Exam (elaborations)
IQM tentamen 2016-1-06
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IQM tentamen vragen en antwoorden
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By: evacabooter • 5 year ago
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Uitwerkingen tentamen IQM – 6 januari 2016Er zijn vaak meerdere manieren om goed antwoord te geven op de opgaven.
Bij de uitwerkingen hieronder geven we steeds één manier.
1a.
Vereenvoudig zo ver mogelijk. Schrijf je eindantwoord met worteltekens i.p.v. gebroken exponenten.
�𝑎𝑎12
2∙ 1
𝑏𝑏 −3
3
𝑎𝑎4
8∙ 1
𝑏𝑏 6
Antwoord:
�𝑎𝑎12 1
2∙ 1 2 1
𝑏𝑏 −13 2 �𝑎𝑎 � 𝑏𝑏 6
2
1 1 3 1 1 1 1 3 2 1 1 1 1 √𝑏𝑏
3 = ∙ ∙ = ∙ 𝑎𝑎4 ∙ 𝑎𝑎 −4 ∙ 𝑏𝑏 3 ∙ 𝑏𝑏 6 = ∙ 𝑎𝑎4−4 ∙ 𝑏𝑏 6+6 = ∙ 𝑎𝑎−2 ∙ 𝑏𝑏 2 =
𝑎𝑎 4 8 𝑏𝑏 −13 𝑎𝑎34 4 4 4 4√𝑎𝑎
8∙ 1
𝑏𝑏 6
1b.
Schrijf als één vereenvoudigde breuk:
𝑥𝑥 + 1 𝑥𝑥 − 1 2𝑥𝑥 2 − 6
+ −
𝑥𝑥 + 3 𝑥𝑥 − 3 2(𝑥𝑥 2 − 9)
Antwoord:
𝑥𝑥 + 1 𝑥𝑥 − 1 2𝑥𝑥 2 − 6
+ −
𝑥𝑥 + 3 𝑥𝑥 − 3 2(𝑥𝑥 2 − 9)
𝑥𝑥 + 1 𝑥𝑥 − 3 𝑥𝑥 − 1 𝑥𝑥 + 3 2(𝑥𝑥 2 − 3)
= ∙ + ∙ −
𝑥𝑥 + 3 𝑥𝑥 − 3 𝑥𝑥 − 3 𝑥𝑥 + 3 2(𝑥𝑥 2 − 9)
𝑥𝑥 2 − 2𝑥𝑥 − 3 𝑥𝑥 2 + 2𝑥𝑥 − 3 𝑥𝑥 2 − 3
= + − 2
𝑥𝑥 2 − 9 𝑥𝑥 2 − 9 𝑥𝑥 − 9
𝑥𝑥 2 − 2𝑥𝑥 − 3 + 𝑥𝑥 2 + 2𝑥𝑥 − 3 − (𝑥𝑥 2 − 3)
=
𝑥𝑥 2 − 9
𝑥𝑥 2 − 3
=
𝑥𝑥 2 − 9
, 2a.
Los het stelsel van vergelijkingen op:
9 + 5𝑦𝑦 = 6𝑥𝑥
�
9𝑥𝑥 + 30𝑦𝑦 = 6
Antwoord:
9 + 5𝑦𝑦 = 6𝑥𝑥 −6𝑥𝑥 + 5𝑦𝑦 = −9
×9 −54𝑥𝑥 + 45𝑦𝑦 = −81
� ⟹ � � ⟹ � 54𝑥𝑥 + 180𝑦𝑦 = 36
�
9𝑥𝑥 + 30𝑦𝑦 = 6 9𝑥𝑥 + 30𝑦𝑦 = 6
×6
225𝑦𝑦 = −45
45 1 8 4
⟹ 𝑦𝑦 = − = − ⟹ 6𝑥𝑥 = 9 + 5𝑦𝑦 = 9 − 1 = 8 ⟹ 𝑥𝑥 = =
225 5 6 3
2b.
Los het stelsel van vergelijkingen op:
𝑥𝑥 + 2𝑦𝑦 = 6
�
2𝑥𝑥(𝑥𝑥 − 𝑦𝑦) = 0
Antwoord:
𝑥𝑥 + 2𝑦𝑦 = 6 ⟹ 𝑥𝑥 = 6 − 2𝑦𝑦
� �
2 ∙ 𝑥𝑥 ∙ (𝑥𝑥 − 𝑦𝑦) = 0 ⟹ 2 ∙ (6 − 2𝑦𝑦) ∙ (6 − 3𝑦𝑦) = 0 ⟹ 𝑦𝑦 = 3 ∨ 𝑦𝑦 = 2
𝑦𝑦 = 3 ⟹ 𝑥𝑥 = 0
⟹ �
𝑦𝑦 = 2 ⟹ 𝑥𝑥 = 2
Dus twee oplossingen: (𝑥𝑥, 𝑦𝑦) = (0, 3) en (𝑥𝑥, 𝑦𝑦) = (2, 2)
3a.
Los de ongelijkheid op:
3 3 3
−𝑥𝑥 + 2 < 2𝑥𝑥 − 1 < 2𝑥𝑥 + 4
Antwoord:
Eerst de linkerongelijkheid:
3 3 3 3 5 5
−𝑥𝑥 + <
2 2
𝑥𝑥 − 1 ⟹ −𝑥𝑥 − 2𝑥𝑥 < −2 − 1 ⟹ −2𝑥𝑥 < −2 ⟹ 𝑥𝑥 > 1
Vervolgens de rechterongelijkheid:
3 3
2
𝑥𝑥 − 1 < 2𝑥𝑥 + 4 ⟹ −1 < 4
-1 is altijd kleiner dan 4, welke x je ook kiest, dus hier geen restrictie voor x
Beide ongelijkheden moeten tegelijk gelden, dus de oplossing is:
𝑥𝑥 > 1
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