$$f\left( x \right) = f\left( {1 - x} \right)$$ for all $$x \in \left[ { - 1,2} \right]$$
Let $${R_1} = \int\limits_{ - 1}^2 {xf\left( x \right)dx,} $$ and $${R_2}$$ be the area of the region bounded by $$y=f(x),$$ $$x=-1,$$ $$x=2,$$ and the $$x$$-axis. Then
$$\,\,\,\,$$ $$\,\,\,\,$$ $$\,\,\,\,$$ Column-$$I$$
(A) $$\,\,\,\,$$If $$\overrightarrow a = \widehat j + \sqrt 3 \widehat k,\overrightarrow b = - \widehat j + \sqrt 3 \widehat k$$ and $$\overrightarrow c = 2\sqrt 3 \widehat k$$ form a triangle, then the internal angle of the triangle between $$\overrightarrow a $$ and $$\overrightarrow b $$ is
(B)$$\,\,\,\,$$ If $$\int\limits_a^b {\left( {f\left( x \right) - 3x} \right)dx = {a^2} - {b^2},} $$ then the value of $$f$$ $$\left( {{\pi \over 6}} \right)$$ is
(C)$$\,\,\,\,$$ The value of $${{{\pi ^2}} \over {\ell n3}}\int\limits_{7/6}^{5/6} {\sec \left( {\pi x} \right)dx} $$ is
(D)$$\,\,\,\,$$ The maximum value of $$\left| {Arg\left( {{1 \over {1 - z}}} \right)} \right|$$ for $$\left| z \right| = 1,\,z \ne 1$$ is given by
$$\,\,\,\,$$ $$\,\,\,\,$$ $$\,\,\,\,$$ Column-$$II$$
(p)$$\,\,\,\,$$ $${{\pi \over 6}}$$
(q)$$\,\,\,\,$$ $${{2\pi \over 3}}$$
(r)$$\,\,\,\,$$ $${{\pi \over 3}}$$
(s)$$\,\,\,\,$$ $$\pi $$
(t) $$\,\,\,\,$$ $${{\pi \over 2}}$$