If $\mathrm{A}=\left[\begin{array}{cc}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right], \mathrm{C}=\mathrm{ABA}^{\mathrm{T}}$ and $\mathrm{X}=\mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{~A}$, then $\operatorname{det} \mathrm{X}$ is equal to :

If $\tan \mathrm{A}=\frac{1}{\sqrt{x\left(x^2+x+1\right)}}, \tan \mathrm{B}=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}$ and

$\tan \mathrm{C}=\left(x^{-3}+x^{-2}+x^{-1}\right)^{1 / 2}, 0<\mathrm{A}, \mathrm{B}, \mathrm{C}<\frac{\pi}{2}$, then $\mathrm{A}+\mathrm{B}$ is equal to :

If $\mathrm{n}$ is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then $\mathrm{n}$ is equal to :

Let $\mathrm{S}=|\mathrm{z} \in \mathrm{C}:| z-1 \mid=1$ and $(\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2} \mid$. Let $z_1, z_2 \in \mathrm{S}$ be such that $\left|z_1\right|=\max\limits_{z \in s}|z|$ and $\left|z_2\right|=\min\limits _{z \in S}|z|$. Then $\left|\sqrt{2} z_1-z_2\right|^2$ equals :