Let $\mathrm{A}=\left\{x \in(0, \pi)-\left\{\frac{\pi}{2}\right\}: \log _{(2 /\pi)}|\sin x|+\log _{(2 / \pi)}|\cos x|=2\right\}$ and $\mathrm{B}=\{x \geqslant 0: \sqrt{x}(\sqrt{x}-4)-3|\sqrt{x}-2|+6=0\}$. Then $\mathrm{n}(\mathrm{A} \cup \mathrm{B})$ is equal to :

The area of the region enclosed by the curves $y=\mathrm{e}^x, y=\left|\mathrm{e}^x-1\right|$ and $y$-axis is :

For some $a, b,$ let $f(x)=\left|\begin{array}{ccc}\mathrm{a}+\frac{\sin x}{x} & 1 & \mathrm{~b} \\ \mathrm{a} & 1+\frac{\sin x}{x} & \mathrm{~b} \\ \mathrm{a} & 1 & \mathrm{~b}+\frac{\sin x}{x}\end{array}\right|, x \neq 0, \lim \limits_{x \rightarrow 0} f(x)=\lambda+\mu \mathrm{a}+\nu \mathrm{b}.$ Then $(\lambda+\mu+v)^2$ is equal to :

If $\alpha>\beta>\gamma>0$, then the expression $\cot ^{-1}\left\{\beta+\frac{\left(1+\beta^2\right)}{(\alpha-\beta)}\right\}+\cot ^{-1}\left\{\gamma+\frac{\left(1+\gamma^2\right)}{(\beta-\gamma)}\right\}+\cot ^{-1}\left\{\alpha+\frac{\left(1+\alpha^2\right)}{(\gamma-\alpha)}\right\}$ is equal to :