### (Solved) : Vii 10 Points Recurrence Equation T N 2t N 2 N T 1 1 Nb Yields Substitution Method T N B A Q35614492

Please need help asap, theseare mcqs!

VII. (10 points) The recurrence equation T(n) 2T[n/2) +n; T(1)-1 (N.B yields, by the substitution method, that T(n) – B.!!) (assuming n is a power of 2), (a) n lg n (b) nlgn+ r (c) lg n (d) lgn + 1 (e) None of the above VIII. (10 points) Applying the Master Theorem to the recurrence equation T(n) 6T(n/4) +4n + 2; T(1) 0 Yields the result that T(n) is in Applying the Master Theorem to the recurrence equation T(n) 3T(n/3)+n: T(1)-1 Yields the result that T(n) is in (a) Θ(n10 g47) (b) e(log n) (c) Θ(nlogn) (d) e(n2) (e) None of the above (a) Θ(n) (b) e(log n) (c) Θ(nlogn) (d) Θ(m) (e) None of the above Show transcribed image text VII. (10 points) The recurrence equation T(n) 2T[n/2) +n; T(1)-1 (N.B yields, by the substitution method, that T(n) – B.!!) (assuming n is a power of 2), (a) n lg n (b) nlgn+ r (c) lg n (d) lgn + 1 (e) None of the above

VIII. (10 points) Applying the Master Theorem to the recurrence equation T(n) 6T(n/4) +4n + 2; T(1) 0 Yields the result that T(n) is in Applying the Master Theorem to the recurrence equation T(n) 3T(n/3)+n: T(1)-1 Yields the result that T(n) is in (a) Θ(n10 g47) (b) e(log n) (c) Θ(nlogn) (d) e(n2) (e) None of the above (a) Θ(n) (b) e(log n) (c) Θ(nlogn) (d) Θ(m) (e) None of the above