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 $f(x) = \left| \begin{array}{ccc} -\sin x & 2 \sin 2x & 4 \cos^2 x \\ \cos x & 4 \sin^2 x & 2 \sin 2x \\ 0 & -\cos x & \sin x \end{array} \right|$, then $f\left(\frac{5\pi}{4}\right) + f'\left(\frac{5\pi}{4}\right) = $

The determinant $\left| \begin{array}{ccc} ^x{C_1} & ^x{C_2} & ^x{C_3} \\ ^y{C_1} & ^y{C_2} & ^y{C_3} \\ ^z{C_1} & ^z{C_2} & ^z{C_3} \end{array} \right|$ equals:

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

Suppose $\alpha, \beta, \gamma$ are the roots of the equation $x^3+qx+r=0$ (where $r \neq 0$) and they are in $A$.$P$. Then the rank of the matrix $\begin{bmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{bmatrix}$ is

If $A = \begin{bmatrix} 1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6 \end{bmatrix}$ and the rank of $A$ is $2$,then the value of $x$ is equal to