If $A = \begin{vmatrix} x & 1 \\ 1 & x \end{vmatrix}$ and $B = \begin{vmatrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{vmatrix}$,then $\frac{dB}{dx}$ is

Let $f$ be a twice differentiable function defined on $R$ such that $f(0)=1$,$f^{\prime}(0)=2$ and $f^{\prime}(x) \neq 0$ for all $x \in R$. If $\left|\begin{array}{ll}f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}\right|=0$ for all $x \in R$,then the value of $f(1)$ lies in the interval:

Let $A=\left[\begin{array}{rrr}-1 & -2 & -3 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right]$,$B=\left[\begin{array}{rr}1 & -2 \\ -1 & 2\end{array}\right]$ and $C=\left[\begin{array}{rrr}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$. If $a, b$ and $c$ respectively denote the ranks of $A, B$ and $C$,then the correct order of these numbers is:

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 $M$ and $m$ respectively be the maximum and the minimum values of $f(x) = \left| \begin{array}{ccc} 1+\sin^2 x & \cos^2 x & 4\sin 4x \\ \sin^2 x & 1+\cos^2 x & 4\sin 4x \\ \sin^2 x & \cos^2 x & 1+4\sin 4x \end{array} \right|$,$x \in R$. Then $M^4 - m^4$ is equal to:

The rank of the matrix $\begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & -1 \\ -1 & 1 & 1 \end{bmatrix}$ is