1. **Problem statement:** We are given a function $f(x)$ defined by a $4\times4$ matrix with entries involving $x$ and constants. We want to find the constant term of $f(x)$, which is $f(0)$, the determinant of the matrix when $x=0$. 2. **Matrix for $f(x)$:** $$\begin{pmatrix} 2x & 5 & 1 & 1 \\ 2 & x & -1 & 2 \\ -2 & 4 & x & 0 \\ x & -3 & 0 & x \end{pmatrix}$$ 3. **Evaluate $f(0)$ by substituting $x=0$:** $$\begin{pmatrix} 0 & 5 & 1 & 1 \\ 2 & 0 & -1 & 2 \\ -2 & 4 & 0 & 0 \\ 0 & -3 & 0 & 0 \end{pmatrix}$$ 4. **Calculate the determinant $f(0)$:** Using cofactor expansion along the last row (4th row): $$f(0) = 0 \cdot M_{41} - 3 \cdot (-1)^{4+2} \cdot M_{42} + 0 \cdot M_{43} + 0 \cdot M_{44} = -3 \cdot (-1)^6 \cdot M_{42}$$ 5. **Calculate minor $M_{42}$:** Remove 4th row and 2nd column: $$M_{42} = \begin{vmatrix} 0 & 1 & 1 \\ 2 & -1 & 2 \\ -2 & 0 & 0 \end{vmatrix}$$ 6. **Calculate $M_{42}$ determinant:** $$= 0 \cdot \begin{vmatrix} -1 & 2 \\ 0 & 0 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 2 \\ -2 & 0 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & -1 \\ -2 & 0 \end{vmatrix}$$ $$= 0 - 1 (2 \cdot 0 - (-2) \cdot 2) + 1 (2 \cdot 0 - (-2) \cdot (-1))$$ $$= 0 - 1 (0 + 4) + 1 (0 - 2) = -4 - 2 = -6$$ 7. **Substitute back:** $$f(0) = -3 \cdot 1 \cdot (-6) = 18$$ --- 8. **Second problem: Determinant $D$ of a $5\times5$ matrix:** $$D = \begin{vmatrix} 0 & a_{12} & a_{13} & 0 & 0 \\ a_{21} & a_{22} & a_{23} & a_{24} & a_{25} \\ a_{31} & a_{32} & a_{33} & a_{34} & a_{35} \\ 0 & a_{42} & a_{43} & 0 & 0 \\ 0 & a_{52} & a_{53} & 0 & 0 \end{vmatrix}$$ 9. **Given options for $D$ are:** A) 1 B) -1 C) 0 D) 2 10. **Observation:** The first row and last two rows have zeros in the 1st, 4th, and 5th columns, indicating linear dependence or zero determinant. 11. **Conclusion:** The determinant $D$ is zero due to the structure of zeros in the matrix. **Final answers:** - Constant term of $f(x)$ is $f(0) = 18$. - Determinant $D = 0$.