If $$y(t)$$ is a solution of $$\left( {1 + t} \right){{dy} \over {dt}} - ty = 1$$ and $$y\left( 0 \right) = - 1,$$ then $$y(1)$$ is equal to

If $$f\left( x \right) = \int\limits_{{x^2}}^{{x^2} + 1} {{e^{ - {t^2}}}} dt,$$ then $$f(x)$$ increases in

The area bounded by the curves $$y = \sqrt x ,2y + 3 = x$$ and
$$x$$-axis in the 1^st quadrant is

If $$\,\left| z \right| = 1$$ and $$\omega = {{z - 1} \over {z + 1}}$$ (where $$z \ne - 1$$), then $${\mathop{\rm Re}\nolimits} \left( \omega \right)$$ is