Answer D
$\text{Option A}$
Connsider a LR(1) DFA with no RR Conflicts. Take two states, say, I3 and I5 in such LR(1) DFA.
I3 : $[A \rightarrow \alpha.\color{red}{, a}, B \rightarrow \beta.\color{blue}{, b}]$ and
I5: $[A \rightarrow \alpha.\color{blue}{, b}, B \rightarrow \beta.\color{red}{, a}]$
Since the core items are same, we will merge $I_3$ and $I_5$ in LALR, say merged state is $I_{35}$
$I_{35}$ : $[A \rightarrow \alpha.\color{red}{, a}\color{blue}{, b} \text{ } B \rightarrow \beta.\color{red}{, a}\color{blue}{, b}]$
A common confusion: $I_{35}$ has RR conflict on $a \text{ and } b$.
$\text{Do } I_{35} \text{ really has any conflict? }$ – Yes.
See one example – here
$\text{Option B}$
Symbol table is accessed among all phases. For example – “int x”, here lexical analyzer will assign 2 tokens, but lexical analyzer won’t know whether x is of type int since it reads int and x as two different tokens. Syntax analyzer will feed type of x to symbol table.
C. It is optional
D. LR(1) = DCFL Ref: https://cs.stackexchange.com/questions/43/language-theoretic-comparison-of-ll-and-lr-grammars