 = í 12 p2 + 4 ytop = í5 geeft í 12 p2 + 4 = í5 í 12 p2 = í9 p 2 = 18 p = 冑18  p = í冑18 p = 3冑2  p = í3冑2 c xtop = íp en ytop = í 12 p2 + 4 invullen bij y = í3x + 8 geeft í 12 p2 + 4 = í3  íp + 8 í 12 p2 + 4 = 3p + 8

d xtop = í

p = íp, dus p = íxtop. 2 Â 12

í 12 p2 í 3p í 4 = 0 p2 + 6p + 8 = 0 (p + 2)(p + 4) = 0 p = í2  p = í4

ytop = 12 xtop2 + pxtop + 4 f ytop = 12  xtop2 + íxtop  xtop + 4 p = íxtop ytop = 12  xtop2 + íxtop2 + 4 ytop = í 12 xtop2 + 4 Dus de formule van de kromme is y = í 12 x2 + 4.

142 Gemengde opgaven

© Noordhoff Uitgevers bv

11

a f í 1(x) = x2 í 4x í 6 í4 xtop = í = 2, dus a = 2. 2Â1 2

x fí1(x)

3

4

5

6

í10 í9

í6

í1

6

en

x

í10 í9 í6 í1 2

f íinv 1 (x)

3

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=

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b De grafieken snijden elkaar op de lijn y = x. f (x) = x2 í 4x í 6 f x2 í 4x í 6 = x y=x x2 í 5x í 6 = 0 (x í 6)(x + 1) = 0 x = 6  x = í1 vold. vold. niet x = 6 geeft y = 6, dus A(6, 6). 4p = í2p, dus p = í 12 xtop. c xtop = í 2Â1 6 6 ytop = xtop2 + 4pxtop + p 1 2 ¶ ytop = xtop + 4  í 2 xtop  xtop + í 1 x 2 top p = í 12 xtop 12 2 2 ytop = xtop í 2xtop í x top

ytop =

íxtop2

12 íx top

Dus de formule van de kromme is y = íx2 í

© Noordhoff Uitgevers bv

12 . x

Gemengde opgaven 143

12

a Voer in y1 = 0,1x3 í 0,2x2 í 2x + 2 en y2 = 0,2x2 í 3. Intersect geeft x § í4,03, x § 2,08 en x § 5,95. y g

2,08

−4,03 O

x

5,95

ƒ

f (x) < g(x) geeft x < í4,03  2,08 < x < 5,95 b Optie zero geeft x § í4,09, x § 0,95 en x § 5,14. f (x) > 0 geeft í4,09 < x < 0,95  x > 5,14. c Voer in y1 = 0 0,2x2 í 3 0 en y2 = 2. Intersect geeft x = í5, x § í2,24, x § 2,24 en x = 5. y y = |0,2x2 − 3|

y=2

−5

−2,24

O

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x

0 g(x) 0 < 2 geeft í5 < x < í2,24  2,24 < x < 5 d Voer in y1 = (0,1x3 í 0,2x2 í 2x + 2)(0,2x2 í 3). Optie zero geeft x § í4,09, x § í3,87, x § 0,95, x § 3,87 en x § 5,14. y y = (0,1x3 − 0,2x2 − 2x + 2)(0,2x2 − 3)

−4,09 −3,87 O

0,95

3,87

5,14

x

f (x)  g(x) > 0 geeft í4,09 < x < í3,87  0,95 < x < 3,87  x > 5,14 144 Gemengde opgaven

© Noordhoff Uitgevers bv

2 De afgeleide functie Bladzijde 189 13

a Voer in y1 = 500x2 /(x2 + 400). Gebruik de optie dy/dx (TI) of d/dx (Casio). De snelheid op t = 15 is c

ds = 15,36 m/s § 55 km/uur. d dt t = 15

De snelheid op t = 30 is c

ds § 7,1 m/s § 26 km/uur. d dt t = 30

500 Â 502 § 431 m. 502 + 400 Gebruik de optie dy/dx (TI) of d/dx (Casio).

b Na 50 seconden is afgelegd s =

De snelheid op t = 50 is c

ds § 2,38 m/s. d dt t = 50

Dus na 1 minuut is afgelegd 431 + 10 Â 2,38 § 455 m. 14

a Stel l: y = ax + b. Voer in y1 = (5x + 6)/冑2x + 9. De optie dy/dx (TI) of d/dx (Casio) geeft c

dy = 19, dus a = 19. d dx x = í4

y = 19x + b f 19 Â í4 + b = í14 f (í4) = í14, dus A(í4, í14) í76 + b = í14 b = 62 Dus l: y = 19x + 62. b Stel k: y = ax + b. B ligt op de y-as, dus xB = 0. dy De optie dy/d x (TI) of d/dx (Casio) geeft c § 1,44, dus a § 1,44. d dx x = 0 y = 1,44x + b fb = 2 f (0) = 2, dus B(0, 2) Dus k: y = 1,44x + 2. c Stel m: y = ax + b. dy De optie d y/d x (TI) of d/d x (Casio) geeft c = 0,632, dus a = 0,632. d dx y=8 y = 0,632x + b f 0,632 Â 8 + b = 9,2 f (8) = 9,2, dus C(8; 9,2) 5,056 + b = 9,2 b = 4,144 Dus m: y = 0,632x + 4,144. m snijden met de x-as geeft 0,632x + 4,144 = 0 0,632x = í4,144 x § í6,56

© Noordhoff Uitgevers bv

Gemengde opgaven 145

15

a

y

y 4

4 3 2

2 1

1 –4 – 3 – 2 –1 O –1

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helling

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x

h

146 Gemengde opgaven

© Noordhoff Uitgevers bv

16

f (x + h) í f (x) h hm0

a f '(x) = lim

3(x + h)2 + 5(x + h) + 6 í (3x2 + 5x + 6) h hm0

= lim

3(x2 + 2xh + h2) + 5x + 5h + 6 í 3x2 í 5x í 6 h hm0

= lim

3x2 + 6xh + 3h2 + 5x + 5h + 6 í 3x2 í 5x í 6 h hm0

= lim

6xh + 3h2 + 5h h hm0

= lim

= lim (6x + 3h + 5) = 6x + 5 hm0

b g'(x) = lim

hm0

g(x + h) í g(x) h

(x + h)3 í 4(x + h) í (x3 í 4x) h hm0

= lim

(x + h)(x2 + 2xh + h2) í 4x í 4h í x3 + 4x h hm0

= lim

x3 + 2x2h + xh2 + x2h + 2xh2 + h3 í 4h í x3 h hm0

= lim

3x2h + 3xh2 + h3 í 4h h hm0

= lim

= lim (3x2 + 3xh + h2 í 4) = 3x2 í 4 hm0

Bladzijde 190 17

a f (x) = íx(2x í 7) = í2x2 + 7x geeft f '(x) = í4x + 7 b f (x) = (x2 í 1)(x í 1) = x3 í x2 í x + 1 geeft f '(x) = 3x2 í 2x í 1 (5 í 2x) Â 2 í (2x í 1) Â í2 10 í 4x + 4x í 2 8 2x í 1 c f (x) = geeft f '(x) = = = 5 í 2x (5 í 2x)2 (5 í 2x)2 (5 í 2x)2 x2 + 8x 1 2 x í 12 x geeft f '(x) = í 18 x í 12 = 7 í 16 16 e f (x) = x(3x + 2)2 = x(9x2 + 12x + 4) = 9x3 + 12x2 + 4x geeft f '(x) = 27x2 + 24x + 4 f f (x) = 8 í (x í 1)2 = 8 í (x2 í 2x + 1) = 8 í x2 + 2x í 1 = íx2 + 2x + 7 geeft f '(x) = í2x + 2 x2 í 2x + x4 geeft g f (x) = x+1

d f (x) = 7 í

f '(x) =

(x + 1) Â (2x í 2) í (x2 í 2x) Â 1 2x2 í 2x + 2x í 2 í x2 + 2x x2 + 2x í 2 3 3 + 4x + 4x + 4x3 = = (x + 1)2 (x + 1)2 (x + 1)2

h f (x) = 3x2 í f '(x) = 6x í

18

4x + 3 geeft 2x í 1 (2x í 1) Â 4 í (4x + 3) Â 2 8x í 4 í 8x í 6 10 = 6x í = 6x + (2x í 1)2 (2x í 1)2 (2x í 1)2

x2 + 3x + 2 (í2)2 + 3 Â í2 + 2 0 = = =0 3 x2 í 1 (í2)2 í 1 x m í2

a lim f (x) = lim x m í2

(x + 1)(x + 2) x2 + 3x + 2 x + 2 í1 + 2 1 = lim = lim = = = í 12 2 í1 í 1 í2 x í1 x m í1 x m í1 (x + 1)(x í 1) x m í1 x í 1

lim f (x) = lim

x m í1

© Noordhoff Uitgevers bv

Gemengde opgaven 147

b Stel k: y = ax + b. f (x) =

(x2 í 1) Â (2x + 3) í (x2 + 3x + 2) Â 2x x2 + 3x + 2 = geeft f '(x) (x2 í 1)2 x2 í 1 2x3 + 3x2 í 2x í 3 í 2x3 í 6x2 í 4x = (x2 í 1)2 í3x2 í 6x í 3 = (x2 í 1)2

a = f '(2) =

í3 Â 22 í 6 Â 2 í 3 í27 = = í3 9 (22 í 1)2

y = í3x + b f í3  2 + b = 4 f (2) = 4, dus door (2, 4) í6 + b = 4 b = 10 Dus k: y = í3x + 10. xí3 xí3 1 c g(x) = 2 mits x  3. = = x í 2x í 3 (x í 3)(x + 1) x + 1 De grafiek van g heeft een perforatie voor x = 3. xí3 xí3 1 lim 2 = lim = lim = 14 x m 3 x í 2x í 3 x m 3 (x í 3)(x + 1) x m3 x + 1 De perforatie is dus (3, 14 ) . x = 3 invullen in k geeft y = í3  3 + 10 = í9 + 10 = 1, dus de perforatie ligt niet op k. 19

a Stel k: y = ax + b. f (x) = (x2 í 9)(x í 1) = x3 í x2 í 9x + 9 geeft f '(x) = 3x2 í 2x í 9 a = f '(2) = 3  22 í 2  2 í 9 = í1 y = íx + b f í 2 + b = í5 f (2) = í5, dus A(2, í5) b = í3 Dus k: y = íx í 3. b Stel m: y = ax + b. B ligt op de y-as, dus xB = 0. a = f '(0) = í9 y = í9x + b f b=9 f (0) = 9, dus B (0, 9) Dus m: y = í9x + 9. c De rc van de raaklijn in C is f '(í1) = 3  (í1)2 í 2  í1 í 9 = í4. rc raaklijn = í4  0, dus de raaklijn in C is niet horizontaal.

20

a Stel k: y = ax + b. f (x) = 13 x3 í 12 x2 í 2x + 1 geeft f '(x) = x2 í x í 2 A ligt op de y-as, dus xA = 0. a = f '(0) = í2 y = í2x + b f b=1 f (0) = 1, dus A(0, 1) Dus k: y = í2x + 1. b Raaklijn horizontaal, dus rc = 0, dus f '(x) = 0. f '(x) = 0 geeft x2 í x í 2 = 0 (x í 2)(x + 1) = 0 x = 2  x = í1 f (2) = í213 en f (í1) = 216 De punten zijn (í1, 216 ) en (2, 213 ) . c rc l = 4 en l is evenwijdig met de raaklijn, dus f '(x) = 4. f '(x) = 4 geeft x2 í x í 2 = 4 x2 í x í 6 = 0 (x + 2)(x í 3) = 0 x = í2  x = 3 xB = í2 en yB = f (í2) = 13 , dus B (í2, 13 ) . xC = 3 en yC = f (3) = í 12 , dus C (3, í 12 ) .

