Find where the signal x(t) = 1/(t^2 – 3t + 2) finds its maximum value between (1.25, 1.75):
A. 1.40
B. 1.45
C. 1.55
D. 1.50
Answer:A
Find where the signal x(t) = 1/(t^2 – 3t + 2) finds its maximum value between (1.25, 1.75):
A. 1.40 B. 1.45 C. 1.55 D. 1.50
Explanation: Differentiate the function for an optima, put it to zero, we will obtain t = 1.5 as the required instant.