math.uncc.edu · PDF fileTitle: Fall2012_1103

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1 Final Exam Math 1103, Fall 2012 1. Find the slope and y-intercept of the line that is parallel to 2 + 3 = 5 and passes through the point (1, βˆ’1) a. = ! ! ; βˆ’ = ! ! b. = βˆ’ ! ! ; βˆ’ = ! ! c. = βˆ’ ! ! ; βˆ’ = βˆ’ ! ! d. = βˆ’ ! ! ; βˆ’ = ! ! e. None of the above 2. Find the domain of the function = ! ! ! !!!! a. (βˆ’βˆž, ∞) b. β‰  1 c. β‰  βˆ’2 d. β‰  2, βˆ’1 e. β‰  βˆ’2,1 3. If the point βˆ’2,1 is on the graph of () and () is known to be odd, what other point must be on the graph of () a. βˆ’2, βˆ’1 b. 2, βˆ’1 c. βˆ’2,1 d. 1, βˆ’1 e. 0, βˆ’1

Transcript of math.uncc.edu · PDF fileTitle: Fall2012_1103

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Final Exam

Math 1103, Fall 2012

1. Find the slope and y-intercept of the line that is parallel to 2π‘₯ + 3𝑦 = 5 and passes through the point (1,βˆ’1) a. π‘†π‘™π‘œπ‘π‘’ = !

!;𝑦 βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘ =   !

!

b. π‘†π‘™π‘œπ‘π‘’ = βˆ’ !

!;𝑦 βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘ =   !

!

c. π‘†π‘™π‘œπ‘π‘’ = βˆ’ !

!;𝑦 βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘ =  βˆ’ !

!

d. π‘†π‘™π‘œπ‘π‘’ = βˆ’ !

!;𝑦 βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘ =   !

!

e. None of the above

2. Find the domain of the function 𝑓 π‘₯ = !

!!!!!!

a. (βˆ’βˆž,∞) b. π‘₯ β‰  1 c. π‘₯ β‰  βˆ’2 d. π‘₯ β‰  2,βˆ’1 e. π‘₯ β‰  βˆ’2,1

3. If the point βˆ’2,1 is on the graph of 𝑓(π‘₯) and 𝑓(π‘₯) is known to be odd, what other point must be on the graph of 𝑓(π‘₯) a. βˆ’2,βˆ’1 b. 2,βˆ’1 c. βˆ’2,1 d. 1,βˆ’1 e. 0,βˆ’1

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4. Find the value of 𝑓 2 βˆ’ 𝑓(0), if

𝑓 π‘₯ = 2βˆ’ π‘₯,                                        π‘₯ < 1  π‘₯! βˆ’ π‘₯ + 1,                    π‘₯ β‰₯ 1

a. 3 b. βˆ’1  c. 2 d. 0 e. 1

5. If 𝑓 π‘₯ = !!+ 1 and 𝑔 π‘₯ = !

!βˆ’ 1, find 𝑓𝑔 (π‘₯).

a. !

!!βˆ’ 1

b. 1

c. 1βˆ’ π‘₯  

 d. !

!!!  

 e. 0

6. The length of a rectangle is 5 units longer than twice its width. Assuming that the width of the rectangle is w and the area is 𝐴, find the area  as a function of the width. a. 𝐴 𝑀 = 𝑀! + 5𝑀 b. 𝐴 𝑀 = 2𝑀! + 5 c. 𝐴 𝑀 = 2𝑀! βˆ’ 5𝑀 d. 𝐴 𝑀 = 2𝑀! + 5𝑀 e. None of the above

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7. 1000 dollars grows to 1 million dollars after 60 years in a bank. If interest is compounded continuously, what is the rate of interest per year? a. 1.83% b. 11.51% c. 28.13% d. 3.84% e. 0.12%

8. Find the sum of all the zeros of the polynomial 𝑓 π‘₯ = π‘₯! + 2π‘₯! βˆ’ 5π‘₯ βˆ’ 6 a. βˆ’5

b. βˆ’2  

c. 0

d. 2

e. 6

9. The graph of 𝑦 = (π‘₯ βˆ’ 4)! + 5 can be obtained by the transformation of 𝑔 π‘₯ = π‘₯!.

Which of the following transformations must be used? I. Move 5 units down. II. Move 5 units up. III. Move 4 units down. IV. Move 4 units left V. Move 4 units right.

a. V, then II b. IV, then II c. III, then I d. II, then III e. III, then II

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10. Which of the following functions represents the inverse of the function 𝑓 π‘₯ = 3𝑒!  . a. 𝑓 π‘₯ = 3𝑒!! b. 𝑓 π‘₯ = !

