JEE Advance - Mathematics (2011 - Paper 2 Offline - No. 11)
$$\,\,\,\,$$ $$\,\,\,\,$$ $$\,\,\,\,$$ 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) $$\,\,\,\,$$ $${{\pi \over 2}}$$
Explanation
(A) We have, $$\overrightarrow a - \overrightarrow b = - 1 + 3 = 2$$, where $$|\overrightarrow a | = 2$$ and $$|\overrightarrow b | = 2$$.
$$\cos \theta = {2 \over {2 \times 2}} = {1 \over 2}$$
$$\theta = {\pi \over 3},\,{{2\pi } \over 3}$$ ; however, it is $${{2\pi } \over 3}$$ as its opposite to side of maximum length.
(B) $$\int\limits_a^b {(f(x) - 3x)dx = {a^2} - {b^2}} $$
$$\int\limits_a^b {f(x)dx = {3 \over 2}({b^2} - {a^2}) + {a^2} - {b^2} = {{ - {a^2} + {b^2}} \over 2} \Rightarrow f(x) = x} $$
Therefore, $$f(\pi /6) = ({\pi ^2}/6)$$
(C) $${{{\pi ^2}} \over {\ln 3}}\left( {{{\ln |(\sec \pi x + \tan \pi x)|_{7/6}^{5/6}} \over \pi }} \right)$$
$$ = {\pi \over {\ln 3}}\left( {\ln \left| {\sec {{5\pi } \over 6} + \tan {{5\pi } \over 6}} \right| - \ln \left| {\sec {{7\pi } \over 6} + \tan {{7\pi } \over 6}} \right|} \right) = \pi $$
(D) Let us consider,
$$u = {1 \over {1 - z}} \Rightarrow z = 1 - {1 \over u}$$
$$|z| = 1 \Rightarrow \left| {1 - {1 \over u}} \right| = 1 \Rightarrow |u - 1| = |u|$$
Hence, the locus of u is perpendicular bisector of line segment joining 0 and 1.
Therefore, the maximum arg(u) approaches $$\pi$$/2, but it will not attain.
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