(A) We have a−b = - 1 + 3 = 2, where ∣a∣ = 2 and b = 2. cos θ = 2×22 = 21 θ = 2π,32π; however , it is 32π as its opposite to side of maximum length. (B) a∫b (f (x) - 3x) dx = a2−b2a∫b f (x) dx = 23(b2−a2)+a2+b2 = 2−a2+b2 ⇒ f (x) = x Therefore, f (π/6) = (π2/6) (C) ln3π2πln(sec(px)+tan(πx))7/65/6 =
ln3π(lnsec65π+tan65π−lnsec67π+tan67π)=π
(D) Let us consider u = 1−z1 ⇒ z = 1 - u1 |z| = 1 ⇒ 1−u1 = 1 ⇒ |u - 1| = |u| Hence, the locus of u is perpendicular bisector of line segment joining 0 and 1. Therefore, the maximum arg(u) approaches π/2, but it will not attain.