If the curve y = y(x) is the solution of the differential equation

$$2({x^2} + {x^{5/4}})dy - y(x + {x^{1/4}})dx = {2x^{9/4}}dx$$, x > 0 which

passes through the point $$\left( {1,1 - {4 \over 3}{{\log }_e}2} \right)$$, then the value of y(16) is equal to :

Two tangents are drawn from a point P to the circle x² + y² $$-$$ 2x $$-$$ 4y + 4 = 0, such that the angle between these tangents is $${\tan ^{ - 1}}\left( {{{12} \over 5}} \right)$$, where $${\tan ^{ - 1}}\left( {{{12} \over 5}} \right)$$ $$\in$$(0, $$\pi$$). If the centre of the circle is denoted by C and these tangents touch the circle at points A and B, then the ratio of the areas of $$\Delta$$PAB and $$\Delta$$CAB is :

Let O be the origin. Let $$\overrightarrow {OP} = x\widehat i + y\widehat j - \widehat k$$ and $$\overrightarrow {OQ} = - \widehat i + 2\widehat j + 3x\widehat k$$, x, y$$\in$$R, x > 0, be such that $$\left| {\overrightarrow {PQ} } \right| = \sqrt {20} $$ and the vector $$\overrightarrow {OP} $$ is perpendicular $$\overrightarrow {OQ} $$. If $$\overrightarrow {OR} $$ = $$3\widehat i + z\widehat j - 7\widehat k$$, z$$\in$$R, is coplanar with $$\overrightarrow {OP} $$ and $$\overrightarrow {OQ} $$, then the value of x² + y² + z² is equal to :

If the Boolean expression $$(p \wedge q) \odot (p \otimes q)$$ is a tautology, then $$ \odot $$ and $$ \otimes $$ are respectively given by :