is the language L={a^nb^n : n>=1} U {b} is determinstic

Question

is the language L={a^nb^n : n>=1} U {b} deterministic

Comments

@akash.dinkar12 union of two dcfl need not be dcfl..

L={$a^{n}b^{n}$: n>=1} U {b} , Here we can deterministically determine string is belong to which part of union by looking at first symbol, if first alphabet is 'a' we will looking for string in the form $a^{n}b^{n}$, else if first alphabet is 'b' we are looking for b only, if anything comes after b we reject the string.

So here we are able to deterministically determine that string is accepted or not, thats why its DCFL