If $D(x) = \begin{vmatrix} x - 1 & (x - 1)^2 & x^3 \\ x - 1 & x^2 & (x + 1)^3 \\ x & (x + 1)^2 & (x + 1)^3 \end{vmatrix}$,then the coefficient of $x$ in $D(x)$ is

If the points $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ are collinear,then the rank of the matrix $\begin{bmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{bmatrix}$ will always be less than

Let $f(x) = \left| \begin{array}{ccc} 1 + \sin^2 x & \cos^2 x & 4 \sin 2x \\ \sin^2 x & 1 + \cos^2 x & 4 \sin 2x \\ \sin^2 x & \cos^2 x & 1 + 4 \sin 2x \end{array} \right|$,then the maximum value of $f(x)$ is:

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

For some $a, b$,let $f(x) = \left|\begin{array}{ccc} a+\frac{\sin x}{x} & 1 & b \\ a & 1+\frac{\sin x}{x} & b \\ a & 1 & b+\frac{\sin x}{x} \end{array}\right|, \quad x \neq 0$. If $\lim_{x \rightarrow 0} f(x) = \lambda + \mu a + \nu b$,then $(\lambda + \mu + \nu)^2$ is equal to:

Let $l, m, n \in R$ and $A = \begin{bmatrix} 1 & r & r^2 & l \\ r & r^2 & 1 & m \\ r^2 & 1 & r & n \end{bmatrix}$. Then the set of all real values of $r$ for which the rank of $A$ is $3$,is