Mat1511 Quiz 3

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Precalculus B
3
QUIZ

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Precalculus B
Due date N/A
Time taken 1 hour
Grade out of 50.00


QUESTION 1
Use Descartes Rule of signs to find the possible number of positive
real zeros of
𝑷(𝒙) = −𝟑𝒙𝟔 + 𝒙𝟓 − 𝟒𝒙𝟒 − 𝒙𝟑 + 𝟕𝒙𝟐 − 𝟐𝒙 − 𝟔 ?
Choose the correct answer:
o a. 6 𝑜𝑟 4 𝑜𝑟 2 𝑜𝑟 0
o b. 4 𝑜𝑟 2 𝑜𝑟 0
o c. 5 𝑜𝑟 3 𝑜𝑟 1
o d. 2 𝑜𝑟 0
o e. None of the options


𝑃(𝑥) = −3𝑥 6 + 𝑥 5 − 4𝑥 4 − 𝑥 3 + 7𝑥 2 − 2𝑥 − 6

Number of sign changes: 4
∴ The number of positive real zeros is: 4 or 2 or 0

QUESTION 2
Solve for 𝒚 in the given matrix equation
4 2 1
𝑑𝑒𝑡 [0 3 𝑦] = 6
2 1 1
Choose the correct answer:
o a. 𝑦 = −1
o b. 𝑦=5
o c. 𝑦 = −3
o d. 𝑦=2
o e. None of the options

N/A
• THERE IS AN ERROR IN THIS QUESTION.
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QUESTION 3
Using the De Moivre’s Theorem to find all the complex roots of
𝒛𝟑 + 𝟐𝟕 = 𝟎 we obtain?
Choose the correct answer:
𝜋 5𝜋
o a. 3∠ ; 3∠𝜋; 3∠
3 3

2𝜋 4𝜋
o b. 3∠0; 3∠ ; 3∠
3 3

𝜋 5𝜋
o c. 27 ∠ ; 27 ∠ 𝜋 ; 27 ∠
3 3

𝜋 5𝜋
o d. 1∠ ; 1∠𝜋; 1∠
3 3

o e. None of the options



𝑧 3 + 27 = 0
𝑧 3 = −27
3
𝑧 = √−27
−𝟐𝟕 is located in directly on the negative real axis of the
complex plane.
𝑟 2 = 𝑎2 + 𝑏 2 𝑏 0
tan 𝜃 = = =0
𝑟 2 = (−27)2 + (0)2 𝑎 −27
𝑟 2 = 272 𝜃 = tan−1 ( 0 )
𝜃 =0+𝜋
𝑟 = √272
𝜃=𝜋
𝑟 = 27

1 𝜃 + 2𝜋𝑘 𝜃 + 2𝜋𝑘
𝑊𝑘 = 𝑟 𝑛 [cos ( ) + 𝑖 sin ( )] for 𝑘 = 0; 1; 2; … ; 𝑛 − 1
𝑛 𝑛
1 𝜋 + 2𝜋𝑘 𝜋 + 2𝜋𝑘
(3)
𝑊𝑘 = (27) [cos ( ) + 𝑖 sin ( )] for 𝑘 = 0; 1; 2
3 3
1 𝜋 + 2𝜋𝑘 𝜋 + 2𝜋𝑘
𝑊𝑘 = (33 )3 [cos ( ) + 𝑖 sin ( )] for 𝑘 = 0; 1; 2
3 3
𝜋 + 2𝜋𝑘 𝜋 + 2𝜋𝑘
𝑊𝑘 = 3 [cos ( ) + 𝑖 sin ( )] for 𝑘 = 0; 1; 2
3 3



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𝜋 + 2𝜋(0) 𝜋 + 2𝜋(0)
∴ 𝑾𝟎 = 3 [cos ( ) + 𝑖 sin ( )]
3 3
𝜋 𝜋
= 3 [cos ( ) + 𝑖 sin ( )]
3 3
𝜋
= 3∠
3

𝜋 + 2𝜋(1) 𝜋 + 2𝜋(1)
∴ 𝑾𝟏 = 3 [cos ( ) + 𝑖 sin ( )]
3 3
3𝜋 3𝜋
= 3 [cos ( ) + 𝑖 sin ( )]
3 3
= 3[cos 𝜋 + 𝑖 sin 𝜋]
= 3∠𝜋

𝜋 + 2𝜋(2) 𝜋 + 2𝜋(2)
∴ 𝑾𝟐 = 3 [cos ( ) + 𝑖 sin ( )]
3 3
𝜋 + 4𝜋 𝜋 + 4𝜋
= 3 [cos ( ) + 𝑖 sin ( )]
3 3
5𝜋 5𝜋
= 3 [cos ( ) + 𝑖 sin ( )]
3 3
5𝜋
= 3∠
3

𝜋 5𝜋
∴ The cube roots of − 27 are: 3∠ , 3∠𝜋 and 3∠
3 3


QUESTION 4
A farmer can cultivate tomatoes and corn on 80 hectares of land.
A hectare of corn yields a profit of R12,000 and requires four
labour days. A hectare of tomatoes yields a profit of R18,000 and
requires six labour-days. The maximum number of labour-days
available to the farmer is 420.

