is this solvable2

consider x to lie in [√n,√(n+1))....nεN
x² ε [n , n+1)
[x²]=n....therefore,{x²}=x²-n

e^{[log{l{x2}l2}]}

e^{[log{l(x2-n)l2}]}.....for x² ε [n,n+1)
|x²-n|≤1.....therefore, {|x²-n|²}=(x²-n)²

hence, f(x)=e^[2ln(x2-n)]

considering the limit at the points where x=√n....it can be easily seen tht the function is discontinuous for x=√n...nεN

the other points of discontinuity can be got by considering the continuity of [2ln(x²-n)]

considering, 2ln(x²-n) ε {(-1,0),(-2,-1),(-3,-2).......},the points of discontinuity can be found out.

the points of discontinuity are,
x=√n....for all nεN
x=\sqrt{e^{\frac{-k}{2}}+n} for all kεN and for any particular nεN

pls correct me if i'm wrong...also help me with the graph