Answer
True
Work Step by Step
Since $Ker(T)$ is one-dimensional, we have: $dim [Ker(T)]=1$ and $T$ is one-to-one, then $Ker(T)=\{0\} \rightarrow T$ is not one-to-one.
Apply Rank Nullity Theorem:
$\dim [Ker(T)]+\dim [Rng(T)]=\dim M_2(R)\\
\rightarrow 1+\dim [Rng(T)]=4\\
\rightarrow \dim [Rng(T)]=3$
$\rightarrow \dim [P_2(R)]=3$
With $Rng(T) \subset P_2(R)$, we know $Rng(T)=P_2(R)$
Hence, $T$ is onto but not one-to-one.
