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1NF
Every table is in 1 NF, no need to check for it.

2NF
Candidate keys are BA, BC, BD, BE. So prime attributes are <A,B,C,D,E>. Since every attributes is prime attribute, no partial dependency is going to exist. That means relation is in 2NF.

3NF
Since all attributes are prime attributes, no transitive dependency will exist, so it satisfy 3NF as well. 

BCNF
Now in BCNF Left hand side need to be superkey, but here some dependency don't have left hand side as superkey. 
So we divide the table as
R1 (ABC)  F1 : (AB -> C)
R2(CD) F2 : (C - > D)
R3(DE) F3 : (D -> E)
R4(EA)  F4 : (E -> A)

For lossless join, common attribute must be a candidate key or some key in the table 
R1 $\cap$ R2 = C [C is key in R2 ]
R2 $\cap$ R3 = D [D is key in R3]
R3 $\cap$ R4 = E [E is key in R4]

Definitely its lossless join.

For Functional Dependency if F1 $\cup$ F2 $\cup$ F3 $\cup$ F4 = F, then we say its dependency preserving, in our case its satisfy this as well. So its dependency preserving as well.

Remember : Its not a rule if relation is in BCNF, then dependency preserving always violates.
Correct statement is :In BCNF lossless join is guaranteed but dependency preserving may or may not be guaranteed.
Refer : https://www.iitg.ernet.in/awekar/teaching/cs344fall11/lecturenotes/september%2006.pdf

For detail explanation about normalization : https://learnwithnitish.wordpress.com/2016/08/12/dbms-normalization-and-normal-forms/

Your Doubt : If i decompose it as R1(CDEA) and R2(CB), what  is the problem in it ?
R1 ( CDEA), F1 ( C ->D, D->E, E-> A)
Here D+ = DEA, its not a super key, so this relation is not in BCNF.

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answer should be C

(ABC)        (CD)        (DE)         (AE)

AB->C      C->D       D->E         E->A

all dependencies are preserved

and in rel (ABC)  and (CD) C is common which is key for (CD) 

and in rel (CD)  and (DE)  D is key for(DE)

and in rel   (DE)  and (AE) E is key for (AE)  so it is  lossless also

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