Dependency Preservation and Normal Forms
Given R(A B C D E F G H I J ) and the FD's specified above
We would get three relations after normalizing it till 2NF:
Since the FD: A->DE is not satisfying 2NF 
Finding A closure we get (ADEIJ).
Since the FD B-> F also does not satisfy 2NF
Finding B Closure we get (BFGH).
Hence we get the relations:
R1(ADEIJ) and R2(BFGH).
A is the Candidate Key for R1; B is the Candidate Key for R2
What remains from R is C.
So we get the third relation R3 (ABC).

This is a dependency preserving and loss less decomposition.