The rank of the matrix $\begin{bmatrix} -1 & 2 & 5 \\ 2 & -4 & a - 4 \\ 1 & -2 & a + 1 \end{bmatrix}$ is

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

Let $f(x) = \begin{vmatrix} \cos x & x & 1 \\ 2 \sin x & x^2 & 2x \\ \tan x & x & 1 \end{vmatrix}$. Then $\lim_{x \to 0} \frac{f'(x)}{x} =$

Let $A = \begin{bmatrix} 1 & 0 & -1 & -3 \\ 0 & 1 & 1 & k-1 \\ 0 & 0 & k-1 & 1 \end{bmatrix}$ and $k \in R$. Then,the value of $k$,if it exists,for which the rank of $A$ is $2$,is

If $f(x) = \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \\ 3 + 2 \cos^4 x & 2 \sin^4 x & \sin^2 2x \\ 2 \cos^4 x & 3 + 2 \sin^4 x & \sin^2 2x \end{vmatrix}$,then $\frac{1}{5} f'(0)$ is equal to: