In a matrix $A$,if all the sub-matrices of order $k$ are singular and there is at least one non-singular sub-matrix of order $r$ $(r < k)$,then the rank $(\rho)$ of the matrix $A$:

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:

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

If $\left| \begin{array}{ccc} 1 + ax & 1 + bx & 1 + cx \\ 1 + a_1x & 1 + b_1x & 1 + c_1x \\ 1 + a_2x & 1 + b_2x & 1 + c_2x \end{array} \right| = A_0 + A_1x + A_2x^2 + A_3x^3$,then $A_1$ is equal to:

Let $f(x) = \left| \begin{array}{ccc} \cos x & x & 1 \\ 2 \sin x & x & 2x \\ \sin x & x & x \end{array} \right|$. Then,$\lim_{x \rightarrow 0} \frac{f(x)}{x^2}$ is

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: