Mat1511 Quiz 2

0




Precalculus B
2
QUIZ

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Precalculus B
Due date N/A
Time limit 1 hours 30 minutes
Grade out of 50.00


QUESTION 1


Find the term containing 𝑥 3 in (𝑥 + 2𝑦)7 :
Choose the correct answer:
o a. 560𝑥3𝑦4
o b. 1120𝑥3𝑦4
o c. 280𝑥3 𝑦4
o d. 2240𝑥3𝑦4
o e. None of the options


𝑥 3 in the expansion of (𝑥 + 2𝑦)7
𝑎𝑟 ……………………(𝑎 + 𝑏)𝑛
➢ 𝑟 = 3 ; 𝑛 = 7 ; 𝑎 = 𝑥 ; 𝑏 = 2𝑦
𝑛
= ( ) 𝑎𝑟 𝑏 𝑛−𝑟
𝑟
7
= ( ) (𝑥)3 (2𝑦)7−3
3
7!
=[ ] ∙ 𝑥 3 (2𝑦)4
3! (7 − 3)!
7 × 6 × 5 × 4!
= ∙ 16𝑥 3 𝑦 4
3 × 2 × 1 × 4!
5
= 2 × × 7 × 16𝑥 3 𝑦 4
2
= 35 × 16𝑥 3 𝑦 4
= 560𝑥 3 𝑦 4

QUESTION 2

According to the Rational Zero Theorem, which list contains all
possible rational zeros of
𝟒𝒙𝟑 − 𝟕𝒙𝟐 + 𝟓𝒙 − 𝟐 ?

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Choose the correct answer:
1 1
o a. ±1; ±2; ± ; ±
2 4
o b. ±1; ±2
1
o c. ± only
4
o d. ±2; ±4
o e. None of the options


The leading coefficient term is 4 and the constant term is
−2.
𝑓𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡𝑒𝑟𝑚
∴ 𝑝𝑜𝑠𝑖𝑏𝑙𝑒 𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑧𝑒𝑟𝑜𝑠 =
𝑓𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 𝑙𝑒𝑎𝑑𝑖𝑛𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡

±1; ±2
=
±1; ±2; ±4
1 1
= ± ; ± ; ±1; ±2
4 2

QUESTION 3

Determine whether the given statement is true?


2 + 4 + 6 + ⋯ + 2𝑛 = 𝑛(𝑛 + 1)
Choose the correct option for the statement:
o a. True for all natural numbers 𝑛
o b. False
o c. True only for even 𝑛
o d. True only for odd 𝑛
o e. None of the options

Let 𝑃(𝑛) denote the statement 2 + 4 + 6 + ⋯ + 2𝑛 = 𝑛(𝑛 + 1)
1. 𝑷(𝟏)
2(1) = (1)[(1) + 1]
2 = 1(2)
2=2
∴ 𝑃(1) is true.

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2. INDUCTION HYPOTHESIS
𝑷(𝒌)
2 + 4 + 6 + ⋯ + 2(𝑘) = 𝑘(𝑘 + 1)
𝑷(𝒌 + 𝟏)
2 + 4 + 6 + ⋯ + 2𝑘 + 2(𝑘 + 1) = (𝑘 + 1)[(𝑘 + 1) + 1]
𝑘(𝑘 + 1) + 2(𝑘 + 1) = (𝑘 + 1)(𝑘 + 2)
(𝑘 + 1)(𝑘 + 2) = (𝑘 + 1)(𝑘 + 2)
𝐿𝐻𝑆 = 𝑅𝐻𝑆
∴ 𝑃(𝑘) and 𝑃(𝑘 + 1) is true.
CONCLUSION: Having proven Step 1 and Step 2, we can conclude by the
Principle of Mathematical Induction that 𝑃(𝑛) is true for all natural numbers
𝑛.

