The rank of the matrix $\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & 1 & -2 \\ 1 & -1 & 4 \\ 2 & 2 & 8\end{array}\right]$ is

If $y = \sin(mx)$,then the value of $\left| \begin{array}{ccc} y & y_1 & y_2 \\ y_3 & y_4 & y_5 \\ y_6 & y_7 & y_8 \end{array} \right|$ (where subscripts of $y$ denote the order of derivative) is:

Let $\Delta = \begin{vmatrix} \sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \cos \theta \cos \phi & \cos \theta \sin \phi & -\sin \theta \\ -\sin \theta \sin \phi & \sin \theta \cos \phi & 0 \end{vmatrix}$. Then:

Let $A=\begin{bmatrix} -1 & -2 & -3 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix}$,$B=\begin{bmatrix} 1 & -2 \\ -1 & 2 \end{bmatrix}$ and $C=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$. If $a, b$ and $c$ respectively denote the ranks of $A, B$ and $C$,then the correct order of these numbers is:

If $f'(x) = \left| \begin{array}{ccc} mx & mx - p & mx + p \\ n & n + p & n - p \\ mx + 2n & mx + 2n + p & mx + 2n - p \end{array} \right|$,then $y = f(x)$ represents

If $f(x) = \left| \begin{array}{ccc} 2\cos^2 2x & \sin 2x & -\sin x \\ \sin 2x & 2\sin^2 x & \cos x \\ \sin x & -\cos x & 0 \end{array} \right|$,then the value of $\int_{0}^{\frac{\pi}{2}} f'(x) \,dx$ is equal to