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IQM – Introduction Quantitative Methods 21 oktober 2015 18:00 – 20:301a. (5)
a, b en c zijn positief. Vereenvoudig tot één breuk zonder gebroken en negatieve exponenten:
1 3
∙ 𝑎𝑎
𝑐𝑐 √ 3 2
2 ∙ �𝑏𝑏 ∙ 2 − 3 ∙ 𝑏𝑏 �
−
𝑎𝑎 3
1 1 2
1 3
∙ 𝑎𝑎
𝑐𝑐 √ 3 2 𝑐𝑐 −1 ∙ 𝑎𝑎3 9 4 𝑎𝑎3 ∙ 𝑎𝑎3 5 5𝑎𝑎𝑎𝑎
2 ∙ �𝑏𝑏 ∙ 2 − 3 ∙ 𝑏𝑏 � = 2 ∙ �6 − 6� ∙ 𝑏𝑏 = ∙ 6 ∙ 𝑏𝑏 =
𝑎𝑎−3 𝑎𝑎−3 𝑐𝑐 6𝑐𝑐
1b. (5)
x, y en z zijn positief en verschillend. Schrijf z als functie van of formule met alleen x en y:
𝑥𝑥(𝑥𝑥 − 𝑧𝑧)
𝑦𝑦 − 2𝑥𝑥 = ⟹ 𝑧𝑧 = ⋯
𝑧𝑧
𝑥𝑥(𝑥𝑥 − 𝑧𝑧)
𝑦𝑦 − 2𝑥𝑥 = ⟹ 𝑦𝑦𝑦𝑦 − 2𝑥𝑥𝑥𝑥 = 𝑥𝑥 2 − 𝑥𝑥𝑥𝑥 ⟹
𝑧𝑧
𝑥𝑥 2
𝑦𝑦𝑦𝑦 − 𝑥𝑥𝑥𝑥 = 𝑥𝑥 2 ⟹ 𝑧𝑧(𝑦𝑦 − 𝑥𝑥) = 𝑥𝑥 2 ⟹ 𝑧𝑧 =
𝑦𝑦 − 𝑥𝑥
2a. (6)
Los het stelsel van vergelijkingen op:
1
𝑦𝑦 = 3𝑥𝑥 − 1
�3𝑦𝑦 − 𝑥𝑥 = −3
2
𝑦𝑦 + 2 = −3𝑥𝑥
(het lijkt er eerst op dat we het snijpunt van drie lijnen zoeken,
maar het blijkt om slechts twee lijnen te gaan)
1
1
𝑦𝑦 = 3𝑥𝑥 − 1 ⎧ 𝑦𝑦 = 3𝑥𝑥 − 1 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 1
⎪ � 𝑦𝑦 = 3𝑥𝑥 − 1
�3𝑦𝑦 − 𝑥𝑥 = −3 ⟹
1
𝑦𝑦 = 3𝑥𝑥 − 1 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 ⟹ � 2
2 ⎨ 𝑦𝑦 = −3𝑥𝑥 − 2
𝑦𝑦 + 2 = −3𝑥𝑥 2
⎪𝑦𝑦 = − 𝑥𝑥 − 2 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
⎩ 3
1 2 4 1
𝑦𝑦 = 𝑦𝑦 ⟹ 3𝑥𝑥 − 1 = −3𝑥𝑥 − 2 ⟹ 𝑥𝑥 = −1 ⟹ 𝑦𝑦 = −3 �= −13�
2b. (6)
Los de ongelijkheid op:
1 4
−4𝑥𝑥 + 4 < 5𝑥𝑥 + 4 < −2𝑥𝑥 + 32
Eerst de linkerongelijkheid:
1 4 1 4 5 16 21
−4𝑥𝑥 +4 < 5
𝑥𝑥 + 4 ⟹ −4𝑥𝑥 < 5𝑥𝑥 ⟹ −20𝑥𝑥 < 20𝑥𝑥 ⟹ −20𝑥𝑥 < 0 ⟹ 𝑥𝑥 > 0
Dan de rechterongelijkheid:
4 4 14
5
𝑥𝑥 + 4 < −2𝑥𝑥 + 32 ⟹ 5
𝑥𝑥 + 2𝑥𝑥 = 5
𝑥𝑥 < 28 ⟹ 𝑥𝑥 < 10
Dus samen:
𝑥𝑥 > 0 ∧ 𝑥𝑥 < 10 ⟹ 0 < 𝑥𝑥 < 10
1
, 3a. (6)
Los het stelsel van vergelijkingen op:
𝑦𝑦 = (𝑥𝑥 − 2)(𝑥𝑥 − 5)
�
𝑦𝑦 = −(𝑥𝑥 − 2)(𝑥𝑥 − 4)
Manier 1:
𝑦𝑦 = (𝑥𝑥 − 2)(𝑥𝑥 − 5) 𝑦𝑦 = 𝑥𝑥 2 − 7𝑥𝑥 + 10
� ⟹ � ⟹
𝑦𝑦 = −(𝑥𝑥 − 2)(𝑥𝑥 − 4) 𝑦𝑦 = −𝑥𝑥 2 + 6𝑥𝑥 − 8
𝑦𝑦 − 𝑦𝑦 = 𝑥𝑥 2 − 7𝑥𝑥 + 10 − (−𝑥𝑥 2 + 6𝑥𝑥 − 8) ⟹
1
0 = 2𝑥𝑥 2 − 13𝑥𝑥 + 18 ⟹ 𝑥𝑥 2 − 62𝑥𝑥 + 9 = 0 ⟹
1 1
𝑥𝑥 2 + �−2 − 42� 𝑥𝑥 + �−2 ∙ −42� = 0 ⟹
(𝑥𝑥 − 2) �𝑥𝑥 − 41� = 0 ⟹ 𝑥𝑥 = 2 ∨ 𝑥𝑥 = 41 �= 9�
2 2 2
Manier 2:
𝑦𝑦 = (𝑥𝑥 − 2)(𝑥𝑥 − 5)
� ⟹
𝑦𝑦 = −(𝑥𝑥 − 2)(𝑥𝑥 − 4)
(𝑥𝑥 − 2)(𝑥𝑥 − 5) = −(𝑥𝑥 − 2)(𝑥𝑥 − 4) ⟹
(𝑥𝑥 − 2)(𝑥𝑥 − 5) + (𝑥𝑥 − 2)(𝑥𝑥 − 4) = 0 ⟹
(𝑥𝑥 − 2)(𝑥𝑥 − 5 + 𝑥𝑥 − 4) = 0 ⟹
(𝑥𝑥 − 2)(2𝑥𝑥 − 9) = 0 ⟹
9 1
𝑥𝑥 = 2 ∨ 2𝑥𝑥 = 9 ⟹ 𝑥𝑥 = 2 ∨ 𝑥𝑥 = 2 �= 42�
3b. (6)
Waar ligt de top van deze parabool? Bepaal a, r en h.
2𝑥𝑥 2 − 10𝑥𝑥 + 13 = 𝑎𝑎(𝑥𝑥 − 𝑟𝑟)2 + ℎ
Manier 1: uitwerken
𝑎𝑎(𝑥𝑥 − 𝑟𝑟)2 + ℎ = 𝑎𝑎(𝑥𝑥 2 − 2𝑟𝑟𝑟𝑟 + 𝑟𝑟 2 ) + ℎ = 𝑎𝑎𝑎𝑎 2 − 2𝑎𝑎𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑟𝑟 2 + ℎ
� ⟹
𝑎𝑎(𝑥𝑥 − 𝑟𝑟)2 + ℎ = 2𝑥𝑥 2 − 10𝑥𝑥 + 13
𝑎𝑎 = 2 𝑎𝑎 = 2 𝑎𝑎 = 2
𝑎𝑎 = 2 5 5
⟹ � 𝑟𝑟 = 2 ⟹ � 𝑟𝑟 = 2
5
� −2𝑎𝑎𝑎𝑎 = −10 ⟹ � 𝑟𝑟 = 2
25 1
𝑎𝑎𝑟𝑟 2 + ℎ = 13 𝑎𝑎𝑟𝑟 2 + ℎ = 13 + ℎ = 13 ℎ=2
2
Dus de top ligt op (𝑟𝑟, ℎ) = �52, 12�
Manier 2: in de top is de afgeleide nul
𝑦𝑦 = 2𝑥𝑥 2 − 10𝑥𝑥 + 13 ⟹ 𝑦𝑦 ′ = 4𝑥𝑥 − 10
10 5
𝑦𝑦 ′ = 0 ⟹ 𝑥𝑥 = 4
=2
Invullen in y geeft de hoogte:
5 2 5 1
𝑦𝑦 = 2 �2� − 10 ∙ 2 + 13 = 2
Top �52, 12� invullen in 𝑎𝑎(𝑥𝑥 − 𝑟𝑟)2 + ℎ geeft:
5 2 1 10 25 1 25 1
𝑎𝑎 �𝑥𝑥 − 2� + 2 = 𝑎𝑎 �𝑥𝑥 2 − 2
𝑥𝑥 + 4 � + 2 = 𝑎𝑎𝑥𝑥 2 − 5𝑎𝑎𝑎𝑎 + 4
𝑎𝑎 +2
25 1
𝑎𝑎𝑥𝑥 2 − 5𝑎𝑎𝑎𝑎 + 4
𝑎𝑎 + 2
= 2𝑥𝑥 2 − 10𝑥𝑥 + 13 ⟹ 𝑎𝑎 = 2
2
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