Calculate and justify the time complexity of the following function:
int f(int n)
{
     int j = 10 ;
     while (j < n) {
          j += sqrt(j);
     }
     return j;
}

We observe that j has to increase from a constant value to n, and in each iteration of the while loop, the value of j increases by at least 1 and by at most sqrt(n).  
Therefore, clearly, the time complexity of this algorithm has to be between O(n) and O(sqrt(n)).
We claim that the time complexity is O(sqrt(n)).
We can analyze separate phases of the while loop in terms of the value of j.
For the value of j to go from n/4 to n, we require at most n/sqrt(n/4)), that is 2 sqrt(n) steps.  
For the value of j to go from n/16 to n/4 , we require at most n/4 sqrt(n/16) , that is, sqrt(n) steps.   
Similarly, for the value of j to go from n/64 to n/16 , we require at most n/16(sqrt(n/64)), that is sqrt(n)/2 steps.
Therefore, overall, for the value of j to go from 1 to n, we require at most sqrt(n) (2 + 1 + 1/2+ 1/4 + ... ) = 4 sqrt(n) steps.
Therefore, the total time taken by the algorithm is O(sqrt(n)).