1. **Stating the problem:** We have a linear transformation $T : \mathbb{P}_3 \to \mathbb{P}_3$ defined by $$T(p(x)) = p'(x) + p(x),$$ where $\mathbb{P}_3$ is the space of all real polynomials with degree at most 3. We want to find the nullity and rank of $T$. 2. **Understanding the space:** The vector space $\mathbb{P}_3$ is 4-dimensional, with basis $\{1,x,x^2,x^3\}.$ Therefore, $$\dim(\mathbb{P}_3) = 4.$$ 3. **Nullity of $T$:** Nullity is the dimension of the kernel (null space) of $T$. We find $p(x)$ such that $$T(p(x)) = 0,$$ i.e., $$p'(x) + p(x) = 0.$$ 4. **Solving the differential equation:** The equation $$p'(x) + p(x) = 0$$ is a first-order linear ODE with general solution $$p(x) = Ce^{-x}$$ for some constant $C$. 5. **Polynomial constraint:** Since $p(x)$ must be a polynomial of degree at most 3, and $e^{-x}$ is not a polynomial, the only polynomial solution is $$p(x) = 0.$$ 6. **Conclusion on nullity:** Since the only solution to $T(p) = 0$ in $\mathbb{P}_3$ is the zero polynomial, the nullity of $T$ is $$\text{nullity}(T) = 0.$$ 7. **Rank of $T$:** By the rank-nullity theorem, $$\dim(\mathbb{P}_3) = \text{rank}(T) + \text{nullity}(T).$$ Substituting, $$4 = \text{rank}(T) + 0,$$ so $$\text{rank}(T) = 4.$$ **Final answer: nullity$(T) = 0$ and rank$(T) = 4$.** Thus, option d: 0 and 4.