If $${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $$,where –1 $$ \le $$ x $$ \le $$ 1, – 2 $$ \le $$ y $$ \le $$ 2, x $$ \le $$ $${y \over 2}$$ , then for all x, y, 4x² – 4xy cos $$\alpha $$ + y² is equal to :

Let y = y(x) be the solution of the differential equation,
$${{dy} \over {dx}} + y\tan x = 2x + {x^2}\tan x$$, $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$, such that y(0) = 1. Then :

The area (in sq.units) of the region bounded by the curves y = 2^x and y = |x + 1|, in the first quadrant is :

If z and w are two complex numbers such that |zw| = 1 and arg(z) – arg(w) = $${\pi \over 2}$$ , then :