148 Gemengde opgaven

© Noordhoff Uitgevers bv

21

a Stel k: y = ax + b. f (x) = (x2 + 2)(1 í x) = x2 í x3 + 2 í 2x = íx3 + x2 í 2x + 2 geeft f '(x) = í3x2 + 2x í 2 a = f '(2) = í3  22 + 2  2 í 2 = í10 y = í10x + b f í10  2 + b = í6 f (2) = í6, dus A(2, í6) í20 + b = í6 b = 14 Dus k: y = í10x + 14. b rc k = í10 en k is evenwijdig met de raaklijn, dus f '(x) = í10. f '(x) = í10 geeft í 3x2 + 2x í 2 = í10 í3x2 + 2x + 8 = 0 D = 22 í 4  í3  8 = 100 í2 í 10 í2 + 10 x= =2x= = í113 í6 í6 Dus xB = í113 .

22

a s = 0,06t3 + 1,2t2 geeft v = 0,18t2 + 2,4t Op t = 4 is de snelheid v = 0,18 Â 42 + 2,4 Â 4 = 12,48 m/s. Op t = 6 is de snelheid v = 0,18 Â 62 + 2,4 Â 6 = 20,88 m/s. 100 b 100 km/uur = m/s 3,6 Voer in y1 = 0,18x2 + 2,4x en y2 = 100/3,6. Intersect geeft x § 7,43. Dus na ongeveer 7,43 seconden is de snelheid 100 km/uur. c Na 8 seconden is s = 0,06 Â 83 + 1,2 Â 82 = 107,52 m. Op t = 8 is de snelheid v = 0,18 Â 82 + 2,4 Â 8 = 30,72 m/s. 300 í 107,52 = 192,48 m 192,48 § 6,3, dus na ongeveer 8 + 6,3 = 14,3 seconden heeft de motor 300 meter afgelegd. 30,72

23

a Stel k: y = ax + b. (x + 3) Â 0 í 4 Â 1 4 f (x) = x í 2 + geeft f '(x) = 1 + x+3 (x + 3)2

Bladzijde 191

4 (x + 3)2 A is het snijpunt met de y-as, dus xA = 0. 4 a = f '(0) = 1 í =5 (0 + 3)2 9 = 1í

k: y = 59 x + b f (0) = í 23 , dus A (0, í 23 )

s

b = í 23

Dus k: y = 59 x í 23 . b Stel l: y = ax + b. a = f '(í2) = í3 y = í3x + b f í3 Â í2 + b = 0 door B(í2, 0) 6+b=0 b = í6 Dus l: y = í3x í 6. Stel m: y = ax + b. a = f '(1) = 34 m: y = 34 x + b f door C(1, 0)

3 4

Â1 + b = 0

b = í 34

© Noordhoff Uitgevers bv

Gemengde opgaven 149

Dus m: y = 34 x í 34 . í3x í 6 = 34 x í 34 í12x í 24 = 3x í 3 í15x = 21 21 x= = í125 geeft y = í3 Â í125 í 6 = í145 í15 Het snijpunt van l en m is ( í125 , í145 ) . 24

a Stel k: y = ax + b. (x í 1)(2x í 2) í (x2 í 2x + 2) Â 1 x2 í 2x + 2 f (x) = geeft f '(x) = xí1 (x í 1)2

a = f '(í2) =

=

2x2 í 2x í 2x + 2 í x2 + 2x í 2 (x í 1)2

=

x2 í 2x (x í 1)2

(í2)2 í 2 Â í2 8 =9 (í2 í 1)2

y = 89 x + b f f (í2) = í313 , dus A (í2, í313 )

8 9

 í2 + b = í313

1 í 16 9 + b = í33

b = í159 Dus k: y = 89 x í 159 . b f '(í1) =

(í1)2 í 2 Â í1 3 = , dus rc l = 34 . 4 (í1 í 1)2

32 í 2 Â 3 3 = , dus rc m = 34 . (3 í 1)2 4 rc l = rc m, dus l en m zijn evenwijdig. f '(3) =

3 Vergelijkingen en herleidingen 25

a 17 í (2x í 1)4 = 1 (2x í 1)4 = 16 2x í 1 = 2  2x í 1 = í2 2x = 3  2x = í1 x = 112  x = í 12 b x6 í 6x3 + 5 = 0 Stel x3 = u. u2 í 6u + 5 = 0 (u í 1)(u í 5) = 0 u=1  u=5 x3 = 1  x3 = 5 35 x = 1  x =冑 4 2 c 10x = 17x + 657 10x4 í 17x2 í 657 = 0 Stel x2 = u. 10u2 í 17u í 657 = 0 D = (í17)2 í 4  10  í657 = 26569 17 í 163 17 + 163 3 u= = í710 =9  u= 20 20 3 x2 = í710  x2 = 9 x = 3  x = í3

150 Gemengde opgaven

© Noordhoff Uitgevers bv

d Stel f (x) = x3 + 5x en g(x) = 6x2. f (x) = g(x) geeft x3 + 5x = 6x2 x3 í 6x2 + 5x = 0 x(x2 í 6x + 5) = 0 x(x í 5)(x í 1) = 0 x=0x=5x=1 y

g

O ƒ

1

5

x

f (x) • g(x) geeft 0 ” x ” 1  x • 5 e (2x2 í 1)2 = x2 4x2 í 4x2 + 1 = x2 4x2 í 5x2 + 1 = 0 Stel x2 = u. D = (í5)2 í 4  4  1 = 9 5í3 1 5+3 u= =4  u= =1 8 8 1 x2 = 4  x2 = 1 x = 12  x = í 12  x = 1  x = í1 Alternatieve oplossing (2x2 í 1)2 = x2  2x2 í 1 = íx 2x2 í 1 = x 2 2x í x í 1 = 0  2x2 + x í 1 = 0 x2 í 12 x í 12 = 0  x2 + 12 x í 12 = 0 (x í 1) (x + 12 ) = 0  (x + 1) (x í 12 ) = 0 x = 1  x = í 12  x = í1  x = 12 f (2x í 1)4 í 5(2x í 1)2 + 4 = 0 Stel (2x í 1)2 = u. u2 í 5u + 4 = 0 (u í 4)(u í 1) = 0 u=4  u=1 (2x í 1)2 = 4  (2x í 1)2 = 1 2x í 1 = 2  2x í 1 = í2  2x í 1 = 1  2x í 1 = í1 2x = 3  2x = í1  2x = 2  2x = 0 x = 112  x = í 12  x = 1  x = 0 g

冑2 í 2x + 2x = 0 冑2 í 2x = í2x

kwadrateren geeft 2 í 2x = (í2x)2 2 í 2x = 4x2 í4x2 í 2x + 2 = 0 2x2 + x í 1 = 0 D = 12 í 4  2  í1 = 9 í1 í 3 í1 + 3 1 x= = í1  x = =2 4 4 x = í1 geeft 冑4 = 2 voldoet x = 12 geeft 冑1 = í1 voldoet niet

© Noordhoff Uitgevers bv

Gemengde opgaven 151

h x3 í 3x冑x í 108 = 0 Stel x冑x = u. u2 í 3u í 108 = 0 (u í 12)(u + 9) = 0 u = 12  u = í9 x冑x = 12  x冑x = í9 geen oplossingen kwadrateren geeft x3 = 144 3 144 x =冑 3 144 geeft x =冑 3 3 144 í 3  冑 144  冑冑 144 í 108 = 0 voldoet.

26

a x5 í 16x3 + 28x = 0 x(x4 í 16x2 + 28) = 0 x = 0  x4 í 16x2 + 28 = 0 Stel x2 = u. x = 0  u2 í 16u + 28 = 0 x = 0  (u í 2)(u í 14) = 0 x = 0  u = 2  u = 14 x = 0  x2 = 2  x2 = 14 x = 0  x = 冑2  x = í冑2  x = 冑14  x = í冑14 b Stel f (x) = x4 en g(x) = x2 + 12. f (x) = g(x) geeft x4 = x2 + 12 x4 í x2 í 12 = 0 Stel x2 = u. u2 í u í 12 = 0 (u + 3)(u í 4) = 0 u = í3  u = 4 x2 = í3  x2 = 4 geen opl. x = 2  x = í2

e 6x5 + 10x2  冑x í 464 = 0 3x5 + 5x2  冑x í 232 = 0 Stel x2  冑x = u. 3u2 + 5u í 232 = 0 D = 52 í 4  3  í232 = 2809 í5 í 53 í5 + 53 u= = í923  u = =8 6 6 x2  冑x = í923  x2  冑x = 8 geen opl. kwadrateren geeft x5 = 64 5 64 x =冑 5 x =冑 64 geeft

f

(冑5 64 )2 Â 冑冑5 64 = 8 voldoet.

冑3x í 2 + 2 = x 冑3x í 2 = x í 2

kwadrateren geeft 3x í 2 = (x í 2)2 3x í 2 = x2 í 4x + 4 íx2 + 7x í 6 = 0 x2 í 7x + 6 = 0 (x í 6)(x í 1) = 0 x=6x=1 x = 6 geeft 冑16 = 4 voldoet.

y

x = 1 geeft 冑1 = í1 voldoet niet. –2

2

x

x4 < x2 + 12 geeft í2 < x < 2 c 0 x4 í 7x2 0 = 18 x4 í 7x2 = 18  x4 í 7x2 = í18 x4 í 7x2 í 18 = 0  x4 í 7x2 + 18 = 0 Stel x2 = u. u2 í 7u í 18 = 0  u2 í 7u + 18 = 0 D = (í 7)2 í 4  1  18 = í 23 (u í 9)(u + 2) = 0 u = 9  u = í2 geen oplossingen x2 = 9  x2 = í2 x = 3  x = í3 2x + 4 12 d = x+1 x (2x + 4)(x + 1) = 12x 2x2 + 4x + 2x + 4 = 12x 2x2 í 6x + 4 = 0 x2 í 3x + 2 = 0 (x í 2)(x í 1) = 0 x=2x=1 vold. vold. 152 Gemengde opgaven

g (2x í 3)(x2 í 3) + 3 = 2x (2x í 3)(x2 í 3) = 2x í 3 2x í 3 = 0  x2 í 3 = 1 2x = 3  x2 = 4 x = 112  x = 2  x = í2 h

x2 í 9 x2 í 9 = 2x + 3 x + 4 2x + 3 = x + 4  x2 í 9 = 0 x = 1  x2 = 9 x = 1  x = 3  x = í3 vold. vold. vold.

© Noordhoff Uitgevers bv

Bladzijde 192 27

a y=

x4 í x2 x2(x2 í 1) x2(x í 1)(x + 1) = = = x2(x + 1) mits x  1 xí1 xí1 xí1

15x 2x(x + 2) 15x í 2x2 í 4x í2x2 + 11x 15x í 2x = í = = x+2 x+2 x+2 x+2 x+2 2x 2x 10 í  (x + 3) 10 í x +3 10(x + 3) í 2x 8x + 30 x+3 c y= mits x  í3 = = = 5x + 17 2 5(x + 3) + 2 2 5+ 5+  (x + 3) x+3 x+3 3t3 + 3t2 í 6t 3t(t2 + t í 2) 3t(t í 1)(t + 2) d N= = = = 3t í 3 mits t  0 œ t  í2 t2 + 2t t(t + 2) t(t + 2) 2a a í 1 a + 2 2a a(a í 1) í 2(a + 2) 2a a2 í 3a í 4 í 2 = =   e K= a + 2 2a a+2 a+2 a 2a2 2a2 b y=

( (

) )

(

=

2a(a2 í 3a í 4)

2a2(a + 2) q2 í2 2 q +1 = f P= q í 2q q2 + 1 28

a e

)

a2 í 3a í 4 a(a + 2) q2 í 2  (q2 + 1) q2 í 2(q2 + 1) q2 + 1 íq2 í 2 q2 + 2 = = = q q í 2q(q2 + 1) í2q3 í q 2q3 + q 2 + 1) í 2q (q  q2 + 1 4x + 6y = 116 geeft e 15x í 6y = 36 + = 152 19x x=8 f 5  8 í 2y = 12 5x í 2y = 12 40 í 2y = 12 í 2y = í28 y = 14 =

( (

2x + 3y = 58 2 ` ` ௘5x í 2y = 12 ௘3

) )

