$\mathrm{cos\theta }=\frac{-\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}}{|-\overrightarrow{\mathrm{a}}|\left|\overrightarrow{\mathrm{b}}\right|}=-\frac{1}{2}\Rightarrow \mathrm{\theta }=\frac{2\mathrm{\pi }}{3}$

(B)  $\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\left(\mathrm{f}\right(\mathrm{x})-(\mathrm{x}\left)\right)\mathrm{dx}={\mathrm{a}}^2-{\mathrm{b}}^2$

differentiating w.r.t (b).

f(b) – 3b = – 2b

f(b) = b

So $\mathrm{f}\left(\frac{\mathrm{\pi }}{6}\right)=\frac{\mathrm{\pi }}{6}$ ; if a = b then any value of f(x) is possible

(C) $\mathrm{I}=\frac{{\mathrm{\pi }}^2}{\mathrm{\ell n}3}\underset{7/6}{\overset{5/6}{\int }}\sec \mathrm{\pi x}\mathrm{dx}$

 $\mathrm{I}=\frac{{\mathrm{\pi }}^2}{\mathrm{\ell n}3}{\left|\mathrm{\ell n}|\mathrm{sec\pi x}+\mathrm{tan\pi x}|\right|}_{7/6}^{5/6}$

 $\mathrm{I}=\frac{\mathrm{\pi }}{\mathrm{\ell n}3}\cdot \mathrm{\ell n}3=\mathrm{\pi }$

(D)  $\therefore$|z| = 1

 $\mathrm{z}=\mathrm{cos\theta }+\mathrm{isin\theta }\forall \mathrm{\theta }\in (-\mathrm{\pi },\mathrm{\pi }]\mathrm{and}\mathrm{\theta }\ne 0$

 $\left|\mathrm{Arg}\frac{1}{(1-\mathrm{z})}\right|=\left|\mathrm{Arg}\left(\frac{1}{1-\mathrm{cos\theta }-\mathrm{isin\theta }}\right)\right|=\left|\mathrm{Arg}(\frac{1}{2}+\frac{\mathrm{icot}\frac{\mathrm{\theta }}{2}}{2})\right|$

 $=\left|\frac{\mathrm{\pi }-\mathrm{\theta }}{2}\right|$      so maximum value is $\mathrm{\pi }$.

*           The most appropriate answer to this question is

            A $\to$ q; B $\to$ p or p, q, r, s & t; C $\to$ s; D $\to$ s

            But because of ambiguity in language, IIT has declared

            A $\to$ q; B $\to$ p or p, q, r, s & t; C $\to$ s; D $\to$ t as correct answer