Match the statements given in Column I with the values given in...

(A) $\therefore|\mathbf{a}|=\sqrt{1+3}=2 |\mathbf{b}|=\sqrt{1+3}=2 |\mathbf{c}|=\sqrt{12}=2 \sqrt{3}$
Using cosine law, \begin{aligned} \cos C & =\frac{|\mathbf{a}|^2+|\mathbf{b}|^2-|\mathbf{c}|^2}{2|\mathbf{a}||\mathbf{b}|} \\ & =\frac{4+4-12}{2 \times 2 \times 2}=\frac{-4}{8}=\frac{-1}{2} \\ \Rightarrow \quad \angle C & =120^{\circ}=\frac{2 \pi}{3} \end{aligned} (B) \begin{aligned} & \int_a^b f(x) d x-3\left(\frac{x^2}{2}\right)_a^b=\left(a^2-b^2\right) \\ & \Rightarrow \int_a^b f(x) d x-\frac{3}{2}\left(b^2-a^2\right)=\left(a^2-b^2\right) \\ & \Rightarrow \int_a^b f(x) d x=\left(a^2-b^2\right)+\frac{3}{2}\left(b^2-a^2\right) \\ & =\frac{b^2-a^2}{2} \\ & \Rightarrow \int_a^b f(x) d x=\frac{b^2-a^2}{2} \\ & \end{aligned}$$f(x)=x \Rightarrow f\left(\frac{\pi}{6}\right)=\frac{\pi}{6}$$ (C) \begin{aligned} & \frac{\pi^2}{\log _e 3} \int_{7/6}^{5 / 6} \sec (\pi x) d x \\ & \Rightarrow \frac{\pi^2}{\log _e 3}\left\{\frac{\log |\sec \pi x+\tan \pi x|}{\pi}\right\}_{7/6}^{5 / 6} \\ & \Rightarrow \frac{\pi}{\log 3}\left\{\log \left|\sec \frac{5 \pi}{6}+\tan \frac{5 \pi}{6}\right|\right. \\ & \left.\quad-\log \left|\sec \frac{7 \pi}{6}+\tan \frac{7 \pi}{6}\right|\right\} \\ & \Rightarrow \frac{\pi}{\log 3}\left\{\log |\sqrt{3}|-\log \left|\frac{1}{\sqrt{3}}\right|\right\} \\ & \Rightarrow \frac{\pi}{\log 3}\{\log 3\}=\pi \end{aligned} (D) \begin{aligned} & \left|\arg \frac{1}{(1-z)}\right|, \text{ for }|z|=1 \\ & \Rightarrow\left|\arg (1-z)^{-1}\right| \\ & \Rightarrow|-\arg (1-z)| \Rightarrow|\arg (1-z)| \end{aligned}
From figure, $\arg (z-1)$ is maximum $=\pi$.