L2={ww$^{R}$ x ∣ w,x ∈ {0,1}$^{*}$}
L2 is regular. Since, w can be ϵ and x∈(0+1)$^{*}$, making L2=(0+1)$^{*}$=Σ$^{*}$. i.e.; the set of strings generated by L2 is {ϵ,0,1,00,01,10,11,000,…}=Σ$^{*}$
Hence, L2 is also DCFL.
if asked for,
L={ww$^{R}$ x ∣ w,x∈ (0+1)$^{+}$}
Here, w cannot be ϵ and hence to accept the string we do need the power of a PDA making L, NCFL (non-determinism is required to guess the start of W).
