Math116Exam2F11Sol

# Math116Exam2F11Sol - Math 116 Second Midterm Name EXAM...

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Math 116 — Second Midterm November 16, 2011 Name: EXAM SOLUTIONS Instructor: Section: 1. Do not open this exam until you are told to do so. 2. This exam has 13 pages including this cover. There are 9 problems. Note that the problems are not of equal difficulty, so you may want to skip over and return to a problem on which you are stuck. 3. Do not separate the pages of this exam. If they do become separated, write your name on every page and point this out to your instructor when you hand in the exam. 4. Please read the instructions for each individual problem carefully. One of the skills being tested on this exam is your ability to interpret mathematical questions, so instructors will not answer questions about exam problems during the exam. 5. Show an appropriate amount of work (including appropriate explanation) for each problem, so that graders can see not only your answer but how you obtained it. Include units in your answer where that is appropriate. 6. You may use any calculator except a TI-92 (or other calculator with a full alphanumeric keypad). However, you must show work for any calculation which we have learned how to do in this course. You are also allowed two sides of a 3 00 × 5 00 note card. 7. If you use graphs or tables to find an answer, be sure to include an explanation and sketch of the graph, and to write out the entries of the table that you use. 8. Turn off all cell phones and pagers , and remove all headphones. Problem Points Score 1 12 2 12 3 11 4 11 5 11 6 8 7 13 8 10 9 12 Total 100

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Math 116 / Exam 2 (November 16, 2011) page 2 1 . [12 points] Indicate if each of the following is true or false by circling the correct answer. No justification is required. a . [2 points] The function y ( t ) = cos 3 t + B sin 3 t + 1 9 t is a solution of y 00 + 9 y = 0 with y (0) = 1. True False Solution: y 00 = - 9 cos 3 t - 9 B sin 3 t 9 y = 9 cos 3 t + 9 B sin 3 t + t Hence y 00 + 9 y = t 6 = 0. b . [2 points] The value of the integral used to compute the area enclosed by a curve r = f ( θ ) given in polar coordinates can be negative if f ( θ ) 0. True False Solution: f ( θ ) 2 0, then A = R b a 1 2 f ( θ ) 2 0. c . [2 points] If f ( x ) is a continuous function such that R 1 f ( x ) dx diverges, then R 1 f ( x ) 2 dx must diverge. True False Solution: If f ( x ) = 1 x , then R 1 1 x dx diverges but R 1 1 x 2 dx converges. d . [2 points] If P ( x ) is a cumulative distribution function for the probability density function p ( x ), then 1 + P ( x ) is also a cumulative distribution function for p ( x ). True False Solution: A cumulative distribution function P ( x ) must satisfy lim x →∞ P ( x ) = 1, but lim x →∞ 1 + P ( x ) = 2 6 = 1. Hence 1 + P ( x ) can’t be a cumulative distribution. e . [2 points] All solutions to the differential equation y 0 = 1 + y 4 are increasing functions. True False Solution: Since 1 + y 4 > 0, then y 0 > 0. Then all the solution curves y must be increasing.
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