!!!

c. 𝑓 π‘₯ = ln  (!!)

d. 𝑓 π‘₯ = !!ln  (π‘₯)

e. 𝑓 π‘₯ = log  (!!)

11. The vertex of the parabola 𝑓 π‘₯ = 2π‘₯! βˆ’ 4π‘₯ + 7 is a. βˆ’1  , 13 b. 1  , 7 c. 2  , 5 d. βˆ’1  , 5 e. 1  , 5

 12. Find the horizontal asymptote (HA) and vertical asymptote (VA) of

𝑓 π‘₯ =π‘₯! βˆ’ 4π‘₯(π‘₯ + 2)

a. HA: 𝑦 = 1 VA: π‘₯ = 0, π‘₯ = βˆ’2

b. HA: 𝑦 = 0 VA: π‘₯ = 0, π‘₯ = βˆ’2 c. HA: 𝑦 = 1 VA: π‘₯ = 0 d. HA: 𝑦 = 0 VA: π‘₯ = 0 e. HA: None VA: π‘₯ = 0, π‘₯ = βˆ’2

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13. Find the oblique asymptote of  

𝑓 π‘₯ =π‘₯! + 1π‘₯ βˆ’ 1

a. 𝑦 = 1 b. 𝑦 = π‘₯ βˆ’ 1 c. 𝑦 = π‘₯! + 1 d. 𝑦 = π‘₯ + 1 e. 𝑦 = 0

14. If 𝑓 π‘₯ = !! and 𝑔 π‘₯ = 1βˆ’ !

!   , find (𝑔 ∘ 𝑓)(π‘₯)

a. 1 b. 1βˆ’ π‘₯ c. βˆ’1+ !

!!

d. !

!βˆ’ !

!!

e. 0

15. What are all the possible rational roots of 𝑓 π‘₯ = 6π‘₯! βˆ’ π‘₯! βˆ’ 4π‘₯! βˆ’ π‘₯ βˆ’ 2? a. Β±1,Β±2,Β±3,Β±6,Β± !

!,Β± !

!

b. Β±1,Β±2,Β± !!,Β± !

!,Β± !

!,Β± !

!

c. βˆ’ !!, !!

d. βˆ’1, !!

e. None of the above

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16. Which of the following intervals represents the solution set to the inequality

!!!!!!

> 3 a. βˆ’4,∞ b. βˆ’βˆž,βˆ’ !"

!

c. βˆ’ !"!,βˆ’4

d. (βˆ’4,1) e. βˆ’βˆž, 3)

 

17. Which of the following statements are true? I. (ln π‘₯)! = 2 ln π‘₯ II. log! 3π‘₯! = 4 log!(3π‘₯) III. log π‘₯ βˆ’ 𝑦 = !"#!

!"#!

IV. log!!!= 2βˆ’ log! 4

V. ln π‘₯! = 2 ln π‘₯ a. I and II only b. I, II, and III only c. I and III only d. IV and V only e. I and IV only

18. Solve π‘™π‘œπ‘”(π‘₯ βˆ’ 1)+ π‘™π‘œπ‘”(π‘₯ + 1) = 0 a. π‘₯ = 2 b. π‘₯ = βˆ’1, π‘₯ = 1 c. π‘₯ =1 d. π‘₯ = βˆ’ 2, π‘₯ = 2 e. π‘₯ = 2

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19. Solve the equation 2!!! = 16! a. π‘₯ = 0 b. π‘₯ = 1 c. π‘₯ = 2 d. π‘₯ = 3 e. π‘₯ = !

!

20. Convert the equation 3!! = !!   to logarithmic form

a. log!(

!!) = βˆ’2

b. log!(βˆ’2) =!!

c. log!!(!!) = 3

d. log!!(3) = βˆ’2

e. log!!(βˆ’2) = 3

21. Find the range of the function 𝑓 π‘₯ = 5 sin[2(π‘₯ + !!)]βˆ’ 4

a.  [βˆ’1,1]

b. [βˆ’ !!, !!]

c. [βˆ’1,9]

d. [βˆ’9,1]

e. None of the above.

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22. Suppose that sinπœƒ = !! and πœƒ is in Quadrant 2. Evaluate cosπœƒ

a. !!!"

b. !

!"

c. βˆ’ !"!

d. !"!

e. !"!

23. Find the quadrant in which the terminal side of πœƒ = 4  π‘Ÿπ‘Žπ‘‘π‘–π‘Žπ‘›π‘     is located a. One b. Two c. Three d. Four e. None of the above.

