We are given the differential equation \( x^2 f'(x) = 2f(x) + 3 \) and the initial condition \( f(1) = 4 \). To solve for \( f(x) \), we first divide both sides of the equation by \( x^2 \): \[ f'(x) = \frac{2f(x) + 3}{x^2}. \] We solve this first-order linear differential equation using the method of integrating factors. After solving, we substitute \( x = 2 \) and calculate \( 2f(2) \).
Final Answer: \( 2f(2) = 29 \).

Let \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).
If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) is equal to: