If $f(x) = \left| \begin{array}{ccc} x^3+x & x+1 & x-2 \\ 2x^3+3x-1 & 3x & 3x-3 \\ x^3+2x+3 & 2x-1 & 2x-1 \end{array} \right|$, then $\frac{d}{dx}(f(x))$ is equal to

The trace of a square matrix is defined as the sum of its diagonal entries. If $A$ is a $2 \times 2$ matrix such that the trace of $A$ is $3$ and the trace of $A^3$ is $-18$,then the value of the determinant of $A$ is:

$A$ value of $b$ for which the rank of the matrix $A = \begin{bmatrix} 1 & 1 & -1 & 0 \\ 4 & 4 & -3 & 1 \\ b & 2 & 2 & 2 \\ 9 & 9 & b & 3 \end{bmatrix}$ is $3$,is

If ${D_p} = \begin{vmatrix} p & 15 & 8 \\ p^2 & 35 & 9 \\ p^3 & 25 & 10 \end{vmatrix}$,then ${D_1} + {D_2} + {D_3} + {D_4} + {D_5} = $

$f(x) = \left| \begin{array}{ccc} x^3 & x^2 & 3x^2 \\ 1 & -6 & 4 \\ p & p^2 & p^3 \end{array} \right|$,where $p$ is a constant,then $\frac{d^3f(x)}{dx^3}$ is

The number of distinct real roots of $\left|\begin{array}{lll}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$ in the interval $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ is