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Exam (elaborations)

IQM tentamen

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IQM tentamen vragen en antwoorden

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  • January 4, 2018
  • 7
  • 2017/2018
  • Exam (elaborations)
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Tentamen IQM 26-10-2016


1a. Vereenvoudig tot één breuk:
2 2

𝑥𝑥 − 𝑦𝑦 𝑥𝑥
Uitwerking:
2 2 2𝑥𝑥 2(𝑥𝑥 − 𝑦𝑦)
− = −
𝑥𝑥 − 𝑦𝑦 𝑥𝑥 𝑥𝑥(𝑥𝑥 − 𝑦𝑦) 𝑥𝑥(𝑥𝑥 − 𝑦𝑦)
2𝑥𝑥 − (2𝑥𝑥 − 2𝑦𝑦)
=
𝑥𝑥(𝑥𝑥 − 𝑦𝑦)
2𝑥𝑥 − 2𝑥𝑥 + 2𝑦𝑦
=
𝑥𝑥(𝑥𝑥 − 𝑦𝑦)
2y
=
𝑥𝑥(𝑥𝑥 − 𝑦𝑦)


1b. Vereenvoudig tot één breuk:
3
√𝑎𝑎2 ∙ √𝑏𝑏 2
6
√𝑎𝑎4 ∙ √𝑏𝑏 3
Uitwerking:
3
2 2 2 0
√𝑎𝑎2 ∙ √𝑏𝑏 2 𝑎𝑎3 ∙ 𝑏𝑏 2 𝑎𝑎3 ∙ 𝑏𝑏 2 1
6 = 4 3 = 2 1 =
√𝑎𝑎4 ∙ √𝑏𝑏 3 𝑎𝑎6 ∙ 𝑏𝑏 2 𝑎𝑎3 ∙ 𝑏𝑏 2 √𝑏𝑏


2a. Lijn l gaat door het punt (4, 4) en staat loodrecht op lijn m met de vergelijking 3𝑥𝑥 − 𝑦𝑦 = −2.
Bepaal de vergelijking van lijn l.
Uitwerking:
𝑚𝑚: 3𝑥𝑥 − 𝑦𝑦 = −2 ⟹ 𝑦𝑦 = 3𝑥𝑥 + 2
⟹ 3 is de richtingscoëfficiënt van lijn m
1
⟹ − is de richtingscoëfficiënt van lijn l
3
1
⟹ 𝑙𝑙: 𝑦𝑦 = − 𝑥𝑥 + 𝑏𝑏
3

Lijn l gaat door (4, 4) dus:
1 1 16 1 1
4 = −3 ∙ 4 + 𝑏𝑏 ⟹ 𝑏𝑏 = 53 = 3
⟹ 𝑙𝑙: 𝑦𝑦 = −3𝑥𝑥 + 53


2b. De parabool p met vergelijking 𝑦𝑦 = 𝑎𝑎 𝑥𝑥 2 + 𝑏𝑏 𝑥𝑥 + 𝑐𝑐 gaat door de punten (–1, 0), (0, 2) en (1, 8).
Bepaal de coëfficiënten a, b en c.
Uitwerking:
Punten (x, y) invullen in 𝑦𝑦 = 𝑎𝑎 𝑥𝑥 2 + 𝑏𝑏 𝑥𝑥 + 𝑐𝑐:

0 = 𝑎𝑎 (−1)2 + 𝑏𝑏 (−1) + 𝑐𝑐 0 = 𝑎𝑎 − 𝑏𝑏 + 𝑐𝑐 𝑐𝑐=2
0 = 𝑎𝑎 − 𝑏𝑏 + 2 −2 = 𝑎𝑎 − 𝑏𝑏
�2 = 𝑎𝑎 02 + 𝑏𝑏 0 + 𝑐𝑐 ⟹ �2 = 𝑐𝑐 ���� � ⟹ �
8 = 𝑎𝑎 + 𝑏𝑏 + 2 6 = 𝑎𝑎 + 𝑏𝑏
8 = 𝑎𝑎 12 + 𝑏𝑏 1 + 𝑐𝑐 8 = 𝑎𝑎 + 𝑏𝑏 + 𝑐𝑐
4 = 2𝑎𝑎
⟹ 𝑎𝑎 = 2 ⟹ 8 = 2 + 𝑏𝑏 + 2 ⟹ 𝑏𝑏 = 4
Dus parabool p: 𝑦𝑦 = 2𝑥𝑥 2 + 4𝑥𝑥 + 2

