If a is given as
= 2 −−−√ + 1 ℎ ( − 1)2a = 2 + 1 the evaluate (a − 1a)2
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(ii) Given. When have,. g(s) = 4s2 − 4s + 1. g(s) = 4s2 − 2s − 2s + 1. g(s) = 2s (2s ... Thus, the zeros of h(s)=(2s-1)(s-√2) are α=12 and β=√2. Now,.
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