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v.(10 points) If g(n)) then itIfafm) E ofn), then it fol follows that If g(n) E on)), then it follows that (a) g(n) E (f(n))
v.(10 points) If g(n)) then itIfafm) E ofn), then it fol follows that If g(n) E on)), then it follows that (a) g(n) E (f(n)) (b) g(n) E 0(f(n) (a)g(n) er/(n)) (c)ftn) E o(g(n) (d) tn) o(g(n)) (e) None of the above (e) None of the above VI. (10 points) 2n lim n n+2 allows us to determine that (a)2" E (n2 + n + 2) 0 2 Eo(n2 + n+ 2) (c) 2n E (n2 + n + 2) (d) (n2 + n + 2) (2n) (e) (n2 + n + 2) E (2n) og 37 allows us to determine that (a) nlo9, E o(n2+n+2) (b) n o9,7 Ew(n2+n+ 2) (c) nwyE (n2 + n + 2) (d) nlog,E (n2 + n + 2) (e) None of the above VII. (10 points) The recurrence equation T[n)-Ttn/5) +3; T(1): 1 (assuming n is a power of 5). yields, by the substitution method, that TIn)- (a) 5 loga n (b) 5 logan+1 (c) 3 logs n (d) 3 logs n+ 1 (e) None of the above
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