Dus (x, y) = (8, 14). b 2x + y = 13 geeft y = í2x + 13 Substitutie van y = í2x + 13 in x2 + y2 = 58 geeft x2 + (í2x + 13)2 = 58 x2 + 4x2 í 52x + 169 = 58 5x2 í 52x + 111 = 0 D = (í52)2 í 4  5  111 = 484 52 í 22 52 + 22 x= =3x= = 725 10 10 x = 3 geeft y = í2  3 + 13 = 7 x = 725 geeft y = í2  725 + 13 = í145 Dus (x, y) = (3, 7)  (x, y) = (725 , í145 ) . c e

2,4x í 1,92y = 12 0,4x í 0,32y = 2 6 ` ` geeft e 2,4x í 1,12y = 20 ௘0,6x í 0,28y = 5 4

í í0,8y = í8 y = 10 f 0,4x í 0,32 Â 10 = 2 0,4x í 0,32y = 2 0,4x í 3,2 = 2 0,4x = 5,2 x = 13

Dus (x, y) = (13, 10).

© Noordhoff Uitgevers bv

Gemengde opgaven 153

d x + 5y = 17 geeft x = í5y + 17. Substitutie van x = í5y + 17 in (2x í 1)2 + (3y í 1)2 = 73 geeft (2(í5y + 17) í 1)2 + (3y í 1)2 = 73 (í10y + 34 í 1)2 + (3y í 1)2 = 73 (í10y + 33)2 + (3y í 1)2 = 73 100y2 í 660y + 1089 + 9y2 í 6y + 1 = 73 109y2 í 666y + 1017 = 0 D = (í666)2 í 4  109  1017 = 144 666 í 12 666 + 12 12 y= =3y= = 3109 218 218 y = 3 geeft x = í5  3 + 17 = 2 12 12 49 y = 3109 geeft x = í5  3109 + 17 = 1109 49 12 Dus (x, y) = (2, 3)  (x, y) = (1109 , 3109 ).

29

a (4x2 í 1)2 í (3x í 1)3 = 16x4 í 8x2 + 1 í (9x2 í 6x + 1)(3x í 1) = 16x4 í 8x2 + 1 í (27x3 í 9x2 í 18x2 + 6x + 3x í 1) = 16x4 í 8x2 + 1 í 27x3 + 9x2 + 18x2 í 6x í 3x + 1 = 16x4 í 27x3 + 19x2 í 9x + 2 2 3 1 (2t í 1)(t + 2) 2t2 + 4t í t í 2 2t2 + 3t í 2 2t2 3t = = = 2+ 2í 2=1+ í 2 b T= 2t2 2t2 2t2 2t 2t 2t 2t t a a í 2 Â (a2 + 1) 2+1 a í 2(a2 + 1) í2 a 2 a +1 = 12a í 6 Â = 12a í 6 Â c B = 12a í 6 Â 5a(a2 + 1) 5a(a2 + 1) 5a

(

= 12a í 6 Â d K= y=

3y í 2 2y í 1 4x xí1

a í 2a2 í 2 5a(a2 + 1)

= 12a +

( ) ( )

)

12a2 í 6a + 12 5a(a2 + 1)

(( ) ) (( ) )

4x 4x í2 3 í 2 Â (x í 1) 3 Â 4x í 2(x í 1) 10x + 2 xí1 xí1 u K= = = = 4x 4x 2 Â 4x í (x í 1) 7x + 1 2 í1 2 í 1 Â (x í 1) xí1 xí1 3

a+b 3 = b+2 a a(a + b) = 3(b + 2) a2 + ab = 3b + 6 ab í 3b = ía2 + 6 (a í 3)b = ía2 + 6 ía2 + 6 b= aí3 3x + 2 6y + 1 = f xí1 y+3 (3x + 2)(y + 3) = (x í 1)(6y + 1) 3xy + 9x + 2y + 6 = 6xy + x í 6y í 1 í3xy + 8x + 8y + 7 = 0 í 3xy + 8y = í8x í 7 í 3xy + 8x = í8y í 7 y(í3x + 8) = í8x í 7 x(í3y + 8) = í8y í 7 í8x í 7 í8y í 7 y= x= í3x + 8 í3y + 8 8x + 7 8y + 7 y= x= 3x í 8 3y í 8

e

154 Gemengde opgaven

© Noordhoff Uitgevers bv

Bladzijde 193 30

31

(í4, 42) invullen geeft 42 = 12  (í4)3 + a  (í4)2 + b  í4 + 6 42 = í32 + 16a í 4b + 6 16a í 4b = 68 (2, 12) invullen geeft 12 = 12  23 + a  22 + b  2 + 6 12 = 4 + 4a + 2b + 6 4a + 2b = 2 16a í 4b = 68 16a í 4b = 68 1 geeft e e ௘4a + 2b = 2 ` 2 ` 8a + 4b = 4 + 24a = 72 a=3 f 4  3 + 2b = 2 4a + 2b = 2 12 + 2b =2 2b = í10 b = í5 Dus a = 3 en b = í5. a px3 + 2px2 + x2 + 214 x = 0 x ( px2 + 2px + x + 214 ) = 0 x = 0  px2 + 2px + x + 214 = 0 x = 0  px2 + (2p + 1)x + 214 = 0 De vergelijking heeft drie oplossingen als px2 + (2p + 1)x + 214 = 0 twee oplossingen heeft. p = 0 geeft x + 214 = 0, dus één oplossing. Voor p  0 is D = (2p + 1)2 í 4  p  214 = 4p2 + 4p + 1 í 9p = 4p2 í 5p + 1 f 4p2 í 5p + 1 > 0 twee oplossingen als D > 0 4p2 í 5p + 1 = 0 D = (í5)2 í 4  4  1 = 9 5í3 1 5+3 =4  p= =1 p= 8 8 D

D = 4p2 – 5p + 1

O

1 4

p 1

De vergelijking heeft drie oplossingen voor p < 0  0 < p < 14  p > 1.

