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 the rank of the matrix $A=\begin{bmatrix} 1 & 2 & 1 & -1 \\ -1 & 2 & 3 & 5 \\ 0 & 1 & k & k \end{bmatrix}$ is $2$ and $k$ is a real number,then $k$ is a root of the following quadratic equation:

The rank of $\left[\begin{array}{ccc}1 & -1 & 1 \\ 1 & 1 & -1 \\ -1 & 1 & 1\end{array}\right]$ is

If $f(x) = \left| \begin{array}{ccc} \cos(2x) & \cos(2x) & \sin(2x) \\ -\cos x & \cos x & -\sin x \\ \sin x & \sin x & \cos x \end{array} \right|$,then:
$A$. $f'(x) = 0$ at exactly three points in $(-\pi, \pi)$
$B$. $f'(x) = 0$ at more than three points in $(-\pi, \pi)$
$C$. $f(x)$ attains its maximum at $x = 0$
$D$. $f(x)$ attains its minimum at $x = 0$

If $A = \begin{bmatrix} 1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6 \end{bmatrix}$ and the rank of $A$ is $2$,then the value of $x$ is equal to