QJEEAdv11Ma1e1a: JEE Advanced 2011 Physics - Class 11 JEEAdv11Ma1e1a Solution: Match the statements given in Column - with the values given in …

(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.