Question: List the constraints’ function and the objective
function?

Choose the correct answer:

2𝑥 ≤ 80
10𝑦 ≤ 420
o a. {
𝑥≥0
𝑦≥0
Objective Function ⇒ 𝑃 = 12 000𝑥 + 18 000𝑦

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𝑥 + 𝑦 ≤ 80
4𝑥 + 6𝑦 ≤ 420
o b. {
𝑥≥0
𝑦≥0
Objective Function ⇒ 𝑃 = 6 000(2𝑥 + 3𝑦)

𝑥 + 𝑦 ≥ 80
2𝑥 + 3𝑦 ≥ 420
o c. {
𝑥≥0
𝑦≥0
Objective Function ⇒ 𝑃 = 12 000𝑥 + 18 000𝑦

4𝑥 + 6𝑦 ≤ 80
𝑥 + 𝑦 ≤ 420
o d. {
𝑥≥0
𝑦≥0
Objective Function ⇒ 𝑃 = 12 000𝑥 − 18 000𝑦

o e. None of the options


Let: • 𝑥 = Corn
• 𝑦 = Tomato

Labour constraint: 4𝑥 + 6𝑦 ≤ 420

Land constraint: 𝑥 + 𝑦 ≤ 80

Corns and Tomatoes cannot be negative: 𝑥≥0 , 𝑦≥0

Profit function: 12 000𝑥 + 18 000𝑦

✓ ∴ THE FINAL ANSWER IS:
𝑥 + 𝑦 ≤ 80
4𝑥 + 6𝑦 ≤ 420
{
𝑥≥0
𝑦≥0
𝑂𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝐹𝑢𝑛𝑐𝑡𝑖𝑜𝑛 ⇒ 𝑃 = 6000(2𝑥 + 3𝑦)


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QUESTION 5
𝜋 𝜋
If 𝑧 = 2 (cos 6 + 𝑖 sin 6 ) , calculate the value of 𝒛𝟔
Choose the correct answer:
o a. 2√3
o b. None of the options
o c. √3
o d. 1 − 2𝑖
o e. −64

De Moivre’s Theorem states that 𝑧 𝑛 = 𝑟 𝑛 (cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃)

𝜋 𝜋
∴ 𝑧 6 = (2)(6) [cos(6) ( ) + 𝑖 sin(6) ( )]
6 6
= 64(cos 𝜋 + 𝑖 sin 𝜋)
= 64[−1 + 𝑖(0)]
= −64

QUESTION 6
1 2 𝑤 𝑥 1 0
If [ ][ ] = [ ] , then 𝑤 = ⋯
3 5 𝑦 𝑧 0 1
Choose the correct answer:
o a. 𝑤 = −1
o b. 𝑤 =3
o c. 𝑤 = −5
1
o d. 𝑤 = −2
o e. None of the options

1(𝑤) + 2(𝑦) 1(𝑥) + 2(𝑧) 1 0
[ ]=[ ]
3(𝑤) + 5(𝑦) 3(𝑥) + 5(𝑧) 0 1
𝑤 + 2𝑦 = 1 3(1 − 2𝑦) + 5𝑦 = 0
3𝑤 + 5𝑦 = 0 3 − 6𝑦 + 5𝑦 = 0
𝑦=3
𝑤 = 1 − 2𝑦 𝑤 = 1 − 2(3)
∴ 𝑤 = −5

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QUESTION 7
𝟐𝝅
Convert the following polar coordinates (−𝟔 ; ) into rectangular
𝟑
coordinates.

Choose the correct answer:
o a. ( 3√3 ; −3 )
o b. ( −2 ; 2√3 )
o c. ( 3 ; −3√3 )
o d. None of the options
e. ( −3 ; −3√3 )


𝑥 = 𝑟 ∙ cos 𝜃 𝑦 = 𝑟 ∙ sin 𝜃
2𝜋 2𝜋
= −6 ∙ cos = −6 ∙ sin
3 3
𝜋 𝜋
= −6 ∙ cos (𝜋 − ) = −6 ∙ sin (𝜋 − )
3 3
𝜋 𝜋
= −6 (− cos ) = −6 (sin )
3 3
1 √3
= −6 × − = −6 ×
2 2
=3 = −3√3


∴ (𝑥; 𝑦) = ( 3 ; −3√3 )

QUESTION 8
State if the following is true or false.
If we perform the row operation, 𝑹𝟐 + 𝑹𝟏 → 𝑹𝟐 ,then 𝑹𝟑 + 𝟐𝑹𝟏 → 𝑹𝟑
on the matrix
1 1 1 −4
(−1 2 −3| 3 )
−2 −3 −1 0
We obtain


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