MATH 115 Precalculus Summer 2018 Final Examination Solved

This document presents a comprehensive solution set for the MATH 115 Precalculus final examination from the Summer of 2018. The exam, designed as an open-book assessment, includes a variety of problem types to evaluate students' understanding of precalculus concepts. The solution provides detailed answers to multiple-choice questions, demonstrating the correct selection and reasoning behind each choice. Additionally, it offers complete solutions to short-answer questions, showcasing the step-by-step work required for problem-solving. The content covers a wide array of topics, including inequalities, polynomial functions, logarithmic expressions, distance formulas, equations of circles, and trigonometric functions and identities. It also addresses inverse functions, angles, parabolas, and vector operations. The solutions are designed to help students understand the problem-solving process and to provide a reliable reference for studying and exam preparation. The document covers topics ranging from solving inequalities and logarithmic expressions to working with trigonometric functions, parabolas, and vectors.

Multiple Choices
Question 1
3−4 x ≤−7∨3−4 x ≥ 7
For 3−4 x ≤−7
−4 x ≤−10
x ≤ 10
4
x ≤ 5
2
3−4 x ≥7
−4 x ≤ 4
x ≤−1
Combining the two , we have : x ≤−1∨x ≥ 5
2
Interval Notation : (−∞ ,−1 ] ∪ [ 5
2 , ∞ )(D)
Question 2
D ¿ f ( x )=−6 x3 + x+ 8
Question 3
3 log( x +2)+log 1−log z
¿ log ( x +2 )3+ log 1
z
¿ log ( x+2 ) 3
z (C)
Question 4
C ¿ ( −∞ , 1 ) ∧(4 , ∞)
Question 5

D ¿ f ( x )=e x+2
Question 6
A ¿ f ( x )=3−cos x
Short Answer
Question 7
a) d= √ ( x1−x2 )2 + ( y1− y2 )2 = √ (−8+2 )2 + ( 5−7 )2=√ ¿
¿ 2 √10
b) Center of the ˚¿ ( x1 + x2
2 , y1 + y2
2 )=(−10
2 , 12
2 )=(−5,6)
c) Equation of a ˚is givenby : ( x−h ) 2 + ( y−k ) 2=r2 ,
Where ( h , k ) is thecenter of the ˚¿
( x +5 ) 2 + ( y−6 ) 2= ( 2 √ 10 ) 2=4∗10=40
x2+ 10 x + y2−12 y=−21
Question 8
log3
1
81
Let x=log3
1
81
3x= 1
81
( 1
3 )
−x
= ( 1
3 )
4
−x=4
x=−4
log3
1
81 =−4
Question 9
Commission Earned=5237.20−1650=3587.20
Thus 7.6 %=3587.20

100 %=?
100
7.6 ∗3587.20
¿ $ 47200
Question 10
T ( t ) =35+ 43 e− ( 0.058 t )
T ( t ) =35+ 43 e− ( 0.058∗12 ) =35+21.4=56.4 ° F
Question 11
f ( x )= 3
8 − 5
6 x
x= 3
8 − 5
6 y
y=−24 x +9
20
Inverse of the function is : y=−24 x +9
20
Question 12
a) Ar =360 °− Ac=360−300=60 °
b) 30 °= π
6
300 °=?
300
30 ∗π
6 = 10 π
6 = 5 π
3 radians
Question 13
a) Period :π
b) Phase Shift : π
2
Question 14
− √2 cos x−1=0

x= 3 π
4 , 5 π
4 .
Thus x=135 ° ,225 °
2 sin x−1= π
6 , 5 π
6 ,
Thus x=30 ° ,150 °
x=135 ° ,225 ° , 30° , 150°
Question 15
a) 0
b) π
3
Question 16
a) Direction of the Parabola: Up
b) Vertex: (5,-3)
c) Focus: (5,-1)
Question 17
a) Domain: (−∞,−3 ) ∪(−3 , ∞)
b) Vertical Asymptotes: x=−3
c) Horizontal Asymptotes: y=−1
d) Graph C
Short Answer: Work Required
Question 18
m= ( y2− y1 )
x2−x1
=−9−3
7−1 =−12
6 =−2
y−3
x−1 =−2
y−3=−2 x+ 2
y=−2 x +5
Question 19

4 x2 =1+12 x
4 x2 −12 x−1=0
x1,2=−b ± √ b2−4 ac
2 a
x= 3+ √10
2 ∨x= 3− √ 10
2
Question 20
a) ( f Ο g ) ( x )=2 ( x −4 )2−9
2 [ x2 −8 x+18 ]=2 x2−16 x +36
( f Ο g ) ( x )=2 x2−16 x +36
b) ( f Ο g ) ( x )=2 x2−16 x +36=2 (−2 )2−16 (−2 ) +36
8+32+36
¿ 76
Question 21
a) h=−16 t2 +72 t+18
At Maximumheight ,
dh
dt =0
dh
dt =−32 t+72=0
−32 t =−72
t=2.25 seconds
b) h=−16 t2 +72 t+18
h=−16∗2.252+72∗2.25+18
¿ 99 feet
Question 22
x+11
x +6 + 60
x2−36 =0
x+11
x +6 + 60
x2−36 = x +11
x +6 ( x +6 ) ( x−6 ) + 60
x2−36 ( x+6 ) ( x−6 )
¿ 0∗( x+6 ) (x−6)
( x +11 ) ( x−6 ) +60=0

x=1 , x=−6
Sincethe equation isundefined at x=−6 , x=1
Question 23
a) sin θ=2
7
( sin θ )2= 4
49
( cos θ )2=1− ( sin θ )2 =1− 4
49 = 45
49
cos θ= √ 45
49 = √ 45
7 = 3 √ 5
7
b) sin 2 θ=2 sin θ cos θ=
2∗2
7 ∗ √45
7 = 4 √45
49 =12 √ 5
49
Question 24
( sin x +cos x ) 2=sin2 x+ 2sin x cos x+ cos2 x
But sin2 x +cos2 x=1∧2sin x cos x=sin 2 x
Thus sin2 x+2sin x cos x+ cos2 x=1+ sin 2 x
Question 25
tan x= Opp
Adj
Thus tan 41.7 °= Height of thetree
55
Height of the Tree=55∗tan 41.7
¿ 49 feet
Question 26
Sine Rule : sin A
a = sin B
b = sin C
c
C=180−48−62=70 °
sin 70
35 =sin 48
a

a= 35∗sin 48
sin 70 =27.68
Question 27
a ¿ :⃗ u .⃗ v=a1 b1 +a2 b2
¿ 9∗4 + (−3 )∗12=36−36=0
b ¿ : Let the angle⃗ u∧⃗ v=x
cos x= (⃗ u⃗ . v )
||⃗u ||.∨¿⃗ v∨¿ = 0
( √92+ (−3 )2 )∗ ( √122+42 ) =0
cos x=¿0
x=90 °
Question 28
a) The major axis is horizontal
b) For an ellipse withmajor axis∥¿ the x−axis ,the foci are defined as
( h+c , k ) , ( h−c , k ) , where c= √a2−b2
is the distance ¿ the center ( h , k ) ¿ a focus .
With the given Equation, Ellipse withcenter ( h , k )= (−1,16 ) ,∧semi−axis
a=8 , b=4
(−1+c , 16 ) , (−1−c ,16 )
c= √ 82−42
¿ 4 √3
( −1+ 4 √ 3 ,16 ) , ( −1−4 √ 3 , 16 )