, 3a. Los op:
2𝑥𝑥 − 2 < 𝑥𝑥 2 − 5 < 𝑥𝑥 + 7
Uitwerking:
• Eerst de linkerongelijkheid: lijn onder een dalparabool 2𝑥𝑥 − 2 < 𝑥𝑥 2 − 5
Snijpunten:
2𝑥𝑥 − 2 = 𝑥𝑥 2 − 5 ⟹ 𝑥𝑥 2 − 2𝑥𝑥 − 3 = 0 ⟹ (𝑥𝑥 − 3)(𝑥𝑥 + 1) = 0 ⟹ 𝑥𝑥 = 3 ∨ 𝑥𝑥 = −1
Ongelijkheid:
2
2𝑥𝑥 − 2 < 𝑥𝑥 − 5 ⟹ 𝑥𝑥 < −1 ∨ 𝑥𝑥 > 3
• Vervolgens de rechterongelijkheid: dalparabool onder een lijn 𝑥𝑥 2 − 5 < 𝑥𝑥 + 7
Snijpunten:
𝑥𝑥 2 − 5 = 𝑥𝑥 + 7 ⟹ 𝑥𝑥 2 − 𝑥𝑥 − 12 = 0 ⟹ (𝑥𝑥 − 4)(𝑥𝑥 + 3) = 0 ⟹ 𝑥𝑥 = 4 ∨ 𝑥𝑥 = −3
Ongelijkheid:
𝑥𝑥 2 − 5 < 𝑥𝑥 + 7 ⟹ −3 < 𝑥𝑥 < 4
• Dus de dubbele ongelijkheid:
−3 < 𝑥𝑥 < −1 ∨ 3 < 𝑥𝑥 < 4


3b. Los op:
𝑥𝑥 3 − 6𝑥𝑥 2 − 3𝑥𝑥 = 0
Uitwerking:
𝑥𝑥 3 − 6𝑥𝑥 2 − 3𝑥𝑥 = 0 ⟹ 𝑥𝑥(𝑥𝑥 2 − 6𝑥𝑥 − 3) = 0 ⟹ 𝑥𝑥 = 0 ∨ 𝑥𝑥 2 − 6𝑥𝑥 − 3 = 0
abc-formule voor 𝑥𝑥 2 − 6𝑥𝑥 − 3 = 0:
𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎 = (−6)2 − 4 ∙ 1 ∙ −3 = 48 > 0 dus twee nulpunten
oplossingen:
−𝑏𝑏 ± √𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎 6 ± √48
𝑥𝑥 = = = 3 ± 2√3
2𝑎𝑎 2
dus: 𝑥𝑥 = 0 ∨ 𝑥𝑥 = 3 − 2√3 ≈ −0.464 ∨ 𝑥𝑥 = 3 + 2√3 ≈ 6.464


4a. Bepaal de afgeleide van 𝑓𝑓(𝑥𝑥) = √𝑥𝑥 2 + 4 en schrijf je antwoord zonder negatieve en gebroken
exponenten
Uitwerking:
1 kettingregel 1 1 𝑥𝑥
𝑓𝑓(𝑥𝑥) = �𝑥𝑥 2 + 4 = (𝑥𝑥 2 + 4)2 ��������� 𝑓𝑓 ′ (𝑥𝑥) = 2(𝑥𝑥 2 + 4)−2 ∙ 2𝑥𝑥 =
√𝑥𝑥 2 +4


5𝑥𝑥 2 +𝑥𝑥
4b. Bepaal de afgeleide en vereenvoudig: 𝑔𝑔(𝑥𝑥) =
3𝑥𝑥 2 +𝑥𝑥

Uitwerking:
Differentiëren en daarna vereenvoudigen:
5𝑥𝑥 2 + 𝑥𝑥
𝑔𝑔(𝑥𝑥) =
3𝑥𝑥 2 + 𝑥𝑥
quotiëntregel (10𝑥𝑥 + 1)(3𝑥𝑥 2 + 𝑥𝑥) − (5𝑥𝑥 2 + 𝑥𝑥)(6𝑥𝑥 + 1)
����������� 𝑔𝑔′ (𝑥𝑥) =
(3𝑥𝑥 2 + 𝑥𝑥)2
(30𝑥𝑥 3 + 13𝑥𝑥 2 + 𝑥𝑥) − (30𝑥𝑥 3 + 11𝑥𝑥 2 + 𝑥𝑥)
=
(3𝑥𝑥 2 + 𝑥𝑥)2
2𝑥𝑥 2 2𝑥𝑥 2 2
= = =
(3𝑥𝑥 2 + 𝑥𝑥)2 𝑥𝑥 2 ∙ (3𝑥𝑥 + 1)2 (3𝑥𝑥 + 1)2

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