Let $A = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{bmatrix}$ and $A^2 = A$. If $r$ is the rank of $A$, then $r + x =$

If $f(x) = \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \\ x - \frac{\pi}{2} & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} \right|$, then $f^{\prime}(\pi)$ is equal to

If $f(x) = \left| \begin{array}{ccc} x^3 - x & a + x & b + x \\ x - a & x^2 - x & c + x \\ x - b & x - c & 0 \end{array} \right|$,then:

If $f(x) = \left| \begin{array}{ccc} \sin(x + \alpha) & \sin(x + \beta) & \sin(x + \gamma) \\ \cos(x + \alpha) & \cos(x + \beta) & \cos(x + \gamma) \\ \sin(\alpha + \beta) & \sin(\beta + \gamma) & \sin(\gamma + \alpha) \end{array} \right|$ and $f(10) = 10$,then $f(\pi)$ is equal to

If $A = [a_{ij}]$,$1 \leq i, j \leq n$ with $n \geq 2$ and $a_{ij} = i + j$ is a matrix,then the rank of $A$ is

If $f(x) = \begin{vmatrix} x & x^2 & x^3 \\ 1 & 2x & 3x^2 \\ 0 & 2 & 6x \end{vmatrix}$,then the ratio $f^{\prime \prime}(x) : f^{\prime}(x) =$