© Noordhoff Uitgevers bv

Gemengde opgaven 155

b 2px4 í px3 + 5x3 + 2x2 = 0 x2(2px2 í px + 5x + 2) = 0 x2 = 0  2px2 í px + 5x + 2 = 0 x = 0  2px2 + (íp + 5)x + 2 = 0 De vergelijking heeft precies één oplossing als 2px2 + (íp + 5)x + 2 = 0 geen oplossingen heeft. p = 0 geeft 5x + 2 = 0, dus één oplossing. Voor p  0 is D = (íp + 5)2 í 4  2p  2 = p2 í 10p + 25 í 16p = p2 í 26p + 25 f p2 í 26p + 25 < 0 geen oplossingen als D < 0 p2 í 26p + 25 = 0 (p í 1)(p í 25) = 0 p = 1  p = 25 D D = p2 – 26p + 25

O 1

p

25

De vergelijking heeft precies één oplossing voor 1 < p < 25. 32

(x2 + 1)  10 í 10x  2x 10x2 + 10 í 20x2 í10x2 + 10 10x '(x) geeft f = = = x2 + 1 (x2 + 1)2 (x2 + 1)2 (x2 + 1)2 2 + 10 4 í10x rc van de raaklijn gelijk aan í 45 geeft = í5 (x2 + 1)2 í5(í10x2 + 10) = 4(x2 + 1)2 50x2 í 50 = 4(x4 + 2x2 + 1) 50x2 í 50 = 4x4 + 8x2 + 4 4x4 í 42x2 + 54 = 0 Stel x2 = u. 4u2 í 42u + 54 = 0 D = 422 í 4  í4  í54 = 900 í42 í 30 í42 + 30 u= =9  u= = 112 í8 í8 x2 = 9  x2 = 112 f (x) =

冑

冑

x = 3  x = í3  x = 112  x = í 112 x = 3  x = í3  x = 冑6  x = í 12 冑6 vold. vold. vold. vold. 1 2

33

a Voer in y1 = (2x í 3)(x í 4)2. Optie maximum geeft x § 2,33 en y § 4,63. Optie minimum geeft x = 4 en y = 0. Dus min. is f (4) = 0 en max. is f (2,33) § 4,63. y b (5, 7)

7 ƒ (2,33; 4,63) (2, 4)

O

2

4

5

x

f (2) = 4 en f (5) = 7, dus B f = 3 0, 7 4 . 156 Gemengde opgaven

© Noordhoff Uitgevers bv

c f (x) = (2x í 3)(x í 4)2 f (2x í 3)(x í 4)2 = 2x í 3 k : y = 2x í 3 2x í 3 = 0  (x í 4)2 = 1 2x = 3  x í 4 = 1  x í 4 = í1 x = 112  x = 5  x = 3 x = 112 geeft y = 2  112 í 3 = 0 x = 5 geeft y = 2  5 í 3 = 7 x = 3 geeft y = 2  3 í 3 = 3 Dus de coördinaten van de snijpunten van de gra¿eken van f en k zijn (112 , 0) , (5, 7) en (3, 3). d g(x) = x2 í 8x + 16 f x2 í 8x + 16 = 2x í 3 k: y = 2x í 3 x2 í 10x + 19 = 0 (x í 5)2 í 25 + 19 = 0 (x í 5)2 = 6 x í 5 = 冑6  x í 5 = í冑6 x = 5 + 冑6  x = 5 í 冑6 x = 5 + 冑6 geeft y = 2  (5 + 冑6 ) í 3 = 10 + 2冑6 í 3 = 7 + 2冑6 x = 5 í 冑6 geeft y = 2  (5 í 冑6 ) í 3 = 10 í 2冑6 í 3 = 7 í 2冑6 Dus de coördinaten van de snijpunten van de gra¿eken van g en k zijn (5 + 冑6, 7 + 2冑6 ) en (5 í 冑6, 7 í 2冑6 ) . e f (x) = (2x í 3)(x í 4)2 ( f 2x í 3)(x í 4)2 = x2 í 8x + 16 g(x) = x2 í 8x + 16 (2x í 3)(x í 4)2 = (x í 4)2 2x í 3 = 1  (x í 4)2 = 0 2x = 4  x í 4 = 0 x=2  x=4 x = 2 geeft y = (2  2 í 3)(2 í 4)2 = 4 x = 4 geeft y = (2  4 í 3)(4 í 4)2 = 0 Dus de coördinaten van de snijpunten van de gra¿eken van g en k zijn (2, 4) en (4,0). 34

2x 24 = xí2 x+1 2x(x + 1) = 24(x í 2) 2x2 + 2x = 24x í 48 2x2 í 22x + 48 = 0 x2 í 11x + 24 = 0 (x í 8)(x í 3) = 0 x=8∨x=3 vold. vold. 2x 24 b f (x)  g(x) = í24 geeft = í24  xí2 x+1 48x í24 = (x í 2)(x + 1) 1 48x = í24(x í 2)(x + 1) í2x = (x í 2)(x + 1) í2x = x2 í x í 2 íx2 í x + 2 = 0 x2 + x í 2 = 0 (x + 2)(x í 1) = 0 x = í2  x = 1 a f(x) = g(x) geeft

© Noordhoff Uitgevers bv

Gemengde opgaven 157

2x 24 í =8 xí2 x+1 8x(x + 1) í 24(x í 2) =8 (x í 2)(x + 1)

c 4 Â f (x) í g(x) = 8 geeft 4 Â

8x2 + 8x í 24x + 48 =8 (x í 2)(x + 1) 8x2 í 16x + 48 =8 (x í 2)(x + 1) x2 í 2x + 6 =1 (x í 2)(x + 1) x2 í 2x + 6 = (x í 2)(x + 1) x2 í 2x + 6 = x2 + x í 2x í 2 íx + 8 = 0 x=8 voldoet (x í 2)  2 í 2x  1 2x í 4 í 2x 2x í4 d f (x) = geeft f '(x) = = = 2 2 xí2 (x í 2) (x í 2) (x í 2)2 g(x) =

