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The Laplace transform of exp(at), where a > 0, is defined only for the Laplace parameter, s > a since | |
A. | the function is exponential. [Wrong Answer] |
B. | the Laplace transform of integral of exp(at) has finite values only for s > a. [Correct Answer] |
C. | the Laplace transform integral of exp(at) has initial values only for s > a. [Wrong Answer] |
D. | the function exp(at) is piece-wise continuous only for s > a. [Wrong Answer] |