24. Find ! !!! !!(!)!

if 𝑓 π‘₯ = π‘₯! + 2π‘₯ βˆ’ 1 a. β„Ž + 6 b. 2 c. 2+ β„Ž d. β„Ž! + 2β„Ž βˆ’ 1 e. 1

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25. Find the exact value of π‘π‘œπ‘‘!!(βˆ’1).

a. βˆ’ !!

b. !!

c. !!!

d. βˆ’ !!!

e. None of the above.

26. Find the inverse of the function 𝑓 π‘₯ = sin(!!), where βˆ’ !

!πœ‹ ≀ π‘₯ ≀ !

!πœ‹,

a. 𝑓!! π‘₯ = !!"#(!!)

b. 𝑓!! π‘₯ = csc(5π‘₯)

c. 𝑓!! π‘₯ = !!sin!!(π‘₯)

d. 𝑓!! π‘₯ = 5sin!!(π‘₯)

e. 𝑓!! π‘₯ = sin!!( !!!)

27. By using sum or difference formulas, cos(!!βˆ’ π‘₯) can be written as

a. βˆ’ cos π‘₯

b. sin π‘₯ c. cos π‘₯ d. βˆ’ sin π‘₯ e. None of the above

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28. Which of the following is an expression for cos(2𝛼) a. 1+ 2cos  !(𝛼) b. βˆ’1+ 2cos! 𝛼 c. 1βˆ’ cos  !(𝛼) d. βˆ’1βˆ’ cos  !(𝛼) e. 2cos(𝛼)

29. A 41 meter guy wire is attached to the top of a 34.6 meter antenna and to a point on the ground. What angle, in degrees, does the guy wire make with the ground?

a. 1Β°

b. 57.55Β°

c. 37.65Β°

d. 45Β°

e. None of the above.

30. Find an angle πœƒ  between 0∘ and 360∘ that is coterminal with βˆ’790∘ a. πœƒ = 90Β° b. πœƒ = 70Β° c. πœƒ = βˆ’70Β° d. πœƒ = 290Β° e. None of the above

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31. The general solution of the equation cos  (2πœƒ) = 1 is

a. π‘˜πœ‹, where π‘˜ is an integer b. 0 c. 2π‘˜πœ‹, where π‘˜ is an integer d. !

!+ 2π‘˜πœ‹ , where π‘˜ is an integer

e. !!!+ 2π‘˜πœ‹, where π‘˜ is an integer

32. Simplify sec π‘₯ βˆ’ sec π‘₯   . sin! π‘₯    

a. 1 b. sec π‘₯   c. sin! π‘₯ d. cos! π‘₯ e. cos π‘₯

33. Find the length of the side 𝑐 in the triangle 𝐴𝐡𝐢 where π‘Ž = 3, 𝑏 = 3 and ∠𝐴𝐢𝐡 = 120Β°

(Note the figure is not drawn to scale)

a. 27

b. 27

c. 18 βˆ’ 9 3

d. 18 + 9 3

e. 10

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34. For what values of π‘₯ in the interval [βˆ’2πœ‹, 2πœ‹] does the graph of 𝑦 = cot(2π‘₯) have a vertical asymptote? (Angles are measured in radians)

a. βˆ’2,βˆ’1, 0, 1, 2

b. βˆ’2πœ‹,βˆ’ !!!,βˆ’πœ‹,βˆ’ !

!, 0, !

!,πœ‹, !!

!, 2πœ‹

c. βˆ’2πœ‹,βˆ’πœ‹, 0,πœ‹, 2πœ‹

d. βˆ’ !!!, !!!

e. βˆ’2πœ‹, 0, 2πœ‹

35. In the figure below, ∠𝐢 = 125Β°  ,𝐴𝐡 = 8.6  π‘–π‘›π‘β„Žπ‘’π‘ ,  and 𝐴𝐢 = 5.7  π‘–π‘›π‘β„Žπ‘’π‘ . Find ∠𝐡 in degrees.

(Note the figure is not drawn to scale)

a. ∠𝐡 = 29.8° b. ∠𝐡 = 32.9° c. ∠𝐡 = 35.7° d. ∠𝐡 = 38.2° e. ∠𝐡 = 30.6°

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SOLUTION KEY

1. c 2. e 3. b 4. e 5. a 6. d 7. b 8. b 9. a 10. c 11. e 12. c 13. d 14. b 15. b 16. c 17. d 18. a 19. e 20. a 21. d 22. c 23. c 24. a 25. c 26. d 27. b 28. b 29. b 30. d 31. a 32. e 33. a 34. b 35. b