Test1Solns - MATH 3705-B TEST 1 SOLUTIONS* January 2007...

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TEST 1 – SOLUTIONS* January 2007 *Note: This version of solutions was edited by Andrey Zakurdaev and differs slightly from the original written by Dr. Elena Devdariani. It contains detailed solutions to MC questions and displays the allocation of marks (in blue) to be used during grading of Section B2 papers. Questions 1-5 are multiple choice. Circle the correct answer. Only the answer will be marked. [12 marks] [No part marks are awarded here. Full marks for correct answer and zero otherwise.] 1. [2] L{ e - 2 t cos(3 t ) } = ( a ) s + 2 ( s + 2) 2 + 9 ( b ) s s 2 + 9 ( c ) s - 2 ( s - 2) 2 + 9 ( d ) None of the above Solution: L{ e - 2 t |{z} a = - 2 cos(3 t ) | {z } f ( t ) } use 1st Shifting Theorem F ( s ) = s s 2 + 3 2 = s s 2 + 9 = F ( s - a ) = F ( s + 2) = s + 2 ( s + 2) 2 + 3 2 = ( a ) 2. [2] L{ t sin(3 t ) } = ( a ) 6 s ( s 2 + 9) 2 ( b ) s 2 - 9 ( s 2 + 9) 2 ( c ) 9 - s 2 ( s 2 + 9) 2 ( d ) None of the above Solution: L{ t |{z} n =1 sin(3 t ) | {z } f ( t ) } use general rule for L{ t n f ( t ) } F ( s ) = s s 2 + 3 2 = s s 2 + 9 = ( - 1) n d n ds n F
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This note was uploaded on 07/16/2009 for the course MATH 3705 taught by Professor Jaberabdualrahman during the Winter '08 term at Carleton CA.

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Test1Solns - MATH 3705-B TEST 1 SOLUTIONS* January 2007...

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