Answer
See below
Work Step by Step
Obtain: $Ker(T)=\{p \in P_1(R): T(p)=0\}\\
=\{ax^2+bx+c: T(ax^2+bx+c)=0\}\\
=\{ax^2+bx+c:(a-b)x+c=0\}$
We have the system: $a-b=0\\
c=0\\
\rightarrow a=b=\\
\rightarrow Ker(T)=\{a(x^2+x): a\in R\}\\
=\{x^2+x\}$
Hence, $T$ is not one-to-one.
Apply Rank Nullity Theorem:
$\dim [Ker(T_2T_1)]+\dim [Rng(T)=\dim P_2(R)\\
1+\dim [Rng(T)]=3\\
\dim [Rng(T)]=2$
$\dim P_1(R)=2=\dim [Rng(T)]\\
\rightarrow Rng=P_1(R)$
$\rightarrow T$ is onto.
Since $T$ is not one-to-one, $T^{-1}$ does not exist.
