If $A = \begin{bmatrix} \sqrt{2020} & \sqrt{2021} & \sqrt{2022} & \sqrt{2023} \\ \sqrt{4040} & \sqrt{4042} & \sqrt{4044} & \sqrt{4046} \\ \sqrt{6060} & \sqrt{6063} & \sqrt{6066} & \sqrt{6069} \\ \sqrt{8080} & \sqrt{8084} & \sqrt{8088} & \sqrt{8092} \end{bmatrix}$,then the rank of $A$ is

Let for $i = 1, 2, 3$,$p_i(x)$ be a polynomial of degree $2$ in $x$,$p'_i(x)$ and $p''_i(x)$ be the first and second order derivatives of $p_i(x)$ respectively. Let $A(x) = \begin{bmatrix} p_1(x) & p'_1(x) & p''_1(x) \\ p_2(x) & p'_2(x) & p''_2(x) \\ p_3(x) & p'_3(x) & p''_3(x) \end{bmatrix}$ and $B(x) = [A(x)]^T A(x)$. Then the determinant of $B(x)$

The rank of the matrix $\begin{bmatrix} 3 & 5 & -1 & 4 \\ 2 & 1 & 3 & -2 \\ 8 & 11 & 1 & 6 \\ -7 & -14 & 6 & -14 \end{bmatrix}$ is

If $A(x) = \left| \begin{array}{ccc} 1 & 2 & 3 \\ x+1 & 2x+1 & 3x+1 \\ x^2+1 & 2x^2+1 & 3x^2+1 \end{array} \right|$, then $\int_0^1 A(x) \, dx$ equals

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 number of real values of $x$ at which the function $f(x) = \left| \begin{array}{ccc} 1 & |x| & x^2 \\ 1 & |x-1| & (x-1)^2 \\ 1 & |x-2| & (x-2)^2 \end{array} \right|$ is not differentiable is