(x + 1) Â 0 í 24 Â 1 24 í24 geeft g'(x) = = x+1 (x + 1)2 (x + 1)2

112 Â f '(x) í g'(x) = 0 geeft 112 Â

í4 í24 í =0 (x í 2)2 (x + 1)2

í 24 í6 = 2 (x + 1)2 (x í 2) í6(x + 1)2 = í24(x í 2)2 (x + 1)2 = 4(x í 2)2 x2 + 2x + 1 = 4(x2 í 4x + 4) x2 + 2x + 1 = 4x2 í 16x + 16 í3x2 + 18x í 15 = 0 x2 í 6x + 5 = 0 (x í 5)(x í 1) = 0 x=5x=1 35

a a + b = 150 en 8,6a + 7,0b = 150 Â 7,9 7 7a + 7b = 1050 a + b = 150 e ` ` geeft e ௘ 8,6a + 7,0b = 1185 1 8,6a + 7b = 1185 í1,6a

í = í135 a = 84,375 f 84,375 + b = 150 a + b = 150 b = 65,625

Dus a = 84,375 en b = 65,625. b Stel hij neemt x ml van de oplossing van 15% en y ml van de oplossing van 30%. Nu moet gelden x + y = 600 en 0,15x + 0,3y = 600 Â 0,22. x + y = 600 3 3x + 3y = 1800 e ` ` geeft e 1,5x + 3y = 1320 ௘ 0,15x + 0,3y = 132 10 í 1,5x = 480 x = 320 f 320 + y = 600 x + y = 600 y = 280 Hij moet 320 ml van de oplossing van 15% en 280 ml van de oplossing van 30% mengen.

158 Gemengde opgaven

© Noordhoff Uitgevers bv

4 Meetkunde Bladzijde 194 36

“BAC = “ EDC (F-hoeken) f +ABC ऴ +DEC “ABC = “ DEC (F-hoeken) AC

AB

CD

DE

geeft

8

5

5

DE

5Â5 = 318 . 8 “ABD = “ BDE (Z-hoeken) f +ABS ऴ +EDS “BAE = “AED (Z-hoeken) Stel DS = x, dus BS = 5 í x. Dus DE =

AB

BS

DE

DS

geeft

5

5íx

318

x

Dus 5x = 318 (5 í x) 5x = 1558 í 318 x 818 x = 1558 x=

1558 = 818

125 8 65 8

8 125 12 = 125 8 Â 65 = 65 = 113

Dus DS = 112 13 . 37

In +ABC is “A = 45°, dus BC = AB = x + x + 2 = 2x + 2 2x + 2 In +BCD is tan(55°) = x+2 Voer in y1 = tan(55) en y2 =

2x + 2 . x+2

Intersect geeft x = 1,49... Dus AB = 2 Â 1,49... + 2 § 4,99. In +BCD is CD2 = (x + 2)2 + (2x + 2)2 = (1,49... + 2)2 + (2 Â 1,49... + 2)2 = 37,18... Dus CD = 冑37,18... § 6,10. 38

“ADC = 180° í 62° í 48° = 70° “BDC = 180° í 70° = 110° “DBC = 180° í 62° í 48° í 25° = 45° AC 6 AD In +ABD is = = sin(62°) sin(48°) sin(70°) geeft AD =

6 sin(70°) 6 sin(48°) = 5,04... en AC = = 6,38... sin(62°) sin(62°)

In +BCD is geeft BD =

BC 6 BD = = sin(45°) sin(25°) sin(110°)

6 sin(110°) 6 sin(25°) = 3,58... en BC = = 7,97... sin(45°) sin(45°)

De omtrek van +ABC = 5,04... + 3,58... + 7,97... + 6,38... § 23,0.

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Gemengde opgaven 159

Bladzijde 195 39 B

Noord

300 m C 100 m 20°

A

In 1 minuut heeft zij 100 m afgelegd. BC2 = 3002 + 1002 í 2 Â 300 Â 100 Â cos(20°) = 43618,44... geeft BC = 208,85... m Maaike heeft 208,85... + 100 í 300 § 9 meter extra afgelegd. 40

In +BCS is BC = 2 en “ BSC = 30°, dus BS = 2  2 = 4 en CS = 2冑3. AS = BS = 4 en DS = CS = 2冑3 In +ABD is AB2 = BD2 + AD2 AB2 = (4 + 2冑3 )2 + 22 AB2 = 16 + 16冑3 + 12 + 4 AB2 = 32 + 16冑3

冑

冑

冑

AB = 32 + 16冑3 = 16(2 + 冑3) = 4 2 + 冑3 “BAS = “SCD (Z-hoeken) f +ABS ऴ +DCS “ABS = “SDC (Z-hoeken) BS

AB

DS

CD

Dus CD = 41

geeft

4 2 + 冑3

2冑3

CD

冑

2冑3 Â 4 2 + 冑3 4

a

冑

4

冑

冑

= 2冑3 Â 2 + 冑3 = 2 6 + 3冑3.

M

10

A

10

N

B

360° = 72°, dus “ AMN = 36°. 5 AN sin(36°) = geeft AN = 10 sin(36°) = 5,87..., dus a = AB = 2  5,87... § 11,8. 10 “ AMB =

160 Gemengde opgaven

© Noordhoff Uitgevers bv

b

M

A

N

2,5

B

2,5

2,5 2,5 geeft AM = = 4,25... AM sin(36°) Oppervlakte gekleurde vlakdelen is ʌ Â 4,25...2 í 5 Â 12 Â 4,25... Â 4,25... Â sin(72° ) § 13,8. sin(36°) =

Bladzijde 196 42

O(ABCD) = DG Â AB = DG Â 14 = 112 geeft DG = 8 In +AGD is AG = 冑102 í 82 = 冑36 = 6, dus BG = 8. In +BDG is BD = 冑82 + 82 = 冑128 = 8冑2. De zijde × hoogte-methode in +ABD geeft BD × AE = AB × DG 8冑2 Â AE = 14 Â 8 14 Â 8 14 冑2 14 AE = = Â = 冑2 = 7冑2 8冑2 冑2 冑2 2 In +ADE is DE =

冑

102

í (7冑

2)2

D

C E

10

A

= 冑2.

F B

G 14

BF = DE = 冑2, dus EF = 8冑2 í 冑2 í 冑2 = 6冑2. 43

D

Stel CS = x. In +BCS is BS2 + CS2 = BC2 BS2 + x2 = 100 BS2 = 100 í x2 BS = 冑100 í x2 O(ABCD) = 12  AC  BD geeft 12  2x  2冑100 í x2 = 60 x冑100 í = 30 kwadrateren geeft x2(100 í x2) = 900 íx4 + 100x2 í 900 = 0 x4 í 100x2 + 900 = 0 Stel x2 = u. u2 í 100u + 900 = 0 (u í 90)(u í 10) = 0 u = 90  u = 10 x2 = 90  x2 = 10 x = 冑90  x = í冑90  x = 冑10  x = í冑10 vold. vold. niet vold. vold. niet x = 3冑10  x = 冑10

10

10

x2

S

A

x

C

x

10

10

De lengten van de diagonalen zijn 6冑10 en 2冑10. B

© Noordhoff Uitgevers bv

Gemengde opgaven 161

44

5

E

N

D

60°

60°

M

60° 60° 5 30°

60° F

K

60°

60°

A

B

C

L

In +MDC is “M = 90° , “D = 60° en DC = 5, dus MD = 212 en MC = 212 冑3. Dus MN = 212 + 5 + 212 = 10 en LM = 212 冑3 + 212 冑3 = 5冑3. Dit geeft O(KLMN) = 10  5冑3 = 50冑3. 45

Stel QR = x, dan is DR + QC = 5 í x f DR = 12 (5 í x) DR = QC AS = QR = x, dus DS = 5 í x. O(+DRS) = 12  DR  DS = 12  12 (5 í x)  (5 í x) = 14 (5 í x)2 O(+CPQ) = O(+DRS) = 14 (5 í x)2 O(ABPQRS) = O(ABCD) í O(+DRS) í O(+CPQ) = 52 í 14 (5 í x)2 í 14 (5 í x)2 = 25 í 12 (5 í x)2 O(ABPQRS) = 15 geeft 25 í 12 (5 í x)2 = 15 10 = 12 (5 í x)2 (5 í x)2 = 20 5 í x = 冑20  5 í x = í冑20 x = 5 í 冑20  x = 5 + 冑20 voldoet niet Dus AS = 5 í 冑20 = 5 í 冑4  5 = 5 í 2冑5.

46

SC // DE en SE // DC, dus SCDE is een parallellogram. CS = ED = 1 f CS = ES = 1 ES = CD = 1 BE = EC = x en ES = 1, dus BS = x í 1 “ESC = “ ASB (overstaande hoeken) f +ESC ऴ +BSA “ECS = “ BAS (Z-hoeken) AB

BS

EC

ES

geeft

1 x

xí1 1

Dus x(x í 1) = 1 x2 í x í 1 = 0 D = (í1)2 í 4  1  í1 = 5 1 í 冑5 1 1 1 + 冑5 1 1 x= = 2 í 2 冑5  x = = 2 + 2 冑5 2 2 vold. niet vold. Dus EC = 12 + 12 冑5.

162 Gemengde opgaven

© Noordhoff Uitgevers bv

Bladzijde 197 47

Stel AD = x, dan is AB = 2x, BD = x冑3 en BC = x冑3.

O(ABCD) = O(+ABD) + O(+BCD) = 12  x  x冑3 + 12  x冑3  x冑3 = 12 x2冑3 + 112 x2 O(ABCD) = 712 geeft 12 x2冑3 + 112 x2 = 712 x2冑3 + 3x2 = 15 x2 (冑3 + 3) = 15 15 x2 = 冑3 + 3 x = 15 BD =  冑3 = 冑3 + 3 48

15 x=í 冑3 + 3

15 冑3 + 3

vold. niet 45 15 en AB = 2 Â 冑3 + 3 冑3 + 3

Het langste lijnstuk binnen een cirkel is een middellijn. Gedurende het rollen is er altijd een middellijn die de afstand bepaalt tot het punt dat op dat moment het verst van de vijfhoek afligt. Zolang de cirkel langs een zijde rolt, wordt de buitengrens dus een lijnstuk evenwijdig aan de zijde. Als de cirkel om een hoekpunt rolt, beschrijft de middellijn een cirkelboog. De oppervlakte van het gebied dat door de cirkel bestreken wordt, is dus gelijk aan vijf vierkantjes met zijde 4 en vijf cirkelsectoren (taartpunten) die samen een cirkel vormen met straal 4. De gevraagde oppervlakte is: 5 Â 42 + ʌ Â 42 = 16(5 + ʌ).

© Noordhoff Uitgevers bv

Gemengde opgaven 163

Verantwoording

Omslagontwerp: In Ontwerp, Assen Ontwerp binnenwerk: Ebel Kuipers, Sappemeer Technisch tekenwerk: OKS, Delhi (India) Lay-out: OKS, Delhi (India)

0 / 15 © 2015 Noordhoff Uitgevers bv, Groningen/Houten, The Netherlands. Behoudens de in of krachtens de Auteurswet van 1912 gestelde uitzonderingen mag niets uit deze uitgave worden verveelvoudigd, opgeslagen in een geautomatiseerd gegevensbestand of openbaar gemaakt, in enige vorm of op enige wijze, hetzij elektronisch, mechanisch, door fotokopieën, opnamen of enige andere manier, zonder voorafgaande schriftelijke toestemming van de uitgever. Voor zover het maken van reprogra¿sche verveelvoudigingen uit deze uitgave is toegestaan op grond van artikel 16h Auteurswet 1912 dient men de daarvoor verschuldigde vergoedingen te voldoen aan Stichting Reprorecht (Postbus 3060, 2130 KB Hoofddorp, www.reprorecht.nl). Voor het overnemen van (een) gedeelte(n) uit deze uitgave in bloemlezingen, readers en andere compilatiewerken (artikel 16 Auteurswet 1912) kan men zich wenden tot Stichting PRO (Stichting Publicatie- en Reproductierechten Organisatie, Postbus 3060, 2130 KB Hoofddorp, www.stichting-pro.nl). All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without prior written permission of the publisher. ISBN 978-90-01-84247-5