You can work on this in groups of two if you want. Ask questions if you're stuck. 1) Recall that the definition of Big-O was : f(n) is O(g(n)) if we can find constants k and no so that for all n > no, f(n) k*g(n). Prove the following statements, by showing appropriate values of k and no. a. n is O(n^) b. n*log n is O(n?) c. 200n3+150n2+75n+2000 is O(n?) d. C*n is O(n) for any real C >0 (In other words, treat C as a constant, and your k should probably be related to C) e. f(n) + g(n) is O(max(f(n),g(n)) (Hint: f(n) + g(n) & v){ for(int i = 0; i& v){ for(int i = 0; i&v){ for(int i = 0; i&v){ for(int i =1;i& v, int x){ // x is initially N if(x==0) return v[x]; else return 2*g(v,x-1); } //show the run times of all 3 functions g. int f(vector& v){ for(int i = 0; i& v, int x){ // x is initially N if(x == 0) return v[x]; else return f(v)*g(v,x-1); } int h(vector& v){ for(int i = 0; i<><><><><><>