The rank of the following matrix $A$ is
$A = \begin{bmatrix} 1 & -2 & 3 & -4 \\ 2 & 9 & 4 & 5 \\ 4 & 5 & 10 & -3 \\ 1 & 11 & -1 & 9 \end{bmatrix}$

The determinant $\left| {\begin{array}{ccc} 4 + {x^2} & -6 & -2 \\ -6 & 9 + {x^2} & 3 \\ -2 & 3 & 1 + {x^2} \end{array}} \right|$ is not divisible by

If $y = \left|\begin{array}{ccc}f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c\end{array}\right|$,then $\frac{dy}{dx}$ is equal to

If $\begin{vmatrix} x^2+x & x+1 & x-2 \\ 2x^2+3x-1 & 3x & 3x-3 \\ x^2+2x+3 & 2x-1 & 2x-1 \end{vmatrix} = xA+B$,where $A$ and $B$ are determinants of order $3$ not involving $x$,then $|A|=$

Let $f(x) = \left| \begin{array}{ccc} 2\cos^2 x & \sin(2x) & -\sin x \\ \sin(2x) & 2\sin^2 x & \cos x \\ \sin x & -\cos x & 0 \end{array} \right|$. Then,evaluate $\int_{0}^{\frac{\pi}{2}} [f(x) + f'(x)] dx$.

If $f(x) = \begin{vmatrix} 1 + \sin x + \sin 2x + \sin 3x & \frac{3 + \sin 2x}{2} & \frac{-2 + \sin 3x}{3} \\ 3 + 4 \sin x & \frac{3}{2} & \frac{4}{3} \sin x \\ 1 + \sin x & \frac{1}{2} \sin x & \frac{1}{3} \end{vmatrix}$,then $\int_0^{\pi / 2} (f(x) + f^{\prime}(x)) dx =$