\[
\text { Match the statements given in Column I with the values given in Column II. }
\]

(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 \(=\pi\).