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Unformatted text preview: Version 106 Homework 01 Radin (58415) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. Welcome to Quest Learning and As sessment. Best of luck this semester. 001 10.0 points Find the value of f (0) when f ( t ) = 4(3 t 1) and f (1) = 5 , f (1) = 2 . 1. f (0) = 2 2. f (0) = 5 3. f (0) = 1 correct 4. f (0) = 4 5. f (0) = 3 Explanation: The most general antiderivative of f has the form f ( t ) = 6 t 2 4 t + C where C is an arbitrary constant. But if f (1) = 5, then f (1) = 6 4 + C = 5 , i.e., C = 3 . From this it follows that f ( t ) = 6 t 2 4 t + 3 , and the most general antiderivative of the latter is f ( t ) = 2 t 3 2 t 2 + 3 t + D , where D is an arbitrary constant. But if f (1) = 2, then f (1) = 2 2 + 3 + D = 2 , i.e., D = 1 . Consequently, f ( t ) = 2 t 3 2 t 2 + 3 t 1 . At x = 0, therefore, f (0) = 1 . 002 10.0 points Consider the following functions: ( A ) F 1 ( x ) = cos 2 x 4 , ( B ) F 2 ( x ) = sin 2 x 2 , ( C ) F 3 ( x ) = cos 2 x 2 . Which are antiderivatives of f ( x ) = sin x cos x ? 1. F 2 only 2. F 1 and F 3 only 3. F 3 only 4. all of them 5. none of them 6. F 1 and F 2 only 7. F 1 only 8. F 2 and F 3 only correct Explanation: By trig identities, cos 2 x = 2 cos 2 x 1 = 1 2 sin 2 x , while sin 2 x = 2 sin x cos x . But d dx sin x = cos x, d dx cos x = sin x . Consequently, by the Chain Rule, ( A ) Not antiderivative. Version 106 Homework 01 Radin (58415) 2 ( B ) Antiderivative. ( C ) Antiderivative. 003 10.0 points Find f ( x ) on ( 2 , 2 ) when f ( x ) = 2 cos x + 4 sec 2 x and f ( 4 ) = 2. 1. f ( x ) = 5 4 tan x + 2 sin x 2. f ( x ) = 7 4 tan x 2 cos x 3. f ( x ) = 4 tan x + 2 sin x 3 correct 4. f ( x ) = 4 tan x 2 sin x 1 5. f ( x ) = 4 tan x 2 cos x 3 Explanation: The most general antiderivative of f ( x ) = 2 cos x + 4 sec 2 x is f ( x ) = 2 sin x + 4 tan x + C with C an arbitrary constant. But if f parenleftBig 4 parenrightBig = 2, then f parenleftBig 4 parenrightBig = 1 + 4 + C = 2 , so C = 3 . Consequently, f ( x ) = 4 tan x + 2 sin x 3 . 004 10.0 points Find the unique antiderivative F of f ( x ) = e 5 x 4 e 2 x + 3 e 3 x e 2 x for which F (0) = 0. 1. F ( x ) = 1 3 e 3 x 4 x 3 5 e 5 x + 4 15 correct 2. F ( x ) = 1 5 e 5 x + 4 x + 1 3 e 3 x 2 5 3. F ( x ) = 1 3 e 3 x 4 x 1 3 e 3 x 4. F ( x ) = 1 5 e 5 x 4 x 3 5 e 5 x 2 5 5. F ( x ) = 1 3 e 3 x + 4 x + 3 5 e 5 x 14 15 6. F ( x ) = 1 3 e 3 x + 4 x + 3 5 e 3 x 4 15 Explanation: After division, e 5 x 4 e 2 x + 3 e 3 x e 2 x = e 3 x 4 + 3 e 5 x ....
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This note was uploaded on 04/29/2008 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas at Austin.
 Spring '08
 RAdin

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