Turing machine that can remember
Given any language $L$ there is a special Turing machine $T$ that decides $L$ in polynomial time.

The Turing machine has 3 states: the initial states $q_0$, the accepting state $q_A$ and the rejecting state $q_R$. 
Let $\Gamma$ be the tape alphabet (including the blank symbol $\varepsilon$), and let $\alpha x$ denote the contents of the tape starting from the left edge to the head position in tape, where $\alpha \in \Gamma^*$ and $x \in \Gamma$. As usual, I am going to assume that the input is written starting from the beginning of the tape and that the head is initially positioned at the beginning of the tape.
The transition function $\delta$ is defined as follows:


$\delta(q_0, \alpha x) = (q_0, x, R)$ if $x \neq \varepsilon$
$\delta(q_0, \alpha x) = (q_A, x, R)$ if $x = \varepsilon$ and $\alpha \in L$
$\delta(q_0, \alpha x) = (q_A, x, R)$ if $x = \varepsilon$ and $\alpha \not\in L$


Since there are undecidable languages for regular Turing Machines, this immediately implies that A and B are false.

Answers D and C are unclear. They talk about "both answers" but you are given $4$ possible answers.