The determinant $\left| \begin{array}{ccc} \cos(\theta + \phi) & -\sin(\theta + \phi) & \cos 2\phi \\ \sin \theta & \cos \theta & \sin \phi \\ -\cos \theta & \sin \theta & \cos \phi \end{array} \right|$ 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:

If the system of equations $a_1 x + b_1 y + c_1 z = 0$,$a_2 x + b_2 y + c_2 z = 0$,and $a_3 x + b_3 y + c_3 z = 0$ has only the trivial solution,then the rank of the matrix $A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$ is:

If $f(x) = \left|\begin{array}{ccc} x^3 & 2x^2+1 & 1+3x \\ 3x^2+2 & 2x & x^3+6 \\ x^3-x & 4 & x^2-2 \end{array}\right|$ for all $x \in R$,then $2f(0) + f'(0)$ is equal to

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:

If $f(x) = \left| \begin{array}{ccc} \cos(2x) & \cos(2x) & \sin(2x) \\ -\cos x & \cos x & -\sin x \\ \sin x & \sin x & \cos x \end{array} \right|$,then:
$A$. $f'(x) = 0$ at exactly three points in $(-\pi, \pi)$
$B$. $f'(x) = 0$ at more than three points in $(-\pi, \pi)$
$C$. $f(x)$ attains its maximum at $x = 0$
$D$. $f(x)$ attains its minimum at $x = 0$