QUESTION 4
= 𝟐 is a zero of 𝑷(𝒙) = 𝒙𝟑 − 𝟓𝒙𝟐 + 𝟐𝒙 + 𝟖, determine the
If 𝒙
remaining zeros of 𝑷(𝒙) ?
Choose the correct answer:
o a. 1 𝑎𝑛𝑑 4
o b. −1 𝑎𝑛𝑑 4
o c. 2 𝑎𝑛𝑑 4
o d. −2 𝑎𝑛𝑑 − 1
o e. None of the options



2 1 -5 2 8
2 -6 -8
1 -3 -4 0
𝑃(𝑥) = (𝑥 − 2)(𝑥 2 − 3𝑥 − 4) = 0
𝑥−2=0 OR 𝑥 2 − 3𝑥 − 4 = 0
𝑥=2 −(−3) ± √(−3)2 − 4(1)(−4)
𝑥=
2(1)
3 ± √25
𝑥=
2
3±5
𝑥=
2
𝑥 = −1 OR 𝑥 = 4



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QUESTION 5

2𝑥+7
Express in partial fractions.
𝑥 2 −2𝑥−3

Choose the correct answer:
7 2𝑥
o a. +
𝑥+1 𝑥−3

−5 13
o b. −
4(𝑥+1) 4(𝑥−3)

−5 13
o c. +
4(𝑥+1) 4(𝑥−3)

o d. None of the options

−7 2𝑥+14
o e. +
𝑥 2 +1 𝑥−3


2𝑥 + 7 𝐴 𝐵
= +
(𝑥 + 1)(𝑥 − 3) 𝑥 + 1 𝑥 − 3
2𝑥 + 7 𝐴(𝑥 − 3) 𝐵(𝑥 + 1)
= +
(𝑥 + 1)(𝑥 − 3) (𝑥 + 1)(𝑥 − 3) (𝑥 − 3)(𝑥 + 1)
2𝑥 + 7 = 𝐴(𝑥 − 3) + 𝐵(𝑥 + 1)
2𝑥 + 7 = 𝐴𝑥 − 3𝐴 + 𝐵𝑥 + 𝐵
2𝑥 + 7 = 𝐴𝑥 + 𝐵𝑥 − 3𝐴 + 𝐵
2𝑥 + 7 = (𝐴 + 𝐵)𝑥 − 3𝐴 + 𝐵

𝐴+𝐵 =2
−3𝐴 + 𝐵 = 7

𝐵 = 2−𝐴 −3𝐴 + (2 − 𝐴) = 7
−4𝐴 + 2 = 7
2 − 7 = 4𝐴
−5 = 4𝐴
5 13 5
∴ 𝐵 = 2 − (− ) = ∴𝐴=−
4 4 4


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5 13
2𝑥 + 7 −
∴ 2 = 4 + 4
𝑥 − 2𝑥 − 3 𝑥 + 1 𝑥 − 3
−5 13
= +
4(𝑥 + 1) 4(𝑥 − 3)


QUESTION 6


Given 𝑷(𝒙) = 𝟐𝒙𝟑 + 𝟓𝒙𝟐 − 𝟖𝒙 − 𝟕 ,using the Upper and Lower
Bounds Theorem to show that the real zeros of 𝑷(𝒙) lie between
−𝟒 and 𝟑 , we get:


A.


3 2 5 -8 -7
-6 -3 -33
2 -1 -11 -40
B.

-4 2 5 -8 -7
-8 12 -16
2 -3 4 -23

Which of the statements are true?
Choose the correct option for the statements:
o a. Only A
o b. Only B
o c. Both A and B
o d. Neither A nor B
o e. None of the options


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•
3 2 5 -8 -7
6 33 75
2 11 25 68
All entries are positive
•
Entries alternate in sign
-4 2 5 -8 -7
-8 +12 -16
2 -3 4 -23
∴ Only B is correct.
QUESTION 7

A boat travels due south at 60 km/h while the current flows due east at
15 km/h.
Determine the true velocity vector?

Choose the correct answer:
o a. 𝑣 = 15𝑖 − 60𝑗
o b. 𝑣 = 60𝑖 + 15𝑗
o c. 𝑣 = 15𝑖 + 60𝑗
o d. 𝑣 = −15𝑖 − 60𝑗
o e. None of the options




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