Correct option: A.$(A) \rightarrow q (B) \rightarrow p\ or \ p, q, r, s \ and \ t (C) \rightarrow s(D) \rightarrow t$
a
$(A)$ $|\vec{a}|=2,|\vec{b}|=2,|\vec{c}|=2 \sqrt{3}$
Angle between $\vec{a}$ and $b$ is
$\cos \theta=\frac{\left|\hat{a}^2+\right| \vec{b}^2-\vec{c}^2}{2|\vec{a}||\vec{b}|}=\frac{4+4-12}{2 \times 2 \times 2}=-\frac{1}{2}$
$\Rightarrow \theta=\frac{2 \pi}{3}$
$(B)$ $\int_a^b(f(x)-3 x) d x=a^2-b^2$
$\Rightarrow \int_a^b f(x) d x-\int_a^b 3 x d x=a^2-b^2$
$\Rightarrow \int_a^b f(x) d x=\frac{b^2-a^2}{2}$
$\Rightarrow f(x)=x$
$\Rightarrow f\left(\frac{\pi}{6}\right)=\frac{\pi}{6}$
$(C)$ $I=\frac{\pi^2}{\ln 3} \int_{7 / 6}^{5 / 6} \sec (\pi x) d x$
$=\frac{\pi^2}{\ln 3}[\ln |(\sec (\pi x)+\tan (\pi x))|]_{7 / 6}^{5 / 6}=\frac{\pi^2}{\ln 3} \times \frac{1}{\pi} \ln 3$
$=\pi$
$(D)$ Let $u=\frac{1}{1-z}$
$\Rightarrow z=1-\frac{1}{u}$
Given that $|z|=1 \Rightarrow\left|1-\frac{1}{u}\right|=1$
$\Rightarrow|u-1|=|u|$
$\therefore$ locus of $u$ is perpendicular bisector of line segment joining $(1,0)$ and $(0,0)$ in complex plane.
$\Rightarrow$ maximum argument approaches $\frac{\pi}{2}$