If $${{dy} \over {dx}} = {{{2^x}y + {2^y}{{.2}^x}} \over {{2^x} + {2^{x + y}}{{\log }_e}2}}$$, y(0) = 0, then for y = 1, the value of x lies in the interval :

An angle of intersection of the curves, $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ and x² + y² = ab, a > b, is :

If $$y{{dy} \over {dx}} = x\left[ {{{{y^2}} \over {{x^2}}} + {{\phi \left( {{{{y^2}} \over {{x^2}}}} \right)} \over {\phi '\left( {{{{y^2}} \over {{x^2}}}} \right)}}} \right]$$, x > 0, $$\phi$$ > 0, and y(1) = $$-$$1, then $$\phi \left( {{{{y^2}} \over 4}} \right)$$ is equal to :

The sum of the roots of the equation

$$x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0$$, is :